Cracking the Mental Math Code : A Repertoire of Number Sense Techniques

Cracking the Mental Math Code: A Repertoire of Number Sense Techniques

By: Yash Chandak

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Special thanks to Mr. Newton, Mom, Dad, and Sneha

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About this Book From those who read a poster asking them to join the school’s Number Sense team to the seasoned competitor eyeing State medals, this book serves as a guide and as inspiration, to not only pique interest but to build a foundation on. You’ll find in this book a step-by-step guide on how-to solve Number Sense problems and prepare you for any mental math competitions. There are over 80 mental math techniques are explained in this book, and over 750 problems to learn, train, and improve with. Just a note: as fun as it is to learn new tricks and techniques, it’s just as important to master them and implement them in full-length tests! I wish you all the best of luck, and happy computing!

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Contents 1

PEMDAS.................................................................................................................................... 8 1.1

Fast Addition and Subtraction ................................................................................. 8

1.2

PEMDAS Approximations ....................................................................................... 10

1.3

Multiply by 11 and like............................................................................................. 12

1.4

Multiplying by 25 ........................................................................................................ 16

1.5

Multiplying by 125 ..................................................................................................... 18

1.6

Multiplying by any Number Ending in 5 ........................................................... 19

1.7

Multiplying by 101 and like ................................................................................... 21

1.8

Multiplying Two Numbers just above 100 ...................................................... 23

1.9

Multiplying Two Numbers just below 100 ...................................................... 26

1.10

Multiplying 2 Numbers with an Even Difference ...................................... 28

1.11 Multiplying 2 Numbers with the Same Tens Digit and Ones Digits that Add to 10 .......................................................................................................................... 30 1.12 2

FOIL .............................................................................................................................. 32

Squares .................................................................................................................................. 35 2.1

Squares............................................................................................................................ 35

2.2

Squares Ending in 5 ................................................................................................... 36

2.3

Sum of Squares in the form a2+(3a)2 .................................................................. 37

2.4

Sum of Two Consecutive Squares ........................................................................ 38

2.5 Sum of Squares when the Outer Digits add to 10 and the Inner Digits are 1 Apart ................................................................................................................................ 40 2.6 3

4

Approximations of Exponents............................................................................... 43 GCD and LCM .................................................................................................................... 44

3.1

How to find GCD .......................................................................................................... 44

3.2

How to find LCM.......................................................................................................... 47

Remainder ............................................................................................................................ 50 4.1

Remainder with Division......................................................................................... 50

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5

6

7

8

9

4.2

Remainder with Operations................................................................................... 53

4.3

Remainder with Exponents .................................................................................... 55

Higher Order Exponents ................................................................................................ 59 5.1

Cubes................................................................................................................................ 59

5.2

Powers of Special Numbers.................................................................................... 59

5.3

Difference of Cubes 1 Apart ................................................................................... 60

5.4

Laws of Exponents ..................................................................................................... 62

Roots ....................................................................................................................................... 64 6.1

Square Roots ................................................................................................................. 64

6.2

Nested Square Roots ................................................................................................. 65

6.3

Other Roots ................................................................................................................... 67

6.4

Approximations of Roots......................................................................................... 68

Primes and Divisors ......................................................................................................... 71 7.1

Primes.............................................................................................................................. 71

7.2

Number of Positive Integral Divisors ................................................................. 72

7.3

Sum of Positive Integral Divisors......................................................................... 74

7.4

Relatively Prime .......................................................................................................... 76

Other Topics ........................................................................................................................ 78 8.1

Additive and Multiplicative Identities and Inverses.................................... 78

8.2

Absolute Value ............................................................................................................. 80

8.3

Roman Numerals ........................................................................................................ 82

8.4

Conversion Factors .................................................................................................... 84

8.5

Complex Numbers ...................................................................................................... 86

Types of Numbers ............................................................................................................. 89 9.1

Polygonal Numbers ................................................................................................... 89

9.2

Deficient, Perfect, and Abundant Numbers ..................................................... 91

9.3

Happy and Unhappy Numbers .............................................................................. 92

9.4

Evil and Odious Numbers........................................................................................ 94

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9.5

Polite Numbers and Politeness ............................................................................. 95

9.6

Frugal, Economical, Equidigital, Wasteful, and Extravagant Numbers 97

10

Factorials and Combinations ..................................................................................... 99

10.1

Factorials .................................................................................................................... 99

10.2

Permutations .......................................................................................................... 100

10.3

Combinations.......................................................................................................... 102

11

Sequences ........................................................................................................................ 104

11.1

Consecutive Integer Sequences ...................................................................... 104

11.2

Arithmetic Sequences ......................................................................................... 107

11.3

Geometric Sequences .......................................................................................... 111

11.4

Fibonacci................................................................................................................... 114

11.5

Other Sequences.................................................................................................... 117

12

Memorization ................................................................................................................. 118

12.1

Fractions and Decimals ...................................................................................... 118

12.2

Approximations ..................................................................................................... 119

13

Adding Fractions........................................................................................................... 121

13.1

Adding Inverses ..................................................................................................... 121

13.2

Fractions in the form a/b+b/(a+b) ............................................................... 124

13.3

Special Fractions ................................................................................................... 126

13.4

Adding Fractions with no Trick ...................................................................... 129

14

Multiplying Fractions.................................................................................................. 130

14.1

Whole Numbers Same and Fractions Add to 1 ........................................ 130

14.2

Multiplying 2 Mixed Numbers with the Same Fraction ........................ 132

14.3

Fractions in the Form a*a/(a+b) .................................................................... 136

14.4

Fractions in the Form a*(a+n)/(a+2n) ........................................................ 138

14.5

Multiplying using Improper Fractions......................................................... 141

14.6

Comparison of Fractions.................................................................................... 142

15

Repeating Decimals ..................................................................................................... 144

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15.1

Repeating Decimals in the Form of .aaa…, .ababab…, or .abcabcabc… 144

15.2

Repeating Decimals in the Form of .abbb…, .abccc…, or .abcbcbc… 147

16

Bases .................................................................................................................................. 151

16.1

Base n to Base 10 .................................................................................................. 151

16.2

Base 10 to Base n .................................................................................................. 153

16.3

Base a to Base a^n ................................................................................................ 155

16.4

Decimals in Other Bases .................................................................................... 159

17

Linear Equations........................................................................................................... 162

17.1

Solving Linear Equations ................................................................................... 162

17.2

Slope and y-intercept .......................................................................................... 164

17.3

Inequalities .............................................................................................................. 167

18

Quadratic Equations.................................................................................................... 168

18.1

Roots........................................................................................................................... 168

18.2

Vieta’s Formulas .................................................................................................... 171

19

Sets...................................................................................................................................... 173

19.1

Subsets ...................................................................................................................... 173

19.2

Proper and Improper Subsets ......................................................................... 174

19.3

Intersection and Union of Sets ........................................................................ 175

19.4

Cartesian Product ................................................................................................. 177

20

Logarithms ...................................................................................................................... 179

20.1

Solving Equations with Logarithms .............................................................. 179

20.2

Nested Logarithms ............................................................................................... 182

21

Matrices ............................................................................................................................ 183

21.1

Adding and Subtracting Matrices................................................................... 183

21.2

Multiplying Matrices............................................................................................ 185

21.3

Determinants .......................................................................................................... 189

Answers ....................................................................................................................................... 191

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1

PEMDAS

1.1

Fast Addition and Subtraction Most all tricks in this book will require you to be quick in basic operations. Given that you only have 7.5 seconds to read, solve, and write the answer for each question, you don’t get much time for the intermediate steps. So, let’s teach you a trick on arguably the simplest topic in mathematics: addition and subtraction. Basic addition involves adding one digit at a time moving right to left. To get the answer faster, however, you can add two digits at a time, write the answer for that, and maintain the carries. Let’s see a couple of examples. 1357 + 2633 = 57 + 33 = 90 13 + 26 = 39 𝐹𝑖𝑛𝑎𝑙 𝑎𝑛𝑠𝑤𝑒𝑟: 3990

2035 − 328 = 35 − 28 = 07 20 − 3 = 17 𝐹𝑖𝑛𝑎𝑙 𝑎𝑛𝑠𝑤𝑒𝑟: 1707

2626 + 262 − 26 = 26 + 62 − 26 = 62 26 + 2 = 28 𝐹𝑖𝑛𝑎𝑙 𝑎𝑛𝑠𝑤𝑒𝑟: 2862 Note: it’s almost always advantageous to subtract before you add. Dealing with smaller numbers is usually easier.

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Practice Problem Set 1.1 1) 2060 + 6020 =

2) 4703 + 3211 =

3) 864 + 579 =

4) 357 + 864 − 135 =

5) 3205 − 3088 =

6) 2024 − 3026 =

7) 2010 + 201 + 20 =

8) 1234 + 123 + 12 + 1 =

9) 2134 − 1123 =

10) 1914 − 3229 =

11) 2468 − 9753 =

12) 2551 + 125 − 51 + 2 =

13) 975 + 318 − 642 =

14) 289 + 33445 − 291 =

15) 4110 − 4128 + 2008 =

16) 16183 − 14135 + 230 =

17) 437 − 734 =

18) 537 − 33 + 284 =

19) 2016 − 424 + 508 =

20) 78 − 3457 + 230 =

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1.2

PEMDAS Approximations Almost every Number Sense test will have an approximation question that is solely addition and subtraction, usually 5 or 6 digits long. Since you can be within 5% of the actual answer and still get the question correct, it is advantageous for you to only do the operations for the highest 2 place values and round appropriately rather than trying to chug through all the digits.

71239 + 47458 − 22334 =

To solve this question, I would use the ten-thousands and thousands digits to determine an approximate value.

71 + 47 − 22 = 71 + 25 = 96 My answer: 96,000 Range: 91,545 to 101,181

Most times, rounding in these questions is beneficial because the ranges are quite large and the problem can become quite easier to solve. However, if you round up on one number a substantial amount (relative to the number of course), it’s good to round down on another to balance out the error. It’s important to be both quick and accurate when rounding. You don’t want to waste time on easy questions trying to choose how to round. Just do what feels right and start calculating!

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Practice Problem Set 1.2 1) 314 + 272 − 31 − 27 = 2) 2101 + 3202 − 2323 = 3) 9876 + 1234 − 50 = 4) 57346 − 45634 + 34365 = 5) 753 + 2468 − 901 + 2005 = 6) 449 ∗ 9 + 451 = 7) 789 − 3120 + 645 = 8) 2345 − 456 + 3298 − 264 = 9) 376 − 1785 + 63 = 10) 2585 + 20345 + 245 + 23 = 11) 4554 − 5665 − 6776 = 12) 74859 − 27384 + 2604 = 13) 3131 − 311 + 133 − 33 = 14) 3091 + 2512 + 7896 =

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1.3

Multiply by 11 and like The one thing more common than addition on a Number Sense test is multiplication. I cannot emphasize how drastically multiplication speed affects your score. It goes without saying that with a lot of multiplication comes a lot of multiplication tricks. The easiest and most common trick is multiplying by 11. I could try to explain this trick in words for you, but that would be quite confusing and lengthy, so let’s use an example.

35 ∗ 11 =

Let’s use the original way of multiplying to see the trick. 3 5 x

1 1 3 5

+ 3 5 0 3 8 5 It looks like all we did was add the two digits (3+5=8) and put that sum in the middle of that number. Let’s see another example.

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57 ∗ 11 = 5 7 x

1 1

1

5 7

+ 5 7 0 6 2 7 In this example, the sum of the two digits was 12, which sadly isn’t just one digit. So, the 2 was put between the two numbers again but the 5 was increased to a 6 because of the carry in 12. Using the previous two examples, we can create the following steps to follow when multiplying by 11: Step 1: write the ones digit as the ones digit of your answer. Step 2: add the ones digit and the tens digit. If the value is less than ten, write the digit as the tens digit of your answer. Else, note the carry and then write one digit. Step 3: write the tens digit (after adding any carries) as the hundreds digit of your answer. Let’s see it pictorially. Each line in this diagram represents the addition for the trick. 0

5

7 Ones Tens Hundreds

This trick even works for numbers with more than 2 digits. We keep adding 2 digits at a time, working from right to left.

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429 ∗ 11 =

Step 1: write the 9. Step 2: add 2 and 9 to get 11. Write 1 and carry 1. Step 3: add 4 and 2 to get 6. Add the carry to get 7. Write 7. Step 4: write 4. Our final answer is 4719. Checking it with basic multiplication, we see not only the answer but the same addition we did to get the answer. 4 2 9 x

1 1 4 2 9

+ 4 2 9 0 4 7 1 9 This trick can also be extended to multiply by 111. When multiplying by 111, you follow the same steps as multiplying by 11 but build up to adding 3 digits at a time.

429 ∗ 111 =

Step 1: write the last digit, a 9. Step 2: add 2 and 9 to get 11. Write the 1 and carry 1. Step 3: this is where it changes. Instead of shifting over to add just 4 and 2, we’re gonna add the 4, 2, and 9 together along with the carry to get 16. We did this because there are 3 ones in the number we are multiplying with. Rule of thumb: the number of ones is the same as the number of digits you have to add. 14 | P a g e

Step 4: add the 4 and 2 as well as the carry to get 7. Step 5: write the 4. Final answer: 47619 4

2

9 Ones Tens Hundreds Thousands Ten thousands

Practice Problem Set 1.3 1) 44 ∗ 11 =

2) 35 ∗ 11 =

3) 73 ∗ 11 =

4) 49 ∗ 11 =

5) 43 ∗ 22 =

6) 187 ∗ 11 =

7) 55 ∗ 46 =

8) 33 ∗ 13 =

9) 789 ∗ 11 =

10) 11 ∗ 258 =

11) 111 ∗ 475 =

12) 214 ∗ 111 =

13) 418 ∗ 111 =

14) 123 ∗ 333 =

15) 111 ∗ 987 =

16) 111 ∗ 45 =

17) 1111 ∗ 111 =

18) 111 ∗ 23 =

19) 121 ∗ 32 =

20) 121 ∗ 14 = 15 | P a g e

1.4

Multiplying by 25 Multiplying by 25 is not only quite common but also quite easy. The easiest way to visualize this trick is with money. It’s known that 4 quarters make a dollar and each quarter is worth 25 cents, so let’s use that. If we multiply 25 by 4, we get 100, an easy number to work with. However, we have to divide the other number by 4 to keep the transformed expression equal. Let’s look at an example.

56 ∗ 25 = 56 ∗ 25 ∗ =

4 4

56 ∗ (25 ∗ 4) 4 = 14 ∗ 100 = 1400

Sometimes, the other number is not divisible by 4. When the number isn’t divisible, imagine the decimal remainder that would occur. When this is subsequently multiplied by 100, shift the decimal place over.

47 ∗ 25 =

47 ∗ (25 ∗ 4) 4

= 11.75 ∗ 100 = 1175

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Practice Problem Set 1.4 1) 24 ∗ 25 =

2) . 12 ∗ 25 =

3) 73 ∗ 25 =

4) 37 ∗ 25 =

5) 75 ∗ 121 =

6) 25 ∗ 46 =

7) 12.8 ∗ 25 =

8) 137 ∗ 25 =

9) 36.4 ∗ 25 =

10) 0.124 ∗ 25 =

11) 25.25 ∗ 25 =

12) 12.016 ∗ 25 =

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1.5

Multiplying by 125 Multiplying by 125 is quite similar to Multiplying by 25 (Trick 1.4). Since 125 ∗ 8 = 1000, we can divide the other number by 8 and then multiply by 1000.

72 ∗ 125 =

72 ∗ 1000 8 = 9000

Practice Problem Set 1.5 1) 24 ∗ 125 =

2) 0.12 ∗ 125 =

3) 73 ∗ 125 =

4) 37 ∗ 125 =

5) 125 ∗ 1.7 =

6) 375 ∗ 48 =

7) 12.8 ∗ 125 =

8) 137 ∗ 125 =

9) 36.4 ∗ 125 =

10) 0.124 ∗ 125 =

11) 25.25 ∗ 125 =

12) 12.016 ∗ 125 =

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1.6

Multiplying by any Number Ending in 5 The number 5 is so versatile. One of the helpful things when conducting mental math is that 5 ∗ 2 = 10. With this obvious fact in mind, let’s look at the double-and-half trick. The name says it all; double the number ending in 5 and half the other one. This will keep the value of the expression the same but make the problem much easier to solve.

32 ∗ 15 = 32 ∗ 15 ∗ =

2 2

32 ∗ (15 ∗ 2) 2 = 16 ∗ 30 = 480

If the other number is odd, it won’t divide evenly by 2. In this case, it is faster to FOIL (Trick 1.13). Practice Problem Set 1.6 1) 45 ∗ 12 =

2) 35 ∗ 16 =

3) 75 ∗ 28 =

4) 78 ∗ 35 =

5) 25 ∗ 46 =

6) 65 ∗ 128 =

7) 789 ∗ 45 =

8) 14 ∗ 95 =

9) 135 ∗ 18 =

10) 42 ∗ 45 = 19 | P a g e

11) 48 ∗ 75 =

12) 85 ∗ 46 =

13) 55 ∗ 72 =

14) 155 ∗ 62 =

15) 95 ∗ 108 =

16) 38 ∗ 95 =

17) 115 ∗ 26 =

18) 135 ∗ 54 =

19) 75 ∗ 120 =

20) 85 ∗ 136 =

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1.7

Multiplying by 101 and like It is very difficult to go wrong when multiplying by 101 if you know what you’re doing. The most important thing to notice is that multiplying by 101 is the same thing as multiplying by 100, multiplying by 1, and adding the two products. Let’s see an example with a 2-digit number.

72 ∗ 101 = 72 ∗ 100 + 72 ∗ 1 = 7200 + 72 = 7272

The quotient is literally the number twice, which makes sense given the operations we did. This works for all two digit numbers. Now let’s up the game.

381 ∗ 101 = 381 ∗ 100 + 381 ∗ 1 = 38100 + 381 = 38481

With three digits, we had to add the hundreds digit from the product with 1 and the ones digit from the product with 100. So, when solving multiplication with 101, be wary of addition when it’s not with a two-digit number. The same theory applies for multiplying by 1001. Just be wary of the place values when thinking of the addition. Rather than writing out an explanation, I’ll let you figure out the place values with practice problems. 21 | P a g e

Practice Problem Set 1.7 1) 101 ∗ 7 =

2) 15 ∗ 101 =

3) 23 ∗ 101 =

4) 123 ∗ 101 =

5) 101 ∗ 258 =

6) 12 ∗ 202 =

7) 7 ∗ 1001 =

8) 14 ∗ 1001 =

9) 135 ∗ 1001 =

10) 1234 ∗ 1001 =

11) 36912 ∗ 1001 =

12) 48653 ∗ 101 =

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1.8

Multiplying Two Numbers just above 100 Let’s prove the trick of how to multiply two numbers above 100.

(100 + 𝑎) ∗ (100 + 𝑏) = 10000 + 100 ∗ 𝑎 + 100 ∗ 𝑏 + 𝑎𝑏 = 10000 + 100(𝑎 + 𝑏) + 𝑎𝑏

Well, that doesn’t look like a trick. But, let’s look at it a bit closer. The 10000 won’t affect any place value below the ten-thousands place and the 100(𝑎 + 𝑏) term won’t affect any place value below the hundreds place. That means that the term 𝑎𝑏 is the only term that affects the tens and ones place. Similarly, the 100(𝑎 + 𝑏) term will affect the thousands and hundreds place (along with carries from the 100(𝑎 + 𝑏) term). Let’s see an example to see this in action.

102 ∗ 103

Step 1: Multiply the amount above 100 for both numbers and make it take up two digits. Multiply 2 and 3 to get 06. Step 2: Add the amount above 100 for both numbers and make it take up two digits. In this problem, add 2 and 3 to get 05. Step 3: Write a 1. Final answer: 10506. Let’s try a more complicated version of this problem.

115 ∗ 109

Step 1: Multiply 15 and 9 to get 135. Write 35 and carry 1. 23 | P a g e

Step 2: Add 15 and 9 along with the carry of 1 to get 25. Step 3: Write a 1. Final answer: 12535. This trick also works for 2 numbers just above any multiple of 100 as long as the multiple is the same.

(ℎ ∗ 100 + 𝑎) ∗ (ℎ ∗ 100 + 𝑏) = ℎ2 ∗ 10000 + ℎ ∗ 100 ∗ 𝑎 + ℎ ∗ 100 ∗ 𝑏 + 𝑎𝑏 = ℎ2 ∗ 10000 + ℎ ∗ 100(𝑎 + 𝑏) + 𝑎𝑏

When comparing this formula to the one for just above 100, we see that ℎ is being multiplied by the term in Step 2 and ℎ2 is the ten thousands place. Let’s do an example.

511 ∗ 506

Step 1: Multiply 11 and 6 to get 66. Write 66. Step 2: Add 11 and 6 to get 17. Multiply this by 5 to get 85. Step 3: calculate 52 to get 25. Final answer: 258566.

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Practice Problem Set 1.8 1) 104 ∗ 102 =

2) 105 ∗ 103 =

3) 103 ∗ 112 =

4) 117 ∗ 104 =

5) 302 ∗ 306 =

6) 114 ∗ 107 =

7) 115 ∗ 113 =

8) 1172 =

9) 107 ∗ 214 =

10) 304 ∗ 317 =

11) 7072 =

12) 212 ∗ 213 =

13) 418 ∗ 111 =

14)921 ∗ 907 =

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1.9

Multiplying Two Numbers just below 100 This trick is very similar to Trick 1.7, from derivation to operations (for obvious reasons). But, since it’s important to know, it’s in this book.

(100 − 𝑎) ∗ (100 − 𝑏) = 100 ∗ 100 − 100 ∗ 𝑎 − 100 ∗ 𝑏 + 𝑎𝑏 = 100 ∗ 100 − 100(𝑎 + 𝑏) + 𝑎𝑏 = 100(100 − 𝑎 − 𝑏) + 𝑎𝑏 = 100([100 − 𝑎] − 𝑏) + 𝑎𝑏 = 100([100 − 𝑏] − 𝑎) + 𝑎𝑏

If you grasped Trick 1.7, the first three steps in this derivation probably make sense. Notice how I didn’t multiply the two integers and rather factored out the 100 from three terms. That allowed me to get to step 4, which seems like a good stopping point. But why go one step further? Look at the original expression. Now, look at what’s in square brackets. See any similarities? The term in the square brackets is one of the numbers you will see on your test. When trying to derive tricks with algebraic terms, it’s important to look back at the original expression to see any similarities or simplifications. Now, what does the last line say? The last two digits of your answer are determined by multiplying the each of the differences between the number and 100. The thousands and hundreds digit are one of the numbers minus the difference between the other number and 100. Let’s now try an example.

26 | P a g e

93 ∗ 96

Step 1: Calculate 𝑎 = 7 and 𝑏 = 4. Step 2: Multiply 7 and 4 to get 28. Step 3: Subtract 4 from 93 (or 7 from 96) to get 89. Final answer: 8928.

88 ∗ 91

Step 1: Calculate 𝑎 = 12 and 𝑏 = 9. Step 2: Multiply 12 and 9 to get 108. Write 08 and carry 1. Step 3: Subtract 12 from 91 (or 9 from 88) to get 79. Add the carry from Step 2 to get 80. Final answer: 8008. Practice Problem Set 1.9 1) 98 ∗ 97 =

2) 97 ∗ 94 =

3) 91 ∗ 96 =

4) 93 ∗ 91 =

5) 96 ∗ 94 =

6) 89 ∗ 92 =

7) 95 ∗ 91 =

8) 88 ∗ 89 =

9) 89 ∗ 95 =

10) 11 ∗ 258 =

11) 88 ∗ 87 =

12) 87 ∗ 94 =

27 | P a g e

1.10

Multiplying 2 Numbers with an Even

Difference This trick is most often seen when two numbers are multiplied that have a difference of two and the middle number ends in 5. Really, it’s using algebra to simplify the difference of two squares as the multiplication of two numbers. For this trick, it’s important to know your squares (which is covered in section 2).

𝑎2 − 𝑏 2 = (𝑎 + 𝑏)(𝑎 − 𝑏)

These problems are given as the right side of the equation and should be converted to a difference of two squares.

22 ∗ 24 = (23 − 1) ∗ (23 + 1) = 232 − 12 = 529 − 1 = 528

To go from step 3 to step 4 will require you to know 232 (and 12 , but if you don’t know that we have bigger problems), but this is much faster than working out the multiplication when you start working with squares. This trick works for any two numbers with an even difference.

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29 ∗ 35 = (32 − 3) ∗ (32 + 3) = 322 − 32 = 1024 − 9 = 1015 Practice Problem Set 1.10 1) 43 ∗ 47 =

2) 35 ∗ 31 =

3) 19 ∗ 21 =

4) 24 ∗ 26 =

5) 89 ∗ 91 =

6) 37 ∗ 43 =

7) 46 ∗ 54 =

8) 36 ∗ 34 =

9) 99 ∗ 103 =

10) 18 ∗ 22 =

11) 41 ∗ 51 =

12) 214 ∗ 111 =

13) 48 ∗ 56 =

14) 33 ∗ 27 =

29 | P a g e

1.11

Multiplying 2 Numbers with the Same Tens

Digit and Ones Digits that Add to 10 I find these problems to be a special case of Trick 1.9. Even though trick 1.9 is fast, this is faster. Let’s see it. Let a represent the tens digit of both numbers, let 𝑏 represent the ones digit of the first number, and let 10 − 𝑏 represent the ones digit of the second number. Then:

(10𝑎 + 𝑏) ∗ (10𝑎 + [10 − 𝑏]) = 100𝑎2 + 10𝑎(10 − 𝑏) + 10𝑎𝑏 + 𝑏(10 − 𝑏) = 100𝑎2 + 100𝑎 − 10𝑎𝑏 + 10𝑎𝑏 + 𝑏(10 − 𝑏) = 100𝑎(𝑎 + 1) + 𝑏(10 − 𝑏)

Translating the last line to words, we see that the last two digits of the answer is simply the product of the two ones digits and the hundreds digit of the answer is the tens digit multiplied by one more than itself.

23 ∗ 27

Step 1: Multiply the ones digits and make it take up two place values. The product of 3 and 7 is 21, so write 21. Step 2: Multiply the tens digit by one more than itself. The product of 2 and 2+1=3 is 6, so write 6. Final answer: 621.

30 | P a g e

121 ∗ 129

Step 1: multiply the ones digits and make it take up two place values. The product of 1 and 9 is 9, so write 09. Step 2: multiply the tens digit by one more than itself. The product of 12 and 12+1=13 is 156, so write 156. Final answer: 15609. Practice Problem Set 1.11 1) 18 ∗ 12 =

2) 84 ∗ 86 =

3) 79 ∗ 71 =

4) 37 ∗ 33 =

5) 43 ∗ 22 =

6) 24 ∗ 26 =

7) 109 ∗ 101 =

8) 162 ∗ 168 =

9) 117 ∗ 113 =

10) 178 ∗ 172 =

31 | P a g e

1.12

FOIL This, hands down, is the greatest and most useful trick in this book. Put in one sentence, this trick makes it possible to multiply any two numbers either 2 or 3 digits in your head. Now you must be asking, “Why mention all the tricks before this if we have FOIL?” Well, every trick before this, based on numerous students as well as personal experience, is faster than FOIL. There are other multiplication tricks you can find online or in other sources, but I found those to be slower or equal to FOIL in terms of time. Let’s see what FOIL says. Let a represent the tens digit and b the ones digit of the first number and c represent the tens digit and d the ones digit of the second number. That means that the numbers can be represented as the following:

"𝑎𝑏" 𝑎𝑠 10𝑎 + 𝑏 𝑎𝑛𝑑 "𝑐𝑑" 𝑎𝑠 10𝑐 + 𝑑

Now, if we want to multiply the two numbers, we can expand it as so:

(10𝑎 + 𝑏)× (10𝑐 + 𝑑) = 100𝑎𝑐 + 10𝑎𝑑 + 10𝑏𝑐 + 𝑏𝑑 = 100(𝑎𝑐) + 10(𝑎𝑑 + 𝑏𝑐) + 𝑏𝑑

What does this mean?

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Step 1: The ones digit of the answer is simply the product of the ones digits of the two numbers multiplied. Step 2: The tens digit of the answer is simply the sum of the product of the Outer digits plus the product of the Inner digits. Step 3: The hundreds digit is simply the product of the tens digits of the two numbers multiplied. All the while, follow carries like every other trick before this. Let’s see FOIL in action. 28 ∗ 52

Step 1: Calculate the ones digit. 8 ∗ 2 = 16; write 6 as ones digit and carry 1. Step 2: Multiply the outer and inner digits to get 2 ∗ 2 = 4 and 8 ∗ 5 = 40. Now add the two products and the carry to get 45. Write 5 as the tens digit and carry 4. Step 3: Multiply the tens digits. 2 ∗ 5 = 10; add the carry to get 14. Final answer: 1456. 18 ∗ 21

Step 1: Calculate the ones digit. 8 ∗ 1 = 8; write 8 as ones digit. Step 2: Multiply the outer and inner digits to get 1 ∗ 1 = 1 and 8 ∗ 2 = 16. Now add the two products and the carry to get 17. Write 7 as the tens digit and carry 1. Step 3: Multiply the tens digits. 1 ∗ 2 = 2; add the carry to get 3. Final answer: 378. This also works for 3 digit multiplication; simply group the tens and hundreds digits as one number and FOIL. There are more calculations that happen so it takes longer to solve, but with practice this multiplication becomes easier.

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273 ∗ 156

Step 1: Calculate the ones digit. 3 ∗ 6 = 18; write 8 as ones digit and carry 1. Step 2: Group the hundreds and tens digit of each number. Multiply the outer and inner numbers to get 27 ∗ 6 = 120 + 42 = 162 and 15 ∗ 3 = 45. Now add the two products and the carry to get 208. Write 8 as the tens digit and carry 20. Step 3: Multiply the outer numbers (use a multiplication trick or FOIL again). 27 ∗ 15 = 405; add the carry to get 425. Final answer: 42588. Practice Problem Set 1.12 1) 44 ∗ 14 =

2) 37 ∗ 76 =

3) 28 ∗ 86 =

4) 83 ∗ 36 =

5) 72 ∗ 46 =

6) 89 ∗ 32 =

7) 76 ∗ 88 =

8) 82 ∗ 34 =

9) 79 ∗ 49 =

10) 58 ∗ 47 =

11) 97 ∗ 78 =

12) 83 ∗ 36 =

13) 61 ∗ 52 =

14) 113 ∗ 124 =

15) 117 ∗ 139 =

16) 131 ∗ 124 =

17) 211 ∗ 315 =

18) 317 ∗ 532 =

19) 453 ∗ 862 =

20) 232 ∗ 874 = 34 | P a g e

2

Squares

2.1

Squares This is a list of squares you should memorize for Number Sense. 112=121

192=361

272=729

352=1225

432=1849

122=144

202=400

282=784

362=1296

442=1936

132=169

212=441

292=841

372=1369

452=2025

142=196

222=484

302=900

382=1444

462=2116

152=225

232=529

312=961

392=1521

472=2209

162=256

242=576

322=1024

402=1600

482=2304

172=289

252=625

332=1089

412=1681

492=2401

182=324

262=676

342=1156

422=1764

502=2500

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2.2

Squares Ending in 5 There’s a neat little trick to find the square of any number ending in 5. (10𝑎 + 5)2 = (10𝑎 + 5) ∗ (10𝑎 + 5) = 100𝑎2 + 50𝑎 + 50𝑎 + 25 = 100𝑎2 + 100𝑎 + 25 = 100 ∗ 𝑎 ∗ (𝑎 + 1) + 25

The algebra indicates that the last two numbers of the square is a 25. The other digits are simply the tens digit multiplied by one more than itself. For example: 352

Step 1: Write 25. Step 2: Multiply 3 by 3+1=4 to get 12. Final answer: 1225. Practice Problem Set 2.2 1) 252 =

2) 352 =

3) 452 =

4) 552 =

5) 652 =

6) 752 =

7) 852 =

8) 952 =

9) 1052 =

10) 1152 =

11) 1252 =

12) 1352 =

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2.3

Sum of Squares in the form a2+(3a)2 The sum of squares in this form is seen on basically every other Number Sense test. It can be solved using simple algebra. 𝑎2 + (3𝑎)2 = 𝑎2 + 9𝑎2 = 10𝑎2 What this means is that you only have to calculate the smaller square and put a 0 after the term. Let’s see an example. 82 + 242 = 10 ∗ 82 = 10 ∗ 64 = 640 Although significantly less common, questions may arise that are in the form of 𝑎2 + (2𝑎)2 , 𝑎2 + (7𝑎)2 , or 𝑎2 + (10𝑎)2 . Each of these cases has a simple trick that I’ll let you derive.

Practice Problem Set 2.3 1) 92 + 272 =

2) 42 + 122 =

3) 132 + 392 =

4) 172 + 512 =

5) 92 + 182 =

6) 872 + 292 =

7) 72 + 492 =

8) 32 + 302 =

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2.4

Sum of Two Consecutive Squares The sum of consecutive squares can often be computed manually by high-scoring individuals. However, this often takes upwards of 10 seconds and breaks tempo. This trick simplifies the process and makes sure you don’t have to calculate any square.

𝑎2 + (𝑎 + 1)2 = 𝑎2 + 𝑎2 + 2𝑎 + 1 = 2𝑎2 + 2𝑎 + 1 = 2𝑎(𝑎 + 1) + 1

Let’s compare the last line to the first. Both numbers in the original expression are present in the final one, multiplied. In words, this trick says, “The sum of consecutive squares is twice the product of the two base numbers plus one.” All you have to do is multiply the two numbers you see on the problem, double it, and add 1. Example time!

252 + 262 = 2 ∗ 25 ∗ 26 + 1 = 13 ∗ 100 + 1 = 1301

More often than not one of the numbers will end in 5, making the multiplication all the more easier (refer to trick 1.5).

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Practice Problem Set 2.4 1) 142 + 152 =

2) 172 + 182 =

3) 202 + 212 =

4) 262 + 252 =

5) 362 + 352 =

6) 742 + 752 =

7) 442 + 452 =

8) 502 + 512 =

9) 132 + 122 =

10) 602 + 592 =

11) 392 + 402 =

12) 1142 + 1152 =

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2.5

Sum of Squares when the Outer Digits add to

10 and the Inner Digits are 1 Apart This sum of squares trick has a complicated derivation (as compared to other tricks) but comes up frequently on tests. Usually, this problem can be identified because the squares are ridiculous by themselves to compute. It is important to note that order matters in this trick as the inner and outer digits would change if the order of the squares change. First, let a represent the tens digit and b the ones digit of the first number. That means that the first number can be represented as the following:

10𝑎 + 𝑏

Now, if the inner digits differ by 1, then the tens digit of the second number is 1 less than the ones digit of the first number. Similarly, if the outer digits add to 10, then the ones digit of the second number is 10 less than the tens digit of the first number. This can be written as the following: 10(𝑏 − 1) + (10 − 𝑎)

Now, we’ll use algebra to square both expressions and add the terms. (10𝑎 + 𝑏)2 + [10(𝑏 − 1) + (10 − 𝑎)]2 = (10𝑎 + 𝑏)2 + (10𝑏 − 10 + 10 − 𝑎)2 = (10𝑎 + 𝑏)2 + (10𝑏 − 𝑎)2 = 100𝑎2 + 20𝑎𝑏 + 𝑏 2 + 𝑎2 − 20𝑎𝑏 + 100𝑏2 = 101𝑎2 + 101𝑏2 = 101(𝑎2 + 𝑏 2 ) 40 | P a g e

As can be seen by the final expression, we will need to multiply by 101 (refer to 1.6 if you need a refresher) and calculate squares of individual digits, more specifically the sum of the squares of the digits of the first number. As said before, order matters; the first number must have a ones digit one greater than the tens digit of the second number. Let’s see an example.

572 + 652

Step 1: make sure that the outer digits add to 10 (5+5) and the inner digits have a difference of 1 (7-6) Step 2: calculate the sum of the squares of the digits of the first number (52 + 72 = 25 + 49 = 74) Step 3: multiply this sum by 101 to get the final answer (7474) It is crucial to note that this trick will not work for 562 + 752 because 6 − 7 = −1. Almost every time the question asks the sum of two large squares, the numbers follow this pattern. On a rare occasion, you may need to reorder the two squares to see the pattern. For example, if this question read 652 + 572 , you would need to switch the position of the two squares before continuing with Step 1.

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Practice Problem Set 2.5 1) 142 + 392 =

2) 172 + 692 =

3) 222 + 182 =

4) 262 + 582 =

5) 362 + 572 =

6) 742 + 332 =

7) 442 + 362 =

8) 582 + 752 =

9) 132 + 292 =

10) 642 + 672 =

11) 392 + 872 =

12) 312 + 942 =

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2.6

Approximations of Exponents Approximating squares and cubes is essential for Number Sense. One thing to look for is a number you know the square to that you can obtain by light rounding. Beware that excessive rounding with base numbers should be avoided because when the number is squared, the minor change becomes major. Sometimes using properties of exponents help make an approximation question manageable. One that comes into use is:

𝑎𝑛 ∗ 𝑏 𝑛 = (𝑎 ∗ 𝑏)𝑛

Hopefully these practice problems help you build speed and see how to manipulate the numbers. Practice Problem Set 2.6 1) 453 ≈

2) 182 ÷ 93 ∗ 36 ≈

3) 272 ÷ 92 ∗ 182 ≈

4) 124 ÷ 83 ∗ 42 ≈

5) 22 ∗ 24 ∗ 26 ≈

6) 153 ÷ 33 ∗ 93 ≈

7) 134 ≈

8) 124 ÷ 63 ∗ 32 ≈

9) 74 ∗ 52 ÷ 36 ≈

10) 244 ≈

11) 502 ∗ 403 ÷ 304 ≈

12) 37 ∗ 38 ∗ 39 ∗ 40 ≈

43 | P a g e

3

GCD and LCM

3.1

How to find GCD The concept of Greatest Common Divisor (or Greatest Common Factor) and Least Common Multiple will require knowledge in prime numbers (covered in Chapter 7). Most students in middle school and higher know the conventional way to calculate the two values: compare the prime factorization of the two numbers. Though accurate, this method isn’t efficient because it requires a student to find the complete prime factorization of both numbers, compare them, and multiply the numbers that are common. Rather, there is simpler way to find GCD when working with 2 numbers. It can be shown that the GCD of two numbers divides their difference. That means that when you find the positive difference of two numbers, the GCD must be a factor of that number. Let 𝑑 be the GCD of 𝑎 and 𝑏, which are defined as positive integers (numbers with no fractional parts) and 𝑎 > 𝑏. Also, let 𝑘 and 𝑙 be positive integers. It can then be written that:

𝑑 = 𝐺𝐶𝐷(𝑎, 𝑏) 𝑎 =𝑑∗𝑘 𝑏 =𝑑∗𝑙

Now, let’s subtract both numbers.

𝑎−𝑏 =𝑑∗𝑘−𝑑∗𝑙 = 𝑑(𝑘 − 𝑙)

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We see that the GCD is a factor of the positive difference. That means that instead of using the traditional way of finding GCD, we can subtract the numbers and use this number to find the GCD. After finding the difference, use simple trial and error starting from the largest factor to find the GCD. Although still a bit of work, this is much easier and faster. Usually the difference doesn’t have many factors and the GCD becomes obvious. Let’s see some examples.

𝐺𝐶𝐷(34, 85) 85 − 34 = 51 51 = 3 ∗ 17 𝑁𝑜𝑡𝑖𝑐𝑒 𝑡ℎ𝑎𝑡 34 = 17 ∗ 2 𝑎𝑛𝑑 85 = 17 ∗ 5 𝐺𝐶𝐷 = 17

I used the prime factorization of 51 to realize that both numbers are a multiple of 17.

𝐺𝐶𝐷(69,115) 115 − 69 = 46 46 = 2 ∗ 23 𝑁𝑜𝑡𝑖𝑐𝑒 𝑡ℎ𝑎𝑡 69 = 23 ∗ 3 𝑎𝑛𝑑 115 = 23 ∗ 5 𝐺𝐶𝐷 = 23

Once again, I used the prime factorization of 46 to analyze the original two numbers and determine the GCD.

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Practice Problem Set 3.1 1) 𝐺𝐶𝐷(12, 16) =

2) 𝐺𝐶𝐷(72, 135) =

3) 𝐺𝐶𝐷(30, 36) =

4) 𝐺𝐶𝐷(36, 90) =

5) 𝐺𝐶𝐷(18, 25) =

6) 𝐺𝐶𝐷(68, 119) =

7) 𝐺𝐶𝐷(32, 50) =

8) 𝐺𝐶𝐷(24, 42) =

9) 𝐺𝐶𝐷(343, 98) =

10) 𝐺𝐶𝐷(16, 36) =

11) 𝐺𝐶𝐷(26, 65) =

12) 𝐺𝐶𝐷(21, 70) =

13) 𝐺𝐶𝐷(24, 30, 42) =

14) 𝐺𝐶𝐷(96, 168) =

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3.2

How to find LCM Another interesting property of GCD and LCM is the following:

𝐺𝐶𝐷(𝑎, 𝑏) ∗ 𝐿𝐶𝑀(𝑎, 𝑏) = 𝑎 ∗ 𝑏

I’ll explain this formula with a Venn diagram.

𝑎

𝑎 𝐺𝐶𝐷

𝑏

𝐺𝐶𝐷

𝑏 𝐺𝐶𝐷

As seen in the diagram above, the two circles represent the two numbers. The intersection would be the GCD as it divides into both numbers. What’s left is each number divided by the GCD. Now, to find the LCM, we need to get the smallest number that divides both numbers. Pictorially, we need to cover each part of the diagram once.

𝐿𝐶𝑀 =

𝑎 𝑏 ∗ 𝐺𝐶𝐷 ∗ 𝐺𝐶𝐷 𝐺𝐶𝐷 =

𝑎∗𝑏 𝐺𝐶𝐷

𝐿𝐶𝑀 ∗ 𝐺𝐶𝐷 = 𝑎 ∗ 𝑏

47 | P a g e

To find LCM, we can find GCD using trick 3.1, find the product of the two numbers, and then divide the latter by the former (although it is advisable to divide before to multiply, so divide one of the numbers by the GCD and then multiply by the other number). Let’s use an example from 3.1. It was determined that:

𝐺𝐶𝐷(34, 85) = 17 𝐿𝐶𝑀(34,85) = =

34 ∗ 85 𝐺𝐶𝐷

34 ∗ 85 17

= 2 ∗ 85 𝐿𝐶𝑀 (34,85) = 170

Additionally, questions sometimes arise that give you 3 of the 4 numbers in the last equation and ask for the fourth one. Knowing this trick, such questions become a breeze.

𝐺𝑖𝑣𝑒𝑛 𝐺𝐶𝐷(24, 𝑥) = 8 𝑎𝑛𝑑 𝐿𝐶𝑀(24, 𝑥) = 120, 𝑓𝑖𝑛𝑑 𝑥:

To find x, simply use the expression:

𝑥=

120 ∗ 8 24

=5∗8 = 40

48 | P a g e

Practice Problem Set 3.2 1)𝐿𝐶𝑀(12, 16) =

2) 𝐿𝐶𝑀(72, 135) =

3) 𝐿𝐶𝑀(30, 36) =

4) 𝐿𝐶𝑀(36, 90) =

5) 𝐿𝐶𝑀(18, 25) =

6) 𝐿𝐶𝑀(68, 119) =

7) 𝐿𝐶𝑀(32, 50) =

8) 𝐿𝐶𝑀(24, 42) =

9) 𝐿𝐶𝑀(343, 98) =

10) 𝐿𝐶𝑀(16, 36) =

11) 𝐿𝐶𝑀(24, 30, 42) =

12) 𝐿𝐶𝑀(96, 168) =

13) 𝐺𝑖𝑣𝑒𝑛 𝐺𝐶𝐷(26, 𝑥) = 13 𝑎𝑛𝑑 𝐿𝐶𝑀(26, 𝑥) = 130, 𝑓𝑖𝑛𝑑 𝑥: 14) 𝐺𝑖𝑣𝑒𝑛 𝐺𝐶𝐷(21, 𝑥) = 7 𝑎𝑛𝑑 𝐿𝐶𝑀(21, 𝑥) = 210, 𝑓𝑖𝑛𝑑 𝑥:

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4

Remainder

4.1

Remainder with Division Number Sense usually asks you to find the remainder when dividing by 2 numbers: 9 and 11. The proofs of each trick involves modular arithmetic and isn’t necessary to know. Instead, let’s focus on the trick itself. The remainder when a number is divided by 9 is the same as the remainder when the sum of the digits of the number is divided by 9. That means you can add the digits of the number and divide that by 9 rather than the original number. Let’s see an example.

24371 ÷ 9 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓 2 + 4 + 3 + 7 + 1 = 17 17 ÷ 9 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓 8 𝐹𝑖𝑛𝑎𝑙 𝑎𝑛𝑠𝑤𝑒𝑟: 8

However, there’s a way to make the thought process easier. Every time you get 9 in your addition, make it 0. In the previous example, you would likely start left to right. 2 + 4 = 6, and 6 + 3 = 9 Now that you have a multiple of 9, make it 0. Moving forward, 7 + 1 = 8 so your final remainder is 8. To find the remainder when dividing by 11, we need to employ a slightly more complex trick. Step 1: start at the ones digit and add every other digit reading backwards (right to left). This will be called 𝑆𝑢𝑚 𝑂𝑛𝑒. Step 2: start at the tens digit and add every other digit reading backwards (right to left). This will be called 𝑆𝑢𝑚 𝑇𝑤𝑜. Step 3: Determine 𝑆𝑢𝑚 𝑂𝑛𝑒 – 𝑆𝑢𝑚 𝑇𝑤𝑜. If this difference is between 0 and 10, write the answer. If the difference is less than 0, add multiples of 11 until you get into this range. If the difference is 50 | P a g e

greater than 10, subtract multiples of 11 until you get into this range. The reason you need to get into the range of 0 to 10 is because when dividing an integer by 11, the only possible remainders are 0 to 10.

296834 ÷ 11 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓 𝑆𝑢𝑚 𝑂𝑛𝑒: 4 + 8 + 9 = 21 𝑆𝑢𝑚 𝑇𝑤𝑜: 3 + 6 + 2 = 11 21 − 11 = 10 𝐹𝑖𝑛𝑎𝑙 𝑎𝑛𝑠𝑤𝑒𝑟: 10

738492 ÷ 11 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓 𝑆𝑢𝑚 𝑂𝑛𝑒: 2 + 4 + 3 = 9 𝑆𝑢𝑚 𝑇𝑤𝑜: 9 + 8 + 7 = 24 9 − 24 = −15 −15 + 11 = −4 −4 + 11 = 7 𝐹𝑖𝑛𝑎𝑙 𝑎𝑛𝑠𝑤𝑒𝑟: 7

51 | P a g e

Practice Problem Set 4.1 1) 717 ÷ 9 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓

2) 717 ÷ 11 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓

3) 423 ÷ 9 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓

4) 423 ÷ 11 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓

5) 276 ÷ 9 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓

6) 276 ÷ 11 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓

7) 3468 ÷ 9 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓

8) 3468 ÷ 11 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓

9) 5863 ÷ 9 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓

10) 5863 ÷ 11 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓

11) 3569 ÷ 9 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓

12) 3569 ÷ 11 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓

13) 7475 ÷ 9 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓

14) 7475 ÷ 11 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓

15) 47347 ÷ 9 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓

16) 47347 ÷ 11 ℎ𝑎𝑠 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓

17) 24842 ÷ 9 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓

18) 24842 ÷ 11 ℎ𝑎𝑠 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓

19) 93757 ÷ 9 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓

20) 93757 ÷ 11 ℎ𝑎𝑠 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓

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4.2

Remainder with Operations This trick is quite useful to drastically simplify questions asking about remainder. When trying to find the remainder of an expression divided by a divisor, instead of using the original terms in the expression, you can use the remainder when each individual term is divided by the divisor. What does that mean? Step 1: Find the remainder of each term when divided by the divisor. Step 2: Plug in each remainder for the original number and keep the operations the same. Step 3: Calculate the final remainder.

(123 + 17 ∗ 4) ÷ 12 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓 123 ÷ 12 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓 3 17 ÷ 12 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓 5 4 ÷ 12 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓 4 (3 + 5 ∗ 4) ÷ 12 = 23 ÷ 12 23 ÷ 12 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓 11 𝐹𝑖𝑛𝑎𝑙 𝑎𝑛𝑠𝑤𝑒𝑟: 11

Note that the quotient of the original expression and the final expression are not the same, only the remainder (which is what we’re looking for).

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Practice Problem Set 4.2 1) (45 ∗ 12 − 13) ÷ 7 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓: 2)(132 + 12 ∗ 143) ÷ 3 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓: 3)(171 ∗ 34 − 54) ÷ 8 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓: 4) (164 + 26 ∗ 34) ÷ 11 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓: 5) (19 ∗ 12 + 13) ÷ 7 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓: 6) ([74 − 29] ∗ 18) ÷ 9 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓: 7) (27 + 223 − 13 ∗ 3) ÷ 5 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓: 8) (765 + 432 ∗ 11) ÷ 6 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓: 9) (2413 ∗ 19 + 36) ÷ 12 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓: 10) (27 + 34 ∗ 727) ÷ 8 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓:

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4.3

Remainder with Exponents Remainders with exponents are quite similar to Trick 4.2. Usually a late fourth column question, these questions can be extremely simple or fairly difficult. The entire trick is attempting to establish a pattern. There are 2 common, easy patterns that I’ll mention, but the others will require you to do multiplication. As a side note, you should be familiar with the Laws of Exponents (Trick 5.4). When determining the remainder of an expression in the form of 𝑎𝑛 ÷ 𝑏: If 𝑎 > 𝑏, first find the remainder of 𝑎 ÷ 𝑏 (let this be 𝑟). The remainder of 𝑟 𝑛 ÷ 𝑏 will have the same remainder as 𝑎𝑛 ÷ 𝑏. This means that you can plug in the remainder rather than the original number and still get the same remainder (just like in Trick 4.2). Pattern 1: When the base of the exponent is one greater than the dividend.

𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑤ℎ𝑒𝑛 126 ÷ 11 12 ÷ 11 ℎ𝑎𝑠 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 1 1 6 ÷ 11 ℎ𝑎𝑠 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 1 𝐹𝑖𝑛𝑎𝑙 𝑎𝑛𝑠𝑤𝑒𝑟: 1

In this pattern, the answer will always be 1. Pattern 2: When the base of the exponent is one less than the dividend. This will take the form of:

𝑎2 ÷ (𝑎 + 1)

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We know for a fact that 𝑎 divided by 𝑎 + 1 will have a remainder of 𝑎 since it will not divide into a number one greater than itself. We also know that:

𝑎2 − 1 (𝑎 + 1) =

(𝑎 + 1)(𝑎 − 1) (𝑎 + 1) =𝑎−1

We can see that 𝑎2 − 1 is divisible by 𝑎 + 1, so 𝑎2 will have a remainder of 1 when divided by 𝑎 + 1.

173 ÷ 18 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓 17 ÷ 18 ℎ𝑎𝑠 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 17 (𝑑𝑜𝑒𝑠 𝑛𝑜 𝑔𝑜𝑜𝑑) 172 ÷ 18 ℎ𝑎𝑠 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 1 (289 ÷ 18 = 16 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 1) 172 ∗ 17 ÷ 18 ℎ𝑎𝑠 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑎𝑠 1 ∗ 17 ÷ 18 1 ∗ 17 ÷ 18 ℎ𝑎𝑠 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 17 𝐹𝑖𝑛𝑎𝑙 𝑎𝑛𝑠𝑤𝑒𝑟: 17

174 ÷ 18 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓 (172 )2 ÷ 18 ℎ𝑎𝑠 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑎𝑠 (1)2 ÷ 18 12 ÷ 18 ℎ𝑎𝑠 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 1 𝐹𝑖𝑛𝑎𝑙 𝑎𝑛𝑠𝑤𝑒𝑟: 1

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Looking at the pattern, if the exponent is even, the remainder will be 1. If the exponent is odd, the remainder will be the base number (in the example it’s 17). Else: If neither of the two conditions are met, keep plugging in the remainder to try to either get a pattern or get to the answer. For example:

1312 ÷ 11 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓 1312 ÷ 11 ℎ𝑎𝑠 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑎𝑠 212 ÷ 11 212 ÷ 11 = (24 )3 ÷ 11 = 163 ÷ 11 163 ÷ 11 ℎ𝑎𝑠 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑎𝑠 53 ÷ 11 53 ÷ 11 = 125 ÷ 11 = 11 𝑤𝑖𝑡ℎ 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 4 𝐹𝑖𝑛𝑎𝑙 𝑎𝑛𝑠𝑤𝑒𝑟: 4

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Practice Problem Set 4.3 1) 1211 ÷ 13 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓: 2)184 ÷ 17 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓: 3)2122 ÷ 22 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓: 4) 148 ÷ 15 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓: 5) 139 ÷ 12 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓: 6) 910 ÷ 11 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓: 7) 67 ÷ 8 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓: 8) 119 ÷ 7 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓: 9) 1413 ÷ 12 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓: 10) 1516 ÷ 8 ℎ𝑎𝑠 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓:

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5

Higher Order Exponents

5.1

Cubes Here are a list of cubes you need to memorize.

5.2

13=1

43=64

73=343

103=1000

133=2197

23=8

53=125

83=512

113=1331

143=2744

33=27

63=216

93=729

123=1728

153=3375

Powers of Special Numbers You also want to memorize these numbers. Note that oftentimes these powers will be combined in approximation questions. 21=2

31=3

51=5

π1≈3.1

e1≈2.7

ϕ1≈1.6

22=4

32=9

52=25

π 2≈9.9

e2≈7.5

ϕ 2≈2.6

23=8

33=27

53=125

π 3≈31

e3≈20

ϕ 3≈4.2

24=16

34=81

54=625

π 4≈97

e4≈55

ϕ 4≈6.8

25=32

35=243

55=3125

π 5≈306

e5≈150

ϕ 5≈11

26=64

36=729

56=15625 π 6≈960

e6≈400

ϕ 6≈18

27=128

π e≈22.5

eπ ≈23

28=256 29=512 210=1024 Oftentimes, approximating π2 to be 10 and π4 to be 100 works if you subtract a little bit at the end. It’s also convenient to know e3 is 20 and work from there.

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5.3

Difference of Cubes 1 Apart Usually the difference of two cubes comes late in the fourth column. As with numerous other tricks, finding the difference of cubes 1 apart will require simple algebra. It’s important to know that if you encounter new variations of questions that you want to derive a trick for, it is best to try to start with algebra and work to the original numbers if possible. Let’s take a look at this derivation.

(𝑎 + 1)3 − 𝑎3 = 𝑎3 + 3𝑎2 + 3𝑎 + 1 − 𝑎3 = 3𝑎2 + 3𝑎 + 1 = 3𝑎(𝑎 + 1) + 1

Looking at the final expression, we can conclude that the positive difference of cubes one apart is one more than the product of the two bases tripled. For example:

163 − 153 = 3 ∗ 16 ∗ 15 + 1 = 3 ∗ 8 ∗ 30 + 1 = 8 ∗ 90 + 1 = 721

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Practice Problem Set 5.3 1) 43 − 33 =

2) 83 − 73 =

3) 73 − 63 =

4) 93 − 83 =

5) 123 − 113 =

6) 93 − 103 =

7) 173 − 163 =

8) 153 − 143 =

9) 303 − 293 =

10) 413 − 403 =

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5.4

Laws of Exponents It’s crucial to know how to work with exponents to simplify a difficult expression. First know that 𝑎0 = 1. When the base is the same, the following statements are true:

𝑎 𝑥 ∗ 𝑎 𝑦 = 𝑎 𝑥+𝑦 (𝑎 𝑥 )𝑦 = 𝑎 𝑥∗𝑦

However, not every question will have numbers with the same base. When this is the case, know the following statements:

𝑎 𝑥 ∗ 𝑏 𝑥 = (𝑎𝑏)𝑥 𝑎𝑥 𝑎 𝑥 =( ) 𝑏𝑥 𝑏 Most every time you must manipulate the numbers given to fit one of the expressions above. For this, use the following fact:

𝑎 𝑥 = 𝑎 ∗ 𝑎 𝑥−1

Let’s work an example now.

124 ÷ 63 = 12 ∗ 123 ÷ 63 = 12 ∗ (12 ÷ 6)3 = 12 ∗ 23 = 12 ∗ 8 = 96 62 | P a g e

Practice Problem Set 5.4 1) 273 ∗ 94 ÷ 316 =

2) 144 ÷ 73 ÷ 28 =

3) 284 ÷ 145 ∗ 72 =

4) 325 ÷ 166 =

5) 523 ÷ 136 ∗ 263 =

6)6254 ÷ 256 ∗ 5 =

7) 185 ÷ 94 =

8) 364 ÷ 184 =

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6

Roots

6.1

Square Roots Square roots are the opposite of squares (much like subtraction is the opposite of addition and division that of multiplication). Number Sense will require you to memorize the following square roots to the thousandths place.

√2 = 1.414

√5 = 2.236

√8 = 2.828

√3 = 1.732

√6 = 2.45

√9 = 3

√4 = 2

√7 = 2.645

√10 = 3.16

There are two types of questions asked: truncate and round. When you truncate, you always round down, no matter if the next place value is 1 or 9. In contrast, when you round, you look at whether the next place value is less than 0.5 or not.

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6.2

Nested Square Roots A square root is a number raised to the ½ power. Nested square roots are intimidating and seemingly complex, but they are simple if you know how to work with the radicals. If given as an expression (as compared to an equation) and asked to find what it equals, work inside-out. For example:

√√16 = √4 = 2 √4 + √25 = √4 + 5 = √9 = 3

However, if given as an equation, you may need to square both sides to de-nest the nested square root. Let’s find X in the following examples.

√9 + √𝑋 = 4 9 + √𝑋 = 16 √𝑋 = 7 𝑋 = 49

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√16 − √12√4 − 𝑥 = 2 16 − √12√4 − 𝑥 = 4 12 = √12√4 − 𝑥 144 = 12√4 − 𝑥 12 = √4 − 𝑥 144 = 4 − 𝑥 𝑥 = −140 Practice Problem Set 6.2

1) √√1296 =

2) √√√256 =

3) √17 + √2 ∗ 32 =

4) √4 ∗ √2 ∗ 8 − √60 − √121 =

5) √24 + 3 ∗ √3𝑋 − 8 = 5. 𝐹𝑖𝑛𝑑 𝑋: 6) √33 − √4 − 𝑋 = 4. 𝐹𝑖𝑛𝑑 𝑋: 7) √14 + √2𝑋 = 5. 𝐹𝑖𝑛𝑑 𝑋:

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6.3

Other Roots Numbers can be raised to other fractional exponents as well. The 1/3 power is called the cube root. These fractional exponents are often simplified using Laws of Exponents (Trick 5.4).

2 83

2

2

= (23 )3 = 23∗3 = 22 = 4

3

3

3

10245 = (210 )5 = 210∗5 = 26 = 64 =

3 5 (4 )5

=

3 5∗ 4 5

= 43 = 64

Note how in the second example there are two paths, both of which get to the correct answer. If you can recognize that 1024 is an even power of 2 and thus is a power of 4, the calculations become a bit simpler and quicker. Practice Problem Set 6.3 2

−1

1) 273 =

2) 64 3 = −2

3

3) 6254 =

4) 125 3 =

2

5

5) 3433 = −2 3

7) 1728 3

9) 2435 =

6) 40966 = =

3 2

8) 169 = 3

10) 21877 =

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6.4

Approximations of Roots The majority of questions with roots on a Number Sense test deal with approximations. There are a handful of strategies to quickly and accurately estimate roots. Step 1: Look for a perfect square that is close to the number under the radical.

√1230 ≈ √1225 = 35

Step 2: Employ the following fact:

√𝑥 ∗ √𝑥 = 𝑥

For example:

√134 ∗ √136 ≈ √135 ∗ √135 = 135

√345 ∗ √1400 ≈ √350 ∗ (√4 ∗ √350) = 2 ∗ 350 = 700

Step 3: For long square roots, only worry about the most important digits; that is, ignore the digits in the tens and ones place. This step is best explained with an example.

√14541 ≈ √14500 = √145 ∗ 100 = 10 ∗ √145

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Note how we can round the tens and ones digits such that we can factor out a 100. Now, we just worry about the 145.

10 ∗ √145 ≈ 10 ∗ √144 = 120 𝑂𝑢𝑟 𝑎𝑛𝑠𝑤𝑒𝑟: 120

Note that the actual answer is 120.58 and the range of accepted answers is 115-127. Let’s try another example.

√1233321 ≈ √1233300 = 10 ∗ √12333

Since the radical is still fairly large, we can factor out one more 10 and not affect the value too much.

10 ∗ √12333 ≈ 10 ∗ √12300 = 100 ∗ √123 100 ∗ √123 ≈ 100 ∗ √121 = 1100 𝑂𝑢𝑟 𝑎𝑛𝑠𝑤𝑒𝑟: 1100

The correct answer is 1110.54 with a range of 1055-1166. In this example, since we rounded 1233321 down to 1210000, it wouldn’t do much harm to add 5 or 10 to our final answer. Either way, we would fall in the range.

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Practice Problem Set 6.4 1) √363 ∗ 189 ≈

2) √120 ∗ √122 ≈

3) √52113 ≈

4) √65748 ≈

5) √1491625 ≈

6) √1730 ∗ √167 ∗ 11 ≈

3

7) √3380 ∗ √223 ∗ 16 ≈ 4

3

3

8) √327 ∗ √397 ∗ √487 ≈

9) √14643 ∗ √1329 ∗ √120 ≈

10) √86420 ≈

11) √262626 ≈

12) √346598 ≈

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7

Primes and Divisors

7.1

Primes A prime number is one that is divisible only by 1 and itself. Every integer greater than 1 can be written as the product of 1 or more primes; this is called a number’s prime factorization. Since prime numbers are essential in competition math (including Number Sense), here is a list of primes under 100 you should be familiar with.

2

13

31

53

73

3

17

37

59

79

5

19

41

61

83

7

23

43

67

89

11

29

47

71

97

Note that neither 0 nor 1 is prime. Also, 2 is the only even prime number as every even integer greater than 2 will be divisible by 2.

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7.2

Number of Positive Integral Divisors We use the prime factorization of a number to determine the number of positive integral divisors (factors) it has. It’s best shown with an example.

𝐻𝑜𝑤 𝑚𝑎𝑛𝑦 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑑𝑖𝑣𝑖𝑠𝑜𝑟𝑠 𝑑𝑜𝑒𝑠 12 ℎ𝑎𝑣𝑒? 12 = 22 ∗ 3

Using the prime factorization, we can see that any combination of 2𝑥 from 𝑥 = 0 to 𝑥 = 2 and 3𝑥 from 𝑥 = 0 to 𝑥 = 1 will create a factor of 12.

20 ∗ 30 = 1

20 ∗ 31 = 3

21 ∗ 30 = 2

21 ∗ 31 = 6

22 ∗ 30 = 4

22 ∗ 31 = 12

In fact, we can derive a formula based on the prime factorization of a number that tells us the number of factors the number has. Since 31 had 2 possible combinations (30 and 31 ) and 22 had 3 possible combinations, we can see that there is one more combination than the exponent of the prime number. If we multiply these combinations, we get the number of factors. If we define 𝑛 to be 𝑛 = 𝐴𝑎 ∗ 𝐵𝑏 ∗ 𝐶 𝑐 … where 𝐴, 𝐵, 𝐶 … are prime, then:

# 𝑜𝑓 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑑𝑖𝑣𝑖𝑠𝑜𝑟𝑠: (𝑎 + 1) ∗ (𝑏 + 1) ∗ (𝑐 + 1) …

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When trying to determine the prime factorization, make sure to always be working with primes (hint: 4 isn’t prime, contrary to the belief of many students in the middle of a Number Sense test). Keep dividing the number by the largest prime you see it is divisible by. Let’s work an example from start to finish. 60 = 5 ∗ 12 = 5 ∗ 2 ∗ 6 = 5 ∗ 2 ∗ 2 ∗ 3 = 5 ∗ 3 ∗ 22 5 ∗ 3 ∗ 22 = 51 ∗ 31 ∗ 22 # 𝑜𝑓 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑑𝑖𝑣𝑖𝑠𝑜𝑟𝑠: (1 + 1) ∗ (1 + 1) ∗ (2 + 1) = 2∗2∗3 = 12 Practice Problem Set 7.2 1) 12 ℎ𝑎𝑠 ℎ𝑜𝑤 𝑚𝑎𝑛𝑦 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑑𝑖𝑣𝑖𝑠𝑜𝑟𝑠: 2)18 ℎ𝑎𝑠 ℎ𝑜𝑤 𝑚𝑎𝑛𝑦 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑑𝑖𝑣𝑖𝑠𝑜𝑟𝑠: 3)24 ℎ𝑎𝑠 ℎ𝑜𝑤 𝑚𝑎𝑛𝑦 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑑𝑖𝑣𝑖𝑠𝑜𝑟𝑠: 4) 26 ℎ𝑎𝑠 ℎ𝑜𝑤 𝑚𝑎𝑛𝑦 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑑𝑖𝑣𝑖𝑠𝑜𝑟𝑠: 5) 32 ℎ𝑎𝑠 ℎ𝑜𝑤 𝑚𝑎𝑛𝑦 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑑𝑖𝑣𝑖𝑠𝑜𝑟𝑠: 6) 37 ℎ𝑎𝑠 ℎ𝑜𝑤 𝑚𝑎𝑛𝑦 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑑𝑖𝑣𝑖𝑠𝑜𝑟𝑠: 7) 45 ℎ𝑎𝑠 ℎ𝑜𝑤 𝑚𝑎𝑛𝑦 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑑𝑖𝑣𝑖𝑠𝑜𝑟𝑠: 8) 72 ℎ𝑎𝑠 ℎ𝑜𝑤 𝑚𝑎𝑛𝑦 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑑𝑖𝑣𝑖𝑠𝑜𝑟𝑠: 9) 144 ℎ𝑎𝑠 ℎ𝑜𝑤 𝑚𝑎𝑛𝑦 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑑𝑖𝑣𝑖𝑠𝑜𝑟𝑠: 10) 195 ℎ𝑎𝑠 ℎ𝑜𝑤 𝑚𝑎𝑛𝑦 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑑𝑖𝑣𝑖𝑠𝑜𝑟𝑠:

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7.3

Sum of Positive Integral Divisors The sum of a number’s factors has a fairly lengthy formula but is faster than figuring out each factor and adding them up. We will stick with our definition in Trick 7.2 of 𝑛 to be 𝑛 = 𝐴𝑎 ∗ 𝐵𝑏 ∗ 𝐶 𝑐 … where 𝐴, 𝐵, 𝐶 … are prime. The formula reads:

∑ 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑑𝑖𝑣𝑖𝑠𝑜𝑟𝑠 = (1 + 𝐴 + 𝐴2 + ⋯ + 𝐴𝑎 ) ∗ (1 + 𝐵 + 𝐵2 + ⋯ + 𝐵𝑏 ) ∗ (1 + 𝐶 + 𝐶 2 + ⋯ + 𝐶 𝑐 ) … 𝐴𝑎+1 − 1 𝐵𝑏+1 − 1 𝐶 𝑐+1 − 1 = ∗ ∗ … 𝐴−1 𝐵−1 𝐶−1 The formula is quite daunting, but is actually repetitive and goes fast when you have memorized the powers listed in Trick 5.2: Powers of Special Numbers. Let’s work a couple of examples.

𝑊ℎ𝑎𝑡 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑑𝑖𝑣𝑖𝑠𝑜𝑟𝑠 𝑜𝑓 12? 12 = 22 ∗ 3 22+1 − 1 31+1 − 1 ∗ ∑ 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑑𝑖𝑣𝑖𝑠𝑜𝑟 = 2−1 3−1 8−1 9−1 = ∗ = 7 ∗ 4 = 28 1 2

If the exponent in the prime factorization is 1, the term

𝐴𝑎+1 −1 𝐴−1

will equal 𝐴 + 1. Also, when the prime number is 2, the denominator will always be 1 and can thus be ignored. This will simplify calculations tremendously.

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𝑊ℎ𝑎𝑡 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑑𝑖𝑣𝑖𝑠𝑜𝑟𝑠 𝑜𝑓 84? 84 = 2 ∗ 42 = 2 ∗ 6 ∗ 7 = 22 ∗ 3 ∗ 7 = 22 ∗ 31 ∗ 71 ∑ 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑑𝑖𝑣𝑖𝑠𝑜𝑟 = (22+1 − 1) ∗ (3 + 1) ∗ (7 + 1) = (8 − 1) ∗ 4 ∗ 8 = 7 ∗ 4 ∗ 8 = 28 ∗ 8 = 160 + 64 = 224 Practice Problem Set 7.3 1) 𝑊ℎ𝑎𝑡 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑑𝑖𝑣𝑖𝑠𝑜𝑟𝑠 𝑜𝑓 12: 2) 𝑊ℎ𝑎𝑡 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑑𝑖𝑣𝑖𝑠𝑜𝑟𝑠 𝑜𝑓 16: 3) 𝑊ℎ𝑎𝑡 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑑𝑖𝑣𝑖𝑠𝑜𝑟𝑠 𝑜𝑓 26: 4) 𝑊ℎ𝑎𝑡 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑑𝑖𝑣𝑖𝑠𝑜𝑟𝑠 𝑜𝑓 30: 5) 𝑊ℎ𝑎𝑡 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑑𝑖𝑣𝑖𝑠𝑜𝑟𝑠 𝑜𝑓 36: 6) 𝑊ℎ𝑎𝑡 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑑𝑖𝑣𝑖𝑠𝑜𝑟𝑠 𝑜𝑓 47: 7) 𝑊ℎ𝑎𝑡 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑑𝑖𝑣𝑖𝑠𝑜𝑟𝑠 𝑜𝑓 148:

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7.4

Relatively Prime Two numbers are relatively prime when their GCD is 1 (or the LCM is the product the two numbers). Almost every question pertaining to relatively prime numbers in Number Sense will read “How many positive integers less than or equal to X are relatively prime to X?” There’s actually a pretty simple formula for this. Sticking with our definition in Trick 7.2 of 𝑛 to be 𝑛 = 𝐴𝑎 ∗ 𝐵𝑏 ∗ 𝐶 𝑐 … where 𝐴, 𝐵, 𝐶 … are prime, the formula reads:

# 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒𝑙𝑦 𝑝𝑟𝑖𝑚𝑒 = 𝑛 ∗

𝐴−1 𝐵−1 𝐶−1 ∗ ∗ ∗… 𝐴 𝐵 𝐶

Notice that this formula doesn’t use the exponents in the prime factorization. The fractions should cancel out and yield an integer, else you probably did something wrong. Let’s see an example:

𝐻𝑜𝑤 𝑚𝑎𝑛𝑦 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑒𝑟𝑠 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 𝑜𝑟 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 42 𝑎𝑟𝑒 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒𝑙𝑦 𝑝𝑟𝑖𝑚𝑒 𝑡𝑜 42? 42 = 6 ∗ 7 = 2 ∗ 3 ∗ 7 2−1 3−1 7−1 ∗ ∗ 2 3 7 1 2 6 = 42 ∗ ∗ ∗ = 12 2 3 7

# 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒𝑙𝑦 𝑝𝑟𝑖𝑚𝑒 = 42 ∗

There are a variety of ways to simplify the multiplication; I’ll leave it up to you to decide what works best for you. Just a word of advice: it’s better to divide or simplify before multiplying.

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Practice Problem Set 7.4 1) 𝐻𝑜𝑤 𝑚𝑎𝑛𝑦 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑒𝑟𝑠 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 𝑜𝑟 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 13 𝑎𝑟𝑒 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒𝑙𝑦 𝑝𝑟𝑖𝑚𝑒 𝑡𝑜 13?

2) 𝐻𝑜𝑤 𝑚𝑎𝑛𝑦 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑒𝑟𝑠 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 𝑜𝑟 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 32 𝑎𝑟𝑒 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒𝑙𝑦 𝑝𝑟𝑖𝑚𝑒 𝑡𝑜 32?

3) 𝐻𝑜𝑤 𝑚𝑎𝑛𝑦 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑒𝑟𝑠 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 𝑜𝑟 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 43 𝑎𝑟𝑒 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒𝑙𝑦 𝑝𝑟𝑖𝑚𝑒 𝑡𝑜 43?

4) 𝐻𝑜𝑤 𝑚𝑎𝑛𝑦 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑒𝑟𝑠 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 𝑜𝑟 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 55 𝑎𝑟𝑒 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒𝑙𝑦 𝑝𝑟𝑖𝑚𝑒 𝑡𝑜 55?

5) 𝐻𝑜𝑤 𝑚𝑎𝑛𝑦 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑒𝑟𝑠 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 𝑜𝑟 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 48 𝑎𝑟𝑒 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒𝑙𝑦 𝑝𝑟𝑖𝑚𝑒 𝑡𝑜 48?

6) 𝐻𝑜𝑤 𝑚𝑎𝑛𝑦 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑒𝑟𝑠 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 𝑜𝑟 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 78 𝑎𝑟𝑒 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒𝑙𝑦 𝑝𝑟𝑖𝑚𝑒 𝑡𝑜 78?

7) 𝐻𝑜𝑤 𝑚𝑎𝑛𝑦 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑒𝑟𝑠 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 𝑜𝑟 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 60 𝑎𝑟𝑒 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒𝑙𝑦 𝑝𝑟𝑖𝑚𝑒 𝑡𝑜 60?

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8

Other Topics

8.1

Additive and Multiplicative Identities and

Inverses The additive identity is the number than when added to any value yields the original value. Thus, the additive identify is 0. Similarly, the multiplicative identity is the number that when multiplied to any value doesn’t change the value. So, the multiplicative identity is 1. We use identities to calculate inverses. A value added to its additive inverse yields the additive identity, or 0. That means that the additive inverse is just the original value with the opposite sign.

𝐴𝑑𝑑𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑣𝑒𝑟𝑠𝑒 𝑜𝑓 4 𝑖𝑠 − 4 𝐴𝑑𝑑𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑣𝑒𝑟𝑠𝑒 𝑜𝑓 − 2⁄3 𝑖𝑠 2⁄3 A value multiplied to its multiplicative inverse yields the multiplicative identity, or 1. That means that the multiplicative inverse is just the reciprocal of the original value.

𝑀𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑐𝑎𝑡𝑖𝑣𝑒 𝑖𝑛𝑣𝑒𝑟𝑠𝑒 𝑜𝑓 4 𝑖𝑠 1⁄4 𝑀𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑐𝑎𝑡𝑖𝑣𝑒 𝑖𝑛𝑣𝑒𝑟𝑠𝑒 𝑜𝑓 − 2⁄3 𝑖𝑠 − 3⁄2

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Practice Problem Set 8.1 1) 𝑊ℎ𝑎𝑡 𝑖𝑠 𝑡ℎ𝑒 𝑎𝑑𝑑𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑣𝑒𝑟𝑠𝑒 𝑜𝑓 12: 4

2)𝑊ℎ𝑎𝑡 𝑖𝑠 𝑡ℎ𝑒 𝑎𝑑𝑑𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑣𝑒𝑟𝑠𝑒 𝑜𝑓 : 3

3)𝑊ℎ𝑎𝑡 𝑖𝑠 𝑡ℎ𝑒 𝑎𝑑𝑑𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑣𝑒𝑟𝑠𝑒 𝑜𝑓 − 4: 3

4) 𝑊ℎ𝑎𝑡 𝑖𝑠 𝑡ℎ𝑒 𝑎𝑑𝑑𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑣𝑒𝑟𝑠𝑒 𝑜𝑓 − : 2

5) 𝑊ℎ𝑎𝑡 𝑖𝑠 𝑡ℎ𝑒 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑐𝑎𝑡𝑖𝑣𝑒 𝑖𝑛𝑣𝑒𝑟𝑠𝑒 𝑜𝑓 12: 4

6)𝑊ℎ𝑎𝑡 𝑖𝑠 𝑡ℎ𝑒 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑐𝑎𝑡𝑖𝑣𝑒 𝑖𝑛𝑣𝑒𝑟𝑠𝑒 𝑜𝑓 : 3

7)𝑊ℎ𝑎𝑡 𝑖𝑠 𝑡ℎ𝑒 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑐𝑎𝑡𝑖𝑣𝑒 𝑖𝑛𝑣𝑒𝑟𝑠𝑒 𝑜𝑓 − 4: 3

8) 𝑊ℎ𝑎𝑡 𝑖𝑠 𝑡ℎ𝑒 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑐𝑎𝑡𝑖𝑣𝑒 𝑖𝑛𝑣𝑒𝑟𝑠𝑒 𝑜𝑓 − : 2

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8.2

Absolute Value The best way to describe absolute value is the distance between the 0 and the number on a number line. Basically, it makes any number positive (and keeps 0 at 0). Number Sense usually asks nested absolute value questions, that is, absolute values inside absolute values. The best way to solve nested absolute value questions is to work inside out. Since most all operations in these questions are single digit addition and subtraction, the number one thing you need to watch are the signs.

2 − |−3 + |−5| − 7| = 2 − |−3 + 5 − 7| = 2 − |−5| = 2−5 = −3

14 + |−12 − |4 + |−4| − 5|| − 2 = 14 + |−12 − |4 + 4 − 5|| − 2 = 14 + |−12 − |−5|| − 2 = 14 + |−12 − 5| − 2 = 14 + |−17| − 2 = 14 + 17 − 2 = 29

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Practice Problem Set 8.2 1) |4 + |3 − 7| − 2| =

2) 4 − |−12 + 4|1 − 2|| =

3) |3|4 − 2| − 8| − 7 =

4) |−2 + |3 + |−7|| + 1| =

5) |4 − |4 − 4 − 4| − 4| − 4 =

6) |−6 + |3 + |−8|| + 5| =

7) |1 + |−2| − 4| − 12 =

8) 2 − |−7 + 2|−5 − 4|| =

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8.3

Roman Numerals It’s important to be able to read Roman Numerals quickly and accurately. First, you need to know the following conversions:

I=1

C = 100

V=5

D = 500

X = 10

M = 1000

L = 50

To represent numbers with Roman Numerals, work from the biggest to smallest value (M to I). Roman Numerals are additive, so you try to find the largest value is less than the number.

1100 = 𝑀𝐶 1200 ≠ 𝐷𝐷𝐶

Also, you generally write the biggest value first and descend accordingly.

1100 = 𝑀𝐶 1100 ≠ 𝐶𝑀

Next, you need to know that there can never be 4 of any letter in a row. Instead, the trick is to subtract to get that value. To write this in a way that indicates subtraction, you write the smaller number before the larger number (this is the exception mentioned before).

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4 = 𝐼𝑉 4 ≠ 𝐼𝐼𝐼𝐼

The best strategy when converting Arabic numerals (normal digits) to Roman numerals is to work one place value at a time.

99 = 90 + 9 = 𝑋𝐶𝐼𝑋 99 ≠ 𝐼𝐶 Practice Problem Set 8.3 1) 𝐶𝑋𝐼 =

2) 𝑀𝑋𝑉𝐼𝐼 =

3) 𝐷𝐿𝑉 =

4) 𝐶𝑋𝑋𝐼𝐼𝐼 =

5) 𝑀𝐷𝐶𝐶𝑋 =

6) 𝑀𝑀𝑋𝑉𝐼𝐼 =

7) 𝑀𝐶𝑀 =

8) 𝐶𝐶𝐼𝑋 =

9) 𝐶𝑀𝑋𝐿𝐼𝑋 =

10) 𝑀𝑀𝑀𝑋𝐶𝑉𝐼𝐼𝐼 =

11) 17 =

12) 35 =

13) 1037 =

14) 765 =

15) 987 =

16) 114 =

17) 432 =

18) 596 =

19) 698 =

20) 999 =

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8.4

Conversion Factors You should know both the metric and customary conversion factors for length, area, volume, weight, and temperature. Distance: 12 inches = 1 foot

36 inches = 3 feet = 1 yard

16.5 feet = 1 rod

5280 feet = 1760 yards =320 rods = 1 mile

Area: 43560 square feet = 1 acre

640 acres = 1 square mile

Volume: 3 teaspoons = 1 tablespoon

2 pints = 1 quart

2 tablespoons = 1 fluid ounce

4 quarts = 1 gallon

8 fluid ounces = 1 cup

2 gallons = 1 peck

2 cups = 1 pint

4 pecks = 1 bushel

Weight: 16 ounces = 1 pound

2000 pounds = 1 ton

Temperature: °F =

9 ∗ °C + 32 5

°C =

5 ∗ (°F − 32) 9

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Metric System Prefixes: Giga:

109

Nano:

10-9

Mega:

106

Micro:

10-6

Kilo:

103

Milli:

10-3

Hecto:

102

Centi:

10-2

Deca:

101

Deci:

10-1

You use the prefixes on base units for every measurement. These include: Length: meter (kilometer, centimeter) Mass: gram (kilogram, milligram) Volume: liter (milliliter) It’s also important for you to know the following conversions as they frequently appear on Number Sense tests. 15 5

𝑚𝑖𝑙𝑒𝑠 𝑓𝑒𝑒𝑡 = 22 ℎ𝑜𝑢𝑟 𝑠𝑒𝑐𝑜𝑛𝑑

𝑓𝑒𝑒𝑡 𝑖𝑛𝑐ℎ =1 𝑚𝑖𝑛𝑢𝑡𝑒 𝑠𝑒𝑐𝑜𝑛𝑑

231 𝑐𝑢𝑏𝑖𝑐 𝑖𝑛𝑐ℎ𝑒𝑠 = 1 𝑔𝑎𝑙𝑙𝑜𝑛

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8.5

Complex Numbers It is well established that square roots can only be taken of positive numbers. But what about the square root of negative numbers? The square root of a negative number doesn’t result in a real number (since no number squared can be negative), so we define the imaginary number 𝑖 to be √−1. That means that: 𝑖 2 = −1 𝑖 3 = 𝑖 ∗ 𝑖 2 = −𝑖 𝑖 4 = (𝑖 2 )2 = (−1)2 = 1 𝑖5 = 𝑖 ∗ 𝑖4 = 𝑖

The cycle continues for every 4 powers. Problems in Number Sense asking about imaginary numbers will be in the following form:

(2 + 5𝑖)(3 + 4𝑖) = 𝑎 + 𝑏𝑖. 𝐹𝑖𝑛𝑑 𝑎

To solve such questions, you will have to expand the expression by multiplying the two binomials.

(𝑎 + 𝑏)(𝑐 + 𝑑) = 𝑎𝑐 + 𝑎𝑑 + 𝑏𝑐 + 𝑏𝑑 (2 + 5𝑖)(3 + 4𝑖) = 2 ∗ 3 + 2 ∗ 4𝑖 + 5𝑖 ∗ 3 + 5𝑖 ∗ 4𝑖 = 6 + 8𝑖 + 15𝑖 + 20𝑖 2 = 6 + 23𝑖 − 20 = −14 + 23𝑖 𝑎 = −14

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Notice how the problem above only asked for 𝑎, or the real number component of the resulting binomial. Thus, we didn’t have to calculate 𝑏. When multiplying two binomials in this form, know the following: 𝑎, or the real number component of the answer, is determined by multiplying the two real numbers in each binomial and adding this to the product of the two imaginary numbers in each binomial.

𝑎 = 2 ∗ 3 + (5𝑖) ∗ (4𝑖) = 6 + 20𝑖 2 = 6 − 20 = −14

𝑏, or the coefficient of the imaginary component of the answer, is determined by multiplying the real number of one binomial by the imaginary component of the other binomial and adding the two products.

𝑏 → 2 ∗ (4𝑖) + 3 ∗ (5𝑖) = 8𝑖 + 15𝑖 = 23𝑖 𝑏 = 23

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Practice Problem Set 8.5 1) (4 + 3𝑖)(2 − 3𝑖) = 𝑎 + 𝑏𝑖. 𝐹𝑖𝑛𝑑 𝑎: 2) (7 − 2𝑖)(4 + 5𝑖) = 𝑎 + 𝑏𝑖. 𝐹𝑖𝑛𝑑 𝑏: 3) (9 − 1𝑖)(2 − 6𝑖) = 𝑎 + 𝑏𝑖. 𝐹𝑖𝑛𝑑 𝑎: 4) (12 + 11𝑖)(10 + 9𝑖) = 𝑎 + 𝑏𝑖. 𝐹𝑖𝑛𝑑 𝑏: 5) (8 − 5𝑖)(4 + 1𝑖) = 𝑎 + 𝑏𝑖. 𝐹𝑖𝑛𝑑 𝑎: 6) (2 − 8𝑖)(4 − 5𝑖) = 𝑎 + 𝑏𝑖. 𝐹𝑖𝑛𝑑 𝑏: 7) (9 + 7𝑖)(2 + 8𝑖) = 𝑎 + 𝑏𝑖. 𝐹𝑖𝑛𝑑 𝑎 + 𝑏: 8) (2 + 1𝑖)(4 − 4𝑖) = 𝑎 + 𝑏𝑖. 𝐹𝑖𝑛𝑑 𝑎 − 𝑏: 9) (8 − 5𝑖)(2 − 8𝑖) = 𝑎 + 𝑏𝑖. 𝐹𝑖𝑛𝑑 𝑎 + 𝑏: 10) (4 + 5𝑖)(2 + 11𝑖) = 𝑎 + 𝑏𝑖. 𝐹𝑖𝑛𝑑 𝑎 − 𝑏:

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9

Types of Numbers

9.1

Polygonal Numbers Polygonal numbers probably the most common type of number on the test. These can be triangular numbers, squares, pentagonal numbers, and so on. For example, 10 is a triangular number because you can construct a triangle with 10 dots like bowling pins. However, 10 isn’t a square number because 10 dots cannot be arranged to make a square. Instead, 9 dots can make a square. Each type of polygonal number has its own sequence and are completely separate from one another. However, the formula for calculating the nth s-gonal number is the same:

𝑛

𝑡ℎ

𝑛2 (𝑠 − 2) − 𝑛(𝑠 − 4) 𝑠 − 𝑔𝑜𝑛𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟: 2

It’s also helpful to be acquainted with triangular numbers as they show up frequently: 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, and so on. The formula for triangular numbers is also useful because it is much simpler than the formula above:

𝑇(𝑛) =

𝑛 ∗ (𝑛 + 1) 2

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Practice Problem Set 9.1 1) 𝑇ℎ𝑒 4𝑡ℎ ℎ𝑒𝑥𝑎𝑔𝑜𝑛𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑖𝑠: 2) 𝑇ℎ𝑒 5𝑡ℎ 𝑛𝑜𝑛𝑎𝑔𝑜𝑛𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑖𝑠: 3) 𝑇ℎ𝑒 7𝑡ℎ ℎ𝑒𝑝𝑡𝑎𝑔𝑜𝑛𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑖𝑠: 4) 𝑇ℎ𝑒 8𝑡ℎ 𝑝𝑒𝑛𝑡𝑎𝑔𝑜𝑛𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑖𝑠: 5) 𝑇ℎ𝑒 6𝑡ℎ 𝑜𝑐𝑡𝑎𝑔𝑜𝑛𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑖𝑠: 6) 𝑇ℎ𝑒 3𝑟𝑑 𝑛𝑜𝑛𝑎𝑜𝑛𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑖𝑠: 7) 𝑇ℎ𝑒 7𝑡ℎ ℎ𝑒𝑥𝑎𝑔𝑜𝑛𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑖𝑠: 8) 𝑇ℎ𝑒 12𝑡ℎ 𝑡𝑟𝑖𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑛𝑢𝑚𝑏𝑒𝑟 𝑖𝑠: 9) 𝑇ℎ𝑒 25𝑡ℎ 𝑡𝑟𝑖𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑛𝑢𝑚𝑏𝑒𝑟 𝑖𝑠: 10) 𝑇ℎ𝑒 3𝑟𝑑 𝑑𝑒𝑐𝑎𝑔𝑜𝑛𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑖𝑠:

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9.2

Deficient, Perfect, and Abundant Numbers To determine whether a number is deficient, perfect, or abundant, you look at the sum of the divisors of the number. Every positive integer is either deficient, where the sum of the factors is less than twice the number; perfect, where the sum of the factors is equal to twice the number; or abundant, where the sum of the factors is greater than twice the number.

12 = 1 ∗ 12 =2∗6 =3∗4 𝑆𝑢𝑚 = 1 + 12 + 2 + 6 + 3 + 4 = 28 28 > 24 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 12 𝑖𝑠 𝑎𝑏𝑢𝑛𝑑𝑎𝑛𝑡

28 = 1 ∗ 28 = 2 ∗ 14 =4∗7 𝑆𝑢𝑚 = 1 + 28 + 2 + 14 + 4 + 7 = 56 56 = 56 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 28 𝑖𝑠 𝑝𝑒𝑟𝑓𝑒𝑐𝑡

Just some notes. The only perfect numbers less than 10,000 are 6, 28, 496, and 8128 (which you should memorize). The smallest abundant number is 12, so every single digit integer is either deficient or perfect. Most abundant numbers are even; the first odd abundant number is 945.

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9.3

Happy and Unhappy Numbers Happy numbers take a fair amount of computation to get to. To determine if a number is happy or not, begin by squaring each digit and adding up the squares. With the number that results, repeat the steps; square each digit and add up the squares. Continue this until one of two things happen: 1. You get to the number 1. If this happens the number is Happy. 2. You get stuck in an endless loop. If this happens the number is Unhappy.

23 → 22 + 32 = 13 13 → 12 + 32 = 10 10 → 12 + 02 = 1

Therefore the number 23 is happy.

25 → 22 + 52 = 29 29 → 22 + 92 = 85 85 → 82 + 52 = 89 89 → 82 + 92 = 145 145 → 12 + 42 + 52 = 42 42 → 42 + 22 = 20 20 → 22 + 02 = 4 20 → 42 = 16 16 → 12 + 62 = 37 37 → 32 + 72 = 58 58 → 52 + 82 = 89 92 | P a g e

Note how 89 shows up again (step 3). That means that we’re going to cycle with the same numbers infinitely if we continue the calculations. That means that 25 is unhappy. Now, it is obviously very tedious to calculate all of those squares and do the addition. That’s why it’s best to memorize some of the unhappy numbers so that as soon as you see them, you can stop and conclude that the number is unhappy.

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9.4

Evil and Odious Numbers To determine if an integer is evil or odious, the first step is to convert the number to base 2, or binary (bases are covered in Chapter 16). After the number is converted to binary, you count the number of ones that appear in the binary form. If the number of ones is even, the number is evil; if odd, then odious (even: evil and odd: odious).

1410 = 11102 𝑆𝑖𝑛𝑐𝑒 𝑡ℎ𝑒𝑟𝑒 𝑎𝑟𝑒 𝑡ℎ𝑟𝑒𝑒 1′𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑏𝑖𝑛𝑎𝑟𝑦 𝑓𝑜𝑟𝑚, 14 𝑖𝑠 𝑜𝑑𝑖𝑜𝑢𝑠

2710 = 110112 𝑆𝑖𝑛𝑐𝑒 𝑡ℎ𝑒𝑟𝑒 𝑎𝑟𝑒 𝑓𝑜𝑢𝑟 1′𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑏𝑖𝑛𝑎𝑟𝑦 𝑓𝑜𝑟𝑚, 27 𝑖𝑠 𝑒𝑣𝑖𝑙

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9.5

Polite Numbers and Politeness A polite number is any number that can be written as the sum of two or more consecutive integers. The most important takeaway is that powers of 2 are impolite; every other positive integer is polite.

7=3+4 18 = 5 + 6 + 7 =3+4+5+6

The number of ways a number can be expressed as the sum of positive integers is called the number’s politeness. The politeness of a number is the number of odd divisors the number has. To determine this, follow the following steps: Step 1: Find the prime factorization of the number Step 2: Add 1 to every power of a prime number greater than 2 Step 3: Multiply each sum Step 4: Subtract 1 from the final product For example:

18 = 2 ∗ 32 𝑃𝑜𝑙𝑖𝑡𝑒𝑛𝑒𝑠𝑠: (2 + 1) − 1 = 2 18 = 5 + 6 + 7 =3+4+5+6

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360 = 23 ∗ 32 ∗ 51 𝑃𝑜𝑙𝑖𝑡𝑒𝑛𝑒𝑠𝑠: (2 + 1) ∗ (1 + 1) − 1 = 5 360 = 119 + 120 + 121 = 70 + 71 + 72 + 73 + 74 = 36 + 37 + ⋯ + 43 + 44 = 17 + 18 + ⋯ + 30 + 31 = 15 + 16 + ⋯ + 29 + 30

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9.6

Frugal, Economical, Equidigital, Wasteful,

and Extravagant Numbers To determine if a number is frugal (or economical, as they are the same thing), equidigital, or wasteful (or extravagant, as they are the same thing), first find its prime factorization. Then count the number of digits in the number’s prime factorization (including exponents besides 1) and compare it to the number of digits in the number itself. If: 𝑑𝑖𝑔𝑖𝑡𝑠𝑝𝑟𝑖𝑚𝑒 𝑓𝑎𝑐𝑡𝑜𝑟𝑖𝑧𝑎𝑡𝑖𝑜𝑛 < 𝑑𝑖𝑔𝑖𝑡𝑠𝑛𝑢𝑚𝑏𝑒𝑟 , 𝑡ℎ𝑒𝑛 𝑓𝑟𝑢𝑔𝑎𝑙 𝑑𝑖𝑔𝑖𝑡𝑠𝑝𝑟𝑖𝑚𝑒 𝑓𝑎𝑐𝑡𝑜𝑟𝑖𝑧𝑎𝑡𝑖𝑜𝑛 = 𝑑𝑖𝑔𝑖𝑡𝑠𝑛𝑢𝑚𝑏𝑒𝑟 , 𝑡ℎ𝑒𝑛 𝑒𝑞𝑢𝑖𝑑𝑖𝑔𝑖𝑡𝑎𝑙 𝑑𝑖𝑔𝑖𝑡𝑠𝑝𝑟𝑖𝑚𝑒 𝑓𝑎𝑐𝑡𝑜𝑟𝑖𝑧𝑎𝑡𝑖𝑜𝑛 > 𝑑𝑖𝑔𝑖𝑡𝑠𝑛𝑢𝑚𝑏𝑒𝑟 , 𝑡ℎ𝑒𝑛 𝑤𝑎𝑠𝑡𝑒𝑓𝑢𝑙

Take a look at the following examples:

125 = 53 → 𝑓𝑟𝑢𝑔𝑎𝑙/𝑒𝑐𝑜𝑛𝑜𝑚𝑖𝑐𝑎𝑙 16 = 24 → 𝑒𝑞𝑢𝑖𝑑𝑖𝑔𝑖𝑡𝑎𝑙 12 = 22 ∗ 3 → 𝑤𝑎𝑠𝑡𝑒𝑓𝑢𝑙/𝑒𝑥𝑡𝑟𝑎𝑣𝑎𝑔𝑎𝑛𝑡

97 | P a g e

Practice Problem Set 9.2 to 9.6 1) 𝑊ℎ𝑖𝑐ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑜𝑙𝑙𝑜𝑤𝑖𝑛𝑔 𝑖𝑠 𝑎 ℎ𝑎𝑝𝑝𝑦 𝑛𝑢𝑚𝑏𝑒𝑟: 13, 11, 9? 2) 𝑊ℎ𝑖𝑐ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑜𝑙𝑙𝑜𝑤𝑖𝑛𝑔 𝑖𝑠 𝑎𝑛 𝑒𝑣𝑖𝑙 𝑛𝑢𝑚𝑏𝑒𝑟: 14, 15, 16? 3) 𝑊ℎ𝑖𝑐ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑜𝑙𝑙𝑜𝑤𝑖𝑛𝑔 𝑖𝑠 𝑎 𝑑𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑛𝑢𝑚𝑏𝑒𝑟: 5, 12, 28? 4) 𝑊ℎ𝑖𝑐ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑜𝑙𝑙𝑜𝑤𝑖𝑛𝑔 𝑖𝑠 𝑛𝑜𝑡 𝑎 𝑝𝑒𝑟𝑓𝑒𝑐𝑡 𝑛𝑢𝑚𝑏𝑒𝑟: 28, 469, 8128? 5) 𝑊ℎ𝑖𝑐ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑜𝑙𝑙𝑜𝑤𝑖𝑛𝑔 𝑖𝑠 𝑎 ℎ𝑎𝑝𝑝𝑦 𝑑𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑛𝑢𝑚𝑏𝑒𝑟: 23, 25, 28? 6) 𝑊ℎ𝑖𝑐ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑜𝑙𝑙𝑜𝑤𝑖𝑛𝑔 𝑖𝑠 𝑎 𝑓𝑟𝑢𝑔𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟: 125, 250, 375? 7) 𝑊ℎ𝑖𝑐ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑜𝑙𝑙𝑜𝑤𝑖𝑛𝑔 𝑖𝑠 𝑎 ℎ𝑎𝑝𝑝𝑦 𝑒𝑣𝑖𝑙 𝑛𝑢𝑚𝑏𝑒𝑟: 7, 10, 13? 8) 𝑊ℎ𝑖𝑐ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑜𝑙𝑙𝑜𝑤𝑖𝑛𝑔 𝑖𝑠 𝑎 𝑑𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑑𝑖𝑜𝑢𝑠 𝑛𝑢𝑚𝑏𝑒𝑟: 26, 28, 30?

98 | P a g e

10

Factorials and Combinations

10.1

Factorials The factorial of a positive integer is the product of every integer from the integer down to 1. Factorials are used when calculating how many possible ways there are to arrange objects, probability, and algebra. It is also worth noting that 0! = 1.

𝑛! = 𝑛 ∗ (𝑛 − 1) ∗ (𝑛 − 2) ∗ … ∗ 1

For example:

5! = 5 ∗ 4 ∗ 3 ∗ 2 ∗ 1 = 120

99 | P a g e

10.2

Permutations The idea behind permutations is that you are ordering all the elements of a set, either with replacement or without, while caring about the order they’re put in. For example, to calculate the number of possible codes of a 3digit lock, you would look at permutations because order matters. If a digit could be used more than once, the permutations would be with replacement (once you use a number, it can be used again). Since there are 10 possible digits to choose from for each slot, there are:

10 ∗ 10 ∗ 10 = 103 = 1000 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑐𝑜𝑑𝑒𝑠

However, if every digit had to be separate, then there would be no replacement (once you use a number, you can’t use it again). There are 10 possible digits to choose from for the first slot. For the second slot, since we can’t choose the number chosen in the first slot, there are only 9 possible digits. Likewise, the third slot only has 8 possible digits. This results in:

10 ∗ 9 ∗ 8 = 720 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑐𝑜𝑑𝑒𝑠

Let’s say we want to arrange all 10 digits to create a 10 digit number. Conventionally, numbers don’t start with 0, but we’ll accept it for this problem. We have 10 options for the first digit of the number, 9 options for the second, and so on and so forth until there’s 1 option for the last digit. The number of possible 10 digit numbers with these conditions is:

10 ∗ 9 ∗ 8 ∗ 7 ∗ 6 ∗ 5 ∗ 4 ∗ 3 ∗ 2 ∗ 1 = 10! = 3628800 100 | P a g e

As for notation, to count the number of orderings for a set of elements r that are chosen from the set n, we say it to be:

𝑃(𝑛, 𝑟)

For instance, the example of creating a 3-digit number would be written 𝑃(10,3). When given in this form, you calculate it with replacement. The general formula is:

𝑃(𝑛, 𝑟) =

𝑛! (𝑛 − 𝑟)!

Practice Problem Set 10.2 1) 𝑃(4,2) =

2) 𝑃(7,7) =

3) 𝑃(6,2) =

4) 𝑃(5,3) =

5) 𝑃(7,4) =

6) 𝑃(9,3) =

7) 𝑃(12,4) =

8) 𝑃(8,4) =

9) 𝑃(10,8) =

10) 𝑃(10,2) =

101 | P a g e

10.3

Combinations Combinations counts how to choose items from a set in such a way that order doesn’t matter. Many times this is phrased as the combination of n things taken k at a time. Like permutations, combinations can be calculated with or without repetition. Assume we have a bag of 6 basketballs labeled 1 to 6. If we were to choose 2 basketballs from the 6 without replacement, we wouldn’t put back the first basketball we chose. That means we have 6 options for the first basketball and 5 for the second one.

6 ∗ 5 = 30

However, the problem doesn’t end there. Choosing basketballs 3 and 5 is the same thing as choosing basketballs 5 and 3, yet they are counted individually so far. Thus, we have to divide our initial product by 2 to get the possible combinations.

30 ÷ 2 = 15

We can continue this process for choosing 3 basketballs from the 6. First, we have 6 basketballs to choose from, then 5, then 4. So the initial product is:

6 ∗ 5 ∗ 4 = 120

Next, we have to remove repeats. Since choosing basketballs 1, 2, and 3 can also be counted as 1-3-2, 2-1-3, 2-3-1, 3-1-2, and 3-21, we are counting 6 times as many combinations as there really are. To remove these, we divide the product by 6:

102 | P a g e

120 = 20 𝑐𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛𝑠 6 We can continue this process for other possible cases. The general formula for choosing r items from a set of n items ends up being:

𝑛! (𝑛 − 𝑟)! ∗ 𝑟!

Note: factorials are easy to cancel out. Always divide before multiplying when working with factorials. For example:

6! 6 ∗ 5 ∗ 4 ∗ 3 ∗ 2 ∗ 1 = = 6 ∗ 5 = 30 4! 4∗3∗2∗1 Practice problems including mix between combinations and permutations. Practice Problem Set 10.3 1) 𝐶(4,2) =

2) 𝐶(7,7) =

3) 𝐶(6,2) =

4) 𝐶(5,3) =

5) 𝐶(7,4) =

6) 𝐶(9,3) =

7) 𝐶(12,4) =

8) 𝐶(8,4) =

9) 𝐶(10,8) =

10) 𝐶(10,2) =

103 | P a g e

11

Sequences

11.1

Consecutive Integer Sequences There are 3 common consecutive integer sequences you should know. The formulas for all three are relatively simple and shouldn’t take much time to calculate.

Sum of consecutive integers starting with 1:

1 + 2 + 3 + ⋯+ 𝑛 𝑛

= ∑𝑖 𝑖=1

=

𝑛 ∗ (𝑛 + 1) 2

Interestingly enough, this formula is identical to triangular numbers. If you research more into triangular numbers, you can see why this is the case.

Sum of consecutive even integers starting with 2:

2 + 4 + ⋯ + 𝑚 = 2 ∗ (1 + 2 + ⋯

𝑚 ) 2

𝑚 2

= 2 ∗ ∑𝑖 𝑖=1

𝑚 𝑚 ∗ ( + 1) 2 =2∗ 2 2 𝑚 𝑚 = ∗ ( + 1) 2 2 104 | P a g e

Sum of consecutive odd integers starting with 1:

1 + 3 + ⋯+ 𝑙 = (2 ∗ 1 − 1) + (2 ∗ 2 − 1) + ⋯ + (2 ∗

𝑙+1 − 1) 2

𝑙+1 2

= ∑ 2𝑖 − 1 𝑖=1

𝑙+1 2 =( ) 2 In words, the equations mean: The sum of consecutive integers from 1 to 𝑛 is the 𝑛𝑡ℎ triangular number. The sum of consecutive even integers from 2 to 𝑚 is half of 𝑚 multiplied by that half plus one. The sum of consecutive odd integers from 1 to 𝑙 is the square of half of one more than 𝑙. Let’s try an example of each. 𝑆𝑢𝑚 𝑜𝑓 𝑓𝑖𝑟𝑠𝑡 10 𝑛𝑎𝑡𝑢𝑟𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑖𝑠: 10 ∗ (10 + 1) 10 ∗ 11 = = 5 ∗ 11 = 55 2 2 𝐹𝑖𝑛𝑎𝑙 𝑎𝑛𝑠𝑤𝑒𝑟: 55

2 + 4 + 6+. . +24 = 𝐹𝑖𝑟𝑠𝑡: 24⁄2 = 12 12 ∗ (12 + 1) = 12 ∗ 13 = 156 𝐹𝑖𝑛𝑎𝑙 𝑎𝑛𝑠𝑤𝑒𝑟: 156

105 | P a g e

1 + 3 + 5 + ⋯ + 17 = 𝐹𝑖𝑟𝑠𝑡:

17 + 1 18 = =9 2 2 92 = 81

𝐹𝑖𝑛𝑎𝑙 𝑎𝑛𝑠𝑤𝑒𝑟: 81 If a problem arises that doesn’t start with 1 (or 2 for consecutive even integers), add as if the sequence did. Then simply subtract out the missing terms.

5 + 7 + 9 + ⋯ + 21 = 𝐹𝑖𝑟𝑠𝑡:

21 + 1 22 = = 11 2 2 112 = 121

121 − (1 + 3) = 117 𝐹𝑖𝑛𝑎𝑙 𝑎𝑛𝑠𝑤𝑒𝑟: 117 Practice Problem Set 11.1 1) 1 + 2 + 3 + ⋯ + 23 =

2) 1 + 2 + 3 + ⋯ + 29 =

3) 1 + 3 + 5 + ⋯ + 27 =

4) 2 + 4 + 6 + ⋯ + 12 =

5) 1 + 3 + 5 + ⋯ + 17 =

6) 1 + 2 + 3 + ⋯ + 24 =

7) 2 + 4 + 6 + ⋯ + 28 =

8) 5 + 6 + 7 + ⋯ + 23 =

9) 5 + 7 + 9 + ⋯ + 23 =

10) 8 + 10 + 12 + ⋯ + 34 =

11) 8 + 9 + 10 + ⋯ + 21 =

12) 10 + 12 + 14 + ⋯ + 40 =

13) 13 + 15 + 17 + ⋯ + 31 =

14) 1 + 2 + 3 + ⋯ + 100 =

106 | P a g e

11.2

Arithmetic Sequences An arithmetic sequence starts with any number (not integer) and progresses such that the difference between consecutive terms is constant. The most important numbers when working with arithmetic sequences are the first term (𝑎0 ) and the common difference (𝑑). Algebraically, an arithmetic sequence can be expressed as:

𝑎0 , 𝑎0 + 𝑑, 𝑎0 + 2𝑑, 𝑎0 + 3𝑑, 𝑎0 + 4𝑑 …

Some questions may ask you to find a certain term of the sequence given the first couple of values. Using the initial term and common difference, to find the 𝑛𝑡ℎ term plug into the equation:

(𝑛 − 1) ∗ 𝑑 + 𝑎0

Let’s see it in action.

𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 12𝑡ℎ 𝑡𝑒𝑟𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑎𝑟𝑖𝑡ℎ𝑚𝑒𝑡𝑖𝑐 𝑠𝑒𝑞𝑢𝑒𝑛𝑐𝑒 2, 5, 8, … 𝑎0 = 2 𝑎𝑛𝑑 𝑛 = 12 𝑎𝑛𝑑 𝑑 = 3 (12 − 1) ∗ 3 + 2 = 11 ∗ 3 + 2 = 33 + 2 = 35

Other questions pertaining to arithmetic sequences will ask you add up a certain number of its terms, starting from a value 𝑎0 and going until 𝑎𝑛 with a common difference 𝑑. To determine the sum, it should first be noted that:

𝑎0 + 𝑎𝑛 = 𝑎1 + 𝑎𝑛−1 = 𝑎2 + 𝑎𝑛−2 = ⋯

107 | P a g e

In words, this says that the sum of the first and last terms is the same as the sum of second and second-to-last terms, which is the same as the sum of the third and third-to-last terms, etc. Because of this property, the average value of the terms in this finite sequence (number of terms can be even or odd) is: 𝑎1 + 𝑎𝑛 2 In this finite sequence there are 𝑛 terms. However in most cases this value isn’t given to us. To find it, use this equation:

𝑛=

𝑎𝑛 − 𝑎1 +1 𝑑

Now that we know both the number of terms and the average value of each term, we can find the sum of the terms in the finite arithmetic sequence by multiplying the two quantities: 𝑛

𝑎𝑛 − 𝑎1 𝑎1 + 𝑎𝑛 + 1) ∗ ( ∑ 𝑎1 + (𝑖 − 1) ∗ 𝑑 = ( ) 𝑑 2 𝑖=1

Since the algebra may look daunting, how about a couple of examples?

108 | P a g e

2 + 5 + 8 + ⋯ + 26 = 𝑎1 = 2 𝑎𝑛𝑑 𝑎𝑛 = 26 𝑎𝑛𝑑 𝑑 = 3 𝑆𝑢𝑚: (

𝐿𝑎𝑠𝑡 𝑡𝑒𝑟𝑚 − 𝐹𝑖𝑟𝑠𝑡 𝑡𝑒𝑟𝑚 𝐹𝑖𝑟𝑠𝑡 𝑡𝑒𝑟𝑚 + 𝐿𝑎𝑠𝑡 𝑡𝑒𝑟𝑚 + 1) ∗ ( ) 𝐶𝑜𝑚𝑚𝑜𝑛 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 2 (

26 − 2 2 + 26 + 1) ∗ ( ) 3 2 =(

24 28 + 1) ∗ ( ) 3 2

= (8 + 1) ∗ 14 = 9 ∗ 14 = 126

Finding the sum of an arithmetic sequence is calculation heavy. It’s always beneficial to divide before you multiply, but the best way to get faster is to work practice problems and understand what works best for you. Let’s work a slightly more complex sequence:

. 4 + 1 + 1.6 + ⋯ + 7.6 = 𝑎1 = .4 𝑎𝑛𝑑 𝑎𝑛 = 7.6 𝑎𝑛𝑑 𝑑 = .6 𝑆𝑢𝑚: (

7.6 − .4 . 4 + 7.6 + 1) ∗ ( ) .6 2 =(

7.2 8 + 1) ∗ ( ) .6 2

= (12 + 1) ∗ 4 = 13 ∗ 4 = 52

109 | P a g e

Practice Problem Set 11.2 1) 3 + 7 + 11 + ⋯ + 39 =

2) 6 + 10 + 14 + ⋯ + 42 =

3) 9 + 14 + 19 + ⋯ + 59 =

4) 4 + 7 + 10 + ⋯ + 46 =

5) 5 + 12 + 19 + ⋯ + 75 =

6) 2 + 7 + 12 + ⋯ + 52 =

7) 14 + 18 + 22 + ⋯ + 70 =

8) 7 + 15 + 23 + ⋯ + 79 =

9) 13 + 18 + 23 + ⋯ + 73 =

10) 8.2 + 8.8 + 9.4 + ⋯ + 14.8 =

11) 4 + 11 + 18 + ⋯ + 60 =

12) 18 + 35 + 52 + ⋯ + 188 =

110 | P a g e

11.3

Geometric Sequences Similar to arithmetic sequences who can have any number as the first term and every consecutive term has a common difference, a geometric sequence can also have any number as the first term and every consecutive term has a common ratio. Instead of adding the common difference to get the next term, you multiply the common ratio. Algebraically, this can be expressed as:

𝑎, 𝑎𝑟, 𝑎𝑟 2 , 𝑎𝑟 3 , 𝑎𝑟 4 …

Note that 𝑎 and 𝑟 can be any number (fractional, negative, irrational, etc.). To find the 𝑛𝑡ℎ term of a geometric sequence, plug into the following equation:

𝑛𝑡ℎ 𝑡𝑒𝑟𝑚 𝑜𝑓 𝑔𝑒𝑜𝑚𝑒𝑡𝑟𝑖𝑐 𝑠𝑒𝑞𝑢𝑒𝑛𝑐𝑒: 𝑎 ∗ 𝑟 𝑛−1

Geometric sequences go on infinitely. When the ratio is greater than 1 (𝑟 > 1), the terms will tend to infinity. However, when the ratio is less than 1 (𝑟 < 1), the terms will tend to 0. In the latter case, there exists a formula for the sum of all terms in this sequence. Given a sequence that starts with a value 𝑎 and has a common ratio 𝑟 with 𝑟 < 1, the sum of all terms in this sequence is:

∑=

𝑎 1−𝑟

111 | P a g e

Let’s work with a couple of sequences.

𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 10𝑡ℎ 𝑡𝑒𝑟𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑒𝑞𝑢𝑒𝑛𝑐𝑒: 3, 6, 12, 24 … 𝑎 = 3 𝑎𝑛𝑑 𝑟 = 2 10𝑡ℎ 𝑡𝑒𝑟𝑚: 3 ∗ 29 = 3 ∗ 512 = 1500 + 36 = 1536

3 3 𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑖𝑛𝑓𝑖𝑛𝑖𝑡𝑒 𝑔𝑒𝑜𝑚𝑒𝑡𝑟𝑖𝑐 𝑠𝑒𝑞𝑢𝑒𝑛𝑐𝑒: 3, , , … 2 4 1 𝑎 = 3 𝑎𝑛𝑑 𝑟 = 2 3 𝑆𝑢𝑚: 1 1− 2 3 = 1 (2) =6

112 | P a g e

𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑖𝑛𝑓𝑖𝑛𝑖𝑡𝑒 𝑔𝑒𝑜𝑚𝑒𝑡𝑟𝑖𝑐 𝑠𝑒𝑞𝑢𝑒𝑛𝑐𝑒:

4 8 16 , , … 7 21 63

4 2 𝑎𝑛𝑑 𝑟 = 7 3 4 (7) 𝑆𝑢𝑚: 2 1− 3 4 (7) = 1 (3) 4 = ∗3 7 12 = 7

𝑎=

Practice Problem Set 11.3 1) 𝑊ℎ𝑎𝑡 𝑖𝑠 𝑡ℎ𝑒 9𝑡ℎ 𝑡𝑒𝑟𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑒𝑞𝑢𝑒𝑛𝑐𝑒 1, 2, 4, 8: 1 1

2) 𝑊ℎ𝑎𝑡 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑒𝑞𝑢𝑒𝑛𝑐𝑒: 1, , , … ? 2 4

3) 𝑊ℎ𝑎𝑡 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑒𝑞𝑢𝑒𝑛𝑐𝑒: 9, 3,1 … ? 7

2

2

7

4) 𝑊ℎ𝑎𝑡 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑒𝑞𝑢𝑒𝑛𝑐𝑒: , 1, , … ? 1

5) 𝑊ℎ𝑎𝑡 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑒𝑞𝑢𝑒𝑛𝑐𝑒: 3, ,

1

6 108

6) 𝑊ℎ𝑎𝑡 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑒𝑞𝑢𝑒𝑛𝑐𝑒: 10, 6,

18 5

,…? ,…?

113 | P a g e

11.4

Fibonacci The Fibonacci sequence is a sequence of numbers where the next number is the sum of the previous two terms. The original sequence starts with 0 and 1. The series begins:

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233 …

It is advisable to be familiar with the first 10 to 15 terms of the sequence. The convention for denoting Fibonacci numbers is 𝐹0 = 0 and 𝐹1 = 1. Fibonacci-like sequences are sequences that have two different numbers to start but follow the same convention of adding the previous two terms to get the third term. A common example is the Lucas sequence:

2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, …

Most questions pertaining to Fibonacci or Fibonacci-like sequences are seen in the fourth column of Number Sense. Rather than asking a specific term from a sequence, the question usually asks the sum of a certain number of terms. This can be solved in a variety of ways. 𝑛

∑ 𝐹𝑖 = 𝐹𝑛+2 − 𝐹2 = 2𝐹𝑛 + 𝐹𝑛−1 − 𝐹2 𝑖=1

114 | P a g e

This formula states that the sum of the first 𝑛 terms of a Fibonacci sequence is the sum of twice the value of the Fibonacci term 2 terms ahead and the Fibonacci term 1 term ahead minus the second Fibonacci term. I understand this may be confusing; maybe an example will help.

2 + 3 + 5 + 8 + ⋯ + 89 + 144 = 𝑇ℎ𝑒 𝑛𝑒𝑥𝑡 𝑡𝑒𝑟𝑚 𝑖𝑛 𝑡ℎ𝑒 𝑠𝑒𝑞𝑢𝑒𝑛𝑐𝑒 𝑖𝑠 89 + 144 = 233 = 𝐹𝑛+1 𝑇ℎ𝑒 𝑡𝑒𝑟𝑚 𝑎𝑓𝑡𝑒𝑟 𝑡ℎ𝑎𝑡 𝑖𝑠 144 + 233 = 377 = 𝐹𝑛+2 𝑆𝑢𝑚: 𝐹𝑛+2 − 𝐹2 = 377 − 3 = 374 𝑇ℎ𝑒 𝑜𝑡ℎ𝑒𝑟 𝑓𝑜𝑟𝑚𝑢𝑙𝑎: 2𝐹𝑛 + 𝐹𝑛−1 − 𝐹2 = 2(144) + (89) − 3 = 288 + 86 = 374

This formula works best when you know the last two numbers in the sequence (as compared to just the last number in the sequence). Other common questions ask you the sum of the first 8, first 9, first 10, or first 11 terms in a Fibonacci-like sequence. There are simple formulas for each: 𝑆𝑢𝑚 𝑜𝑓 𝐹𝑖𝑟𝑠𝑡 𝐸𝑖𝑔ℎ𝑡: 4𝐹7 + 𝐹3 𝑆𝑢𝑚 𝑜𝑓 𝐹𝑖𝑟𝑠𝑡 𝑁𝑖𝑛𝑒: 7𝐹7 − 𝐹4 𝑆𝑢𝑚 𝑜𝑓 𝐹𝑖𝑟𝑠𝑡 𝑇𝑒𝑛: 11𝐹7 𝑆𝑢𝑚 𝑜𝑓 𝐹𝑖𝑟𝑠𝑡 𝐸𝑙𝑒𝑣𝑒𝑛: 18𝐹7 − 𝐹3

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Practice Problem Set 11.4 1) 𝑇ℎ𝑒 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 9 𝑡𝑒𝑟𝑚𝑠 𝑜𝑓 𝑡ℎ𝑒 𝐹𝑖𝑏. 𝑠𝑒𝑞𝑢𝑒𝑛𝑐𝑒 1,1,2,3,5 … 𝑖𝑠: 2) 𝑇ℎ𝑒 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 12 𝑡𝑒𝑟𝑚𝑠 𝑜𝑓 𝑡ℎ𝑒 𝐹𝑖𝑏. 𝑠𝑒𝑞𝑢𝑒𝑛𝑐𝑒 1,2,3,5,8,13,21 … 𝑖𝑠: 3) 𝑇ℎ𝑒 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 10 𝑡𝑒𝑟𝑚𝑠 𝑜𝑓 𝑡ℎ𝑒 𝐹𝑖𝑏. 𝑠𝑒𝑞𝑢𝑒𝑛𝑐𝑒 3,4,7,11,18 … 𝑖𝑠: 4) 𝑇ℎ𝑒 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 9 𝑡𝑒𝑟𝑚𝑠 𝑜𝑓 𝑡ℎ𝑒 𝐹𝑖𝑏. 𝑠𝑒𝑞𝑢𝑒𝑛𝑐𝑒 3,5,8,13,21 … 𝑖𝑠: 5) 𝑇ℎ𝑒 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 8 𝑡𝑒𝑟𝑚𝑠 𝑜𝑓 𝑡ℎ𝑒 𝐹𝑖𝑏. 𝑠𝑒𝑞𝑢𝑒𝑛𝑐𝑒 4,6,10,16,26 … 𝑖𝑠: 6) 𝑇ℎ𝑒 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝐹𝑖𝑏. 𝑠𝑒𝑞𝑢𝑒𝑛𝑐𝑒 2,5,7,12, … 131,212 𝑖𝑠: 7) 𝑇ℎ𝑒 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝐹𝑖𝑏. 𝑠𝑒𝑞𝑢𝑒𝑛𝑐𝑒 2,1,3,4, … 123,199 𝑖𝑠:

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11.5

Other Sequences The following sequences seldom show up but are worth knowing. Squares

1, 4, 9, 16, 25, 36, 49, 64 … 𝑛

∑ 𝑥2 = 𝑥=1

𝑛 ∗ (𝑛 + 1) ∗ (2𝑛 + 1) 6

Cubes

1, 8, 27, 64, 125, 216, 343, 512 … 𝑛

2

𝑛 ∗ (𝑛 + 1) ∑ 𝑥3 = ( ) = (𝑇𝑛 )2 2

𝑥=1

𝑇𝑛 𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡𝑠 𝑡ℎ𝑒 𝑛𝑡ℎ 𝑡𝑟𝑖𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑛𝑢𝑚𝑏𝑒𝑟

Triangular Numbers

1, 3, 6, 10, 15, 21, 28, 36 … 𝑛

∑ 𝑇𝑛 = 𝑥=1

𝑛 ∗ (𝑛 + 1) ∗ (𝑛 + 2) 6

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12

Memorization

12.1

Fractions and Decimals It is crucial you know the following table so that you can recognize special numbers in problems and convert from one form to another and simplifying the problem.

Fraction

Decimal

Fraction

Decimal

1⁄ 2 3⁄ 4 3⁄ 8 7⁄ 8 1⁄ 3 1⁄ 6

.5

1⁄ 4 1⁄ 8

.25

1⁄ 12 1⁄ 5 1⁄ 11 1⁄ 14 1⁄ 20

.75 .375 .875 . 3̅ . 16̅ . 083̅ .2 ̅̅̅̅ . 09 ̅̅̅̅̅̅̅̅̅̅ . 0714285 .05

5⁄ 8 1⁄ 16 2⁄ 3 5⁄ 6 11⁄ 12 1⁄ 9 1⁄ 7 1⁄ 15 1⁄ 40

.125 .625 .0625 . 6̅ . 83̅ . 916̅ . 1̅ ̅̅̅̅̅̅̅̅̅̅ . 142857 . 3̅ .025

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12.2

Approximations Fractions, decimals, and percents are often seen in questions that ask you to approximate the answer; most times they will simplify with other numbers in the expression. The values usually aren’t the exact fraction or decimal but are close enough that you can round and still fall in the range (granted that everything else goes right). The key is to be able to recognize the fraction, convert it to the form you need it to be, and then simplify it with other numbers in the expression. Here are some examples.

375 ÷ 833 ∗ 555 3 5 5 ≈ ( ∗ 1000) ÷ ( ∗ 1000) ∗ ( ∗ 1000) 8 6 9 =

3 6 5 ∗ ∗ ∗ 1000 8 5 9 1 = ∗ 1000 4 = 250

𝐶𝑜𝑟𝑟𝑒𝑐𝑡 𝑟𝑎𝑛𝑔𝑒: 238 𝑡𝑜 262

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Practice Problem Set 12.2 7

1) 142.857 ∗ 428.571 ≈

2) 888 ∗ 87.5% ÷

3) 62.5 ÷ 83.3 ∗ 888 ≈

4) 678 ∗ 12.6% ÷ .5 ≈

1

5) 106 % ∗ 799 ∗ .125 ≈ 4

2

7) 429 ∗ 8.6 ∗ ≈

5 12

≈

∗ .196 ≈

8) 285.714 ∗ 209 ≈

5

2

2

9) 654 ∗ ∗ 16.7% ≈

10) 3333 ÷ 66 % ∗ 3.6 =

3

11) 8888 ∗ 62.5% ∗

6) 833 ÷

11

5 11

3

=

12) 428.571 ∗ 87.5 =

120 | P a g e

13

Adding Fractions

13.1

Adding Inverses Operations with fractions are a big aspect of Number Sense. Luckily, most all questions asked have simple tricks with recurring patterns and simple algebraic proofs, so fear not! Fraction questions are quick and simple, not cumbersome nor lengthy. The first trick involves adding two fractions, one of which is the multiplicative inverse of the other. First, let’s see the trick:

𝑎 𝑏 + 𝑏 𝑎 𝑎2 𝑏 2 = + 𝑎𝑏 𝑎𝑏 𝑎2 + 𝑏 2 = 𝑎𝑏 𝑎2 + 𝑏 2 + (2𝑎𝑏 − 2𝑎𝑏) = 𝑎𝑏 𝑎2 − 2𝑎𝑏 + 𝑏 2 + 2𝑎𝑏 = 𝑎𝑏 (𝑎 − 𝑏)2 + 2𝑎𝑏 = 𝑎𝑏 (𝑎 − 𝑏)2 2𝑎𝑏 = + 𝑎𝑏 𝑎𝑏 (𝑎 − 𝑏)2 =2+ 𝑎𝑏 Note: in the fourth line, we added the value of 0 in the form of (2𝑎𝑏 − 2𝑎𝑏). Since adding 0 does not change the value of the numerator, we can use this to help simplify the expression.

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Let’s break down the algebraic terms into steps: Step 1: write the whole number 2. Step 2: find the difference between the two numbers and square it (since the difference is squared, it will always be positive). Note this as the numerator. Step 3: multiply the two numbers together; this becomes the denominator. Simplify with the numerator if possible; write the final fraction next to the whole number. Time for an example.

7 3 + 3 7 (7 − 3)2 =2+ 7∗3 16 =2 21 4 9 + 9 4 (9 − 4)2 =2+ 4∗9 =2

25 36

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Practice Problem Set 13.1 3

4

4

3

4

7

7

4

5

4

4

5

3

8

8

3

6

5

5

6

1) + = 3) + = 5) + = 7) + − 2 = 9) + + 1 =

2

5

5

2

7

2

2

7

9

4

4

9

2) + = 4) + = 6) + = 8)

13 7

10)1

+ 11 14

7 13

+

= 14 11

=

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13.2

Fractions in the form a/b+b/(a+b) Oftentimes fractions in this form are difficult to recognize and are simple enough to cross multiply (refer to Trick 13.4). However, the trick does save time and thus should be implemented whenever possible.

𝑎 𝑏 + 𝑏 𝑎+𝑏 𝑎 ∗ (𝑎 + 𝑏) 𝑏2 = + 𝑏 ∗ (𝑎 + 𝑏) 𝑏 ∗ (𝑎 + 𝑏) 𝑎2 + 𝑎𝑏 + 𝑏 2 = 𝑏 ∗ (𝑎 + 𝑏) 𝑏 ∗ (𝑎 + 𝑏) + 𝑎2 = 𝑏 ∗ (𝑎 + 𝑏) 𝑏 ∗ (𝑎 + 𝑏) 𝑎2 = + 𝑏 ∗ (𝑎 + 𝑏) 𝑏 ∗ (𝑎 + 𝑏) 𝑎2 =1+ 𝑏 ∗ (𝑎 + 𝑏)

Let’s break this down. Step 1: write the number 1 as the whole number. Step 2: square the 𝑎 term. Remember this to be the numerator. Step 3: multiply the denominators of the two fractions to get the denominator of the answer. Simplify with the numerator from Step 2 if possible. Write the final fraction as the fraction next to the 1.

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Here’s an example.

2 5 + 5 7 22 =1+ 5∗7 4 =1 35 Practice Problem Set 13.2 3

4

4

7

1) + = 4

7

7

11

5

4

4

9

3

8

8

11

3) +

=

5) + = 7) +

−2=

3

5

5

8

7

2

2

9

2) + = 4) + = 4

9

9

13

6) + 8)

7 13

+

=

13 20

=

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13.3

Special Fractions Sometimes in the middle of a Number Sense test you come across a difference of fractions that seems to have no trick. The numbers are large, there’s heavy computation, and time’s dwindling. Luckily, nine times out of ten the question is a special form that makes the calculation significantly easier. These questions ask about a difference of two fractions where the numerator of the second fraction is a multiple of the numerator of the first fraction plus a certain value. The denominator of the second fractions is also the same multiple of the denominator of the first fraction minus that same certain value used in the numerator. Hopefully the algebra makes a bit more sense. We’ll use 𝑘 to represent the multiple and 𝑑 to represent that certain value. In that case, we can represent the question as the following:

𝑎 𝑘∗𝑎+𝑑 − 𝑏 𝑘∗𝑏−𝑑 With this, we can derive a trick using some cross multiplication.

𝑎 𝑘∗𝑎+𝑑 − 𝑏 𝑘∗𝑏−𝑑 =

𝑎 ∗ (𝑘 ∗ 𝑏 − 𝑑) 𝑏 ∗ (𝑘 ∗ 𝑎 + 𝑑) − 𝑏 ∗ (𝑘 ∗ 𝑏 − 𝑑) 𝑏 ∗ (𝑘 ∗ 𝑏 − 𝑑) =

𝑘𝑎𝑏 − 𝑎𝑑 − (𝑘𝑎𝑏 + 𝑏𝑑) 𝑏 ∗ (𝑘 ∗ 𝑏 − 𝑑)

=

𝑘𝑎𝑏 − 𝑎𝑑 − 𝑘𝑎𝑏 − 𝑏𝑑 𝑏 ∗ (𝑘 ∗ 𝑏 − 𝑑) =

−𝑎𝑑 − 𝑏𝑑 𝑏 ∗ (𝑘 ∗ 𝑏 − 𝑑)

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=

−𝑑(𝑎 + 𝑏) 𝑏 ∗ (𝑘 ∗ 𝑏 − 𝑑)

Let’s use the formula in a handful of examples.

7 29 − 13 51 Step 1: realize the following:

7 29 7 4∗7+1 − = − 13 51 13 4 ∗ 13 − 1 Step 2: calculate 𝑑. In this example 𝑑 = 1. Step 3: Add the numerator and denominator of the first fraction, then multiply by 𝑑. This will be the numerator of your answer. In this example, we see−𝑑 ∗ (𝑎 + 𝑏) = −1 ∗ (7 + 13) = −20. Step 4: Multiply the denominators of the two fractions to get the denominator of your final answer. For this 13 ∗ 51 = 13 ∗ 50 + −20 13 ∗ 1 = 650 + 13 = 663. Final answer: . 663

5 29 − 12 73 5 5∗6−1 = − 12 12 ∗ 6 + 1 −(−1) ∗ (5 + 12) 12 ∗ 73 17 = 876

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3 23 − 7 47 3 3∗7+2 = − 7 3∗7−2 −2 ∗ (3 + 7) 7 ∗ 47 −20 = 329 Practice Problem Set 13.3 5

24

8

41

1) − 3) 5)

8 11 7 20

− −

87 122 55 161

2

101

3

149

7) − 9)

=

17 22

−

35 43

8

87

9

100

2) − =

4)

=

6)

11 16 8 11

− −

32 49 31 45

4

21

7

34

=

8) −

=

10)

11 12

−

= = =

= 32 37

=

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13.4

Adding Fractions with no Trick There come times where no trick is available to add or subtract the fractions given. When this situation arises, you have to add fractions the good old-fashioned way: with least common denominators. The numerators and denominators are usually manageable. Practice is what will make these problems fast.

Practice Problem Set 13.4 3

4

5

7

1) + = 4

9

7

11

3) 3 −

=

1

3

4

7

7

2

8

11

2

7

3

8

5) 1 − 2 = 7) 4 + 2

2

3

5

8

2) 1 + =

=

9) 7 + 7 =

4

5

5

7

4) 7 + 3 = 3

1

7

4

6) − 3 = 3

3

8

11

8) 1 − 2

=

1

13

5

15

10) 5 − 15

=

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14

Multiplying Fractions

14.1

Whole Numbers Same and Fractions Add to 1 This is probably the simplest trick when multiplying fractions. It will ask you to multiply two mixed numbers, the whole numbers of which are the same and the fractions add to 1. To derive the trick, let’s represent the whole number as 𝑎 and the fraction of the first mixed number as 𝑓. In that case, the fraction of the second fraction will be 1 − 𝑓. Now it’s algebra time!

(𝑎 + 𝑓) ∗ (𝑎 + [1 − 𝑓]) = (𝑎 + 𝑓) ∗ (𝑎 + 1 − 𝑓) = 𝑎2 + 𝑎 − 𝑎𝑓 + 𝑎𝑓 + 𝑓 − 𝑓 2 = 𝑎2 + 𝑎 + 𝑓 − 𝑓 2 = 𝑎 ∗ (𝑎 + 1) + 𝑓 ∗ (1 − 𝑓)

Although it doesn’t look like much, the last line is a real game changer. It tells us the following things: 1) The fraction of the answer is the product of the two fractional components of the mixed numbers. 2) The whole number of the answer is the product of the whole number in the mixed numbers multiplied by one more than itself. An example is in order now. 2 5 4 ∗4 7 7 2 5 = 4 ∗ (4 + 1) + ( ) ∗ ( ) 7 7 = 20

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Practice Problem Set 14.1 3

1

4

4

3

4

7

7

6

1

7

7

1) 4 ∗ 4 =

2) 5

3) 7 ∗ 7 =

4

9

9

2 11

∗ 12

9 13

1

2

3

3

3

2

5

5

=

6) 9 ∗ 9 =

7) 11 ∗ 11 = 9) 12

13

∗5

4) 3 ∗ 3 =

5) 6 ∗ 6 = 5

4

9 11

1

4

5

5

8) 10 =

11) 25 ∗ 25 =

3 10

10) 15

∗ 10

7 12

7 10

∗ 15

=

5 12

=

1 2

12) (50 ) = 2

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14.2

Multiplying 2 Mixed Numbers with the Same

Fraction Multiplication of two such numbers can be simple in itself, but it can go even faster if you understand FOIL with fractions. Let 𝑎1 and 𝑎2 represent the two whole numbers and 𝑓 represent the fraction shared by both proper numbers. In that case, we can see that:

(𝑎1 + 𝑓) ∗ (𝑎2 + 𝑓) = 𝑎1 ∗ 𝑎2 + 𝑎1 ∗ 𝑓 + 𝑓 ∗ 𝑎2 + 𝑓 2 = 𝑎1 ∗ 𝑎2 + 𝑓 ∗ (𝑎1 + 𝑎2 ) + 𝑓 2 In most cases, the term 𝑓 ∗ (𝑎1 + 𝑎2 ) results in an integer. This means that the only fraction in the answer is 𝑓 2 . When working with such questions, follow these steps: Step 1: Add the two whole numbers and multiply the sum by the fraction shared by both terms. Step 2: If the result of Step 1 is an integer, add it to the product of the two whole numbers and write this as the whole number of the answer. If the result of Step 1 is a mixed number, add the whole number component to the product of the two whole numbers. Wait to write the whole number until after Step 3 because the result of Step 3 may be greater than 1. Step 3: Square the fraction. If from Step 2 there is a fractional component still left, add this to the square of the fraction. Now let’s work some examples to get a feel for the different cases.

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1 1 6 ∗9 3 3 𝑊ℎ𝑜𝑙𝑒 𝑛𝑢𝑚𝑏𝑒𝑟: 6 ∗ 9 + (6 + 9) ∗

1 3

= 54 + 5 = 59 1 2 𝐹𝑟𝑎𝑐𝑡𝑖𝑜𝑛: ( ) 3 =

1 9

𝐹𝑖𝑛𝑎𝑙 𝑎𝑛𝑠𝑤𝑒𝑟: 59

1 9

3 3 4 ∗2 7 7 𝑊ℎ𝑜𝑙𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 = 4 ∗ 2 + (4 + 2) ∗

3 7

18 7 4 = 8+2 7 4 = 10 → 10 7 =6+

3 2 4 𝐹𝑟𝑎𝑐𝑡𝑖𝑜𝑛: ( ) + 7 7 =

9 28 + 49 49 37 = 49

𝐹𝑖𝑛𝑎𝑙 𝑎𝑛𝑠𝑤𝑒𝑟: 10

37 49 133 | P a g e

4

9 9 ∗7 10 10

𝑊ℎ𝑜𝑙𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 = 4 ∗ 7 + (4 + 7) ∗

9 10

99 10 9 = 20 + 9 10 9 = 29 → 29 10 = 28 +

9 2 9 𝐹𝑟𝑎𝑐𝑡𝑖𝑜𝑛: ( ) + 10 10 =

81 90 + 100 100 71 =1 100

𝐹𝑖𝑛𝑎𝑙 𝑎𝑛𝑠𝑤𝑒𝑟: 30

71 100

Now I understand the second and third examples look quite computational and time-consuming, but it should be noted that most all questions on Number Sense with the same fraction will follow the first example.

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Practice Problem Set 14.2 3

3

4

4

1) 4 ∗ 8 = 3

3

7

7

2) 4

3) 7 ∗ 14 = 6

6

7

7

5

9

9

2

13

1

1

3

3

3

3

5

5

2

8) 10

3 10

=

10) 15

11) 25 ∗ 50 =

12) 23

11

∗ 10

4

=

6) 9 ∗ 6 =

7) 11 ∗ 7 = 9) 12

13

∗9

4) 3 ∗ 9 =

5) 6 ∗ 8 = 5

4

11

1

1

5

5

∗ 20

7 12 13 17

∗9

3 10 7 12

∗ 11

= =

13 17

=

135 | P a g e

14.3

Fractions in the Form a*a/(a+b) When people usually see fractions in this form, they quickly square the numerator and then go about trying to simplify it with the denominator to get the final answer into a mixed number. Although this may be fast on certain questions, the following trick is sure to be faster (with a bit of practice): 𝑎 𝑎∗ 𝑎+𝑏 𝑎2 = 𝑎+𝑏 𝑎2 + (𝑏 2 − 𝑏 2 ) = 𝑎+𝑏 𝑎2 − 𝑏 2 + 𝑏 2 = 𝑎+𝑏 (𝑎 + 𝑏) ∗ (𝑎 − 𝑏) + 𝑏 2 = 𝑎+𝑏 =

(𝑎 + 𝑏) ∗ (𝑎 − 𝑏) 𝑏2 + 𝑎+𝑏 𝑎+𝑏 𝑏2 = (𝑎 − 𝑏) + 𝑎+𝑏

A simple analysis of the final algebraic expression results in the following steps to follow: Step 1: Calculate the value of 𝑏 Step 2: Find 𝑎 − 𝑏; this is the whole number of the answer. Step 3: Square 𝑏; this becomes the numerator of the fraction. The denominator is simply the denominator of the fractional term. Let’s see it in action!

136 | P a g e

13 ∗

13 17

𝑏=4 42 (13 − 4) + 17 16 =9 17

17 ∗

17 15

𝑏 = −2 (−2)2 (17 − [−2]) + 17 4 = 19 17 Practice Problem Set 14.3 1) 12 ∗ 3) 9 ∗ 5) 8 ∗

12 17

9 13 8 11

7) 17 ∗ 9) 19 ∗

=

2) 7 ∗

7 12

=

4) 11 ∗

=

6) 15 ∗

17 13 19 14

=

11 7 15 8 23

=

8) 23 ∗

=

10) 30 ∗

25

= = =

30 31

=

137 | P a g e

14.4

Fractions in the Form a*(a+n)/(a+2n) Here’s another common, easy fraction’s trick that shows up semi-frequently. Questions in this form are fairly easy to recognize and evaluate. Let’s take a look at the algebra below.

𝑎∗ =

𝑎+𝑛 𝑎 + 2𝑛

𝑎 ∗ (𝑎 + 𝑛) 𝑎 + 2𝑛

𝑎2 + 𝑎𝑛 = 𝑎 + 2𝑛 𝑎2 + 𝑎𝑛 + (2𝑛2 − 2𝑛2 ) = 𝑎 + 2𝑛 𝑎2 + 𝑎𝑛 − 2𝑛2 + 2𝑛2 = 𝑎 + 2𝑛 (𝑎 − 𝑛) ∗ (𝑎 + 2𝑛) + 2𝑛2 = 𝑎 + 2𝑛 (𝑎 − 𝑛) ∗ (𝑎 + 2𝑛) 2𝑛2 = + 𝑎 + 2𝑛 𝑎 + 2𝑛 2𝑛2 = (𝑎 − 𝑛) + 𝑎 + 2𝑛 Skipping down to the last line, we can extract these steps when solving questions in this form: Step 1: Determine 𝑛. Step 2: Square 𝑛 and multiply by 2 to get the numerator of the fraction. Simplify with the denominator if possible and write the fraction as the fractional part of the answer. If the numerator turns out to be greater than the denominator, make it into a mixed number and carry any values. 138 | P a g e

Step 3: Subtract 𝑛 from 𝑎 and add any carries to get the whole number of the final answer. Seems simple enough! The number one part that messes people up is forgetting to double the numerator and then simplify. Since you can’t mark over or erase on tests, you should simplify the fraction in your mind before writing the answer. Let’s see a couple of examples.

12 ∗

13 14

𝑛=1 2 ∗ 12 14 2 = 11 + 14 1 = 11 7

(12 − 1) +

13 ∗

17 21

𝑛=4 2 ∗ 42 (13 − 4) + 21 32 =9+ 21 11 = 10 21

139 | P a g e

17 ∗

15 13

𝑛 = −2 2 ∗ (−2)2 (17 − [−2]) + 13 8 = 19 13 Practice Problem Set 14.4 6

1) 5∗ =

2) 7 ∗

7

3) 9 ∗ 5) 8 ∗

11 13 11 14

7) 17 ∗ 9) 19 ∗

9 11

=

9

=

4) 11 ∗ =

=

6) 15 ∗

15 13 21 23

7

14 13 17

=

8) 12 ∗

=

10) 30 ∗

22

= =

37 44

=

140 | P a g e

14.5

Multiplying using Improper Fractions More like a technique rather than a trick, multiplying using improper fractions is easier than multiplying mixed numbers for the following reasons: 1) Oftentimes numerators and denominators will cancel out. 2) It is easier to multiply two integers than FOIL two mixed numbers. Thus, when you see a question which asks the product of two mixed numbers and there seems to be no trick, try converting the numbers to improper fractions. More than likely something will cancel.

4 2 1 ∗2 5 9 9 20 = ∗ 5 9 =4 Practice Problem Set 14.5 3

1

4

7

1) 1 ∗ 1 = 4

7

7

9

3

4

8

5

3) 2 ∗ = 5) 4 ∗ = 2

2

8

7

7) 5 ∗ 2 =

2

5

5

9

2) 5 ∗ = 4) 5

7 10

∗

4 19

1

11

6

13

6) 2 ∗ 1 8) 3

=

6 13

∗4

=

12 17

=

141 | P a g e

14.6

Comparison of Fractions Nearly every other test will ask you to compare two improper fractions. The general technique is to find the least common denominator and then find the numerator for both fractions; the fraction with the larger numerator is the larger fraction. Instead of following this lengthy trick, we will employ the same principles but simplify calculations. This is done using crossmultiplication. The basic idea goes that instead of trying to find the least common denominator, we just find a common denominator, namely the product of the two denominators. But we don’t evaluate this. Instead, we just determine the numerators if this were the denominator and compare those numerators. In other words, we’re going to multiply the denominator of one fraction by the numerator of the other and then the denominator of the other fraction by the numerator of the first (hence, cross multiply). 𝑎 𝑏

𝑐 𝑑

𝑁𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟 1 = 𝑎 ∗ 𝑑 𝑁𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟 2 = 𝑏 ∗ 𝑐

Let’s see a few examples to make sense of this all.

142 | P a g e

𝑊ℎ𝑖𝑐ℎ 𝑖𝑠 𝑙𝑎𝑟𝑔𝑒𝑟,

4 13 𝑜𝑟 ? 7 24

4 ∗ 24 = 96 13 ∗ 7 = 91 𝑆𝑖𝑛𝑐𝑒 96 > 91, 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝐹𝑖𝑛𝑎𝑙 𝑎𝑛𝑠𝑤𝑒𝑟:

𝑊ℎ𝑖𝑐ℎ 𝑖𝑠 𝑠𝑚𝑎𝑙𝑙𝑒𝑟,

4 13 > 7 24

4 7

−4 −24 𝑜𝑟 ? 9 53

(−4) ∗ 53 = −212 9 ∗ (−24) = −216 𝑆𝑖𝑛𝑐𝑒 − 212 > −216, 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝐹𝑖𝑛𝑎𝑙 𝑎𝑛𝑠𝑤𝑒𝑟:

−4 −24 > 9 53

−24 53

Practice Problem Set 14.6 3

6

7

13

1) 𝑊ℎ𝑖𝑐ℎ 𝑖𝑠 𝑙𝑎𝑟𝑔𝑒𝑟, 𝑜𝑟 3) 𝑊ℎ𝑖𝑐ℎ 𝑖𝑠 𝑙𝑎𝑟𝑔𝑒𝑟, 5) 𝑊ℎ𝑖𝑐ℎ 𝑖𝑠 𝑙𝑎𝑟𝑔𝑒𝑟,

12 17

19

18

𝑜𝑟

9) 𝑊ℎ𝑖𝑐ℎ 𝑖𝑠 𝑠𝑚𝑎𝑙𝑙𝑒𝑟,

:

3 11

8

31

−10 17

4) 𝑊ℎ𝑖𝑐ℎ 𝑖𝑠 𝑙𝑎𝑟𝑔𝑒𝑟,

−2

3

7) 𝑊ℎ𝑖𝑐ℎ 𝑖𝑠 𝑠𝑚𝑎𝑙𝑙𝑒𝑟, 𝑜𝑟

𝑜𝑟

4

25

7

48

2) 𝑊ℎ𝑖𝑐ℎ 𝑖𝑠 𝑠𝑚𝑎𝑙𝑙𝑒𝑟, 𝑜𝑟

13

𝑜𝑟

−12

:

:

−5 11

6) 𝑊ℎ𝑖𝑐ℎ 𝑖𝑠 𝑠𝑚𝑎𝑙𝑙𝑒𝑟,

:

𝑜𝑟

4 13

:

−11

𝑜𝑟

2

1

9

5

24 5

:

:

17

8) 𝑊ℎ𝑖𝑐ℎ 𝑖𝑠 𝑙𝑎𝑟𝑔𝑒𝑟, 𝑜𝑟 :

−14 23

:

10) 𝑊ℎ𝑖𝑐ℎ 𝑖𝑠 𝑙𝑎𝑟𝑔𝑒𝑟,

−6 13

𝑜𝑟

−13 27

:

143 | P a g e

15

Repeating Decimals

15.1

Repeating Decimals in the Form of .aaa…,

.ababab…, or .abcabcabc… The two most important pieces of information to know before starting repeating decimals are the following:

1 = 0.1 10 1 = 0.111 … = 0. 1̅ 9 Note: the bar above the 1 means that the pattern of digits under the bar repeats continuously. When you have a repeating decimal where every digit is the 1 same, the fraction will be a multiple of . For example: 9

. 4̅ =

4 9

6 2 1. 6̅ = 1 = 1 9 3 When you have a repeating decimal in the form of . 𝑎𝑏𝑎𝑏𝑎𝑏 … = . ̅̅̅ 𝑎𝑏, the fraction will have a denominator of 99 and a numerator of 𝑎𝑏. For example:

̅̅̅̅ = . 43

43 99 144 | P a g e

̅̅̅̅ = 1 1. 57

57 19 =1 99 33

̅̅̅̅̅ Lastly, repeating decimals in the form of . 𝑎𝑏𝑐𝑎𝑏𝑐𝑎𝑏𝑐 … =. 𝑎𝑏𝑐 will have a denominator of 999 and a numerator of 𝑎𝑏𝑐.

̅̅̅̅̅ = . 274

1. ̅̅̅̅̅ 123 = 1

274 999

123 41 =1 999 333

The process can also be reversed, where you’re given a fraction and asked to write the first handful of digits of the decimal expansion. This can go like this:

𝑊𝑟𝑖𝑡𝑒 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 4 𝑑𝑖𝑔𝑖𝑡𝑠 𝑜𝑓

14 99

14 ̅̅̅̅ → .1414 = . 14 99

𝑊𝑟𝑖𝑡𝑒 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 4 𝑑𝑖𝑔𝑖𝑡𝑠 𝑜𝑓

4 111

4 36 ̅̅̅̅̅ → .0360 = = . 036 111 999

145 | P a g e

Practice Problem Set 15.1 1) . 444 … =

2) . 232323 … =

3) . 454545 … =

4) . 159159159 … =

5) . 987987987 … =

6) . 374374374 … = 7

7) 𝑊𝑟𝑖𝑡𝑒 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 4 𝑑𝑖𝑔𝑖𝑡𝑠 𝑜𝑓 : 9

8) 𝑊𝑟𝑖𝑡𝑒 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 4 𝑑𝑖𝑔𝑖𝑡𝑠 𝑜𝑓 9) 𝑊𝑟𝑖𝑡𝑒 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 4 𝑑𝑖𝑔𝑖𝑡𝑠 𝑜𝑓

13 99

:

146 999

10) 𝑊𝑟𝑖𝑡𝑒 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 4 𝑑𝑖𝑔𝑖𝑡𝑠 𝑜𝑓

8

: :

37

146 | P a g e

15.2

Repeating Decimals in the Form of .abbb…,

.abccc…, or .abcbcbc… Every fraction has a numerator and a denominator; we’ll learn to determine each value separately. Determining the denominator of these fractions uses the two pieces of information highlighted at the beginning of Trick 15.1. The number of repeating digits is the decimal is the number of 9’s in the denominator, and the number of digits to the right of the decimal point but do not repeat are the number of 0’s in the denominator (before simplification). Let’s look at the three cases: . 𝑎𝑏𝑏𝑏 … →. 𝑎𝑏̅ 1 𝑟𝑒𝑝𝑒𝑎𝑡𝑖𝑛𝑔 𝑑𝑖𝑔𝑖𝑡, 1 𝑛𝑜𝑛 − 𝑟𝑒𝑝𝑒𝑎𝑡𝑖𝑛𝑔 𝑑𝑖𝑔𝑖𝑡 𝐷𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟: 90

. 𝑎𝑏𝑐𝑐𝑐 … →. 𝑎𝑏𝑐̅ 1 𝑟𝑒𝑝𝑒𝑎𝑡𝑖𝑛𝑔 𝑑𝑖𝑔𝑖𝑡, 2 𝑛𝑜𝑛 − 𝑟𝑒𝑝𝑒𝑎𝑡𝑖𝑛𝑔 𝑑𝑖𝑔𝑖𝑡𝑠 𝐷𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟: 900 ̅̅̅ . 𝑎𝑏𝑐𝑏𝑐𝑏𝑐 … →. 𝑎𝑏𝑐 2 𝑟𝑒𝑝𝑒𝑎𝑡𝑖𝑛𝑔 𝑑𝑖𝑔𝑖𝑡𝑠, 1 𝑛𝑜𝑛 − 𝑟𝑒𝑝𝑒𝑎𝑡𝑖𝑛𝑔 𝑑𝑖𝑔𝑖𝑡 𝐷𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟: 990

Now for the numerator. The first step is to imagine that the decimal is written with a vinculum (the overhead bar). Next, take out the decimal point. Finally, subtract the non-repeating part of the number from the entire number to get the numerator. Let’s see the three cases again:

147 | P a g e

. 𝑎𝑏𝑏𝑏 … →. 𝑎𝑏̅ 𝑁𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟: 𝑎𝑏 − 𝑎

. 𝑎𝑏𝑐𝑐𝑐 … →. 𝑎𝑏𝑐̅ 𝑁𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟: 𝑎𝑏𝑐 − 𝑎𝑏 ̅̅̅ . 𝑎𝑏𝑐𝑏𝑐𝑏𝑐 … →. 𝑎𝑏𝑐 𝑁𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟: 𝑎𝑏𝑐 − 𝑎

Now that we know the formulas, let’s work some problems!

. 90444 … 𝑡𝑜 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 . 90444 … = .904̅ 𝐷𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟: 900 𝑁𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟: 904 − 90 = 814 814 407 = 900 450 407 𝐹𝑖𝑛𝑎𝑙 𝑎𝑛𝑠𝑤𝑒𝑟: 450

𝐹𝑟𝑎𝑐𝑡𝑖𝑜𝑛:

148 | P a g e

. 1363636 … 𝑡𝑜 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 ̅̅̅̅ . 1363636 … = .136 𝐷𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟: 990 𝑁𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟: 136 − 1 = 135 135 27 9 3 = = = 990 198 66 22 3 𝐹𝑖𝑛𝑎𝑙 𝑎𝑛𝑠𝑤𝑒𝑟: 22

𝐹𝑟𝑎𝑐𝑡𝑖𝑜𝑛:

1.4222 … 𝑡𝑜 𝑖𝑚𝑝𝑟𝑜𝑝𝑒𝑟 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 1.4222 … = 1.42̅ 𝐷𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟: 90 𝑁𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟: 42 − 4 = 38 38 19 = 90 45 19 64 𝐹𝑖𝑛𝑎𝑙 𝑎𝑛𝑠𝑤𝑒𝑟: 1 + = 45 45 𝐹𝑟𝑎𝑐𝑡𝑖𝑜𝑛:

149 | P a g e

Practice Problem Set 15.2 1) . 2444 … =

2) . 1232323 … =

3) . 3454545 … =

4) . 41555 … =

5) . 9878787 … =

6) . 37444 … =

7) . 3777 … =

8) . 5464646 … =

9) 2.2464646 … =

10) 1.35777 … =

11) 3.3696969 … =

12) 5.12333 … =

13) 𝑊𝑟𝑖𝑡𝑒 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 4 𝑑𝑖𝑔𝑖𝑡𝑠 𝑜𝑓 14) 𝑊𝑟𝑖𝑡𝑒 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 4 𝑑𝑖𝑔𝑖𝑡𝑠 𝑜𝑓 15) 𝑊𝑟𝑖𝑡𝑒 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 4 𝑑𝑖𝑔𝑖𝑡𝑠 𝑜𝑓 16) 𝑊𝑟𝑖𝑡𝑒 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 4 𝑑𝑖𝑔𝑖𝑡𝑠 𝑜𝑓 17) 𝑊𝑟𝑖𝑡𝑒 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 4 𝑑𝑖𝑔𝑖𝑡𝑠 𝑜𝑓 18) 𝑊𝑟𝑖𝑡𝑒 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 4 𝑑𝑖𝑔𝑖𝑡𝑠 𝑜𝑓

7

:

90

23 90

:

143 990 411 900

: :

12

:

45 7

:

55

150 | P a g e

16

Bases

16.1

Base n to Base 10 A base is the number of digits that can be used to represent numbers. For example, most all computation is in base 10, called decimal, because we use the digits 0-9 to represent numbers. Other common bases include base 2, or binary, and base 16, or hexadecimal. We can only use 0’s and 1’s to represent numbers in binary. However, we need more characters than 0-9 to represent numbers in hexadecimal. Thus, we use the alphabet: A=10, B=11, C=12 … F=15. For example, a number in base 16 could read 3D2 or A7. The best way to understand numbers in other bases is through place value. In base 10 (what you’re likely used to), there is the ones place value, then tens, then hundreds, etc. Each subsequent place value is 10 times the previous, where the first place value is 1.

15610 = 1 ∗ 102 + 5 ∗ 101 + 6 ∗ 1

The same pattern occurs in numbers in other bases, where each time you increase a place value you multiply by the base you’re in, starting from 1. Basically, you replace the 10 in the expression above with the base you’re working in. This is how you convert numbers in other bases to base 10.

2𝐴16 = ______10 2𝐴16 = 2 ∗ 161 + 10 ∗ 1 = 32 + 10 = 4210

151 | P a g e

1𝐷214 = ______10 1𝐷214 = 1 ∗ 142 + 13 ∗ 141 + 2 ∗ 1 = 196 + 182 + 2 = 38010

2044 = ______10 2044 = 2 ∗ 42 + 0 ∗ 41 + 4 ∗ 1 = 32 + 0 + 4 = 3610 Practice Problem Set 16.1 1) 247 = ______________10

2) 449 = ______________10

3) 1𝐴13 = ______________10

4) 3𝐹16 = ______________10

5) 1245 = ______________10

6) 3078 = ______________10

7) 1101012 = ______________10

8) 1200213 = ______________10

9) 12345 = ______________10

10) 1𝐶𝐵15 = ______________10

11) 3𝐴412 = ______________10

12) 50037 = ______________10

13) 𝐴𝐵𝐶14 = ______________10

14) 2𝐶615 = ______________10

152 | P a g e

16.2

Base 10 to Base n Converting a number from base 10 to another base has a completely different system. Here are the steps to follow: Step 1: Divide the number you want to convert by the desired base you want to convert to. Step 2: Remember the quotient and write the remainder as the last digit in the answer (if no remainder, write 0). Step 3: Divide the quotient from Step 2 by the desired base. Write the remainder as the next-to-last digit of the answer. Step 4: Continue this process until the quotient and remainder are 0. Let’s see some examples.

𝐸𝑥𝑎𝑚𝑝𝑙𝑒: 1810 = ____4 18 ÷ 4 = 4 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 2 4 ÷ 4 = 1 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 0 1 ÷ 4 = 0 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 1 𝐹𝑖𝑛𝑎𝑙 𝑎𝑛𝑠𝑤𝑒𝑟: 1024

𝐸𝑥𝑎𝑚𝑝𝑙𝑒: 20410 = ____12 204 ÷ 12 = 17 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 0 17 ÷ 12 = 1 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 5 1 ÷ 12 = 0 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 1 𝐹𝑖𝑛𝑎𝑙 𝑎𝑛𝑠𝑤𝑒𝑟: 15012 𝐸𝑥𝑎𝑚𝑝𝑙𝑒: 57710 = ____14 577 ÷ 14 = 41 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 3 41 ÷ 14 = 2 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 13 → 𝐷 153 | P a g e

2 ÷ 14 = 0 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 2 𝐹𝑖𝑛𝑎𝑙 𝑎𝑛𝑠𝑤𝑒𝑟: 2𝐷314 Practice Problem Set 16.2 1) 2410 = ______________4

2) 4410 = ______________8

3) 1410 = ______________7

4) 3910 = ______________12

5) 24310 = ______________8

6) 30710 = ______________11

7) 18210 = ______________12

8) 32110 = ______________6

9) 123410 = ______________16

10) 151110 = ______________15

11) 35410 = ______________5

12) 50310 = ______________9

13) 72910 = ______________2

14) 23610 = ______________13

154 | P a g e

16.3

Base a to Base a^n There comes a special property of numbers when you have to convert from, say base 2 to base 4 or base 8, base 3 to base 9, etc. The idea is grouping digits. This is best explained with an example. Let’s say you want to solve the following question:

101102 = ____ 4

Since 4 is the second power of 2, we will be taking groups of two (the power). For each grouping, starting from the right and working left, convert the number to base 4. This can be done by using the place value idea in 16.1 to convert to base 10, and then to the desired base (Note: when converting to bases under 10, you can imagine converting to base 10 as the number will be the same). For our example, this will be a digit between 0 and 3. This becomes the last digit of the number in base 4. We continue to convert these groupings to convert the number. So, we’ll start with 10110. We can see that 102 = (1 ∗ 21 + 0 ∗ 20 )10 = 210 = 24 . This means that our last digit is 2. (Note: we can just convert the number to base 10 since 4