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Tutorials - Arithmetic Algebra Arithmetic Geometry Pages: 1 2 3 4 5 1) POWER CYCLE Main Application to find the last digit of a power expression b

The last digit of a number of the form a falls in a particular sequence or order depending on the unit digit of the number (a) and the power the number is raised to (b). The power cycle of a number thus depends on its unit digit. Consider the power cycle of 2 1

2

3

4

5

6

7

8

2 =2, 2 =4, 2 =8, 2 =16, 2 =32, 2 =64, 2 =128, 2 =256 th

As it can be observed, the unit digit gets repeated after every 4 power of 2. Hence, we can say that 2 has a power cycle of 2,4,8,6 with frequency 4. This means that, a number of the form 4k+1

will have the last digit as 2

4k+2

will have the last digit as 4

4k+3

will have the last digit as 8

4k+4

will have the last digit as 6 (where k=0, 1, 2, 3…)

2

2 2 2

This is applicable not only for 2, but for all numbers ending in 2. ( eg 12

32

123

,3452

)

Therefore to find the last digit of a number raised to any power, we just need to know the power cycle of digits from 0 to 9, which is given below Unit digit 0 1

Power cycle 0 1

Frequency 1 1

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2,4,8,6 3,9,7,1 4,6 5 6 7,9,3,1 8,4,2,6 9,1

4 4 2 1 1 4 4 2

APPLICATIONS OF POWER CYCLE I. DIRECT QUESTIONS TO FIND THE UNIT DIGIT OF A POWER EXPRESSION OF THE FORM a

b

55

Eg 1) Find the last digit of 4

55 can be written as 4k+3 In the cycle of 4, in the third position we get 4 itself. Last digit of the expression is 4 itself 34

Eg 2) Find the last digit of 123457

Since 7 is the units digit write 34=4k+2 nd

In the cycle of 7, in the 2

position = 9. last digit of the expression=9

II. REMAINDER QUESTIONS WHERE THE DIVISIBILITY OF THE DIVISOR DEPENDS ONLY ON ITS UNIT DIGIT (eg: when the divisors are 2,5,10) Eg 3) Find the remainder when 3

75

is divided by 5.

1) Express the power in the form, 4k+x where x=1, 2, 3, 4. In this case 75 = 4k+3. 2) Take the power cycle of the base (3) which is 3,9,7,1. Since the form is 4k+3, take the third digit in the cycle, which is 7. Any number divided by 5, the remainder will be that of the unit digit divided by 5. Hence the remainder is 2. 66

Eg 4) Find the remainder when 44

is divided by 10?

1) Express the power in the form, 4k+x where x=1, 2, 3, 4. In this case 66 = 4k+2. 2) Take the power cycle of the base (4) which is 4,6,4,6. Since the form is 4k+2, take the second digit in the cycle, which is 6. Any number divided by 10, the remainder will be that of the unit digit divided by 10. Hence the remainder is 6. Sometimes, you may get a question in the term of variables, where you need to substitute values to get the answer in the fastest way possible. For example, Eg 5) Find the unit digit of 7

3^4n 3^4

Put n=1, the problem reduces to 7

, which is 7

81.

Since 81=4k+1, take the first digit in the power cycle of 7, which is 7. What is the logic behind this? Since “n” is a variable, whichever value of “n” you substitute; you have to get the same answer. We substitute n=1 for ease of calculation. You can try substituting n=2 as well, you will get the answer as 7. III. POWER CYCLE QUESTIONS OCCURRING IN GROUPS In some questions, given a sequence of numbers; there will be a pattern in the unit digit which will be followed with a particular frequency. Based on this frequency, we can form groups of the sequence of numbers. In those cases, we need to find out the frequency and we can easily evaluate the unit digit of the entire group based on one single group. Let us understand this concept with the help of a typical example Eg 6) The sum of the third powers of the first 100 natural numbers will have a unit’s digit of a) 3 b) 6 c) 9 d) 0 3

3

3

3

The question asks us to find 1 +2 +3 +…….100

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3

3

3

3

3

3

3

3

If you notice, the unit digit of 1 +2 +..10 is the same as the unit digit of 11 +12 ….+20 and so on till 91 +92 +….100 3

3

3

Hence it is is sufficient to find the unit digit of the first group and then multiply by 10 to get the unit digit of 1 +2 +3 +…….100

3

Let us find out the unit digit of each element in the first group. Using the concept of power cycle, it can be found as Element

Unit digit

3

1

1

3

2

8

3

3

7

3

4

4

3

5

5

3

6

6

3

7

3

3

8

2

3

9

9 3

10

0

?

..5 3

3

Therefore units digit of 1 ….100 = ..5*10= 0 Shortcut:- since there are 10 groups involved, the unit digit of the first group has to be multiplied by 10 to get the answer. There is absolutely no need to even find the unit digit of the first group as any number multiplied by 10 gives 0 as the unit digit!! IV. TO FIND THE RIGHT MOST NON ZERO INTEGER 343

Eg 7) Find the rightmost non-zero integer of the expression 1430

367

+1470

?

a) 3 b) 9 c) 7 d) 1 367

It can be easily observed that 1470

343

has more number of zeroes as compared to 1430 343

integer will depend on the unit digit of only 1430

. hence, the rightmost non-zero 343

, which in turn will depend on the unit digit of 3

343 can written as 4k+3 In the power cycle of 3 (3,9,7,1), the third digit is 7. Hence, the last non-zero integer in the expression is 7. Useful technique to find the last 2 digits of any expression of the form a

b

Depending on the last digit of the number in question, we can find the last two digits of that number. We can classify the technique to be applied into A) Odd Numbers B) Even Numbers A) Odd Numbers TYPE METHOD EXAMPLES 1) Numbers ending in 1

nd

67

The last digit is always 1. The 2 last digit = product 21 =__41 (2 * 7=__4) of tens digit of base * unit digit of the power. 87 9) 41 =__ 81

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67

For example, In 21 ; 2 is the tens digit of base and 7 is the unit digit of power

( 4 *7= __8) 10) 1261

167

=__21

( 6 * 7= __2) 124

11) 31

=__ 21

( 3 * 4=__2) 2) Number ending with 5 Tens digit of base power Last digit Second last digit

Odd

odd

5

7

Odd

Even 5

2

Even

Odd

5

2

even

Even 5

2

3) Numbers ending in 3, 7, Change the power so that the base ends with 1 and 9 then use the same technique as for those numbers 4

4

12) 1555

34

= __ 25

image004

2

ending with 1. eg) 3 , 7 &9 all will end in1. B) For even Number It is based on the cycle of 2 i.e. 2

10

10

raised to even power ends with 76 and

10

2 raised to odd power ends with 24. This can be used for all even numbers 4) For even numbers (2,4,6,8)

Use the pattern of the number 1024 10

10

=2 i.e.*2 raised to even power ends with 76 and

image004

10

* 2 raised to odd power ends with 24. Note:76 x 04=04 76 x 08=08 76 x 16 =16 76 x 32= 32 76 x 64= 64 76 x 28= 28 76 x 56 = 56…… ILLUSTRATIONS 234

Eg 14) Find the last two digits of 8 234

8

=2

10 70

3(234)

=2

702

2

(2 ) *2 = 76 x 04 = __04 Eg 15) What are the last two digits of 475

23

?

a) 25 b) 50 c) 75 d) None of these

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Ans – (c ) If the digit in the tens place is odd and the power is odd, then any number ending with 5, always have its last two digits as 75. 288

Eg 16) Find the ten’s digit of 74 (a) 3 (b) 6 (c) 7 (d) 8 Ans. (c) 288

= (37×2)

288

x2

74 37

288

288

4 72

x (2)

2

2 –5672

(37 )

288

(37 x37 )

x(__56)

72

(69 x 69) x(__56) 72

(__61) x(__56) (__21)x(__56) = __76. Answer is 7 2

Shortcut of last two digit of squares:- in the above question to get the last two digits of 37 as 69;we do not need to calculate manually. There is a pattern which will be very useful here 2

2

2

2

2

2

2

2

2

The last two digits of x , (50-x) , (50+x) , (100-x) will always be the same. For example 12 , 38 , 62 , 88 , 112 …. will all be the same. That is, the last two digits of these numbers will always be 44 2

2

2

2

2

2

2

2

Also, last two digits of 11 =39 =61 =89 =111 =139 =161 =189 and so on 1084

Eg 17) Find the remainder when (148)

is divided by 100.

a) 06 b) 66 c) 56 d) 76 When the question is about finding the remainder when you divide a number by 100, what we need to find out is the last two digits (148)

1084

= (37)

4 271

= (37 ) 2

1084

x (2)

2 271

=(37 x 37 )

x (4)

1084

2168

x …56 2

2

(The last two digits of 37 will be the same as the last two digits of 13 =69) =(..69 x ..69) =(61)

271

271

x …56

x …56 = …21 x …56= …76

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