05 Number system

Actual CAT Problems 1998-2006

Number System

CAT 1998 1.

2.

n3 is odd. Which of the following statement(s) is(are) true? I. n is odd. II. n2 is odd. III. n2 is even. a. I only b. II only c. I and II

d. I and III

(BE)2 = MPB, where B, E, M and P are distinct integers. Then M = a. 2 b. 3 c. 9

d. None of these

3.

Five-digit numbers are formed using only 0, 1, 2, 3, 4 exactly once. What is the difference between the maximum and minimum number that can be formed? a. 19800 b. 41976 c. 32976 d. None of these

4.

A certain number, when divided by 899, leaves a remainder 63. Find the remainder when the same number is divided by 29. a. 5 b. 4 c. 1 d. Cannot be determined

5.

A is the set of positive integers such that when divided by 2, 3, 4, 5, 6 leaves the remainders 1, 2, 3, 4, 5 respectively. How many integers between 0 and 100 belong to set A? a. 0 b. 1 c. 2 d. None of these

6.

How many five-digit numbers can be formed using the digits 2, 3, 8, 7, 5 exactly once such that the number is divisible by 125? a. 0 b. 1 c. 4 d. 3

7.

What is the digit in the unit’s place of 251? a. 2 b. 8

8.

c. 1

d. 4

A number is formed by writing first 54 natural numbers in front of each other as 12345678910111213 ... Find the remainder when this number is divided by 8. a. 1 b. 7 c. 2 d. 0

CAT 1999 9.

Let a, b, c be distinct digits. Consider a two-digit number ‘ab’ and a three-digit number ‘ccb’, both defined under the usual decimal number system, if (ab)2 = ccb > 300, then the value of b is a. 1 b. 0 c. 5 d. 6

10.

The remainder when 784 is divided by 342 is a. 0 b. 1

Number System

c. 49

d. 341

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11.

If n = 1 + x where x is the product of four consecutive positive integers, then which of the following is/are true? A. n is odd B. n is prime C. n is a perfect square a. A and C only b. A and B only c. A only d. None of these

12.

For two positive integers a and b define the function h(a,b) as the greatest common factor (G.C.F) of a, b. Let A be a set of n positive integers. G(A), the G.C.F of the elements of set A is computed by repeatedly using the function h. The minimum number of times h is required to be used to compute G is a.

13.

1 n 2

b. (n – 1)

If n2 = 12345678987654321, what is n? a. 12344321 b. 1235789

c. n

d. None of these

c. 111111111

d. 11111111

Directions for questions 14 to 16: Answer the questions based on the following information. There are 50 integers a1, a2 … a50, not all of them necessarily different. Let the greatest integer of these 50 integers be referred to as G, and the smallest integer be referred to as L. The integers a1 through a24 form sequence S1, and the rest form sequence S2. Each member of S1 is less than or equal to each member of S2. 14.

All values in S1 are changed in sign, while those in S2 remain unchanged. Which of the following statements is true? a. Every member of S1 is greater than or equal to every member of S2. b. G is in S1. c. If all numbers originally in S1 and S2 had the same sign, then after the change of sign, the largest number of S1 and S2 is in S1. d. None of the above

15.

Elements of S1 are in ascending order, and those of S2 are in descending order. a24 and a25 are interchanged. Then which of the following statements is true? a. S1 continues to be in ascending order. b. S2 continues to be in descending order. c. S1 continues to be in ascending order and S2 in descending order. d. None of the above

16.

Every element of S1 is made greater than or equal to every element of S2 by adding to each element of S1 an integer x. Then x cannot be less than a. 210 b. the smallest value of S2 c. the largest value of S2 d. (G – L)

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Number System

CAT 2000 17.

Let D be a recurring decimal of the form D = 0. a1 a2 a1 a2 a1 a2 ..., where digits a1 and a2 lie between 0 and 9. Further, at most one of them is zero. Which of the following numbers necessarily produces an integer, when multiplied by D? a. 18 b. 108 c. 198 d. 288

18.

Consider a sequence of seven consecutive integers. The average of the first five integers is n. The average of all the seven integers is a. n b. n + 1 c. k × n, where k is a function of n

19.

2 d. n +   7

Let S be the set of integers x such that I. 100 ≤ x ≤ 200 , II. x is odd and III. x is divisible by 3 but not by 7. How many elements does S contain? a. 16 b. 12

c. 11

d. 13

20.

Let x, y and z be distinct integers, that are odd and positive. Which one of the following statements cannot be true? a. xyz2 is odd b. (x – y)2 z is even 2 c. (x + y – z) (x + y) is even d. (x – y)(y + z)(x + y – z) is odd

21.

Let S be the set of prime numbers greater than or equal to 2 and less than 100. Multiply all elements of S. With how many consecutive zeros will the product end? a. 1 b. 4 c. 5 d. 10

22.

What is the number of distinct triangles with integral valued sides and perimeter 14? a. 6 b. 5 c. 4 d. 3

23.

Let N = 1421 × 1423 × 1425. What is the remainder when N is divided by 12? a. 0 b. 9 c. 3 d. 6

24.

The integers 34041 and 32506, when divided by a three-digit integer n, leave the same remainder. What is the value of n? a. 289 b. 367 c. 453 d. 307

25.

Each of the numbers x1, x 2 , L, x n , n ≥ 4, is equal to 1 or –1. Suppose x1x 2 x 3 x 4 + x 2 x 3 x 4 x 5 + x 3 x 4 x 5 x 6 + L + x n −3 x n − 2 x n −1x n + x n −2 x n −1x n x1 + x n −1x n x1x 2 + x n x1x 2 x 3 = 0, then a. n is even b. n is odd c. n is an odd multiple of 3 d. n is prime

Number System

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26.

Sam has forgotten his friend’s seven-digit telephone number. He remembers the following: the first three digits are either 635 or 674, the number is odd, and the number 9 appears once. If Sam were to use a trial and error process to reach his friend, what is the minimum number of trials he has to make before he can be certain to succeed? a. 10,000 b. 2,430 c. 3,402 d. 3,006

27.

Let N = 553 + 173 – 723. N is divisible by a. both 7 and 13 b. both 3 and 13

28.

c. both 17 and 7

Convert the number 1982 from base 10 to base 12. The result is a. 1182 b. 1912 c. 1192

d. both 3 and 17

d. 1292

CAT 2001 29.

Let x, y and z be distinct integers. x and y are odd and positive, and z is even and positive. Which one of the following statements cannot be true? a. y(x – z)2 is even b. y2(x – z) is odd c. y(x – z) is odd d. z(x – y)2 is even

30.

In a four-digit number, the sum of the first 2 digits is equal to that of the last 2 digits. The sum of the first and last digits is equal to the third digit. Finally, the sum of the second and fourth digits is twice the sum of the other 2 digits. What is the third digit of the number? a. 5 b. 8 c. 1 d. 4

31.

Anita had to do a multiplication. In stead of taking 35 as one of the multipliers, she took 53. As a result, the product went up by 540. What is the new product? a. 1050 b. 540 c. 1440 d. 1590

32.

x and y are real numbers satisfying the conditions 2 < x < 3 and – 8 < y < –7. Which of the following expressions will have the least value? a. x2y b. xy2 c. 5xy d. None of these

33.

In a number system the product of 44 and 11 is 3414. The number 3111 of this system, when converted to the decimal number system, becomes a. 406 b. 1086 c. 213 d. 691

34.

All the page numbers from a book are added, beginning at page 1. However, one page number was added twice by mistake. The sum obtained was 1000. Which page number was added twice? a. 44 b. 45 c. 10 d. 12

35.

If a, b, c and d are four positive real numbers such that abcd = 1, what is the minimum value of (1 + a)(1 + b)(1 + c)(1 + d)? a. 4 b. 1 c. 16 d. 18

36.

A set of consecutive positive integers beginning with 1 is written on the blackboard. A student came along and erased one number. The average of the remaining numbers is 35 number erased? a. 7

Page 4

b. 8

c. 9

7 . What was the 17

d. None of these

Number System

37

Let b be a positive integer and a = b2 – b. If b ≥ 4 , then a2 – 2a is divisible by a. 15 b. 20 c. 24 d. All of these

38.

In some code, letters a, b, c, d and e represent numbers 2, 4, 5, 6 and 10. We just do not know which letter represents which number. Consider the following relationships: I. a + c = e, II. b – d = d and III. e + a = b Which of the following statements is true? a. b = 4, d = 2 b. a = 4, e = 6 c. b = 6, e = 2 d. a = 4, c = 6

39.

Let n be the number of different five-digit numbers, divisible by 4 with the digits 1, 2, 3, 4, 5 and 6, no digit being repeated in the numbers. What is the value of n? a. 144 b. 168 c. 192 d. None of these

CAT 2002 40.

If there are 10 positive real numbers n1 < n2 < n3 ... < n10 , how many triplets of these numbers

(n1, n2 ,n3 ), (n2 ,

n3 ,n4 ), ... can be generated such that in each triplet the first number is always

less than the second number, and the second number is always less than the third number? a. 45 b. 90 c. 120 d. 180 41.

Number S is obtained by squaring the sum of digits of a two-digit number D. If difference between S and D is 27, then the two-digit number D is a. 24 b. 54 c. 34 d. 45

42.

A rich merchant had collected many gold coins. He did not want anybody to know about him. One day, his wife asked, " How many gold coins do we have?" After a brief pause, he replied, "Well! if I divide the coins into two unequal numbers, then 48 times the difference between the two numbers equals the difference between the squares of the two numbers." The wife looked puzzled. Can you help the merchant's wife by finding out how many gold coins the merchant has? a. 96 b. 53 c. 43 d. None of these

43.

A child was asked to add first few natural numbers (i.e. 1 + 2 + 3 + …) so long his patience permitted. As he stopped, he gave the sum as 575. When the teacher declared the result wrong, the child discovered he had missed one number in the sequence during addition. The number he missed was a. less than 10 b. 10 c. 15 d. more than 15

44.

When 2256 is divided by 17, the remainder would be a. 1 b. 16 c. 14

45.

d. None of these

At a bookstore, ‘MODERN BOOK STORE’ is flashed using neon lights. The words are individually 1 1 1 s, 4 s and 5 s respectively, and each word is put off after a second. 2 4 8 The least time after which the full name of the bookstore can be read again is a. 49.5 s b. 73.5 s c. 1744.5 s d. 855 s

flashed at the intervals of 2

Number System

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46.

1 3 1 lb, 6 lb and 7 lb respectively are to be divided into parts of 2 4 5 equal weight. Further, each part must be as heavy as possible. If one such part is served to each guest, then what is the maximum number of guests that could be entertained? a. 54 b. 72 c. 20 d. None of these

Three pieces of cakes of weights 4

47.

After the division of a number successively by 3, 4 and 7, the remainders obtained are 2, 1 and 4 respectively. What will be the remainder if 84 divides the same number? a. 80 b. 75 c. 41 d. 53

48.

If u, v, w and m are natural numbers such that true? a. m ≥ min(u, v, w) c. m < min(u, v, w)

49.

50.

um + vm = w m , then which one of the following is b. m ≥ max(u, v, w) d. None of these

76n – 66n , where n is an integer > 0, is divisible by a. 13 b. 127 c. 559

d. All of these

How many numbers greater than 0 and less than a million can be formed with the digits 0, 7 and 8? a. 486 b. 1,084 c. 728 d. None of these

CAT 2003 (Leaked Paper) 51.

How many even integers n, where 100 ≤ n ≤ 200 , are divisible neither by seven nor by nine? a. 40 b. 37 c. 39 d. 38

52.

A positive whole number M less than 100 is represented in base 2 notation, base 3 notation, and base 5 notation. It is found that in all three cases the last digit is 1, while in exactly two out of the three cases the leading digit is 1. Then M equals a. 31 b. 63 c. 75 d. 91

Directions for question 53: Each question is followed by two statements, A and B. Answer each question using the following instructions. Choose (a) if the question can be answered by one of the statements alone but not by the other. Choose (b) if the question can be answered by using either statement alone. Choose (c) if the question can be answered by using both the statements together, but cannot be answered by using either statement alone. Choose (d) if the question cannot be answered even by using both the statements together. 53.

Is a44 < b11, given that a = 2 and b is an integer? A. b is even B. b is greater than 16

54.

How many three digit positive integers, with digits x, y and z in the hundred's, ten's and unit's place respectively, exist such that x < y, z < y and x ≠ 0 ? a. 245 b. 285 c. 240 d. 320

Page 6

Number System

55.

If the product of n positive real numbers is unity, then their sum is necessarily a. a multiple of n

56.

b. equal to n +

1 n

c. never less than n

d. a positive integer

The number of positive integers n in the range 12 ≤ n ≤ 40 such that the product (n − 1)(n − 2)K 3.2.1 is not divisible by n is a. 5 b. 7 c. 13 d. 14

CAT 2003 (Re test) Directions for questions 57 to 59: Answer the questions on the basis of the information given below. The seven basic symbols in a certain numeral system and their respective values are as follows: I = 1, V = 5, X = 10, L = 50, C = 100, D = 500 and M = 1000 In general, the symbols in the numeral system are read from left to right, starting with the symbol representing the largest value; the same symbol cannot occur continuously more than three times; the value of the numeral is the sum of the values of the symbols. For example, XXVII = 10 + 10 + 5 + 1 + 1 = 27. An exception to the left-to-right reading occurs when a symbol is followed immediately by a symbol of greater value; then the smaller value is subtracted from the larger. For example, XLVI = (50 – 10) + 5 + 1 = 46. 57.

58.

59.

The value of the numeral MDCCLXXXVII is a. 1687 b. 1787

c. 1887

d. 1987

The value of the numeral MCMXCIX is a. 1999 b. 1899

c. 1989

d. 1889

Which of the following represent the numeral for 1995? I. MCMLXXV II. MCMXCV III. MVD

IV. MVM

a. Only I and II

d. Only IV

b. Only III and IV

c. Only II and IV

60.

What is the sum of all two-digit numbers that give a remainder of 3 when they are divided by 7? a. 666 b. 676 c. 683 d. 77

61.

An intelligence agency forms a code of two distinct digits selected from 0, 1, 2, …, 9 such that the first digit of the code is non-zero. The code, handwritten on a slip, can however potentially create confusion when read upside down — for example, the code 91 may appear as 16. How many codes are there for which no such confusion can arise? a. 80 b. 78 c. 71 d. 69

62.

What is the remainder when 496 is divided by 6? a. 0 b. 2 c. 3

63.

d. 4

Let n (>1) be a composite integer such that n is not an integer. Consider the following statements: A: n has a perfect integer-valued divisor which is greater than 1 and less than n B: n has a perfect integer-valued divisor which is greater than n but less than n a. Both A and B are false b. A is true but B is false c. A is false but B is true d. Both A and B are true

Number System

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64.

If a, a + 2 and a + 4 are prime numbers, then the number of possible solutions for a is a. one b. two c. three d. more than three

65.

Let a, b, c, d and e be integers such that a = 6b = 12c, and 2b = 9d = 12 e. Then which of the following pairs contains a number that is not an integer?

 a c b.  ,   36 e 

 a b a.  ,   27 e 

 a bd  c.  ,   12 18 

a c d.  ,   6 d

CAT 2004 66.

On January 1, 2004 two new societies s1 and s2 are formed, each n numbers. On the first day of each subsequent month, s1 adds b members while s2 multiplies its current numbers by a constant factor r. Both the societies have the same number of members on July 2, 2004. If b = 10.5n, what is the value of r? a. 2.0 b. 1.9 c. 1.8 d. 1.7

67.

Suppose n is an integer such that the sum of digits on n is 2, and 1010 < n 10n. The number of different values of n is a. 11 b. 10 c. 9 d. 8

68.

Let y =

1 2+

1 3+

1 2+

1 3 + ...

What is the value of y? a. 69.

11 + 3 2

b.

11 − 3 2

c.

15 + 3 2

The reminder, when (1523 + 2323) is divided by 19, is a. 4 b. 15 c. 0

d.

15 − 3 2

d. 18

CAT 2005 70.

If R =

3065 – 2965

3064 + 2964 a. 0 < R ≤ 0.1

, then b. 0 < R ≤ 0.5

c. 0.5 < R ≤ 0.1

d.R > 1.0

71.

If x = (163 + 173 + 183 + 193), then x divided by 70 leaves a remainder of a. 0 b. 1 c. 69 d. 35

72.

let n! = 1 × 2 × 3 × … × n for integer n ≥ 1 . If p = 1! + (2 × 2!) + (3 × 3!) + … + (10 × 10!), then p + 2 when divided by 11! Leaves a remainder of a.10 b. 0 c. 7 d. 1

Page 8

Number System

73.

The digits of a three-digit number A are written in the reverse order to form another three-digit number B. If B > A and B-A is perfectly divisible by 7, then which of the following is necessarily true? a. 100 < A < 299 b. 106 < A < 305 c. 112 < A < 311 d. 118< A < 317

74.

The rightmost non-zero digits of the number 302720 is a. 1 b. 3 c. 7

d. 9

75.

For a positive integer n, let pn denote the product of the digits of n and sn denote the sum of the digits of n. The number of integers between 10 and 1000 for which pn + sn = n is a.81 b. 16 c. 18 d. 9

76.

Let S be a set of positive integers such that every element n of S satisfies the conditions (a) 1000 ≤ n ≤ 1200 (b) every digit in n is odd Then how many elements of S are divisible by 3? a. 9 b. 10 c. 11 d. 12

CAT 2006 77.

If x = – 0.5, then which of the following has the smallest value? 1.

78.

1 2x

Which among 1. 21/2

2.

1 2 2

,

1 x

1 , 3 3

3.

1 44

2. 31/3

,

1 6 6

and

1 x

1 12 12

2

4. 2X

5.

1 −x

is the largest?

3. 41/4

4. 61/6

5. 12/12

79.

A group of 630 children is arranged in rows for a group photograph session. Each row contains three fewer children than the row in front of it. What number of rows is not possible? 1. 3 2. 4 3. 5 4. 6 5. 7

80.

The sum of four consecutive two-digit odd numbers, when divided by 10, becomes a perfect square. Which of the following can possibly be one of these four numbers? 1. 21 2. 25 3. 41 4. 67 5. 73

81.

When you reverse the digits of the number 13, the number increases by 18. How many other twodigit numbers increase by 18 when their digits are reversed? 1. 5 2. 6 3. 7 4. 8 5. 10

Number System

Page 9

Answers and Explanations CAT 1998 1. c

If n3 is odd, then n should also be odd. Hence, n2 should also be odd. And n2 will again be odd and not even. So only I and II are true.

2. b

Since MPB is a three-digit number, and also the square of a two-digit number, it can have a maximum value of 961 viz. 312. This means that the number BE should be less than or equal to 31. So B can only take the values 0, 1, 2 and 3. Since the last digit of MPB is also B, it can only be 0 or 1 (as none of the squares end in 2 or 3). The only squares that end in 0 are 100, 400 and 900. But for this to occur the last digit of BE also has to be 0. Since E and B are distinct integers, both of them cannot be 0. Hence, B has to be 1. BE can be a number between 11 and 19 (as we have also ruled out 10), with its square also ending in 1. Hence, the number BE can only be 11 or 19. 112 = 121. This is not possible as this will mean that M is also equal to 1. Hence, our actual numbers are 192 = 361. Hence, M = 3.

3. c

The maximum and the minimum five-digit numbers that can be formed using only 0, 1, 2, 3, 4 exactly once are 43210 and 10234 respectively. The difference between them is 43210 – 10234 = 32976.

4. a

The best way to solve this question is the method of simulation, i.e. take a number which when divided by 899 gives a remainder of 63. The smallest such number is (899 + 63) = 972. 972, when divided by 29 gives a remainder of 5. Hence, the answer is 5. Students, please note that 899 itself is divisible by 29. Hence, the required remainder is the same as obtained by dividing 63 by 29, i.e. 5. Shortcut: Since 899 is divisible by 29, so you can directly divide

6. c

Let us find some of the smaller multiples of 125. They are 125, 250, 375, 500, 625, 750, 875, 1000 ... A five-digit number is divisible by 125, if the last three digits are divisible by 125. So the possibilities are 375 and 875, 5 should come in unit’s place, and 7 should come in ten’s place. Thousand’s place should contain 3 or 8. We can do it in 2! ways. Remaining first two digits, we can arrange in 2! ways. So we can have 2! × 2! = 4 such numbers. There are: 23875, 32875, 28375, 82375.

7. b

Since 2 has a cyclicity of 4, i.e. 21 = 2, 22 = 4, 23 = 8, 24 = 16, 25 = 32, 26 = 64 ..., the last digits (2, 4, 8, 6) are in four cycles. 51

∴ On dividing 4 , we get the remainder as 3. ∴ The last digit has to be 23 = 8 Shortcut: Since cyclicity of the power of 2 is 4, so 251 can be written in 24(12) + 3 or unit digit will be 23 = 8. 8. a

CAT 1999 9. a

(ab)2 = ccb, the greatest possible value of ‘ab’ to be 31. Since 312 = 961 and since ccb > 300, 300 < ccb < 961, so 18 < ab < 31. So the possible value of ab which statisfies (ab)2 = ccb is 21. So 212 = 441, ∴ a = 2, b = 1, c = 4.

10. b

Note: 342 = 73 – 1. On further simplification we get,

Note that the difference between the divisors and the remainders is constant. 2–1=3–2=4–3=5–4=6–5=1 In such a case, the required number will always be [a multiple of LCM of (2, 3, 4, 5, 6) – (The constant difference)]. LCM of (2, 3, 4, 5, 6) = 60 Hence, the required number will be 60n – 1. Thus, we can see that the smallest such number is (60 × 1) – 1 = 59 The second smallest is (60 × 2) – 1 = 119 So between 1 and 100, there is only one such number, viz. 59.

Number System

(73 )28 34328 (342 + 1) = = 342 342 342

28

=

63 will give 5 as a the remainder of 63 by 29, so 29 remainder, option (a).

5. b

The number formed by the last 3 digits of the main number is 354. The remainder is 2 if we divide 354 by 8. So the remainder of the main number is also 2 if we divide it by 8.

342 N + 1 1 = 342 342 Hence, remainder = 1, optiion (b).

=

11. a

Use the method of simulation, viz. take any sample values of x and verify that n is both odd as well as a perfect square.

12. b

If there are n numbers, the function h has to be performed one time less.

13. c

The square root is 11111111.

14. d

None of the statements are true.

Page 1

15. a

S1 remains the same, but S2 changes.

16. d

x must be equal to the greatest difference in the value of numbers of S1 and S2.

26. c

There are two possible cases. The number 9 comes at the end, or it comes at position 4, 5, or 6. For the first case, the number would look like:

9 ... 9. 674 In both these cases, the blanks can be occupied by any of the available 9 digits (0, 1, 2, ..., 8). Thus, total possible numbers would be 2 × (9 × 9 × 9) = 1458. For the second case, the number 9 can occupy any of the given position 4, 5, or 6, and there shall be an odd number at position 7. Thus, the total number of ways shall be 2[3(9 × 9 × 4)] = 1944. Hence, answer is 3402. 635 ...

CAT 2000 17. b

18. b

a1 a2 . So D must be multiplied 99 by 198 as 198 is a multiple of 99. 99 × D = a1a2. Hence, D=

Use any 7 consecutive numbers to check the answers. n=

(1 + 2 + 3 + 4 + 5 ) = 3 ,

k=

(1 + 2 + 3 + 4 + 5 + 6 + 7 ) = 4 .

average of 7 integers is

5

7 So k = n + 1. Alternately, the average of the first 5 terms is the middle term which is third term, and the average of the first 7 terms is the middle term which is the fourth term. Hence, it is one more than the previous average.

19. d

Numbers which are divisible by 3 (between 100 and 200) are 33. Numbers which are divisible by 21, i.e. LCM of 7 and 3 (between 100 and 200) are 5. Out of the 33 numbers divisible by 3, 17 are even and 16 are odd. Out of the 5 numbers divisible by 7, three are odd. Hence, the number of odd numbers divisible by 3 but not by 7 is (16 – 3) = 13.

20. d

Take any three odd and positive numbers and check this out.

21. a

There is only one 5 and one 2 in the set of prime numbers. Hence, there would be only one zero at the end of the resultant product.

22. c

If the sides of the triangle are a, b and c, then a + b > c. Given a + b + c = 14. Then the sides can be (4, 4, 6), (5, 5, 4), (6, 5, 3) and (6, 6, 2). Hence, four triangles are possible.

23. c

N = 1421 × 1423 × 1425. When divided by 12, it shall (1416 + 5 ) × (1416 + 7 ) × (1416 + 9 ) . look like  12 Now the remainder will be governed by the term 5 × 7 × 9, which when divided by 12 leaves the remainder 3.

24. d

25. a

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Let r be the remainder. Then 34041 – r and 32506 – r are perfectly divisible by n. Hence, their difference should also be divisible by the same. (34041 – r) – (32506 – r) = 1535, which is divisible by only 307. Each term has to be either 1 or –1. Hence, if the sum of n such terms is 0, then n is even.

27. d

N can be written either (54 + 1)3 + (18 – 1)3 – 723 or (51+ 4 )3 + 173 – (68 + 4)3 . The first form is divisible by 3, and the second by 17.

28. c 1 2 19 82 1 2 16 5 – 2 1 2 13 – 9 1 – 1

The answer is 1192.

CAT 2001 29. a

Use the answer choices and the fact that: Odd × Odd = Odd Odd × Even = Even Even × Even = Even

30. a

Let the four-digit number be abcd. a+b=c+d ... (i) b + d = 2(a + c) ... (ii) a+d=c ... (iii) From (i) and (iii), b = 2d From (i) and (ii), 3b = 4c + d ⇒ 3(2d) = 4c + d ⇒ 5d = 4c

5 d 4 Now d can be 4 or 8. But if d = 8, then c = 10 not possible. So d = 4 which gives c = 5. ⇒ c=

31. d

Let the number be x. Increase in product = 53x – 35x = 18x ⇒ 18x = 540 ⇒ x = 30 Hence new product = 53 × 30 = 1590.

32. c

The value of y would be negative and the value of x would be positive from the inequalities given in the question. Therefore, from (a), y becomes positive. The value of xy2 would be positive and will not be the minimum. From (b) and (c), x2y and 5xy would give negative values but we do not know which would be the minimum.

Number System

c = 6, c = 8 (Not possible) c = 10, e = 10 (Not possible since both c and e cannot be 10) From (ii), we have c = 2, 4, 6, 10. For c = 2, e = 3 (Not possible) c = 4, e = 4 (Not possible) c = 6, e = 5 (Possible) c = 10, e = 7 (Not possible) Considering the possibility from B that c = 6 and e = 5 means e + a = 4 ⇒ a = –1 (Not possible) Hence, only possibility is b = 10, d = 5, c = 2, e = 6. e + a = 10 ⇒ a = 4

On comparing (a) and (c), we find that x2 < 5x in 2 < x < 3.

∴ x2 y > 5xy [Since y is negative.] ∴ 5xy would give the minimum value. 33. a

The product of 44 and 11 is 484. If base is x, then 3411 = 3x3 + 4 x2 + 1x1 + 4 × x0 = 484

⇒ 3 x3 + 4 x2 + x = 480 This equation is satisfied only when x = 5. So base is 5. In decimal system, the number 3111 can be written 39. c

3 × 53 + 1 × 5 2 + 1 × 51 + 1 × 50 = 406 34. c

Remaining digits can be chosen in

Let the total number of pages in the book be n.

∑i + x = 1000 i =1

n(n + 1) + x = 1000 2 Thus,

40. c

n(n + 1) = 990 (for n = 44). 2 Hence x = 10. Since

Take a = b = c = d = 1.

36. a

Let the highest number be n and x be the number erased.

P3 = 24 ways.

CAT 2002

n(n + 1) ≤ 1000 gives n = 44 2

35. c

4

Hence, total number of such five-digit numbers = 24 × 8 = 192.

n

Let page number x be repeated. Then

The last two digits can be 12, 16, 24, 32, 36, 52, 56, and 64, i.e. 8 possibilites

Total possible arrangements = 10 × 9 × 8 Now 3 numbers can be arranged among themselves in 3! ways = 6 ways Given condition is satisfied by only 1 out of 6 ways. Hence, the required number of arrangements =

10 × 9 × 8 = 120 6

41. b

Check choices Choice (b) 54 ⇒ S = (5 + 4)2 = 81 ⇒ D – S = 81 – 54 = 27. Hence, the number = 54

42. d

Let the number of gold coins = x + y 48(x – y) = X2 – Y2 48(x – y) = (x – y)(x + y) ⇒ x + y = 48 Hence the correct choice would be none of these.

43. d

575 =

n(n + 1) –x 7 602 . 2 Then = 35 = (n – 1) 17 17

Hence, n = 69 and x = 7 satisfy the above conditions. 37. d

38. b

a = b2 – b, b ≥ 4 a2 – 2a = (b2 – b)2 – 2(b2 – b) = (b2 – b)(b2 – b – 2) Using different values to b ≥ 4 and we find that it is divisible by 15, 20, 24. Hence all of these is the right answer. From II, b = 2d Hence, b = 10, d = 5 or b = 4, d = 2 From III, e + a = 10 or e + a = 4 From I, a + c = e or e – a = c From III and I, we get 2e = 10 + c or 2e = 4 + c ⇒ e=5+

c 2

Number System

1150 = n2 + n – 2x n(n + 1) ≥ 1150 n2 + n ≥ 1150 The smallest value for it is n = 34. For n = 34 40 = 2x ⇒ x = 20

44. a

(2 ) 4

64

= (17 – 1)64 = 17n + (−1)64 = 17n + 1

Hence, remainder = 1

... (i)

c ... (ii) 2 From (i), we can take c = 2, 4, 6, 10. For c = 2, e = 6 c = 4, e = 7 (Not possible)

or e = 2 +

n2 + n –x 2

45. b

Because each word is lit for a second, 17 41  5  7 21 49  LCM  + 1, , + 1, + 1 = LCM  ,  4 8 2  2 4 8  LCM(7, 21, 49) 49 × 3 = = 73.5 s HCF ( 2, 4, 8) 2

Page 3

46. d

 9 27 36  HCF(9, 27, 36) = 9 HCF  , , lb  = LCM (2, 4, 5) 20 2 4 5  = Weight of each piece Total weight = 18.45 lb

Maximum number of guests = 47. d

48. d

49. d

18.45 ×20 = 41 9

54. c

If y = 2 (it cannot be 0 or 1), then x can take 1 value and z can take 2 values. Thus with y = 2, a total of 1 × 2 = 2 numbers can be formed. With y = 3, 2 × 3 = 6 numbers can be formed. Similarly checking for all values of y from 2 to 9 and adding up we get the answer as 240.

55. c

The best way to do this is to take some value and verify.

3(4(7x + 4) + 1) + 2 = 84x + 53 Therefore, remainder is 53.

um + vm = w m

1 and 1. Thus, n = 3 and the sum of the three 2 numbers = 3.5. Thus options 1, 2 and 4 get eliminated.

u2 + v 2 = w 2 Taking Pythagorean triplet 3, 4 and 5, we see m < min (u, v, w) Also 1' + 2' = 3' and hence m ≤ min (u, v, w)

Alternative method: Let the n positive numbers be a1, a2, a3 … an a1, a2, a3 … an = 1 We know that AM ≥ GM

E.g. 2,

1 (a1 + a2 + a3 + … + an ) ≥ (a1a2 …an )1/ n n or (a1 + a2 + a3 + … an) ≥ n

76n – 66n Put n = 1

Hence

76 – 66 = (73 – 63 )(73 + 63 )

This is a multiple of 73 – 63 = 127 and 73 + 63 = 559 and 7 + 6 = 13

56. b

From 12 to 40, there are 7 prime number, i.e. 13, 17, 19, 23, 29, 31, 37, which is not divisible by (n–1)!

CAT 2003 (Retest) 50. c

Number of ways for single digit = 2 2 digits = 2 × 3 = 6 3 digits = 2 × 3 × 3 = 18 4 digits = 2 × 3 × 3 × 3 = 54 5 digits = 2 × 3 × 3 × 3 × 3 = 162 6 digits = 2 × 3 × 3 × 3 × 3 × 3 = 486 Total = 728

57. b

MDCCLXXXVII = 1000 + 500 + 100 + 100 + 50 + 10 + 10 + 10 + 5 + 1 + 1 = 1787

58. a

MCMXCIX = 1000 + (1000 – 100) + (100 – 10) + (10 – 1) = 1000 + 900 + 90 + 9 = 1999

59. c

(I) MCMLXXV = 1000 + (1000 – 100) + 50 + 10 + 10 + 5 = 1975 (II) MCMXCV = 1000 + (1000 – 100) + (100 – 10) + 5 = 1995 (III) MVD = 1000 + (500 – 5) = 1495 (IV) MVM = 1000 + (1000 – 5) = 1995 Therefore, the answer is (II) and (IV), i.e. option (c).

60. b

Such numbers are 10, 17, …, 94. These numbers are in AP. There are 13 numbers.

CAT 2003 (leaked) 51. c

52. d

There are 101 integers in all, of which 51 are even. From 100 to 200, there are 14 multiples of 7, of which 7 are even. There are 11 multiples of 9, of which 6 are even. But there is one integer (i.e. 126) that is a multiple of both 7 and 9 and also even. Hence the answer is (51 – 7 – 6 + 1) = 39 Since the last digit in base 2, 3 and 5 is 1, the number should be such that on dividing by either 2, 3 or 5 we should get a remainder 1. The smallest such number is 31. The next set of numbers are 61, 91. Among these only 31 and 91 are a part of the answer choices. Among these, (31)10 = (11111)2 = (1011)3 = (111)5 Thus, all three forms have leading digit 1. Hence the answer is 91.

53. a

Page 4

Solution cannot be found by using only Statement A since b can take any even number 2, 4, 6…. But we can arrive at solution by using statement B alone. If b > 16, say b = 17 Hence 244 < (16 + 1)11 244 < (24 + 1)11

∴ Sum =

10 + 94

× 13 2 = 52 × 13 = 676

61. c

Total codes which can be formed = 9 × 9 = 81. (Distinct digit codes) The digits which can confuse are 1, 6, 8, 9, from these digit we can form the codes = 4 × 3 = 12 Out of these 12 codes two numbers 69 and 96 will not create confusion. Therefore, (12 – 2) = 10 codes will create a confusion. Therefore, total codes without confusion = 81 – 10 = 71.

Number System

62. d

496 6 Let’s come down to basic property of dividing the power of 4 by 6, i.e.

67. a

We have (1) 1010 < n < 1011 (2) Sum of the digits for 'n' = 2 Clearly(n)min = 10000000001 (1 followed by 9 zeros and finally 1) Obviously, we can form 10 such numbers by shifting '1' by one place from right to left again and again. Again, there is another possibility for 'n' n = 20000000000 So finally : No. of different values for n = 10 + 1 = 11 ans.

68. d

y=

Remainder when

41 =4 6 42 =4 6 43 =4 6 44

=4

6 Hence, any power of 4 when divided by 6 leaves a remainder of 4. 63. d

65. d

=

n.

As any prime number greater than 3 can be expressed in the form 6n ± 1 , minimum difference between three consecutive prime numbers will be 2 and 4. The values that satisfy the given conditions are only 3, 5 and 7, i.e. only one set is possible.

–6 ± 36 + 24 4

–6 ± 60 –3 ± 15 = 4 2

Since 'y' is a +ve number, therefore: y=

69. c

a = 6b = 12c and 2b = 9d = 12e. Dividing the first equations by 12 and second by 36,

15 –3 ans. 2

1523 = (19 – 4)23 = 19x + (–4)23 where x is a natural

number. 2323 = (19 + 4)23 = 19y + (4)23 where y is a natural

a b c b d e = = = = we get and 12 2 1 18 4 3 i.e.

3+y 7 + 2y

⇒y=

n > 2 >1.

Therefore, statement B is true. 64. a

1 3+ y

⇒ 2y 2 + 6y – 3 = 0

n = 6 ≈ 2.4 Therefore, divisors of 6 are 1, 2, 3. Therefore,

If we take 3 as divisor, then 6 > 3 > 2.4, i.e. n > 3 >

2+

⇒y=

Let n = 6

If we take 2 as divisor, then Statement A is true.

1

number.

1523 + 2323 = 19 ( x + y ) + 423 + (–4)23

a b c b d e = = = = and 108 18 9 18 4 3

= 19 ( x +y)

a b c d e = = = = 108 18 9 4 3 ∴ a : b : c : d : e = 108 : 18 : 9 : 4 : 3. ∴

∴

c 9 = is not an integer. d 4

CAT 2005

70. d

There will be an increase of 6 times. No. of members s1 will be in A.P. On July 2nd , 2004, s1 will have n + 6 b members = n + 6 × 10.5 n = 64n No. of members in s2 will be in G.P On July 2nd, 2004 Number of members in s2 = nr6 They are equal, Hence 64 n = nr6

⇒ 64 = r 6 ⇒ r = 2

Number System

3064 + 2964

>1

as 3065 – 2965 > 3064 + 2964 3064 (30 – 1) > 2964 (29 + 1) 3064 × 29 > 2964 × 30 3063 > 2963 Hence option d.

CAT 2004 66. a

3065 – 29 65

71. a

x = 163 + 173 + 183 + 193 is even number Therefore 2 divides x. a3 + b3 = (a + b) (a2 – ab – b2) ⇒ a + b always divides Therefore 163 + 193 is divisible by 35 183 + 173 is divisible by 35 Hence x is divisible by 70. Hence option a.

Page 5

72. d

If p = 1! = 1 Then p + 2 = 3 when divided by 2! remainder will be 1. If p = 1! + 2 × 2! = 5 Then p + 2 = 7 when divided by 3! remainder is still 1. Hence p = 1! + (2 × 2!) + (3 × 3!) + … + (10 × 10!) when divided by 11! leaves remainder 1 Alternative method: P = 1 + 2.2! + 3.3!+ ….10.10! = (2 –1)1! + ( 3 – 1)2! + (4 – 1)3! + ….(11 – 1)10! =2! – 1! + 3! – 2! + ….. 11! –10! = 1 + 11! Hence the remainder is 1.

73. b

74. a

−1/ 2 = (4) 2

78. 2

2

79. 4

4 680

The 100th and 1000th position value will be only 1 . Now the possibility of unit and tens digits are (1, 3), (1, 9), (3, 1), (3, 7), (5, 5), (7, 3), (7, 9), (9, 1), (9, 7).

CAT 2006 77. 2

Go by option, put x = −2 (1) 2 =

(2)

12 3

4

12 2

6

12

121

Let the no. of students in front row be x. So, the no. of students in next rows be x – 3, x - 6, x – 9…. so on If n i.e. no. of rows be 3 then no. of students x + (x – 3) + (x – 6) = 630 3x = 639 x = 213 So possible similarly n = 4 x + (x – 3) + (x – 6) + (x - 9) = 630 4x – 18 = 630

648 = 162 4 If n = 5 (4x – 18) + (x - 12) = 630 5x – 30 = 630 x = 120 Again possible. If n = 6 (5x - 30) + (x - 15) = 630 6x - 45 = 630 6x = 675 x ≠ Integer Hence n ≠ 6 80. 3

By options checking option (3), four consecutive odd numbers are 37, 39, 41 and 43. The sum of these 4 numbers is 160. When divided by 10, we get 16 which is a perfect square. ∴ 41 is one of the odd numbers.

81. 2

Let the number be 10x + y so when number is reversed the number because 10y + x. So, the number increases by 18 Hence (10y + x) - (10x + y) = 9 (y - x) = 18 y-x=2 So, the possible pairs of (x, y) is (3, 1) (4, 2) (5, 3) (6, 4), (7, 5) (8, 6) (9, 7) But we want the number other than 13 so, there are 6 possible numbers are there i.e. 24, 35, 46, 57, 68, 79. So total possible numbers are 6.

⇒ bc = 99 + 9

76. a

3

x=

((30) )

b a But the maximum value for bc = 81 And RHS is more than 99. Hence no such number is possible. Hence option d.

12 4

∴ 34 is greatest Note: n1/n is maximum when n = e (2.718). Among the options n = 3 is closest to the value of e.

Let A = 100 x + 10y + z ⇒ B = 100z + 10y + x B - A = 99(z - x) For B - A to be divided by 7, z - x has to be divisible by 7. Only possibility is z = 9, x = 2. Biggest number A can be 299 Option b.

10 < n < 1000 Let n is two digit number. n = 10a + b ⇒ pn = ab, sn = a + b Then ab + a + b = 10a + b ⇒ ab = 9a ⇒ b = 9 There are 9 such numbers 19, 29, 33, … 99 Then Let n is three digit number ⇒ n = 100a + 10b + c ⇒ pn = abc, sn= a + b + c then abc + a + b + c = 100a + 10b + c ⇒ abc = 99a + 9b

LCM of 2, 3, 4, 6, 12 = 12 12 6

Hence the right most non-zero digit is 1. 75. d

1 2

−1 2

1 4

1 1 ⇒ = −2 x −1/ 2

1 1 ⇒ =4 (3) x 2 ( −1/ 2 )2

Page 6

Number System