Divisibility Rules & Remainder Theorem Questions and Answers

Q: The number 897159*2 is divisible by 4. What is the sum of all possible values of *?

Correct answer: 25A number is divisible by 4 when the number formed by its last two digits is divisible by 4. Here those digits are *2, so we need the two-digit endings 02, 12, 22, ... , 92 that 4 divides.\[12 = 4\times 3,\quad 32 = 4\times 8,\quad 52 = 4\times 13,\quad 72 = 4\times 18,\quad 92 = 4\times 23\]The remaining endings - 02, 22, 42, 62 and 82 - are each 2 more than a multiple of 4, so they fail. The admissible values of * are therefore 1, 3, 5, 7 and 9, and their sum is \(1+3+5+7+9 = 25\).

Q: What is the remainder when the product \(121 \times 122 \times 123\) is divided by 7?

Correct answer: 3Reduce each factor modulo 7 before multiplying, rather than forming the product first.\[121 = 7\times 17 + 2,\qquad 122 = 7\times 17 + 3,\qquad 123 = 7\times 17 + 4\]So the product is congruent to \(2\times 3\times 4 = 24\), and \(24 = 7\times 3 + 3\).\[121\times 122\times 123 \equiv 3 \pmod 7\]

Important Instructions

41.

A number when divided by 296 gives a remainder of 75. When the same number is divided by 37, what is the remainder?

A.1
B.2
C.8
D.11

Answer: A

Let the number be N. We have N = 296k + 75. We need to find N mod 37. First, check if the first divisor (296) is a multiple of the second divisor (37). \(296 = 37 \times 8\). Yes, it is. In this case, the new remainder is simply the old remainder divided by the new divisor. So, we find the remainder of \(75 \div 37\). \(75 = 2 \times 37 + 1\). The remainder is 1.

Discuss in Forum

42.

Which of the following numbers is divisible by 11?

A.235641
B.245642
C.315624
D.415624

Answer: D

A number is divisible by 11 if the difference between the sum of digits at odd places and the sum of digits at even places is 0 or a multiple of 11.

A) (1+6+3) - (4+5+2) = 10 - 11 = -1

B) (2+6+4) - (4+5+2) = 12 - 11 = 1

C) (4+6+1) - (2+5+3) = 11 - 10 = 1

D) (4+6+1) - (2+5+4) = 11 - 11 = 0. Since the difference is 0, 415624 is divisible by 11.

Discuss in Forum

43.

The number 897159*2 is divisible by 4. What is the sum of all possible values of *?

A.15
B.18
C.20
D.25

Answer: C

Correct answer: 25

A number is divisible by 4 when the number formed by its last two digits is divisible by 4. Here those digits are *2, so we need the two-digit endings 02, 12, 22, ... , 92 that 4 divides.

\[12 = 4\times 3,\quad 32 = 4\times 8,\quad 52 = 4\times 13,\quad 72 = 4\times 18,\quad 92 = 4\times 23\]

The remaining endings - 02, 22, 42, 62 and 82 - are each 2 more than a multiple of 4, so they fail. The admissible values of * are therefore 1, 3, 5, 7 and 9, and their sum is \(1+3+5+7+9 = 25\).

Discuss in Forum

44.

The number 897159*4 is divisible by 4. What is the sum of all possible values of *?

A.15
B.18
C.20
D.25

Answer: C

For a number to be divisible by 4, the number formed by its last two digits must be divisible by 4. The last two digits form the number '*4'.

We need to find the values of * such that '*4' is divisible by 4. Let's check single digits for *:

04 is divisible by 4 (so * can be 0).

14 is not divisible by 4.

24 is divisible by 4 (so * can be 2).

34 is not divisible by 4.

44 is divisible by 4 (so * can be 4).

54 is not divisible by 4.

64 is divisible by 4 (so * can be 6).

74 is not divisible by 4.

84 is divisible by 4 (so * can be 8).

The possible values for * are 0, 2, 4, 6, 8. The sum of these values is \(0+2+4+6+8 = 20\).

Discuss in Forum

45.

What is the remainder when the product \(121 \times 122 \times 123\) is divided by 7?

Answer: A

Correct answer: 3

Reduce each factor modulo 7 before multiplying, rather than forming the product first.

\[121 = 7\times 17 + 2,\qquad 122 = 7\times 17 + 3,\qquad 123 = 7\times 17 + 4\]

So the product is congruent to \(2\times 3\times 4 = 24\), and \(24 = 7\times 3 + 3\).

\[121\times 122\times 123 \equiv 3 \pmod 7\]

Discuss in Forum

46.

What is the remainder when the product \(121 \times 122 \times 124\) is divided by 7?

Answer: A

We find the remainder of each number when divided by 7 individually and then multiply the remainders.

Remainder of \(121 \div 7\) is 2 (since \(121 = 17 \times 7 + 2\)).

Remainder of \(122 \div 7\) is 3 (since \(122 = 17 \times 7 + 3\)).

Remainder of \(124 \div 7\) is 5 (since \(124 = 17 \times 7 + 5\)).

The remainder of the overall product is the remainder of \((2 \times 3 \times 5) \div 7\).

\(2 \times 3 \times 5 = 30\).

Now, we find the remainder of \(30 \div 7\).

\(30 = 4 \times 7 + 2\).

The remainder is 2.

Discuss in Forum

47.

What is the remainder when \(9^{19} + 6\) is divided by 8?

Answer: D

We can solve this in two parts. First, find the remainder of \(9^{19} \div 8\). Since \(9 = 8+1\), the expression is \((8+1)^{19} \div 8\). The remainder will be \(1^{19}\), which is 1. Second, find the remainder of \(6 \div 8\), which is 6. The final remainder is the sum of these remainders, divided by 8 if necessary. Remainder = \((1 + 6) \div 8 = 7 \div 8\). The remainder is 7.

Discuss in Forum

48.

Find the largest 4-digit number that is exactly divisible by 88.

A.9944
B.9900
C.9988
D.9966

Answer: A

The largest 4-digit number is 9999. To find the largest 4-digit number divisible by 88, we divide 9999 by 88 and find the remainder.

\(9999 \div 88\).

\(9999 = 113 \times 88 + 55\).

The remainder is 55. This means 9999 is 55 more than a multiple of 88. To get the required number, we subtract the remainder from 9999.

Required Number = \(9999 - 55 = 9944\).

Discuss in Forum

49.

What is the remainder when \(67^{67} + 67\) is divided by 68?

A.1
B.66
C.67
D.0

Answer: B

Let's analyze the expression in two parts modulo 68. Part 1: \(67^{67} \pmod{68}\). We can write 67 as (68 - 1). So, \((68-1)^{67} \pmod{68}\). This is equivalent to \((-1)^{67} \pmod{68}\), which is -1. Part 2: \(67 \pmod{68}\) is 67. Adding the remainders: \(-1 + 67 = 66\). The remainder is 66.

Discuss in Forum

50.

The number 58129745812974 is divided by a certain number. The quotient is 5812974 and the remainder is 0. The divisor is:

A.10
B.11
C.101
D.10000001

Answer: D

The dividend is a number formed by repeating the block '5812974' twice. Let X = 5812974. The dividend is \(X \times 10^7 + X = X(10^7 + 1)\). The quotient is X and the remainder is 0. According to the division algorithm, Dividend = Divisor × Quotient + Remainder. So, \(X(10^7+1) = Divisor \times X + 0\). Dividing by X, we get Divisor = \(10^7 + 1 = 10000001\).

Discuss in Forum

Loading…

Topics

Other Subjects

Tags: Divisibility Rules Questions and Answers || Remainder Theorem MCQ Questions and Answers || Divisibility and Remainder Theorem GK Questions || GK on Divisibility Rules || Quantitative Aptitude Questions and Answers || Divisibility Theorem MCQ

Divisibility Rules & Remainder Theorem Questions and Answers

Important Instructions

Topics

Other Subjects

Tags: Divisibility Rules Questions and Answers || Remainder Theorem MCQ Questions and Answers || Divisibility and Remainder Theorem GK Questions || GK on Divisibility Rules || Quantitative Aptitude Questions and Answers || Divisibility Theorem MCQ

About Us

Stay informed about all important Dates & Notifications