Divisibility Rules & Remainder Theorem Questions and Answers

Answer: B

A number is divisible by 3 if the sum of its digits is a multiple of 3.

A) 2+3+4+5+6 = 20 (Not divisible by 3)

B) 5+4+3+2+1 = 15 (Divisible by 3, since 15 = 3 × 5)

C) 7+8+9+0+1 = 25 (Not divisible by 3)

D) 1+3+5+7+9 = 25 (Not divisible by 3)

Answer: D

To find the remainder when a number is divided by 5, we only need to look at its last digit. The remainder is the same as the remainder of the last digit when divided by 5.

The last digit is 4. When 4 is divided by 5, the remainder is 4.

Answer: D

We check the number formed by the last three digits for divisibility by 8.

A) 780 ÷ 8 = 97.5 (Not divisible)

B) 345 ÷ 8 = 43.125 (Not divisible)

C) 764 ÷ 8 = 95.5 (Not divisible)

D) 672 ÷ 8 = 84 (Divisible). So, 45672 is divisible by 8.

Answer: C

For a number to be divisible by 8, the number formed by its last three digits must be divisible by 8. Here, the number is 6*2.

Let's check values for *:

If * = 1, we have 612. \(612 \div 8 = 76.5\) (Not divisible)

If * = 2, we have 622. \(622 \div 8 = 77.75\) (Not divisible)

If * = 3, we have 632. \(632 \div 8 = 79\) (Divisible)

The least value is 3.

Answer: A

We can use the remainder theorem. \(17\) can be written as \((18 - 1)\). So, we need to find the remainder of \((18-1)^{200} \div 18\).

Using the binomial expansion, every term except the last one will have 18 as a factor. The last term is \((-1)^{200}\).

\((-1)^{200} = 1\).

So the remainder is 1.

Answer: B

For a number to be divisible by 6, it must be divisible by both 2 and 3.

1. Divisibility by 2: The number ends in 4, which is even. So, it is divisible by 2.

2. Divisibility by 3: The sum of its digits must be divisible by 3. Sum = \(9+8+7+*+5+4 = 33 + *\).

We need \(33 + *\) to be a multiple of 3. Since 33 is already a multiple of 3, * must also be a multiple of 3. The possible single-digit values are 0, 3, 6, 9. From the options, the value is 3.

Answer: C

We look at the pattern of remainders of powers of 2 when divided by 5.

\(2^1 \div 5 \rightarrow rem=2\)

\(2^2 \div 5 \rightarrow rem=4\)

\(2^3 \div 5 \rightarrow rem=3\)

\(2^4 \div 5 \rightarrow rem=1\)

The cycle of remainders is (2, 4, 3, 1), which has a length of 4. To find the remainder of \(2^{31}\), we find the remainder of the power when divided by the cycle length: \(31 \div 4 \rightarrow rem=3\). The required remainder is the 3rd in the cycle, which is 3.

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.

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.

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 \(1+3+5+7+9 = 25\).