Divisibility Rules & Remainder Theorem Questions and Answers

Answer: C

Correct answer: 3

The powers of 2 modulo 5 repeat in a cycle of four.

\[2^1 \equiv 2, \quad 2^2 \equiv 4, \quad 2^3 \equiv 3, \quad 2^4 \equiv 1 \pmod 5\]

Since \(31 = 4 \times 7 + 3\), the exponent falls on the third entry of the cycle, so the remainder is 3.

Answer: B

We can write 15 as \(17-2\). So we need \((17-2)^{37} \pmod{17}\), which is equivalent to \((-2)^{37} \pmod{17}\). Since the power is odd, this is \(-(2^{37}) \pmod{17}\). By Fermat's Little Theorem, since 17 is prime, \(2^{16} \equiv 1 \pmod{17}\). We can write \(2^{37} = 2^{16} \times 2^{16} \times 2^5 \equiv 1 \times 1 \times 32 \pmod{17}\). The remainder of \(32 \div 17\) is 15. So we have \(-(15) \pmod{17}\), which is \(-15 \equiv 2 \pmod{17}\). The remainder is 2.

Answer: B

The smallest 4-digit number is 1000. We divide 1000 by 7 to find the remainder. \(1000 = 142 \times 7 + 6\). The remainder is 6. To find the smallest 4-digit number divisible by 7, we need to add a number to 1000 to make it a multiple of 7. The number to be added is \(7 - \text{remainder} = 7 - 6 = 1\). The required number is \(1000 + 1 = 1001\).

Answer: C

According to the Polynomial Remainder Theorem, if a polynomial P(x) is divided by \((x-a)\), the remainder is P(a). Here, the divisor is \((x-2)\), so a=2. We need to calculate P(2). \(P(2) = (2)^3 + 2(2)^2 - 5(2) + 7 = 8 + 2(4) - 10 + 7 = 8 + 8 - 10 + 7 = 16 - 10 + 7 = 6 + 7 = 13\). The remainder is 13.

Answer: D

A number formed by repeating a digit 6 times is always divisible by 3, 7, 11, and 13. This is because such a number is a multiple of 111111, and \(111111 = 3 \times 7 \times 11 \times 13 \times 37\). Let's check for this specific number. Divisibility by 3: sum of digits = 24 (divisible). Divisibility by 11: (4+4+4)-(4+4+4)=0 (divisible). Divisibility by 7 and 13: This is a property of such numbers. The number is 4 x 111111, so it is divisible by all prime factors of 111111.

Answer: A

First, let's find the remainder when the base, 143, is divided by 11. Sum of alternate digits: (3+1) - 4 = 0. Since the difference is 0, 143 is divisible by 11. The remainder is 0. Therefore, \(143^3\) will also be divisible by 11, and the remainder will be 0.

Answer: C

Correct answer: 1

Since \(88 = 8 \times 11\), both tests must pass.

Divisibility by 8. The last three digits are 46B, that is \(460 + B\). Because \(460 = 8\times 57 + 4\), we need \(4 + B \equiv 0 \pmod 8\), so \(B = 4\).

Divisibility by 11. With \(B = 4\), take the alternating sum of 8A5146B from the right:

\[(4 + 4 + 5 + 8) - (6 + 1 + A) = 21 - (7 + A) = 14 - A\]

This must be 0 or a multiple of 11. Only \(14 - A = 11\) leaves \(A\) a digit, so \(A = 3\).

The number is 8351464, and \(8351464 \div 88 = 94903\). Hence \(B - A = 4 - 3 = 1\).

Answer: C

For a number to be divisible by 88, it must be divisible by both 8 and 11.

1. Divisibility by 8: The number formed by the last three digits, 47B, must be divisible by 8. Let's test values for B. If B=2, we have 472. \(472 \div 8 = 59\). So, B=2 is a valid value.

2. Divisibility by 11: The difference between the sums of alternate digits must be 0 or a multiple of 11. The number is 8A51472. Sum of digits at odd places: \(2+4+5+8 = 19\). Sum of digits at even places: \(7+1+A = 8+A\). The difference is \(19 - (8+A) = 11-A\). For this to be a multiple of 11, the only possibility for a single-digit A is when \(11-A=0\), which gives A=11 (not a digit), or when \(11-A=11\), which gives A=0. So, A=0.

We have A=0 and B=2. The value of B-A = 2-0=2.

Answer: D

Correct answer: 2

Reduce each term modulo 11 separately.

Since \(13 \equiv 2 \pmod{11}\) and Fermat’s little theorem gives \(2^{10} \equiv 1 \pmod{11}\), write \(73 = 10\times 7 + 3\):

\[13^{73} \equiv 2^{73} = \left(2^{10}\right)^{7}\times 2^{3} \equiv 1\times 8 = 8 \pmod{11}\]

And \(14 \equiv 3\), so \(14^{3} \equiv 3^{3} = 27 \equiv 5 \pmod{11}\).

Adding the two parts, \(8 + 5 = 13 \equiv 2 \pmod{11}\).

Answer: D

We evaluate each term modulo 11 separately.

Part 1: \(13^{74} \pmod{11}\). First, simplify the base: \(13 \equiv 2 \pmod{11}\). The problem becomes \(2^{74} \pmod{11}\). By Fermat's Little Theorem (since 11 is prime), we know \(a^{p-1} \equiv 1 \pmod{p}\), so \(2^{10} \equiv 1 \pmod{11}\). We can rewrite the power: \(2^{74} = (2^{10})^7 \times 2^4\). So, \(2^{74} \equiv (1)^7 \times 2^4 \equiv 16 \pmod{11}\). The remainder of \(16 \div 11\) is 5.

Part 2: \(14^3 \pmod{11}\). Simplify the base: \(14 \equiv 3 \pmod{11}\). The problem becomes \(3^3 = 27 \pmod{11}\). The remainder of \(27 \div 11\) is 5.

Total Remainder = \((5 + 5) \pmod{11} = 10 \pmod{11}\). The final remainder is 10.