Divisibility Rules & Remainder Theorem Questions and Answers

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.

Answer: B

\[ 100 = 3 \times 33 + 1 \quad\Longrightarrow\quad 100 \equiv 1 \pmod{3} \]

\[ 100^{100} \equiv 1^{100} = 1 \pmod{3} \]

Equivalently, the digit sum of \( 100^{100} \) is 1.

Answer: C

Let the number be N and the divisor be D. We have \(N = Dk + 24\). Now consider twice the number, 2N. \(2N = 2(Dk + 24) = 2Dk + 48\). When 2N is divided by D, the remainder is the same as the remainder of \(48 \div D\). We are given that this remainder is 11. So, \(48 = Dm + 11\) for some integer m. This means \(D \times m = 37\). Since 37 is a prime number, its only factors are 1 and 37. The divisor D must be greater than the remainder (24), so D must be 37.

Answer: C

Correct answer: 548

A number leaving the same remainder 8 under every one of these divisors has the form (a common multiple of them) \(+\ 8\). The least such number uses the LCM.

\[12 = 2^2\times 3,\quad 15 = 3\times 5,\quad 20 = 2^2\times 5,\quad 54 = 2\times 3^3\]

Taking the highest power of each prime, \(\text{LCM} = 2^2\times 3^3\times 5 = 540\).

\[N = 540 + 8 = 548\]

Strictly, 8 itself also leaves remainder 8 under each of these divisors, since it is smaller than all of them. The intended reading, standard for this question type, is the least number larger than the divisors, which is 548.