Divisibility Rules & Remainder Theorem Questions and Answers

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.

Answer: C

Correct answer: 12

Divide and take the remainder, since subtracting it leaves an exact multiple.

\[5000 = 43 \times 116 + 12\]

So 12 must be subtracted, giving \(4988 = 43 \times 116\).

Answer: C

This is a direct application of Wilson's Theorem, which states that for any prime number p, \((p-1)! \equiv -1 \pmod{p}\). Here, p = 31, which is a prime number. We need to find the remainder of \((31-1)!\) when divided by 31. According to the theorem, \(30! \equiv -1 \pmod{31}\). A remainder of -1 is equivalent to a remainder of \(-1+31=30\). So the remainder is 30.

Answer: B

A number divisible by 3 and 5 is divisible by their LCM, which is 15. The number of multiples of 15 up to 300 is \(\lfloor 300/15 \rfloor = 20\). Now we need to subtract the numbers that are also divisible by 7. A number divisible by 15 and 7 is divisible by their LCM, which is 105. The number of multiples of 105 up to 300 are \(\lfloor 300/105 \rfloor = 2\) (which are 105 and 210). So, the required number of numbers is \(20 - 2 = 18\).

Answer: C

First, find the remainder of the base. \(5 \div 3\) leaves a remainder of 2. So the problem is equivalent to finding the remainder of \(2^5 \div 3\). \(2^5=32\). Now, find the remainder of \(32 \div 3\). The sum of digits is 5, and \(5 \div 3\) leaves a remainder of 2. So the remainder is 2. Alternatively, \(2 \equiv -1 \pmod{3}\), so \(2^5 \equiv (-1)^5 = -1 \equiv 2 \pmod{3}\).

Answer: C

Correct answer: 7

Work out how quickly the powers of 7 repeat modulo 100.

\[7^1 = 7,\quad 7^2 = 49,\quad 7^3 = 343 \equiv 43,\quad 7^4 = 2401 \equiv 1 \pmod{100}\]

The cycle length is 4, so only the exponent modulo 4 matters. Since \(129 = 4\times 32 + 1\),

\[7^{129} = \left(7^{4}\right)^{32}\times 7 \equiv 1^{32}\times 7 = 7 \pmod{100}\]

Answer: C

We need to find the remainder of \(7^{131} \pmod{100}\). We can find the cycle of remainders for powers of 7.

\(7^1 \equiv 7 \pmod{100}\)

\(7^2 \equiv 49 \pmod{100}\)

\(7^3 = 343 \equiv 43 \pmod{100}\)

\(7^4 = 7^2 \times 7^2 = 49 \times 49 = 2401 \equiv 1 \pmod{100}\)

The cycle of remainders is (7, 49, 43, 1), and the length of the cycle is 4.

To find the remainder for \(7^{131}\), we divide the power (131) by the cycle length (4).

\(131 \div 4 = 32\) with a remainder of 3.

The required remainder is the 3rd term in our cycle, which is 43.

Answer: A

The expression can be factored as \(n^2(n-1)(n+1)\), or \(n \times n \times (n-1) \times (n+1)\). This can be rearranged as \((n-1)n(n+1)n\). The term \((n-1)n(n+1)\) is the product of three consecutive integers, which is always divisible by \(3! = 6\). So the expression is divisible by 6. We also need to check for divisibility by other factors. The expression is divisible by 3 and 2. We need to check for 4. If n is even, let n=2k. Then \(n^2=4k^2\), so the expression is divisible by 4. If n is odd, let n=2k+1. Then n-1 and n+1 are consecutive even numbers, \((2k)(2k+2) = 4k(k+1)\). This product is divisible by 4. So the expression is always divisible by 3 and 4, which are co-prime. Therefore, it is always divisible by \(3 \times 4 = 12\).