Divisibility Rules & Remainder Theorem Questions and Answers

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 \(3 \times 4 = 12\).

Answer: B

First, find the smallest number that leaves a remainder of 7 when divided by 9, 10, and 15. This number is of the form \(k \times \text{LCM}(9,10,15) + 7\). LCM of 9, 10, 15 is 90. The numbers are 97, 187, 277, etc. We need to find a number of this form that is just less than 1936. Let's divide 1936 by 90. \(1936 = 21 \times 90 + 46\). The nearest multiple of 90 below 1936 is \(21 \times 90 = 1890\). The required number is \(1890+7=1897\). The number that must be subtracted from 1936 to get 1897 is \(1936 - 1897 = 39\).

Answer: C

Let the number be N. Let the quotient when divided by 3 be q. So, \(N = 3q+1\). We are given that when q is divided by 2, the remainder is 1. So, \(q = 2k+1\) for some integer k. Substitute this expression for q into the first equation: \(N = 3(2k+1)+1 = 6k+3+1 = 6k+4\). This shows that when the number N is divided by 6, the remainder is 4.

Answer: A

We can write the base 9 as \(8+1\). The expression becomes finding the remainder of \((8+1)^{100} \div 8\). According to the remainder theorem, this is equivalent to finding the remainder of \(1^{100} \div 8\). Since \(1^{100} = 1\), the remainder is 1.