Divisibility Rules & Remainder Theorem Questions and Answers

Answer: B

Correct answer: 2

The doubling rule for 7 tests divisibility but does not hand you the remainder, so divide directly.

\[23473 = 7\times 3353 + 2\]

Since \(7\times 3353 = 23471\), the remainder is \(23473 - 23471 = 2\).

As a check, \(23471\) is exactly divisible by 7, and 23473 sits two above it.

Answer: B

To find the remainder, we can perform long division.

\(23 \div 7 = 3\) with a remainder of 2. Carry over 2.

\(24 \div 7 = 3\) with a remainder of 3. Carry over 3.

\(37 \div 7 = 5\) with a remainder of 2. Carry over 2.

\(23 \div 7 = 3\) with a remainder of 2.

The final remainder is 2. Alternatively, \(23473 = 21000 + 2100 + 350 + 21 + 2\). All terms except the last one are divisible by 7. So the remainder is 2.

Answer: A

First, we divide 1056 by 23 to find the remainder. \(1056 = 45 \times 23 + 21\). The remainder is 21. The number that must be added to make the sum divisible by 23 is \(\text{Divisor} - \text{Remainder} = 23 - 21 = 2\).

Answer: A

We can express the base 17 as \(16+1\). So, we need to find the remainder of \((16+1)^{23} \div 16\). According to the remainder theorem, this is equivalent to finding the remainder of \(1^{23} \div 16\). Since \(1^{23} = 1\), the remainder is 1.

Answer: D

Correct answer: 13

List the multiples of 13 in order and test each against the second condition.

\[13 \div 5 = 2 \text{ remainder } 3\]

The very first multiple already satisfies it, so 13 is the smallest such number. For comparison, 26 leaves remainder 1, 39 leaves 4 and 52 leaves 2, so none of the larger options qualifies as the smallest.

Answer: A

Correct answer: 1

\[1057 = 23 \times 45 + 22\]

The remainder is 22, so the next multiple of 23 is one step away.

\[23 - 22 = 1, \qquad 1057 + 1 = 1058 = 23 \times 46\]

Note the contrast with subtraction, where the remainder itself is removed.

Answer: D

Correct answer: 3

Here \(82 = 2\times 41\) and 3 shares no factor with either, so Euler’s theorem applies.

\[\phi(82) = \phi(2)\,\phi(41) = 1\times 40 = 40 \quad\Longrightarrow\quad 3^{40} \equiv 1 \pmod{82}\]

Since \(81 = 2\times 40 + 1\),

\[3^{81} = \left(3^{40}\right)^{2}\times 3 \equiv 1^{2}\times 3 = 3 \pmod{82}\]

Answer: D

\[ 3^{3} = 27 \]

Since \( 27 < 82 \), the division gives quotient 0 and the whole number is the remainder:

\[ 27 \]

Answer: D

A number is divisible by 18 if it is divisible by its co-prime factors, 2 and 9.

Option D, 65043, ends in 3, so it is an odd number and not divisible by 2. Therefore, it cannot be divisible by 18.

Let's check the others to be sure. A) 34056: Ends in 6 (div by 2). Sum is 18 (div by 9). So divisible by 18. B) 54036: Ends in 6 (div by 2). Sum is 18 (div by 9). So divisible by 18. C) 50436: Ends in 6 (div by 2). Sum is 18 (div by 9). So divisible by 18.

Answer: A

The sum of the first n cubes is given by the formula \(S_n = [\frac{n(n+1)}{2}]^2\). Here, n=100. Sum = \([\frac{100(101)}{2}]^2 = (50 \times 101)^2 = (5050)^2\). We need to find the remainder of \(5050^2 \div 4\). First, find the remainder of \(5050 \div 4\). The number formed by the last two digits is 50. \(50 \div 4\) gives a remainder of 2. So, we need to find the remainder of \(2^2 \div 4 = 4 \div 4\). The remainder is 0.