Divisibility Rules & Remainder Theorem Questions and Answers

Answer: B

There are 8 such numbers.

A divisor d leaves remainder 28 on 1458 exactly when d divides \(1458 - 28 = 1430\) and d is greater than 28 (a remainder must be smaller than the divisor).

\[ 1430 = 2 \times 5 \times 11 \times 13 \]

1430 has 16 divisors: 1, 2, 5, 10, 11, 13, 22, 26, 55, 65, 110, 130, 143, 286, 715 and 1430. Those greater than 28 are 55, 65, 110, 130, 143, 286, 715 and 1430, which is 8 numbers. Answers of 9 or 10 come from including divisors that are not greater than 28, such as 26 or 22, which actually leave remainders of 2 and 6.

Answer: D

The remainder is 0.

The remainder on division by 100 is the last two digits of the product. Two of the factors are 60, and \(60 \times 60 = 3600\), which already ends in 00.

Multiplying a number ending in 00 by any whole numbers keeps the last two digits 00, so the whole product \(86 \times 60 \times 12 \times 60 \times 94\) is divisible by 100 and leaves no remainder. Any of the other remainders would need a product not divisible by 100, which the two factors of 60 rule out.

Answer: C

The number divisible by 6 is 9n.

The single-digit numbers greater than 2 that are even are 4, 6 and 8. Of these, 4 and 8 are divisible by 4, so n = 6.

\[ \begin{aligned} 9n &= 54 = 6 \times 9 \\ 9n + 4 &= 58,\ \ 6n + 7 = 43,\ \ 5n + 3 = 33 \end{aligned} \]

Only 54 is divisible by 6. 58 and 43 are not divisible by 3 (43 is not even even), and 33 is odd, so it fails the test for 2.

Answer: D

The remainder is 2.

The remainder on division by 10 is the units digit of the product, and only the units digits of the factors matter: 3, 7, 7, 8 and 2.

\[ \begin{aligned} 3 \times 7 &= 21 \to 1 \\ 1 \times 7 &= 7 \\ 7 \times 8 &= 56 \to 6 \\ 6 \times 2 &= 12 \to 2 \end{aligned} \]

So the product ends in 2. 6 is the units digit one step before the last factor, a sign of dropping the 52. 3 and 8 come from slips in the chain.

Answer: C

Statement I is true but Statement II is false.

Statement I: every multiple of 4 is \(4k = 2(2k)\), so it is also a multiple of 2. True.

Statement II is the converse, and it fails: 2, 6 and 10 are divisible by 2 but not by 4. A rule of the form 'divisible by m implies divisible by n' holds only when n is a factor of m, so the step from 4 to 2 works but the step from 2 to 4 does not.

That rules out 'both true', 'both false' and the reversed option.

Answer: B

Only two pairs are correctly matched.

Pair 1 is right: 2 and 3 are coprime, so a number divisible by both is divisible by their product, 6.

Pair 2 is wrong: 4, 12 and 20 are divisible by 4 but not by 8. Divisibility passes down to factors, not up to multiples.

Pair 3 is right: the condition already includes divisibility by 10, and divisibility by 5 adds nothing, because 5 is a factor of 10. Note that the rule in pair 1 does not extend to non-coprime pairs: divisible by 2 and 4 does not mean divisible by 8.

Answer: C

The value of a is 3.

A number is divisible by 11 when the difference between the sum of its digits in odd places and the sum in even places is 0 or a multiple of 11. For 5a108, the odd-place digits are 5, 1 and 8 and the even-place digits are a and 0:

\[ \begin{aligned} (5 + 1 + 8) - (a + 0) &= 14 - a \\ 14 - a = 11 &\Rightarrow a = 3 \end{aligned} \]

Check: 53108 = 11 × 4828. With a = 1, 2 or 4 the difference is 13, 12 or 10, none of which is a multiple of 11.

Answer: A

The remainder is 1.

Work with powers of 2 modulo 17. Since \(2^4 = 16 = 17 - 1\):

\[ \begin{aligned} 2^{4} &\equiv -1 \pmod{17} \\ 2^{8} &\equiv 1 \pmod{17} \\ 2^{256} &= (2^{8})^{32} \equiv 1 \pmod{17} \end{aligned} \]

16 would be the remainder for \(2^4\), or any power \(2^{4k}\) with \(k\) odd; 256 is a multiple of 8, so the result is 1. Fermat's little theorem gives the same, since \(2^{16} \equiv 1\) and 256 is a multiple of 16.

Answer: C

The remainder is 5.

First reduce the base modulo 19: 2019 = 19 × 106 + 5, so 2019 leaves remainder 5. Hence \(2019^{19}\) leaves the same remainder as \(5^{19}\).

Since 19 is prime and does not divide 5, Fermat's little theorem gives \(5^{18} \equiv 1 \pmod{19}\). Therefore:

\[ \begin{aligned} 5^{19} &= 5^{18} \times 5 \\ &\equiv 1 \times 5 = 5 \pmod{19} \end{aligned} \]

More generally \(a^{p} \equiv a \pmod{p}\) for a prime p, so the remainder is simply that of 2019 itself. The options 2, 3 and 7 do not satisfy this.