Divisibility Rules & Remainder Theorem Questions and Answers

Answer: D

The number that is not a factor is 9.

Factorise 3630:

\[ \begin{aligned} 3630 &= 10 \times 363 = 2 \times 5 \times 3 \times 121 \\ &= 2 \times 3 \times 5 \times 11^{2} \end{aligned} \]

So 11, 3 and 5 are factors. 3 appears only once, so 9 = 3 x 3 is not. The divisibility test confirms it: the digit sum of 3630 is 3 + 6 + 3 + 0 = 12, which is divisible by 3 but not by 9.

Answer: C

The numbers divisible by 2, 3 and 11 are A and D: 8448 and 9636.

Divisibility by 2, 3 and 11 means divisibility by 66. Check each rule:

• 8448: even; digit sum 24, divisible by 3; alternating sum 8 - 4 + 4 - 8 = 0, divisible by 11. (8448 = 66 × 128.)
• 9812: digit sum 20, not divisible by 3.
• 9126: even and digit sum 18, but alternating sum 9 - 1 + 2 - 6 = 4, not divisible by 11.
• 9636: even; digit sum 24; alternating sum 9 - 6 + 3 - 6 = 0. (9636 = 66 × 146.)

Options including B or C therefore fail one of the tests.

Answer: B

The value of \((2x + 3y)\) is 26.

A number divisible by 88 is divisible by both 8 and 11.

Divisibility by 8: the last three digits, 74y, must be divisible by 8. Only 744 = 8 × 93 works, so y = 4.

Divisibility by 11: in 5 6 5 x 9 2 7 4 4, the digits in odd places sum to 5 + 5 + 9 + 7 + 4 = 30 and those in even places to 6 + x + 2 + 4 = 12 + x. Their difference, 18 - x, must be a multiple of 11, so x = 7.

\[ 2x + 3y = 2(7) + 3(4) = 14 + 12 = 26 \]

Answer: B

The value of (3x + y) is 5.

88 = 8 × 11, and 8 and 11 have no common factor, so the number must be divisible by both.

By 8: the last three digits, 67y, must be divisible by 8. Among 670 to 679 only 672 works, so y = 2.

By 11: the difference between the sums of alternate digits of 3 0 x 0 8 6 7 2 must be a multiple of 11:

\[ \begin{aligned} &(3 + x + 8 + 7) - (0 + 0 + 6 + 2) \\ &= 18 + x - 8 = 10 + x \end{aligned} \]

With x a single digit, 10 + x = 11 gives x = 1. Then 3x + y = 3 + 2 = 5. The number is 30108672 = 88 × 342144.

Answer: B

The difference is 24, which lies between 20 and 25.

Find each remainder:

\[ \begin{aligned} 2002 &= 101 \times 19 + 83 \\ 2201 &= 102 \times 21 + 59 \\ 83 - 59 &= 24 \end{aligned} \]

Check: \(101 \times 19 = 1919\) and \(2002 - 1919 = 83\), which is less than 101; \(102 \times 21 = 2142\) and \(2201 - 2142 = 59\), which is less than 102.

24 is in the range 20 to 25, so the other ranges are wrong. Using quotients 20 and 22 by mistake would give negative remainders, a sign the division has gone one step too far.

Answer: B

The value of \(3x + 2\) is 17.

A number is divisible by 36 when it is divisible by both 4 and 9. The last two digits, 92, are divisible by 4, so that condition holds for any \(x\). For 9, the digit sum must be a multiple of 9:

\[ \begin{aligned} 7 + 2 + x + 2 + 9 + 2 &= 22 + x \\ 22 + x = 27 &\Rightarrow x = 5 \\ 3x + 2 &= 3(5) + 2 = 17 \end{aligned} \]

The next multiple, 36, would need \(x = 14\), which is not a digit. 16, 20 and 15 correspond to \(x\) values that do not make 72x292 divisible by 9.

Answer: D

The sum of the possible values of \(x\) is 18.

A number is a multiple of 3 when the sum of its digits is a multiple of 3. For 3x15:

\[ 3 + x + 1 + 5 = 9 + x \]

Since 9 is already a multiple of 3, \(x\) itself must be a multiple of 3. As a single digit, \(x\) can be 0, 3, 6 or 9, and

\[ 0 + 3 + 6 + 9 = 18 \]

Leaving out 9 gives 9, and leaving out 0 changes nothing, so 9, 6 and 12 all miss at least one valid digit.

Answer: D

The number not divisible by 72 is 95142.

Since 72 = 8 × 9 and 8 and 9 have no common factor, a number is divisible by 72 when it is divisible by both 8 (last three digits divisible by 8) and 9 (digit sum divisible by 9).

95142: the last three digits 142 are not divisible by 8, and the digit sum 21 is not divisible by 9, so it fails.

The others pass both tests: 88848 (848 = 8 × 106, digit sum 36), 88704 (704 = 8 × 88, digit sum 27) and 95112 (112 = 8 × 14, digit sum 18).

Answer: B

The value of \((x - y)\) is 2.

\(88 = 8 \times 11\), and 8 and 11 are co-prime, so the number must be divisible by both.

Divisible by 8: the last three digits 91y must be divisible by 8. Among 910 to 919 only 912 is (\(912 = 8 \times 114\)), so \(y = 2\).

Divisible by 11: the difference between the sums of alternate digits of 179x0912 must be a multiple of 11.

\[ \begin{aligned} (1 + 9 + 0 + 1) &= 11 \\ (7 + x + 9 + 2) &= 18 + x \\ (18 + x) - 11 &= 7 + x \end{aligned} \]

For \(7 + x\) to be a multiple of 11 with x a single digit, \(x = 4\). So \(x - y = 4 - 2 = 2\). The value 4 is x itself, not \(x - y\).

Answer: C

The statement that is not true is x = 9 and y = 5.

72 = 8 × 9, so the number must be divisible by both 8 and 9. For 8, the last three digits 9y2 must be a multiple of 8: 912, 952 and 992 work, so y is 1, 5 or 9. For 9, the digit sum must be a multiple of 9:

\[ 9 + 4 + 7 + 1 + x + 9 + y + 2 = 32 + x + y \]

so x + y must be 4 or 13. The pairs (8, 5), (4, 9) and (3, 1) give 13, 13 and 4, but (9, 5) gives 14, which makes the digit sum 46, not a multiple of 9.