Divisibility Rules & Remainder Theorem Questions and Answers

Answer: C

The value of 2x + y is 8.

72 = 8 × 9, and 8 and 9 have no common factor, so the number must be divisible by both.

By 8: the last three digits, 58y, must be divisible by 8. Among 580 to 589 only 584 is, so y = 4.

By 9: the digit sum must be a multiple of 9.

\[ \begin{aligned} 1+3+4+x+5+8+4 &= 25 + x \\ 25 + x = 27 &\Rightarrow x = 2 \end{aligned} \]

(x = 11 is impossible for a digit.) So 2x + y = 4 + 4 = 8, and the number is 1342584. The values 6, 7 and 9 come from a wrong digit for x or y.

Answer: B

The value is 7. A number divisible by 3, 7 and 11 is divisible by their product 231. Look for a multiple of 231 between 89800 and 89899. Since 231 x 389 = 89859, the number is 89859, so x = 5 and y = 9. Check: the digit sum 8 + 9 + 8 + 5 + 9 = 39 is divisible by 3, and 231 x 389 is divisible by 7 and 11 by construction. Then \(5x - 2y = 25 - 18 = 7\). The nearby multiples 231 x 388 = 89628 and 231 x 390 = 90090 fall outside the range 898xy, so 89859 is the only number.

Answer: C

The digit sum is 9. The divisors minus the remainders are 14 - 5 = 9, 16 - 7 = 9, 20 - 11 = 9 and 35 - 26 = 9, so x + 9 is a common multiple of 14, 16, 20 and 35. Their LCM is 560. Between 5000 and 5500, x + 9 must lie between 5009 and 5509, and the only multiple of 560 there is 560 x 9 = 5040. So x = 5031. Check: 5031 divided by 14 leaves 5, by 16 leaves 7, by 20 leaves 11 and by 35 leaves 26. The digit sum is 5 + 0 + 3 + 1 = 9.

Answer: A

The remainder is 0. Multiply the given expression by 2: \[2(7x + 2y - 5z) = 14x + 4y - 10z\] Reducing the coefficients modulo 11 gives 14 = 3 + 11, 4 stays 4 and -10 = 1 - 11, so \[14x + 4y - 10z \equiv 3x + 4y + z \pmod{11}\] Now -7y is congruent to 4y and 12z to z modulo 11, so \(3x - 7y + 12z \equiv 3x + 4y + z\). Therefore 3x - 7y + 12z is congruent to twice a multiple of 11, and the remainder is 0. The options 1, 5 and 7 would need a different relationship between the two expressions.

Answer: C

546714 is not divisible by 4. A number is divisible by 4 if its last two digits form a number divisible by 4.

467540 ends in 40 and 367328 ends in 28 and 556824 ends in 24, all multiples of 4, while 546714 ends in 14, which is not, so it is the answer.

The reason the last-two-digits test works is that 100 is a multiple of 4, so only the last two digits decide whether the number is divisible. The other numbers can also be checked by halving twice: 467540 / 2 = 233770 and 233770 / 2 = 116885, a whole number, which agrees with the test.

Answer: D

The statement that is not true is x = 7 and y = 3. A number is divisible by 72 when it is divisible by both 8 and 9. For divisibility by 8, the last three digits 9y2 must be divisible by 8, which holds for y = 1, 5 and 9 (912, 952 and 992 are divisible by 8) and not for y = 3 (932 is not).

For divisibility by 9 the digit sum must be a multiple of 9:

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

So x + y = 4 or 13. This gives x = 3, y = 1; x = 8, y = 5; and x = 4, y = 9, which are the first three options. The pair x = 7, y = 3 fails both tests, so the fourth option is the one that is not true.

Answer: A

When different numbers leave the same remainder y on division by x, x divides the differences between the numbers. The differences are:

\[ 3481 - 2629 = 852 = 2^2 \times 3 \times 71 \]

\[ 4972 - 3481 = 1491 = 3 \times 7 \times 71 \]

The common factors are 3 and 71 (their product, 213, is the HCF). The two-digit divisor is therefore x = 71. The remainder is 2629 = 71 x 37 + 2, so y = 2.

\[ 2x - 5y = 142 - 10 = 132 \]

The options 133, 131 and 129 differ by one or three from 132, and come from a wrong remainder.

Answer: C

The sum of the digits of \(x\) is 16. The least number divisible by 11, 21, 24 and 28 is their LCM: \[ \text{LCM}(21,24,28) = 168,\quad 168 \times 11 = 1848 \] So \(x = 1848k + 10\). Since \(1848 = 37\times 49 + 35\), we have \(1848 \equiv -2 \pmod{37}\), and \(x\) divisible by 37 needs \(10 - 2k \equiv 0\), so \(k = 5\) (the least positive value). \[ x = 1848\times 5 + 10 = 9250 = 37\times 250 \] \[ 9 + 2 + 5 + 0 = 16 \] The options 13, 15 and 17 do not match the digit sum of 9250.

Answer: B

The factor is \((x-1)\).

By the factor theorem, \((x - a)\) is a factor exactly when \(f(a) = 0\):

\[ f(1) = 2(1)^{7} + 3(1) - 5 = 2 + 3 - 5 = 0 \]

Check the others: \(f(-1) = -2 - 3 - 5 = -10\), so \((x+1)\) is not a factor. For \((2x-5)\), \(f(2.5)\) is large and positive. For \(x^{3}+8\), its real root is \(x = -2\), and \(f(-2) = -256 - 6 - 5 \ne 0\).

Answer: D

The expression divisible by 11 is 4x − 9y.

Look for a multiple of 4x − 9y that differs from a multiple of 3x + 7y by a multiple of 11.

\[ \begin{aligned} 3(4x - 9y) &= 12x - 27y \\ &= 4(3x + 7y) - 55y \end{aligned} \]

Both 4(3x + 7y) and 55y are multiples of 11, so 3(4x − 9y) is too. Since 11 is prime and does not divide 3, it must divide 4x − 9y.

A check with numbers: x = 5, y = 1 gives 3x + 7y = 22. Then 4x − 9y = 11, while 4x + 6y = 26, x + y + 4 = 10 and 9x + 4y = 49, none of which is a multiple of 11.