Multiple choice

A divisor divides a number and gives the remainder 21. When thrice the original number is again divided by the same number, the remainder changes to 9. What will be the divisor to give the remainders of 21 and 9 in this condition?

  1. 15

  2. 22

  3. 27

  4. 31

  5. 37
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C Correct answer
Explanation

Let the divisor be D. N = qD + 21. 3N = q'D + 9. Since 3N = 3(qD + 21) = 3qD + 63, the remainder when 3N is divided by D is the remainder of 63 divided by D. 63 = kD + 9, so kD = 54. Possible divisors are factors of 54 greater than 21. 27 is a factor of 54 and greater than 21.

AI explanation

Let the original number be N and the divisor be D, meaning N equals k times D plus 21 for some integer k. Multiplying by three gives 3N equals 3k times D plus 63. Since dividing 3N by D leaves a remainder of 9, dividing 63 by D must also leave a remainder of 9. This requires D to be greater than 21 and to divide 54 exactly, making the divisor 27.

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