IMC 2019 Q24
24. When 5655 is divided by a two-digit positive integer \(N\), the remainder is 11. When 5879 is divided by the same positive integer \(N\), the remainder is 14.
What is the sum of the digits of \(N\)?
- A 6
- B 7
- C 8
- D 9
- E 10
Intermediate Mathematical Challenge 2019, Question 24. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: C 8
We are given that \(5655 = N \times j + 11\) and \(5879 = N \times k + 14\).
Subtracting the first equation from the second, we obtain, \(224 = N \times (k - j) + 3\). Therefore \(221 {=} N \times (k - j)\).
It follows that \(N\) is a positive factor of 221, so that \(N\) is equal to 1, 13, 17 or 221. But one of the remainders is 14, so \(N\) is equal to 17 or 221. However, 221 does not have two digits.
Therefore \(N = 17\) and the sum of its digits is 1 + 7 = 8.
Solution © UK Mathematics Trust, from the IMC 2019 solutions. Longer versions with investigations: Solutions and investigations (PDF).