IMC 2018 Q17
17. How many three-digit numbers are increased by 99 when their digits are reversed?
- A 4
- B 8
- C 10
- D 80
- E 90
Intermediate Mathematical Challenge 2018, Question 17. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: D 80
The value of ‘abc’ equals \(100a + 10b + c\). We need to find integers of the form ‘abc’ such that ‘cba’ = ‘abc’ + 99.
So \(100c + 10b + a = 100a + 10b + c + 99\), that is \(99c = 99a + 99\). This condition simplifies to \(c = a + 1\). So there are 10 such integers of the form ‘1b2’, 10 of the form ‘2b3’, 10 of the form ‘3b4’, …, and 10 of the form ‘8b9’. Therefore the required number is 8 × 10 = 80.
Solution © UK Mathematics Trust, from the IMC 2018 solutions. Longer versions with investigations: Solutions and investigations (PDF).