PK 2020 Q24
24. Eight consecutive three-digit positive integers have the following property: each of them is divisible by its last digit. What is the sum of the digits of the smallest of these eight integers?
- A 9
- B 10
- C 11
- D 12
- E 13
Pink Kangaroo 2020, Question 24. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: E 13
Since we cannot divide by zero, the eight numbers must have the form \(ABn\) where \(n\) runs from 1 to 8 or from 2 to 9. However if \(ABn\) is divisible by \(n\) for each \(n\) from 2 to 9, it is also divisible by \(n\) for each \(n\) from 1 to 8 and \(AB1\) is smaller than \(AB2\). Hence our solution will use the digits 1 to 8. Note that \(ABn\) is divisible by \(n\) if and only if \(AB0\) is divisible by \(n\). So we require the smallest number \(AB0\) which is divisible by 1, 2, 3, 4, 5, 6, 7, 8. The LCM of these 8 numbers is 8 × 7 × 5 × 3 = 840. The eight numbers are then 841, 842, 843, 844, 845, 846, 847, 848. The digit sum of 841 is 13.
Solution © UK Mathematics Trust, from the PK 2020 solutions.