PK 2016 Q22
22. Jakob wrote down four consecutive positive integers. He then calculated the four possible totals made by taking three of the integers at a time. None of these totals was a prime. What is the smallest integer Jakob could have written?
- A 12
- B 10
- C 7
- D 6
- E 3
Pink Kangaroo 2016, Question 22. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: C 7
Let \(n\) be the smallest of the integers. The four consecutive integers have sum \(n + n + 1 + n + 2 + n + 3 = 4n + 6\). Then the four possible sums formed by taking three of these at a time are
\(4n + 6 - n = 3n + 6\) (divisible by 3 so not prime)
\(4n + 6 - (n + 1) = 3n + 5\) (if \(n\) is odd then this is even, so is not prime)
\(4n + 6 - (n + 2) = 3n + 4\)
\(4n + 6 - (n + 3) = 3n + 3\) (divisible by 3 so not prime)
Hence we are looking for the smallest \(n\) for which neither \(3n + 4\) nor \(3n + 5\) is prime.
| n | \(3n + 4\) | \(3n + 5\) |
| 1 | 7 prime | 8 not prime |
| 2 | 10 not prime | 11 prime |
| 3 | 13 prime | 14 not prime |
| 4 | 16 not prime | 17 prime |
| 5 | 19 prime | 20 not prime |
| 6 | 22 not prime | 23 prime |
| 7 | 25 not prime | 26 not prime |
When \(n = 7\), none of the four sums is prime.
Solution © UK Mathematics Trust, from the PK 2016 solutions.