Maclaurin 2016 Q1
1. The positive integer \(N\) has five digits.
The six-digit integer \(P\) is formed by appending the digit 2 to the front of \(N\).
The six-digit integer \(Q\) is formed by appending the digit 2 to the end of \(N\).
Given that \(Q = 3P\), what values of \(N\) are possible?
Maclaurin Olympiad 2016, Question 1. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: 85 714
We have \(P = N + 200\,000\) and \(Q = 10N + 2\). Hence
\(\begin{aligned}10N + 2 &= 3 \times (N + 200\,000)\\[6pt] &= 3N + 600\,000.\end{aligned}\)
Therefore, subtracting \(3N + 2\) from each side, we get
\(7N = 599\,998\)
and so, dividing each side by 7, we obtain
\(N = 85\,714.\)
Hence the only possible value of \(N\) is 85 714.
Solution © UK Mathematics Trust, from the Maclaurin Olympiad 2016 solutions.