Hamilton 2020 Q2
2. In the UK, 1p, 2p and 5p coins have thicknesses of 1.6 mm, 2.05 mm and 1.75 mm respectively.
Using only 1p and 5p coins, Joe builds the shortest (non-empty) stack he can whose height in millimetres is equal to its value in pence. Penny does the same but using only 2p and 5p coins.
Whose stack is more valuable?
Hamilton Olympiad 2020, Question 2. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: Penny’s (135p, compared with Joe’s 125p)
Say Joe has \(a\) 1p coins and \(b\) 5p coins. Then he wants the minimal \(a\) and \(b\) such that
This simplifies to
Since \(a\) and \(b\) are integers, and 12 and 65 have no factors in common (other than 1), \(a\) must be a multiple of 65 and \(b\) must be a multiple of 12. Clearly the smallest (positive) \(a\) and \(b\) which satisfy this equation are \(a = 65\) and \(b = 12\). So Joe’s stack comprises sixty-five 1p coins and twelve 5p coins, giving it a value of 125p.
Now say Penny has \(c\) 2p coins and \(d\) 5p coins. Then similarly, she wants the minimal \(c\) and \(d\) such that
which simplifies to
The smallest positive \(c\) and \(d\) which satisfy this equation are \(c = 65\) and \(d = 1\). So Penny’s stack comprises sixty-five 2p coins and one 5p coin, giving it a value of 135p.
Hence Penny’s stack is more valuable.
Solution © UK Mathematics Trust, from the Hamilton Olympiad 2020 solutions.