(a - b)

^{2}+ (b – c)

^{2}+ (c – d)

^{2}+ (d – e)

^{2}+ (e – f)

^{2}

over all possible arrangements of (a, b, c, d, e, f) of the six numbers 2, 3, 5, 12, 15, 16.

On the face of it, this seems like a long hard slog of enumerating 6! = 720 different arrangements of the six numbers. However, there is a shortcut!

This question has been adapted from a UKMT Cayley paper. This means it is aimed at the best lower secondary mathematicians in the UK – only the top 500 are invited to participate – so I think it should be solvable by upper secondary students. I’d rate this at about 10-15 minutes of writing and thinking time.

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