(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.
Feel free to comment, ask questions and even check your answer in the comments box below powered by
If you enjoy using this website then please consider making a donation - every little helps :-)
You can receive these questions directly to your email box or read them in an RSS reader. Subscribe using the links on the right.
Don’t forget to follow Gifted Mathematics on Google+, Facebook or Twitter. You may add your own interesting questions on our Google+ Community and Facebook..
You can also subscribe to our Bookmarks on StumbleUpon and Pinterest. Many resources never make it onto the pages of Gifted Mathematics but are stored in these bookmarking websites to share with you.