Enigma 1703

Spoiler for New Scientist Enigma 1703:

I was staying at my sister’s house when my niece Amy came home from school feeling special. The class had been shown how to split a whole number, T, into whole number parts in such a way that the product of the parts was the greatest, G, that could be obtained for that T. For instance, she explained, 10 could be split into ten ones, or 2 and 4 and 4, or 5 and 5, and so on, which would yield products of 1, 32, and 25 respectively. But, she warned, G exceeds 32 for T=10.

Why did she feel special? Well, each pupil in the class had been given a different number in the range 20-50 inclusive for their personal T, and she had noted that, when she added the digits of her G together, the sum was exactly half of her T, and no one else in the class had T and G with this property.

What value of T was Amy given?

I couldn’t do this without a good deal of computer help. I wrote a Perl script to find G(T) for small values of T; the number of possible partitions of T becomes too large to test all the numbers out to 50, but it did show me the pattern, which once you know it can be proved by induction:

  • For T = 3k, G(T) = 3k.
  • For T = 3k+1, G(T) = 4 x 3k–1.
  • For T = 3k+2, G(T) = 2 x 3k.

For instance, for T = 3, k = 1 and G(3) = 3; for T = 4, k = 1 and G(4) = 1 x 3 = 3; for T = 5, k = 1 and G(3) = 2 x 3 = 6; for T = 6, k = 2 and G(3) = 3 x 3 = 9, and so on.

That’s about as interesting as this enigma gets. After that you pretty much just have to calculate G(T) for all even T from 20 to 50 and find the one whose digits sum to half of T. More computer work; I wrote a spreadsheet. The answer is T = 36, G(T) = 312 = 531441 with digit sum 18.


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