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Programming Enigma Puzzles

3 July 2012

Posted by on **From New Scientist #2694, 7th February 2009** [link]

I have written down three 3-figure numbers which, overall, use no digit more than once. One of the numbers is a perfect square and the two other numbers are primes. Their total is 2009.

With a little logic it is possible to calculate the three numbers very quickly.

What (in increasing order) are they?

[enigma1531]

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Here’s my original Perl solution. It runs in 19ms.

Solution:The numbers are 401, 625 and 983.And here’s a similar Python solution. It runs in 44ms.

I couldn’t resist having a go at this one as well:

I think I’ve worked out the little logic:

The digital root of 2009 is 2. The digits 0 to 9 sum to 45,

so we must omit 7 to leave a total 38 with digital root 2.

The primes are necessarily odd, so the square must be too, to give an odd total.

Odd 3-digit squares with no repeated digit are 169, 289, 361, 529, 625, 841, 961

(omitting 729).

If the square were to end in 1, the primes would have to end in 1 and 7, but that repeats the 1;

so 361, 841, and 961 are excluded.

If the square were to end in 9, the primes would have to end in 1 and 9 (duplication!) or 3 and 7, but we’ve excluded 7 from the start.

Therefore the square is 625 and the primes end in 1 and 3.

They have to sum to 1384, and must not start with 3, so 401 is the smallest possible.

By chance we’ve hit on the solution: 401 and 983.