Re: Ask EU: Mathematical puzzle

BrritSki wrote:
New Scientist Feb 7th Enigma 1531 reads as follows:
"The Year In Question.
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 other two numbers are primes. Their total is 2009.
With a little logic it is possible to calculate the three numbers. What (in increasing order) are they ?"

So I tried with the following logic:

Primes can only end in 1, 3, 7 & 9, so the possible ending combinations for the 2 primes are 1&3 1&7 1&9 3&7 3&9 & 7&9.

Summing each of those pairs and subtracting from 9 (or 19) gives the following results:
Sum 1 3 4 5 1 7 8 1 Not allowed as repeats a digit
1 9 10 9 Not allowed as repeats a digit 3 7 10 9 3 9 12 7 No - squares must end 0,1,4,9,6,5
7 9 16 3 No - squares must end 0,1,4,9,6,5

So the square must end in either:
5 so will have a value of 225 (not allowed as repeats a digit)
or 625.
9 so will have a value of 361 (not allowed as repeats a digit already in one of the primes) or 841.

Taking the first example with the square being 625 and the primes ending in 1 & 3, the remaining digits are 4 7 8 & 9, which means you must have the following combinations (forgetting for a moment about whether they combine to form primes in some way):
100s 10s Sum + 629
4 7 8 9 1899
4 8 7 9 1989
4 9 7 8 2079
7 8 4 9 2259
7 9 4 8 2349
8 9 4 7 2439

Similarly for the 2nd example with the square being 841 and the primes ending in 3 & 7, the remaining digits are 2 5 6 & 9, which means you must have the following combinations (forgetting for a moment about whether they combine to form primes in some way):
100s 10s Sum + 851
2 5 6 9 1701
2 6 5 9 1791
2 9 5 6 2061
5 6 2 9 2061
5 9 2 6 2331
6 9 2 4 2411

HELP, none of these answers is 2009, so where have I gone wrong ?????????

TIA

I don't have time to work through the whole thing, but I think you've made an invalid assumption at the very top - there's no reason that the two primes cannot end with the same digit.

--
David
.

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