31329 is a 5-digit perfect square. It happens that its digits can be used successively, with a couple of dividing spaces, to form three prime numbers: two with two digits each, and one with only one digit:

31 3 29

It's also possible to do this subdividing another way: 3 13 29.

Several other 5-digit squares can be divided in this manner into two 2-digit primes and one 1-digit prime. But the goal of this particular puzzle is to find three such squares so that a total of nine different primes are formed by the subdivisions, being of course six 2-digit primes and three 1-digit primes due to the nature of how they were formed.

There are no leading zeros for any of the primes or squares. I'll tell you this: 31329 is unique in having two ways of being split in this manner. All the rest of the squares for which this is possible have only one way of being validly split.

By squaring all 3 digit numbers from 100 to 317 in Excel I found the following set of squares which can be broken into two 2-digit primes and a one digit prime:

171 29241     29  2  41
177 31329     31  3  29
231 53361     53  3  61
267 71289     71  2  89
269 72361      7 23  61
277 76729      7 67  29
281 78961      7 89  61

We need to eliminate 31329 as that was the given example.
In the set 53361 has 53 and 3 unique to itself.  Since 61 appears in it also we can elimate all other squares which feature 61. That leaves us with 76729 [7 67 29] which begin with 7.  Since that number contains a 29 we can further eliminate 29241 [29 2 41] which leaves 71289 [71 2 89].

The three squares therefore which can give us six 2-digit primes and three 1-digit primes are thus:
          53361     53  3  61
          71289     71  2  89  
 and  76729      7 67  29

Posted by brianjn on 2010-10-19 21:16:24