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.
The easiest approach was to square numbers from 100 to 316 (thus limiting examinations to squares), and then to generate for each the three divisions a-bc-de, ab-c-de, ab-cd-e. The last of these does not produce any triads of primes.
a-bc-de allows 3-13-29, 7-23-61, 7-67-29, 7-89-61
ab-c-de allows 29-2-41, 31-3-29, 53-3-61, 71,2,89
From these: 3-13-29 , 7-23-61, and 71-2-89 give nine different primes.
If we are not to use the sample 31329, then we could also have: 71-2-89, 53-3-61, 7-67-29 (if "such squares" means "other").
Edited on October 19, 2010, 12:54 pm
Edited on October 19, 2010, 12:55 pm
solution
