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.

1. The last 3 digits can quickly be ascertained from a list of squares mod 1000. They are{129,161,241,289,329,361,529,561,729,761,929,961}

2. Call each of these N and solve for integer x,y{x^2-N = 1000y, 0<y<100} to obtain candidates:

129:y=15,16  (1,61,29)
161:y=14,17  (1,41,61)(1,71,61)
241:y=29      (29:2:41)
289:y=54,71  (71:2:89)  
329:y=31      (31:3:29)(3:13:29)  
361:y=53,72  (53:3:61)(7:23:61)  
529:y=51,74  (no solutions)  
561:y=28       (no solutions)  
729:y=49,76  (7:67:29)  
761:y=32      (no solutions)
929:y= 29     (no solutions)
961:y=47,78  (7,89,61)

(Note: the solutions starting with a 1 are treated as not qualifying as 1 is not strictly a prime number. Otherwise, the candidate (1,41,61) could replace any of those numbers having 61 as a component, below.)

3. Since 9 different primes are required:231^2=(53:3:61),267^2=(71:2:89) and 277^2=(7:67:29).

4.In addition, there are two more solutions if the sample square given in the problem is used:(71:2:89)(31:3:29)(7:23:61) and (71:2:89)(3:13:29)(7:23:61).

Edited on October 19, 2010, 2:09 pm

Posted by broll on 2010-10-19 13:40:02