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Going pandigital with P and Q (Posted on 2008-10-11) Difficulty: 2 of 5
Determine all possible pair(s) (P, Q) of 5-digit perfect squares, such that P and Q together contain each of the digits 0 to 9 exactly once. Neither P nor Q can contain any leading zeroes.

See The Solution Submitted by K Sengupta    
Rating: 1.6667 (3 votes)

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mathematia's answer | Comment 1 of 2

wrote up a quick optimized search in mathematica and tested all 5 digit pandigital pairs (P,Q) for pairs where both are squares.  and got the following list

 ( 15876,23409 ) = ( 126^2,153^2)

 ( 15876,39204 ) = ( 126^2,198^2)

 ( 20736,51984 ) = ( 144^2,228^2)

 ( 20736,95481 ) = ( 144^2,309^2)

 ( 23409,15876 ) = ( 153^2,126^2)

 ( 30276,51984 ) = ( 174^2,228^2)

 ( 30276,95481 ) = ( 174^2,309^2)

 ( 38025,47961 ) = ( 195^2,219^2)

 ( 39204,15876 ) = ( 198^2,126^2)

 ( 47961,38025 ) = ( 219^2,195^2)

 ( 51984,20736 ) = ( 228^2,144^2)

 ( 51984,30276 ) = ( 228^2,174^2)

 ( 63504,71289 ) = ( 252^2,267^2)

 ( 71289,63504 ) = ( 267^2,252^2)

 ( 95481,20736 ) = ( 309^2,144^2)

 ( 95481,30276 ) = ( 309^2,174^2)

only caveat being that it is not stated wether or not (P,Q) and (Q,P) are unique so I included them just in case

**oops, messed up on the spelling of Mathematica in the title**

Edited on October 11, 2008, 12:04 pm

Edited on October 11, 2008, 12:05 pm
  Posted by Daniel on 2008-10-11 12:03:01

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