Determine all possible 7digit decimal (base 10) perfect square(s), each of whose digits is nonzero and even.
Note: While a solution is trivial with the aid of a computer program, show how to derive it without one.
The root of a 7digit perfect square is 4 digits (inclusively between 1000 to 3162). Representing the digits of the root as ABCD, ABCD
^{2} is equal to
10
^{6}*(A
^{2}) +
10
^{5}*(2AB) +
10
^{4}*(2AC + B
^{2}) +
10
^{3}*(2AD + 2BC) +
10
^{2}*(2BD + C
^{2}) +
10
^{1}*(2CD) +
10
^{0}*(D
^{2})
As the final digit must be nonzero and even, D must be 2 or 8; and the 10
^{0}digit digit must be 4. Proceeding to the next, if D was 2 then C must be 2 or 6 with respective values for the 10
^{1}digit as 8 or 4; and if D was 8 then C must be 0, 3, 6 or 7 with respective values for the 10
^{1}digit as 6, 4, 2 or 8. By substituting each of the ten possible digits for B and one of the three (13) for A, eliminating any number where any one of the digits is zero or odd, one can find each digit of the 7digit perfect square and it's root, ABCD. Only three solutions are found:
2862864 (1692
^{2}),
8282884 (2878
^{2}), and
8868484 (2978
^{2}).
Using the above procedure manually may be onerous, but it may be done using simple math and without the aid of a computer program.

Posted by Dej Mar
on 20081027 04:12:05 