Determine all possible 7-digit 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 7-digit perfect square is 4 digits (inclusively between 1000 to 3162). Representing the digits of the root as ABCD, ABCD2
is equal to
*(2AC + B2
*(2AD + 2BC) +
*(2BD + C2
As the final digit must be non-zero and even, D must be 2 or 8; and the 100
-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 101
-digit as 8 or 4; and if D was 8 then C must be 0, 3, 6 or 7 with respective values for the 101
-digit as 6, 4, 2 or 8. By substituting each of the ten possible digits for B and one of the three (1-3) for A, eliminating any number where any one of the digits is zero or odd, one can find each digit of the 7-digit perfect square and it's root, ABCD. Only three solutions are found:2862864
), and 8868484
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 2008-10-27 04:12:05