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.

(In reply to re: solution TRUE BUT WHY by Ady TZIDON)

As the 4-digit number squared, ABCD², is
10⁶*(A²) + 10⁵*(2AB) + 10⁴*(2AC + B²) + 10³*(2AD + 2BC) + 10²*(2BD + C²) + 10¹*(2CD) + 10⁰*(D²), the last digit is the right-most digit of D².

D D²
0 00
1 01
2 04    <--
3 09
4 16
5 25
6 36
7 49
8 64    <--
9 81

As we exclude numbers with a zero or an odd digit, D can not be 0, 1, 3, 5, 7, or 9. Both the squares of 4 and 6 will have an odd carry to the 10¹-digit. As the 10¹-digit is the right-most digit of 2*C*D plus any carry from the result of D², adding any odd-digit to the even product of 2*C*D [any integer multiplied by 2 is even] will result in an odd number. Thus, D can be neither 4 nor 6, leaving only 2 and 8 as possible values for D. In both cases, 2² (4) and 8² (64), the right-most digit results in a 4.  The carry for either to be added to 2*C*D, to find the 10¹-digit, is, respectively, 0 and 6 -- both even numbers.

 

Edited on October 28, 2008, 1:30 am

Posted by Dej Mar on 2008-10-27 06:33:41