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.
(In reply to
re: solution TRUE BUT WHY by Ady TZIDON)
As the 4digit number squared, ABCD^{2}, is
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}), the last digit is the rightmost digit of D^{2}.
D D^{2}
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 10^{1}digit. As the 10^{1}digit is the rightmost digit of 2*C*D plus any carry from the result of D^{2}, adding any odddigit 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, 2^{2} (4) and 8^{2} (64), the rightmost digit results in a 4. The carry for either to be added to 2*C*D, to find the 10^{1}digit, is, respectively, 0 and 6  both even numbers.
Edited on October 28, 2008, 1:30 am

Posted by Dej Mar
on 20081027 06:33:41 