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Even Digit Perfect Square(s) (Posted on 2008-10-25) Difficulty: 3 of 5
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.

See The Solution Submitted by K Sengupta    
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re(2): solution TRUE BUT WHY Comment 10 of 10 |
(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

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