Saurav's Square #2
Number Theory
Level
3
Let ABCD be a 4 digit number such that (ABCD)^2 = WXYZABCD. Find the digit sum of ABCD.
Details and Assumptions : WXYZABCD is a 8 digit number.
The answer is 25.
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1 solution
If the digits must all be different, then: A>=3 (as first digit W of square will form from A^2 + (some carry over from right hand side)
D = 0,1,5,or 6 (no other digit yields the same last digit as itself in its own square)
CD = 00,01,25,51, or 76 --> D can't be 0 (00 not allowed as final number has last 4 digits as ABCD and not 0000)
BCD -> can't end in 0 or 1; 625 or 376
ABCD = 0625 or 9376 (0625 will not give a full 8 digit square)
So, 9376^2 = 87909376