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January 31st, 2015, 06:48 PM   #1
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Can you do this without a complete bash?

If abcd = (ab+cd)^2 and c is the only digit that can be 0, find the sum of all possible abcd.

Note that abcd is a four digit number, not a*b*c*d (same with ab and cd, but they are 2 digit numbers).

Last edited by skipjack; February 1st, 2015 at 09:10 AM.
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February 1st, 2015, 09:50 AM   #2
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Start by considering possible values of a.

There are very few candidates. As only c can be zero, abcd = 3025 is ruled out.
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February 1st, 2015, 11:05 AM   #3
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but how will you be able to tell that a number abcd works without testing it (i.e. squaring)?

can you eliminate some squares right away (besides the one that have b=0 or d=0)
like 46^2 for example? (without squaring 46 and seeing if ab+cd=46)
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February 1st, 2015, 04:50 PM   #4
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As abcd = (ab + cd)², its digital root is 1 or 9.
As (ab + cd)² ends in d, bd = 81, 44, 49, 84, 86 or 89.
As (ab + cd)² begins with a,
if a < 5, c = 1 or 2,
if a = 5 or 6, c = 1,
if a = 7, c = 0 or 1,
and if a = 8 or 9, c = 0.
(The above deductions about c require knowledge of some squares or square roots.)

The above leaves less than ten possible values for ab + cd, and 46 isn't one of them.
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