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January 31st, 2015, 06:48 PM  #1 
Member Joined: Jan 2015 From: Orlando, Florida Posts: 92 Thanks: 10  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. 
February 1st, 2015, 09:50 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 19,190 Thanks: 1649 
Start by considering possible values of a. There are very few candidates. As only c can be zero, abcd = 3025 is ruled out. 
February 1st, 2015, 11:05 AM  #3 
Member Joined: Jan 2015 From: Orlando, Florida Posts: 92 Thanks: 10 
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) 
February 1st, 2015, 04:50 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 19,190 Thanks: 1649 
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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