A square (Posted on 2004-06-25)

	Submitted by Ady TZIDON
Rating: 3.1429 (7 votes)
Solution:	(Hide)
Answer: 7744=88*88 It is easy as soon as you perceive that A+B=11 Remark: clearly AABB is divisible by 11, hence by 121. so A0B is also divisible by 11. For any number to be divisible by 11 , one evaluates two sums: SO - the sum of all the digits found on the odd-numbered places , and SE -the sum of all the digits found on the even-numbered places . The absolute value of SO-SE has to be divisible by 11. Hence A+B=11. There are only 2 possibilities for a square to be terminated by two identical digits ==> BB IS EITHER 00 or 44. Hence the answer 7744=88^2 ady

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