A number AABB is the square of an integer. Find this integer, aided by pen and paper. No other calculating aids allowed.
(In reply to
Puzzle Solution by K Sengupta)
Let AABB = x^2 (say)
Then, 1< = x^2 < = 9999, so that:
1<= x < 100, so that the number of digits of x is 2.
Now, we know that if M = 50*P +/- N, then:
M^2 (Mod 100) = N^2
Hence, for the minimum value of N, having the above property, we merely need to check for N = 1 to 25
Thus, by inspecting the minimum values of N = 1 to 25, for which the last two digits of N^2 are alike occur at N = 00, 12. But, if B=0, then AABB = 1100*A = 100*11*A, and so the minimum value of A is 11, and at A =11, we have: AABB = 12100, which is a contradiction. Hence N = 12, and accordingly:
x = 50* P +/- 12. Since the number of digits in x is 2, we have:
x = 12, 38, 62, 88 as the only possibilities.....(i)
But, we note that AABB is divisible by 11, so that x must be divisible by 11.
The only value in (i) divisible by 11 occurs at x = 88, giving:
x^2 = (100 - 12)^2
= 100^2 - 24*100 + 144
= 100(100-24) + 144
= 7600 + 144
= 7744
Consequently, the required integer is 7744.
Edited on September 4, 2008, 12:14 pm
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