Question on general analysis of Number and Perfect squares
As @lulu pointed it out, the sum of number and its reverse would be of the form $ 11*(a+b) $. For this to be a perfect square, (a+b) must be of the form $ 11*c $, where c is also a perfect square.
On solving it under the given constraints, we find that c must be 1.
So, a+b must be 11, we have
Thus, all the 8 possible two digit numbers are : 29, 38, 47, 56, 65, 74, 83 and 92.
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mnulb 16 days
How many two-digit positive integers N have the property that the sum of N and the number obtained by reversing the order of the digits of N is a perfect square? Answer given is 8.
lulu over 6 yearsHint: if $a,b$ are the digits then your sum is $11(a+b)$.