There is at least one two-digit number such that when it is added to the two-digit number having the same digits in reverse order, the sum is a perfect square. Find the sum of all such two-digit numbers.
Here's how Alan may have found his answers....
[And then again, maybe not !!! ]
Let the first number be 10 a + b
And let the other number be 10 b + a
And we know that
[10 a + b] + [10 b + a] = N^2
10a + a + 10b + b = N^2
11a + 11b = N^2
11 ( a + b) = N^2
But..... if the right side is a perfect square then ( a + b) must equal 11
So.....the possibilities are
And subbing these into 10a + b and 10b + a will generate Alan's two digit numbers