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.

benjamingu22 Jul 20, 2017

#2**+1 **

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

9,2

8,3

7,4

6,5

And subbing these into 10a + b and 10b + a will generate Alan's two digit numbers

CPhill Jul 20, 2017