Question

how many two digit positive integers N have the property that the sum of N and the number obtained reversing the order of the digits of N is a perfect square ?

Answer 1

There are 8 integers who have the property.

Let the two digit positive integer be 10x + y.

N = 10x + y

Then the number obtained by reversing the order of digits will be 10y + x.

Take the sum of both.

10x + y + 10y + x = 11x + 11y = 11(x + y)

If 11(x + y) is a perfect square, x + y should be equal to 11.

If x + y = 11, 11(x + y) = 11 × 11 = 121 (Perfect square)

Let's find how many types 11 can be written as sum of 2 one-digit positive integers.

11 = 5 + 6 → Type 1

11 = 4 + 7 → Type 2

11 = 3 + 8 → Type 3

11 = 2 + 9 → Type 4

There are 4 types. From these types, N = {56, 47, 38, 29}.

These 4 numbers can obtain new 4 numbers by reversing order of digits.

So there are a total of 8 two-digit positive integers.

And that integers are 56, 65, 47, 74, 38, 83, 29 and 92.

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