Question

When a certain 2 digit number is added to another two digit number having same digits in reverse order, the sum is perfect square. how many such two digit numbers are there?

Answer 1

let no 10x + y
a/q 10x + y + 10y + X = 11x + 11y
= 11(x + y) is perfect square
there for X + y = 11
then X = 2. y = 9
X = 3, y = 8
X = 4, y = 7
X = 5, y = 6
no. are 29, 92
38, 83
47, 74
56, 65

Answer 2

There are 4 pairs of such digits.

GIVEN: 2-digit number is added to another two-digit number having the same digits in reverse order, the sum is a perfect square.
TO FIND: how many such two-digit numbers are there
SOLUTION:

As we are given in the question,

We need to find the two-digit number that is having same digits in reverse order and the sum is a perfect square.

Now, let us assume the number to be  10x + y

Now,

According to the question,

10x + y + 10y + X

= 11x + 11y

= 11(x + y) is perfect square

Using the trial and error method we get the following answers.

Implying that,

X = 2. y = 9

X = 3, y = 8

X = 4, y = 7

X = 5, y = 6

Therefore,

The numbers are 29, 92,38, 83,47, 74,56, 65

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