Question

Digits of a two digit number are interchanged. Sum of new number obtained and originalnumber is a perfect square. If tens place digit is larger than units place digit, how manysuch two digit numbers exist?​

Answer 1

Answer: No Number

Step-by-step explanation:

Let the digit be 10x+y

Then the interchanged digit will be 10y+x

So their sum, is 11x+11y which is a perfect square.

Tens place digit i.e x>y

x and y are single digit whole numbers as the form a whole number.

Now , x>y

Hence , 11x>11y

Let 11x+11y = 11(x+y) = 121

Here x = y = 11 but x>y

We want a value of x and y unequal and adding up to a perfect square.

The numbers divisible by 11 which is perfect square are: 121, (121)^2, (121)^3...

As x and y are single digit no. x=y=11

And by the question, x>y plus x<=9 This is contradictory.

Hence such number doesn't exist.