**Q) **The sum of a two digit number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2, find the number. How many such numbers are there?

**Ans: **Let’s consider X and Y are the digits of the given number.

Hence the given number is 10 X + Y

By reversing the digits, we will get new number as: 10 Y + X

Given that the sum of new number and original number is 66, therefore:

(10 Y + X) + (10 X + Y) = 66

11 X + 11 Y = 66

X + Y = 6 …………… (i)

Given that the digits differ by 2, hence,

X – Y = 2 ………….. (ii)

and

Y – X = 2 ……….. (iii)

Since, it is not given if 1st digit is larger than or smaller than the 2nd digit; hence, X – Y = 2 and Y – X = 2 both are valid possibilities.

Next, let’s solve the equations (i) & (ii), we get:

X = 4 and Y = 2. Hence the original number is 10 X + Y = 42.

By solving the equations (i) & (iii), we get:

X = 2 and Y = 4. Hence the original number is 10 X + Y = 24.

**Therefore, there are two such possible numbers i.e. 42 and 24.**