The sum of two–digit number and the number formed by reversing the order of digits is 66. If the two digits differ by 2, find the number. How many such numbers are there?

To find: the number. How many such numbers are there.

Solution:

Let the one’s digit be ‘a’ and ten’s digit be ‘b’.

The 2 digit number is formed as ( 10×number on tens' digit + number on one's digit )

So the number is 10 b + a,

If we reverse the digits one's digit will be 'b' and tens' digit will be 'a'.

So number after reversing will become 10 a + b.

As it is given that sum of two–digit number and the number formed by reversing the order of digits is 66.

⇒ 10 b + a + 10 a + b = 66

⇒ 11 b + 11 a = 66

⇒ a + b = 6 .... (1)

Also, digits differ by 2.

⇒ a – b = 2 ...... (2)

or b – a = 2 ..... (3)

Adding eq 1 and 2 we get

a + b + a – b = 6 + 2

⇒ 2a = 8

⇒ a = 4

Putting value of a in 1 we get,

4 + b = 6

⇒ b = 6 – 4

⇒ b = 2

Adding eq 1 and 3 we get,

a + b + b – a = 6 + 2

⇒ 2b = 8

⇒ b = 4

Putting value of a in 1 we get,

a + 4 = 6

⇒ a = 6 – 4

⇒ a = 2

Thus,

a = 4 and b = 2

or a = 2 and b = 4

Now 2 digit number is 10 b + a

For a = 4 and b = 2

The 2 digit number is 10 (2) + 4 = 20 + 4 = 24

For a = 2 and b = 4

The 2 digit number is 10 (4) + 2 = 40 + 2 = 42

Hence Numbers can be 24 or 42.