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

A two-digit number has 2 digits, one number at the units place and one number at the tens place.

## Answer: There are 2 such numbers, that are, 42 and 24.

The general form of a two-digit number with 'a' as the digit at the units place and 'b' as the digit at the tens place is:

10 × b + a = 10b + a

**Explanation:**

Let the digit at the unit's place be 'a' and the digit at the ten's place be 'b',

So, the original number = 10b + a.

Therefore, when the number obtained when the order of the digits are reversed = 10a + b.

Since, the sum of a two-digit number and the number formed by reversing the order of its digits is 66:

(10b + a) + (10a + b) = 66

or, 10b + a + 10a + b = 66

or, 11a + 11b = 66

or, a + b = 6

Also, if the two digits differ by 2, then we get two conditions:

Either (a - b) = 2 or (b - a) = 2

Now let's solve these linear equations to get the values of a and b by substitution method:

Since, a + b = 6

or, a = 6 - b

Taking (a - b) = 2, we get:

a = 2 + b

or, 6 - b = 2 + b

or, 2b = 6 - 2

or, 2b = 4

or, b = 2

Now taking (b - a) = 2, we get:

b = 2 + a

or, b = 2 + 6 - b

or, 2b = 8

or, b = 4

Now if b = 2, then a = 6 - b = 6 - 2 = 4, and the original number = 10a + b = 10 × 4 + 2 = 40 + 2 = 42

Also, if b = 4, then a = 6 - b = 6 - 4 = 2, and the original number = 10a + b = 10 × 2 + 4 = 20 + 4 = 24