# The sum of a 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?

Let’s assume the digit at unit’s place as x and ten’s place as y. Thus from the question, the number needed to be found is 10y + x.

From the question it’s told as, the two digits of the number are differing by 2. Thus, we can write

x – y = ±2………….. (i)

Now after reversing the order of the digits, the number becomes 10x + y.

Again from the question it’s given that, the sum of the numbers obtained by reversing the digits and the original number is 66. Thus, this can be written as;

(10x+ y) + (10y+x) = 66

⇒ 10x + y + 10y + x = 66

⇒ 11x +11y = 66

⇒ 11(x + y) = 66

⇒ x + y = 66/11

⇒ x + y = 6………….. (ii)

Now, we have two sets of systems of simultaneous equations

x – y = 2 and x + y = 6

x – y = -2 and x + y = 6

Let’s first solve the first set of system of equations;

x – y = 2 …………. (iii)

x + y = 6 ………….. (iv)

On adding the equations (iii) and (iv), we get;

(x – y) + (x + y) = 2+6

⇒ x – y + x + y = 8

⇒ 2x =8

⇒ x = 8/2

⇒ x = 4

Putting the value of x in equation (iii), we get

4 – y = 2

⇒ y = 4 – 2

⇒ y = 2

Hence, the required number is 10 × 2 +4 = 24

Now, let’s solve the second set of system of equations,

x – y = -2 …………. (v)

x + y = 6 ………….. (vi)

On adding the equations (v) and (vi), we get

(x – y)+(x + y )= -2 + 6

⇒ x – y + x + y = 4

⇒ 2x = 4

⇒ x = 4/2

⇒ x = 2

Putting the value of x in equation 5, we get;

2 – y = -2

⇒ y = 2+2

⇒ y = 4

Hence, the required number is 10×4+ 2 = 42

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

#### 1 Comment

1. kumbuka paterne

good