Question

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

Answer 1

$\large{\pmb{\sf{\underline{Question}}}}$

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?

$\large{\pmb{\sf{\underline{Required\: solution...}}}}$

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.

Step:01

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)

Step:02

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

Step:03

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)

Step:04

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)

Step:05

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

Step:06

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

Step:07

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

Step:08

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.

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