Question

A two-digit number is such that the product of its digit is 35. If 18 is added to the number, the digits interchange their places. Find the number.

Answer 1

Let the tens and the units digits of the required number be x and y​, respectively.
Then, we have:
xy = 35 ....(i)
Required number = (10x + y)
Number obtained on reversing its digits = (10y + x)
∴ (10x + y) + 18 = 10y + x
⇒ 9x − 9y = −18
⇒ 9(y − x) = 18
⇒ y − x = 2 ....(ii)

We know:
(y + x)² − (y − x)² = 4xy
⇒ $\left(y+x\right)=\pm \sqrt{{\left(y-x\right)}^2+4xy}$
⇒ $\left(y+x\right)=\pm \sqrt{4+4\times 35}=\pm \sqrt{144}=\pm 12$
∴ y + x = 12 .....(iii) (∵ x and y cannot be negative)
On adding (ii) and (iii), we get:
2y = 2 + 12 = 14
⇒ y = 7
On substituting y = 7 in (ii), we get:
7 − x = 2
⇒ x = (7 − 2) = 5
∴ Number = (10x + y) = 10 × 5 + 7 = 50 + 7 = 57
Hence, the required number is 57.