Question

A two-digit number is such that the product of its digit is 18. When 63 is subtracted from 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 = 18 ....(i)
Required number = (10x + y)
Number obtained on reversing its digits = (10y + x)
∴ (10x + y) − 63 = 10y + x
⇒ 9x − 9y = 63
⇒ 9(x − y) = 63
⇒ x − y = 7 ....(ii)

We know:
(x + y)² − (x − y)² = 4xy
⇒ $\left(x+y\right)=\pm \sqrt{{\left(x-y\right)}^2+4xy}$
$$\begin{gathered} \Rightarrow \left(x+y\right)=\pm \sqrt{49+4\times 18} \\ =\pm \sqrt{49+72} \\ =\pm \sqrt{121}=\pm 11 \end{gathered}$$
∴ x + y = 11 ....(iii) (∵ x and y cannot be negative)
On adding (ii) and (iii), we get:
2x = 7 + 11 = 18
⇒ x = 9
On substituting x = 9 in (ii), we get:
9 − y = 7
⇒ y = (9 − 7) = 2
∴ Number = (10x + y) = 10 × 9 + 2 = 90 + 2 = 92
Hence, the required number is 92.