    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.

Open in App
Solution

## 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 ⇒ $\left(x+y\right)=±\sqrt{{\left(x-y\right)}^{2}+4xy}$ $⇒\left(x+y\right)=±\sqrt{49+4×18}\phantom{\rule{0ex}{0ex}}=±\sqrt{49+72}\phantom{\rule{0ex}{0ex}}=±\sqrt{121}=±11$ ∴ 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.  Suggest Corrections  1      Similar questions  Related Videos   Basics Revisted
MATHEMATICS
Watch in App  Explore more