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.

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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.

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