Question

A two-digit number is such that the product of its digits 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`
`⇒ (x + y) = ± sqrt((x−y)2+4xy)`
`⇒ (x + y) = ± sqrt(49+4 ×18)`
                `= ± sqrt(49+72)`
                 `= ± 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 = 9in (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.