Question

A two-digit number is such that the product of its digits 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`
`⇒ (y + x) = ± sqrt((y−x)2+4xy)`
`⇒ (y + x) = ± sqrt(4+4 ×35 )= ± sqrt(144 )= ±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 = 7in (ii) we get
7 – x = 2
⇒ x = (7 – 2) = 5
∴ The number = (10x + y) = 10 × 5 + 7 = 50 + 7 = 57
Hence, the required number is 57.