Question

The product of the digits of a two digit number is one third that digit. If we as 18 to the required number we get the number consisting of same digits in reverse order. Find the digits

Answer 1

Answer:

The sum should read as "The product of the digits of a two digit number is one third that NUMBER....."

Let digit in 10s place of original number be x

let digit in 1s place of original number be y

The number obtained is 10x + y. We are given that product of digits is 1/3rd the number. Therefore,

xy = 1/3*(10x + y)

3xy = 10x + y ... Eqn 1

We are also given that if 18 is added to the original number, we get a number with digits reversed.

Number with digits reversed will be 10y + x

Hence,

10x + y + 18 = 10y + x

On simplification, we get:

9x - 9y = -18

y - x = 2

y = x + 2 ...Eqn 2

Substituting for y in Eqn 1, we get:

3x(x + 2) = 10x + x + 2

3x^2 + 6x = 11x + 2

3x^2 - 5x -2 = 0

(3x + 1)(x - 2) = 0 ......(on factorization)

Solutions are x = -1/3 or x = 2. As x is a digit, -1/3 is not admissible,

Hence x = 2.

Using Eqn 2, we get y = 2 + 2 = 4

The digits are 2 and 4

Hence, the original number is 24 and the reversed number is 42.

Verify:

Product of digits of original number is 2*4=8 which is 1/3rd of the number. (8 is one third of 24)

If we add 18 to the original number (24), we get 42, which is the original number with digits reversed.

Step-by-step explanation: