Question

A two digit number is such that the product of the digits is 8. When 18 is added to the number they interchange their places. Determine the number.

Answer 1

Let the two-digit number = “ xy ”

Where x takes the “tens” place and y, the unit. Hence you can write the two-digit no. “ xy” as “(10x + y)”.

Product of digits equals 8; x(y) = 8 …….(1)

Adding 18 to the number reverses the digits;

xy + 18 = yx Or (10x + y) + 18 = (10y + x)

This simplifies to; 9x + 18 = 9y …….(2)

Multiplying eq/n (2) through by “y” gives; 9x(y) + 18(y)= 9y(y)……(3).

Substituting in x(y) = 8 from eq/n (1) into (3) gives; 9(8) + 18y = 9y^2, which reduces to 8 + 2y = y^2. This is a quadratic eq/n whose solution for y equals 4 & -2. For y = 4, x = 2 by substitution into eq/n (1).

Hence the two-digit no. xy = 24

Answer 2

Answer:

Required number = 24

Step-by-step explanation:

Let ten's place digit = x ,

Unit place digit = y

The number = 10x + y ----( 1 )

If interchange the places then

the new number so formed

= 10y + x ----( 2 )

Given ,

xy = 8 ---( 3 )

And

10x + y + 18 = 10y + x

=> 10x - x -10y + y + 18 = 0

=> 9x - 9y + 18 = 0

=> x - y + 2 = 0

=> x - y = -2 ---( 4 )

( x + y )² = ( x - y )² + 4xy

=> ( x + y )² = ( - 2 )² + 4 × 8

= 4 + 32

= 36

=> ( x + y ) = √36

=> x + y = 6 ---( 5 )

Add ( 4 ) and ( 5 ), we get

2x = 4

=> x = 4/2 = 2

Put x = 2 in equation ( 5 ), we get

y = 6 - 2 = 4

Therefore ,

Required number = 10x + y

= 10 × 2 + 4

= 20 + 4

= 24

••••