A two-digit number is 4 more than 6 times the sum of its digits. If 18 is subtracted from the number, the digits are reversed. Find the number.

Given :

A two-digit number is 4 more than 4 times the sum of its digits. If 18 is subtracted from the number, the digits are reversed.

To find :

We have to find the number.

Solution :

Let the number be $10x+y$.

According to the question,

$10x+y=6(x+y)+4$

$10x+y=6x+6y+4$

$10x-6x=6y-y+4$

$4x=5y+4$

$5y=4x-4$

$5y=4(x-1)$....(i)

It is also given that if 18 is subtracted from the number the digits gets reversed.

Therefore,

$10x+y-18 = 10y+x$

$10x+y-10y-x = 18$

$10x-x+y-10y= 18$

$9x-y(10-1)=18$

$9x-9y=18$

$9(x-y)=18$

$x-y=\frac{18}{9}$

$x-y=2$

$y=x-2$....(ii)

Substituting equation (ii) in equation (i), we get,

$5(x-2)=4(x-1)$

$5x-10=4x-4$

$5x-4x=10-4$

$x=6$

This implies,

$y=x-2=6-2=4$

The original number is $10(6)+4=60+4=64$.

Therefore, the number is 64. 

Tutorialspoint

Simply Easy Learning

Updated on: 10-Oct-2022

2K+ Views

Kickstart Your Career

Get certified by completing the course

Get Started