A two-digit number is 3 more than 4 times the sum of its digits. If 18 is added to the number, the digits are reversed. Find the number. 

Given :

A two-digit number is 3 more than 4 times the sum of its digits. If 18 is added to 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,

$10x+y=4(x+y)+3$

$10x+y=4x+4y+3$

$10x-4x=4y-y+3$

$6x=3y+3$

$6x=3(y+1)$

$2x=y+1$

$y=2x-1$....(i)

It is also given that if 18 is added to the number the digits gets reversed.

Therefore,

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

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

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

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

$9y-9x=18$

$9(y-x)=18$

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

$y-x=2$

$2x-1-x=2$   (From (i))

$x=2+1$

$x=3$

This implies,

$y=2x-1=2(3)-1=6-1=5$

The original number is $10(3)+5=30+5=35$.

Therefore, the number is 35.  

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Updated on: 10-Oct-2022

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