The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

Given :

The sum of the digits of a two-digit number is 9.

Nine times this number is twice the number obtained by reversing the order of the digits.

To do :

We have to find the given number.

Solution :

 Let the two-digit number be $10x+y$.

$x + y = 9$

$x=9-y$.....(i)

The number formed on reversing the digits is $10y+x$.

Therefore,

$9(10x+y) = 2(10y+x)$

$90x+9y=20y+2x$

$90x-2x+9y-20y=0$

$88x-11y=0$

$11(8x-y) = 0$

$8x-y = 0$

$y=8x$

Substituting $y = 8x$ in equation (i), we get,

$x =9-8x $

$x+8x = 9$

$9x = 9$

$x=1$

This implies,

$y = 8x = 8(1)=8$

The original number is $10(1)+8 = 10+8 = 18$.

The original number is 18. 

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Updated on: 10-Oct-2022

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