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Question

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.


Solution

Let the ten's digit no. be x and one's digit no. be y.

So the no. will be = $$10x +y$$.

Given :  $$x + y = 9 $$-----(I)

             $$9( 10x + y ) = 2( 10y + x )$$ $$\Rightarrow 88x - 11y =0$$ -----(II)

On solving I and II simultaneously you will get $$x = 1$$ and $$y =8$$.
Therefore your desired no. is 18.

Maths

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