The digits of a two-digit number differ by 5. If the digits are interchanged and the resulting number is added to the original number then we get 99. Find the original number.

Given :

The digits of a two-digit number differ by 5.

If the digits are interchanged and the resulting number is added to the original number, we get 99.

To do :

We have to find the original number.

Solution :

Let the two-digit number be $10x+y$.

$x-y=5$ or $y-x=5$

If the digits are interchanged and the resulting number is added to the original number, we get 99.

This implies,

$(10x+y)+(10y+x) = 99$

$10x+x+10y+y=99$

$11x+11y=99$

$11(x+y)=99$

$x+y=\frac{99}{11}$

$x+y=9$

If $x-y=5$ and $x+y=9$

Adding both, we get,

$x-y+x+y=5+9$

$2x=14$

$x=7$

$7+y=9$

$y=9-7=2$

$x = 7, y=2$

Then the original number is $10x+y=10(7)+2=72$

If $y-x=5$,

$x+y=9$

$y-x+x+y=5+9$

$2y=14$

$y=\frac{14}{2}$

$y=7$

$x+7=9$

$x=9-7=2$

Then the original number is $10x+y=10(2)+7=27$

Therefore, the original number is 72 or 27. 

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Updated on: 10-Oct-2022

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