A number consist of two digits whose sum is five.


A number consist of two digits whose sum is five. When the digits are reversed, the number becomes greater by nine. Find the number.


Let the digits at units and tens place of the given number be $x$ and $y$ respectively. Thus, the number is $10 y+x$.

The sum of the digits of the number is 5 . Thus, we have $x+y=5$

After interchanging the digits, the number becomes $10 x+y$.

The number obtained by interchanging the digits is greater by 9 from the original number. Thus, we have

$10 x+y=10 y+x+9$

$\Rightarrow 10 x+y-10 y-x=9$

$\Rightarrow 9 x-9 y=9$

$\Rightarrow 9(x-y)=9$

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

$\Rightarrow x-y=1$

So, we have two equations



Here x and y are unknowns. We have to solve the above equations for x and y.

Adding the two equations, we have


$\Rightarrow x+y+x-y=6$

$\Rightarrow 2 x=6$

$\Rightarrow x=\frac{6}{2}$

$\Rightarrow x=3$

Substituting the value of in the first equation, we have


$\Rightarrow y=5-3$


$\Rightarrow y=2$

Hence, the number is $10 \times 2+3=23$.

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