The sum of digits of a two number is 15.


The sum of digits of a two number is 15. The number obtained by reversing the order of digits of the given number exceeds the given number by 9. Find the given 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 15 . Thus, we have $x+y=15$

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

The number obtained by interchanging the digits is exceeding 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


$\Rightarrow x+y+x-y=16$

$\Rightarrow 2 x=16$

$\Rightarrow x=\frac{16}{2}$

$\Rightarrow x=8$

Substituting the value of in the first equation, we have


$\Rightarrow y=15-8$


$\Rightarrow y=7$

Hence, the number is $10 \times 7+8=78$.

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