The sum of digits of a two number is 15.

Solution:

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

$x+y=15$

 

$x-y=1$

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

Adding the two equations, we have

$(x+y)+(x-y)=15+1$

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

$\Rightarrow 2 x=16$

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

$\Rightarrow x=8$

Substituting the value of x in the first equation, we have

$8+y=15$

$\Rightarrow y=15-8$

 

$\Rightarrow y=7$

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