Question

The sum of the digits of a $3$ digit number is subtracted from the number. The resulting number is always____.

Answer 1

Explanation.

Let the number consist of $\mathrm{a},\mathrm{b}\mathrm{and}\mathrm{c}$ in its one's tens and hundred places respectively.

So the number is $\mathrm{cba}$ which can be expressed as follow:

$\mathrm{cba}=100\mathrm{c}+10\mathrm{b}+\mathrm{a}$

The Sum of digits can be written as$\mathrm{c}+\mathrm{b}+\mathrm{a}$

When the sum of the digits of a $3$ digit number is subtracted from the number:

$$\begin{gathered} \mathrm{Difference}=\mathrm{cba}-\left(\mathrm{c}+\mathrm{b}+\mathrm{a}\right) \\ =100\mathrm{c}+10\mathrm{b}+\mathrm{a}-\left(\mathrm{c}+\mathrm{b}+\mathrm{a}\right) \\ =100\mathrm{c}-\mathrm{c}+10\mathrm{b}-\mathrm{b} \\ =99\mathrm{c}+9\mathrm{b} \\ =9\left(11\mathrm{c}+\mathrm{b}\right) \\ =9k,\mathrm{where}\mathrm{k}=11\mathrm{c}+\mathrm{b} \end{gathered}$$

Thus the resulting difference is always divisible by $9$.

Hence, the sum of the digits of a $3$ digit number is subtracted from the number. The resulting number is always divisible by $9$.