Question

If sum of the digits of any integer lying in between $100$ and $1000$ is subtracted from that integer, then the result is always divisible by which number?

• A. $7$
• B. $6$
• C. $5$
• D. $9$

Answer 1

The correct option is D $9$

let the three digit number be $abc$

where $abc=100a+10b+c$

Therefore, $(100a+10b+c)-(a+b+c)=99a+9b$

$=9(11a+b)$ which is divisible by $9$

Therefore, when $a+b+c$ is subtracted from the integer, the result is always divisible by $9$