Question

What is the sum of all $2$-digit numbers.which when divided by $3$ leave $2$ as remainder ?

• A. $1565$
• B. $1585$
• C. $1635$
• D. $1655$

Answer 1

Answer: (C)

The first and the last $2$-digit number which leaves $2$ as remainder when divided by $3$ are $11$ and $98$.

Let there are $n$ number between $11$ and $98$ which leaves $2$ as remainder when divided by $3$.

Then $11+(n-1)3=98$

or, $(n-1)3=87$

or, $n=30$.

$\therefore$ the sum of all these numbers $=\frac{30}{2}(11+98)=15\times 109=1635$.