Question

The sum of all two digit numbers which leave remainder $1$ when divided by $3$ is

• A. $1616$
• B. $1602$
• C. $1605$
• D. None of these

Answer 1

The correct option is C $1605$

Clearly, the two digits numbers which leave remainder $1$ when divided by $3$ are $10,13,16,...,97$.

This is an AP with first term $a=10$, common difference $d=3$ and last term $l=97$.
Let there be $n$ terms in this AP, then
$a_n=97=a+(n-1)d$
$\therefore 10+(n-1)\times 3=97$

$10+3n-3=97$

$3n=97+3-10$

$3n=90$

$\therefore n=30$
$\therefore$ Required sum $=\frac{n}{2}[a+l]=\frac{30}{2}[10+97]=15\times 107=1605$
Option C is correct.