Question

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

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

Answer 1

The correct option is C $1635$

All the numbers which when divided by ${}^{\prime }{3}^{\prime }$ leaves remainder ${}^{\prime }{2}^{\prime }$ will be of form $=3k+2,k=[3,32]\to$ for all two digit numbers

All two digit nos/ $=11,14,.....,98$

This is a A.P. with $a=11d=3$

$\therefore Tn=a+(n-1)d$

$\to 98=11+(n-1)\left(3\right)\to 87=(n-1)\left(3\right)$

$\to 29=n-1\to n=30$

$Sn=$ Sum of $AP=\left(\frac{n}{2}\right)(2a+(n-1\left)d\right)$

$Sn=\left(\frac{30}{2}\right)[22+(29\left)\right(3\left)\right]$

$=\left(15\right)(22+87)=1635$.