Question

The sum of all two digit numbers which when divided by $4$, yield unity as remainder, is

• A. 1100
• B. 1200
• C. 1210
• D. None of these

Answer 1

Answer: (C)

The first two digit number which when divided by $4$ leaves remainder $1$ is $4\cdot 3+1=13$ and last is $4\cdot 24+1=97.$
Thus, we have to find the sum
$13+17+21+\dots +97$
which is an A.P.
$\therefore 97=13+(n-1)\cdot 4$
$\Rightarrow n=22$
and $S_n=[a+l]=11\times [13+97]$
$=11\times 110=1210$