The sum of all two digit numbers which when divided by 4, yield unity as remainder is,
A. 1012
B. 1201
C. 1212
D. 1210

Hint: Sometimes language of the question can be very confusing like our problem. Here in the question I want to say is find the sum of those digit numbers which have 1 as a remainder when that number is divided by 4.

Complete step by step answer:
Suppose N is a number which has 1 as a remainder when divided by 4.
Then, we can write the number N = 4M +1 form.
Using some basic knowledge we can say 13 is first ${{N}_{1}}$ where M will be 3.
So, ${{N}_{1}}=13$ when M = 3
${{N}_{2}}=17\ when\ M=4........{{N}_{22}}=97\ when\ M=24$
Take the last double digit number 99 and divide by 4 to get the last digit of our N series, when we do this we get 3 as a remainder but we know 1 as a remainder. So, we subtract (3 – 1) from 99 to get the last two digit number of our N number series.
So, 99 – (3 – 1) = 97 will be the last number of our N number series.
Put N = 97 in N = 4M + 1 to get M,
$\Rightarrow M=\dfrac{97-1}{4}=24$
Now, N: 13, 17, 21, ....97
When we observe this series, we can see the series is an arithmetic progression with the first term as 13 and last term as 97 and the number of terms are 22 with a common difference between terms is 4.
Now, we know the sum of A.P.
So,
$sum\ of\ n\ terms\ of\ A.P.\ {{S}_{n}}=\dfrac{n}{2}\left( {{a}_{1}}+{{a}_{n}} \right);\ where\ {{a}_{1}}=first\ term\ and\ {{a}_{n}}=\ last\ term$
$\begin{align}
  & {{S}_{n}}=\dfrac{22}{7}\left( 13+97 \right) \\
 & {{S}_{n}}=11\times 110=1210 \\
\end{align}$
Hence, option (D) is the correct answer.

Note: Catch the pattern of any series is very helpful for using direct formulas of sum or find the terms, don’t go directly to sum the whole terms of series one by one which takes too much time and hard work.