Find the sum of all two-digit numbers that give a remainder of 3 when they are divided by 7.

Correct option is C

Given: 
Divisor = 7
Remainder = 3
Concept used:
A, (A + D), (A + 2D), ...., Nth term The series mentioned above is in Arithmetic Progression.
Number of terms in the series (N) =

Sum of all N terms =

Solution:
Numbers from 1 to 100 that are when divided by 7 leave 0 as a remainder are as follows:
7, 14, 21, ...., 84, 91, 98
Now, two-digit numbers that are when divided by 7 leave 3 as a remainder are as follows:
(7 + 3), (14 + 3), (21 + 3), ...., (84 + 3), (91 + 3)
=> 10, 17, 24, ...., 87, 94
As we can see, the series mentioned above is in Arithmetic Progression, 
where 10 is the first term and (17 - 10), i.e., 7 is the common difference.
Number of terms in the series = (94−10)/7+1 = 13
Now, the sum of the terms of the series
= (94+10)/2×13
= 52 × 13
= 676
The sum of all two-digit numbers that give a remainder of 3 when they are divided by 7 is 676.