# Find the sum of all such integers less than 100, which leave a remainder 1 when divided by 3 and leave a remainder 2, when divided by 4?

Jan 30, 2017

Sum of all such integers less than $100$ would be $416$.

#### Explanation:

Integers which leave a remainder $1$ when divided by $3$ are

$\left\{4 , 7 , 10 , 13 , 16 , 19 , 22 , \ldots \ldots\right\}$

and of these, those which leave a remainder $2$ when divided by $4$ are

$\left\{10 , 22 , 34. \ldots . .\right\}$

this is an arithmetic sequence, with first term as ${a}_{1} = 10$ and common difference as $d = 12$ (note it is LCM of $3$ and $4$). As $\frac{100 - 10}{12} = 7 + \ldots .$, the last number in the sequence, up to $100$ should be $7 + 1$ or ${8}^{t h}$ term.

and sum of the arithmetic sequence given by S_n=n/2(2a(n-1)d)) would be

$\frac{8}{2} \left(2 \times 10 + \left(8 - 1\right) \times 12\right) = 4 \left(20 + 84\right) = 4 \times 104 = 416$