Question

What is the sum of all 3 digit numbers that leave a remainder of '2' when divided by 3?

• A. 897
• B. 164850
• C. 164749
• D. 149700

Answer 1

The correct option is B

164850

The smallest 3 digit number that will leave a remainder of 2 when divided by 3 is 101.
The next number that will leave a remainder of 2 when divided by 3 is 104, 107, ....
The largest 3 digit number that will leave a remainder of 2 when divided by 3 is 998.

So, it is an AP with the first term being 101 and the last term being 998 and common difference being 3.

Sum of an AP = $\frac{n}{2}(a+l)$

we know that in an A.P., the nth term $a_n=a_1+(n-1)\times d$

In this case, therefore,

998 = 101 + (n - 1) $\times$ 3
$\Rightarrow 897=(n-1)\times 3$
$\Rightarrow 299=(n-1)$
$\Rightarrow n=300$

Sum of the AP will therefore, be $\frac{(101+998)}{2}$$\times 300=164,850$