Question

Find the sum of all 3-digit numbers which leave the remainder 2, when divided by 3.

Answer 1

Step1: Obtaining the common difference, first term, and last term.

The set of 3-digit numbers which leave the remainder 2, when divided by 3 forms an A.P. series having common difference $d=3$.

The smallest such number is $101$. Thus, the first term of the A.P. is $a=101$

The largest such number is $998$. Thus, the last term of the A.P. is$l=998$.

Let there be $n$ such numbers.

Step2: Calculation of the numbers $n$.

The last term of the series A.P. is given by $l=a+\left(n-1\right)d$, where $a$ is the first term and $d$ is the common difference.

Substitute 998 for $l$, 101 for $a$ and 3 for $d$ in the formula $l=a+\left(n-1\right)d$.

$$\begin{gathered} \Rightarrow 998=101+(n-1)3 \\ \Rightarrow 998-101=(n-1)3(subtracting101frombothsides) \\ \Rightarrow 897=(n-1)3 \\ \Rightarrow \frac{897}{3}=n-1(dividingbothsidesby3) \\ \Rightarrow 299=n-1 \\ \Rightarrow 299+1=n(adding1tobothsides) \\ \therefore 300=n \end{gathered}$$

Thus, there are 300, 3-digit numbers that leave the remainder 2, when divided by 3.

Step3: Obtaining the sum of 3-digit numbers which leave the remainder 2, when divided by 3.

The sum of first $n$ terms of an A.P. is given by the formula $S_n=\frac{n}{2}\left[a+l\right]$, where $a$ is the first term and $l$ is the last term.

Thus, the sum all 300 terms of the A.P. is given by:

$$\begin{gathered} \Rightarrow S_{300}=\frac{300}{2}\left[101+998\right] \\ \Rightarrow S_{300}=150\left[1099\right] \\ \therefore S_{300}=164850 \end{gathered}$$

Final Answer: The sum of 3-digit numbers which leave the remainder 2, when divided by 3 is $164850$.