Question

How many three-digit numbers are divisible by 3?

• A. 300
• B. 150
• C. 200
• D. 250

Answer 1

The correct option is A

300

We have an AP starting at 102 because it is the first three-digit number divisible by 3.

AP will end at 999 because it is the last three digit number divisible by 3.

Therefore, we have an AP 102,105,108, ..., 999

First term, $a_0$ = 102

Common difference, $d$ = 105 - 102 = 3

Using formula $a_n=a_0+(n-1)d$, to find $n^{th}$ term of arithmetic progression, we can say that

$999=102+(n-1)\left(3\right)$

$\Rightarrow 999=102+(n-1)\left(3\right)$

$\Rightarrow 897=3(n-1)$

$\Rightarrow n-1=\frac{897}{3}\Rightarrow n=299+1=300$

It means 999 is the ${300}^{th}$ term of AP. Therefore, there are 300 terms in AP. In other words, we can also say that there are 300 three digit numbers divisible by 3.