Question

How many three digit numbers are divisible by 6?
(a) 102
(b) 150
(c) 151
(d) 966

Answer 1

Hint: Here, we will first try to form an arithmetic progression of the three digit numbers which are divisible by 6. Then, we will use the formula ${{a}_{n}}=a+\left( n-1 \right)d$ to calculate the value of n which is the total number of terms of the AP.

Complete step-by-step answer:
We know that the first three digit number which is divisible by 6 is 102.
In order, to get the numbers divisible by 6 after 102, we can keep adding 6 again and again.
The number which is divisible by 6 and nearest to 102 is 108. Similarly the number divisible by 6 and nearest to 108 is 114.
So, the sequence of three digits numbers divisible by 6 is given as:
102, 108, 114, ……….
Now, we will try to find the last term of this sequence.
We know that the greatest 3 digit number is 999. On dividing 999 by 6 we get remainder equal to 3.
Therefore, the greatest three digit number which is divisible by 6 is 996.
Also, the sequence obtained here is an arithmetic progression, because the difference between the two consecutive terms is always 6 which is a constant. This constant is the common difference of this AP.
Now, we know that in an AP any general term is given as:
${{a}_{n}}=a+\left( n-1 \right)d........\left( 1 \right)$
Here, ${{a}_{n}}$ is any general term of an AP, a is the first term and d is the common difference of the AP.
We have, a = 102, d = 6 and ${{a}_{n}}$ = 996.
On putting these values in equation (1), we get:
$\begin{align}
  & 996=102+\left( n-1 \right)6 \\
 & \Rightarrow 996-102=6n-6 \\
 & \Rightarrow 894+6=6n \\
 & \Rightarrow n=\dfrac{900}{6} \\
 & \Rightarrow n=150 \\
\end{align}$
So, the value of n comes out to be 150.
It means that 996 is the 150th term of this AP. Therefore, there are 150 three digit numbers divisible by 6.
Hence, option (b) is the correct answer.

Note: Students should note here that the sequence obtained here is an AP, because the difference between any two consecutive terms of this sequence is a constant. Students should remember the formula for the nth term of an AP and do the calculations properly to avoid unnecessary mistakes.