Question

How many $3$-digit natural numbers are completely divisible by $6$?

Answer 1

Step 1: Condition for divisibility by $6$

A number is said to be divisible by $6$ when the number is divisible by both $2$ and $3$.

For example, let us consider a random number $468$.

The last digits of this numbers is even, hence it is divisible by $2$.

The sum of the digits $=4+6+8=18$.

We know that $6\times 3=18$, hence $18$ is divisible by $3$.

Therefore $468$ is divisible by $6$.

Step 2: Estimate the count of natural numbers

Let us enlist such $3$-digit natural:

$102,108,114,120,126......972,978,984,990,996$

This is an Arithmetic Progression where

$a=102$

$d=6$

$t=996$.
where $a=$ First Number, $t=$Last Number, and $d=$ difference of two consecutive numbers.
Let the number of terms be $n$.

Use the formula for $n$ terms of arithmetic progression.
$\therefore a+(n–1)xd=996$
$\Rightarrow$.$102+(n–1)x6=996$
$\Rightarrow$ $6(n–1)=894$
$\Rightarrow$ $(n–1)=149$
$\Rightarrow$ $n=150$

There are a total of $150$ digits in this range.

Hence, the number of $3$-digit natural numbers that are completely divisible by $6$ is $150$.