Question

How many three-digit numbers are divisible by 7?

Answer 1

First three-digit number that is divisible by $7 = 105$
Next number $= 105 + 7 = 112$
Therefore, $105, 112, 119, ….$
All are three digit numbers which are divisible by 7 and thus, all these are terms of an A.P. having first term as $105$ and common difference as $7$.
The maximum possible three-digit number is $999$.
When we divide it by 7, the remainder will be $5$.
Clearly, $999 − 5 = 994$ is the maximum possible three-digit number that is divisible by $7$.
The series is as follows.
$105, 112, 119, …, 994$
Let $994$ be the nth term of this A.P.
$a = 105$, $d = 7$ and $a_n = 994$, $n = ?$
$a_n = a + (n − 1) d$
$994 = 105 + (n − 1)7$
$889 = (n − 1)7$
$n − 1 = 127$
$n = 128$

Therefore, $128$ three-digit numbers are divisible by $7$.