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 \(7\).