Question

How many three-digit numbers are divisible by 7?

Answer 1

Hint: - Here we go through by first finding the smallest three digit number which is divisible by 7 and the largest three digit number which is divisible by 7 and we can see that these numbers form an AP then find out the total number of terms of AP.

Complete step-by-step answer:
As we know the smallest 3 digit number which is divisible by 7 is 105 i.e.$7 \times 15$.
And the largest 7 digit number can be found by dividing the largest 3 digit number by 7 and subtract the remaining remainder from the largest 3 digit number. i.e. $999 \div 7 = 7 \times 142 + 5$. Here 5 is the remainder so subtract 5 from 999 we get, the largest number which is divisible by 7 is 994.
The first three-digit number which is divisible by 7 is 105 and the last three-digit number which is divisible by 7 is 994.
Now form an AP with first term 105 an with the common difference of 7 because by adding 7 in first the number comes out is also divisible by 7 by continuing the step we form an AP
This is an A.P in which a=105, d=7 and l=994.
Let the number of terms be n then ${T_n} = 994$.
We know that ${n}^{th}$ term of AP is written as ${T_n} = a + (n - 1)d$
By putting the values we get,
$\ \Rightarrow 994 = 105 + (n - 1)7 \\\ \Rightarrow 994 - 105 = (n - 1)7 \\\ \Rightarrow (n - 1) = \dfrac{{889}}{7} = 127 \\\ \Rightarrow n = 127 + 1 = 128 \\\$
$\therefore n = 128$
Therefore there are 128 three-digit numbers which are divisible by 7.

Note: - Whenever we face such a type of question the key concept for solving the question is first find out the smallest and the largest number of a given digit which is divisible by the given number and then find out the total number in between them by the help of making a series of AP. As we can say dividend is the common difference. And by the formula of the ${n^{th}}$ term of AP we will find out the total numbers.