Question

# How many three-digit numbers are divisible by 7?

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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 make this term as AP then find out the total number of terms by the rule of AP.

As we know the smallest 3 digit number which is divisible by 7 is 105 i.e.$7 \times 15$.
And the largest 3 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.
Let the number of terms be n then ${T_n} = 994$.
We know that nth term of AP is written as ${T_n} = a + (n - 1)d$
$\ \Rightarrow 994 = 105 + (n - 1)7 \\ \Rightarrow 994 - 105 = (n - 1)7 \\ \Rightarrow (n - 1) = \dfrac{{889}}{7} = 127 \\ \Rightarrow n = 127 + 1 = 128 \\$
$\therefore n = 128$