How many three digit numbers are divisible by 7?

In this problem, we need to find out how many numbers of three digits are divisible by 7.

So, we know that the first three digit number that is divisible by 7 is 105 and the last three digit number divisible by 7 is 994. Also, all the terms which are divisible by 7 will form an A.P. with the common difference of 7.

So here,

First term (a) = 105

Last term (an) = 994

Common difference (d) = 7

So, let us take the number of terms as n

Now, as we know,

$a_{n}=a+(n-1) d$

So, for the last term,

$994=105+(n-1) 7$

$994=105+7 n-7$

$994=98+7 n$

$994-98=7 n$

Further simplifying,

$896=7 n$

$n=\frac{896}{7}$

$n=128$

Therefore, the number of three digit terms divisible by 7 is 128 .