I agree with aryabatta.
Since it asks that how many integrers are divisible by 3 between 100 and 200.the first step is to find out the first number and the last number divisible by 3 between 100 and 200.
These are 102 and 198 respectively.
Now since these numbers are in an arithmetic progression with the first term 102 ,last term 198 and difference 3 we can find the number of terms in this AP by(using AP formula T(n)=T(1)+(n-1)d )
198 =102+(n-1)3
solving this
n=33