divisibility by 21 It's a simple problem but I am stuck.
The multiple of 21 between 700 and 950 are 714, 735, 756, 777, 798, 819, 840,861,882,903,924,945.
So, there are 12 multiples of 21 between 700 and 950 but 21 *12=252.
So 12 multiples of 21 should take an interval of 252 , and there are only  250 numbers between 700 and 950, So there should be only 11 multiples .
Where am i making a wrong argument ?
Please help.
Thanks. 
 A: Let's look at multiples of $3$ between $2$ and $13$. There are $3, 6, 9, 12$, which is four of them. But $4 \times 3 = 12$, and between $2$ and $13$ is only $11$ numbers. 
The problem is that the number $4 \times 3$ doesn't represent the length of the interval containing the four multiples. There are only three gaps of length $3$, plus one more for the last number. That's a length of ten, which fits easily inside the $11$ spaces you've got. 
A: You are wrong when you claim that “$21\times12=252$, so $12$ multiples of $21$ should take an interval of $252$”. The numbers $12$ and and $24$ are two consecutive multiples of $12$, but they take an interval of $13$ numbers only. 
A: If you consider, for example, the setof multiples of $15$ between $14$ and $94$ you obtain that there are only $6$ numbers but if you compute $6\cdot15=90$ that is different from $94-14$. You don't have to consider the first of this multiples. In this case, in fact, your method works: $94-14\geq(6-1)\cdot15.$
A: $21\times12$=252 , so it seems that we need 2 more numbers but when we observe that the the first multiple of 21 in this interval is 714 , we have a surplus of 7 numbers , leaving a net of 7-2=5 numbers extra(from 945 to 950) after 12 multiples. 
Thanks to everyone for helping me out.
