A "fast" way for finding sum of the numbers from $1$ to $100$,which are not divisible by $3$ and $5$ 
Find sum of numbers from $1$ to $100$ which are not divisible by $3$
  and $5$?

I can understand that the here we require to sum up these numbers:
$1+2+4+7+8+11+13+14+16+17+19+22+23+26+28+29+31+32+34$ $+37+38+41+43+44+46+47+49+52+53
+56+58+59+61+62+64+67+68+71$ $+73+74+76+77+79+82+83+86+88+89+91+92+94+97+98$
Which gives the sum $= 2632$,but I any didn't relation among the numbers in this series,any ideas? Also,I am supposed to do this under a mint so please hint/answer accordingly.
 A: The sum of the numbers which are divisible by $3$ in this range is
$$
3+6+\cdots+99=3(1+2+\cdots+33)
$$
and likewise the sum of those divisible by $5$ is
$$
5+10+\cdots+100=5(1+2+\cdots+20).
$$
You should be aware of a quick formula to add these two sums. Be careful to add back all the numbers that are divisible by $15$, as they have been subtracted twice. Thanks to Gauss, we know that the sum of the first $100$ integers is $5050$, and after doing the arithmetic you get your answer.
A: Hint: First try finding the sum of numbers which are divisible by $3$ or $5$
A: There's something called inclusion-exclusion you wanna use : sum all integers up to 100, remove those divisible by 3, remove those divisible by 5, then add those divisible by 15. =)
A: This is a combinatorial argument called Inclusion-exclusion principle:
$$X=1+\ldots+100 - 3(1+\ldots+33) - 5(1+\ldots+20) + 15(1+\ldots+6)$$
First we add all the numbers, then we remove all those which are divisible by $3$, and those divisible by $5$. But wait a minute! What about $15$? We removed that number twice! We therefore need to add it once, as well the rest of its multiples below $100$.
Now using the summation formula: $1+\ldots+n = \frac{n(n+1)}{2}$ the answer follows.
