Find the sum of the multiples of $3$ and $5$ below $709$? I just cant figure this question out:
Find the sum of the multiples of $3$ or $5$ under $709$
For example, if we list all the natural numbers below $10$ that are multiples of $3$ or $5$, we get $3$, $5$, $6$ and $9$. The sum of these multiples is $23$.

There are $\displaystyle\left\lfloor\frac{709}{3}\right\rfloor = 236$ multiples of $3$ below $709$, $\displaystyle\left\lfloor\frac{709}{5}\right\rfloor = 141$ multiples of $5$ below $709$ and $\displaystyle\left\lfloor\frac{709}{15}\right\rfloor = 47$ multiples of $15$ below $709$.
By the inclusion/exclusion principle:
$$\mathop{\sum_{(n|3\ \vee\ n|5)} n}_{ n \leq 709} = \mathop{\sum_{n|3} n}_{ n \leq 709}  + \mathop{\sum_{n|5} n}_{ n \leq 709}  - \mathop{\sum_{n|15} n}_{ n \leq 709} $$
Note that:
$$\mathop{\sum_{n|3} n}_{ n \leq 709} = 3 + 6 + \cdots + 3·236 = 3(1+2+\cdots + 236) = 3\frac{236·237}{2} = 83898$$
Similarly, 
$$\mathop{\sum_{n|5} n}_{ n \leq 709} = 5 + 10 + \cdots + 5·141 = 5(1+2+\cdots + 141) = 5\frac{141·142}{2} = 50055$$
$$\mathop{\sum_{n|15} n}_{ n \leq 709} = 15 + 30 + \cdots + 15·47 = 15(1+2+\cdots + 47) = 15\frac{47·48}{2} = 16920$$
So the total sum is: $83898 + 50055 - 16920 = \boxed{117033}$
