The multiples of 7 are added one after another: first 7, then 7 + 14, then 7 + 14 + 21, and so on.
How many multiples must be added before the running total first passes 500?

The total of the first $n$ multiples of 7 is 7(1 + 2 + ... + n) = $\frac{7n(n+1)}{2}$. For $n=11$ this is $\frac{7(11)(12)}{2}=462$, still below 500, and for $n=12$ it is $\frac{7(12)(13)}{2}=546$, which passes 500. So 12 multiples are needed. 11 falls short, while 13 (total 637) and 15 (total 840) are past the first crossing.