Find the number of ordered triplets $(a, b, c)$ of positive integers such that $30a + 50b + 70c < 343$ 
Find the number of ordered triplets $(a, b, c)$ of positive integers such that $30a + 50b + 70c < 343$

My Attempt:
$c$ cannot be $5$ since $70 \times 5 > 343$. 
It can't be $4$ either since if we put $c = 4$, we get $30a + 50b < 63$, which would mean there are no positive integer solutions for at least one of $a$ or $b$.
So, there are 3 options for $c$.
Proceeding similarly, we will get that $b = 1,2,3,4$ or there are $4$ options for b.
Similarly, there are $7$ options for $c$. 
Since any value of $a$ can be paired with any value of $b$ and $c$, we get  $3 \times 4 \times 7=84$ total triplets which is not the answer              

$(x^3+x^6+x^9+x^{12}+...x^{33})\cdot (x^5+x^{10}+x^{15}+x^{20}+x^{30}) \cdot (x^7+x^{14}+x^{21}+x^{28})$
Expand the above using geometric progression sum, and then find the sum of coefficients of $x^k$ where $k\leq 34$
