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 $30a + 50b + 70c &lt; 343$


My Attempt:

$c$ cannot be $5$ since $70 \times 5 &gt; 343$. 

It can't be $4$ either since if we put $c = 4$, we get $30a + 50b &lt; 63$, 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
The problem is that if $c=3$ you need $30a+50b \lt 133$.  In that case $b$ can only be $1$ or $2$.  If $b=2$ you must have $a=1$, while if $b=1$ you can have $a=1$ or $2$.  The choices are not independent, so you cannot multiply.  

You can just continue the casework.  There are not too many cases for $b,c$.
$(x^3+x^6+x^9+x^{12}+...x^{33})\cdot (x^5+x^{10}+x^{15}+x^{20}+x^{30}) \cdot (x^7+x^{14}+x^{21}+x^{28})$

Expand the above using geometric progression sum, and then find the sum of coefficients of $x^k$ where $k\leq 34$