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              
 A: 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 $b,c$.
A: You are correct in saying that $c=1,2,3$ but this is where you then made the mistake. We need to consider each of these cases separately
When $c=3$, we have $30a+50b<133$. Here, $b$ cannot be greater than $3$ as $50\times3=150>133$. So we have $b=1,2$. 
When $b=1$, then we have $30a<83$ meaning $a<2.766\ldots$ so $a=1,2$ are the integer solutions
When $b=2$, then we have $30a<33$ meaning $a<1.1$ so $a=1$ is the only integer solution
So, for $c=3$, we have $3$ possible solutions:
\begin{align}(a,b,c)&=(1,2,3)\\
&=(1,1,3)\\
&=(2,1,3)\end{align}
Now we can do similar things for $c=2$ and $c=1$
A: $(x^3+x^6+x^9+x^{12}+...x^{33})\cdot (x^5+x^{10}+x^{15}+x^{20}+x^{30}) \cdot (x^7+x^{14}+x^{21}+x^{28})$
Expand the above using geometric progression sum, and then find the sum of coefficients of $x^k$ where $k\leq 34$
