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
Best 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 $b,c$.