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

  • $\begingroup$ What is the answer in your book? $\endgroup$
    – lioness99a
    Mar 10, 2020 at 15:47
  • 2
    $\begingroup$ "Since any value of a can be paired with any value of b and c" this right here is the mistake. If we have $c=3$ we reduce to $30a + 50b < 133$. However we need to now be careful we don't choose too large a value for $b$. In your attempt you assume $b=4$ works but this isn't so $\endgroup$
    – WaveX
    Mar 10, 2020 at 15:49
  • 1
    $\begingroup$ @lioness99a, The solution requires that $a,b$ and $c$ be positive. If the OP's claim that there is no positive solution for either a or b were true, then that would be impossible. But the OP's claim is not true. There are positive solutions for $a$ and there are positive solutions for $b$. So to show this is impossible the OP needs to make a different claim. The different claim should have been "There are no simultaneous positive solutions for both $a$ and $b$". Language is important. $\endgroup$
    – fleablood
    Mar 10, 2020 at 16:08
  • 2
    $\begingroup$ @fleablood I think we all knew what OP meant... $\endgroup$
    – lioness99a
    Mar 10, 2020 at 16:11
  • 1
    $\begingroup$ "I think we all knew what OP meant... " Of course we did. But that doesn't mean the OP wasn't being careless with numbers. And carelessness leads to errors as WaveX points out. $\endgroup$
    – fleablood
    Mar 10, 2020 at 16:23

3 Answers 3


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 $b,c$.


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 $c=1$


$(x^3+x^6+x^9+x^{12}+...x^{33})\cdot (x^5+x^{10}+x^{15}+x^{20}+x^{30}) \cdot (x^7+x^{14}+x^{21}+x^{28})$

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


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