# 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

• What is the answer in your book? Mar 10, 2020 at 15:47
• "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 Mar 10, 2020 at 15:49
• @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. Mar 10, 2020 at 16:08
• @fleablood I think we all knew what OP meant... Mar 10, 2020 at 16:11
• "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. Mar 10, 2020 at 16:23

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 $$k\leq 34$$