A group of college students is split into two teams, red shirts, and blue shirts. They participate in a contest. Each contestant plays each other contestant only once. There are 10x as many reds as there are blues. The red shirts win 4.5x as many games as the blue shirts. How many blue shirts are there?

This seems to be a combination of combinatorics, algebra, and logic (maybe?)... but I have no idea how to solve this.

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    $\begingroup$ I assume each match in the contest consists of one red shirted player against one blue shirted player? It's not specified as far as I can see ... edit: also, in these matches the winner earns 1 point for winning? does the loser lose a point or just get no points? it might be obvious but it's best to make sure $\endgroup$ – Zubin Mukerjee Sep 24 '18 at 22:21
  • $\begingroup$ Wasnt specified in the question either. I think it's just student vs student. Wish I could help more :( $\endgroup$ – QuantumHoneybees Sep 24 '18 at 22:24

Let's consider the case where all contestants play against each other. There are $b$ blue shirts and $10b$ red shirts.

There are $\binom b2$ games between two blue shirts, which are guaranteed to give the blue team a point. There are $\binom{10b}{2}$ games between two red shirts, which are guaranteed to give the red team a point. Finally, there are $10b^2$ games between a blue shirt and a red shirt, which could go either way.

So the blue team can earn at most $\binom b2 + 10b^2$ points, if they win all of those games; the red team must earn at least $\binom{10b}{2}$ points.

If the red team earns $4.5$ times as many points as the blue team, then we must have $$ \binom{10b}{2} \le \text{red score} = 4.5(\text{blue score}) \le 4.5 \left(\binom b2 + 10b^2\right). $$ Expanding this inequality gives us $$ 50b^2 - 5b \le 4.5 \left(\frac{21}{2}b^2 -\frac12 b\right) \iff 2.75b^2 - 2.75b \le 0 $$ which holds when $0 \le b \le 1$.

Ignoring the $b=0$ case where there are no participants at all, we must have $b=1$. In this case, there is $1$ blue shirt and $10$ red shirts. If the blue shirt wins all $10$ of their games while the red shirts get $45$ points from games with each other, then the conditions of the problem are satisfied. From the inequality above, we know that this is the only possible case.


Let the number of blue-shirted players be $b$.

The number of red-shirted players is therefore $10b$.

Each blue-shirted player plays $10b$ matches (one against each red-shirted player).

This means that the total number of matches is $10b^2$.

This is the same as the total number of points scored in those matches.

Let the total number of points scored by the blue-shirts be $p$.

The total number of points scored by the red-shirts is therefore $4.5p$.

We've calculated the total number of points scored in two different ways, but they must be equal:

$$ \begin{align*} p+4.5p &= 10b^2\\\\ 5.5p&=10b^2\\\\ 11p&=20b^2 \tag{potato} \end{align*} $$

The above equation tagged $\textbf{potato}$ is what is known as a Diophantine equation, as we are looking for integer solutions $(p,b)$.

Note that the left hand side of $\textbf{potato}$ is divisible by $11$. This means that the right hand side must also be divisible by $11$. Since $\gcd(11,20) = 1$, we know that $b^2$ must be a multiple of $11$. But $11$ is prime, so we find that $$b \equiv 0\pmod{11} \,\,\,\,\text{and}\,\,\,\,b^2 \equiv 0 \pmod{121}$$

If $20b^2$ is divisible by $121$, then the left hand side of $\textbf{potato}$ must also be divisible by $121$. So we have

$$p \equiv 0 \pmod{11}$$

This information yields an infinite set of solutions for the number of points scored by blue-shirts $p$ and the number of blue-shirts $b$:

$$(p,b) \,\,\,=\,\,\, (220,11) \,\,\,\text{or}\,\,\,(880,22) \,\,\,\text{or}\,\,\,(1980,33) \,\,\,\text{or}\,\,\,(3520,44)\,\,\, \cdots$$

So, I think all we can say about the number of blue-shirts is that it's a multiple of $11$.

$$\boxed{\,b \in \left\{11k \,\mid k \in \mathbb{N}\,\right\}\,\,}$$

  • $\begingroup$ If we assume that each contestant plays every other contestant regardless of shirt color, does that help give a single solution? $\endgroup$ – QuantumHoneybees Sep 24 '18 at 22:48
  • $\begingroup$ I'd guess that would still lead to infinite solutions*, but you could work it out. In that case, the number of matches (and also the total number of points scored) would be the number of ways to choose two players from the entire group: $$\binom{11b}{2}$$ $\endgroup$ – Zubin Mukerjee Sep 24 '18 at 22:51
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    $\begingroup$ Ok.. I guess that's what I thought. Thanks! $\endgroup$ – QuantumHoneybees Sep 24 '18 at 22:53
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    $\begingroup$ If all players play each other, then things become more interesting: we are forced to take $b=1$, because otherwise the required number of points ($\frac{2}{11}$ of the total number of points) is impossible to achieve even if all the blue shirts beat all the red shirts. $\endgroup$ – Misha Lavrov Sep 24 '18 at 22:56
  • $\begingroup$ @MishaLavrov Ah, the red-shirts get so many guaranteed points just from playing against each other! Interesting $\endgroup$ – Zubin Mukerjee Sep 24 '18 at 23:01

Each member of blue faces off against each member of red.

There are $b(10b) = 10b^2$ total matches

Suppose red wins $p$ matches blue wins $q$ matches.

$p,q,b$ are integers and

$\frac {p}{q} = \frac {9}{2}\\ p+q = 10b^2\\ \frac {11}{2}q = 10b^2\\ 11q = 20b^2$

$b$ must be divisible by $11$

But, there is not enough information beyond that.


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