Clare had birthday party and invited 9 of her closest friends. Clare had her 13th Birthday party at a pizzeria. Clare invited $9$ of her closest friends, $6$ girls and $3$ boys. Four pizzas were ordered and the boys and girls had a competition to see who ate the most pizza. The boys ate one pizza while the girls ate the other $3$ pizzas All pizzas were cut into the same amount of equal pieces. No girl ate the same amount of pizza as any other girl and no boy ate the same amount as any other boy. Who won the eating competition?
I am $99\%$ the girls won, But I have a heavy feeling there is a trick to this question and I got it wrong. Could you please explain how you got your answer if you do end up answering? I would like to know how to solve it as well.
Thanks
 A: I will assume that the amount of slices is not known(I have never ordered pizza). In that case:
The problem is undefined. Let $x$ be the amount of slices of pizzas in which the pizzas where cut. We have 7 girls and 3 boys. So, 3 boys ate $x$ slices of pizzas, and 7 girls ate $3x$ slices of pizza.
The minimum amount of slices of pizzas girls could it is $$0+1+...+6=21.$$ Hence $3x\geq 21,$ from which $x\geq 7.$ 
The maximum amount of slices one boy could eat happens when the other two eats 0 and 1 slices, and hence, the quantity is $x-1.$ 
The maximum amount of slices a girl could eat happens when the other girls eat 0, 1, 2, 3, 4 and 5 slices, and hence the quantity is $3x-15.$  (Note that since $x\geq 7,$ we have $3x-15\geq 6>5$)
We claim that depending on how the girls ate, the competition could be won by boys, girls or to end in a tie. 
If $x= 7,$ then $x-1= 3x-15,$ and there would be a tie. 
On the other hand, if $x>7,$ we will have $3x-15> x-1,$ and in this case one of the girls would win. 
The previous analysis was under a fixed distribution on the slices, such that one girl and one boy could ate the most they can. But this is not the only possible distribution. If we change this, then boys could also win. 
In order to see this, assume $x>7$ is divisible by 7. Then, a possible distribution for the girls is 
$$\frac{3x}{7}+3, \frac{3x}{7}+2, \ldots, \frac{3x}{7}-3,$$ on which the winner among girls ate $\frac{3x}{7}+3$ slices. The above distribution for the boys($0, 1, x-1$) will give us now 
$$x-1 > \frac{3x}{7}+3,$$ and then boys would win.
A: There are $4n$ slices of pizza available. The boys ate $n$ slices. While the girls ate $3n$ slices.
Since the boys and the girls have been segregated we will denote the amount of slices ate by guy $ i $ or by girl $j$ as :
$n_i \forall \ i \in \{1,2,3\} \\ n_j \forall \ j \in \{4,..,10\}$
You can compute numerically the highest possible slices eaten because you know no one ate the same amount of slices as anyone else in their own group. 
So that : $n_1 \ne n_2 \ne n_3 \ne n_1$ and $n_t \ne n_u $ for all $ \ t\ne u \ \ $ with  $t,u \in \{4,5,...,10\} $
The minimal amount of slices one can eat is $0$. So that the maximum amount achievable for the boys group is : $n_3=n-1-0$.
For the girls the maximum is : $ n_{10} = 3n - \sum_{k=0}^5 k$.
In the end $ $
$\begin{cases}
    n_3 = n-1,& \\
    n_{10}=3n - 15              & 
\end{cases}$
What matters now is the conditions on $4n$ the amount of total slices, $3n \ge 21 \Rightarrow 4n \ge 28 \Rightarrow n \ge 7$ again because of the initial condition on the $n_j$ being different for every girl. 
For the girls to win you need :
$$\begin{align} n_{10} &>n_{3}  \\
 3n-15 & > n - 1 \\
n & > 7
\end{align}$$
But this is only valid if boys and girls use the best strategy, aka starving most of the group so that one can eat more. 
