Extra Condition to solve Three Conferences with constant success

Question

A cycle of three conferences had constant success, that is: in each session, there were the same number of assistants. However, half of the ones who attended the first conference didn't come back, a third of the ones who attended the second conference only watched this one and a fourth of the ones who attended the third conference didn't watch neither the first nor the second conferences. Knowing that there were a total of $300$ registered participants and that each participant attended to at least one conference, answer: How many people attended to each conference? How many people attended all the three conferences?

So, I tried setting a constant for the total of participants of each conference and then working up the equations of the intersections of the conferences and also of the ones who only attended one conference, but in the end I missed the fact that the numbers of participants should always be a positive integer, which would give a numeric range to solve the equations. In the end, the real difficulty of the problem was the not so obvious nor that explicit extra condition(at least for me as I'm not really used to this kind of approach and I don't recall ever using it for this type of problem), which was brought up by 57Jimmy and Christian Blatter. Thanks for the effort.

Answer 1

Warning: this is probably the most inelegant solution one could provide. But it is a solution. The difficulty lies in the fact that many constraints are not explicit: any number of people must be a nonnegative integer, for instance.

Let $x$ be the total number of people who were present at one of the three conferences. Denote by $N_i$ the number of people that attended exactly $i$ conferences. We have

$$ N_1 = x/2 + x/3 + x/4 = 13x/12$$ $$ N_2 = (300-x-x/2-x/3)+(300-x-x/2-x/4)+(300-x-x/3-x/4) = $$ $$ = 900 - 3x - 2x/2 - 2x/3 - 2x/4 = 900 - 31x/6$$ $$ N_3 = 300 - N_1 - N_2 = 49x/12 - 600$$

We do not have a lot more information, but we know that $0 \le N_i < 300$ for all $i\in \{1,2,3\}$ and that $x$ is divisible by $12$. Thus we get conditions

$$ 0 \le N_2 < 300 \Longrightarrow 117 \le x \le 174$$ $$ 0 \le N_3 < 300 \Longrightarrow 147 \le x \le 220$$

Hence $x \in \{156, 168\}$ and, correspondingly, $N_3 \in \{37, 86\}$. Now one should check whether both cases are actually possible, distinguishing in a finer way how many people took part in exactly what conferences. Hence we distinguish numbers $M_1, M_2, M_3, M_{1,2}, M_{1,3}, M_{2,3}, M_{1,2,3}$, according to what conferences they visited. The first three numbers and the last are given, so we must check if there are solutions for the three numbers of people that went to two conferences, under the constraints

$$ N_2 = M_{1,2} + M_{1,3} + M_{2,3}$$ $$ x = M_1 + M_{1,2} + M_{1,3} + M_{1,2,3} = M_2 + M_{1,2} + M_{2,3} + M_{1,2,3} = M_3 + M_{1,3} + M_{2,3} + M_{1,2,3}$$

Equivalently, we must have

$$ N_2 = M_{1,2} + M_{1,3} + M_{2,3}$$ $$ x - N_2 - M_{1,2,3} = M_1 - M_{2,3} = M_2 - M_{1,3} = M_3 - M_{1,2} $$

In the first case $( x = 156)$, from the second line we get

$$ 25 = 78 - M_{2,3} = 52 - M_{1,3} = 39 - M_{1,2} $$

and the numbers obtained also satisfy the first line, so that we have a solution. In the second case $(x = 168)$ we do not get a solution: we obtain

$$ 50 = 84 - M_{2,3} = 56 - M_{1,3} = 42 - M_{1,2} $$

but then $M_{1,2}$ would be negative, which is not allowed! Hence the only possible solution is

$$ x = 156, N_3 = 37.$$

Answer 2

Some hints:

You got $6$ equations for the seven unknowns $x_1$, $x_2$, $x_3$, $x_{12}$, $x_{13}$, $x_{23}$, $x_{123}$. Consider $x_1$ as given, and solve for the remaining unknowns in terms of $x_1$. Look at the result, and get the necessary extra conditions for $x_1$ from the knowledge that all $x_\iota\in{\mathbb N}_{\geq0}$. In the end everything will be determined uniquely.