is the answer 21??

Yeah even i got 21. But my solution is entirely different than anyone at my center. It uses summations and bounding

@above me too!

Please post your solutions, and then I'll post my entirely different solution. Most people whom I could hear were saying something like $48x+45y=972$

Math-wiz wrote: Please post your solutions, and then I'll post my entirely different solution. Most people whom I could hear were saying something like $48x+45y=972$ Exactly!!!!

An example of my stupidity in the exam- I solved P6 but when I checked them in the original eqn, I did a calculation mistake in verifying and ended up writing that there are no solutions.

Here's a summary of how I solved it. Hope y'all like it. Denote $a_i, b_j$ as the number of students who did not solve $O_i, F_j$ respectively. Then, $\sum a_i+\sum b_j=n$ And, $a_{ij}=n-a_i-b_j$, where $n$ is total number of students. Summing both sides with $1\leq i\leq 9, 1\leq j\leq 6$, $972=54n-6\sum a_i-9\sum b_j=48n-3\sum b_j$ So, $\sum b_j=16n-324$ Now, $0\leq \sum b_j\leq n$ So, $\frac{324}{16}\leq n \leq\frac{324}{15}$ This gives $n=21$

Math-wiz wrote: Here's a summary of how I solved it. Hope y'all like it. Denote $a_i, b_j$ as the number of students who did not solve $O_i, F_j$ respectively. Then, $\sum a_i+\sum b_j=n$ And, $a_{ij}=n-a_i-b_j$, where $n$ is total number of students. Summing both sides with $1\leq i\leq 9, 1\leq j\leq 6$, $972=54n-6\sum a_i-9\sum b_j=48n-3\sum b_j$ So, $\sum b_j=16n-324$ Now, $0\leq \sum b_j\leq n$ So, $\frac{324}{16}\leq n \leq\frac{324}{15}$ This gives $n=21$ I wish I could give more than 1 upvote!.Cleanest solution ever

Even i liked my solution too much. But, everyone at my center ruined the fun with some other method. Even the dumbest ones solved this one Ohh, great @below. You're one of the few rare ones i am the only one out of 50 students who have solved it this way

I did the same as math wiz

Ok, two people asked me to post the diophantine solution. In my opinion it's actually the most straightforward (I got this in like 5 minutes). Let \[S = \sum_{i=1}^9 \sum_{j=1}^6 a_{ij}\] Every student has either (a) attempted everything except \(O_i\) for some \(i\), or (b) attempted everything except \(F_i\) for some \(i\) In the first case, observe that this person is counted exactly \(8\cdot 6 = 48\) times. In the second case, this person has been counted exactly \(9\cdot 5 = 45\) times in \(S\). As an example, if a person attempted everything except \(O_9\), then he is counted once for each of \[a_{1j}, a_{2j}, \ldots, a_{8j}\]for every \(j\) (since they attempted all FITB questions). So, if \(a\) denotes the number of people accounted for in case (a), and \(b\) denotes the number of people in case (b), then the condition \(S=972\) becomes \[48a + 45b = 972\]Now just check that the only positive integer solution to this is \((12,9)\), which gives \(a+b = \boxed{21}\).