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I don't know why they deleted their own problem. It's a good challenge, and someone may find it educational. Paraphrased: Forty students took a test that had three difficult questions - #1 on geometry, #2 on probability, and #3 on modular equations. 4 students solved only #1. 3 students solved only #2. 5 students solved only #3. 19 students solved #1 and at least one other question. 20 students solved #2 and at least one other question. 21 students solved #3 and at least one other question. How many students solved all three questions?

Bafs
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  • Did you solve it? If so IMO you should say that in the question, and maybe explain whether you're looking for other solutions. If not it would be good to show your attempt here. – coffeemath Jun 02 '22 at 01:24
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    If you want discussion about why it was deleted, that belongs on Meta, not here. If you already know the solution, this might be better on puzzling.stackexchange.com than here. – aschepler Jun 02 '22 at 01:26
  • I'll go to puzzling, now that I know about it. Had to edit, anyway. – Bafs Jun 02 '22 at 01:33
  • You could always post a Question and also provide the best answer in order to leave a good depository for someone who might need it next time. – algevristis Jun 03 '22 at 13:51

2 Answers2

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Let $a$, $b$, $c$, $d$, and $e$ represent the number of people who solved exactly 1 and 2, 1 and 3, 2 and 3, all, and none, respectively. Then we have a system of equations $$a+b+d=19,$$ $$a+c+d=20,$$ $$b+c+d=21,$$ $$a+b+c+d+4+3+5+e=40.$$ Reducing, we see $c=b+1=a+2$, and $2a+d=18$, thus $a+(2a+d)=3a+d=25-e$, so $a=7-e$. Then, unless I'm missing something, for each possible value of $a$, $0\leq a\leq7$, we have $d=18-2a$, and the corresponding values for $b$, $c$, and $e$ will satisfy the system and be nonnegative integers, and thus satisfy all the parameters of the question. Thus, the number solving all questions, $d$, may be $4$, $6$, $8$, $10$, $12$, $14$, $16$, or $18$.

Alex
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  • Your answer starts off good, but goes off track at a + b + d = 19. If you model this on paper, using a Venn-like picture, it doesn't add to 19. – Bafs Jun 03 '22 at 15:21
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    How so? The students who solved 1 and and least one other question should be those who solved just 1 and 2, just 1 and 3, and all 1, 2, and 3, correct? Further, we should be able to add as such sets of people are disjoint by definition. – Alex Jun 03 '22 at 15:25
  • It could be the variables you chose, not sure. I started with a: solved only #1, b: solved only #2, c: solved only #3, d: solved #1 and one other, e: solved #2 and one other, f: solved #3 and one other, g: solved all 3 questions. Then I drew a 3-pointed star, and placed the variables between the points (in sections for each question). You need to represent d,e, and f twice each, and represent g in all three sections. – Bafs Jun 03 '22 at 15:51
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    Let A be the set of people who answered one and at least one other problem correctly. Let B be those who answered 1 and 2 correctly and 3 wrong, let C be those who answered 1 and 3 correctly and 2 wrong, and let D be those who answered all correctly. Then B, C, and D are in A, form a partition of A, and have orders a, b, and d respectively (as I defined them in the answer), thus $a+b+d$ is precisely the order of A, which is 19. The other equations in the system are similarly justified. Also, I have represented the values you mention multiple times as the variables appear in several equations. – Alex Jun 03 '22 at 16:08
  • You are forgetting those who answered #2 and #3, and got #1 wrong. Furthermore, in my picture I separated d, e, and f into actual students represented by uppercase alphabet letters, and then lowercase, to make 40. All of this was to verify the math, which gives only one solution. – Bafs Jun 03 '22 at 16:14
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    That value is c in my answer, which is used in the last equation to make 40. I did not forget c, because the students who got 1 wrong do not factor into those who got 1 and at least one other question right, so should not be considered in adding up to 19. You should post your solution then so I can learn from it. Perhaps I am missing something obvious. – Alex Jun 03 '22 at 16:20
  • I'm sorry. You were correct from the beginning. I just worked through this since finding the problem yesterday. Your technique makes multiple solutions more obvious. There are eight possible solutions, which you identified: 4, 6, 8, 10, 12, 14, 16, and 18. I'll edit my answer. – Bafs Jun 03 '22 at 19:30
  • Let us continue this discussion in chat. – Alex Jun 04 '22 at 00:11
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My variables were a: solved only #1, b: solved only #2, c: solved only #3, d: solved #1 and one other, e: solved #2 and one other, f: solved #3 and one other, g: solved all three. Then 40 students were represented by upper and lower case letters, spread across three areas of a diagram. Variables d, e, and f were shown twice, in two areas. Variable g was shown in each of the three areas. I will clarify that I divided each of the variables d, e, and f into two groups of students (alphabet letters) - those also solving the question to their left (clockwise), and the rest solving the question to their right (counter-clockwise). Here's the logic of the math: Out of 40 students, 12 solved only one question. a+b+c=12. That leaves 40-12=28 who solved exactly two or three questions. What makes this a pigeonhole problem is that there are 19+20+21 = 60 "pigeonholes", or questions on all of the tests, where these 28 students solved correctly. 28 times 2 gives 56, the first and second questions answered by d, e, and f. That leaves 60-56 = 4 additional (third) questions solved by: 4 students / perfect scores. However, there are more solutions, accounting for students who solved none of the questions. If 1 solved none, 6 solved three. 2 solving none gives 8. For every score of zero, there are two more perfect scores, until 7 solving none gives 18 solving all three questions. The math is in the comment section. Thank you, Alex.

Bafs
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