Perfect Score question, deleted by poster, earlier today

Question

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?

Answer 1

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 $18$.

Answer 2

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.