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There are 3 questions in a test and the full mark of each question is 7. Each question can only be marked with integers: $1, 2, 3, \cdots, 7$. We know that the product of everyone's marks of the 3 questions is 36 (i.e., the product of the sums of the marks of the three questions across everyone is 36) and that the marks of any two people are not exactly the same. How many people did the test at most?

What I have done:

Assume there are $n$ people, let $x_i$ be the sum of the three questions for person $i$ where $i = 1, 2, \cdots, n$. We know that $x_1x_2x_3 \cdots x_n=36$ and that $x_i \neq x_j$ for $i \neq j$. How can we find the largest possible $n$?

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  • $\begingroup$ @GoodDeeds Sorry, I have not read this carefully enough. $\endgroup$ – almagest May 5 '16 at 10:00
  • $\begingroup$ So 10 people might score: (1,1,1),(1,1,2),(1,2,1),(2,1,1),(3,1,1),(1,3,1),(1,1,3), (2,1,3),(1,3,2),(3,2,1). $\endgroup$ – almagest May 5 '16 at 10:08
  • $\begingroup$ Note that $36=2 \times 2 \times 3 \times 3$. Also, $x_i>1$. $\endgroup$ – SinTan1729 May 5 '16 at 10:15
  • $\begingroup$ @TrueTears I think you have misunderstood the question. The mark 1 can appear multiple times. The requirement is that if A has mark $(a,b,c)$ on the three questions and B has mark $(a',b',c')$ then we cannot have $a=a',b=b'$ and $c=c'$. So the example I gave above fulfils the conditions. $\endgroup$ – almagest May 5 '16 at 10:42
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We can write, $$36=1^n\times2^2\times3^2$$ where $n$ can be any integer. We would like to maximize the number of factors in the product. However, no factor can repeat. Hence, $1$ can appear at most once. But if it appears, the only way it could have been formed is if the score in one test had been $1$, and $0$ in the rest. As a score of $0$ is not allowed, $1$ cannot appear. Similarly, $2$ cannot appear.

The rest of the factors are $2,2,3,3$. Again, no two of them can repeat. The ways of factorizing then are $$4\times9$$ $$12\times3$$ $$18\times2$$

$$36$$

The maximum number of factors that appear is $2$. The maximum number of students who could have appeared is $2$.

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  • $\begingroup$ Thanks, however you can't score zero for a question, so across 3 questions, there's no possible way to score a "1"? $\endgroup$ – TeTs May 5 '16 at 9:59
  • $\begingroup$ @TrueTears Sorry, I missed that. Hope it is correct now? $\endgroup$ – GoodDeeds May 5 '16 at 10:04

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