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Sorry for the title, but I couldn't think of something else, its not actually homework, but rather a question from a maths question book I am currently stuck on, but still I've tagged it in homework.

The question is as follows:

An institute holds 32 mock tests, students have the option to appear for any number of mocks, even 0. Student A gives 16, student B gives 18, student C gives 20. What is the minimum number of mocks which were written by more than one among A, B, and C.

The way I approached was as follows:

Since total mocks by A, B, and C if we sum them are 54, and the total mocks held by the institute were 32, so atleast 54 - 32 = 22 mocks could have been written by more than one.
The problem is I am stuck here, and can't move ahead.

I think this problem maybe solved by Set theory, but being an aptitude problem, I think there's more to it and it can be solved logically without going into much mathematics.

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    Suppose André is right (I haven't checked his arithmetic). Then what you need to prove is twofold: (a) that there _is_ a way for the three students to take their tests such that exactly$11$tests were taken by more than one of them, and (b) that there _is no way_ they could take the tests with _fewer than_ 11 tests being attended by two or more. The proof of part (a) could simply be by showing a concrete example of how it is possible. For part (b) you'll need an abstract argument, probably along the lines André suggests.2012-06-30

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We will minimize the number of common exams written if we make sure that each exam written by at least two people is written by all three. (Of course it is not yet clear that this can be arranged).

If it can be arranged, let $n$ be the number of exams written by all three. Then A will write $16-n$ exams that no one else writes, B will write $18-n$, and C will write $20-n$. Thus $n+(16-n)+(18-n)+(20-n)\le 32.$ Solve. We get $n\ge 11$.

We can arrange for $n$ to be exactly $11$ by giving the first $11$ exams to everybody. In addition, A writes exams $12$ to $16$, B writes $17$ to $23$, and C writes $24$ to $32$.

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    Take $n=11$. Give these to A, plus $5$ new. Give them to B, plus $7$ new. Give them to C, plus $9$ new. There were $5+7+9$ "news" (that is $21$). Added to our $11$, that's the full $32$. And everybody got the right number.2012-06-30