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December 27th, 2016, 10:32 AM   #1
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Work in 7 weeks

I have a problem that I can't solve. It says:

A person work at least 1 hour every day during 7 weeks, at most 11 hours per week. Prove that there is a period of consecutive days in which he will work exactly 20 hours. (The number of hours he work is an integer number).

I just know, for the pigeonhole principle, that there are at least 3 days every week in which he works exactly 1 hour. But I don't know how to solve the problem with that information.

Thanks a lot!
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December 27th, 2016, 05:09 PM   #2
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Let $x_n$ be the number of hours worked on the $n$'th day, $1 \le n \le 49$.

Consider the partial sums

$s_1 = x_1$
$s_2 = x_1 + x_2$
...
$s_{49} = x_1 + \ldots + x_{49}$

If you can show that $s_j - s_i = 20$, for some $i,j$, then you're done.

Suppose there is no such pair $i,j$ (assumption *).

But the sequence $s_1,s_2,\ldots,s_{49}$ has 49 terms, hence there must be 3 terms $s_i,s_j,s_k$, say, with $i < j < k$, such that $s_i,s_j,s_k$ are congruent (mod 20).

By assumption *, $s_j - s_i > 20$ and $s_k - s_j > 20$.

Then $s_k - s_i = (s_k - s_j) + (s_j - s_i) \ge 40 + 40 = 80$, but that implies $s_k > 80$, contrary to the stated requirement that the person works at most 11 hours per week.
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Last edited by quasi; December 27th, 2016 at 05:17 PM.
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December 28th, 2016, 06:43 AM   #3
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Thanks you very much quasi, I have understood your proof, but just I'm not sure how can I prove that there are 3 terms congruent (mod 20). The sequence has 49 terms, but they are not consecutive.

Thanks againç

Edit: In fact, if I prove that there are 3 terms congruent (mod 20), the difference between two of them must be 20 as we can prove, right?

Last edited by relativo94; December 28th, 2016 at 06:55 AM.
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December 28th, 2016, 05:32 PM   #4
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There are only 20 distinct residue classes, mod 20. Since 49 is more than twice 20, there can't be at most 2 terms in each residue class. It follows that some residue class has at least 3 distinct terms. Thus, there must be 3 distinct terms which are congruent (mod 20).
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