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 December 27th, 2016, 10:32 AM #1 Newbie   Joined: Dec 2016 From: España Posts: 11 Thanks: 0 Math Focus: Logic 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!
 December 27th, 2016, 05:09 PM #2 Member   Joined: Dec 2016 From: USA Posts: 46 Thanks: 11 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. Thanks from relativo94 Last edited by quasi; December 27th, 2016 at 05:17 PM.
 December 28th, 2016, 06:43 AM #3 Newbie   Joined: Dec 2016 From: España Posts: 11 Thanks: 0 Math Focus: Logic 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.
 December 28th, 2016, 05:32 PM #4 Member   Joined: Dec 2016 From: USA Posts: 46 Thanks: 11 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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