# Pigeonhole principle problem

The problem I'm working on says:

A basketball player has been training for 112 hours during 12 days. He has trained an integer number of hours every day. Prove that there was two consecutive days where he has trained for at least 19 hours.

I'm following this rationale to try prove it:

As we are interested in two consecutive days we split the 12 days into 6 pairs: {1,2}, {3,4}, {5,6}, {7,8}, {9,10}, {11, 12}

Using a corollary of the Pigeonhole principle we know that the sum of one the pairs is greater than 18, but not 19.

Does this mean that I can't prove it this way?

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Firstly your grouping is erroneous. You need to look at the following groups $\{1,2\}, \{2,3\}, \{3,4\}, \cdots, \{11, 12 \}, \{12, 1\}$. If in each pair, time spent was less than 19, say only 18, then you would have $18 \times 12=216 < 2 \times 112 = 224$. Hence a contradiction. – user21436 Dec 28 '11 at 16:25
Wouldn't we need to leave {12,1} out? – Javi Dec 28 '11 at 16:41
Yeah, as your question stands, I shouldn't have added $\{12,1\}$. Then you are in a little odd shape. Because, there's actually nothing that tells you the number of hours he practised together on the $12^{th}$ day and $1^{st}$ day. – user21436 Dec 28 '11 at 16:48
On a totally unrelated note, this seems to be your first post. And your comments seem to not use $\LaTeX$. So, Please use it for subsequent posts, be it questions, answers or comments. You'll help the community a bunch. – user21436 Dec 28 '11 at 16:50
OK. I think I got it now. We know that at least one pair is > 18. And as the problem states that he trained an integer number of hours, it has to be, at least, 19 hours. The grouping issue that Kannappan points out, can be solved if we group it like this: {1,12}, {2,3}, {4,5},{6,7},{8,9},{10,11}. This way we cover all the possible pairs and we can apply the same principle. – Javi Dec 28 '11 at 16:51

This can be proved easily. Suppose the statement is NOT true. Then any given pair of consecutive days he trains for at most 18 hours. So he trains a total of at most 18 * 6 = 108 hours. This is a contradiction since we are told that he trains a total of 112 hours.

I hope this helps! :)

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The 6 is there multiplying because you have exactly 6 pairs of consecutive days – chango Dec 28 '11 at 16:54
That's very simple! I was trying to fit in the pigeonhole principle as I could, but this is even more obvious. – Javi Dec 28 '11 at 17:11
Javi, cuando la matemática es intuitiva es mejor no complicarse al divino cohete. Saludos y suerte con tus estudios! – chango Dec 28 '11 at 17:32

Your grouping works just fine. The average over the $6$ groups is $\dfrac{112}{6}$, which is greater than $18$. So at least one of the group sums is greater than $18$. Since all group sums are integers, at least one of these must be $\ge 19$.

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Thank you so much! I only used the integer clue to assume that every day he trained a positive integer number of hours. – Javi Dec 28 '11 at 17:03
@Javi: There is no need to assume positive integer number of hours, $\ge 0$ is fine. Actually, negative integer also allows the proof to go through. If you have a sequence $a_1,a_2, \dots, a_{12}$ of $12$ integers, with sum $112$, then some two consecutive elements of the sequence add up to $19$ or more. But negative integers would not make much sense for basketball pracice. – André Nicolas Dec 28 '11 at 17:13