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I have two Pigeon Hole Principle questions that I am having trouble with.

First Question: In a room there are 10 people, none of whom are older than 60, but each of whom is at least 1 year old. Prove that one can always find two groups of people (with no common person) the sum of whose ages is the same.

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So I let [itex]a_i[/itex] = age of person i, i = 1,2,...,10.

Thus [itex]1 \leq a_i \leq 60[/itex] for i = 1,2,...,10

we can also assume that [itex]a_i \neq a_j \forall i \neq j[/itex] since if there were two (or more) that were equal then we would be done.

The book has the following hint: The number of sums that can be formed with 10 numbers is [itex]2^{10} - 1[/itex]. No sum can exceed 600

Now I understand how they got these hints. For the first, you can use the idea of a Powerset, and the number of elements in P(S) = 2^n (2^n - 1 if you subtract the empty set, n is the number of elements in S). And 600 is obvious since 60x10 = 600 the max (though we could do more to show that the sum is actually lower; it is easy to show that the sum cannot exceed 555 (and it is even lower than that) and also that the sum must exceed 55, (and is even higher than that)).

However, I am not sure of how to use the first hint. There is obvious overcounting, and also we are interested in two groups (or subsets if we want to stick with that idea) whose sums are equal. The 2^n - 1 includes the sum of all 10, which should not be in there, it also includes things that repeat, like a1+a2 = a1, a1+a2+a3 = a1+a2, etc. Do I need to then count these things (and then subtract them from 2^10 - 1), and if so what would be an efficient way (I can think of the brute force way, but I would not want to have to do that )? Any ideas?

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Second Question: Use the pigeonhole principle to prove that the decimal expansion of a rationl number m/n eventually is repeating.

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Now I am guessing they are also including decimal expansions that terminate (for example 3/4 = .75)

Let us again look at the book's hint: What are the possible remainders when an integer is divided by n?

The answer to this question is: 1, 2, 3, ...., n-1

So we have n-1 possible remainders, but I don't yet see what we can do with this.

So how about the following question: If I have m/n, how do I get the decimal expansion of this number? (I am guessing we will keep on dividing somewhere and that is where the remainders will come into play, but I don't know how to get a decimal expansion.) Any help would be appreciated, thanks!

First Question: In a room there are 10 people, none of whom are older than 60, but each of whom is at least 1 year old. Prove that one can always find two groups of people (with no common person) the sum of whose ages is the same.

-------

So I let [itex]a_i[/itex] = age of person i, i = 1,2,...,10.

Thus [itex]1 \leq a_i \leq 60[/itex] for i = 1,2,...,10

we can also assume that [itex]a_i \neq a_j \forall i \neq j[/itex] since if there were two (or more) that were equal then we would be done.

The book has the following hint: The number of sums that can be formed with 10 numbers is [itex]2^{10} - 1[/itex]. No sum can exceed 600

Now I understand how they got these hints. For the first, you can use the idea of a Powerset, and the number of elements in P(S) = 2^n (2^n - 1 if you subtract the empty set, n is the number of elements in S). And 600 is obvious since 60x10 = 600 the max (though we could do more to show that the sum is actually lower; it is easy to show that the sum cannot exceed 555 (and it is even lower than that) and also that the sum must exceed 55, (and is even higher than that)).

However, I am not sure of how to use the first hint. There is obvious overcounting, and also we are interested in two groups (or subsets if we want to stick with that idea) whose sums are equal. The 2^n - 1 includes the sum of all 10, which should not be in there, it also includes things that repeat, like a1+a2 = a1, a1+a2+a3 = a1+a2, etc. Do I need to then count these things (and then subtract them from 2^10 - 1), and if so what would be an efficient way (I can think of the brute force way, but I would not want to have to do that )? Any ideas?

-------------------

Second Question: Use the pigeonhole principle to prove that the decimal expansion of a rationl number m/n eventually is repeating.

------

Now I am guessing they are also including decimal expansions that terminate (for example 3/4 = .75)

Let us again look at the book's hint: What are the possible remainders when an integer is divided by n?

The answer to this question is: 1, 2, 3, ...., n-1

So we have n-1 possible remainders, but I don't yet see what we can do with this.

So how about the following question: If I have m/n, how do I get the decimal expansion of this number? (I am guessing we will keep on dividing somewhere and that is where the remainders will come into play, but I don't know how to get a decimal expansion.) Any help would be appreciated, thanks!

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