## Math Puzzles

December 31, 2007 by philo

I’ve returned from my conference equipped with a few new ideas, and three interesting mathematical puzzles:

- Show that, in any community of finitely many people, at least two people have the same number of friends. (Assume that no one is friend of himself.)
- In a certain park, there are some trees, and in those trees are some birds. Each tree has the same number of birds as every other tree. The total number of birds is between 200 and 300. Jane, on learning how many birds were in the park, was able to deduce how many birds were in each tree. How many trees are in the park?
- What is the next number in the sequence 1, 11, 21, 1211, …?

Now the humbling part: The first two are questions from math competitions for *middle school students*.

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Posted in mathematics | Tagged mathematics, puzzles | 7 Comments

on December 31, 2007 at 2:11 pm |jd27181. This is nice, in that it requires no knowledge outside the problem (compare my: “show that there are at least two (non-bald) people in New York City with exactly the same number of hairs on their head” which requires the knowledge that no one has 8 million hairs on their head.)

2. I haven’t seen this before, but also like it. Are we clear that “some” means “more than one”? That’s not how it is normally used in mathematics (and you’ve labeled these as math puzzles, which puts the definition in play)

3. I hate this problem. I think it is a pretty cheap trick.

Jonathan

on December 31, 2007 at 3:52 pm |philoGood point: I’m taking ‘some trees’ and ‘some birds’ to mean ‘at least two trees’ and ‘at least two birds’—the ‘at least two’ coming from the plural, not the ‘some’. I should have made that more explicit.

The third problem is indeed a cheap trick.

on December 31, 2007 at 4:28 pm |Matt2. SPOILER ALERT!

I think there are 17 trees with 17 birds in each, total birds of 289. Largest prime number squared whose product is greater than 200 but less than 300. Probably grackles or starlings. Yuck.

Other in process, but not progressing.

on December 31, 2007 at 5:04 pm |MattPuzzle 1. Kinda like the Y2K dilemma. Was year 2000 the end of the century or the beginning of the next? For the friends, I think you have to assume each community member has at least 1 friend, otherwise this doesn’t work. If you were to assign a whole number to each person in the group > 0, and the largest number was the population less 1 (can’t be a friend to yourself) then you cannot assign a unique number to everyone, so conversely some pair has the same number of friends.

on December 31, 2007 at 6:19 pm |jd2718Matt,

#1 doesn’t need additional conditions. If we have n people, the smallest number of possible friends is 0, the largest is n-1, but, and here’s a big but, is someone has 0 friends, then the guy with the most can’t have n-1, he could have at most n-2. If he does have n-1 friends, than he is friends with everyone, and no one has 0.

It’s easier with numbers. If there’s 3 of us, the largest number of friends would be 2, the smallest would be 0. But if I’ve got no friends, you’ve got at most 1. And if you are friends with both of the others (including me), than I have one friend.

We get forced back to the pigeon hole, no matter what. By the way, there is a related post with video in the newest carnival of mathematics.

on January 1, 2008 at 10:17 am |MattJD I believe you are correct, I hadn’t thought of it that way. I didn’t consider that friendship must be reflected back. Could A consider B to be a friend yet B not consider A a friend? It happens in plot devices in movies quite often.

#3 has me beat. I’ve tried base 3, mirrors, dance steps, etc

on January 1, 2008 at 12:02 pm |philoMatt,

Good point; we have to assume that friendship is symmetric.

On #3, here’s a hint: say it out loud, digit by digit.