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Old 10-13-2017, 06:14 AM
CairoCarol CairoCarol is offline
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Did anyone ever participate in the Putnam math competition? What'd you think?

I'd never heard of this competition until my son participated. (I wasn't a math geek, so it's not too surprising, though I spent so much time at MIT mooning after nerds you'd think I would have at least known about it.) My son enjoyed it and will keep doing it, even though his score wasn't high enough to confer any bragging rights.

Seems like it could be fun for the right kind of person. Anyone on the SD ever do it?
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  #2  
Old 10-13-2017, 06:30 AM
Bullitt Bullitt is offline
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Back when I was in college I heard about it. Thought about it, but never did it. But my college had an annual mathematics competition and I did that twice. Fun and interesting. I placed 3rd my first year, and then 2nd the next year, and won a little money. Math competitions are fun because you use your bag of tricks to solve problems, and the problems can be solved using different methods. You got points for elegance, too. It was a great experience. I was a bit of a math geek back then. Not especially good at it, but I enjoyed it.

Sounds like the CairoKid likes it! It's something he can put on his resumé.

Last edited by Bullitt; 10-13-2017 at 06:33 AM.
  #3  
Old 10-13-2017, 06:47 AM
Shalmanese Shalmanese is offline
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The Putnam is generally accepted among math people as having the most nerd cred. If your son has any plans of going into a STEM career, a good Putnam score could give him a leg up. Even if he doesn't, as long as he's enjoying it, it should be encouraged. The Putnam is one of the few remaining academic tests where even the top scorers get a miserable low mark and I feel tests like that are important for young people to experience.
  #4  
Old 10-13-2017, 06:58 AM
CairoCarol CairoCarol is offline
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Quote:
Originally Posted by Shalmanese View Post
The Putnam is one of the few remaining academic tests where even the top scorers get a miserable low mark and I feel tests like that are important for young people to experience.
I hadn't thought of that, but you make a very good point.
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Old 10-13-2017, 07:13 AM
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My nephew does the Scottish maths challenge and does well at it.
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  #6  
Old 10-13-2017, 08:53 AM
Wesley Clark Wesley Clark is offline
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I didn't participate but I knew people who did in college. I think they all got 5 points or less.
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Last edited by Wesley Clark; 10-13-2017 at 08:54 AM.
  #7  
Old 10-13-2017, 03:16 PM
Shalmanese Shalmanese is offline
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Originally Posted by Wesley Clark View Post
I didn't participate but I knew people who did in college. I think they all got 5 points or less.
Considering the median score is 1 (out of 120), a 5 is a very good score!
  #8  
Old 10-13-2017, 05:25 PM
Wesley Clark Wesley Clark is offline
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Originally Posted by Shalmanese View Post
Considering the median score is 1 (out of 120), a 5 is a very good score!
I thought the median score was 3.

Edit, nope. median score is 0-1. I can't remember what my friends and acquaintances got, but they seemed to get 0 or 1 on most of the questions. No 10s for any of them for what I remember.
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Last edited by Wesley Clark; 10-13-2017 at 05:26 PM.
  #9  
Old 10-13-2017, 07:03 PM
puzzlegal puzzlegal is offline
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I took it twice. I scored 6 or 7 points the first time, and the secretary who gave me my score sort of apologized, but I was happy. The second time, my boyfriend coached me a little about the importance of elegance (or at least, not leaving loose ends all over the place) and I doubled my score.

It was a good experience.
  #10  
Old 10-13-2017, 10:24 PM
Left Hand of Dorkness Left Hand of Dorkness is offline
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How is it scored?

Also, I found some sample questions here. My high school/college-level math concepts and vocabulary are frankly pitiful, so I don't even understand how to start most of them. But I think I have one of them (okay, in typing it out I realize I don't understand it as well as I thought and would love some help):
Quote:
Consider the following game played with a deck of 2n cards numbered from 1to 2n. The deck is randomly shuffled and n cards are dealt to each of two players.

Beginning with A, the players take turns discarding one of their remaining cards and announcing its number. The game ends as soon as the sum of the numbers on the discarded cards is divisible by 2n + 1. The last person to discard wins the game. Assuming optimal strategy by both A and B, who wins?
My answer:
SPOILER:
B wins, and here's why. 2n+1 is 1 higher than the highest card in the deck. A sequence from 1 to 2n contains an even number of cards, and the cards can be rearranged into pairs with the value 2n+1. For example, if n=5, the cards are from 1-10, 2n+1=11, and you can arrange the cards into (1, 10); (2,9); (3,8); (4,7); and (5,6).

Player A may have some of these pairs, and optimal play will involve playing members of a complete pair. Continuing the example, if Player A has a hand of (1, 3, 4, 6, 8), the only complete pair is (3, 8); she should play one of these cards, knowing that player B won't have the other card to win the game.

The problem is that for every pair Player A has, Player B necessarily has a pair. In this example, Player B must have a hand of (2, 5, 7, 9, 10), with the matching pair (2, 9). Optimal play will look something like this:
A: 3
B: 2

A knows not to complete her match with (8), because then B completes his with (9) and wins. But if A plays anything that's not from a match, B can then win.

Huh. I'm losing my train here. Basically, I think B's ability to choose to play from matches or not matches, in response to A's plays, creates an unbeatable advantage. Can anyone smooth out my thinking on this, using very small words?
  #11  
Old 10-13-2017, 11:51 PM
DPRK DPRK is offline
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Hmm, I found this page containing a whole bunch of problems as well as solutions. It says students have 6 hours to solve 12 problems, and can get up to 10 points per problem. Anyway the score results are also listed.
  #12  
Old 10-14-2017, 03:04 AM
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Quote:
Originally Posted by Left Hand of Dorkness View Post
The second problem on that page seems very easy.
SPOILER:

If neither T nor U is closed, then ab = z and xy = c for some a,b,c in T and x,y,z in U.
So abc = xyz. But we are given that abc is in T, xyz is in U and T,U are disjoint.


The third problem has an easy solution I think:
SPOILER:
Clearly B wins if the game lasts until the end. Can B be zugzwanged earlier? No! B will always have a safe move by a simple parity argument.

If K (modulo 2n+1) is the local goal for a pair, B need only avoid moves x where A can take K-x and win. All the pairs (x,K-x) are therefore off-limits to B. But B will always have an odd number of choices, and the elements of pairs (x,K-x) will have an even count. This would break down if (x, x+2n+1, K-x) were all available -- two pairs would actually be a triplet, but that is impossible: the starting numbers are all within 2n-1 of each other, so x+2n+1 cannot be in the set if x is.

Once upon a time I did very well on tests like this. It's refreshing to see that this increasingly idiotic idiot-savant still retains a tinge of the savant!

  #13  
Old 10-14-2017, 08:49 AM
Hari Seldon Hari Seldon is offline
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I never took it. I was supposed to but the professor who was supposed to apply for us missed the deadline. But I have looked over the questions regularly and they are very hard. Even as a professional mathematician, I doubt I would get a very high score. For example, I just read the question posed by LHoD and I would not have the foggiest idea how to get started. I think his solution is correct, although I have not thought it through thoroughly.

There are two parts. The first part consists of questions that many will get. I can usually do about half of them. The second part is much harder and I cannot usually do any of them.

The team that won the very first Putnam in the 1930s was from U. of Toronto and, over the years, I met all three members. One was an absolutely brilliant mathematician, arguably the top American mathematician of his era (he spent his entire career in the US). The second member was a decent enough mathematician, nothing spectacular, and the third was a cipher. So it is hard to know what the significance of winning is. I guess it rewards a certain kind of quickness, rather than the steady growth of understanding that always characterized my research.
  #14  
Old 10-14-2017, 11:38 AM
DPRK DPRK is offline
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Quote:
Originally Posted by Left Hand of Dorkness View Post
My answer:
SPOILER:
B wins, and here's why. 2n+1 is 1 higher than the highest card in the deck. A sequence from 1 to 2n contains an even number of cards, and the cards can be rearranged into pairs with the value 2n+1. For example, if n=5, the cards are from 1-10, 2n+1=11, and you can arrange the cards into (1, 10); (2,9); (3,8); (4,7); and (5,6).

Player A may have some of these pairs, and optimal play will involve playing members of a complete pair. Continuing the example, if Player A has a hand of (1, 3, 4, 6, 8), the only complete pair is (3, 8); she should play one of these cards, knowing that player B won't have the other card to win the game.

The problem is that for every pair Player A has, Player B necessarily has a pair. In this example, Player B must have a hand of (2, 5, 7, 9, 10), with the matching pair (2, 9). Optimal play will look something like this:
A: 3
B: 2

A knows not to complete her match with (8), because then B completes his with (9) and wins. But if A plays anything that's not from a match, B can then win.

Huh. I'm losing my train here. Basically, I think B's ability to choose to play from matches or not matches, in response to A's plays, creates an unbeatable advantage. Can anyone smooth out my thinking on this, using very small words?
SPOILER:

I think the simplest way to put it is: whenever it is A's turn, no matter what card A plays B can respond so as to prevent A from winning on the next turn.

Whatever A chooses to play, it will be B's turn; A has k-1 cards in hand and B has k. That means B can leave k different totals while at most k-1 of these allow A to win.
  #15  
Old 10-15-2017, 02:41 PM
Left Hand of Dorkness Left Hand of Dorkness is offline
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Quote:
Originally Posted by DPRK View Post
SPOILER:

I think the simplest way to put it is: whenever it is A's turn, no matter what card A plays B can respond so as to prevent A from winning on the next turn.

Whatever A chooses to play, it will be B's turn; A has k-1 cards in hand and B has k. That means B can leave k different totals while at most k-1 of these allow A to win.
That explanation makes total sense to me, and is pretty cool. Thanks!
  #16  
Old 10-15-2017, 02:57 PM
hogarth hogarth is online now
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I did the Putnam once or twice. I thought it was kind of fun, although I didn't distinguish myself (I can't remember my scores, but probably less than 10). It eats up a whole day, though!

My math contest skills peaked in grade 11 when I did well in the Fermat competition.
  #17  
Old 10-16-2017, 03:25 PM
Some Call Me... Tim Some Call Me... Tim is offline
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Quote:
Originally Posted by Left Hand of Dorkness View Post
My answer:
SPOILER:
...Player A has a hand of (1, 3, 4, 6, 8)
...
Optimal play will look something like this:
A: 3
B: 2
...
That wouldn't be optimal play by B...
SPOILER:

A: 6
Running sum is 11, which is divisible by 2n+1. A wins.
  #18  
Old 10-16-2017, 03:50 PM
That Don Guy That Don Guy is offline
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I am in the "I heard about the Putnam, but never actually tried it in college" group.

If calculus and really high-level stuff throws you, try the USA Mathematical Olympiad questions - six questions in two groups of three, with 4 1/2 hours allowed per group, and no calculus.
Archive of past USAMO questions

For those of us who do much better with number-crunching problems than proofs, the next level down is the American Invitational Mathematics Exam (in order to be eligible to take it, you need to score 100 or more on the AMC 12 exam, formerly known as the AHSME, but my high school teachers always called it "the MAA exam")
Archive of past AIME questions
  #19  
Old 10-16-2017, 10:29 PM
Left Hand of Dorkness Left Hand of Dorkness is offline
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Quote:
Originally Posted by Some Call Me... Tim View Post
That wouldn't be optimal play by B...
SPOILER:

A: 6
Running sum is 11, which is divisible by 2n+1. A wins.
So yeah: no, I never entered the Putnam Math Competition.
  #20  
Old 10-17-2017, 10:01 AM
CalMeacham CalMeacham is offline
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I hadn't heard of it.

In high school we had a Math League. Once a month we went to another high school and took a one hour test with ten questions. You had to give them precisely the correct answer they wanted. Questions were Putnam-like, but a couple steps down in difficulty. I don't think I ever scored above 5 or 6.

One time they gave an advanced math test at my high school. I think it was on a state level. There were more problems than at a Math League meet, and you got more than an hour. I don't know which test it was.
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  #21  
Old 10-17-2017, 12:22 PM
Bullitt Bullitt is offline
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Quote:
Originally Posted by Left Hand of Dorkness View Post
How is it scored?

Also, I found some sample questions here. My high school/college-level math concepts and vocabulary are frankly pitiful, so I don't even understand how to start most of them. But I think I have one of them (okay, in typing it out I realize I don't understand it as well as I thought and would love some help):
Quote:
Consider the following game played with a deck of 2n cards numbered from 1to 2n. The deck is randomly shuffled and n cards are dealt to each of two players.

Beginning with A, the players take turns discarding one of their remaining cards and announcing its number. The game ends as soon as the sum of the numbers on the discarded cards is divisible by 2n + 1. The last person to discard wins the game. Assuming optimal strategy by both A and B, who wins?
My answer:
SPOILER:
B wins, and here's why. 2n+1 is 1 higher than the highest card in the deck. A sequence from 1 to 2n contains an even number of cards, and the cards can be rearranged into pairs with the value 2n+1. For example, if n=5, the cards are from 1-10, 2n+1=11, and you can arrange the cards into (1, 10); (2,9); (3,8); (4,7); and (5,6).

Player A may have some of these pairs, and optimal play will involve playing members of a complete pair. Continuing the example, if Player A has a hand of (1, 3, 4, 6, 8), the only complete pair is (3, 8); she should play one of these cards, knowing that player B won't have the other card to win the game.

The problem is that for every pair Player A has, Player B necessarily has a pair. In this example, Player B must have a hand of (2, 5, 7, 9, 10), with the matching pair (2, 9). Optimal play will look something like this:
A: 3
B: 2

A knows not to complete her match with (8), because then B completes his with (9) and wins. But if A plays anything that's not from a match, B can then win.

Huh. I'm losing my train here. Basically, I think B's ability to choose to play from matches or not matches, in response to A's plays, creates an unbeatable advantage. Can anyone smooth out my thinking on this, using very small words?
Quote:
Originally Posted by DPRK View Post
SPOILER:

I think the simplest way to put it is: whenever it is A's turn, no matter what card A plays B can respond so as to prevent A from winning on the next turn.

Whatever A chooses to play, it will be B's turn; A has k-1 cards in hand and B has k. That means B can leave k different totals while at most k-1 of these allow A to win.


For DPRK, k = (n/2), and I'll tweak it just a little:
SPOILER:
I think the simplest way to put it is: whenever it is A's turn, no matter what card A plays B can respond so as to either win immediately, or prevent A from winning on the next turn.


But I'm not sure if DPRK's is a definitive proof. Is it? For all n? The following shows that B always wins for n = 1 and for n = 2. Furthermore, for n = 1 and n = 2, it is impossible for A to win. B must be the winner, even if B can misplay its hand.


For n = 1, regardless of what A plays, either a 1 or a 2 (the only cards in the deck), B's play results in Σ = 3 and B wins. Trivial, but important to show. If it is true that the same player must win regardless of n, then this proves B always wins for any n, because B wins for n = 1, and the proof is complete.


Continuing, for n = 2, A can have one of six possible hands: (1,2), (1,3), (1,4), (2,3), (2,4), and (3,4).
  • If A(1,2) plays either 1 or 2, respectively, then B(3,4) plays either 4 or 3, respectively. Σ = 5 and B wins.
  • If A(1,3) plays either 1 or 3, respectively, then B(2,4) plays either 4 or 2, respectively. Σ = 5 and B wins.

A(1,4) is just a little more complicated...
  • If A(1,4) plays 1, then regardless of what B(2,3) plays, either the Σ = 3 or Σ = 4. A then plays 4. Still no winner. With B's remaining play, Σ = 10 and B wins.
  • If A(1,4) plays 4, then regardless of what B(2,3) plays, either the Σ = 6 or Σ = 7. A then plays 1. Still no winner. With B's remaining play, Σ = 10 and B wins.

Similarly for A(2,3), A(2,4), and A(3,4), B always wins:
  • Note that for {A(2,3), B(1,4)} each of their hands sum to the winning total, as for {A(1,4), B(2,3)} above -- B can only win after all four cards are played, and as above it is impossible for A to win. B must win.
  • Also note that for {A(2,4), B(1,3)} and for {A(3,4), B(1,2)}, as for A(1,2) and A(1,3) above, regardless of what A plays, B wins on the very next play. Even if B misplays, it is impossible for A to win. Again, B must win

Therefore, for n = 2, B always wins. Not only that, but A can never win -- B must win, even if B misplays the game.

This is a brute force approach. There's probably a more elegant proof!
  #22  
Old 10-17-2017, 05:11 PM
Left Hand of Dorkness Left Hand of Dorkness is offline
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My current understanding:

The way the game is written, for any particular nonwinning sum at the end of my turn, my opponent has one theoretical card that can be played to win. If there are ten cards, if I end my turn at 11, I win; but if I end it at 12, my opponent must play a 10 to win, and if I end at 13, they must play a 9, and so on.

Call the sum at the end of my turn the "end-sum"; the card I play is the "sum determiner"; the card they must play to win is the "winning card."

Each card is unique. I can choose among my cards which sum determiner to play, giving a unique end-sum for each choice, and requiring a unique winning card.

I have one more sum determiner than they have potential winning cards. If I have five cards left in my hand, that's five different possible end sums. And they only have four cards. I can choose the end sum for which they lack the winning card.
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