First Test

Today, I took the first day of the 1992 Putnam Exam. I haven't taken a 3 hour math exam in some time, so I had to readjust. I began to lose concentration around the 1.5 hours mark, and was completely useless after an additional hour, so I basically ended up giving up 20 minutes early, because I just couldn't bring myself to do anything more. This will, I hope, improve with time and effort. I took it pretty casually, with a lot of bathroom/water breaks, and I didn't formally write up my solutions. I will be more rigorous with conditions as this process goes on.

The results themselves were actually relatively good. I solved the first and fourth problems, and basically solved the third, but made a stupid in error in getting the final answer. I will post about how Putnam is evaluated later, but I give myself 2.5 problems, which is 25 points. Had I bothered to check that error (which I would have in real competition settings), and written down some preliminary work in the other 3 problems, I would have gotten a 33. Looking at the competition statistics, this is already enough for top 200, (cutoff was 32), despite the fact that I only did the first half of the competition! This is pretty awesome, but unfortunately, it doesn't mean all that much.

The only real conclusion I can draw here is that the competition got significantly more difficult from 1992 to 2013. (In general, competitions get harder as time goes by, because more training materials are developed, and people put a lot more effort into practicing). I should probably concentrate on exams after the year 2000.

Sample Problem:


Prove that f(n) = 1 − n is the only integer-valued function defined on the
integers that satisfies the following conditions:
(i) f(f(n)) = n, for all integers n;
(ii) f(f(n + 2) + 2) = n for all integers n;
(iii) f(0) = 1.

Our approach will be to prove any function, f(n), with these properties must be equal to 1-n.

We are going to use a form of induction, similar to what was described in the previous section.

First, we'll prove it for even positive integers, then odd positive integers.

Since f(f(n))=n, f is it's own inverse, so
applying i) to ii), we get:
f(n+2)+2=f(n)

We are now going to use induction, except adding 2 instead of 1, to prove the statement for the even numbers.
Base case:
f(0)=1=1-0

if f(n)=1-n
f(n+2)=f(n)-2=1-n-2=1-(n+2)
So it is true for 0, therefore it is true for 2,4,etc.

Applying i) to iii), we get f(1)=0, so we can make the same argument for odd positive  integers. (f(f(0)=f(1))

For negative integers, note that when n is positive f(n+1)=1-(n+1)=-n, so f(-n)=f(f(n+1))=n+1=1-(-n), so it is true for negative integers, and we are done.