**Below are some strategies / useful bits of information for Putnam problems.
This page is VERY much a work in progress - submissions, comments, addendums
welcome!**

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Induction**

Mathematical induction is a powerful way to prove statements. There are many problems that can be solved with it. Click here for some notes I wrote on the subject. In addition to the standard problems there, another fun instance is the Towers of Hanoi problem. For another problem, see A5 of 2004. Sometimes it helps to do the problem more generally; for this one, imagine you have a rectangle with 2r rows and 2c columns. If r=1 the problem is clear, and thus we need only show that if we can do 2r by 2c we can do 2r+2 by 2c.

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**Combinatorics**

Combinatorics is a vast and important subject. If you click here you can find some notes of mine on the binomial theorem; this, however, is but the first of many wonderful results in the subject. Some of my favorite have applications in mathematical statistics. See the first several pages of this handout (as well as this more basic handout) from my mathstats course at Brown. It includes differentiating identies (ie, how to pass from knowing Sum_n x^n to knowing Sum_n n x^n) to matching coefficients (which allows us to compute Sum_{k=0 to n} (n choose k)^2). See B2 of 2004 for a nice instance where a binomial coefficient is hidden.

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Inequalities**

Inequalities have been called the lifeblood of much of mathematics; one of the most important is that for real x, x^2 >= 0. The arithmetic mean - geometric mean (see also the Wikipedia entry) is another very useful one, as is the Cauchy-Schwarz inequality and Jensen's inequality. One example is given a, b, c > 0 prove (a/b) + (b/c) + (c/a) >= 3. MANY such problems: A2 of 1985, B2 of 1988, B1 of 1993 (mediant property), B5 of 1994, B2 of 1996 (integral test, Stirling's formula).

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In****clusion /
Exclusion**

Inclusion / exclusion is a powerful way to calculate complex probabilities from knowing simpler ones (it's also a great way to mistakenly think you've proved big results like the Riemann Hypothesis!). Here's a nice example: consider n people, each wearing a hat. If we randomly permute the order of the hats (ie, all n! possible permutations are equally likely), what is the probability that at least one person ends up with their hat? We might as well label the hats 1, 2, ..., n; the question is now equivalent to the probability a permutation has at least one fixed element. Let E(i1,i2,...,ik) (i1 < i2 < ... < ik) be the probabilities that a random permutation leaves hats i1, i2, ..., ik alone; note E(i1,...,ik) = E(j1,...,jk) -- all that matters is the number of hats left alone. The probability that at least one person has their hat is Sum_i E(i) - Sum_{i<j} E(i,j) + Sum_{i<j<k} E(i,j,k) - .... This is because the first sum gives us the probability that each person has their hat returned; however, if people i and j both get their hat back, we've counted that twice, and hence must subtract E(i,j). But if three people, say i,j,k, all get their hats back, then we've subtracted that too often: we added it three times from the first sum, and subtracted it three times from the second (from E(i,j), E(i,k) and E(j,k)). Thus we need to add it back once so that it is counted. We can simplify the expressions, and note it is Sum_i E(1) - Sum_{i<j} E(1,2) + Sum_{i<j<k} E(1,2,3) - ..., which is E(1) Sum_i 1 - E(1,2) Sum_{i<j} 1 + E(1,2,3) Sum_{i<j<k} 1 - .... A little thought gives E(1,2,...,m) = (n-m)!/n! (as the first m elements of our permutation are fixed) and Sim_{i1 < i2 < ... < im} 1 = (n choose m) = n! / m!(n-m)!. Thus our probability is (1/n)*(n/1!) - (1/n(n-1))*(n(n-1)/2!) + (1/n(n-1)(n-2))*(n(n-1)(n-2)/3!) - ... = 1/1! - 1/2! +1/3! - ... = 1 - 1 + 1/1! - 1/2! +1/3! - ... = 1 - (1 - 1/1! + 1/2! - 1/3! + ...), which for large n looks like 1 - 1/e.

ASIDE: for many probability problems, it is often easier to calculate the probability the event doesn't happen, and then subtract that from 1; if we did that here, we would find the probability that no one gets their hat is about 1/e for large n -- such a permutation is called a derangement.There are many other fun and important inclusion - exclusion problems, ranging from showing that the probability a number is square-free is about 6/pi^2 to some properties about the density of twin primes. One useful variant of this is that (1-1)^n = 0. A good example of this problem (complete with a derangement) is B5 of 1988.

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**Splitting into cases**

Oftentimes it is easier to solve a problem by breaking it into cases, as this allows us to use more information to solve each part. A great example is the following problem (I unfortunately cannot find where it is again): given Sum_n a_n converges with each a_n > 0, show Sum_n a_n^(n/(n+1)) converges. For each a_n consider two possibilities. First, consider the case when a_n < 1/2^{n+1}. In this case, a_n^(n/(n+1)) < 1/2^n, and by the geometric series formula and the comparison test, the sum of these terms converges. Thus we are left to consider the sums of terms with a_n >= 1/2^{n+1}.

.Thus a_n^{n/(n+1)} equals a_n / a_n^{1/(n+1)}. As a_n >= 1/2^{n+1} we see a_n^{1/(n+1)} >= 1/2, so 1/a_n^{1/(n+1)} <= 2. Thus a_n^{n/(n+1)} <= 2a_n, and as we have a convergent series the sum of these terms also converges. Equivalently, we can say a_n^{n/(n+1)} <= 2a_n + 1/2^n, and thus the sum still converges.Whenever you see something like n/(n+1), it's not bad to think about writing it as 1 - 1/(n+1); this is another common trick

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(Dirichlet's) Pigeon-hole Principle)**

Dirichlet's pigeon-hole principle (see also the Wikipedia entry) is another useful technique. Examples abound: B3 of 1988, A3 of 1990, A4 of 1993, A4 and A6 of 1994, B6 of 1995, B6 of 1997, A5 of 1999, B1 and B6 of 2000.