Everything the Putnam Tests

Almost all scores on a problem are $0$, $1$, $9$, or $10$ — graders reward COMPLETE, rigorous solutions and give almost nothing for partial ideas. A-1 and B-1 are accessible to anyone who has finished this course; A-6 and B-6 are research-adjacent. The winning strategy is not to attempt everything: it is to find the two or three problems you can finish, write them so cleanly that a skeptical grader cannot object, and treat the rest as bonus.

Analysis: sequences and series, limits as Riemann sums or derivatives, the intermediate and mean value theorems, Taylor with remainder, and — the Putnam signature — definite integrals with no antiderivative. Linear algebra: structured determinants, eigenvalues and traces, rank, Cayley–Hamilton, matrices over finite fields. Algebra and polynomials: Vieta and Newton, roots of unity, interpolation, functional equations, the definitions of groups and rings. Number theory: orders, valuations, quadratic residues, Diophantine equations, divisor sums. Combinatorics and probability: bijections, generating functions, recurrences, inclusion–exclusion, expectation by indicators, random walks. Geometry: convexity, lattice points, complex numbers, vectors.

Every unit ends with a practice pool of hand-verified Putnam-level problems; the integral units share the non-elementary pool with the Calculus course. The Linear Algebra and Multivariable courses on this site are the prerequisites this course assumes.

Most lost points on the easier problems are not wrong ideas but unfinished contracts. A "find all" answer that verifies the solutions but never excludes others proves half the statement, and graders score half a proof close to zero. The first minute on a problem is well spent rewriting the question as the two or three statements you must prove, each of which will become a paragraph of the write-up.

The contract: show that the listed values solve the equation, and that no other real $x$ does.

Step 1 — $x = 1$ works: $2 + 3 = 5$.

Step 2 — divide by $5^x > 0$: the equation is equivalent to $g(x) = \left(\tfrac25\right)^x + \left(\tfrac35\right)^x = 1$.

Step 3 — each term is an exponential with base less than $1$, hence strictly decreasing, so $g$ is strictly decreasing and takes the value $1$ at most once.

Step 4 — so $x = 1$ is the only solution. $\blacksquare$

How to find it. Guessing $x = 1$ takes a second; the work is the uniqueness half, and dividing by the largest term is the standard way to turn a sum of exponentials into something monotone.

The contract: answer yes or no, and prove it. The answer is no.

Step 1 — $f$ is injective: if $f(a) = f(b)$ then $-a = f(f(a)) = f(f(b)) = -b$.

Step 2 — a continuous injective function on an interval is strictly monotone (if it went up and then down, the IVT would give two points with the same value).

Step 3 — if $f$ is increasing, $f \circ f$ is increasing; if $f$ is decreasing, $f \circ f$ is increasing too (two reversals). Either way $f(f(x))$ is increasing, but $-x$ is decreasing: contradiction. $\blacksquare$ The hypothesis "continuous" is essential: without it such $f$ exist, e.g. $f(0) = 0$, $f(x) = x + 1$ on each $(2k, 2k + 1]$ and $f(x) = -(x - 1)$ on each $(2k + 1, 2k + 2]$ for $k \ge 0$, and $f(x) = -f(-x)$ for $x < 0$.

How to find it. Try to build one and watch where it fails: $f$ must permute the reals in cycles $x \to f(x) \to -x \to -f(x) \to x$ of length $4$, and a continuous map cannot do that because it cannot reverse the order of the line an odd number of times in two steps.

The contract: show the constants work, and that nothing else does. Constants clearly satisfy the equation.

Step 1 — $f$ is even: $f(-x) = f(x^2) = f(x)$.

Step 2 — for $x > 0$, apply the equation to $\sqrt x, \sqrt[4]x, \ldots$: $f(x) = f\!\left(x^{1/2}\right) = f\!\left(x^{1/4}\right) = \cdots = f\!\left(x^{1/2^n}\right)$ for every $n$. Since $x^{1/2^n} \to 1$ and $f$ is continuous at $1$, $f(x) = \lim_n f\!\left(x^{1/2^n}\right) = f(1)$.

Step 3 — by evenness $f(x) = f(1)$ for $x < 0$, and by continuity at $0$, $f(0) = \lim_{x\to0^+}f(x) = f(1)$. So $f$ is constant. $\blacksquare$

How to find it. The equation links $x$ to $x^2$, so iterate it in the direction that CONVERGES: squaring sends points toward $0$ or $\infty$, but square roots send every positive number toward $1$, where continuity can be applied.

The six pools share almost nothing in content, but each has a small number of opening moves that crack its accessible problems. The three problems below are A1/B1-level representatives; notice that in each, the first move is routine and the solution's length is spent on the finish — the part a grader actually reads.

Step 1 — row operations. For $i = 1, \ldots, n - 1$ replace row $i$ by (row $i$) $-$ (row $i + 1$), using the ORIGINAL rows (do it from the top down). The new row $i$ has entries $|i - j| - |i + 1 - j|$, which is $-1$ for $j \le i$ and $+1$ for $j \ge i + 1$. The last row stays $(n - 1, n - 2, \ldots, 1, 0)$. The determinant is unchanged.

Step 2 — column operations: add the last column to each of the others. In rows $1, \ldots, n - 1$ the last entry is $+1$, so the entries become $0$ for $j \le i$ and $2$ for $i < j < n$, with $1$ in the last column; the last row, whose last entry is $0$, is unchanged.

Step 3 — the first column is now zero except for the entry $n - 1$ at the bottom. Expanding along it gives $(-1)^{n+1}(n - 1)$ times the minor on rows $1, \ldots, n - 1$ and columns $2, \ldots, n$, which is upper triangular with diagonal $2, 2, \ldots, 2, 1$ ($n - 2$ twos).

Step 4 — $\det A = (-1)^{n-1}(n - 1)2^{n-2}$. Check: $n = 2$ gives $\det\begin{pmatrix}0&1\\1&0\end{pmatrix} = -1$, and $n = 3$ gives $4$.

How to find it. Entries depending on $|i - j|$ change by $\pm1$ from one row to the next, so differences of adjacent rows are made of $\pm1$ — far simpler than the rows themselves. That is the row-operation reflex from the determinants topic.

Step 1 — write $X = \sum_iI_i$, where $I_i$ is $1$ if $i$ is fixed. Each $E[I_i] = \tfrac1n$, so $E[X] = 1$ by linearity.

Step 2 — square the sum: $X^2 = \sum_iI_i^2 + \sum_{i\ne j}I_iI_j = X + \sum_{i\ne j}I_iI_j$, using $I_i^2 = I_i$.

Step 3 — for $i \ne j$, $E[I_iI_j] = P(i \text{ and } j \text{ fixed}) = \tfrac{(n-2)!}{n!} = \tfrac{1}{n(n-1)}$, and there are $n(n - 1)$ ordered pairs, so $E\left[\sum_{i\ne j}I_iI_j\right] = 1$.

Step 4 — $E[X^2] = 1 + 1 = 2$, and the variance is $2 - 1^2 = 1$, for every $n \ge 2$. The case $n = 4$ in the figure confirms both numbers.

How to find it. Linearity handles $E[X]$; for $E[X^2]$, expand the square of the indicator sum into pairs. Pairs of indicators are the standard second move, and they turn a variance into a count of pairs.

Step 1 — claim: $F_0F_1\cdots F_{n-1} = F_n - 2$. For $n = 1$: $F_0 = 3 = 5 - 2$. Inductively, $(F_n - 2)F_n = \left(2^{2^n} - 1\right)\left(2^{2^n} + 1\right) = 2^{2^{n+1}} - 1 = F_{n+1} - 2$.

Step 2 — for $m < n$, any common divisor $d$ of $F_m$ and $F_n$ divides the product $F_0\cdots F_{n-1}$ (which contains $F_m$) and $F_n$, hence divides their difference $2$. The $F$'s are odd, so $d = 1$.

Step 3 — each $F_n$ has a prime factor, and by Step 2 these prime factors are all different: infinitely many primes. $\blacksquare$

How to find it. Coprimality questions ask for a relation that expresses one number in terms of the others plus a small constant. Products of Fermat numbers telescope through $(a - 1)(a + 1) = a^2 - 1$, which is the relation.