USAMO Practice Tests & Write-Up Playbook

Treat each set as a session: $4.5$ hours for six problems, no calculator. After finishing, read every solution in full and compare the write-up to yours line by line: did you justify the step the solution justifies? Did you handle the equality/edge case? The gap between "I saw the idea" and "I wrote a 7" is the entire game.

Graders read for GAPS: an unproved claim, a case not covered, a "clearly" that is not clear. Proof by contradiction must state the negated assumption precisely; induction must state the hypothesis and prove the base. Diagrams help but are not proofs — every geometric fact you read off a picture must be justified (angle chase, similar triangles, power of a point) in words. Numbered steps are your friend.

Graders read with a rubric that lists the key steps and the points each earns; a step counts only if it is proved on the page. So the most valuable habit is to write the solution as a sequence of labelled claims, each followed by its proof. That also protects you when the full solution fails: a clearly stated and proved claim earns its points even if the assembly is missing, while the same idea buried in a paragraph of exploration usually earns nothing.

Answer: $n = 1, 2, 4$.

Claim 1. If $n$ is odd, $v_2(3^n - 1) = 1$. Proof. $3^n - 1 = (3 - 1)(3^{n-1} + \cdots + 1)$ and the second factor is a sum of $n$ odd terms, hence odd.

Claim 2. If $n$ is even, $v_2(3^n - 1) = v_2(n) + 2$. Proof. Lifting the exponent for $p = 2$ and even $n$: $v_2(3^n - 1) = v_2(3 - 1) + v_2(3 + 1) + v_2(n) - 1 = 1 + 2 + v_2(n) - 1$.

Conclusion. For odd $n$ we need $1 \ge n$, so $n = 1$. For even $n$ we need $v_2(n) + 2 \ge n$. Since $2^{v_2(n)} \le n$, this gives $2^{n-2} \le 2^{v_2(n)} \le n$, and $2^{n-2} \le n$ holds only for $n \le 4$ (at $n = 5$, $8 > 5$, and the left side doubles while the right grows by $1$). Among even $n \le 4$: $n = 2$ ($v_2(8) = 3 \ge 2$) and $n = 4$ ($v_2(80) = 4 \ge 4$) both work. So the answer is $n = 1, 2, 4$. $\blacksquare$

How to find it. Divisibility by a prime power is a valuation inequality, and LTE computes the valuation exactly; the rest is comparing a logarithmic quantity with a linear one.

What the grader sees: a correct answer (typically worth $0$ or $1$ point on its own), a verification that $1, 2, 4$ work (this half is complete), and a claim about “large $n$” with no proof — which is the entire problem. The phrase “grows much faster” names the idea without computing the power of $2$ in $3^n - 1$; nothing on the page bounds it. Checking $n \le 12$ does not help, because a finite check never covers “for all $n$”. The fix is exactly Claims 1 and 2 of Example 1: two lines each, and they turn “much faster” into the inequality $v_2(n) + 2 \ge n$.

Claim. $2$, $3$ and $5$ each divide $n^5 - n$; since they are pairwise coprime, so does $30$.

Mod 5. By Fermat's little theorem $n^5 \equiv n \pmod 5$ for every integer $n$ (for $5 \mid n$ both sides are $0$).

Mod 3. $n^5 - n = n(n^4 - 1) = n(n^2 - 1)(n^2 + 1) = (n - 1)n(n + 1)(n^2 + 1)$, and one of three consecutive integers is divisible by $3$.

Mod 2. The same factorization contains the consecutive integers $n$ and $n + 1$, one of which is even. $\blacksquare$

The grading point: a write-up saying “$n^5 - n$ is always divisible by $2, 3, 5$ by Fermat” is incomplete — Fermat's theorem with exponent $5$ speaks about the prime $5$ only; the factors $2$ and $3$ need their own line, and the step from three primes to $30$ needs the word coprime.

How to find it. Split a composite modulus into prime powers and treat each separately; a factorization of the polynomial usually handles the small primes at once.

Every gap on this list has the same shape: a sentence that is true in the case the writer had in mind and false in general. The self-check is to read each sentence of your solution and ask “for which cases did I actually prove this?” The three examples below each contain one such sentence; the solutions repair them, and the repaired versions are what a 7 looks like.

The step uses $2n^2 \ge (n + 1)^2$, i.e. $n^2 \ge 2n + 1$, which is false for $n = 1, 2$ (it holds from $n = 3$ on). So the chain breaks between the base case and the first step where the step is valid — and indeed the claim fails: $2^2 = 4 = 2^2$, $2^3 = 8 < 9$, $2^4 = 16 = 4^2$. The correct theorem is $2^n > n^2$ for $n \ge 5$: base $32 > 25$, and the step is valid for all $n \ge 5$. The general lesson: find the smallest $n$ for which the step is valid, and make that (or later) the base.

The bound $\tfrac{2}{\sqrt x}$ depends on $x$, so it is not a lower bound for the minimum at all. AM–GM gives a constant only when the product of the terms is constant; split $x$ so that it is: $\tfrac x2 + \tfrac x2 + \tfrac1{x^2} \ge 3\sqrt[3]{\tfrac x2\cdot\tfrac x2\cdot\tfrac1{x^2}} = \tfrac{3}{\sqrt[3]4}$. Equality needs $\tfrac x2 = \tfrac1{x^2}$, i.e. $x = \sqrt[3]2$, which is allowed, so the minimum is $\tfrac{3}{\sqrt[3]4} = \tfrac{3\sqrt[3]2}{2} \approx 1.890$. Two checks every AM–GM step needs: the right side is constant, and the equality case is attainable.

“Additive functions are linear” needs a regularity hypothesis (continuity, monotonicity, boundedness on an interval); the problem gives none. The repair derives monotonicity from the second equation.

Step 1 — $f(t^2) = f(t)^2 \ge 0$, so $f \ge 0$ on $[0, \infty)$.

Step 2 — for $x \le y$, $f(y) - f(x) = f(y - x) \ge 0$ by additivity: $f$ is non-decreasing.

Step 3 — an additive non-decreasing function is linear: $f(q) = qf(1)$ for rational $q$, and squeezing any real between rationals gives $f(x) = xf(1)$.

Step 4 — then $f(1) = f(1)^2$, so $f \equiv 0$ or $f(x) = x$, and both satisfy the equations. The two-line gap was worth most of the problem.