The American Invitational Mathematics Examination (AIME) runs each February for students who cleared the AMC threshold. It's a completely different game: you have three hours to produce fifteen integer answers in the range \(000\) to \(999\), with no multiple choice and no partial credit. Fifteen binary outcomes.
Every AIME answer is an integer from 000 to 999. You bubble three digits (with leading zeros as needed) on the answer sheet. No fractions, no decimals, no proofs — just the number.
This constraint shapes the problems profoundly. If a problem asks "find the probability," it's phrased so the answer is \(\frac{m}{n}\) with \(\gcd(m,n)=1\), and you're asked for \(m+n\). If a geometry problem's answer is \(\sqrt{3}+2\), you're asked for \(\lfloor 100(\sqrt{3}+2)\rfloor\) or similar. Every answer is forced to be an integer \(\le 999\).
\( \text{AIME score} = (\text{number of correct answers out of 15}) \)
That's it. Each correct answer gives \(+1\) point. Wrong answers and blanks are equivalent — both give \(0\). No guess penalty, but random guessing among \(1000\) integers nets an expected \(15 \cdot \tfrac{1}{1000} = 0.015\) points. So: educated guessing is essentially worthless. Only answer when you have real confidence.
Problem: A student scores 7 on AIME. What was their AMC 12 + AIME index if their AMC 12 was \(114\)?
Step 1 — Apply the Index formula.
\( \text{Index} = 114 + 10 \cdot 7 = 184 \)
Step 2 — Contextualize. USAMO cutoffs have typically lived in \(220\)–\(235\). USAJMO cutoffs roughly \(210\)–\(225\). Index \(184\) is below both — not a USAMO year, but a strong signal (AIME \(7\) is roughly top \(25\%\) of AIME takers).
Step 3 — Plan. The AIME is the 10x lever in the index. Pushing AIME from \(7\) to \(10\) alone raises the index by \(30\) — more than pushing AMC from \(114\) to \(144\). For most students, improving AIME is the fastest path to USAMO/USAJMO.
Takeaway: An AIME point is worth ten AMC points. Train accordingly.
Problem: You've solved a problem and gotten the answer \(\frac{27}{125}\), but the problem asks for \(m+n\) where \(\frac{m}{n}\) is the probability in lowest terms. What do you bubble?
Step 1 — Verify lowest terms. \( \gcd(27, 125) = 1 \) since \(27 = 3^3\) and \(125 = 5^3\). Already in lowest terms. ✓
Step 2 — Compute the target quantity.
\( m + n = 27 + 125 = 152 \)
Step 3 — Bubble the form. On the AIME answer sheet, you bubble 152 as three digits: "1", "5", "2". (If the answer had been \(42\), you'd bubble 042.)
Important warning: Forgetting to simplify \(\frac{m}{n}\) before adding is the most common AIME error. \( 27/125 \) is already reduced — but if the problem had given \( 54/250 \), blindly computing \(54 + 250 = 304\) would be wrong.
Like AMC, AIME problems ramp. A useful mental model in 5-problem blocks:
Historically, AIME median score has hovered around \(\mathbf{5}\) correct — so scoring \(8\) puts you well above the median, and \(10+\) is USAMO territory.
AMC + 10 × AIME. Historical approximate cutoffs:
Concrete targets assuming AMC \(126\), AMC \(132\), AMC \(138\):
\( 126 + 10a \geq 225 \Rightarrow a \geq 9.9 \Rightarrow a \geq 10 \)
\( 132 + 10a \geq 225 \Rightarrow a \geq 9.3 \Rightarrow a \geq 10 \)
\( 138 + 10a \geq 225 \Rightarrow a \geq 8.7 \Rightarrow a \geq 9 \)
Reading: A solid AMC \(130+\) plus an AIME \(9\)–\(10\) is the USAMO target zone.
A student scores 9 on AIME and 120 on AMC 12. What is their USAMO qualification index?