Complete Syllabus & Gap Fillers

Study Sheet

Complete Syllabus & Gap Fillers

Everything the AIME tests, with the pieces that were missing

What the AIME Tests

Concept
The checklist

Integer answers 000000–999999, fifteen problems, three hours. Number theory (CRT, orders, totients, Legendre/Kummer, base representations, digit sums, LTE-lite); algebra (Vieta and Newton sums, symmetric substitutions, functional equations, floor/fractional-part equations, logs and exponentials); counting (recursion, bijections, inclusion–exclusion with structure, expected value, generating functions in easy cases, roots-of-unity filter); geometry (coordinates, complex numbers, mass points, Ceva/Menelaus, power of a point, trig in triangles, 3D solids); sequences (telescoping, periodic recurrences, characteristic equations); trigonometry (product-to-sum, roots of unity products like ∏sin⁡kπn\prod\sin\tfrac{k\pi}{n}).

Reminder — Vieta's formulas:  r1+r2=−ba,r1r2=car_1+r_2=-\frac{b}{a},\qquad r_1 r_2=\frac{c}{a}

Gap Fillers

Tip
Floor functions and base representations

⌊x⌋\lfloor x\rfloor problems are block counts: ∑k=12026⌊k7⌋\sum_{k=1}^{2026}\lfloor\tfrac k7\rfloor groups kk by quotient; "nn divisible by ⌊n⌋\lfloor\sqrt n\rfloor" gives exactly three per square block. Base-bb digit conditions are bijections with strings: base-3 numbers with no digit 22 are binary strings read in base 33.

Tip
Iterated functions and periodicity

Digit sums preserve residues mod 99 and shrink fast — f(f(f(72026)))f(f(f(7^{2026}))) is a single digit found by size bounds plus mod 99. Rational recurrences like xn+1=xn+11−xnx_{n+1} = \tfrac{x_n + 1}{1 - x_n} are tangent-addition in disguise and have small periods; compute one cycle and reduce the index.

Tip
Generating functions, the easy cases

The coefficient of xnx^n in (1+x)a(1+x)b(1 + x)^a(1 + x)^b is Vandermonde (∑(ak)(bn−k)=(a+bn)\sum\binom ak\binom{b}{n-k} = \binom{a+b}{n}; ∑(10k)2=(2010)\sum\binom{10}{k}^2 = \binom{20}{10}). Stars and bars is the coefficient in (1−x)−k(1 - x)^{-k}. Differentiating ∑xn\sum x^n evaluates ∑kxk\sum kx^k — how ∑k(20k)=20⋅219\sum k\binom{20}{k} = 20\cdot2^{19} and ∑k2k=2\sum\tfrac{k}{2^k} = 2 both arise.