What Each Problem Position Tests
Based on 1300 problems from 52 contests. Topic shares use each problem's primary topic.
Problems 1-10
Must-solve target for an AIME qualifier: All ten. A qualifier has zero wrong answers here; spend the last five minutes of the exam re-checking them.
Median difficulty 2.0, typical solve time 1.9 minutes.
| Category | share |
|---|---|
| Algebra | 49.6% |
| Number Theory | 16.5% |
| Counting & Probability | 7.7% |
| Geometry | 19.4% |
| Logic, Statistics & Miscellaneous | 6.7% |
Most frequent topics:
- Linear Equations & Word Problems (75 problems) — topic page
- Ratios, Percents & Averages (63 problems) — topic page
- Exponents, Logarithms & Radicals (25 problems) — topic page
- Sequences & Series (24 problems) — topic page
- Triangles: Area & Pythagorean (23 problems) — topic page
- Circles (22 problems) — topic page
- Algebraic Manipulation (22 problems) — topic page
- Divisibility & Factors (21 problems) — topic page
Most useful techniques: direct_computation (364), plug_in_values (51), algebraic_identity (40), bounding_estimation (39), brute_force_enumeration (32), extremal_principle (23)
Representative problems:
- 2025 AMC 10A #7 — The Remainder Theorem turns the two divisions into P(1) = 4 and P(2) = 6, a two-by-two linear system in a and b.
- 2024 AMC 10A #8 — Amy, Bomani and Charlie pack 10 per 3 minutes for all 105 minutes (350 packages); Daria's remaining 100 at 5 per 4 minutes take 80 minutes, ending at 2:45.
- 2023 AMC 10A #8 — A linear scale preserves proportions: 200 is 90/240 = 3/8 of the way from 110 to 350, so it is 3/8 of the way from 0 to 100.
- 2022 AMC 10B #7 — By Vieta the roots multiply to 36 and k is minus their sum; list the factor pairs of 36, drop 6·6, and remember the negative pairs.
- 2021 AMC 10A #6 — Name the half-trail length d; Chantal's three legs take d/4 + d/2 + d/3 = 13d/12 hours, during which Jean covers exactly d.
- 2020 AMC 10A #7 — Whatever the arrangement, the five rows together contain every number once, so the common sum is (total of -10 through 14)/5 = 50/5 = 10.
- 2019 AMC 10B #9 — For x at least 0 both terms agree, giving 0; for a negative non-integer the floor rounds away from zero, so the difference is exactly -1.
- 2018 AMC 10A #8 — Express every count in terms of the number of nickels x: dimes x + 3, quarters 20 - 2x; the value equation then has one unknown.
- 2017 AMC 10A #9 — Compute each leg as distance over speed; Penny rides the loop backwards, so her uphill leg is the 15 km stretch and her downhill leg is the 10 km one.
- 2016 AMC 10A #8 — Reverse each crossing: add the 40-coin toll back, then halve; three reversals from 0 give 20, 30, 35.
Problems 11-15
Must-solve target for an AIME qualifier: All five, or four with one blank. These are one-idea problems; the idea is almost always on the topic pages linked below.
Median difficulty 3.0, typical solve time 3.4 minutes.
| Category | share |
|---|---|
| Algebra | 30.4% |
| Number Theory | 18.1% |
| Counting & Probability | 15.8% |
| Geometry | 28.8% |
| Logic, Statistics & Miscellaneous | 6.9% |
Most frequent topics:
- Linear Equations & Word Problems (22 problems) — topic page
- Circles (22 problems) — topic page
- Sequences & Series (15 problems) — topic page
- Triangles: Area & Pythagorean (15 problems) — topic page
- Basic Probability (14 problems) — topic page
- Ratios, Percents & Averages (12 problems) — topic page
- Algebraic Manipulation (10 problems) — topic page
- Diophantine Equations (10 problems) — topic page
Most useful techniques: direct_computation (125), bounding_estimation (34), algebraic_identity (33), casework (23), geometric_decomposition (22), symmetry (22)
Representative problems:
- 2025 AMC 10A #11 — The geometric sequence starts at 1, so z = p^3; the arithmetic one forces 3 | z - 1, and p^3 = p mod 3 makes p = 4 smallest.
- 2024 AMC 10B #13 — sqrt(1183) = 13 sqrt(7) forces sqrt(x) = m sqrt(7), sqrt(y) = n sqrt(7), m + n = 13; then 7(m^2 + n^2) is least at 6, 7.
- 2023 AMC 10B #12 — P is positive for x > 10 and its sign flips only when crossing a root of odd multiplicity; track the flips at 9, 7, 5, 3, 1.
- 2022 AMC 10B #15 — With d = 2, S_n = n(a + n − 1), so S_{3n}/S_n is constant only when a = 1; then S_n = n² and S_20 = 400.
- 2021 AMC Fall 10A #11 — Measure in Emily's steps: walking with the ship she gains 210 - 210k, against it 42 + 42k, where k is the ship's speed ratio; both equal the length.
- 2020 AMC 10A #14 — Combine the middle fractions into (x^5 + y^5)/(xy)^2 and build x^5 + y^5 from the power sums using x + y = 4 and xy = -2.
- 2019 AMC 10A #15 — Flip the recursion: 1/a_n = 2/a_(n-1) - 1/a_(n-2), so the reciprocals form an arithmetic sequence with first term 1 and common difference 4/3.
- 2018 AMC 10A #14 — The numerator equals 81 times the denominator minus 65 * 2^96; that leftover is positive yet smaller than the denominator, so the quotient is just below 81.
- 2017 AMC 10A #14 — Turn the percent statements into 5m = A - s and 20s = A - m; solving gives m = 19A/99 and s = 4A/99, total 23A/99.
- 2016 AMC 10A #14 — In 2a + 3b = 2016 the number of threes must be even, so b = 2k with 0 <= k <= 336, giving 337 solutions.
Problems 16-20
Must-solve target for an AIME qualifier: About three of five. This band decides qualification. Pick the two you recognize, solve them fully, and leave the rest blank unless you can eliminate choices.
Median difficulty 3.0, typical solve time 4.5 minutes.
| Category | share |
|---|---|
| Algebra | 18.5% |
| Number Theory | 15.8% |
| Counting & Probability | 23.8% |
| Geometry | 38.8% |
| Logic, Statistics & Miscellaneous | 3.1% |
Most frequent topics:
- Circles (23 problems) — topic page
- Basic Probability (16 problems) — topic page
- Quadrilaterals & Polygon Areas (15 problems) — topic page
- Solid Geometry (14 problems) — topic page
- Sequences & Series (14 problems) — topic page
- Coordinate Geometry (13 problems) — topic page
- Bases & Digits (12 problems) — topic page
- Basic Counting (12 problems) — topic page
Most useful techniques: direct_computation (96), symmetry (56), geometric_decomposition (42), casework (39), auxiliary_lines (31), algebraic_identity (28)
Representative problems:
- 2025 AMC 10A #18 — The harmonic mean needs only the reciprocals' sum, and Vieta on kx^2 - 4x - 3 gives 1/p + 1/q = -4/3 for every k.
- 2024 AMC 10B #17 — Split by the size k of the single tie: choose the tied snails, then order the 6-k blocks: C(5,k)(6-k)!, plus 5! untied.
- 2023 AMC 10B #17 — The diagonal is sqrt(a^2+b^2+c^2) = sqrt((a+b+c)^2 - 2(ab+bc+ca)); the edge sum and face sum give both pieces without finding a, b, c.
- 2022 AMC 10A #16 — The new volume is (r+2)(s+2)(t+2): expand it as rst + 2(rs+st+tr) + 4(r+s+t) + 8 and read every symmetric sum from Vieta.
- 2021 AMC Fall 10A #20 — Both discriminants must be nonpositive: b^2 <= 4c and c^2 <= 4b, which force b, c <= 4; checking the 16 small pairs leaves 6.
- 2020 AMC 10A #17 — P(n) is negative exactly when an odd number of factors are negative, i.e. n lies strictly between k^2 and (k+1)^2 for odd k; each such gap holds 2k integers.
- 2019 AMC 10B #18 — At the limit the two turning points reproduce each other: A = B/4 and B = A + (3/4)(2 - A); solving gives A = 2/5, B = 8/5.
- 2018 AMC 10B #20 — Adding consecutive recursions gives f(n) = f(n-6) + 6, so f(2018) = f(2) + 6 * 336 = 2017.
- 2017 AMC 10B #20 — 21! has 2^18 in its factorization, so the exponent of 2 in a divisor takes 19 equally sized values, and odd divisors are the one value 0.
- 2016 AMC 10B #16 — With ar = 1 the sum is S = 1/(r(1-r)); the denominator peaks at 1/4 when r = 1/2, so S is at least 4.
Problems 21-25
Must-solve target for an AIME qualifier: Zero or one. Attempt only with time left after checking 1–15; a correct guess after eliminating two choices is worth trying.
Median difficulty 4.0, typical solve time 6.6 minutes.
| Category | share |
|---|---|
| Algebra | 13.1% |
| Number Theory | 22.7% |
| Counting & Probability | 28.1% |
| Geometry | 33.5% |
| Logic, Statistics & Miscellaneous | 2.7% |
Most frequent topics:
- Circles (26 problems) — topic page
- Basic Probability (16 problems) — topic page
- Solid Geometry (16 problems) — topic page
- Basic Counting (15 problems) — topic page
- Diophantine Equations (13 problems) — topic page
- Sequences & Series (12 problems) — topic page
- Modular Arithmetic (11 problems) — topic page
- Divisibility & Factors (11 problems) — topic page
Most useful techniques: direct_computation (73), casework (55), symmetry (46), auxiliary_lines (41), algebraic_identity (39), bounding_estimation (36)
Representative problems:
- 2025 AMC 10B #23 — Both fillings are linear in the row and column indices, so setting 91(r-1)+c equal to 141(c-1)+r reduces to 9r = 14c - 5 and hence c = 1 mod 9.
- 2024 AMC 10A #23 — Subtracting pairs of equations factors: (b−1)(a−c) = 13, so b ∈ {2, 0, 14, −12}; only b = −12 works, giving (−9, −12, −8) and 276.
- 2023 AMC 10B #23 — The true sum is 221 or 223, so n divides 442 or 446; d >= 2 forces n <= 14, leaving n = 13, d = 2, a = 5.
- 2022 AMC 10B #21 — Write P = (x²+1)(ax+b) + 2x+1, reduce modulo x²+x+1 with x³ = 1, and match to x+2: a = 1, b = 2.
- 2021 AMC Fall 10A #23 — Since f_1(n) = 2d(n) <= 20 here, iterating small even numbers shows only 12, 18, 20 reach 12; so count n <= 50 with d(n) in {6, 9, 10}.
- 2020 AMC 10B #22 — The divisor times 2^101 - 2^51 + 1 equals 2^202 + 1 by difference of squares, so 2^202 is -1 modulo the divisor.
- 2019 AMC 10A #23 — Tadd's turns are rounds 1, 4, 7, ..., so after m turns he has said m(3m - 1)/2 numbers; his 2019th number falls in his 37th turn, round 109.
- 2018 AMC 10B #21 — The next divisor d must share a factor with 323 = 17 * 19, since otherwise 323d > 9999 divides n; the smallest such d is 340.
- 2017 AMC 10A #24 — Three shared distinct roots force f(x) = g(x)(x - r); the x coefficient gives r = -90, then a = -89 and f(1) = g(1)(1 - r).
- 2016 AMC 10A #23 — Substituting c = b and then b = a in the rules shows a diamond b = a/b, so the equation is 336x = 100.