2024 AMC 10B
All 25 problems, 75 minutes, scored 6 / 1.5 / 0. The answer key and the idea behind each problem are on this page too, folded away until you ask for them.
The problems
Show the answer key, topics and key insights
This gives away all 25 answers and the idea behind each one. Sit the paper first if you mean to.
| # | Answer | Topic | Difficulty | Key insight |
|---|---|---|---|---|
| 1 | B | Basic Counting | Position from the left plus position from the right counts the same person twice, so the line has 1013 + 1010 - 1 people. | |
| 2 | B | Algebraic Manipulation | Since 10 * 9 * 8 = 720 = 6!, the expression is 7! * 720 minus 7! * 720, which is zero. | |
| 3 | E | Absolute Value & Inequalities | 7 pi is just under 22, so |x| <= 10.99, giving the 21 integers from -10 to 10. | |
| 4 | D | Sequences & Series | Step n ends at triangular number n(n+1)/2; since 2016 < 2024 <= 2080, ball 2024 is in step 64, and 64 mod 5 = 4 means bin D. | |
| 5 | B | Sequences & Series | The total is 2500; flipping terms subtracts twice their sum, so flipped terms must exceed 1250, and the largest k odd numbers sum to k(100 - k). | |
| 6 | B | Divisibility & Factors | 2024 = 45^2 - 1 = 44 * 46, a factor pair as close to sqrt(2024) as integers allow, so the perimeter is 2(44 + 46) = 180. | |
| 7 | A | Modular Arithmetic | Factor out 7^2024 to leave 1 + 7 + 49 = 57 = 3 * 19, so the whole sum is a multiple of 19. | |
| 8 | D | Divisibility & Factors | Divisors pair up into products equal to 42, and 42 has 8 divisors, so N = 42^4, whose units digit is that of 2^4 = 16. | |
| 9 | A | Algebraic Manipulation | Square the sum: (a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca), so 0 = 30 + 2(ab+bc+ca) gives ab+bc+ca = -15, mean -5. | |
| 10 | A | Similar & Congruent Triangles | Triangles AFE and CFB are similar with ratio 1:2, so [AFE] = 1 gives [CFB] = 4, and every other region follows from area ratios. | |
| 11 | C | Similar & Congruent Triangles | Equal areas force XM = 2 ZA; the right angle at M makes WXM and MYA similar, pinning ZA = 1; subtract three corner triangles from 32. | |
| 12 | A | Basic Counting | Students correspond to distinct k-subsets of n languages; the most such subsets is C(n, n/2), and C(8,4) = 70 < 100 <= 126 = C(9,4). | |
| 13 | B | Exponents, Logarithms & Radicals | 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. | |
| 14 | B | Geometric Probability | The target is the annulus 18 <= x^2 + y^2 <= 32, area 14 pi, fully inside the diamond of area 128 (inradius sqrt32). | |
| 15 | C | Statistics & Data | The median must be one of x, y, z; the mean forces x+y+z = 11.2 or 20.2; the range forces x = 0, z = 8, or z = x+7. | |
| 16 | A | Games & Processes | Each move removes three numbers, so the count stays 2 mod 3, and odd numbers never exceed the initial 1012; both limits leave 1010. | |
| 17 | D | Arrangements with Restrictions | 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. | |
| 18 | B | Modular Arithmetic | Multiples of 5 give remainder 0; otherwise n^4 = 1+5k and two fifth powers lift this to n^100 = 1 (mod 125). | |
| 19 | C | Coordinate Geometry | Two lattice points force a rational slope, and one lattice point plus a rational slope forces infinitely many; so each row allows exactly two of the four counts. | |
| 20 | A | Arrangements with Restrictions | An L beside an R must be its mate, so no shoe touches two opposite-type shoes; 8 L/R patterns survive, each boundary locking a pair. | |
| 21 | C | Circles | Circles of radii r, s tangent to each other and the floor touch it 2sqrt(rs) apart; the third pipe fits between (1/9) or beyond (1). | |
| 22 | A | Basic Counting | Line up 16 people in four blocks with chair and secretary first: count is 16!/(4! 2^4), with 3^6 in 16! and 3^1 in 4!. | |
| 23 | B | Sequences & Series | The ratios F_{2n}/F_n are the Lucas numbers 1, 3, 4, 7, 11, ... with the same recurrence, so just add ten small terms. | |
| 24 | E | Modular Arithmetic | Over denominator 8, P(m) is an integer iff 8 divides 4m+2m^2+m^4+m^8; even m is trivial, odd m gives 4(m+1). | |
| 25 | E | Systems of Equations | 2p or 7p never equals 3p+1, so each new dimension comes from a different edge: 7p = 3q+1, 2q = 3r+1, 2r = 3p+1. |
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