2024 AMC 10A
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 | A | Algebraic Manipulation | Peel off a power of ten: 9901·101 = 990100 + 9901 = 1000001 and 99·10101 = 1010100 − 10101 = 999999, so the difference is 2. | |
| 2 | B | Linear Equations & Word Problems | Subtracting the two 69-minute equations gives a = 1000b, so b = 0.03 and a = 30; then evaluate 4.2a + 4000b. | |
| 3 | B | Primes | Using 2 makes the sum even, so take five odd primes; 3+5+7+11+13 = 39 is composite, 41 is unreachable, and 3+5+7+11+17 = 43 is prime. | |
| 4 | B | Number Properties | Twenty two-digit numbers total at most 20·99 = 1980 < 2024, while twenty 99s plus one 44 reach 2024 exactly, so 21 is the least. | |
| 5 | D | Primes | 2024 = 2³·11·23, so n! needs the prime factor 23; 23! already contains 2³ and 11, hence n = 23. | |
| 6 | D | Games & Processes | Each adjacent swap reverses the order of exactly one pair of letters, and all C(6,2) = 15 pairs must be reversed, so 15 swaps are necessary and sufficient. | |
| 7 | B | Divisibility & Factors | Allow two negative factors: (−1)(−6)(10) = 60 has sum 3, and checking the factor triples shows a positive sum of 1 or 2 is impossible. | |
| 8 | A | Linear Equations & Word Problems | 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. | |
| 9 | B | Basic Counting | Split the juniors into three unordered pairs (15 ways), the seniors likewise (15 ways), then match junior pairs with senior pairs (3! ways): 15·15·6 = 1350. | |
| 10 | C | Sequences & Series | Iterate: 100→110→120→40→50→60→20→30→10→20→…; from step 6 the values cycle 20, 30, 10 with period 3, and step 100 ≡ step 7 gives 30. | |
| 11 | D | Diophantine Equations | Square to get n² − m² = (n−m)(n+m) = 49 with m ≥ 0; the factor pairs give (n, m) = (±25, 24) and (±7, 0), four pairs. | |
| 12 | E | Statistics & Data | Track deviations from 1700: the six scores are 1700 + (0, 80, −10, −20, 40, 0), and the deviations average 90/6 = 15, so the mean is 1715. | |
| 13 | C | Transformations & Symmetry | Write each map as a formula on (x, y): the dilation commutes with the rotation and the reflection, the translation commutes only with the reflection; the other three pairs fail. | |
| 14 | D | Circles | The circle nestles in the 120° angle between the triangle's side and ℓ; the vertex-to-center kite has area 48√3 and the removed sector is 60°, i.e. 24π. | |
| 15 | E | Diophantine Equations | Subtracting the two squares gives (b−a)(b+a) = 2560; M is largest when b − a is smallest (2), so a = 639 and M = 639² − 1213. | |
| 16 | D | Similar & Congruent Triangles | With long-to-short ratio r, a rectangle of area S has short side √(S/r); rectangle 8 spanning 9 and 1 forces r = √2, then AB² = 200r. | |
| 17 | E | Basic Probability | Summing the three winning scenarios, (2/3)p + (2/3)(1−p)p + (1/3)p² = 1/2 gives 2p² − 8p + 3 = 0, so p = (4 − √10)/2. | |
| 18 | D | Bases & Digits | Because 2024_b = 2(b³+b+2), the condition is b³+b+2 ≡ 0 (mod 8), which holds exactly when b ≡ 3, 6, 7 (mod 8). | |
| 19 | E | Sequences & Series | The ratio 720/a = b/720 = p/q in lowest terms forces p and q to divide 720; the smallest such ratio above 1 is 16/15, giving b = 768. | |
| 20 | C | Number Properties | Ten consecutive integers hold at most 3 elements (four would be spaced exactly 3 apart, putting two odds 6 apart); the pattern 1, 4, 8 per block gives 608. | |
| 21 | C | Sequences & Series | Rows and columns all arithmetic makes the entry a bilinear function A + Bi + Cj + Dij; the four given entries fix the constants, and (1,2) gives 29. | |
| 22 | B | Triangles: Area & Pythagorean | AB = 3 + 3 is the sum of two bases of isosceles triangles with legs √3 and apex 120°, and C lies 3√3/2 above AB, giving area 9√3/2. | |
| 23 | D | Systems of Equations | 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. | |
| 24 | B | Basic Probability | Four steps trace distinct edges of one cube iff consecutive steps use different axes and repeated axes alternate sign: 24 axis patterns, 2^(axes used) sign choices, total 168/1296. | |
| 25 | C | Paths & Grids | A toothpick on a middle cell's vertical side forces a full-height crossing; the loop has exactly two crossings, and each remaining middle cell independently chooses top or bottom. |
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