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.
The same key as a PDF 134 KB — a separate file from the paper, so one can be handed over without the other.
| # | Answer | Topic | Difficulty | Key insight |
|---|---|---|---|---|
| 1 | A | Algebraic Manipulation | Multiplying by 101 means times 100 plus one more copy, and multiplying by 99 means times 100 minus one copy. Both products then come out in one line. | |
| 2 | B | Linear Equations & Word Problems | Both trails take 69 minutes, so subtract one equation from the other. That gives a simple link between a and b: a is 1000 times b. | |
| 3 | B | Primes | If 2 is one of the five primes, the total is even and cannot be prime. So use five odd primes, and try the smallest totals first. | |
| 4 | B | Number Properties | The biggest two-digit number is 99. Twenty of them make only 1980, which is short of 2024, but twenty 99s plus a 44 make exactly 2024. | |
| 5 | D | Primes | Break 2024 into primes: 2 times 2 times 2 times 11 times 23. The factorial must include the prime 23, and 23! already includes everything else. | |
| 6 | D | Games & Processes | All 15 pairs of letters must switch order, and one swap of neighbours switches only one pair, so at least 15 swaps are needed, and 15 are enough. | |
| 7 | B | Divisibility & Factors | The word is integers, so negatives are allowed. Two negative numbers keep the product positive but make the sum small: (−1)(−6)(10) = 60 with sum 3. | |
| 8 | A | Linear Equations & Word Problems | Amy, Bomani and Charlie work the whole 105 minutes and pack 350 packages. Daria packs the other 100, which takes her 80 minutes, so she started 80 minutes before 2:45. | |
| 9 | B | Basic Counting | Split the juniors into three pairs (15 ways), split the seniors into three pairs (15 ways), then match each junior pair with a senior pair (6 ways). | |
| 10 | C | Sequences & Series | Work out the first ten or so steps. From step 6 the values repeat 20, 30, 10, so step 100 matches step 7, which is 30. | |
| 11 | D | Diophantine Equations | Square both sides and factor: (n − m)(n + m) = 49. Then list the ways to split 49 into two integer factors. | |
| 12 | E | Statistics & Data | Turn the day-to-day changes into actual scores with a running total. Then average how far each score is above 1700 instead of adding six big numbers. | |
| 13 | C | Transformations & Symmetry | Write each move as a rule for where (x, y) goes, then do each pair in both orders. Only three pairs give the same point both ways. | |
| 14 | D | Circles | The circle sits in the 120 degree corner outside the triangle. The region is a kite minus a 60 degree slice of the circle. | |
| 15 | E | Diophantine Equations | The two squares differ by 2560 = (b − a)(b + a). M is biggest when a is biggest, which needs the smallest gap b − a. | |
| 16 | D | Similar & Congruent Triangles | Every rectangle has the same long-to-short ratio r. Where rectangle 8 sits exactly on top of rectangles 9 and 1, the widths must match, and that gives r = √2. | |
| 17 | E | Basic Probability | List the three ways Team A can win: win-win, win-lose-win, lose-win-win. Add their chances, set the total to one half, and solve the quadratic for p. | |
| 18 | D | Bases & Digits | In base b, 2024 means 2(b³ + b + 2). So b³ + b + 2 must be a multiple of 8, which depends only on b's remainder mod 8. | |
| 19 | E | Sequences & Series | The middle term squared equals the product of the outer terms, so ab = 720². Make a the largest divisor of 720² below 720. | |
| 20 | C | Number Properties | Ten numbers in a row hold at most 3 members; four would force two odd members 6 apart. The pattern 1, 4, 8 in each ten reaches 608. | |
| 21 | C | Sequences & Series | Measure from the bottom-right 0. Column 5 goes up in equal steps, and the row steps themselves change by equal amounts, so each entry is P·u + v·(Q + R·u). | |
| 22 | B | Triangles: Area & Pythagorean | Every kite corner is 60, 90 or 120 degrees, so all lengths come from 30-60-90 triangles: AB = 3 + 3 and C is 3√3/2 above AB. | |
| 23 | D | Systems of Equations | Subtract two equations and factor: (a − c)(b − 1) = 13. Since 13 is prime there are only four cases, and only b = −12 works. | |
| 24 | B | Basic Probability | Each roll must use a new direction, and a direction used before must be reversed. Counting roll by roll gives 6 · 4 · 7 = 168. | |
| 25 | C | Paths & Grids | An upright middle-row toothpick forces a full top-to-bottom line. A loop has exactly two such lines, and every other middle square picks top or bottom. |
2024 AMC 10A. Problems © Mathematical Association of America, American Mathematics Competitions (maa.org/amc), licensed CC BY-NC-SA 4.0. Typeset and annotated by AMC 10 Step by Step (amcstep.com) and shared under the same licence. Not endorsed by the MAA.