2023 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 | E | Linear Equations & Word Problems | Riders approaching each other close the gap at the combined speed 30 mph, so they meet after 1.5 hours; Alicia covers 18 times 1.5. | |
| 2 | A | Linear Equations & Word Problems | Move the pizza terms to one side and the orange terms to the other: 5/12 of a pizza weighs the same as 3 cups, which is 3/4 pound. | |
| 3 | A | Number Properties | A square is divisible by 5 exactly when its root is, so count multiples of 5 whose square stays below 2023, i.e. up to 40. | |
| 4 | D | Geometric Optimization | Any side of a polygon is shorter than the sum of the others, so the longest side is less than 26/2 = 13, and 12 is achievable. | |
| 5 | E | Exponents, Logarithms & Radicals | Rewrite in primes: 2^15 times 5^15 times 3^5 equals 243 followed by fifteen zeros. | |
| 6 | D | Basic Counting | Each vertex lies on 3 edges and each edge on 2 faces, so every vertex is counted 6 times: the cube's value is 6 times 21. | |
| 7 | B | Basic Probability | The running total hits 3 only through the first one, two or three rolls; list the sequences 3, 1+2, 2+1, 1+1+1. | |
| 8 | D | Linear Equations & Word Problems | 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. | |
| 9 | E | Bases & Digits | 2023 already has one 0 and one 3, so MMDD must contain exactly one 0, one 3 and a matching pair of some other digit. | |
| 10 | D | Ratios, Percents & Averages | Write total = n times mean; both scenarios give linear equations in n and the mean, which reduce to m + n = 10 and 3m + 2n = 27. | |
| 11 | C | Triangles: Area & Pythagorean | The legs satisfy a + b = sqrt 3 and a^2 + b^2 = 2, so ab = 1/2 and the ratio r satisfies r + 1/r = 4. | |
| 12 | B | Divisibility & Factors | The reversed number ends in N's hundreds digit, which must be 5; then just count multiples of 7 from 500 to 599. | |
| 13 | C | Circles | Points seeing AC at 60 degrees lie on a fixed circle through A and C, so AB is longest when it is a diameter. | |
| 14 | B | Conditional Probability & States | Only the 9 multiples of 11 can work, and for each, exactly half its divisors are multiples of 11 since 11 appears once. | |
| 15 | E | Sequences & Series | Each shaded ring between radii 2k-1 and 2k has area (4k-1)pi, so 2n circles shade (2n^2 + n)pi; find the least n with 2n^2 + n >= 2023. | |
| 16 | B | Games & Processes | With 3L players the game count is C(3L,2), and wins split 5:7, so the total must be a multiple of 12; only 36 fits both. | |
| 17 | A | Triangles: Area & Pythagorean | Leg 30 with the other leg at most 28 forces the 16-30-34 triple; then leg 12 offers 5, 9, 16, and only 9 leaves 21-28-35 integer. | |
| 18 | D | Solid Geometry | Twelve rhombi give 24 edges, Euler's formula gives 14 vertices, and 3a + 4b = 48 with a + b = 14 yields a = 8. | |
| 19 | E | Coordinate Geometry | The rotation center is equidistant from A and A' and from B and B', so it is where the two perpendicular bisectors meet: (7/2, 9/2). | |
| 20 | D | Paths & Grids | The center is in every 2x2 block; the four edge-middle squares form a cycle colored with the other 3 colors (18 ways), and the corners are then forced. | |
| 21 | D | Polynomials | Each clue is a value of P: P(1)=1 and P(0)=P(9)=P(4)=0; the cubic gives P(1)=24, so a fourth root r satisfies 24(1-r)=1. | |
| 22 | D | Circles | C3 is centered at the midpoint with radius 3/4; C4's center is on the perpendicular bisector, so a right triangle with legs 1/4, 3/4+r and hypotenuse 1-r gives r. | |
| 23 | C | Diophantine Equations | Complete the square: 4N + 400 and 4N + 529 are both perfect squares differing by 129 = 3 times 43, which factors in only two ways. | |
| 24 | C | Quadrilaterals & Polygon Areas | The 3/7 gap and the gap on the other side of a block sum to 2 because neighboring blocks share a side line, so the frame has side 3. | |
| 25 | A | Basic Probability | By symmetry greater and less are equally likely, so find the tie probability: from R, 5 vertices lie at distance 1, 5 at distance 2, 1 at distance 3. |
Problems © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. The topic tags, difficulty ratings and key insights on this page are original to this site. No problem statements are reproduced here — each links to its own page.