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2019 AMC 10A — answer key
The answer to each problem, with the topic it tests and the idea it turns on.
2019 AMC 10A — answer key
amcstep.com| # | Answer | Topic and the idea it turns on |
|---|---|---|
| 1 | C | Exponents, Logarithms & Radicals — Work each tower from the top down. The first becomes 0 to the power 1, which is 0, then 2 to the power 0, which is 1. |
| 2 | A | Divisibility & Factors — 15! already contains three 5s and plenty of 2s, so it is a multiple of 1000. So is 20!, and so is their difference. |
| 3 | D | Linear Equations & Word Problems — With Bonita b and Ana b squared, last year's condition is b squared minus 1 = 5(b minus 1). Cancel the shared factor b minus 1. |
| 4 | B | Sets, Estimation & Miscellaneous — Build the worst draw with no 15 of a color: 14 of each big color plus every small-color ball, 75 in all. One more forces it. |
| 5 | D | Number Properties — The integers may be negative. A run from -44 up to 44 adds to 0, so adding 45 on the end gives sum 45 with 90 numbers in the run. |
| 6 | C | Circles — A point equally far from all four corners is the center of a circle through them. That needs opposite angles adding to 180 degrees. |
| 7 | C | Coordinate Geometry — Write the two lines through (2,2), find where each meets x + y = 10, then use the side on that line as the base. |
| 8 | C | Transformations & Symmetry — Watch the little tails. A mirror flip changes which way a tail slants; a half-turn does not. The lower row looks exactly like the upper row turned halfway around. |
| 9 | B | Divisibility & Factors — Dividing n! by n(n+1)/2 leaves 2 times (n-1)! over n+1. This fails to be whole only when n+1 is a prime, and the largest prime below 1000 is 997. |
| 10 | C | Paths & Grids — Each grid line crossed means a new tile. The bug crosses 16 lines one way and 9 the other, never two at once, so it visits 26 tiles. |
| 11 | C | Divisibility & Factors — 201 = 3 times 67, so each divisor is a power of 3 times a power of 67. Count squares (25) and cubes (16), then remove the 4 counted twice. |
| 12 | E | Statistics & Data — The modes are 1 to 28, so d = 14.5. The 183rd value is 16, so M = 16. Missing large dates pull the mean just below 16. |
| 13 | D | Circles — Any point on a circle sees the diameter at a right angle, so BD and CE are both heights of the triangle. Then triangle BFC has angles 50 and 20. |
| 14 | D | Angles & Polygons — Four lines give at most 6 crossing points. Parallel lines and three lines through one point reduce that. Every count from 0 to 6 can happen except 2. |
| 15 | E | Sequences & Series — Flip every term upside down. The flipped terms go up by the same amount each time: 1, 7/3, 11/3, 15/3, and so on. |
| 16 | A | Circles — Join the centers of touching circles; each join is 2 long. The top center is 2 times root 3 above the middle, so the big radius is that plus 1. |
| 17 | D | Basic Counting — Stack all 9 cubes and then take off the top one. Every 8-cube tower comes from exactly one 9-cube stack, so just count 9-cube stacks. |
| 18 | D | Bases & Digits — In base ten 0.2323... is 23/99; in base k it is (2k + 3) over (k squared minus 1). So 51 must divide k squared minus 1. |
| 19 | B | Algebraic Manipulation — Pair outside and inside factors: both give x squared plus 5x, plus a number. Call that u; the product is (u + 4)(u + 6), never below minus 1. |
| 20 | B | Basic Probability — Only odd or even matters. The five odd numbers must fill one whole row and one whole column, a cross shape. There are 9 crosses out of 126 odd-even patterns. |
| 21 | D | Solid Geometry — The sphere cuts the triangle's plane in a circle that touches all three sides, which is the inscribed circle, radius 4. Then use a right triangle with hypotenuse 6. |
| 22 | B | Conditional Probability & States — Each number is 0 (chance 1/4), 1 (chance 1/4) or a random point in between (chance 1/2). Split into cases by type and add: 1/8 + 1/4 + 1/16. |
| 23 | C | Sequences & Series — Tadd speaks in every third turn, saying 1, 4, 7, ... numbers. Add these to find which turn holds his 2019th number, then find where that turn starts. |
| 24 | B | Polynomials — Clear the fractions and plug in each root: 1/A is the product of p minus each other root. Adding the three needs only the sum and pair-products of the roots. |
| 25 | D | Divisibility & Factors — Counting ways to split n squared people into n teams shows the fraction is a whole number times (n-1)!/n. Only primes and n = 4 fail. |
2019 AMC 10A — answer key. 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.