2019 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 | C | Exponents, Logarithms & Radicals | Evaluate the towers from the top down: 1^9 = 1 makes 0^1 = 0, so the first term is 2^0 = 1; the second is a power of 1. | |
| 2 | A | Divisibility & Factors | Both 15! and 20! contain the factors 2, 5, 10 and more, so each is a multiple of 1000; their difference ends in three zeros. | |
| 3 | D | Linear Equations & Word Problems | With Bonita's age b, Ana is b^2 and last year b^2-1 = 5(b-1); cancel the common factor b-1 to get b+1 = 5. | |
| 4 | B | Sets, Estimation & Miscellaneous | Worst case: draw 14 of each color that has at least 14 balls and every ball of the small colors, 75 in all; the next ball forces a 15th. | |
| 5 | D | Number Properties | Negative integers are allowed: the run from -44 through 45 cancels in pairs and leaves 45, using 90 integers. | |
| 6 | C | Circles | A point equidistant from all four vertices is a circumcenter, so ask which quadrilaterals are cyclic: opposite angles must add to 180 degrees. | |
| 7 | C | Coordinate Geometry | The other vertices are (6,4) and (4,6); that side has length 2 sqrt 2 and the apex (2,2) is 3 sqrt 2 from the line x+y=10. | |
| 8 | C | Transformations & Symmetry | Lower squares are offset with reversed tails, so a half-turn about a point of the line and a translation by one period work; both reflections fail. | |
| 9 | B | Divisibility & Factors | n(n+1)/2 divides n! exactly when n+1 divides 2(n-1)!, which fails only if n+1 is prime; the largest three-digit n with n+1 prime is 996. | |
| 10 | C | Paths & Grids | Each time the diagonal crosses a grid line it enters a new tile; it crosses 16 vertical and 9 horizontal lines, never at a corner since gcd(10,17) = 1. | |
| 11 | C | Divisibility & Factors | Since 201 = 3 x 67, divisors are 3^a 67^b; squares need both exponents even (25), cubes need multiples of 3 (16), overlap is sixth powers (4). | |
| 12 | E | Statistics & Data | Modes are 1 through 28, so d = 14.5; the 183rd value is 16, so M = 16; missing 29s, 30s, 31s pull the mean just below 16. | |
| 13 | D | Circles | Angles inscribed in the semicircle on BC are right angles, so BD and CE are altitudes and F is the orthocenter; angle BFC = 180 - angle A = 110. | |
| 14 | D | Angles & Polygons | Six pairs of lines give at most 6 points; parallel pairs and concurrent triples remove points. Every count from 0 to 6 is achievable except 2. | |
| 15 | E | Sequences & Series | Flip the recursion: 1/a_n = 2/a_(n-1) - 1/a_(n-2), so the reciprocals form an arithmetic sequence with first term 1 and common difference 4/3. | |
| 16 | A | Circles | Connect centers: two tangent steps of length 2 bending 120 degrees put the top center 2 sqrt 3 from the middle, so R = 2 sqrt 3 + 1. | |
| 17 | D | Basic Counting | Arrange all 9 cubes in a line, 9!/(2! 3! 4!) = 1260 ways, and remove the top one; each tower of 8 arises from exactly one such line. | |
| 18 | D | Bases & Digits | A two-digit repeating block in base k equals (2k+3)/(k^2-1); setting this to 7/51 forces 51 to divide k^2-1, which only k = 16 satisfies. | |
| 19 | B | Algebraic Manipulation | Pair outer and inner factors: (x+1)(x+4) = u+4 and (x+2)(x+3) = u+6 with u = x^2+5x, so the product is (u+5)^2 - 1, at least -1. | |
| 20 | B | Basic Probability | Only parity matters: the five odd numbers must fill one full row and one full column (a cross), 9 placements out of C(9,5) = 126. | |
| 21 | D | Solid Geometry | The sphere meets the triangle's plane in a circle tangent to all three sides, the incircle of radius 4; then distance = sqrt(6^2 - 4^2). | |
| 22 | B | Conditional Probability & States | Each number is 0, 1, or uniform with probabilities 1/4, 1/4, 1/2; split on the types of x and y and add 1/8 + 1/4 + 1/16. | |
| 23 | C | Sequences & Series | Tadd's turns are rounds 1, 4, 7, ..., so after m turns he has said m(3m - 1)/2 numbers; his 2019th number falls in his 37th turn, round 109. | |
| 24 | B | Polynomials | Clear denominators and set s = p: 1/A = (p - q)(p - r); the sum of the three such products simplifies to (p+q+r)^2 - 3(pq+qr+rp) = 484 - 240. | |
| 25 | D | Divisibility & Factors | Since (n^2)!/((n!)^n n!) counts partitions into n groups of n, the quotient equals an integer times (n-1)!/n; it fails exactly for primes and n = 4. |
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