2016 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 | B | Algebraic Manipulation | Factor 10! out of the numerator: 11! - 10! = 10!(11 - 1), and 10!/9! = 10. | |
| 2 | C | Exponents, Logarithms & Radicals | Rewrite every base as a power of 10 so the equation becomes 10^{5x} = 10^{15}, then match exponents. | |
| 3 | C | Ratios, Percents & Averages | David spends 3/4 of what Ben spends, so the $12.50 gap is 1/4 of Ben's total; Ben spent $50, David $37.50. | |
| 4 | B | Number Properties | The quotient (3/8)/(-2/5) = -15/16 has floor -1, not 0, so rem = 3/8 + (-2/5) = -1/40. | |
| 5 | D | Number Properties | Sides k, 3k, 4k give volume 12k^3, so the volume must be 12 times a perfect cube; only 96 = 12 * 8 qualifies. | |
| 6 | D | Bases & Digits | Each replaced units digit 2 lowers the sum by 1 and each replaced tens digit 2 lowers it by 10; count the digit positions, not the numbers. | |
| 7 | D | Statistics & Data | The mean condition alone gives (540 + x)/7 = x, so x = 90; then check that 90 is also the median and mode. | |
| 8 | C | Linear Equations & Word Problems | Reverse each crossing: add the 40-coin toll back, then halve; three reversals from 0 give 20, 30, 35. | |
| 9 | D | Sequences & Series | The rows total N(N+1)/2, so N(N+1) = 4032 = 63 * 64, giving N = 63 and digit sum 9. | |
| 10 | B | Quadrilaterals & Polygon Areas | With inner length x, the three region areas are x, 2x + 6 and 2x + 14; equal gaps force x + 6 = 8, so x = 2. | |
| 11 | D | Quadrilaterals & Polygon Areas | The two slanted segments are the diagonals of a parallelogram, so the shaded bow-tie is half that parallelogram plus two tiny corner triangles. | |
| 12 | A | Basic Probability | Odd product needs three odd picks; without replacement each later odd pick is slightly less likely than 1/2, so p is below 1/8. | |
| 13 | B | Logic Puzzles | Seat numbers always total 15 and the other four moved a net +1, so Ada moved one seat left, from seat 2 to end seat 1. | |
| 14 | C | Diophantine Equations | In 2a + 3b = 2016 the number of threes must be even, so b = 2k with 0 <= k <= 336, giving 337 solutions. | |
| 15 | A | Circles | The dough's radius is 3 (center cookie radius plus one outer cookie's diameter), so the scrap area is 9pi - 7pi = 2pi and its radius is sqrt(2). | |
| 16 | D | Transformations & Symmetry | Track (x, y): the reflection gives (x, -y), the rotation then gives (y, x), so the combined map is reflection over y = x, its own inverse. | |
| 17 | A | Basic Probability | Only the red ball's position matters; with N = 5m it fails exactly when 2m+1 to 3m-1 greens lie to its left, so P(N) = 1 - (m-1)/(5m+1). | |
| 18 | C | Arrangements with Restrictions | Equal face sums force each antipodal pair to differ by one constant d, and 1 through 8 split into equal-difference pairs only for d = 1, 2, 4. | |
| 19 | E | Similar & Congruent Triangles | Similar triangles cut BD in ratios BE : AD = 1 : 3 and BF : AD = 2 : 3, so BP = BD/4 and BQ = 2BD/5. | |
| 20 | B | Distributions & Stars and Bars | All-four-variables terms are solutions of i+j+k+l+m = N with i,j,k,l >= 1 and m >= 0 for the 1; stars and bars gives C(N,4). | |
| 21 | D | Circles | Tangent circles on a common line have horizontal center separations sqrt((r1+r2)^2 - (r1-r2)^2): here 2sqrt2 and 2sqrt6, so the centers are (-2sqrt2, 1), (0, 2), (2sqrt6, 3) and shoelace finishes. | |
| 22 | D | Divisibility & Factors | In 110n^3 the primes 2, 5, 11 give divisor-count factors 3e+2 and other primes give 3e+1, so 110 = 2*5*11 forces exponents {0, 1, 3}. | |
| 23 | A | Functions | Substituting c = b and then b = a in the rules shows a diamond b = a/b, so the equation is 336x = 100. | |
| 24 | E | Circles | Law of cosines at the center gives cos(theta) = 3/4 for a 200-chord, so each diagonal is 100sqrt14; Ptolemy then yields the fourth side. | |
| 25 | A | GCD & LCM | Work one prime at a time: each lcm condition fixes the maximum of two exponents, and the counts for primes 2, 3, 5 (5, 3, 1) multiply. |
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.