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2016 AMC 10A — answer key
The answer to each problem, with the topic it tests and the idea it turns on.
2016 AMC 10A — answer key
amcstep.com| # | Answer | Topic and the idea it turns on |
|---|---|---|
| 1 | B | Algebraic Manipulation — 11! is just 11 times 10!, so the top is 10 copies of 10!. Then 10! divided by 9! is 10, giving 10 times 10. |
| 2 | C | Exponents, Logarithms & Radicals — Write 100 and 1000 as powers of 10. The left side becomes 10 to the power 5x and the right side 10 to the power 15, so 5x = 15. |
| 3 | C | Ratios, Percents & Averages — David spends 75 cents for each dollar Ben spends, so the 12.50 dollar gap is one quarter of Ben's total. Ben spent 50 dollars and David 37.50. |
| 4 | B | Number Properties — The division 3/8 divided by -2/5 is -15/16, a bit above -1. Rounding down gives -1, not 0, so the answer is 3/8 - 2/5 = -1/40. |
| 5 | D | Number Properties — Call the sides k, 3k and 4k. The volume is 12 times k cubed, so it must be 12 times a perfect cube. Only 96 = 12 times 8 works. |
| 6 | D | Bases & Digits — A 2 turned into 1 costs 1 in the ones place and 10 in the tens place. Count the 2s in each place separately. |
| 7 | D | Statistics & Data — Only the mean gives an equation: 540 + x must equal 7x, so x = 90. Then check that 90 is also the median and mode. |
| 8 | C | Linear Equations & Word Problems — Work backwards from 0 coins. To undo a crossing, add back the 40 coins and then cut the amount in half. Three undo steps give 20, then 30, then 35. |
| 9 | D | Sequences & Series — The rows add up to N(N+1)/2 coins, so N times (N+1) is 4032. Since 63 squared is close to 4032, try 63 times 64, which is exactly 4032. |
| 10 | B | Quadrilaterals & Polygon Areas — With inner length x, the colored areas are x, 2x + 6 and 2x + 14. Equal gaps need x + 6 = 8, so x = 2. |
| 11 | D | Quadrilaterals & Polygon Areas — Turning the picture half a turn about the rectangle's center leaves it unchanged, so the slanted lines cross at the center and the two shaded pieces are equal. |
| 12 | A | Basic Probability — All three numbers must be odd. The first pick is odd half the time, each later pick slightly less, so p is below 1/8. |
| 13 | B | Logic Puzzles — The seat numbers always add to 15, so all moves cancel. Bea +2, Ceci -1, the swap 0: Ada moved one seat left. |
| 14 | C | Diophantine Equations — The number of threes must be even, so it can be 0, 2, 4, up to 672. That is 337 choices. |
| 15 | A | Circles — The dough's radius is 3: the center cookie's radius plus an outer cookie's width. The scrap is 9 pi - 7 pi = 2 pi. |
| 16 | D | Transformations & Symmetry — Follow the corners: every point (x, y) ends at (y, x). That is a reflection over y = x, and repeating it undoes it. |
| 17 | A | Basic Probability — Only the red ball's spot matters; all N + 1 spots are equally likely. For N = 5m, exactly m - 1 middle spots fail. |
| 18 | C | Arrangements with Restrictions — Equal face sums force opposite corners to differ by one fixed amount. The numbers 1 to 8 pair up that way only for differences 1, 2 and 4. |
| 19 | E | Similar & Congruent Triangles — Because BC is parallel to AD, lines from A cut BD in the ratios 1 to 3 and 2 to 3. Use twentieths. |
| 20 | B | Distributions & Stars and Bars — Each term records how many brackets give a, b, c, d or 1. With every letter used at least once, stars and bars gives N choose 4. |
| 21 | D | Circles — For two touching circles on a line, the centers are the radius-sum apart and differ in height by the radius difference, so the Pythagorean theorem gives the sideways gap. |
| 22 | D | Divisibility & Factors — The primes 2, 5 and 11 each add a factor of at least 2 to the divisor count 110 = 2 times 5 times 11, leaving nothing for other primes. |
| 23 | A | Functions — Put c = b so that b diamond b = 1 appears, then put b = a. The operation turns out to be division. |
| 24 | E | Circles — Three equal sides make an isosceles trapezoid. Find the diagonal from the center using areas and the Pythagorean theorem, then compare two right triangles to get the fourth side. |
| 25 | A | GCD & LCM — The lcm takes the larger power of each prime, so treat 2, 3 and 5 separately: 5, 3 and 1 choices, and 15 in all. |
2016 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.