2021 AMC Fall 10B
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 | Bases & Digits | Each of the digits 1, 2, 3, 4 appears exactly once in every place value, so the total is (1+2+3+4) times 1111. | |
| 2 | B | Quadrilaterals & Polygon Areas | The shaded arrowhead is the tall triangle with vertices (1,0), (5,0), (3,5) minus the short triangle (1,0), (5,0), (3,2), both sharing base 4. | |
| 3 | E | Algebraic Manipulation | Over the common denominator the numerator is 2021^2 - 2020^2 = (2021-2020)(2021+2020) = 4041, and 4041 shares no factor with 2020 or 2021. | |
| 4 | C | Linear Equations & Word Problems | After the changes the gap is N - 8, and 'differ by 2' means N - 8 = ±2, so N is 10 or 6. | |
| 5 | E | Exponents, Logarithms & Radicals | Convert everything to base 2: 8^2022 = 2^6066, dividing by 4 gives 2^6064, and 6064/2 = 3032 turns it into a power of 4. | |
| 6 | B | Divisibility & Factors | 2021 = 43 * 47, so the exponents are 46 and 42; put the bigger exponent on the smaller prime to get 2^46 * 3^42 = 16 * 6^42. | |
| 7 | C | Fractions & Decimals | Reduce the 14 special fractions; two of them sum to an integer only if both are integers or both have the same denominator 2 or 4. | |
| 8 | C | Primes | 16383 = 2^14 - 1 = (2^7 - 1)(2^7 + 1) = 127 * 129 = 3 * 43 * 127, and 127 is prime. | |
| 9 | C | Ratios, Percents & Averages | Take 42 knights: 12 red, 30 blue, 7 magical; if a fraction x of blue knights is magical, then 12(2x) + 30x = 7. | |
| 10 | A | Logic Puzzles | Alice's uncertainty rules out 1 and 40; Bob then knows only if he holds 1, 2, 39, or 40; prime forces 2, and 200 + a = 225. | |
| 11 | B | Circles | Each reflected arc bulges into the hexagon, so the region is the hexagon minus six circular segments, each a 60-degree sector minus an equilateral triangle. | |
| 12 | D | Algebraic Manipulation | The left side equals half the sum (x-y)^2 + (y-z)^2 + (z-x)^2, which is 1 exactly when two of the integers are equal and the third differs by 1. | |
| 13 | B | Similar & Congruent Triangles | The triangle's width shrinks linearly with height: it is 3 at height 3 and 2 at height 5, so it is 9/2 at height 0 and 0 at height 9. | |
| 14 | C | Basic Probability | The product misses a factor of 4 only if all six dice are odd, or exactly one die shows 2 or 6 and the rest are odd. | |
| 15 | D | Similar & Congruent Triangles | A quarter turn of the square carries BP onto CQ, so CQ = 13; then BR is the altitude to the hypotenuse of right triangle QBC. | |
| 16 | D | Expected Value | By linearity, the answer is 5 times the chance one fixed ball ends home: untouched twice (3/5 * 3/5) or swapped away and swapped straight back (2/5 * 1/5). | |
| 17 | D | Coordinate Geometry | Two reflections in lines through the origin make a rotation by twice their angle; P to P'' is a 90-degree clockwise turn, so m is l turned 45 degrees clockwise. | |
| 18 | E | Quadrilaterals & Polygon Areas | The 24-gon has 12-fold symmetry: its 24 vertices alternate between square corners (every 30 degrees) and side intersections exactly halfway between, so it splits into 24 congruent triangles. | |
| 19 | A | Exponents, Logarithms & Radicals | Write N as 7.77...*10^312 (or 777.7...*10^310 when 5 must divide the exponent) and compare the leading factor with small r-th powers of integers. | |
| 20 | C | Conditional Probability & States | P(Hugo wins) is 1/4 by symmetry, and given a first roll of 5 he wins with probability sum over k other 5s of C(3,k)(1/6)^k(4/6)^(3-k)/(k+1). | |
| 21 | E | Basic Counting | Each side of the polygon with fewer sides crosses the other polygon exactly twice, so polygons with m < n sides meet in 2m points; sum over pairs. | |
| 22 | B | Modular Arithmetic | S_n = ((sum)^2 - sum of squares)/2 = (n-1)n(n+1)(3n+2)/24, and since 3 never divides 3n+2, S_n is a multiple of 3 exactly when 9 divides (n-1)n(n+1). | |
| 23 | D | Basic Probability | With no one-color triangle, every vertex has exactly 2 red and 2 blue edges, so the red edges form a 5-cycle; only 12 of 1024 colorings. | |
| 24 | A | Arrangements with Restrictions | Classify by the most white cubes on one face: 4 gives 1 shape, 3 gives 4 shapes (including two mirror-image spirals), and 2 on every face gives 2 shapes. | |
| 25 | E | Quadrilaterals & Polygon Areas | The 1x3 rectangle and unit square force tan(theta) = 1/3 and side sqrt(10); then R's tilt satisfies a quadratic with two valid roots. |
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.