2022 AMC 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 | A | Functions | Evaluate the innermost diamond first in each half: 1◇(2◇3) = 1◇1 = 0 while (1◇2)◇3 = 1◇3 = 2. | |
| 2 | D | Quadrilaterals & Polygon Areas | All sides of a rhombus are equal, so AB = AD = 5, and the 3-4-5 triangle ABP gives height BP = 4; area = base times height = 20. | |
| 3 | D | Basic Counting | Fix the first two digits; among the ten possible units digits exactly five flip the parity of the even-digit count, so exactly half of all 900 numbers work. | |
| 4 | A | Clocks, Calendars & Time | The 700th hiccup is 699 gaps after the first, so it occurs 699 × 5 = 3495 seconds, i.e. 58 minutes 15 seconds, after 4:00. | |
| 5 | B | Algebraic Manipulation | Each 1 − 1/k² factors as (1 − 1/k)(1 + 1/k), so the radicand is N times its mirror M and the whole expression equals √(N/M) = 2. | |
| 6 | A | Primes | Each term factors as a repunit times 10^n + 1: 121 = 11·11, 11211 = 111·101, 1112111 = 1111·1001, so none is prime. | |
| 7 | B | Quadratics | By Vieta the roots multiply to 36 and k is minus their sum; list the factor pairs of 36, drop 6·6, and remember the negative pairs. | |
| 8 | B | Divisibility & Factors | Every block of 10 holds one or two multiples of 7; with 142 multiples spread over 100 blocks, the number of two-multiple blocks is 142 − 100 = 42. | |
| 9 | D | Sequences & Series | Write k/(k+1)! as 1/k! − 1/(k+1)!; the sum telescopes to 1 − 1/2022!. | |
| 10 | D | Statistics & Data | Mode above median forces the top two values to equal the mode; the two smallest are then distinct with sum 2m − 10, so the mean m ≥ 7. | |
| 11 | B | Logic Puzzles | Rewrite the statement as 'if a school is bigger, then it did not sell more' and take the contrapositive: 'if a school sold more, then it is not bigger'. | |
| 12 | C | Basic Probability | A sum of 7 has probability 1/6 per roll, so avoid it n times with probability (5/6)^n; find the first n with (5/6)^n below 1/2. | |
| 13 | E | Primes | Expand (p+2)³ − p³ = 6p² + 12p + 8 and complete the square: (p+1)² = 5184 = 72², so the primes are 71 and 73. | |
| 14 | B | Number Properties | {13, …, 25} works (smallest sum 26); if M is the largest element, pairing k with M − k caps the size at (M+1)/2 ≤ 13. | |
| 15 | D | Sequences & Series | With d = 2, S_n = n(a + n − 1), so S_{3n}/S_n is constant only when a = 1; then S_n = n² and S_20 = 400. | |
| 16 | D | Quadrilaterals & Polygon Areas | The square is tilted along a 3-4-5 triangle, so the overlap is the square minus a right triangle with legs 5 and 15/4. | |
| 17 | C | Modular Arithmetic | Factoring identities (a ± b divides aⁿ ± bⁿ) eliminate (A), (B), (D), (E) via 3 or 5; only 2^607 − 1 survives. | |
| 18 | B | Systems of Equations | A nonzero solution exists iff the rows are dependent: a zero row, a repeated row, or three distinct rows with one equal to the sum of the others. | |
| 19 | C | Paths & Grids | Only 2 or 3 outer cells can be filled; translate 'everything else ends empty' into adjacency rules on the 8-cell ring and count by how many corners are used. | |
| 20 | D | Angles & Polygons | Extend BE to meet AD at G; D is the midpoint of hypotenuse AG, so DF = DA = DC and A, F, G, C are concyclic about D. | |
| 21 | E | Polynomials | Write P = (x²+1)(ax+b) + 2x+1, reduce modulo x²+x+1 with x³ = 1, and match to x+2: a = 1, b = 2. | |
| 22 | E | Circles | Tangency to both concentric circles forces radius 3 (center at distance 5) or radius 5 (center at distance 3); each case meets the third circle's tangency condition at four centers. | |
| 23 | C | Geometric Probability | She always takes two steps and stops there iff t₁ + t₂ > 1 (probability 1/2); positions are independent of times, so P = ½·½ + ½·⅚. | |
| 24 | B | Functions | Use f(300) = f(900) as a shortcut: |f(800) − f(400)| ≤ 50 + 50 = 100, then halve once more for 50, attained by a piecewise-linear f. | |
| 25 | A | Modular Arithmetic | S_n is the inverse of 7 mod 2^n; because 7 divides 2^{3m} − 1, the bits are 1 then 1,1,0 repeating: x_k = 0 iff 3 divides positive k. |
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.