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2022 AMC 10B — answer key
The answer to each problem, with the topic it tests and the idea it turns on.
2022 AMC 10B — answer key
amcstep.com| # | Answer | Topic and the idea it turns on |
|---|---|---|
| 1 | A | Functions — The brackets decide the order. Work each half from the inside out, and do not expect the two groupings to give the same result. |
| 2 | D | Quadrilaterals & Polygon Areas — All four sides of a rhombus are equal, so AB = 5. That makes triangle ABP a 3-4-5 right triangle, and its side BP = 4 is the height. |
| 3 | D | Basic Counting — Whatever the first two digits are, exactly 5 of the 10 possible last digits make the count of even digits odd. So exactly half of all 900 three-digit numbers work. |
| 4 | A | Clocks, Calendars & Time — From the 1st hiccup to the 700th there are only 699 gaps of 5 seconds, so it happens 3495 seconds (58 minutes 15 seconds) after 4:00. |
| 5 | B | Algebraic Manipulation — Each 1 − 1/k² splits into (1 − 1/k)(1 + 1/k). Then the top of the fraction cancels with part of the square root, leaving a small product. |
| 6 | A | Primes — Every term is a product: 121 = 11 × 11, 11211 = 111 × 101, 1112111 = 1111 × 1001. The pattern never stops, so no term is prime. |
| 7 | B | Quadratics — The roots multiply to 36 and add to minus k. List the pairs that multiply to 36, drop 6 × 6, and include the negative pairs. |
| 8 | B | Divisibility & Factors — Every set of 10 holds one or two multiples of 7. There are 142 multiples in 100 sets, so 142 − 100 = 42 sets hold two. |
| 9 | D | Sequences & Series — Each term k/(k+1)! equals 1/k! minus 1/(k+1)!. Adding them, almost everything cancels, leaving 1 − 1/2022!. |
| 10 | D | Statistics & Data — The mode is above the median, so it fills the top two spots. The two smallest numbers must differ, which forces the mean to be at least 7. |
| 11 | B | Logic Puzzles — The statement says a school cannot be both bigger than Euclid and a better seller than Euclid. Saying it in the other order gives the same meaning. |
| 12 | C | Basic Probability — Work out the chance of never rolling a 7. That is 5/6 multiplied by itself once per roll; find the first number of rolls where it drops below 1/2. |
| 13 | E | Primes — Call the primes p and p + 2. Expanding the cubes leads to (p + 1)² = 5184 = 72², so the primes are 71 and 73. |
| 14 | B | Number Properties — The numbers 13 to 25 work, since every sum is at least 26. Pairing smaller numbers that add to the largest number shows 14 is impossible. |
| 15 | D | Sequences & Series — The ratio stays the same for every n only if the first term is 1. Then the terms are the odd numbers, and their sums are perfect squares. |
| 16 | D | Quadrilaterals & Polygon Areas — The square's side from the bottom edge to the top edge is 5 long and rises 4, so it is tilted like a 3-4-5 triangle. Subtract the corner sticking out. |
| 17 | C | Modular Arithmetic — Knock out four choices with quick remainder checks: (A) and (D) are multiples of 3, (B) and (E) end in 5. Only 2^607 − 1 is left. |
| 18 | B | Systems of Equations — A nonzero solution exists exactly when one equation adds nothing new: it is all zeros, it repeats another, or it is the sum of the other two. |
| 19 | C | Paths & Grids — Only the center's eight neighbours matter, and only two or three of them can be filled. Then check, cell by cell, that nothing else ends up with the wrong count. |
| 20 | D | Angles & Polygons — Extend BE until it meets line AD at G. Then D is the midpoint of AG, and A, F, G, C all lie on one circle centred at D. |
| 21 | E | Polynomials — For the remainder on dividing by x² + 1, replace x² by −1. For x² + x + 1, replace x² by −x − 1 and x³ by 1. |
| 22 | E | Circles — Touching both circles centred at the origin forces radius 3 or radius 5. Each size can then touch the third circle in four positions. |
| 23 | C | Geometric Probability — The times only decide whether she takes 2 or 3 steps, each with chance 1/2. Then find the chance that the step lengths add to more than 1. |
| 24 | B | Functions — Since f(300) = f(900), the values f(400) and f(800) can differ by at most 50 + 50 = 100. Applying f again halves that, so at most 50. |
| 25 | A | Modular Arithmetic — Write the first few S_n in binary. After the first 1 the bits repeat 1, 1, 0, so bits at multiples of 3 are 0. |
2022 AMC 10B — 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.