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.
The same key as a PDF 127 KB — a separate file from the paper, so one can be handed over without the other.
| # | Answer | Topic | Difficulty | Key insight |
|---|---|---|---|---|
| 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. 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.