2000 AMC 10
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 | Divisibility & Factors | Factor 2001 = 3 * 23 * 29; with the product fixed, the sum is largest when two factors are as small as possible, namely 1 and 3. | |
| 2 | A | Exponents, Logarithms & Radicals | Multiplying by one more factor of 2000 raises the exponent by one: 2000^1 times 2000^2000 is 2000^2001. | |
| 3 | B | Ratios, Percents & Averages | Eating 20% leaves 80%, so two days scale the count by (4/5)^2 = 16/25, and the original amount is 32 divided by 16/25, which is 50. | |
| 4 | D | Linear Equations & Word Problems | The January bill differs from December only by one extra unit of connect-time cost, so the difference of the bills is the variable part. | |
| 5 | B | Triangles: Area & Pythagorean | P slides along a line parallel to AB, so its height above AB never changes; MN is always half of AB, so only the perimeter can vary. | |
| 6 | C | Modular Arithmetic | Only units digits matter: add consecutive last digits mod 10 and tick off each digit on first appearance; 6 is ticked last. | |
| 7 | B | Triangles: Area & Pythagorean | Trisecting the right angle at D makes ADP and ADB 30-60-90 triangles sharing the known leg AD = 1, so every side of triangle BDP is a 30-60-90 ratio. | |
| 8 | D | Ratios, Percents & Averages | Equal contestant counts give (2/5)f = (4/5)s, so f = 2s: the class with the smaller participation fraction must be twice as big. | |
| 9 | C | Absolute Value & Inequalities | Since x<2, the quantity x-2 is negative, so |x-2| equals 2-x; hence x=2-p and x-p=2-2p. | |
| 10 | D | Triangles: Area & Pythagorean | The triangle inequality confines both x and y to the open interval (2, 10), so |x - y| can be anything below 8 but can never reach 8. | |
| 11 | C | Primes | Both primes are odd, so the product is odd and the sum is even, making pq - (p + q) odd; a size bound then leaves only 119. | |
| 12 | C | Sequences & Series | Figure n has n^2 + (n+1)^2 squares (two rotated grids interleaved), so figure 100 has 100^2 + 101^2 = 20201. | |
| 13 | B | Arrangements with Restrictions | Five yellow pegs need five distinct rows and columns on a board with exactly five of each, so their spots are forced, and so is every other color. | |
| 14 | C | Modular Arithmetic | Integer averages mean the first k scores sum to a multiple of k; work backwards from the total 400 using mod 4, then mod 3. | |
| 15 | E | Algebraic Manipulation | Combine a/b + b/a into (a^2+b^2)/(ab), then use a^2 + b^2 = (a-b)^2 + 2ab and the given ab = a - b to make everything cancel. | |
| 16 | B | Coordinate Geometry | The lattice gives coordinates: intersect lines AB and CD, then AE is the fraction x_E/x_B of length AB, namely 5/9. | |
| 17 | D | Modular Arithmetic | Quarter-to-nickels and nickel-to-pennies preserve value; penny-to-quarters adds exactly 124 cents, so every reachable total is 1 more than a multiple of 124. | |
| 18 | C | Quadrilaterals & Polygon Areas | The visible set is a 1 km band around the square: 5x5 minus 3x3, four 5x1 rectangles, four quarter-discs. | |
| 19 | D | Similar & Congruent Triangles | Scale the square to side 1; the small hypotenuses share one line, so legs 2m and 1 pair with legs 1 and 1/(2m). | |
| 20 | C | Algebraic Manipulation | Adding A + M + C + 1 = 11 completes the expression to (A+1)(M+1)(C+1); with the factors summing to 13, the product peaks at 4, 4, 5. | |
| 21 | B | Logic Puzzles | Only statement II follows: the creepy crawlers that are alligators are automatically ferocious; I and III each fail in a small made-up world. | |
| 22 | C | Linear Equations & Word Problems | Angela's cup is average-sized, so her share 1/p of the whole lies strictly between 1/6 (coffee) and 1/4 (milk): p=5. | |
| 23 | E | Statistics & Data | The mode is 2 and the mean is (25+x)/7; split on whether the median is 2, x, or 4, then force an arithmetic progression. | |
| 24 | B | Functions | Rewrite f in terms of its own input by substituting x = 3t, then the question becomes a quadratic in z whose root sum Vieta gives directly. | |
| 25 | A | Clocks, Calendars & Time | The gap between the two dates is 265 or 266 days depending on a leap day; only the multiple of 7 fits, revealing which year is leap. |
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