2004 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 | C | Basic Counting | Rows 12 through 22 inclusive are 22 - 12 + 1 = 11 rows, not 10, so the reserved block is 11 times 33. | |
| 2 | B | Basic Counting | Count the two-digit numbers with no 7 at all (8 choices for the tens digit, 9 for the units) and subtract from 90. | |
| 3 | A | Sequences & Series | Doubling four times separates practice 1 from practice 5, so the first count is 48 divided by 2^4. | |
| 4 | B | Divisibility & Factors | P equals 720 divided by the hidden number, so the guaranteed divisor is the gcd of 720/1, ..., 720/6, which is 720/lcm(1,...,6) = 12. | |
| 5 | D | Exponents, Logarithms & Radicals | The 0 must be d, and then 1 * 3^2 = 9 beats 2^3 = 8 and every arrangement that wastes the 1 as base or multiplier. | |
| 6 | C | Number Properties | n! (n+1)! = (n!)^2 (n+1), so a product of two consecutive factorials is a perfect square exactly when the larger index is a perfect square; 100 works. | |
| 7 | A | Ratios, Percents & Averages | Exchanging d dollars yields 10d/7 Canadian; setting 10d/7 - 60 = d gives d = 140, and the question asks for the digit sum 5. | |
| 8 | A | Triangles: Area & Pythagorean | Southwest and southeast are perpendicular directions, so the two cities and the airport form a right triangle with legs 8 and 10 and hypotenuse sqrt(164) ≈ 12.8. | |
| 9 | B | Circles | The circle's center is a corner of the square, so exactly a quarter of the disk overlaps the square; union = square + three-quarters of the disk. | |
| 10 | D | Sequences & Series | The rows hold 1, 3, 5, ..., and the sum of the first n odd numbers is n^2, so n^2 = 100 gives n = 10. | |
| 11 | C | Basic Probability | mn > m + n is the same as (m-1)(n-1) > 1, which fails only when a die shows 1 or both show 2: 16 of 64 pairs. | |
| 12 | A | Circles | A radius meets a tangent at a right angle, so a^2 + c^2 = b^2 and the annulus area pi(b^2 - c^2) equals pi a^2. | |
| 13 | B | Modular Arithmetic | Every thickness is 0.15 mm above a multiple of 0.20 mm, so n coins total 0.15n mod 0.20; hitting 14 exactly forces 4 | n, leaving n = 8. | |
| 14 | C | Ratios, Percents & Averages | Only the last snapshot matters: blue is 1/5 of the bag, so blue : other = 1 : 4, and doubling blue makes it 2 : 4, i.e. one third. | |
| 15 | A | Linear Equations & Word Problems | Swapping adds 5 cents per nickel and removes 5 per dime, so 5(n - d) = 70; with n + d = 20 that is 17 nickels and 3 dimes. | |
| 16 | D | Circles | The three small centers form an equilateral triangle of side 2 whose center is 2/sqrt(3) from each vertex; the big radius is that plus 1. | |
| 17 | B | Bases & Digits | Writing the ages as 10a + b and 10b + a, the five-years condition becomes 8a = 19b + 5, whose only digit solution is (3, 1). | |
| 18 | E | Triangles: Area & Pythagorean | Each corner triangle shares an angle with ACE and uses 1/4 and 3/4 of the enclosing sides, so each is 3/16 of the area. | |
| 19 | C | Sequences & Series | The rule says a_{n+2} + a_{n+3} = a_n + a_{n+1}, so consecutive-pair sums repeat every two steps and the even-indexed terms drop by exactly 2 each time. | |
| 20 | D | Triangle Centers & Cevians | Mass points: A = 5, D = 15 and B = 4, E = 16 both balance at T = 20; then C = 11 and CD/BD = 4/11. | |
| 21 | A | Sequences & Series | Shared terms are 16 mod 21 and run only up to 6010, the slower sequence's last term, so there are 286 of them. | |
| 22 | D | Triangle Centers & Cevians | In a right triangle the circumcenter is the hypotenuse midpoint and the incenter is (r, r) with r = 2; use the distance formula. | |
| 23 | B | Basic Probability | A coloring works iff the minority color covers 0 faces, 1 face, or one opposite pair: 20 of 64 colorings. | |
| 24 | B | Circles | Equal inscribed angles at A give BD = CD; Ptolemy on cyclic ABDC reads 15 CD = 9 AD, so AD/CD = 5/3. | |
| 25 | B | Circles | The shaded region is the lens where the two big disks overlap minus the unit disk; the lens is two 120-degree circular segments, each 4pi/3 - sqrt(3). |
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.