2006 AMC 10A
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 | Linear Equations & Word Problems | Cost is quantity times unit price for each item, then add: 5 times 3 plus 8 times 2. | |
| 2 | C | Functions | Evaluate the inner operation first, h ⊗ h = h^3 − h, then feed that result in as the second input of the outer operation. | |
| 3 | B | Ratios, Percents & Averages | A 3:5 ratio with the 5-part equal to 30 makes each part 6, so the 3-part is 18. | |
| 4 | E | Bases & Digits | Maximize the hour digits and the minute digits separately: hour 9 beats 10, 11, 12 in digit sum, and minutes top out at 59. | |
| 5 | D | Linear Equations & Word Problems | Price per slice: plain slices cost 1 dollar, the anchovy half costs 4 + 2 = 6 for four slices; Dave pays 7, Doug pays 3. | |
| 6 | B | Exponents, Logarithms & Radicals | Write (7x)^14 as ((7x)^2)^7; equal seventh powers force 49x^2 = 14x, and the nonzero root is x = 2/7. | |
| 7 | A | Quadrilaterals & Polygon Areas | Rearranging pieces preserves area, so the square has side 12; the long edge 18 − y of each hexagon must become a full side of that square. | |
| 8 | E | Quadratics | Both points have y = 3, so 2 and 4 are roots of x^2 + bx + (c − 3) = 0; Vieta gives c − 3 = 8. | |
| 9 | C | Number Properties | Sum equals (number of terms) times (average), and the average of consecutive integers is the middle value, so just test lengths 2, 3, 4, 5 against 15. | |
| 10 | E | Number Properties | Set 120 − √x = k^2 with k a nonnegative integer; k runs from 0 to 10, and each k gives exactly one x. | |
| 11 | C | Coordinate Geometry | Expanding leaves xy = 0, which holds exactly when x = 0 or y = 0: the two coordinate axes. | |
| 12 | C | Circles | Both setups give a half-disk of radius 8; in II the rope also bends around the nearby corner with 4 feet to spare, adding a quarter-disk of radius 4. | |
| 13 | D | Expected Value | Winning needs an even first roll (1/2) and a matching second roll (1/6), so P(win) = 1/12 and the fair prize is 5 × 12 = 60. | |
| 14 | B | Sequences & Series | Each ring hangs from the bottom of the one above, so the holes stack end to end: total = 1 + (18 + 17 + ... + 1) + 1. | |
| 15 | D | Linear Equations & Word Problems | Both runners take 2π/5 minutes per lap, so together they close a full turn every π/5 minutes; count how many of those fit in 30 minutes. | |
| 16 | D | Similar & Congruent Triangles | From the apex the circles are similar in ratio 1:2, so AO_2 = 2 AO_1 with AO_2 − AO_1 = 3; hence AO_1 = 3, height 8. | |
| 17 | A | Quadrilaterals & Polygon Areas | With AH = AC = 2 and unit trisection spacing, every drawn line has slope ±1, so WXYZ is a square with horizontal and vertical diagonals of length 1. | |
| 18 | C | Basic Counting | Glue the two adjacent letters into one block: it can start in any of 5 positions among the 6 slots, times 10^4 digit strings times 26^2 letter pairs. | |
| 19 | C | Angles & Polygons | Three angles in arithmetic progression summing to 180 have middle term 60, so the triangle is determined by the common difference d with 1 ≤ d ≤ 59. | |
| 20 | E | Modular Arithmetic | Six integers but only five residues mod 5, so two must share a residue and their difference is a multiple of 5: the event is certain. | |
| 21 | E | Basic Counting | Count the complement: four-digit numbers using neither 2 nor 3 number 7 · 8 · 8 · 8 = 3584; subtract from the 9000 four-digit numbers. | |
| 22 | C | Diophantine Equations | Payments are integer combinations 300p + 210g, which are exactly the multiples of gcd(300, 210) = 30; three goats for two pigs realizes 30. | |
| 23 | B | Circles | Radii to the tangent points are perpendicular to CD, so ACE and BDE are similar right triangles in ratio 3 : 8, and CE = 4 from the 3-4-5 triangle. | |
| 24 | B | Solid Geometry | Split the octahedron into two square pyramids: the base is the square through the four side-face centers, with diagonal 1 and area 1/2, and each apex is 1/2 above it. | |
| 25 | C | Basic Probability | Fix the first two moves by symmetry (3 · 2 ways); only 3 of the remaining continuations visit every vertex, so 18 good paths out of 3^7 equally likely ones. |
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.