2006 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 | Sequences & Series | Powers of -1 alternate between -1 and 1, so consecutive terms cancel in pairs and 2006 terms form 1003 pairs summing to zero. | |
| 2 | A | Functions | The operation is just x^2 - y^2, so evaluate the inner one first: 4 spade 5 = -9, then 3 spade (-9) = 9 - 81. | |
| 3 | A | Linear Equations & Word Problems | Sum and difference are given, so the smaller score is half of (sum minus difference): (34 - 14)/2 = 10. | |
| 4 | D | Circles | Areas scale with the square of the diameter, so the big circle is 9 times the small one and the ring is 9 - 1 = 8 times. | |
| 5 | B | Geometric Optimization | Total area 18 rules out a 4 by 4 square; placing the rectangles side by side (widths 2 + 3, heights 3 and 4) fits in 5 by 5. | |
| 6 | D | Circles | Each side of the square is a diameter, so each semicircular arc has length (pi/2) times the side, which is exactly 1; four arcs give 4. | |
| 7 | A | Exponents, Logarithms & Radicals | The denominator 1 - (x-1)/x collapses to 1/x, so the radicand is x^2 and the square root is |x|, which equals -x for negative x. | |
| 8 | B | Circles | Join the center to a top corner of the square: the radius is the hypotenuse of a right triangle with legs s and s/2, so r^2 = (5/4)s^2 = 50. | |
| 9 | B | Ratios, Percents & Averages | The whole batch is 600 grams with 25 + 386 = 411 calories, and 200 grams is one third of it, so 411/3 = 137. | |
| 10 | A | Triangles: Area & Pythagorean | With sides x, 3x, 15 the binding constraint is x + 15 > 3x, so x < 7.5; x = 7 gives perimeter 43. | |
| 11 | C | Modular Arithmetic | Every factorial from 10! on contains both 2*5 and 10, hence ends in 00; only 7! + 8! + 9! affects the tens digit. | |
| 12 | E | Coordinate Geometry | The intersection point lies on both lines, so substituting x = 1, y = 2 into each equation solves for a and b immediately. | |
| 13 | E | Ratios, Percents & Averages | Joe keeps all 2 ounces of cream; JoAnn's cup is 2/14 cream when she drinks, so she loses 1/7 of her cream and keeps 12/7. | |
| 14 | D | Quadratics | q is the product of the new roots, and expanding (a + 1/b)(b + 1/a) = ab + 2 + 1/(ab) needs only ab = 2, never m. | |
| 15 | C | Quadrilaterals & Polygon Areas | Diagonal BD is the short diagonal of ABCD but the long diagonal of BFDE, so the similarity ratio is 1/sqrt3 and the area ratio 1/3. | |
| 16 | E | Clocks, Calendars & Time | Each ordinary year shifts the weekday by 1, each leap year by 2; 12 + 2*4 = 20 is 6 mod 7, so Sunday becomes Saturday. | |
| 17 | D | Basic Probability | Whatever color Alice sends, Bob's bag holds six balls with that color doubled, and the bags match exactly when Bob sends back one of those two. | |
| 18 | E | Sequences & Series | Computing a few terms shows the sequence cycles with period 6, so a_2006 equals a_2 since 2006 leaves remainder 2 on division by 6. | |
| 19 | A | Circles | Angle DOE is 30 degrees; the shaded region is that sector of radius 2 minus triangles OBD and OBE, each with base sqrt3 - 1 and height 1. | |
| 20 | E | Coordinate Geometry | AB has displacement (2000, 200) and AD, perpendicular with x-change 2, must be (2, -20): AD is AB rotated and scaled by 1/100, so the area is AB^2/100. | |
| 21 | C | Basic Probability | Face k has probability k/21, so a total of 7 has probability (1*6 + 2*5 + 3*4 + 4*3 + 5*2 + 6*1)/21^2 = 56/441. | |
| 22 | D | Diophantine Equations | In cents, N(4B+5J) = 253 = 11*23; the per-sandwich cost is at least 9 and N > 1, so N = 11, B = 2, J = 3. | |
| 23 | D | Triangle Centers & Cevians | Join the top vertex to the cevian intersection; equal-altitude triangles have areas proportional to bases, giving two linear equations in the two pieces. | |
| 24 | B | Circles | Radii to tangent points are perpendicular to the tangent, so OADP is a right trapezoid with bases 2 and 4 and height 4*sqrt(2). | |
| 25 | B | Divisibility & Factors | If 5 were an age the even number would end in 0, forcing the form d0d0 with d = 9, but 9090 is divisible by neither 4 nor 8. |
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.