2018 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 | A | Basic Counting | Each side of the pan holds a whole number of 2-inch pieces: 20/2 = 10 across and 18/2 = 9 down, so 10 times 9 pieces. | |
| 2 | D | Linear Equations & Word Problems | Half an hour at 60 mph is 30 miles and at 65 mph is 32.5 miles, leaving 33.5 miles for the last half hour, i.e. 67 mph. | |
| 3 | B | Basic Counting | The value depends only on how the four digits are split into two pairs, and there are just three pairings: 14, 11 and 10. | |
| 4 | B | Systems of Equations | The three distinct face areas are XY, YZ, ZX; their product is (XYZ)^2 = 288^2, so each edge is 288 divided by one face area. | |
| 5 | D | Basic Counting | Count all 2^8 subsets and subtract the 2^4 subsets built only from the four non-primes 4, 6, 8, 9. | |
| 6 | D | Basic Probability | Exactly three draws happen when the first two chips sum to at most 4, which means the unordered pair is {1,2} or {1,3}: 2 of 10 pairs. | |
| 7 | D | Circles | With small radius r the large radius is Nr, so A:B = N : (N^2 - N) = 1 : (N-1), giving N - 1 = 18. | |
| 8 | C | Sequences & Series | Counting rows of horizontal and vertical toothpicks gives n(n+3) for an n-step staircase; check 3*6 = 18, then solve n(n+3) = 180 to get n = 12. | |
| 9 | D | Basic Probability | Replacing each face k by 7 - k is a bijection, so a sum of 10 is exactly as likely as 49 - 10 = 39. | |
| 10 | E | Solid Geometry | Base BCHE is a 1-by-sqrt(13) rectangle and M is as far from its plane as F, namely the altitude 6/sqrt(13) of right triangle EFB, so V = 2. | |
| 11 | C | Primes | For any prime p other than 3, p^2 is 1 mod 3, so p^2 + 26 is divisible by 3; p = 3 gives 35 = 5*7. | |
| 12 | C | Triangle Centers & Cevians | The centroid is one third of the way from the center O to C, so it traces a circle of radius 4 and area 16 pi, about 50. | |
| 13 | C | Modular Arithmetic | Terms are 10^k + 1 for k = 2 to 2019; since 100 is -1 mod 101, divisibility by 101 happens exactly when k is 2 mod 4. | |
| 14 | D | Statistics & Data | Every non-mode value appears at most 9 times, so the other 2008 entries need at least ceil(2008/9) = 224 values, plus the mode makes 225. | |
| 15 | A | Solid Geometry | Along a diagonal of the sheet, the half-diagonal runs w/2 to the box's side, h up the side, and w/2 across the top, so the side is sqrt(2)(w+h). | |
| 16 | E | Modular Arithmetic | n^3 - n = (n-1)n(n+1) is always divisible by 6, so the cube sum is congruent mod 6 to the plain sum 2018^2018, which is 4 mod 6. | |
| 17 | B | Triangles: Area & Pythagorean | With side s, the corner legs are (8-s)/2 and (6-s)/2, and the Pythagorean theorem gives s^2 + 14s - 50 = 0, so s = -7 + 3 sqrt(11). | |
| 18 | D | Arrangements with Restrictions | Siblings sharing a row forces a front-back sibling pair, so each row has one child per family: 2^3 * 3! front rows, then 2 derangements behind. | |
| 19 | E | Divisibility & Factors | Chloe's age is a multiple of Zoe's exactly when Zoe's age divides Chloe's age minus 1, so that difference has 9 divisors: 36, making Chloe 37 and Joey 38. | |
| 20 | B | Sequences & Series | Adding consecutive recursions gives f(n) = f(n-6) + 6, so f(2018) = f(2) + 6 * 336 = 2017. | |
| 21 | C | Divisibility & Factors | The next divisor d must share a factor with 323 = 17 * 19, since otherwise 323d > 9999 divides n; the smallest such d is 340. | |
| 22 | C | Geometric Probability | With 1 the longest side, the conditions are x + y > 1 and x^2 + y^2 < 1: a quarter disk minus a triangle, area pi/4 - 1/2. | |
| 23 | B | GCD & LCM | Write a = gx, b = gy with x, y coprime; then 63 = g(xy(20 - g) + 12), so g divides 63 and only g = 21 works. | |
| 24 | C | Quadrilaterals & Polygon Areas | Each side of XYZ cuts a corner off ACE through the midpoint of one side and the quarter point of another, removing 1/8 of ACE three times. | |
| 25 | C | Number Properties | Write x = n + f; then (n+f)^2 = 10000f has exactly one root f in [0,1) precisely when (n+1)^2 < 10000, giving n = -100 to 98. |
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.