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.
The same key as a PDF 128 KB — a separate file from the paper, so one can be handed over without the other.
| # | Answer | Topic | Difficulty | Key insight |
|---|---|---|---|---|
| 1 | A | Basic Counting | Count how many 2-inch pieces fit along each side of the pan, 10 along one side and 9 along the other, then multiply. | |
| 2 | D | Linear Equations & Word Problems | Each 30 minutes is half an hour. Find the miles driven in the first two half hours, subtract from 96, and double what is left. | |
| 3 | B | Basic Counting | The value only depends on which two digits get multiplied together. There are just three ways to split 1, 2, 3, 4 into two pairs. | |
| 4 | B | Systems of Equations | Opposite faces of a box match, so the three different areas 24, 48, 72 are the three products of two edges. Find three edges that make them. | |
| 5 | D | Basic Counting | Count all subsets, then take away the ones with no prime at all, which are the subsets made only from 4, 6, 8, 9. | |
| 6 | D | Basic Probability | Three draws are needed exactly when the first two chips add to 4 or less, which only happens for the pairs 1 and 2 or 1 and 3. | |
| 7 | D | Circles | If the big radius is N times the small one, the big semicircle holds N squared small areas, so the leftover is N squared minus N. | |
| 8 | C | Sequences & Series | Half the toothpicks are horizontal. Column by column the horizontal ones number 2, 3, 4, and so on, so find how many columns make 90. | |
| 9 | D | Basic Probability | Turn every die upside down: a face k becomes 7 minus k. A roll with sum 10 becomes a roll with sum 49 minus 10, which is 39. | |
| 10 | E | Solid Geometry | The base is a slanted 1-by-square-root-of-13 rectangle. The height is the distance from M to that slanted rectangle, which you can measure from F in the front face. | |
| 11 | C | Primes | The square of any prime other than 3 is one more than a multiple of 3, so adding 26 always gives a multiple of 3. The prime 3 gives 35. | |
| 12 | C | Triangle Centers & Cevians | The centroid sits one third of the way from the circle's center to C. So it moves on a circle of radius 4, with area 16 pi, about 50. | |
| 13 | C | Modular Arithmetic | The terms are 1 followed by zeros and a 1. Every fourth term, starting with 101 itself, is a multiple of 101, so count 1st, 5th, 9th, and so on. | |
| 14 | D | Statistics & Data | The mode is the only value that appears 10 times, so every other value appears at most 9 times. Fill the other 2008 spots using as few values as possible. | |
| 15 | A | Solid Geometry | Follow the paper from its center to a corner: half the width across the bottom, up the side, half the width on top. That is w plus h. | |
| 16 | E | Modular Arithmetic | A number and its cube always leave the same remainder when divided by 6, so the sum of cubes acts like the plain sum. | |
| 17 | B | Triangles: Area & Pythagorean | The octagon is the rectangle with four matching right triangles cut off its corners. Write both legs using the side length, then use the Pythagorean theorem. | |
| 18 | D | Arrangements with Restrictions | Siblings can never share a row, so each row has one child per family. Choose and order the front row; then only 2 back rows work. | |
| 19 | E | Divisibility & Factors | Chloe's age is a multiple of Zoe's exactly when Zoe's age divides their age gap. So the gap has 9 divisors, which makes it 36, Chloe 37 and Joey 38. | |
| 20 | B | Sequences & Series | Work out the first dozen values. Every 6 steps the value goes up by exactly 6, so jump from f(2) to f(2018) in steps of 6. | |
| 21 | C | Divisibility & Factors | The next divisor must share the factor 17 or 19 with 323; otherwise n would be a multiple of their product, far too big. | |
| 22 | C | Geometric Probability | Side 1 is longest. A triangle needs x + y > 1 and an obtuse one needs x squared + y squared < 1: a quarter circle minus a triangle. | |
| 23 | B | GCD & LCM | Write a = gx and b = gy, with g the gcd and x, y sharing no factor. Then 63 equals g times a whole number, so g divides 63. | |
| 24 | C | Quadrilaterals & Polygon Areas | Each side of XYZ cuts a corner off ACE through the middle of one side and a quarter of the way along another, removing 1/8 each time. | |
| 25 | C | Number Properties | Split x into whole part n and decimal part. Each interval from n to n + 1 has one solution when (n + 1) squared is below 10,000. |
2018 AMC 10B. Problems © Mathematical Association of America, American Mathematics Competitions (maa.org/amc), licensed CC BY-NC-SA 4.0. Typeset and annotated by AMC 10 Step by Step (amcstep.com) and shared under the same licence. Not endorsed by the MAA.