2013 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 127 KB — a separate file from the paper, so one can be handed over without the other.
| # | Answer | Topic | Difficulty | Key insight |
|---|---|---|---|---|
| 1 | C | Fractions & Decimals | Work out the small sums first. The problem becomes 12/9 minus 9/12, which is 4/3 minus 3/4. | |
| 2 | A | Sets, Estimation & Miscellaneous | Change both sides from steps to feet first, giving 30 feet by 40 feet. Find the area in square feet, then take half of it. | |
| 3 | C | Linear Equations & Word Problems | If two numbers average 3, they add up to 6. The high is 16 more than the low, so the low is 8 below the average. | |
| 4 | D | Number Properties | Counting down from 201 to 53 counts both end numbers, so the count is 201 minus 53, plus 1. | |
| 5 | B | Algebraic Manipulation | Write 2a - ab as a(2 - b). Use b = 5 to make (2 - b) as negative as possible, then the largest a, which is 5. | |
| 6 | C | Statistics & Data | Turn each average back into a total: the kids' ages add to 363 and the parents' ages to 1815. Then divide the grand total by all 88 people. | |
| 7 | B | Triangles: Area & Pythagorean | The triangle with three different sides uses two neighbors and the point opposite one of them. Its long side is a diameter, so it has a right angle. | |
| 8 | B | Ratios, Percents & Averages | Pick an easy distance, like 40 miles each. Ray uses 1 gallon and Tom uses 4, so 80 miles on 5 gallons is 16 miles per gallon. | |
| 9 | D | Divisibility & Factors | Relatively prime numbers share no prime, so all the 2s go in one number, all the 3s in another, all the 5s in the third: 8, 27, 125. | |
| 10 | C | Linear Equations & Word Problems | Let x be the three-point tries; then there are 1.5x two-point tries. Points are 1.5x + 1.2x = 2.7x, and 2.7x = 54 gives x = 20. | |
| 11 | B | Quadratics | Complete the square: the equation becomes (x-5)^2 + (y+3)^2 = 0. Squares are never negative, so both must be 0. | |
| 12 | B | Basic Probability | A regular pentagon has 5 sides of one length and 5 diagonals of another length. Whatever segment you pick first, 4 of the other 9 match it. | |
| 13 | E | Sequences & Series | Turn k says k numbers, so after 9 turns 45 numbers have been said. The 53rd number is the 8th number of turn 10, which is 8. | |
| 14 | E | Algebraic Manipulation | The equation becomes 2xy(x - y) = 0, which is true when x = 0, when y = 0, or when x = y: three lines. | |
| 15 | B | Quadrilaterals & Polygon Areas | Write each area using its side length: the triangle's side is a/3 and the hexagon's side is b/6. The hexagon is six small equilateral triangles. | |
| 16 | B | Triangle Centers & Cevians | 1.5, 2, 2.5 is half of a 3-4-5 triangle, so the medians cross at a right angle. Doubling PD and PE gives the other pieces of each median. | |
| 17 | E | Games & Processes | Every trade gives one silver token, so count trades. Trading must stop at 1 red and 2 blue; two equations then give 59 + 44 = 103 trades. | |
| 18 | D | Bases & Digits | For numbers 1bcd, the last digit is 1 + b + c, so b + c is at most 8: that gives 45 numbers. Add 2002. | |
| 19 | D | Quadratics | Exactly one root means b^2 = 4ac, and b is halfway between a and c. Take a = 1, solve for c, and the root is -b/(2a). | |
| 20 | B | Primes | 2013 has the prime factor 61, so the top needs 61!. That brings in the prime 59, so the bottom needs 59! to cancel it. | |
| 21 | C | Sequences & Series | The seventh term is 5a + 8b. Two different starts give the same seventh term only if one start is 8 more in a and 5 less in b. | |
| 22 | C | Arrangements with Restrictions | J is on every line, so the lines have equal sums exactly when the four opposite pairs have equal sums. The eight outer numbers add to 45 - J. | |
| 23 | B | Circles | Right angles at D and F both face AB, so A, F, D, B lie on one circle. Then angle AFE equals angle B, giving EF = 4. | |
| 24 | A | Divisibility & Factors | A number with exactly four divisors is p^3 or p times q for primes. The divisor sums are (1+p)(1+p^2) or (1+p)(1+q); only 2016 works. | |
| 25 | E | Bases & Digits | The last digits in base 5 and base 6 must match, so N is 30m plus a remainder from 0 to 4. The tens digits then depend only on m. |
2013 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.