2018 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 | B | Exponents, Logarithms & Radicals | A negative-one exponent just flips the fraction; work from the innermost parentheses outward, flipping and adding 1 three times. | |
| 2 | A | Ratios, Percents & Averages | Give Jacqueline a convenient amount such as 100; then Liliane has 150 and Alice 125, and 150/125 = 1.2 means 20% more. | |
| 3 | E | Clocks, Calendars & Time | A day is 86400 seconds; cancel prime factors against 10! to get exactly 42 days, and 42 days after noon January 1 is February 12. | |
| 4 | E | Arrangements with Restrictions | Only four sets of three non-adjacent periods exist among six; each set can be filled with the three courses in 3! = 6 orders. | |
| 5 | D | Absolute Value & Inequalities | Negate each claim: d < 6, d > 5, d > 4; the intersection is the open interval (5, 6). | |
| 6 | B | Ratios, Percents & Averages | If 65% of the votes are likes then 35% are dislikes, so the score is 30% of the total; 30% of T equals 90. | |
| 7 | E | Exponents, Logarithms & Radicals | Write 4000 = 2^5 * 5^3; positive n costs factors of 5 (at most 3), negative n costs factors of 2 (at most 5), and n = 0 works. | |
| 8 | C | Linear Equations & Word Problems | Express every count in terms of the number of nickels x: dimes x + 3, quarters 20 - 2x; the value equation then has one unknown. | |
| 9 | E | Similar & Congruent Triangles | Segment DE is four small bases long, so triangle ADE is the small triangle scaled by 4 and has area 16; the trapezoid is 40 - 16. | |
| 10 | A | Algebraic Manipulation | The two radicands differ by 24, so (sum)(difference) = 24 by difference of squares, and the difference is given as 3. | |
| 11 | E | Distributions & Stars and Bars | Every die shows at least 1, so subtract 1 from each: distribute the 3 leftover pips among 7 dice, which is C(9,3) = 84 with no cap ever reached. | |
| 12 | C | Absolute Value & Inequalities | On the line x = 3 - 3y the sign of x is fixed by y, so only three sign regions remain, each giving two linear equations. | |
| 13 | D | Transformations & Symmetry | Folding A onto B makes the crease the perpendicular bisector of AB; it meets AB at its midpoint and forms a right triangle with angle A whose tangent is 3/4. | |
| 14 | A | Exponents, Logarithms & Radicals | The numerator equals 81 times the denominator minus 65 * 2^96; that leftover is positive yet smaller than the denominator, so the quotient is just below 81. | |
| 15 | D | Circles | Tangency points lie on the lines of centers, so A and B extend OP and OQ from length 8 to 13; triangle OAB is triangle OPQ scaled by 13/8. | |
| 16 | D | Triangles: Area & Pythagorean | Sliding along AC, the distance from B falls from 20 to the altitude 420/29 (about 14.5), then rises to 21, hitting each integer once per side. | |
| 17 | C | Divisibility & Factors | A minimum of 2 or 3 bans too many numbers (all evens, or 6, 9, 12 plus half of 4/8 and 5/10), capping the set at 5; {4,5,6,7,9,11} works. | |
| 18 | D | Bases & Digits | Digits -1, 0, 1 in base 3 (balanced ternary) give every integer from -3280 to 3280 exactly once; by symmetry the nonnegative ones number (3^8 + 1)/2. | |
| 19 | E | Modular Arithmetic | Only the units digit of m matters: 1 always works, 5 never, 9 needs even n, 3 or 7 need 4 | n; count within the block of 20. | |
| 20 | B | Paths & Grids | Full square symmetry means the code is determined by one-eighth of the grid: a triangular wedge of 10 cells, so 2^10 colorings minus the two monochrome ones. | |
| 21 | E | Coordinate Geometry | Substituting x^2 = y + a gives a quadratic in y whose root y = -a is the lone axis point; the other root needs 2a - 1 > 0. | |
| 22 | D | GCD & LCM | Exponents of 2 and 3 through the gcd chain force gcd(d,a) = 6k with k coprime to 6; between 70 and 100 only 78 = 6 * 13 fits. | |
| 23 | D | Triangles: Area & Pythagorean | Cut the field into the square, a base-5 height-2 triangle on the hypotenuse, and two right triangles along the legs; total area 6 forces s = 2/7. | |
| 24 | D | Triangle Centers & Cevians | The bisector splits BC and DE in the ratio 5:1, so [ABG] = (5/6)(120) = 100 and [ADF] = (5/6)(30) = 25; the quadrilateral is their difference, 75. | |
| 25 | D | Bases & Digits | In terms of the repunit R the equation reads R(9c - a^2) = b - 2c; two values of n force both sides to vanish. |
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.