Four topic areas, 25 questions, rising difficulty

The AMC 8 doesn't test anything your child hasn't seen the pieces of. There's no calculus, no trigonometry, no algebra beyond what a strong pre-algebra student knows. What makes it hard is that the pieces get combined in unfamiliar ways, and you have 40 minutes for 25 questions. This guide walks through the four main areas, shows one original worked problem in each, and then describes how the test typically gets harder from start to finish.

The problems below are ones we wrote for this article, not real contest problems. Each answer was checked two different ways, and we show both so you can see how a fast method and a safe method compare. Try each one yourself before reading the solution.

Number theory

Number theory on the AMC 8 means the properties of whole numbers: factors, multiples, primes, divisibility, remainders, and the greatest common divisor and least common multiple. It's the area where a little technique pays off most, because a student who knows the tools turns a long search into a short one. Some problems hide a number theory idea inside a word problem.

Problem. A school band has between 40 and 80 members. When they line up 6 to a row, 2 people are left over. When they line up 7 to a row, 4 people are left over. How many members does the band have?

Method 1: list and match. Start with the more restrictive condition. Numbers that leave remainder 4 when divided by 7 are 4, 11, 18, and so on, adding 7 each time. Between 40 and 80 those are 46, 53, 60, 67, and 74. Now test each against the other condition (remainder 2 when divided by 6). 46 leaves 4. 53 leaves 5. 60 leaves 0. 67 leaves 1. 74 leaves 2. Only 74 works.

Method 2: build the answer and check uniqueness. A number that leaves remainder 2 on division by 6 and remainder 4 on division by 7 will repeat every 6 × 7 = 42, because 42 is a multiple of both. Check 74 directly: 74 = 6 × 12 + 2 and 74 = 7 × 10 + 4. Both conditions hold. The next solution down is 74 − 42 = 32, which is below 40, and the next one up is 116, above 80. So 74 is the only one in range.

Answer: 74. The lesson is the repeat length. Whenever two remainder conditions are combined, the solutions come around every LCM of the divisors, and that tells you how many to expect in a range.

Counting and probability

Counting problems ask how many ways something can happen. Probability problems ask what fraction of the possible outcomes meet a condition. At the AMC 8 level, almost everything comes down to the multiplication principle, careful casework, and writing probability as favorable over total. The danger isn't advanced technique; it's double counting or missing a case.

Problem. Two fair six-sided dice are rolled. What is the probability that the sum is a multiple of 4?

Method 1: list by sum. There are 6 × 6 = 36 equally likely outcomes. A sum of 4 can happen as (1,3), (2,2), (3,1): 3 ways. A sum of 8 can happen as (2,6), (3,5), (4,4), (5,3), (6,2): 5 ways. A sum of 12 happens only as (6,6): 1 way. No other sum between 2 and 12 is a multiple of 4. That's 3 + 5 + 1 = 9 favorable outcomes, so the probability is 9/36 = 1/4.

Method 2: fix the first die. For each value of the first die, count the values of the second die that make the sum a multiple of 4. If the first die shows 1, only a 3 works (sum 4). Shows 2: a 2 or a 6 (sums 4, 8). Shows 3: a 1 or a 5 (sums 4, 8). Shows 4: only a 4 (sum 8). Shows 5: only a 3 (sum 8). Shows 6: a 2 or a 6 (sums 8, 12). That's 1 + 2 + 2 + 1 + 1 + 2 = 9 again, and 9/36 = 1/4.

Answer: 1/4. Two different organizing schemes gave the same count, which is the best check available in counting. If the schemes disagree, you've double counted or missed something. Students who get into the habit of counting two ways, even when time allows only a rough second pass, catch most of their errors.

Geometry

Geometry on the AMC 8 is mostly area, perimeter, angles, the Pythagorean theorem, similar triangles, and simple circle and solid-shape facts. The good news is that you're rarely asked to prove anything. The bad news is that the diagrams are often not drawn to scale, and many problems need one extra line or a clever subtraction.

Problem. A rectangle has a perimeter of 34 and a diagonal of length 13. What is its area?

Method 1: algebra with a square. Call the sides a and b. The perimeter tells us a + b = 17. The diagonal and the Pythagorean theorem tell us a² + b² = 169. Square the first equation: (a + b)² = a² + 2ab + b² = 289. Subtract the second equation: 2ab = 289 − 169 = 120, so ab = 60. The area is 60, and we never needed the sides themselves.

Method 2: spot the triple. If a + b = 17 and the diagonal is 13, try the Pythagorean triple 5, 12, 13. The sides 5 and 12 add to 17 and give a diagonal of 13, so they fit both conditions. The area is 5 × 12 = 60. To make sure that's the only rectangle, note that the algebra above gave ab = 60 directly, so any rectangle meeting the conditions has area 60.

Answer: 60. Notice that Method 2 is quick but leans on luck, while Method 1 always works. A good habit is to try the quick idea first, confirm it with the reliable one if you have time, and trust the reliable one if they disagree.

Algebra and logic

This bucket holds linear equations, ratios, rates, averages, percentages, simple sequences, and word problems that need careful reading. It also includes pure logic, such as ordering puzzles and "who is telling the truth" problems. The algebra rarely goes past one or two unknowns. The skill is translating a paragraph into equations without losing a condition. This area shows up everywhere in the test, from problem 1 to problem 25, and it's where careless misreads cost the most points.

Problem. Ana, Ben, and Cy have 49 stickers altogether. Ana has 5 more than Ben, and Cy has twice as many as Ben. How many stickers does Cy have?

Method 1: one variable. Let Ben have b stickers. Then Ana has b + 5 and Cy has 2b. The total is (b + 5) + b + 2b = 4b + 5 = 49, so 4b = 44 and b = 11. Cy has 2 × 11 = 22.

Method 2: a picture. Draw one box of equal size for Ben, one for Ana, and two for Cy. That's four equal boxes, plus Ana's extra 5. Take the extra 5 out and the four boxes hold 44, so each box is 11. Cy has two boxes, 22. As a final check on the whole answer, Ana has 16, Ben has 11, Cy has 22, and 16 + 11 + 22 = 49.

Answer: 22. Always substituting back into the original statement is the cheapest check in the contest. It catches misread conditions, which are the most common algebra error at this level.

How difficulty ramps from question 1 to 25

The MAA builds the AMC 8 roughly in order of difficulty, and that order shapes how you should spend your time. The exact boundary moves from year to year, but the pattern is consistent enough to plan around.

• Questions 1 to 8. Quick and mostly direct. These test fluent arithmetic, fractions, reading a chart, basic area, and simple counting. A prepared student should get nearly all of them, and each takes under a minute.
• Questions 9 to 16. Still accessible, but they start to need a short idea: a remainder, a small casework count, a similar-triangle or area-by-subtraction trick, a rate problem with a twist. This is where most students separate themselves.
• Questions 17 to 21. Multi-step. A problem here usually combines two topics or needs a clever setup, and a wrong first approach can eat several minutes.
• Questions 22 to 25. The hardest. They reward real insight and care, and many strong students don't finish all four. Nothing here requires more than middle school math, but you may need to try more than one approach.

Two practical points follow. First, every question is worth the same one point, so the early ones are the best value per minute and shouldn't be rushed into mistakes. Second, a student aiming at an award will need some of the final problems, while a student aiming for a solid first score is usually better off making the first 15 or 16 airtight. Our pacing guide turns this ramp into a time budget.

How topics mix in practice

Real problems cross the lines. A counting problem may need divisibility to decide which cases are allowed. A geometry problem may need an equation to find an unknown length. A probability problem may be a number theory problem in disguise, as our dice example partly is: "multiple of 4" is a divisibility condition on the sum. That's why we recommend practicing by topic early on, then switching to mixed sets, as laid out in our AMC 8 preparation guide.

Where to go from here

A good test of your own readiness is to sort your recent mistakes by topic. If one area accounts for most of them, that's where the next two weeks of practice should go. A full-length timed test makes this easy: our free full-length mock AMC 8 comes with a score report that breaks your result down by topic and shows, for each missed question, a faster method. That's a quick way to find out whether your weak spot is number theory, counting, geometry, or algebra, before you plan anything else.