AMC 8 Number Theory: Primes, Divisors and the Factor Trick

If you watch a strong student attempt an AMC 8 number theory question, the first thing they write is almost never an answer. It is a prime factorisation. They do it before they have finished reading, because experience has taught them that whatever the question turns out to be asking, the factorisation will be needed and the answer usually falls out of it.

That single habit — factorise first, then look at the question again — is most of what separates a student who finds these questions hard from one who finds them quick.

Why prime factorisation is trustworthy

The reason the habit works is a theorem so fundamental it is named for that: the fundamental theorem of arithmetic. It says that every whole number greater than 1 can be written as a product of primes in exactly one way, apart from the order of the factors.

The word that matters is exactly. It means 360 = 2³ × 3² × 5 is not one possible description of 360 among several; it is the description. Every fact about 360 that concerns multiplication or division is encoded in those three numbers, so a question about divisors, or common factors, or whether something is a perfect square, is really a question about the exponents 3, 2 and 1.

Primes are the atoms. The factorisation is the chemical formula. Once you have it, you are no longer guessing.

Counting divisors without listing them

"How many positive divisors does 72 have?" is a standard AMC 8 question, and it is a trap for students who answer it by listing. Listing works for 72 and fails for 720, and it is slow and error-prone even when it works.

Here is the method. Factorise: 72 = 2³ × 3². Any divisor of 72 must be built from those same primes, using no more of each than 72 itself has. So a divisor looks like 2^a × 3^b, where a is 0, 1, 2 or 3, and b is 0, 1 or 2.

That is 4 choices for a and 3 for b, independently, so 4 × 3 = 12 divisors. The rule in general: add one to each exponent and multiply.

The "add one" catches students out every time, so it is worth being clear about where it comes from. An exponent of 3 gives four choices, not three, because using none of that prime is allowed — that is the choice a = 0, and it is what produces the divisor 1.

Once you trust the rule, questions that look much harder become the same question. The number of divisors of 720 = 2⁴ × 3² × 5 is 5 × 3 × 2 = 30, and you would not want to list those.

GCD and LCM, read off the page

Most students first meet greatest common divisor and least common multiple as procedures — listing factors, or long division. Both are slow. From prime factorisations, both are immediate.

Take 48 = 2⁴ × 3 and 180 = 2² × 3² × 5.

For the GCD, take each prime that appears in both and keep the smaller exponent. Both have 2 and 3; the smaller exponents are 2 and 1. So the GCD is 2² × 3 = 12. The prime 5 contributes nothing, because 48 has none of it and a common divisor cannot use a prime that one of the numbers lacks.

For the LCM, take every prime that appears in either and keep the larger exponent: 2⁴ × 3² × 5 = 720.

Smaller for GCD because a common divisor must fit inside both. Larger for LCM because a common multiple must contain both. Stated that way it is hard to mix them up, which is the usual error.

There is also a relationship worth memorising, because it converts a whole class of puzzle into one line:

GCD(a, b) × LCM(a, b) = a × b

Check it: 12 × 720 = 8640, and 48 × 180 = 8640. When a question gives you three of these four quantities, this identity hands you the fourth without any factorising at all.

Perfect squares, and why the test is about evenness

A perfect square is a number whose prime exponents are all even. That is not a coincidence, it is what squaring does: (2³ × 5)² = 2⁶ × 5², and every exponent has doubled.

This reframing answers questions that otherwise need trial and error. What is the smallest positive integer k such that 540k is a perfect square?

Factorise: 540 = 2² × 3³ × 5. The exponent of 2 is already even. The exponent of 3 is 3, odd, so it needs one more 3. The exponent of 5 is 1, odd, so it needs one more 5. Therefore k = 3 × 5 = 15.

No searching, no guessing. The question was really "which exponents are odd?" and the factorisation answered it directly.

A related fact that makes a good competition shortcut: a number has an odd number of divisors precisely when it is a perfect square. Divisors normally pair up — for 72, the pair 8 and 9 multiply to 72 — and the pairing is what makes the count even. Only a perfect square has a divisor paired with itself, the square root, leaving an odd total.

Putting it together

How many positive divisors of 1800 are perfect squares?

This looks like two hard ideas at once. It is not, because both reduce to reading exponents.

Factorise: 1800 = 2³ × 3² × 5². A divisor is 2^a × 3^b × 5^c with a at most 3, b at most 2, c at most 2. To be a perfect square, all of a, b, c must be even.

So a can be 0 or 2 — two choices. b can be 0 or 2 — two choices. c can be 0 or 2 — two choices. That gives 2 × 2 × 2 = 8.

The same structure as simple divisor counting, with "even" added as a filter. Students who learned divisor counting as a rule to memorise find this question new and difficult; students who learned why the rule works find it a small variation.

Practising AMC 8 number theory properly

Factorise before you think. On any question involving divisors, multiples, common factors or squares, write the prime factorisation first. It costs ten seconds and it frequently makes the question answer itself.

Know the small primes cold. Every prime under 50, instantly. Most AMC 8 numbers are built from 2, 3, 5, 7, 11 and 13, and hesitating over whether 51 is prime — it is not, it is 3 × 17 — wastes time you do not have.

Together these habits cover most of what AMC 8 number theory actually asks for.

Understand the "add one", do not memorise it. A student who knows it comes from allowing exponent 0 can adapt it to perfect-square divisors, odd divisors, or divisors greater than some bound. A student who memorised it is stuck the moment the question varies.

Registering for the AMC 8

The AMC 8 is 25 multiple-choice questions in 40 minutes, for students in grade 8 and below, with no penalty for a wrong answer. Number theory reliably supplies several questions on every paper, and they are among the most learnable marks available — the methods above are finite and they do not require insight on the day.

Families are often surprised that students cannot register themselves. The competition is ordered and administered by a school, university, math circle or learning centre through its Competition Manager. If your child's school does not participate, you will need to find a local centre that does, and it is worth doing that early, since registration closes well before the competition itself.

Current dates, deadlines and fees are published at maa.org/amcreg and change each cycle. Our AMC 8 preparation course teaches number theory as a connected method rather than a list of rules, and the AMC 8 masterclass adds timed competition practice. Students continuing after Class 8 generally move to AMC 10 preparation, where the same factorisation habits carry directly over.