1. Factors and multiples
Two words for the same relationship, viewed from different sides.
A whole number $a$ is a factor (or divisor) of a whole number $b$ if $b \div a$ leaves no remainder. Equivalently, $b = a \cdot k$ for some whole number $k$.
A whole number $b$ is a multiple of $a$ if $b = a \cdot k$ for some whole number $k$. The list of multiples of $a$ is $a, 2a, 3a, 4a, \ldots$ — every number on $a$'s "times table."
These are the same statement read two ways. "$6$ is a factor of $24$" and "$24$ is a multiple of $6$" describe the identical fact: $24 = 6 \cdot 4$. Pick whichever phrasing feels natural for the situation.
A few clean facts to keep in your head:
- $1$ is a factor of every whole number. (Take $k = b$ in $b = 1 \cdot k$.)
- Every whole number is a factor of itself. (Take $k = 1$.)
- $0$ is a multiple of every whole number, but it is not a factor of anything except itself — you can't divide by zero.
- The factors of a number always come in pairs whose product is the number: if $a$ is a factor of $24$, so is $24/a$. The pairs of $24$ are $(1,24), (2,12), (3,8), (4,6)$ — and then they start repeating.
2. Divisibility rules
Before factoring a number, it helps to know which small numbers are likely to divide it. The divisibility rules let you check this almost instantly without doing the long division.
| Divisor | Test | Reason in one sentence |
|---|---|---|
| $2$ | Last digit is $0, 2, 4, 6, 8$. | The last digit determines the parity; the rest of the number is already a multiple of $10$, hence of $2$. |
| $3$ | Sum of digits is divisible by $3$. | Because $10 \equiv 1 \pmod{3}$, every place value contributes its digit value modulo $3$. |
| $4$ | The last two digits form a number divisible by $4$. | Everything past the tens place is a multiple of $100 = 4 \cdot 25$, hence of $4$. |
| $5$ | Last digit is $0$ or $5$. | Same reasoning as for $2$ — the last digit decides. |
| $6$ | Divisible by both $2$ and $3$. | $6 = 2 \cdot 3$ and these two factors are coprime, so a number divisible by both is divisible by their product. |
| $9$ | Sum of digits is divisible by $9$. | Same trick as for $3$: $10 \equiv 1 \pmod{9}$. |
| $10$ | Last digit is $0$. | Trivially — base $10$. |
The digit-sum tests for $3$ and $9$ feel like magic the first time, but they're a direct consequence of place value. Write $3{,}471$ as $3 \cdot 1000 + 4 \cdot 100 + 7 \cdot 10 + 1$. Modulo $9$, every power of $10$ becomes $1$, so the number is congruent to $3 + 4 + 7 + 1 = 15$. And $15$ is divisible by $3$ (so $3{,}471$ is), but not by $9$ (so $3{,}471$ isn't).
When trying to find factors of a number, hit it with the divisibility tests before reaching for long division. Most composite numbers under $1000$ reveal at least one small prime factor instantly.
3. Prime numbers
Some whole numbers can be split further; some can't.
- $12 = 2 \cdot 6 = 3 \cdot 4$ — splits into smaller factors.
- $7$ — its only factors are $1$ and $7$ itself; no smaller piece works.
Numbers like $7$ are the building blocks of multiplication. They're called primes.
A whole number greater than $1$ whose only positive factors are $1$ and itself. A number greater than $1$ that has any other factor is called composite.
The first dozen primes are $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, \ldots$. Note three things about this list:
- $2$ is the only even prime. Every other even number has $2$ as a factor.
- The list thins out as you go but never stops. Euclid proved roughly $2{,}300$ years ago that there are infinitely many primes — see the historical note below.
- $1$ is missing. By the modern convention, $1$ is not prime — for good reason, explained next.
Why $1$ doesn't count as prime
By the literal definition above, $1$ would qualify: its only positive factor is itself (which is also $1$). So why exclude it? Because if $1$ were prime, the next big theorem we'll meet — the uniqueness of prime factorization — would fail. We could write $12 = 2 \cdot 2 \cdot 3$, but also $12 = 1 \cdot 2 \cdot 2 \cdot 3$, or $12 = 1 \cdot 1 \cdot 1 \cdot 2 \cdot 2 \cdot 3$, and so on forever. To make the theorem true as stated, $1$ has to be excluded. Convention won out over a literal reading.
Around 300 BCE, Euclid showed there are infinitely many primes. Suppose the list ended at some largest prime $p$. Multiply every prime together and add $1$: the result $N = 2 \cdot 3 \cdot 5 \cdots p + 1$ leaves remainder $1$ when divided by any prime in the list. So either $N$ itself is a new prime, or its prime factors are all primes not in our list. Either way, the list wasn't complete — contradiction. It's a one-paragraph proof that has stood unchanged for over two millennia.
4. The fundamental theorem of arithmetic
This is the big result that justifies caring about primes in the first place:
Every whole number greater than $1$ can be written as a product of primes in exactly one way, up to the order of the factors.
Two halves to that statement:
- Existence. Every number bigger than $1$ has at least one prime factorization. (If it's prime, that's the factorization; if it's composite, factor it once, then factor each piece, and so on. Numbers don't shrink forever, so the process stops.)
- Uniqueness. No number has two genuinely different prime factorizations. $12 = 2 \cdot 2 \cdot 3$ is the only way to write it as primes — you might shuffle the order ($3 \cdot 2 \cdot 2$), but the multiset of primes is fixed.
The uniqueness piece is the surprising one. It says that primes are not just some way to break a number down; they are the way. Every whole number has a fingerprint — its list of prime factors with multiplicities — and the fingerprint is uniquely its own.
In symbols, every whole number $n > 1$ has a unique expression of the form
$$ n = p_1^{a_1} \cdot p_2^{a_2} \cdot \ldots \cdot p_k^{a_k} $$where $p_1 < p_2 < \ldots < p_k$ are distinct primes and the $a_i$ are positive whole numbers. For example, $360 = 2^3 \cdot 3^2 \cdot 5$ — and that's the only way.
Almost every fact you learn about whole numbers eventually traces back to prime factorization. GCD and LCM are read off the factorizations; tests for perfect squares and cubes are read off the exponents; the structure of fractions in lowest terms is determined by them; and modern cryptography (RSA) relies on the fact that finding the factorization of a large number is, as far as we know, computationally hard.
5. Finding prime factorizations
The practical procedure: peel off small prime factors one at a time. A "factor tree" is the standard picture — split the number into any two factors, then split each of those, and continue until every leaf is prime.
The tree starts with $360$ at the top. At each step, we pick any factorization (we picked $4 \cdot 90$, but $8 \cdot 45$ or $6 \cdot 60$ would have worked just as well — they'd lead to a different-looking tree with the same leaves, by the uniqueness theorem). Composite numbers continue branching; primes (in orange) become leaves.
Read off the leaves and you have the prime factorization: $2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 5$, which is $2^3 \cdot 3^2 \cdot 5$.
A more systematic alternative — useful for big numbers — is to peel off the smallest prime factor at each step:
$$ 360 \xrightarrow{\div 2} 180 \xrightarrow{\div 2} 90 \xrightarrow{\div 2} 45 \xrightarrow{\div 3} 15 \xrightarrow{\div 3} 5 \xrightarrow{\div 5} 1. $$The primes you divided by, in order, are the prime factorization: $2, 2, 2, 3, 3, 5$. Same answer, no branching. Pick whichever method you prefer.
6. GCD and LCM
Two natural questions about a pair of whole numbers:
- What's the largest number that divides both? Their greatest common divisor, GCD$(a, b)$.
- What's the smallest positive number that both divide into? Their least common multiple, LCM$(a, b)$.
Once you have prime factorizations, both answers fall out without effort. Write both numbers as products of the same set of primes (with exponent $0$ where a prime is absent):
$$ 12 = 2^2 \cdot 3^1 \cdot 5^0, \qquad 18 = 2^1 \cdot 3^2 \cdot 5^0. $$- GCD takes the minimum exponent in each prime: $\gcd(12, 18) = 2^1 \cdot 3^1 = 6$.
- LCM takes the maximum exponent in each prime: $\mathrm{lcm}(12, 18) = 2^2 \cdot 3^2 = 36$.
One nice consequence — and a great identity to keep in mind:
$$ \gcd(a, b) \cdot \mathrm{lcm}(a, b) = a \cdot b. $$For $12$ and $18$: $6 \cdot 36 = 216 = 12 \cdot 18$ ✓. This identity follows directly from the min-and-max rule: $\min(x, y) + \max(x, y) = x + y$, applied exponent by exponent.
GCD is what you need to reduce a fraction to lowest terms ($\tfrac{12}{18} = \tfrac{2}{3}$ after dividing both by $\gcd(12,18) = 6$). LCM is what you need to find a common denominator when adding fractions. Both appear in the next topic, where they earn their keep.