Factors, Multiples and Number Theory

Updated 4 October 2026

Once you can divide, one question follows immediately: which divisions come out exactly? Everything on this page is an answer to that question, from a seven-year-old sorting counters into equal rows to a proof that the primes never run out.

It is also the most practical branch of arithmetic that does not look practical. Simplifying a fraction is a factor question. Two buses that leave together and meet again is a multiple question. Whether a decimal stops or repeats forever is settled entirely by the prime factors of the denominator.

What is a factor, and what is a multiple?

A factor of a number divides it exactly, and a multiple is that number times a whole number. The factors of 20 are 1, 2, 4, 5, 10 and 20, and the list stops, since no factor is bigger than the number. The multiples of 20 are 20, 40, 60, 80 and so on, and that list never ends. The two words face opposite ways: 4 × 5 = 20, so 4 is a factor of 20 and 20 is a multiple of 4. Factors come in pairs that multiply to the number, so search upward from 1 and stop when the pairs start to repeat. A square number has one pair made of a number times itself, as 36 = 6 × 6, so it has an odd number of factors. The divisibility rules save trying every number. Multiplying each pair back checks the list: 1 × 20, 2 × 10 and 4 × 5 all make 20.

Dividing Exactly is where "exactly" gets pinned down — no remainder, nothing left over — and Factors and Multiples name the two directions. Factor Pairs is the one that makes finding them reliable: factors come in pairs that multiply to the number, so hunting for 12 means checking 1 × 12, 2 × 6, 3 × 4 and then stopping, because after that the pairs repeat in the other order.

Divisibility without dividing

Divisibility Tests collects the checks that answer "does this go?" faster than the division would.

Every test, with the reason it works: the divisibility rules.

The digit-sum tests are the interesting ones, and they are not coincidence: every power of ten leaves a remainder of 1 when divided by 9, so a number and its digit sum leave the same remainder. That fact is worth more than the test — Casting Out Nines to Check Arithmetic turns it into a way of checking a long multiplication in a couple of seconds, and Modular Arithmetic on a Clock Face is the general idea behind it: arithmetic where only the remainder is kept, which is how clocks, calendars and checksums all work.

What is a prime number?

A prime number is a whole number greater than 1 whose only factors are 1 and itself, so it has exactly two factors. The first primes are 2, 3, 5, 7, 11 and 13. Other whole numbers above 1 are composite, products of smaller numbers, as 15 = 3 × 5. To test whether a number is prime, try dividing it by the primes in order, and stop once the prime squared is bigger than the number. For 97, none of 2, 3, 5 or 7 divides it, and 11 × 11 = 121 is more than 97, so 97 is prime. You can stop there because factors come in pairs, and the smaller factor in a pair, multiplied by itself, is never bigger than the number. The number 1 is not prime, since it has only one factor, and 2 is the only even prime. A number such as 91 looks prime until you try 7: 91 = 7 × 13.

The Sieve of Eratosthenes is the oldest way to find them, and it is a genuinely good algorithm rather than a classroom curiosity: write the numbers out, keep 2 and strike out every later multiple of 2, keep 3 and strike out its multiples, and continue. What survives is prime, and you never test a single number for divisibility.

Prime Factorization is the payoff. Every whole number above 1 breaks into primes in exactly one way — $180=2^2\times 3^2\times 5$180 = 2² × 3² × 5 — and that unique fingerprint is what makes the next section work.

HCF or LCM — which do you need?

Decide whether the question breaks things into equal groups or waits for things to line up. Equal groups need the highest common factor (HCF), the biggest number that divides both. Lining up needs the lowest common multiple (LCM), the smallest number both divide into. Suppose 24 red beads and 36 blue beads go into identical bags with none left over. The most bags you can fill is the HCF, 12, and each bag holds 2 red and 3 blue. If one bus leaves every 6 minutes and another every 8 minutes, they next leave together after the LCM, 24 minutes. Keywords can mislead, so ask whether the answer must fit into both numbers or be reached by both. When two numbers share no factor, the HCF is 1 and the LCM is their product, so 8 and 9 have an LCM of 72. The HCF times the LCM equals the two numbers multiplied, which checks both: 2 × 24 = 48 = 6 × 8.

Common Factors and The Highest Common Factor take the first direction: the largest number that divides both. 12 and 18 share 1, 2, 3 and 6, so the HCF is 6 — which is the largest tray you can pack both piles into without a gap.

Common Multiples and The Lowest Common Multiple take the other: the smallest number both divide into. 12 and 18 first meet at 36 — the first time a 12-minute bus and an 18-minute bus leave together again.

Choosing Between HCF and LCM exists because the arithmetic is rarely the hard part; picking the right one from a word problem is. The prime factorizations answer both at once: take the lowest power of each shared prime for the HCF, the highest power of every prime for the LCM.

HCF and LCM from the prime factorizations

For numbers too big to list factors for, the factorizations do the work. Write both as products of primes, then read off both answers from the same pair of rows.

$84=2^2\times 3\times 7$84 = 2² × 3 × 7 and $90=2\times 3^2\times 5$90 = 2 × 3² × 5.

• HCF: the lowest power of each prime they share — 2 × 3 = 6. Only primes appearing in both count, which is why 5 and 7 are ignored.
• LCM: the highest power of every prime that appears in either — $2^2\times 3^2\times 5\times 7=1$2² × 3² × 5 × 7 = 1,260.

A useful check falls out of this: multiply the HCF by the LCM and you get the product of the two original numbers, every time. 6 × 1,260 = 7,560 = 84 × 90. Each prime is being counted once at its lower power and once at its higher, which between them is exactly the two numbers' worth.

Do the primes ever run out?

No, and the proof is short. Euclid’s Proof That Primes Never Run Out starts by supposing someone has a complete list of every prime. Multiply all the primes on the list together and add 1. The new number leaves a remainder of 1 when divided by any prime on the list, so none of them is a factor. But every whole number above 1 has a prime factor, so some prime is missing from the list, and no finite list can be complete. With the list 2, 3, 5 and 7, the new number is 2 × 3 × 5 × 7 + 1 = 211, which is itself prime. The new number need not be prime: 2 × 3 × 5 × 7 × 11 × 13 + 1 = 30,031 = 59 × 509. Its prime factors are still missing from the list. And 13 × 2,310 = 30,030 confirms the remainder of 1 when dividing by 13.

Which fractions have decimals that stop?

A fraction has a decimal that stops when its denominator, in lowest terms, has no prime factors other than 2 and 5. Why Every Fraction Terminates or Recurs gives the reason: decimals count in tenths, hundredths and thousandths, and every power of ten is built only from 2s and 5s, since 10 = 2 × 5. For $\frac{7}{20}$7/20, the denominator is 2 × 2 × 5, so multiply top and bottom by 5 to get $\frac{35}{100}$35/100, which is 35 hundredths. A denominator with any other prime factor can never reach a power of ten, so the division never ends, and the digits repeat instead. $\frac{1}{8}=0.125$1/8 = 0.125 stops; $\frac{1}{3}=0.333\dots$1/3 = 0.333… cannot. Always simplify first, because a factor that cancels does not count: $\frac{6}{15}$6/15 has a 3 in its denominator, but $\frac{6}{15}=\frac{2}{5}$6/15 = 2/5, which stops. To check a stopping decimal, write it as a fraction over a power of ten: $\frac{1}{8}=\frac{125}{1000}$1/8 = 125/1000.

The repeat is also guaranteed rather than merely likely: dividing by 7 can only ever leave remainders 1 to 6, so within seven steps a remainder must repeat, and once a remainder repeats the digits do too.

Turning Recurring Decimals into Fractions runs the process backwards with one trick — multiply by a power of ten to line the repeating block up with itself, then subtract. x = 0.272727…, so 100x = 27.2727…, and 99x = 27, giving $x=\frac{27}{99}=\frac{3}{11}$x = 27/99 = 3/11.

The same subtraction settles the argument people have about Why 0.999… Equals Exactly One: x = 0.999…, 10x = 9.999…, so 9x = 9 and x = 1. Not almost one. There is no number in between, and two numbers with nothing between them are one number.

What is an irrational number?

An irrational number is a number that cannot be written as a fraction with whole numbers on the top and bottom. Rational and Irrational Numbers draws the line using decimals: every fraction has a decimal that either stops or repeats, so a decimal that goes on forever without repeating cannot be a fraction. The best-known examples are $\sqrt{2}$√2 and $\pi$π. No fraction squares to exactly 2. You can get close, since ${\left(\frac{7}{5}\right)}^2=\frac{49}{25}$(7/5)² = 49/25, which is just under 2, and ${\left(\frac{3}{2}\right)}^2=\frac{9}{4}$(3/2)² = 9/4, which is over 2, but a proof shows that no fraction lands exactly on it. The square root of a perfect square is a whole number, so $\sqrt{49}$√49 is 7, and the square root of $\frac{9}{4}$9/4 is $\frac{3}{2}$3/2. A long decimal on a calculator does not prove a number is irrational either, because the screen stops after a few digits. What decides it is whether the number can be written exactly as one whole number divided by another.

The Square Root of Two Is Irrational proves that at least one such number exists, and does it by contradiction again. Write $\sqrt{2}=\frac{a}{b}$√2 = a/b in lowest terms. Squaring gives $a^2=2b^2$a² = 2b², so $a^2$a² is even, so a is even. Write a = 2c and the same equation becomes $b^2=2c^2$b² = 2c², so b is even too. But a fraction in lowest terms cannot have both parts even. The assumption is impossible, so no such fraction exists.

Two older ideas round the branch out. Egyptian Fractions and the Greedy Algorithm looks at a notation that only allowed unit fractions — $\frac{3}{7}=\frac{1}{3}+\frac{1}{11}+\frac{1}{231}$3/7 = 1/3 + 1/11 + 1/231 — and at the greedy method that always terminates, which is a real theorem hiding in a curiosity. Continued Fractions writes a number as a nest of reciprocals and produces the best possible fractional approximations: it is where $\frac{355}{113}$355/113 comes from, a value for $\pi$π correct to seven digits with three-digit numbers.

The three mistakes worth naming

• Calling 1 prime. It is excluded on purpose: if 1 were prime, prime factorization would stop being unique, since you could paste on any number of 1s.
• Confusing the two lists. Factors are finite and sit below the number; multiples are infinite and sit above it.
• Reading 0.999… as "just under 1". A recurring decimal is not a number creeping toward a limit; it is the limit, written down.

Where this leads next

Factors are the engine under fraction arithmetic — see fractions, equivalence and mixed numbers — and the terminating-or-recurring question hands straight over to the decimals guide. Powers, roots and surds continue the irrational thread in algebraic expressions, indices and surds.