Pick a number — say, a class of 12 pupils, or a $36 pizza bill to split. Two questions come up all the time: how many ways can I split this evenly, and what do all the ways add up to? In number theory this is: how many divisors does n have, and what do they sum to? Both questions have surprisingly clean answers, and both answers are multiplicative functions — meaning they fall straight out of the prime factorisation without ever having to list a single divisor by hand.

These two functions have their own names and symbols: \tau(n) (or d(n)) counts the divisors, and \sigma(n) sums them. They look like small bookkeeping tools, but as you'll see on the next page, one of them is the exact ingredient behind a 2000-year-old mystery.

A school scheduler asks a version of the first question every term: "into how many equal-sized groups can I split 24 pupils?" — that's really asking for the divisors of 24. A warehouse manager stacking 60 boxes into equal rows asks the same question in disguise. Neither of them wants to try every possible group size by hand — and neither will we.

Counting divisors: \tau(n)

Write \tau(n) for the number of positive divisors of n. Suppose n = p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k} is its prime factorisation. Building a divisor means choosing, independently for each prime p_i, an exponent somewhere from 0 up to a_i — that's a_i + 1 independent choices for each prime, and the choices multiply (a basic counting rule: independent choices combine by multiplying the number of options):

Nothing here needs you to actually list the divisors — the exponents in the factorisation tell you everything. Watch it happen concretely for n = 12 = 2^2 \cdot 3^1: the exponent of 2 can be 0, 1, or 2 (three choices), and the exponent of 3 can be 0 or 1 (two choices). Every one of the 3 \times 2 = 6 pairings gives a different divisor — 2^0 3^0{=}1,\ 2^1 3^0{=}2,\ 2^2 3^0{=}4,\ 2^0 3^1{=}3,\ 2^1 3^1{=}6,\ 2^2 3^1{=}12 — and that list is exactly the six divisors of 12, with nothing missing and nothing repeated.

Summing divisors: \sigma(n)

Write \sigma(n) for the sum of the positive divisors (including n itself). On a single prime power the divisors are 1, p, p^2, \ldots, p^a — a geometric series with a tidy closed form:

Because \sigma is multiplicative, the value on a general n is just the product of its values on each prime-power piece of the factorisation:

The closed form for a single prime power is worth trusting rather than memorising blindly: for p = 2, a = 2 it gives \sigma(4) = \tfrac{2^3-1}{2-1} = 7, which matches 1+2+4=7 directly; for p=3, a=1 it gives \sigma(3) = \tfrac{3^2-1}{3-1} = 4, matching 1+3=4. The formula is just a shortcut for adding up a geometric series — it never disagrees with counting by hand.

Worked examples

The whole point of the formulas is that they match brute-force listing — but arrive there without the listing. Three examples, checked both ways.

Example 1: n = 12.

Direct: the divisors of 12 are 1, 2, 3, 4, 6, 12 — that's 6 of them, summing to 1+2+3+4+6+12 = 28.

By formula: 12 = 2^2 \cdot 3, so

Both methods agree, exactly as they must.

Example 2: n = 100.

100 = 2^2 \cdot 5^2, two primes with matching exponents:

Listing all nine divisors (1,2,4,5,10,20,25,50,100) and adding them by hand gives the same 217 — but the formula got there in two multiplications. Since 217 > 200 = 2 \cdot 100, 100 is abundant too — round numbers with lots of small prime factors tend to be.

Example 3: a prime, n = 13.

A prime is its own one-term factorisation, 13 = 13^1, so \tau(13) = (1+1) = 2 and \sigma(13) = 1 + 13 = 14. This is really just the definition of "prime" wearing a formula: a prime's only divisors are 1 and itself.

Seeing the pairs

Divisors always come in pairs that multiply back to n: if d divides n, so does n / d, and d \cdot (n/d) = n. Below, each divisor of n is placed on a scale where equal steps mean doubling (so multiplying two numbers means the steps simply add). On a scale like that, a pair d and n/d always sits mirrored around \sqrt{n} — the dashed line. Try a few values of n and watch the mirror hold.

For n=12 the pairs are 1{\times}12,\ 2{\times}6,\ 3{\times}4 — three pairs, six divisors total, matching \tau(12) = 6 exactly. The doubling scale is chosen on purpose: multiplying two numbers becomes adding their positions on it, so a pair that multiplies to n always lands the same total distance apart, symmetric about the midpoint \sqrt{n}. When n is a perfect square, one divisor — \sqrt{n} itself — sits exactly on the mirror line with no partner, which is precisely why perfect squares are the only numbers with an odd \tau(n).

Reading a number's character

\tau and \sigma reveal a number's "shape" at a glance. A prime has \tau(p) = 2 — the smallest possible value for anything bigger than 1, since every number has at least the two divisors 1 and itself. A perfect square has an odd \tau, because its square-root divisor is unpaired while every other divisor pairs off with a distinct partner. And comparing \sigma(n) against 2n sorts every number into one of three camps: deficient (\sigma(n) < 2n), abundant (\sigma(n) > 2n), or — vanishingly rarely — exactly balanced.

Take 12 from the worked examples above: \sigma(12) = 28 and 2 \cdot 12 = 24, and 28 > 24, so 12 is abundant — it has more "divisor weight" than its own size. Compare that with 15: its divisors 1, 3, 5, 15 sum to 24, and 24 < 30 = 2 \cdot 15, so 15 is deficient. Most numbers you meet are deficient — abundant numbers are a minority, and the exactly-balanced ones are rarer still.