Math Core

"How many positive divisors does $720$ have?" Listing them one by one is slow and easy to botch. With the prime factorization in hand, you can count them in one line and answer related questions too: how many are odd, how many are perfect squares, and what they add up to.

The divisor-counting formula

Take $72 = 2^3 \cdot 3^2$ . Any divisor of $72$ can use only the primes $2$ and $3$ , and no more of each than $72$ has. So every divisor looks like $2^a \cdot 3^b \quad \text{with } a \in \{0, 1, 2, 3\} \text{ and } b \in \{0, 1, 2\}.$ There are $4$ choices for $a$ and $3$ for $b$ , and each pair gives a different divisor (by unique factorization). So $72$ has $4 \times 3 = 12$ divisors.

Number of divisors

If $n = p_1^{e_1} \cdot p_2^{e_2} \cdots p_k^{e_k}$ , then the number of positive divisors of $n$ is $(e_1 + 1)(e_2 + 1) \cdots (e_k + 1).$

Common mistake

The " $+1$ " is there because an exponent of $0$ is allowed. Students who multiply the exponents themselves ( $3 \times 2 = 6$ for $72$ ) get the wrong answer. Also make sure you've factored all the way into primes: writing $72 = 8 \cdot 9$ and computing $2 \times 2$ is wrong.

Worked example: Count them all

How many positive divisors does $720$ have?

$720 = 2^4 \cdot 3^2 \cdot 5$ , so it has $(4+1)(2+1)(1+1) = 5 \cdot 3 \cdot 2 = 30$ divisors.

Divisors with a condition

Restrict the choice of exponents.

Odd divisors: force the exponent of $2$ to be $0$ .
Even divisors: total minus odd, or force the exponent of $2$ to be at least $1$ .
Multiples of $m$ : force each exponent to be at least what $m$ needs.
Perfect-square divisors: allow only even exponents.

Worked example: Even divisors

How many positive divisors of $720$ are even?

Odd divisors of $720 = 2^4 \cdot 3^2 \cdot 5$ use $2^0$ only: $1 \cdot 3 \cdot 2 = 6$ of them. So there are $30 - 6 = 24$ even divisors. (Or directly: the exponent of $2$ is one of $1, 2, 3, 4$ , giving $4 \cdot 3 \cdot 2 = 24$ .)

Sum of divisors

Multiply out $(1 + 2 + 4)(1 + 3 + 9)$ . Each term of the expansion is one choice from each bracket, like $4 \cdot 3 = 12$ . So you get every divisor of $2^2 \cdot 3^2 = 36$ exactly once, and the product is their sum.

Worked example: Sum of divisors

What is the sum of the positive divisors of $180$ ?

$180 = 2^2 \cdot 3^2 \cdot 5$ , so the sum is $(1 + 2 + 4)(1 + 3 + 9)(1 + 5) = 7 \cdot 13 \cdot 6 = 546$ .

Odd numbers of divisors

Divisors come in pairs: $d$ pairs with $\dfrac{n}{d}$ . For $36$ : $1 \cdot 36$ , $2 \cdot 18$ , $3 \cdot 12$ , $4 \cdot 9$ , and then $6 \cdot 6$ , which pairs $6$ with itself. A number has an odd number of divisors exactly when it is a perfect square. This matches the formula: all $e_i$ even means every factor $e_i + 1$ is odd.

Tip

Numbers with exactly $3$ divisors are squares of primes ( $4, 9, 25, 49, \dots$ ), since $3 = 2 + 1$ forces $n = p^2$ . In general, factor the target count to see what shapes $n$ can take: exactly $6$ divisors means $n = p^5$ or $n = p^2 q$ .