Counting Factors of a Number

Learning outcome

Count positive divisors from prime factorisation and apply the method when divisors must satisfy a simple restriction.

Concepts and reasons

In this lesson, factor counting means counting distinct positive divisors of a positive integer. Negative divisors are excluded, and the method is not applied to zero. The integer 1 has exactly one positive divisor: 1.

For n > 1, write n as a product of powers of distinct primes. If n = p^a × q^b, where a and b are positive integers, its number of positive divisors is (a + 1)(b + 1). Include one such factor for every distinct prime.

Why does this work? The number contains a copies of p. A divisor may include none, one, two, and so on up to a copies: a + 1 choices. Independently, it has b + 1 choices for copies of q. Every combination produces one divisor, so multiply the choice counts rather than adding them.

Choosing no copies of a prime contributes a factor of 1, not 0. A divisor cannot introduce a new prime or use more copies than n contains. Unique prime factorisation makes different choices give different divisors.

In a perfect square, prime exponents are even, so every choice count is odd. The total divisor count is therefore odd, matching the square-root factor pairing with itself.

Worked examples

Example 1. Since 72 = 2^3 × 3^2, its divisor count is (3 + 1)(2 + 1) = 12. The choices for copies of 2 are 0, 1, 2, 3; for copies of 3, they are 0, 1, 2.

Example 2. Since 100 = 2^2 × 5^2, its divisor count is 3 × 3 = 9. This odd count matches the fact that 100 is a perfect square.

Example 3. Count divisors of 120 divisible by 6. Write 120 = 2^3 × 3 × 5. Include at least one 2: three choices. Include the 3: one choice. Include or omit 5: two choices. Total = 3 × 1 × 2 = 6.

Common traps

Use distinct prime bases, not composite factors. Combine repeated prime factors first. Count 1 and the number itself. Restrictions change the available choices, not the multiplication principle.

Practice questions

1. How many positive divisors does 180 have?
2. How many positive divisors do 1 and 13 have?
3. How many positive divisors of 200 are odd?

Answers and explanations

1. Since 180 = 2^2 × 3^2 × 5, the primes allow 3, 3 and 2 choices respectively. Thus the count is 3 × 3 × 2 = 18.
2. The number 1 has one divisor: 1. The prime 13 has two: 1 and 13.
3. Since 200 = 2^3 × 5^2, an odd divisor must omit every factor 2. Choosing 0, 1 or 2 copies of 5 gives three divisors: 1, 5, 25.