Number of Factors: Total, Even, Odd, Perfect-Square and Sum

Every question on this topic begins the same way: break the number into primes and write it with exponents. Once you have, say, N = 2ᵃ × 3ᵇ × 5ᶜ, every factor of N is built by choosing a power of 2 from 2⁰ to 2ᵃ, a power of 3 from 3⁰ to 3ᵇ, and a power of 5 from 5⁰ to 5ᶜ. Counting factors is therefore counting choices.

Divide by small primes in order, 2, then 3, then 5, then 7, until you reach 1. Write the result neatly; a wrong exponent at this stage spoils every later step.

For example, 360 = 2³ × 3² × 5. This single line is the input for every worked example below.

There are (a + 1) choices for the power of 2, because the exponent can be 0, 1, …, a. Likewise (b + 1) for 3 and (c + 1) for 5. Multiply the choices:

A useful consequence: the number of ways to write N as a product of two factors is half the number of factors, because factors pair up as a and N ÷ a. If N is a perfect square, the square root pairs with itself, so add 1 before halving.

Restrict the choices to match the condition, then multiply as before.

• Odd factors contain no 2, so drop the 2 entirely and use the other brackets.
• Even factors contain at least one 2, so the power of 2 has only a choices (1 to a). Alternatively, subtract the odd factors from the total.
• Perfect-square factors need every exponent even. For exponent a, the even choices are 0, 2, 4, …, a count of ⌊a ÷ 2⌋ + 1.
• Factors divisible by k are k times a factor of N ÷ k, so count the factors of N ÷ k.

The odd factors are therefore 24 − 18 = 6, which matches (2 + 1)(1 + 1) = 6 when the 2 is dropped.

Each bracket in the factor formula becomes a sum of powers. For N = 2ᵃ × 3ᵇ × 5ᶜ:

Multiplying out the brackets produces every possible combination of one power from each bracket, which is exactly the list of factors, so the product is their sum. The same reasoning explains why the counting formula works.

• ✘ Forgetting the "+1" for the zero exponent. 2³ gives 4 choices, not 3.
• ✘ Factorising carelessly. Check by multiplying the primes back together.
• ✘ Counting even factors with (a + 1) choices for 2. An even factor needs at least one 2.
• ✘ Halving an odd count when pairing factors of a perfect square. Add 1 first.

Factor counts connect to HCF and LCM questions, which use the same prime factorisations, and to the divisibility rules that speed up factorising. For a quick formula sheet, see the Quant cheat code.

Do I count 1 and the number itself as factors?

Yes. The formula counts every factor, including 1 and the number itself. If a question asks for proper factors or factors other than 1 and the number, subtract the ones it excludes from the total.

Why does a perfect square have an odd number of factors?

Factors pair up as a and N ÷ a. For a perfect square, the square root pairs with itself, so one factor has no partner and the total is odd. Every exponent in its factorisation is even, so every bracket (a + 1) is odd.

All problems and solutions on this page are original. No outside facts are relied on; every numeric answer is recomputed by software before publishing.

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