Factors, prime factorisation & trailing zeros — notes, question types & practice · Number System

01

Overview

A factor of a number divides it exactly. The factors of $12$ are $1, 2, 3, 4, 6, 12$ . A multiple is the reverse idea: $12, 24, 36, \ldots$ Every question here starts from the prime factorisation, so that skill comes first.

02

Prime factorisation

Keep dividing by the smallest prime that fits. $720 = 2 \times 360 = 2 \times 2 \times 180 = \cdots = 2^4 \times 3^2 \times 5$ . Write the result in power form $N = p^a \times q^b \times r^c$ ; every method below reads the exponents.

03

Counting factors

A factor of $2^4 \times 3^2 \times 5$ chooses one power of each prime. The power of $2$ : $0, 1, 2, 3$ or $4$ , so five choices. The power of $3$ : $0, 1$ or $2$ , so three choices. The power of $5$ : $0$ or $1$ , so two choices.

Rule: Number of factors $= (a+1)(b+1)(c+1)$ . For $720$ : $5 \times 3 \times 2 = 30$ factors.

04

Odd, even and special factors

Odd factors: drop the power of $2$ completely. For $3600 = 2^4 \times 3^2 \times 5^2$ : odd factors $= 3 \times 3 = 9$ .
Even factors: total minus odd, here $45 - 9 = 36$ .
Factors that are multiples of $k$ : divide $N$ by $k$ and count factors of the result. For $1440$ with $k = 12$ : $1440 \div 12 = 120 = 2^3 \times 3 \times 5$ , giving $16$ .
Perfect-square factors: every exponent may take even values only. Cube factors: $0, 3, 6, \ldots$

05

Sum of factors

Write one bracket per prime and multiply.

\text{sum} = (1 + p + \cdots + p^a)(1 + q + \cdots + q^b) \cdots

For $360 = 2^3 \times 3^2 \times 5$ : $(1 + 2 + 4 + 8)(1 + 3 + 9)(1 + 5) = 15 \times 13 \times 6 = 1170$ . For the sum of even factors only, start the bracket of $2$ at $2$ : $(2 + 4 + 8) \times \cdots$

Tip: The bracket form also answers sum of odd factors: just delete the bracket of $2$ .

06

Product of factors and factor pairs

Factors pair as $(1, N), (2, N/2), \ldots$ and each pair multiplies to $N$ . So the product of all factors of $N$ is $N^{d/2}$ where $d$ is the factor count: for $12$ , with $6$ factors, the product is $12^3 = 1728$ .

The same pairing counts the ways to write $N$ as a product of two factors: $d/2$ ways when $N$ is not a perfect square, $(d+1)/2$ when it is. For two co-prime factors the answer is $2^{k-1}$ , with $k$ the number of distinct primes.

07

Trailing zeros of factorials

A trailing zero needs a $10 = 2 \times 5$ . Factorials hold more twos than fives, so count the fives: divide by $5$ , then the quotient by $5$ again, and add.

Rule: Zeros in $n! = \left\lfloor \dfrac{n}{5} \right\rfloor + \left\lfloor \dfrac{n}{25} \right\rfloor + \left\lfloor \dfrac{n}{125} \right\rfloor + \cdots$ . For $125!$ : $25 + 5 + 1 = 31$ zeros.

08

Highest power of a prime in a factorial

Same successive division by the prime. The power of $3$ in $100!$ is $33 + 11 + 3 + 1 = 48$ . For a composite such as $12 = 2^2 \times 3$ , find each prime's power, divide the power of $2$ by $2$ , and take the smaller result. In $50!$ : twos give $47$ , so $23$ twos-pairs, and threes give $22$ ; the answer is $22$ .

09

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common2 practice Q

Number of factors: total, odd, even

How to spot it:

The question asks how many factors or divisors a number has, or how many of them are odd or even.

N = p^a q^b r^c \ \Rightarrow\ d(N) = (a + 1)(b + 1)(c + 1)

Method

Prime-factorise $N$ .
Total factors: multiply (each exponent $+ 1$ ).
Odd factors: ignore the power of $2$ entirely.
Even factors: total minus odd.

Why it works:

Each factor picks one power of every prime, and the choices are independent.

Try this

How many factors does 420 have?

Show solution

$420 = 2^2 \times 3 \times 5 \times 7$ .
Total: $(2+1)(1+1)(1+1)(1+1) = 3 \times 2 \times 2 \times 2$ .
$= 24$ factors.

Answer

24

Type 2common2 practice Q

Special factors: squares, cubes, multiples of k

How to spot it:

The question asks how many factors of N are perfect squares, perfect cubes, or divisible by a given number.

\#\{\text{factors divisible by } k\} = d(N / k)

Method

For square factors, allow only even exponents $0, 2, 4, \ldots$ for every prime.
For cube factors, allow only $0, 3, 6, \ldots$
For multiples of $k$ , divide $N$ by $k$ and count factors of the quotient.

Why it works:

A square factor carries even powers only; a factor divisible by k is k times a factor of N/k.

Try this

How many factors of 2 to the power 6 x 3 to the power 4 are perfect squares?

Show solution

Powers of $2$ available to a square: $0, 2, 4, 6$ , that is $4$ choices.
Powers of $3$ : $0, 2, 4$ , that is $3$ choices.
$4 \times 3 = 12$ square factors.

Answer

12

Type 3common2 practice Q

Sum of factors

How to spot it:

The question asks for the sum of all factors of N, or the sum of only the odd or only the even factors.

\sigma(N) = (1 + p + \cdots + p^a)(1 + q + \cdots + q^b) \cdots

Method

Prime-factorise $N$ .
Write one bracket per prime, running from $1$ to the full power.
Multiply the brackets.
Odd sum: delete the bracket of $2$ . Even sum: start that bracket at $2$ .

Why it works:

Expanding the bracket product produces every factor exactly once.

Try this

Find the sum of all factors of 90.

Show solution

$90 = 2 \times 3^2 \times 5$ .
Brackets: $(1 + 2)(1 + 3 + 9)(1 + 5)$ .
$= 3 \times 13 \times 6 = 234$ .

Answer

234

Type 4very common2 practice Q

Trailing zeros in a factorial or product

How to spot it:

The question asks how many zeros end a factorial like 90!, or end a written-out product.

Z(n!) = \left\lfloor \dfrac{n}{5} \right\rfloor + \left\lfloor \dfrac{n}{25} \right\rfloor + \cdots

Method

Divide $n$ by $5$ and keep the whole part.
Divide that quotient by $5$ again, and keep going.
Add all the quotients.
For a plain product, count the twos and the fives and take the smaller count.

Why it works:

Each trailing zero needs one two and one five, and fives are the scarce partner.

Try this

How many zeros are at the end of 90 factorial?

Show solution

$90 \div 5 = 18$ .
$18 \div 5 = 3$ .
Zeros: $18 + 3 = 21$ .

Answer

21

Type 5common2 practice Q

Highest power of a prime or composite in a factorial

How to spot it:

The question asks for the largest n such that 7 to the n divides 150!, or the highest power of 12 in 50!.

E_p(n!) = \sum_{k \ge 1} \left\lfloor \dfrac{n}{p^k} \right\rfloor

Method

For a prime $p$ : divide $n$ by $p$ , then the quotient again, and add the quotients.
For a composite, split into prime powers.
Divide each prime's count by the power required.
Take the smallest result.

Why it works:

Floor division counts multiples of p, then the extra p carried by multiples of its square, and so on.

Try this

Find the highest power of 7 that divides 150 factorial.

Show solution

$150 \div 7 = 21$ .
$21 \div 7 = 3$ .
Total: $21 + 3 = 24$ .

Answer

7 to the power 24

Type 6occasional3 practice Q

Product of factors and factor pairs

How to spot it:

The question asks for the product of all factors of N, or in how many ways N can be written as a product of two factors.

\prod \text{factors} = N^{d/2}, \qquad \text{pairs} = \dfrac{d}{2} \ (N \text{ not a square})

Method

Count the factors $d$ .
Product of factors: raise $N$ to $d/2$ .
Two-factor products: $d/2$ ways, or $(d+1)/2$ for a perfect square.
Co-prime pairs: $2^{k-1}$ with $k$ distinct primes.

Why it works:

Factors pair as f with N over f, and each pair multiplies to N.

Try this

In how many ways can 48 be written as a product of two factors?

Show solution

$48 = 2^4 \times 3$ , so $d = (4+1)(1+1) = 10$ factors.
$48$ is not a perfect square.
Ways: $10 \div 2 = 5$ .

Answer

5

10

Formula sheet

Number of factors

d(N) = (a + 1)(b + 1)(c + 1)

Sum of factors

\sigma(N) = (1 + p + \cdots + p^a)(1 + q + \cdots + q^b) \cdots

Even factors

a \times (b + 1)(c + 1)

when 2 has exponent a

Product of all factors

N^{d(N)/2}

Trailing zeros in n factorial

\left\lfloor \dfrac{n}{5} \right\rfloor + \left\lfloor \dfrac{n}{25} \right\rfloor + \left\lfloor \dfrac{n}{125} \right\rfloor + \cdots

Highest power of a prime in n factorial

\sum_{k \ge 1} \left\lfloor \dfrac{n}{p^k} \right\rfloor

11

Shortcuts that save time

⚡ Even factors: force one two

If N has two to the power a in it, even factors equal a times the factor count of the odd part.

Example

How many even factors does 360 have?

Show solution

$360 = 2^3 \times 3^2 \times 5$ .
Even factors $= 3 \times (2 + 1) \times (1 + 1) = 3 \times 3 \times 2$ .
$= 18$ .

Answer

18

⚡ Successive division by five

For trailing zeros, keep dividing n by 5 and add the quotients until the quotient is 0.

Example

How many zeros end 1000!?

Show solution

$1000 \to 200 \to 40 \to 8 \to 1$ .
Sum: $200 + 40 + 8 + 1$ .
$249$ zeros.

Answer

249

⚡ Sum of factors as brackets

One bracket per prime, each running from 1 up to the full power. Multiply the brackets.

Example

Find the sum of all factors of 72.

Show solution

$72 = 2^3 \times 3^2$ .
Brackets: $(1 + 2 + 4 + 8)(1 + 3 + 9) = 15 \times 13$ .
$= 195$ .

Answer

195

12

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Counting only n/5 for trailing zeros.

Add the extra fives from 25, 125 and higher powers of 5 as well.

Mistake 02

Adding exponents instead of multiplying (a+1)(b+1)(c+1).

The choices per prime multiply, they never add.

Mistake 03

Forgetting 1 and N as factors.

Both count; the formula (a+1)(b+1)(c+1) already includes them.

Mistake 04

Adding the zero counts for a sum like 100! + 200!.

The sum keeps only the smaller count of trailing zeros.

Mistake 05

Testing for a prime with squares of primes only.

Divide by every prime up to the square root of N.

13

Quick revision

Read this the night before the exam.

Factor count: $(a+1)(b+1)(c+1)$ ; odd factors: drop the $2$ .
Multiples of $k$ among the factors: count factors of $N/k$ .
Square factors: even exponents only; cube factors: $0, 3, 6, \ldots$
Sum: one bracket per prime, multiply the brackets.
Product of all factors: $N^{d/2}$ .
Zeros in $n!$ : keep dividing $n$ by $5$ and add quotients.
Composite power in $n!$ : split into prime powers and take the smallest.

14

Practice: 17 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 10 questions

Suggested time 8 min · wrong answers go to your mistake notebook automatically.