Number System

high importance~2 Q in Tier 133 formulas⚡ 18 shortcuts6 subtopics

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Subtopic 5 of 6·← Unit digit & cyclicity·Fractions, decimals & recurring decimals →

Factors, prime factorisation & trailing zeros

Write $N = p^a q^b r^c$ (prime factorisation). Everything about factors comes from the exponents.

Number of factors $= (a+1)(b+1)(c+1)$
Odd factors: ignore the power of 2. Even factors = total − odd (or give 2 an exponent from 1 to its max).
Perfect-square factors: each exponent can only take even values 0, 2, 4, …
Sum of factors: multiply $(1 + p + \dots + p^a)$ for each prime.

Trailing zeros of a product = number of (2, 5) pairs = number of 5s (5s are scarcer). In n!, count 5s by successive division by 5.

📖 Notes 🧩 6 question types ⚡ Formulas 🎯 17 practice Q

Detailed notes

Factors and multiples

A factor (divisor) of a number divides it exactly. Factors of 12 are 1, 2, 3, 4, 6, 12. A multiple is the opposite: 12, 24, 36 … are multiples of 12. Every question on "how many factors" starts with prime factorisation — writing the number as a product of primes.

Prime factorisation

Keep dividing by the smallest prime that works. $720 = 2 \times 360 = 2 \times 2 \times 180 = \dots = 2^4 \times 3^2 \times 5$ . Write it in power form: $N = p^a \times q^b \times r^c$ .

Number of factors

A factor of $2^4 \times 3^2 \times 5$ can take the power of 2 as 0, 1, 2, 3 or 4 (5 choices), the power of 3 as 0, 1 or 2 (3 choices), and the power of 5 as 0 or 1 (2 choices). $\text{Number of factors} = (a+1)(b+1)(c+1)$ For 720: $5 \times 3 \times 2 = 30$ factors.

Odd, even and special factors

Odd factors: ignore the 2 completely. $3600 = 2^4 \times 3^2 \times 5^2$ → odd factors $= 3 \times 3 = 9$ .
Even factors = total − odd. For 3600: $45 - 9 = 36$ .
Factors that are multiples of $k$ : divide $N$ by $k$ and count the factors of $N/k$ . For 1440 and $k = 12$ : $1440/12 = 120 = 2^3 \times 3 \times 5$ → $4 \times 2 \times 2 = 16$ .
Perfect-square factors: each power can only be even (0, 2, 4 …). Perfect-cube factors: powers 0, 3, 6 …

Sum of factors

$\text{Sum} = (1 + p + \dots + p^a)(1 + q + \dots + q^b)\dots$ For $360 = 2^3 \times 3^2 \times 5$ : $(1+2+4+8)(1+3+9)(1+5) = 15 \times 13 \times 6 = 1170$ . Sum of even factors: drop the 1 from the bracket of 2: $(2+4+8) \times \dots$

Product of factors

If $N$ has $d$ factors, the product of all its factors is $N^{d/2}$ . Factors pair up as $(1, N), (2, N/2), \dots$ — each pair multiplies to $N$ .

Writing N as a product of two factors

Number of ways $= d/2$ (if $N$ is not a perfect square), or $(d+1)/2$ if it is (the square root pairs with itself).
As a product of two co-prime factors: $2^{k-1}$ , where $k$ = number of different primes in $N$ .

Trailing zeros

A zero at the end comes from a factor 10 = 2 × 5. In $n!$ there are always more 2s than 5s, so count the 5s: $\text{zeros in } n! = \left\lfloor \frac{n}{5} \right\rfloor + \left\lfloor \frac{n}{25} \right\rfloor + \left\lfloor \frac{n}{125} \right\rfloor + \dots$ 125! → 25 + 5 + 1 = 31 zeros. Quick way: keep dividing by 5 and add the quotients. Careful with products like $5 \times 10 \times 15 \times \dots \times 100$ : here 5s are plenty and 2s are fewer, so count both and take the smaller.

Highest power of a prime in n!

Same successive division, but by the prime $p$ : power of 3 in 100! $= 33 + 11 + 3 + 1 = 48$ . For a composite like 12 = $2^2 \times 3$ : find the power of 2 (divide by 2 because of the square) and the power of 3; the answer is the smaller.

Quick revision

Factors: $(a+1)(b+1)(c+1)$ ; odd factors → drop 2.
Multiples of $k$ among factors → factors of $N/k$ .
Sum: product of $(1 + p + \dots + p^a)$ brackets.
Product of factors $= N^{d/2}$ .
Zeros in $n!$ = successive division by 5.
Highest power of $p$ in $n!$ = successive division by $p$ .

Types of questions asked

Every way this subtopic shows up in exams — how to recognise it, the formula or logic to use, and a solved example.

Type 1: Number of factors (total, odd, even)very common2 practice Q

How to spot it:

'How many factors / divisors does 720 have?', 'how many odd (or even) factors …'.

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

Prime-factorise $N$ .
Total: multiply (each power + 1).
Odd: leave out the power of 2. Even: total − odd.

Why: each factor is built by choosing a power for every prime independently.

Example: How many factors does 252 have?

$252 = 2^2 \times 3^2 \times 7$ → $3 \times 3 \times 2 = 18$ factors.

Type 2: Special factors: perfect squares/cubes, multiples of kcommon2 practice Q

How to spot it:

'How many factors of N are perfect cubes?', 'how many factors of 1440 are divisible by 12?'.

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

Perfect squares: count only even powers for each prime (0, 2, 4, …). Cubes: 0, 3, 6, …
Multiples of $k$ : divide $N$ by $k$ and count the factors of the quotient.

Why: a factor divisible by $k$ is $k$ times a factor of $N/k$ .

Example: How many factors of $2^5 \times 3^3$ are perfect squares?

Powers of 2: 0, 2, 4 (3 choices). Powers of 3: 0, 2 (2 choices). Answer $3 \times 2 = 6$ .

Type 3: Sum of factors (all, odd or even)common2 practice Q

How to spot it:

'Find the sum of all factors of 360', 'sum of even factors of 72'.

\sigma(N) = \prod \frac{p^{a+1}-1}{p-1} = (1+p+\dots+p^a)(1+q+\dots+q^b)

Prime-factorise.
Write a bracket $(1 + p + p^2 + \dots)$ for each prime and multiply.
Even factors: bracket for 2 starts from 2. Odd factors: drop the bracket of 2.

Why: expanding the product gives every factor exactly once.

Example: Find the sum of all factors of 60.

$60 = 2^2 \times 3 \times 5$ → $(1+2+4)(1+3)(1+5) = 7 \times 4 \times 6 = 168$ .

Type 4: Trailing zeros in a factorial or productvery common2 practice Q

How to spot it:

'How many zeros are there at the end of 125!?' or at the end of a product like $5 \times 10 \times \dots \times 100$ .

Z(n!) = \left\lfloor \tfrac{n}{5} \right\rfloor + \left\lfloor \tfrac{n}{25} \right\rfloor + \left\lfloor \tfrac{n}{125} \right\rfloor + \dots

For $n!$ : divide $n$ by 5, divide the quotient by 5 again, … and add all quotients.
For other products: count 2s and 5s separately; zeros = the smaller count.

Why: each zero needs one 2 and one 5.

Example: How many zeros are at the end of 80!?

$80 \div 5 = 16$ , $16 \div 5 = 3$ . Zeros $= 16 + 3 = 19$ .

Type 5: Highest power of a prime (or composite) dividing n!common2 practice Q

How to spot it:

'Find the largest n such that $3^n$ divides 100!', or 'highest power of 12 in 50!'.

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

For a prime $p$ : successive division of $n$ by $p$ , add quotients.
For a composite, split into prime powers ( $12 = 2^2 \times 3$ ), find each exponent, divide by the power needed, take the smallest.

Why: $\lfloor n/p \rfloor$ counts multiples of $p$ , $\lfloor n/p^2 \rfloor$ adds the extra $p$ from multiples of $p^2$ , and so on.

Example: Highest power of 7 in 100!?

$100 \div 7 = 14$ , $14 \div 7 = 2$ . Answer $14 + 2 = 16$ .

Type 6: Product of factors; ways to write N as a product of two factorsoccasional3 practice Q

How to spot it:

'Find the product of all factors of 36', 'in how many ways can 360 be written as a product of two (co-prime) factors?'.

\prod d = N^{d(N)/2},\quad \text{pairs} = \frac{d(N)}{2},\quad \text{co-prime pairs} = 2^{k-1}

Count factors $d(N)$ .
Product of factors $= N^{d/2}$ .
Ways as a product of two factors $= d/2$ ( $(d+1)/2$ for a perfect square).
Co-prime pairs: each distinct prime goes wholly to one side → $2^k$ ordered splits, $2^{k-1}$ unordered.

Why: factors pair up as $(f, N/f)$ .

Example: Find the product of all factors of 12.

$12$ has 6 factors → product $= 12^{6/2} = 12^3 = 1728$ . Check: $1 \times 2 \times 3 \times 4 \times 6 \times 12 = 1728$ ✓.

Formulas

Number of factors

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

Sum of factors

\sigma(N) = \frac{p^{a+1}-1}{p-1} \cdot \frac{q^{b+1}-1}{q-1} \cdot \frac{r^{c+1}-1}{r-1}

Even factors

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

when p = 2 with exponent a

Product of all factors

N^{d(N)/2}

Trailing zeros in n!

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

Highest power of prime p in n!

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

Shortcut tricks

⚡ Even factors: force one 2

If $N = 2^a \times (\text{odd part})$ , even factors $= a \times$ (factors of odd part).

Example: How many even factors does 360 have?

$360 = 2^3 \times 3^2 \times 5$ ⇒ even factors $= 3 \times 3 \times 2 = 18$ .

⚡ Successive division by 5

Keep dividing by 5 (dropping remainders) and add the quotients.

Example: How many zeros are at the end of 1000!?

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

⚡ Sum of factors as bracket product

Write one bracket per prime: (1 + p + p² + …). Multiply.

Example: Find the sum of all factors of 72.

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

Where students lose marks

Counting only n/5 for trailing zeros and missing the extra 5s from 25, 125, …
Adding exponents instead of multiplying (a + 1)(b + 1)(c + 1).
Forgetting that 1 and N are themselves factors.
For trailing zeros of a SUM (e.g. 100! + 200!) adding the zeros — the sum has the smaller count.

Practice sets — 17 questions

Sets of 10, mixed across the question types above. Each answer comes with a step-by-step explanation.

Topic test · 10 questions

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