AFCAT Numerical Ability · Number System

Prime Factorisation, and Counting Divisors from It

Write a number as a product of powers of primes, then read off its distinct primes, its number of divisors, its odd and even divisors and the sum of its divisors.

All Number System notes

Why this matters

This is the largest page of the chapter, and half of it asks for a factorisation in powers of primes. The options differ by a single exponent, so multiply your answer back before you choose. The rest reads facts off the factorisation, above all how many divisors a number has.

Concept 1 of 4: Writing a prime factorisation

Divide by the smallest prime that goes in, again and again: 2 until the number is odd, then 3, then 5, then 7. When what is left is 1, the primes you divided by are the factorisation. Splitting into known pieces first (a power of 10, a square) is often faster.

Definition

Every whole number above 1 has exactly one prime factorisation, written as powers: $2^a \times 3^b \times 5^c \times \cdots$
Use the divisibility rules from the last page to see which prime divides next.
Known pieces help: $100 = 2^2 \times 5^2$ , $1000 = 2^3 \times 5^3$ , $2^{10} = 1024$ .
Check by multiplying back. The options usually differ by one exponent.

Prime factorisation

N = p_1^{a_1} \times p_2^{a_2} \times \cdots \times p_k^{a_k} \quad (p_i \text{ prime})

Worked example

Find the prime factorisation of 1,260.

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionHard

Example 1 · Number System · Prime Factorization

Find the prime factorization of

1440

Keep going until every factor is prime

2^2 \times 3 \times 15

is not finished:

15 = 3 \times 5

, so the factorisation is

2^2 \times 3^2 \times 5

Add exponents when pieces share a prime

3600 = 36 \times 100

has

2^2

from each piece, so the power of 2 is

2^4

, not

2^2

Drill 15 more on writing a prime factorisation

Concept 2 of 4: Reading a factorisation: distinct primes, repeats and the number itself

Once the number is written as powers of primes, many questions are a glance: the bases are the distinct primes, the exponents count the repeats, and multiplying out gives the number back.

Definition

Distinct prime factors: the bases. $2^3 \times 5^2$ has two, 2 and 5.
Prime factors counted with repeats: the sum of the exponents. $2^3 \times 5^2$ has five: 2, 2, 2, 5, 5.
Product of the distinct primes: multiply the bases once each.
Exponent form: $2 \times 2 \times 7$ is $2^2 \times 7$ .
A product is a prime factorisation only if every base is prime; a 9, 15 or 21 hides more primes.

For N = pᵃ × qᵇ × rᶜ

\text{distinct primes} = 3,\qquad \text{primes with repeats} = a + b + c

Worked example

2700 = 2^2 \times 3^3 \times 5^2

. Find its number of distinct prime factors, its prime factors counted with repeats, and the product of its distinct primes.

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionModerate

Example 2 · Number System · Prime Factorization

How many total prime factors (with multiplicity) does

72

have?

Distinct is not counted with repeats

2^3 \times 5

has 2 distinct prime factors but 4 counted with repeats. Read which one the question asks for.

A bare exponent counts once

2^4 \times 3 \times 5^2

the 3 has exponent 1. Leaving it out makes the sum of exponents 6 instead of 7.

Drill 7 more on reading a factorisation: distinct primes, repeats and the number itself

Concept 3 of 4: How many divisors a number has

A divisor of

2^3 \times 5^2

takes some power of 2 (from

2^0

2^3

, four choices) and some power of 5 (from

5^0

5^2

, three choices). Every pair of choices is a different divisor, so there are

4 \times 3 = 12

Definition

If $N = p^a \times q^b \times r^c$ , the number of divisors is $(a + 1)(b + 1)(c + 1)$ .
The count includes 1 and $N$ itself.
A prime has 2 divisors; $p^a$ has $a + 1$ .
Only perfect squares have an odd number of divisors, because only there does one divisor pair with itself.

Number of divisors

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

Worked example

How many divisors has 450?

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionModerate

Example 3 · Number System · Prime Factorization

What is the number of factors of

720

Add 1 to each exponent first

2^3 \times 3^2

has

(3 + 1)(2 + 1) = 12

divisors. Multiplying the exponents themselves gives 6, which is wrong.

Factorise fully before counting

Written as

4 \times 25

, 100 seems to give

(1 + 1)(1 + 1) = 4

divisors. Only prime powers work:

2^2 \times 5^2

gives 9.

Drill 3 more on how many divisors a number has

Concept 4 of 4: Odd divisors, even divisors and the sum of the divisors

An odd divisor takes no factor of 2 at all, so count the divisors of the number with its power of 2 removed. Every other divisor is even. The sum of the divisors multiplies out the same choices, adding the powers of each prime instead of counting them.

Definition

For $N = 2^a \times q^b \times r^c$ with $q, r$ odd primes: odd divisors $(b + 1)(c + 1)$ .
Even divisors = all divisors minus odd ones = $a(b + 1)(c + 1)$ .
Sum of divisors: $(1 + p + \cdots + p^a)(1 + q + \cdots + q^b)\cdots$
An odd number has no even divisors at all.

Odd and even divisors of 2ᵃ × qᵇ × rᶜ

\text{odd} = (b + 1)(c + 1),\qquad \text{even} = a(b + 1)(c + 1)

Worked example

How many odd and how many even divisors has 1,200?

Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionHard

Example 4 · Number System · Prime Factorization

How many even divisors does

360

have?

Even divisors are not half the total

2^3 \times 3 = 24

has 8 divisors: 2 odd (1 and 3) and 6 even. Split the count with the formula, never in half.

Drill 2 more on odd divisors, even divisors and the sum of the divisors

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (4)

Writing a prime factorisation
Prime factorisation
$N = p_1^{a_1} \times p_2^{a_2} \times \cdots \times p_k^{a_k} \quad (p_i \text{ prime})$
Reading a factorisation: distinct primes, repeats and the number itself
For N = pᵃ × qᵇ × rᶜ
$\text{distinct primes} = 3,\qquad \text{primes with repeats} = a + b + c$
How many divisors a number has
Number of divisors
$N = p^a q^b r^c \;\Rightarrow\; d(N) = (a + 1)(b + 1)(c + 1)$
Odd divisors, even divisors and the sum of the divisors
Odd and even divisors of 2ᵃ × qᵇ × rᶜ
$\text{odd} = (b + 1)(c + 1),\qquad \text{even} = a(b + 1)(c + 1)$

Watch out for (7)

Keep going until every factor is prime
→ Writing a prime factorisation
Add exponents when pieces share a prime
→ Writing a prime factorisation
Distinct is not counted with repeats
→ Reading a factorisation: distinct primes, repeats and the number itself
A bare exponent counts once
→ Reading a factorisation: distinct primes, repeats and the number itself
Add 1 to each exponent first
→ How many divisors a number has
Factorise fully before counting
→ How many divisors a number has
Even divisors are not half the total
→ Odd divisors, even divisors and the sum of the divisors

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