Sec 1 Primes, Factors and Multiples: Examples & Guide

Secondary 1 Mathematics · G3 focus · Study guide

Build prime factorisations, choose HCF or LCM for a situation, and use prime powers to understand squares, cubes and roots.

Concept → worked examples → practice15 min read · practice linked below

Factors divide; multiples are made by multiplying

A factor of a positive integer divides it exactly. Since $3\times8=24$ , both 3 and 8 are factors of 24, and 24 is a multiple of each. List factors in pairs to avoid omissions: the positive factors of 24 are $1,2,3,4,6,8,12,24$ . Positive multiples continue without end: $24,48,72,\ldots$ .

A prime number is a whole number greater than 1 with exactly two positive factors: 1 and itself. The first primes are $2,3,5,7,11,13$ . A composite number has more than two positive factors. The number 1 is neither prime nor composite; 2 is the only even prime.

A quick divisibility check

A number is divisible by 2 if its last digit is even, by 5 if its last digit is 0 or 5, and by 3 or 9 if its digit sum is divisible by 3 or 9 respectively. These checks help you begin a factorisation. To test whether a number is prime, test prime divisors up to its square root: any composite number has a factor no greater than its square root.

Prime factorisation and index notation

Break a number into factors until every remaining factor is prime. The order of the factors does not matter. Index notation records repeated multiplication: $2^3=2\times2\times2$ , not $2\times3$ .

Worked example 1: factorise 360

Express 360 as a product of prime factors in index notation.

Start with an easy factor pair: $360=36\times10$ .
Break down each factor: $36=6\times6=2\times3\times2\times3$ , and $10=2\times5$ .
Collect equal primes: $\boxed{360=2^3\times3^2\times5}.$

Check: $8\times9\times5=360$ . Stop only when every factor is prime; $6^2\times10$ is a factorisation, but not a prime factorisation.

Different factor trees give the same prime factors, apart from their order.

Why HCF uses smaller powers and LCM uses larger powers

The highest common factor (HCF) is the largest positive integer dividing all the numbers. It cannot contain more copies of a prime than any of the numbers contains. Use the smallest power of each common prime.

The lowest common multiple (LCM) is the smallest positive integer divisible by all the numbers. It must contain enough copies of every prime to include each number. Use the largest power of every prime present. If a prime is absent, its power is 0, and $p^0=1$ .

Worked example 2: HCF and LCM of 72 and 120

First write $72=2^3\times3^2$ and $120=2^3\times3\times5$ .

For the HCF, both numbers contain three 2s and at least one 3. Only 120 contains 5, so 5 is excluded: $\boxed{\operatorname{HCF}=2^3\times3=24}.$
For the LCM, take three 2s, two 3s and one 5: $\boxed{\operatorname{LCM}=2^3\times3^2\times5=360}.$

Check: $72\div24=3$ and $120\div24=5$ , while $360\div72=5$ and $360\div120=3$ . For two positive integers, $\text{HCF}\times\text{LCM}$ equals their product: $24\times360=72\times120$ . Do not assume this product rule works unchanged for three numbers.

Each entry represents one copy of a prime factor. Shared copies appear once in the overlap, so they are not counted twice in the LCM.

Explore the prime powers

Choose two positive whole numbers and compare the prime powers. Predict which powers belong to their HCF and LCM before reading the last two columns.

36 = 2² × 3²; 60 = 2² × 3 × 5.

Choose an exponent for each prime
Prime	Power in 36	Power in 60	HCF: smaller	LCM: larger
2	2	2	2	2
3	2	1	1	2
5	0	1	0	1

HCF = 12; LCM = 180. The HCF divides both numbers; the LCM is divisible by both.

Choose HCF or LCM from the situation

Look at what your answer must do, rather than relying on the word “largest” or “smallest” alone. A common piece length must divide every original length. A common repeat time must be a multiple of every interval.

Worked example 3: equal ribbon pieces

Two ribbons are 84 cm and 126 cm long. Cut both into the longest possible equal pieces, with none left over. Find the piece length and total number of pieces.

The piece length must divide both 84 and 126. We therefore need their HCF.
Factorise: $84=2^2\times3\times7$ and $126=2\times3^2\times7$ . Thus $\operatorname{HCF}=2\times3\times7=42$ .
Each piece is 42 cm. The number of pieces is $84\div42+126\div42=2+3=\boxed{5}.$

Check: two 42 cm pieces reconstruct the first ribbon, and three reconstruct the second. Do not give 42 as the number of pieces: it is a length.

Worked example 4: events repeating together

Two lights flash every 18 seconds and 24 seconds. Both flash at 10:00:00. When do they next flash together?

The waiting time must be a multiple of both intervals. We need the LCM.
Since $18=2\times3^2$ and $24=2^3\times3$ , $\operatorname{LCM}=2^3\times3^2=72$ seconds.
Convert 72 seconds to 1 minute 12 seconds. The next common flash is at 10:01:12.

Check: $72=4\times18=3\times24$ . “Next” excludes the shared flash at the starting time.

Squares, cubes and the missing prime factors

A perfect square has an even power of every prime: its prime factors can be paired. A perfect cube has powers divisible by 3: its factors can be grouped in triples. For example, $144=2^4\times3^2$ , so $\sqrt{144}=2^2\times3=12$ . Also $1728=2^6\times3^3$ , so $\sqrt[3]{1728}=2^2\times3=12$ .

The symbol $\sqrt{144}$ means the non-negative square root, 12. The equation $x^2=144$ has two solutions, $x=12$ and $x=-12$ . A cube root can be negative: $\sqrt[3]{-125}=-5$ .

Worked example 5: make 540 into a square or cube

Find the smallest positive integer to (a) multiply 540 by to obtain a perfect square, (b) multiply it by to obtain a perfect cube, and (c) divide it by to obtain a perfect square.

Start from $540=2^2\times3^3\times5$ .

For a square, the powers 3 and 1 are odd. Add one 3 and one 5: $k=3\times5=\boxed{15},\qquad540k=8100=90^2.$
For a cube, raise the powers of 2 and 5 to 3. The power of 3 is already 3: $k=2\times5^2=\boxed{50},\qquad540k=27000=30^3.$
When dividing to leave a whole-number square, remove one 3 and one 5. This leaves even powers: $d=3\times5=\boxed{15},\qquad540\div15=36=6^2.$

These values are smallest because each adds or removes only the prime factors needed to fix the powers. For a number to be both a square and a cube, every prime power must be a multiple of 6.

Common mistakes and next steps

Calling 1 prime: it has only one positive factor.
Swapping HCF and LCM: ask whether the answer divides the given numbers or is divisible by them.
Multiplying the two numbers to get the LCM: that gives a common multiple, but shared factors may make it larger than necessary.
Adjusting the number rather than the prime powers: for a square, pair every prime factor; for a cube, group every prime factor in threes.

When you are ready to work independently, use the primes, factors and multiples question bank. Begin with prime factorisation, then move to HCF/LCM situations and perfect-power questions.