LCM — prime factorization method, relationship HCF × LCM = product: Worked examples

Before we begin, a one-line reminder of the two words we keep using:

A multiple of $n$ is any number you get by multiplying $n$ by a whole number: $n, 2 n, 3 n, \dots$ — think "footsteps of size $n$ ".
A factor (or divisor) of $n$ is a number that divides $n$ with no remainder — think "a brick $n$ is built from".

The scenario matrix

Every LCM problem falls into one of these cells. The worked examples below are each tagged with the cell(s) they cover, and together they hit all of them.

Cell	Situation	What's tricky about it	Example
A	Two numbers share some primes (overlap)	pick max carefully per prime	Ex 1
B	Two numbers coprime (share nothing, HCF $= 1$ )	LCM is just the product	Ex 2
C	One number divides the other	LCM $=$ the bigger one; HCF $=$ smaller	Ex 3
D	Both are powers of the same prime	only one prime — max wins outright	Ex 4
E	Degenerate: LCM with $1$ ; role of $0$	identity element, and why $0$ breaks LCM	Ex 5
F	Reverse problem: given HCF & LCM, find missing number	use $HCF \times LCM = a \times b$	Ex 6
G	Three numbers	product rule FAILS — must use per-prime min/max	Ex 7
H	Word problem (bells / cycles syncing)	translate real world → LCM	Ex 8
I	Exam twist: HCF and LCM given, must be consistent	HCF must divide LCM, else impossible	Ex 9

Worked example Ex 1 · LCM and HCF of

$24$ and $36$ Forecast: Guess the LCM before reading on. Both are "smallish" — will the LCM be bigger or smaller than $24 \times 36 = 864$ ? (Hint: they overlap a lot, so it'll be much smaller.)

Step 1 — Prime-factorize each. $24 = 2^{3} \cdot 3^{1}, 36 = 2^{2} \cdot 3^{2}$ Why this step? By the Fundamental Theorem of Arithmetic every number has ONE recipe of primes; we need each recipe to compare ingredient-by-ingredient.

Step 2 — List every prime that appears. Here: $2$ and $3$ . Why? A prime only matters for LCM/HCF if it shows up in at least one number.

Step 3 — LCM: take the MAX power of each prime. $2^{m a x (3, 2)} \cdot 3^{m a x (1, 2)} = 2^{3} \cdot 3^{2} = 8 \cdot 9 = 72$ Why max? To be divisible by $24$ you need at least $2^{3}$ ; to be divisible by $36$ you need at least $3^{2}$ . The bigger demand for each prime wins so both are satisfied.

Step 4 — HCF: take the MIN power of each prime. $2^{m i n (3, 2)} \cdot 3^{m i n (1, 2)} = 2^{2} \cdot 3^{1} = 12$ Why min? A common divisor can only use what BOTH have — you can't take $2^{3}$ from a number that only owns $2^{2}$ .

Verify: $72/24 = 3$ ✓, $72/36 = 2$ ✓ (LCM divides cleanly). $24/12 = 2$ , $36/12 = 3$ ✓ (HCF divides both). And $HCF \times LCM = 12 \times 72 = 864 = 24 \times 36$ ✓.

Look at the figure below: the two prime-recipe columns, with the LCM reaching up to the taller bar of each prime, the HCF staying at the shorter bar.

Figure — LCM — prime factorization method, relationship HCF × LCM = product

Cell B — coprime numbers (no shared primes)

Worked example Ex 2 · LCM of

$8$ and $15$ Forecast: $8$ and $15$ look unrelated. What do you think their HCF is? If HCF $= 1$ , the identity $HCF \times LCM = a \times b$ says something very clean about the LCM…

Step 1 — Factorize. $8 = 2^{3}$ , $15 = 3 \cdot 5$ . Why? Same reason always: we compare per prime.

Step 2 — Primes present: $2, 3, 5$ . Notice no prime is shared. Why it matters? When numbers share no primes we call them coprime (HCF $= 1$ ).

Step 3 — HCF (min per prime). Each prime is missing from one number, so its min power is $0$ : $2^{m i n (3, 0)} \cdot 3^{m i n (0, 1)} \cdot 5^{m i n (0, 1)} = 2^{0} \cdot 3^{0} \cdot 5^{0} = 1$ Why $0$ powers? A prime absent from a number contributes power $0$ there, and $min (something, 0) = 0$ .

Step 4 — LCM (max per prime). $2^{3} \cdot 3^{1} \cdot 5^{1} = 8 \cdot 3 \cdot 5 = 120$ Why this equals the product? With nothing shared, nothing is "saved" — LCM just piles all bricks together.

Verify: $8 \times 15 = 120$ , and $HCF \times LCM = 1 \times 120 = 120$ ✓. Rule of thumb confirmed: coprime ⟹ LCM = product.

Recall When is LCM

$(a, b)$ exactly $a \times b$ ? When ::: $a$ and $b$ are coprime (HCF $= 1$ , no shared prime).

Cell C — one number divides the other

Cell D — powers of the same prime

Cell E — degenerate inputs ( $1$ and $0$ )

Worked example Ex 5 · LCM

$(1, 7)$ , and why LCM $(0, 7)$ is a trap Forecast: $1$ has no prime bricks at all. What does the LCM of $1$ and any number equal? And can $0$ even have an LCM?

Part 1 — LCM $(1, 7)$ . $1$ has empty prime recipe. $7 = 7^{1}$ . Max per prime: only $7$ appears, so $LCM = 7$ . Why? The multiples of $1$ are all whole numbers, so the smallest common multiple is just $7$ itself. $1$ is the identity: LCM $(1, n) = n$ , HCF $(1, n) = 1$ .

Part 2 — the $0$ trap. The LCM is defined as the smallest positive multiple shared by both. But $0$ is a multiple of every number, and it's smaller than any positive number — so if we allowed $0$ , the "least" common multiple would collapse to $0$ for anything. To keep LCM meaningful, LCM is defined only for positive integers; $LCM (0, n)$ is either undefined or taken as $0$ by convention. Why flag this? Examiners like the sneaky " $0$ " input. Recognise it as degenerate, don't blindly plug into $\frac{a \times b}{HCF}$ .

Verify: $HCF (1, 7) \times LCM (1, 7) = 1 \times 7 = 7 = 1 \times 7$ ✓.

Cell F — reverse problem (find the missing number)

Worked example Ex 6 · Given HCF

$= 6$ , LCM $= 72$ , and $a = 24$ . Find $b$ . Forecast: You're handed three of the four quantities in $HCF \times LCM = a \times b$ . Which single formula pops $b$ straight out?

Step 1 — Use the identity $HCF \times LCM = a \times b$ (valid for two numbers). Why this tool and not prime factorization? You aren't given $b$ 's factors — you can't factorize what you don't know. The product identity turns the unknown into simple arithmetic.

Step 2 — Rearrange for $b$ . $b = \frac{HCF \times LCM}{a} = \frac{6 \times 72}{24} = \frac{432}{24} = 18$ Why divide by $a$ ? $a \times b$ equals the known left side; dividing by $a$ isolates $b$ .

Step 3 — Consistency check (don't skip!). Does $b = 18$ actually give HCF $= 6$ , LCM $= 72$ with $a = 24$ ? $24 = 2^{3} \cdot 3$ , $18 = 2 \cdot 3^{2}$ . HCF $= 2^{1} \cdot 3^{1} = 6$ ✓, LCM $= 2^{3} \cdot 3^{2} = 72$ ✓.

Verify: $6 \times 72 = 432 = 24 \times 18$ ✓, and the factor check above confirms it's genuinely consistent (not just arithmetically forced).

Cell G — three numbers (product rule FAILS)

Worked example Ex 7 · LCM and HCF of

$12, 18, 30$ Forecast: Tempting to write " $HCF \times LCM = 12 \times 18 \times 30$ ". Resist it. The product identity is a two-number law only. We go back to per-prime min/max.

Step 1 — Factorize all three. $12 = 2^{2} \cdot 3, 18 = 2 \cdot 3^{2}, 30 = 2 \cdot 3 \cdot 5$

Step 2 — List primes: $2, 3, 5$ .

Step 3 — LCM: max power of each across ALL three. $2^{m a x (2, 1, 1)} \cdot 3^{m a x (1, 2, 1)} \cdot 5^{m a x (0, 0, 1)} = 2^{2} \cdot 3^{2} \cdot 5 = 180$ Why over all three? To be a common multiple of three numbers, each prime must be tall enough for the tallest demand among them.

Step 4 — HCF: min power of each across all three. $2^{m i n (2, 1, 1)} \cdot 3^{m i n (1, 2, 1)} \cdot 5^{m i n (0, 0, 1)} = 2^{1} \cdot 3^{1} \cdot 5^{0} = 6$

Step 5 — Show the product rule breaks. $HCF \times LCM = 6 \times 180 = 1080$ , but $12 \times 18 \times 30 = 6480$ . They differ ( $1080 \neq = 6480$ ). Why does it fail? Shared primes get "double-counted" across three factors, so no clean min $+$ max cancellation exists.

Verify: LCM $180$ is divisible by $12, 18, 30$ ( $180/12 = 15$ , $180/18 = 10$ , $180/30 = 6$ ) ✓; HCF $6$ divides all three ✓; and $6 \times 180 \neq = 12 \times 18 \times 30$ ✓ (rule correctly fails).

Cell H — word problem (syncing cycles)

Worked example Ex 8 · Three bells ring every

$9$ , $12$ , and $15$ minutes, all together at $9 : 00$ . When next together? Forecast: "Together again" = the first common multiple of all three intervals = LCM. Guess: will it be under an hour or over?

Step 1 — Translate to maths. "Ring together again" means a time that is a multiple of $9$ and $12$ and $15$ . Smallest such $= LCM (9, 12, 15)$ . Why LCM? Each bell hits multiples of its own interval; they coincide at the smallest shared multiple.

Step 2 — Factorize. $9 = 3^{2}$ , $12 = 2^{2} \cdot 3$ , $15 = 3 \cdot 5$ .

Step 3 — LCM (max per prime, three numbers → Cell G method). $2^{m a x (0, 2, 0)} \cdot 3^{m a x (2, 1, 1)} \cdot 5^{m a x (0, 0, 1)} = 2^{2} \cdot 3^{2} \cdot 5 = 180$

Step 4 — Interpret with units. $180$ minutes $= 3$ hours. So the bells ring together next at $9 : 00 + 3 : 00 = 12 : 00$ (noon). Why convert? The answer is a time; leaving it as " $180$ " without units loses the meaning.

Verify: $180/9 = 20$ , $180/12 = 15$ , $180/15 = 12$ — all whole, so all three bells complete a whole number of cycles ✓. No smaller common multiple (since $180$ is the LCM) ✓.

Cell I — exam twist (impossible data)

Worked example Ex 9 · "Two numbers have HCF

$= 8$ and LCM $= 60$ ." Find them — or show it's impossible. Forecast: Feels routine. But check one thing first: must the HCF divide the LCM?

Step 1 — Recall the structural fact. The HCF is a common divisor of both numbers, and each number divides the LCM. So HCF divides each number, which divides the LCM ⟹ HCF must divide LCM. Why this tool? It's a cheap consistency filter that catches impossible data before you waste effort.

Step 2 — Test it. Does $8$ divide $60$ ? $60/8 = 7.5$ — no. What it looks like: $8 = 2^{3}$ demands $2^{3}$ inside the LCM, but $60 = 2^{2} \cdot 3 \cdot 5$ only owns $2^{2}$ . The HCF asks for more $2$ s than the LCM contains — a contradiction.

Step 3 — Conclude. No such pair of numbers exists. The data is inconsistent. Why not just force it? $b = \frac{8 \times 60}{a}$ would spit out numbers, but they'd fail the factor check — a classic trap answer.

Verify: $60 mod 8 = 4 \neq = 0$ , confirming $8 ∤ 60$ , so the scenario is genuinely impossible ✓.

Recall ladder

Recall Coprime numbers: what is their LCM?

Their product ::: (since HCF $= 1$ , so $LCM = \frac{a \times b}{1} = a \times b$ ).

Recall If

$a$ divides $b$ , what are LCM and HCF? LCM $= b$ , HCF $= a$ ::: the bigger and smaller numbers respectively.

Recall Why can't we use HCF × LCM = product for three numbers?

Because ::: min+max cancellation only pairs two exponents; with three, shared primes are miscounted (e.g. $2, 4, 6$ : $2 \times 12 = 24 \neq = 48$ ).

Recall Given HCF, LCM, and one number

$a$ , formula for $b$ ? $b = \frac{HCF \times LCM}{a}$ ::: valid for exactly two numbers — always re-check consistency.

Connections

LCM — prime factorization method, relationship HCF × LCM = product (parent)
HCF — prime factorization and Euclid's algorithm
Fundamental Theorem of Arithmetic
Prime Numbers and Factorization
Divisibility Rules
Applications — bells ringing, tiling, syncing cycles

Worked examples — LCM — prime factorization method, relationship HCF × LCM = product