Visual walkthrough — LCM — prime factorization method, relationship HCF × LCM = product

Before we begin, three plain-word promises about the symbols:

• $a$ and $b$ are just the two numbers we start with (e.g. $60$ and $72$).
• A prime is a number with no divisors except $1$ and itself: $2,3,5,7,11,\dots$ (see Prime Numbers and Factorization).
• HCF = biggest number dividing both; LCM = smallest number both divide into. That's it.

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Step 1 — Every number is a tower of prime blocks

WHAT. Take one number, say $60$, and keep splitting it until only primes remain: $60=2\times 2\times 3\times 5=2^2\cdot 3\cdot 5$.

WHY. The Fundamental Theorem of Arithmetic promises this splitting is unique — there is one and only one recipe of primes for each number. Because the recipe is unique, we can compare two numbers ingredient by ingredient with no ambiguity. That comparison is the whole engine of this page.

PICTURE. Read the number as towers, one tower per prime. The height of a tower = the exponent (how many copies of that prime).

$60=\underset{\text{prime}}{\underbrace{2}}{}^{\stackrel{\text{exponent = tower height}}{\overbrace{2}}}\cdot 3^1\cdot 5^1$

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Step 2 — Line up the two numbers, prime by prime

WHAT. Put $a=60$ and $b=72$ side by side and write every prime that appears in either, even if one has height $0$.

$60=2^2\cdot 3^1\cdot 5^1,72=2^3\cdot 3^2\cdot 5^0$

WHY. Writing the missing prime as $5^0$ (a tower of height $0$ — an empty spot) lets both numbers use the same list of primes. Now a comparison of heights is a fair, column-by-column game.

PICTURE. Two rows of towers over the same three columns ($2$, $3$, $5$). Notice $72$ has an empty column at $5$.

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Step 3 — What does "divides both" demand? (the HCF rule)

WHAT. A number $d$ divides $a$ if $d$'s recipe fits inside $a$'s recipe — every one of $d$'s towers must be no taller than $a$'s tower for that prime.

WHY. You cannot pull $2^3$ out of a number that only has $2^2$ inside it — there simply aren't enough copies of $2$. So to divide both $a$ and $b$, each of $d$'s towers must fit under ${\alpha }_p$ and under ${\beta }_p$. The tallest tower that fits under both is the shorter of the two — the $\min$.

$\text{HCF: height at prime }p=\underset{\text{the shorter of the two towers}}{\underbrace{\min ({\alpha }_p,{\beta }_p)}}$

PICTURE. For each column, the HCF slices off at the height of the shorter tower (highlighted).

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Step 4 — What does "multiple of both" demand? (the LCM rule)

WHAT. A number $m$ is a multiple of $a$ if $a$ fits inside $m$ — now $m$'s towers must be at least as tall as $a$'s. To be a multiple of both, each tower must clear ${\alpha }_p$ and ${\beta }_p$, so it must reach the taller one — the $\max$. Taking exactly the max (not more) keeps $m$ smallest.

$\text{LCM: height at prime }p=\underset{\text{the taller of the two towers}}{\underbrace{\max ({\alpha }_p,{\beta }_p)}}$

WHY $\max$, not $\min$? "Least" in Least Common Multiple describes the finished number being smallest — but to be divisible by both you still need enough of each prime, so you must reach the taller requirement. Choosing $\min$ here would leave one number unable to divide in.

PICTURE. For each column, the LCM rises to the taller tower (highlighted).

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Step 5 — The magic identity: min + max = the two heights

WHAT. For any two heights $\alpha$ and $\beta$: $\min (\alpha ,\beta )+\max (\alpha ,\beta )=\alpha +\beta$

WHY. Whichever tower is shorter is picked by $\min$; the other one is picked by $\max$. Between them the two picks use up both towers exactly once — nothing double-counted, nothing dropped. This is the hinge of the whole proof.

PICTURE. Stack the "short pick" on top of the "tall pick": their combined height equals the two original towers stacked. Compare against $\alpha +\beta$ — same total.

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Step 6 — Multiply HCF by LCM, one column at a time

WHAT. When you multiply HCF and LCM, the tower heights add column by column (because $p^x\cdot p^y=p^{x+y}$). So at each prime the combined height is $\min ({\alpha }_p,{\beta }_p)+\max ({\alpha }_p,{\beta }_p)$ — which Step 5 says equals ${\alpha }_p+{\beta }_p$.

$\text{HCF}\times \text{LCM}={\prod }_p{p}^{\min ({\alpha }_p,{\beta }_p)+\max ({\alpha }_p,{\beta }_p)}={\prod }_p{p}^{{\alpha }_p+{\beta }_p}$

WHY. And $p^{{\alpha }_p+{\beta }_p}=p^{{\alpha }_p}\cdot p^{{\beta }_p}$ splits back into "the $a$ part" times "the $b$ part": ${\prod }_p{p}^{{\alpha }_p+{\beta }_p}=\left({\prod }_p{p}^{{\alpha }_p}\right)\left({\prod }_p{p}^{{\beta }_p}\right)=a\times b$

PICTURE. Column by column, HCF-tower + LCM-tower rebuilds $a$-tower + $b$-tower. Summed over all columns, HCF × LCM reassembles exactly $a\times b$.

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Step 7 — The edge cases (never leave a scenario unshown)

WHAT & WHY. We must confirm the tower argument survives every boundary.

Case A — a prime missing from one number (${\beta }_p=0$). Empty tower. $\min$ picks $0$ (not shared), $\max$ picks ${\alpha }_p$ (LCM must still carry it). Still $0+{\alpha }_p={\alpha }_p+0$. ✓ (This was the $5$ column above.)

Case B — one number divides the other, e.g. $a=8,b=24$. $8=2^3$, $24=2^3\cdot 3$. Every $a$-tower is $\le$ its $b$-tower, so $\text{HCF}=8$ (the smaller), $\text{LCM}=24$ (the larger). Check $8\times 24=192=8\times 24$. ✓ The identity is trivially true here.

Case C — coprime numbers, e.g. $a=8,b=9$. No column is shared: every $\min =0$, so $\text{HCF}=1$. Then $\text{LCM}=a\times b=72$, and $1\times 72=72$. ✓ (More on coprimes: HCF — prime factorization and Euclid's algorithm.)

Case D — the "three numbers" trap. The min+max pairing works for two heights only. With three towers there is a middle one that neither min nor max touches, so the identity breaks. Example $2,4,6$: $\text{HCF}=2$, $\text{LCM}=12$, but $2\times 12=24\ne 2\cdot 4\cdot 6=48$.

PICTURE. Four mini-panels, one per case, each with its towers.

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The one-picture summary

WHAT. One figure holds the whole story: two towers per prime → $\min$ builds HCF, $\max$ builds LCM → stacking them equals the two originals stacked → HCF × LCM $=a\times b$.

Recall Feynman retelling — say it to a friend

Break each number into piles of prime blocks — one pile per prime. Line up the two numbers so their piles sit in the same columns; if a prime is missing, its pile is empty. Now play two games in every column. In the shared game you keep only the shorter pile — that's what both numbers truly have in common, and stacking all the shorter piles gives you the HCF. In the covering game you keep the taller pile — just enough for both numbers to fit inside — and stacking all the taller piles gives the LCM. Here's the punchline: in any column, the shorter plus the taller pile is just the two piles put together — you used up both, exactly once. So when you multiply HCF by LCM (which stacks all their piles together), you've secretly stacked every original pile of $a$ and every original pile of $b$ — that's $a\times b$. Shared part plus leftover part = the whole thing, no block wasted. That's why $\text{HCF}\times \text{LCM}=a\times b$.

Recall Predict then check

If $\text{HCF}(a,b)=6$, $\text{LCM}=72$, and $a=24$, find $b$. $b=\frac{\text{HCF}\times \text{LCM}}{a}=\frac{6\times 72}{24}=\frac{432}{24}=18$. ✓

Connections

• Parent: LCM & HCF × LCM = product
• Fundamental Theorem of Arithmetic
• Prime Numbers and Factorization
• HCF — prime factorization and Euclid's algorithm
• Divisibility Rules
• Applications — bells ringing, tiling, syncing cycles