Divisibility properties

A number $a$ is divisible by $d$ (equivalently, $d$ divides $a$, written $d \mid a$) when $a = d \times k$ for some whole number $k$. Here $d$ is called a factor of $a$, and $a$ is a multiple of $d$. Every divisibility property below is really the same trick played out with algebra: replace a number by $d \times (\text{its quotient})$, do the arithmetic, and read off the new quotient.

Sums and differences carry the common factor forward. If $d \mid a$ and $d \mid b$, write $a = dk_1$ and $b = dk_2$. Then

where: $d$ is the common factor, $k_1, k_2$ are the (whole-number) quotients of $a$ and $b$ by $d$. Because $k_1 + k_2$ and $k_1 - k_2$ are themselves whole numbers, $d$ divides both the sum and the difference — no new remainder can appear from adding or subtracting two exact multiples.

Multiplying by anything preserves divisibility. If $d \mid a$, then $d \mid (a \times m)$ for any whole number $m$, even one $d$ does not divide: $a = dk \Rightarrow am = d(km)$. This is stronger than the sum rule — it needs only one of the two numbers to carry the factor $d$.

A mismatch breaks the sum. If $d \mid a$ but $d \nmid b$, then $d$ cannot divide $a + b$ or $a - b$ either — if it did, subtracting the multiple of $d$ back out would force $d \mid b$, a contradiction.

Primes have a special veto power. If a prime $p$ divides a product $a \times b$, then $p$ must divide $a$ or divide $b$ (or both). No composite divisor can guarantee this — for instance $6 \mid (4 \times 9=36)$, yet $6$ divides neither $4$ nor $9$. This one property of primes is exactly what makes prime factorisation a trustworthy tool for HCF and LCM.

The goal of this derivation is to show, using nothing beyond prime factorisation and the rules for adding exponents, why $\text{HCF}(a,b) \times \text{LCM}(a,b)$ always equals $a \times b$ for any two whole numbers $a$ and $b$, and why this is not a coincidence but a direct consequence of how the "lowest power" and "highest power" rules for HCF and LCM are built.

Step 1. Write both numbers in prime-factorised form, listing every prime $p$ that appears in either number, using exponent $0$ where a prime is absent. For a single prime $p$, say $a$ contains $p^{x}$ and $b$ contains $p^{y}$ (with $x, y \ge 0$).

Step 2. Recall the definitions of HCF and LCM in terms of exponents. HCF takes the lowest power of each common prime, so its exponent for $p$ is $\min(x, y)$. LCM takes the highest power of every prime appearing in either number, so its exponent for $p$ is $\max(x, y)$.

Step 3. Add the two exponents contributed by $p$ to $\text{HCF} \times \text{LCM}$: it is $\min(x,y) + \max(x,y)$. Whichever of $x, y$ is smaller and whichever is larger, this sum always equals $x + y$ — the smaller and larger of two numbers always add up to their ordinary sum, since one of $\min(x,y)$ or $\max(x,y)$ literally equals $x$ and the other equals $y$.

Step 4. Since $x + y$ is exactly the exponent of $p$ in the product $a \times b$ (exponents add when you multiply), the exponent of every prime $p$ in $\text{HCF}(a,b) \times \text{LCM}(a,b)$ matches its exponent in $a \times b$. Two numbers with identical prime factorisations are equal, so

where: $a, b$ are the two whole numbers, $\text{HCF}(a,b)$ is their highest common factor, $\text{LCM}(a,b)$ is their lowest common multiple. This is why knowing any three of $a$, $b$, $\text{HCF}$, $\text{LCM}$ always lets you compute the fourth by simple division — the relation holds exactly, not approximately, for every pair of whole numbers.

HCF of two numbers 84 and 108 is found to be 12. Find their LCM without listing multiples, then verify by prime factorisation, and check the sum property on the pair.

Step 1. Apply the relation $\text{HCF} \times \text{LCM} = a \times b$ with $a = 84$, $b = 108$: $12 \times \text{LCM} = 84 \times 108 = 9072$, so $\text{LCM} = 9072 / 12 = 756$.

Step 2. Verify by prime factorisation: $84 = 2^2 \times 3 \times 7$ and $108 = 2^2 \times 3^3$. HCF uses the lowest power of each common prime, $2^2 \times 3^1 = 4 \times 3 = 12$, matching the given HCF. LCM uses the highest power of every prime present, $2^2 \times 3^3 \times 7 = 4 \times 27 \times 7 = 756$, matching Step 1.

Step 3. Check the sum property on this pair: since $12 \mid 84$ and $12 \mid 108$, property one guarantees $12 \mid (84+108)$. Indeed $84 + 108 = 192 = 12 \times 16$, confirmed. It also guarantees $12 \mid (108-84)$, and $108 - 84 = 24 = 12 \times 2$, confirmed.

LCM of 84 and 108 is 756, and both the product relation and the sum property check out exactly.

Competitive stretch

If instead only the LCM ($756$) and one number ($84$) were given, the same relation rearranges to find the other: $\text{HCF} \times 756 = 84 \times b$. Combined with $\text{HCF} \mid 84$, this lets you narrow down $b$ without ever listing multiples of $756$ by hand.

• Believing $d \mid (a+b)$ forces $d$ to divide $a$ and $b$ separately — false: $3 \mid (2+7=9)$, yet $3$ divides neither $2$ nor $7$.
• Extending the prime "veto" rule to composite divisors — $d \mid (ab)$ does not imply $d \mid a$ or $d \mid b$ unless $d$ is prime (recall $6 \mid 36$ but $6 \nmid 4$ and $6 \nmid 9$).
• Swapping HCF (always $\le$ both numbers) and LCM (always $\ge$ both numbers) when applying $\text{HCF} \times \text{LCM} = a \times b$.
• Applying the two-number relation $\text{HCF}(a,b) \times \text{LCM}(a,b) = a \times b$ directly to three or more numbers — it only holds for a pair at a time.

Ask the tutor: Want to work through a proof of one of these divisibility properties step by step, or try a few more HCF–LCM problems using the product relation?