Fundamental Theorem of Arithmetic

Pick any whole number bigger than $1$ — say $360$ — and start breaking it into factors any way you like: $8 \times 45$ , or $4 \times 90$ , or a patient factor tree one branch at a time. No matter which route you take, you always land on the same handful of primes: $2, 2, 2, 3, 3, 5$ . That is not a coincidence for $360$ alone — it is a guarantee for every composite number, and it is the Fundamental Theorem of Arithmetic. This single guarantee of uniqueness is what makes prime factorisation trustworthy enough to build an entire method for finding HCF and LCM, the two most-tested skills in the Class 10 Real Numbers chapter.

The core insight

Every composite number has exactly one set of prime factors (order aside), so writing two numbers in prime-power form lets you read off their HCF and LCM directly, and the two are always linked by $\text{HCF}(a,b) \times \text{LCM}(a,b) = a \times b$ .

The core idea

The theorem itself. Every composite number can be expressed as a product of primes, and this factorisation is unique, apart from the order of the factors. "Unique" is the whole point: it means prime factorisation is not just a way to break a number down — it is the way, so any calculation built on it (like HCF or LCM) will always give the same answer no matter how you arrived at the factor list.

Why exponent form is useful. Once a number is written as a product of primes, group repeated primes into powers, for example $360 = 2^3 \times 3^2 \times 5^1$ . This exponent form turns "find the common factors" into simple exponent comparisons between two numbers' prime lists.

Figure

Two different factor trees for 360 both terminating in the same prime leaves 2 2 2 3 3 5

HCF and LCM from prime powers. Write each number in prime-power form. Then:

\text{HCF} = \text{product of the smallest power of each common prime factor}

\text{LCM} = \text{product of the greatest power of each prime factor appearing in any of the numbers}

where: "common prime factor" means a prime present in every number's factorisation, and "smallest/greatest power" compares the exponents of that prime across the numbers being combined. A prime that appears in only one number still enters the LCM (at its own power) but never enters the HCF.

The link between them. For any two positive integers $a$ and $b$ :

\text{HCF}(a,b) \times \text{LCM}(a,b) = a \times b

This relation holds for exactly two numbers only — it does not extend directly to three or more. Euclid's Division Algorithm gives the HCF of two numbers faster than factorising for large values, but prime factorisation is the only method here that also hands you the LCM, which is why examiners favour it for this chapter.

Derivation AHCF times LCM equals the product of the two numbers

The goal is to show, using nothing beyond the Fundamental Theorem of Arithmetic and the rules for combining exponents of the same base, why $\text{HCF}(a,b) \times \text{LCM}(a,b)$ must always equal $a \times b$ for any two positive integers $a$ and $b$ , so that the relation is not a coincidence but a direct consequence of how prime powers multiply.

Step 1. Write both numbers over the same list of primes. Suppose the primes involved across $a$ and $b$ are $p_1, p_2, \ldots, p_k$ (including, for each number, primes it does not actually contain, written with exponent $0$ ). Then $a = p_1^{x_1} p_2^{x_2} \cdots p_k^{x_k}$ and $b = p_1^{y_1} p_2^{y_2} \cdots p_k^{y_k}$ , where each $x_i, y_i \geq 0$ . The Fundamental Theorem of Arithmetic guarantees these exponent lists are unique, so this is a fair, fixed comparison.

Step 2. Recall the definitions in exponent language. For each prime $p_i$ , the HCF takes the smaller of $x_i$ and $y_i$ (call it $\min(x_i, y_i)$ ), and the LCM takes the larger (call it $\max(x_i, y_i)$ ). So $\text{HCF}(a,b) = \prod_i p_i^{\min(x_i,y_i)}$ and $\text{LCM}(a,b) = \prod_i p_i^{\max(x_i,y_i)}$ .

Step 3. Multiply the HCF and LCM together, prime by prime. For a single prime $p_i$ , its contribution to the product $\text{HCF} \times \text{LCM}$ is $p_i^{\min(x_i,y_i)} \times p_i^{\max(x_i,y_i)} = p_i^{\min(x_i,y_i) + \max(x_i,y_i)}$ . But for any pair of numbers, the smaller plus the larger always equals the sum of the two original numbers: $\min(x_i,y_i) + \max(x_i,y_i) = x_i + y_i$ . So each prime's contribution becomes $p_i^{x_i + y_i} = p_i^{x_i} \times p_i^{y_i}$ .

Step 4. Multiply this out over every prime in the list. $\text{HCF}(a,b) \times \text{LCM}(a,b) = \prod_i p_i^{x_i} \times p_i^{y_i} = \left(\prod_i p_i^{x_i}\right) \times \left(\prod_i p_i^{y_i}\right) = a \times b$ . The regrouping works only because each prime's minimum-plus-maximum trick applies independently, prime by prime, which is exactly what unique factorisation lets us do. Check with numbers: $a = 96 = 2^5 \times 3^1$ , $b = 404 = 2^2 \times 101^1$ . For prime $2$ : $\min(5,2) + \max(5,2) = 2 + 5 = 7 = 5 + 2$ . For prime $3$ : $\min(1,0) + \max(1,0) = 0 + 1 = 1 = 1 + 0$ . For prime $101$ : $\min(0,1) + \max(0,1) = 0 + 1 = 1 = 0 + 1$ . Every prime's exponents add up correctly, confirming Step 3's claim.

Worked example

Find the HCF and LCM of $96$ and $404$ using prime factorisation, verify $\text{HCF} \times \text{LCM} = 96 \times 404$ , and then use that HCF to find a third number that has the same HCF with $96$ but a different LCM.

Step 1. Prime factorise each number. $96 = 2^5 \times 3^1$ and $404 = 2^2 \times 101^1$ .

Step 2. Find the HCF. The only common prime is $2$ ; the smaller power between $2^5$ and $2^2$ is $2^2$ . So $\text{HCF}(96,404) = 2^2 = 4$ .

Step 3. Find the LCM. Take the greatest power of every prime seen anywhere ( $2$ , $3$ , $101$ ): $\text{LCM}(96,404) = 2^5 \times 3 \times 101 = 32 \times 3 \times 101 = 9696$ .

Step 4. Verify the relation. $\text{HCF} \times \text{LCM} = 4 \times 9696 = 38784$ , and $96 \times 404 = 38784$ . Both sides match.

Step 5. Now find a third number, $c$ , with $\text{HCF}(96, c) = 4$ but a different LCM from before. Take $c = 100 = 2^2 \times 5^2$ . Comparing with $96 = 2^5 \times 3^1$ : the only common prime is $2$ , and the smaller power is $2^2$ , so $\text{HCF}(96, 100) = 4$ — the same HCF as before, as required. But the LCM now uses $2^5$ , $3^1$ , and $5^2$ : $\text{LCM}(96,100) = 32 \times 3 \times 25 = 2400$ , which is different from $9696$ . This shows the HCF alone does not fix the LCM — the LCM depends on every prime present, including ones that never entered the HCF calculation. The HCF and LCM of $96$ and $404$ are $4$ and $9696$ ; a number like $100$ can share the HCF of $4$ with $96$ while having an entirely different LCM.

Common traps

Assuming $\text{HCF} \times \text{LCM} = a \times b$ works for three or more numbers — it holds only for exactly two.
Mixing up "smallest power" and "greatest power": HCF uses the smallest power of only the common primes; LCM uses the greatest power of every prime seen in any number.
Forgetting to carry a prime that appears in only one of the numbers into the LCM (the $101$ in $404$ must still appear in the LCM even though it is not common to both).
Treating the theorem as being only about proving irrationality — its main syllabus use here is prime-factorisation-based HCF/LCM; irrationality proofs are a separate, related application.

Ask the tutor: Want a fresh pair of numbers to prime-factorise and check the HCF-LCM relation yourself, or a trickier "find the missing number" problem? Ask the AI tutor for one.

Why does prime factorisation being unique matter so much for finding HCF and LCM?

If two different people could factorise the same number into two genuinely different sets of primes, then "the HCF" or "the LCM" would not be a single well-defined answer — it would depend on which factorisation you happened to use. The Fundamental Theorem of Arithmetic rules that out: no matter how you split a number apart, you always reach the same prime list. That guarantee is what lets us safely compare exponents across two numbers and trust the result.

Does the HCF times LCM equals a times b relation work for three numbers, like 12, 18 and 24?

No, not directly. The relation $\text{HCF}(a,b) \times \text{LCM}(a,b) = a \times b$ is proven using only two exponent lists at a time (the min and max of a pair), and that pairing argument breaks down with three or more numbers — the product $a \times b \times c$ does not equal $\text{HCF}(a,b,c) \times \text{LCM}(a,b,c)$ in general. For three or more numbers, find the HCF and LCM directly from all their prime factorisations at once, using the smallest and greatest powers across every number, rather than trying to force the two-number formula.

If I already know the HCF of two numbers, can I work out the LCM without factorising again?

Only if you also know both numbers (or their product). Rearranging the relation gives $\text{LCM}(a,b) = \dfrac{a \times b}{\text{HCF}(a,b)}$ , which is genuinely useful — Worked Example 2 in the Pass-1 material uses exactly this trick to find a missing number. But knowing the HCF alone, without the numbers, tells you nothing about the LCM, because the LCM also depends on every prime that is not common to both numbers, which the HCF never sees.

Fundamental Theorem of Arithmetic

The core idea

Worked example

Common traps

Before this

Where this leads