Find HCF and LCM of 12 15 and 21 using prime factorisation

Question

Find the HCF and LCM of 12, 15, and 21 using prime factorisation. Verify the relationship between HCF and LCM.

Solution — Step by Step

12 = 2^2 \times 3 = 2 \times 2 \times 3

15 = 3 \times 5

21 = 3 \times 7

HCF = product of common prime factors with the lowest power in which they appear.

Looking at all three factorizations:

Factor 2: appears in 12 only (not in 15 or 21) → NOT common to all three
Factor 3: appears in all three → $3^1$ (minimum power = 1)
Factor 5: appears in 15 only → NOT common to all three
Factor 7: appears in 21 only → NOT common to all three

\text{HCF}(12, 15, 21) = 3

LCM = product of all prime factors (from all numbers) with the highest power.

Factor 2: appears with highest power $2^2$ (in 12)
Factor 3: appears with highest power $3^1$ (all have at most $3^1$ )
Factor 5: appears with highest power $5^1$ (in 15)
Factor 7: appears with highest power $7^1$ (in 21)

\text{LCM}(12, 15, 21) = 2^2 \times 3 \times 5 \times 7 = 4 \times 3 \times 5 \times 7 = 420

Check that 420 is divisible by 12, 15, and 21:

$420 \div 12 = 35$ ✓

$420 \div 15 = 28$ ✓

$420 \div 21 = 20$ ✓

All divide evenly. And 420 is the smallest such number.

HCF = 3, LCM = 420

Why This Works

HCF finds the largest number that divides all given numbers — we take only factors that ALL numbers share, and use the minimum power (since we need it to divide ALL of them).

LCM finds the smallest number divisible by all given numbers — we take ALL factors from ALL numbers, using the maximum power (to ensure divisibility by each).

Note: the product formula $\text{HCF} \times \text{LCM} = a \times b$ works for exactly TWO numbers, not three. So we can’t verify as $12 \times 15 \times 21 = 420 \times 3$ here (that would be $3780 \neq 1260$ ). The relationship for three numbers is more complex.

Alternative Method — Successive Division (for HCF)

For two numbers: divide the larger by the smaller, take the remainder, repeat until remainder is 0. Last non-zero divisor is the HCF (Euclidean algorithm).

For three numbers: first find HCF of any two, then find HCF of that result with the third.

$\text{HCF}(12, 15) = 3$ , then $\text{HCF}(3, 21) = 3$ . So $\text{HCF}(12, 15, 21) = 3$ . ✓

Common Mistake

For LCM, students include a prime factor only once, even if it appears in multiple numbers with different powers. The rule is: take each prime factor with its HIGHEST power across all numbers. For example, factor 2 appears as $2^2$ in 12 — the LCM must include $2^2$ , not $2^1$ , even though 15 and 21 have no factor of 2 at all.

Question

Solution — Step by Step

Why This Works

Alternative Method — Successive Division (for HCF)

Common Mistake

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