1 markMCQSection A

The HCF of 960 and 432 is : (a) 48 (b) 54 (c) 72 (d) 36

Q: 1. The HCF of 960 and 432 is : (a) 48 (b) 54 (c) 72 (d) 36

(a) 48 — $HCF(960, 432) = 2^4 \times 3 = 48$, the product of the lowest common powers of 2 and 3 in their prime factorisations.

Q: Can I use Euclid's division algorithm instead?

Yes. $960 = 2\times432 + 96$; $432 = 4\times96 + 48$; $96 = 2\times48 + 0$. The last non-zero remainder, 48, is the HCF — the same answer as prime factorisation.

Real Numbers

HCF by prime factorisation

Options

(A)48

(B)54

(D)36

Official Answer

(a) 48 — $HCF(960, 432) = 2^4 \times 3 = 48$ , the product of the lowest common powers of 2 and 3 in their prime factorisations.

HCFprime factorisationFundamental Theorem of Arithmeticcommon prime factorslowest power482^4 x 3greatest common divisor

Marking Scheme

11 mark: correct option (a) 48.
2Full credit for the correct answer; working not required for an MCQ but $960 = 2^6\times3\times5$ and $432 = 2^4\times3^3$ giving $HCF = 2^4\times3 = 48$ is the expected reasoning.

Hint

Prime-factorise both numbers and multiply the SMALLEST power of each COMMON prime (2 and 3).

Quick Oral Answer

Factorise: $960 = 2^6\times3\times5$ and $432 = 2^4\times3^3$ ; the HCF takes the lowest power of the common primes 2 and 3, giving $2^4\times3 = 48$ .

Analysis & Explanation

This MCQ checks whether HCF can be found via prime factorisation (Fundamental Theorem of Arithmetic).

Concept

HCF = product of the smallest power of each common prime factor.
$960 = 2^6 \times 3 \times 5$ ; $432 = 2^4 \times 3^3$ .

Key points

Common primes are 2 and 3; smallest powers are $2^4$ and $3^1$ .
$HCF = 2^4 \times 3 = 16 \times 3 = 48$ .

Common mistakes

Taking the highest power instead of lowest (gives wrong values like 72 or 36).
Including 5 or 3³, which are not common to both numbers (gives 54).
Check: $960 \div 48 = 20$ , $432 \div 48 = 9$ , and $\gcd(20,9) = 1$ , confirming 48 is correct.

Common Mistakes

1Taking the HIGHEST power of each common prime (which gives a wrong large value like 72) — that is the rule for LCM, not HCF.
2Multiplying all prime factors including non-common ones such as 5, instead of only the common primes 2 and 3.
3Arithmetic slip in factorising 960 (writing $2^5$ or $3^2$ instead of $2^6 \times 3$ ) leading to a wrong power of 2.

Interesting Facts

The Fundamental Theorem of Arithmetic — that every integer > 1 has a unique prime factorisation — was first rigorously proved by Carl Friedrich Gauss in his 1801 work Disquisitiones Arithmeticae.

HCF × LCM = product of the two numbers; here HCF = 48 and the product $960 \times 432 = 414720$ , so the LCM must be $414720 \div 48 = 8640$ , which you can verify as $2^6 \times 3^3 \times 5$ .

Euclid's division algorithm gives the same HCF without factorising: $960 = 2\times432 + 96$ , $432 = 4\times96 + 48$ , $96 = 2\times48 + 0$ → HCF = 48.

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Frequently Asked Questions

How do you find the HCF using prime factorisation?

Write both numbers as products of primes, identify the primes common to both, take the smallest power of each common prime, and multiply them. For $960 = 2^6\times3\times5$ and $432 = 2^4\times3^3$ , the common primes are 2 and 3 with lowest powers $2^4$ and $3^1$ , so $HCF = 2^4\times3 = 48$ .

What is the difference between the HCF and the LCM rule here?

For HCF you multiply the LOWEST powers of only the COMMON primes; for LCM you multiply the HIGHEST powers of ALL primes appearing in either number. Confusing the two is the most common error — it would turn 48 into 72 in this question.

Can I use Euclid's division algorithm instead?

Yes. $960 = 2\times432 + 96$ ; $432 = 4\times96 + 48$ ; $96 = 2\times48 + 0$ . The last non-zero remainder, 48, is the HCF — the same answer as prime factorisation.