Question

The HCF of 960 and 432 is :

• A. 48
• B. 54
• C. 72
• D. 36

Hint:

For larger numbers, you can also use Euclid's Division Algorithm to quickly find the HCF.
Divide the larger number by the smaller number and write down the remainder.
Repeat the process with the divisor and the remainder until the remainder becomes zero.
$960 = 432 \times 2 + 96$
$432 = 96 \times 4 + 48$
$96 = 48 \times 2 + 0$
The last non-zero remainder is 48, which is the HCF.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The topic of this question is Real Numbers, specifically finding the Highest Common Factor (HCF) of two positive integers.
The Highest Common Factor is the largest positive integer that divides each of the given integers without leaving a remainder.
To find this value, we can analyze the common divisibility of 960 and 432.

Step 2: Key Formula or Approach:
We can utilize the Prime Factorization Method.
In this approach, we express each number as a product of its prime factors.
The HCF is then determined by taking the product of the terms containing the lowest powers of the common prime factors.
Mathematically, for any two numbers $A$ and $B$:
\[ \text{HCF}(A, B) = \prod p_i^{\min(a_i, b_i)} \]
where $p_i$ represents the common prime factors, and $a_i, b_i$ are their respective exponents in the prime factorizations.

Step 3: Detailed Explanation:

• First, let us find the prime factorization of 960.
Divide 960 successively by prime numbers:
$960 \div 2 = 480$
$480 \div 2 = 240$
$240 \div 2 = 120$
$120 \div 2 = 60$
$60 \div 2 = 30$
$30 \div 2 = 15$
$15 \div 3 = 5$
$5 \div 5 = 1$
Therefore, the prime factorization of 960 is:
\[ 960 = 2^6 \times 3^1 \times 5^1 \]

• Next, let us find the prime factorization of 432.
Divide 432 successively by prime numbers:
$432 \div 2 = 216$
$216 \div 2 = 108$
$108 \div 2 = 54$
$54 \div 2 = 27$
$27 \div 3 = 9$
$9 \div 3 = 3$
$3 \div 3 = 1$
Therefore, the prime factorization of 432 is:
\[ 432 = 2^4 \times 3^3 \]

• Now, identify the common prime factors of 960 and 432.
The common prime factors are 2 and 3.

• Find the lowest power of each common prime factor:
For prime factor 2, the powers are $2^6$ and $2^4$. The minimum power is $2^4$.
For prime factor 3, the powers are $3^1$ and $3^3$. The minimum power is $3^1$.

• Compute the product of these lowest powers to find the HCF:
\[ \text{HCF} = 2^4 \times 3^1 \]
\[ \text{HCF} = 16 \times 3 = 48 \]


Step 4: Final Answer:
The Highest Common Factor (HCF) of 960 and 432 is 48, which corresponds to option (A).