What is the HCF of `2^(3) xx 7^(2) and 2^(5) xx 3^(1)` ?

Options

A. `2^(3) xx 3^(1)`
B. ` 2^(3) xx 7^(2) xx 3^(1)`
C. `2^(3)`
D. ` 3^(1) xx 7^(2)`

Answer

To find the Highest Common Factor (HCF) of the numbers \(2^3 \times 7^2\) and \(2^5 \times 3^1\), we can follow these steps:

### Step 1: Identify the prime factorization of both numbers.
- The first number is \(2^3 \times 7^2\).
- The second number is \(2^5 \times 3^1\).

### Step 2: List the prime factors and their powers.
- For \(2^3 \times 7^2\):
- The prime factors are \(2\) and \(7\).
- The powers are \(3\) for \(2\) and \(2\) for \(7\).

- For \(2^5 \times 3^1\):
- The prime factors are \(2\) and \(3\).
- The powers are \(5\) for \(2\) and \(1\) for \(3\).

### Step 3: Determine the HCF for each prime factor.
- For the prime factor \(2\):
- The minimum power is \(\min(3, 5) = 3\).

- For the prime factor \(7\):
- It appears only in the first number, so its minimum power is \(\min(2, 0) = 0\) (since \(7\) does not appear in the second number).

- For the prime factor \(3\):
- It appears only in the second number, so its minimum power is \(\min(0, 1) = 0\) (since \(3\) does not appear in the first number).

### Step 4: Write the HCF using the minimum powers.
The HCF is given by multiplying the prime factors raised to their respective minimum powers:
\[ \text{HCF} = 2^{\min(3, 5)} \times 7^{\min(2, 0)} \times 3^{\min(0, 1)} = 2^3 \times 7^0 \times 3^0 \]

### Step 5: Simplify the HCF.
Since any number raised to the power of \(0\) is \(1\), we can simplify:
\[ \text{HCF} = 2^3 \times 1 \times 1 = 2^3 \]

### Final Answer:
Thus, the HCF of \(2^3 \times 7^2\) and \(2^5 \times 3^1\) is:
\[ \text{HCF} = 2^3 = 8 \]

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