Find the H.C.F. of `(2^(3)xx3xx5) and (2^(4)xx5^2xx17)`

Answer

To find the H.C.F. (Highest Common Factor) of the two numbers \( (2^3 \times 3^1 \times 5^1) \) and \( (2^4 \times 5^2 \times 17^1) \), we will follow these steps:

### Step 1: Identify the prime factors
The prime factorization of the two numbers is:
- Number 1: \( 2^3 \times 3^1 \times 5^1 \)
- Number 2: \( 2^4 \times 5^2 \times 17^1 \)

### Step 2: List the common prime factors
The common prime factors between the two numbers are \( 2 \) and \( 5 \). The prime factor \( 3 \) is only present in the first number, and \( 17 \) is only present in the second number.

### Step 3: Determine the lowest power of each common prime factor
- For the prime factor \( 2 \):
- In Number 1: \( 2^3 \)
- In Number 2: \( 2^4 \)
- The lowest power is \( 2^3 \).

- For the prime factor \( 5 \):
- In Number 1: \( 5^1 \)
- In Number 2: \( 5^2 \)
- The lowest power is \( 5^1 \).

### Step 4: Write the H.C.F. using the lowest powers
Now, we can write the H.C.F. as:
\[ \text{H.C.F.} = 2^3 \times 5^1 \]

### Step 5: Calculate the H.C.F.
Now, we calculate the value:
\[ 2^3 = 8 \]
\[ 5^1 = 5 \]
\[ \text{H.C.F.} = 8 \times 5 = 40 \]

### Final Answer
The H.C.F. of \( (2^3 \times 3^1 \times 5^1) \) and \( (2^4 \times 5^2 \times 17^1) \) is \( 40 \).

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