If `a = 3^(129) xx 5^128 xx 7^22`
`b = 3^128 xx 5^129 xx 7^21`
`c = 3^127 xx 5^128 xx 7^24`, then HCF of a,b, and c is

Options

A. `3^(128) xx 5^128 xx 7^21`
B. `3^(127) xx 5^128 xx 7^21`
C. `3^(127) xx 5^128 xx 7^22`
D. `3^(127) xx 5^128 xx 7^24`

Answer

To find the HCF (Highest Common Factor) of the numbers \( a \), \( b \), and \( c \), we will follow these steps:

1. **Identify the prime factorization of each number:**
- \( a = 3^{129} \times 5^{128} \times 7^{22} \)
- \( b = 3^{128} \times 5^{129} \times 7^{21} \)
- \( c = 3^{127} \times 5^{128} \times 7^{24} \)

2. **Find the HCF by taking the lowest power of each prime factor from \( a \), \( b \), and \( c \):**
- For the prime factor \( 3 \):
- The powers are \( 129 \) (from \( a \)), \( 128 \) (from \( b \)), and \( 127 \) (from \( c \)).
- The lowest power is \( 127 \).
- For the prime factor \( 5 \):
- The powers are \( 128 \) (from \( a \)), \( 129 \) (from \( b \)), and \( 128 \) (from \( c \)).
- The lowest power is \( 128 \).
- For the prime factor \( 7 \):
- The powers are \( 22 \) (from \( a \)), \( 21 \) (from \( b \)), and \( 24 \) (from \( c \)).
- The lowest power is \( 21 \).

3. **Combine the lowest powers to find the HCF:**
- Therefore, the HCF of \( a \), \( b \), and \( c \) is:
\[ HCF(a, b, c) = 3^{127} \times 5^{128} \times 7^{21} \]

### Final Answer:
The HCF of \( a \), \( b \), and \( c \) is \( 3^{127} \times 5^{128} \times 7^{21} \).

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