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 ^(23)`
`d = 3 ^(127) xx 5 ^(128) xx 7 ^(24) ,` then HCF of a,b,c and d is

Options

A. `3 ^(128) xx 5 ^(128) xx7 ^(21)`
B. `3 ^(127) xx 5 ^(128) xx 7 ^(21)`
C. `3 ^(127) xx 5 ^(128) xx 7 ^(21)`
D. `3 ^(127) xx 5 ^( 128) xx 7 ^(24)`

Answer

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

1. **Identify the prime factorization of each number:**
- \( a = 3^{129} \times 5^{128} \times 7^{22} \)
- \( b = 3^{128} \times 5^{129} \times 7^{21} \)
- \( c = 3^{127} \times 5^{128} \times 7^{23} \)
- \( d = 3^{127} \times 5^{128} \times 7^{24} \)

2. **Determine the minimum power for each prime factor across all numbers:**
- For the base \( 3 \):
- Powers: \( 129 \) (from \( a \)), \( 128 \) (from \( b \)), \( 127 \) (from \( c \)), \( 127 \) (from \( d \))
- Minimum power: \( 127 \)
- For the base \( 5 \):
- Powers: \( 128 \) (from \( a \)), \( 129 \) (from \( b \)), \( 128 \) (from \( c \)), \( 128 \) (from \( d \))
- Minimum power: \( 128 \)
- For the base \( 7 \):
- Powers: \( 22 \) (from \( a \)), \( 21 \) (from \( b \)), \( 23 \) (from \( c \)), \( 24 \) (from \( d \))
- Minimum power: \( 21 \)

3. **Combine the minimum powers to express the HCF:**
- HCF = \( 3^{127} \times 5^{128} \times 7^{21} \)

Thus, the HCF of \( a \), \( b \), \( c \), and \( d \) is \( 3^{127} \times 5^{128} \times 7^{21} \).