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If a1,a2,a3,.....,a(n+1) be (n+1) diffe...

If `a_1,a_2,a_3,.....,a_(n+1)` be `(n+1)` different prime numbers, then the number of different factors (other than1) of `a_1^m.a_2.a_3...a_(n+1)`, is

A

(a) `m+1`

B

(b) `(m+1)2^(n)`

C

(c) `m*2^(n)+1`

D

(d) None of these

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To find the number of different factors (other than 1) of the expression \( a_1^m \cdot a_2^1 \cdot a_3^1 \cdots a_{n+1}^1 \), where \( a_1, a_2, a_3, \ldots, a_{n+1} \) are different prime numbers, we can follow these steps: ### Step-by-Step Solution: 1. **Identify the Prime Factorization**: The given expression can be represented as: \[ N = a_1^m \cdot a_2^1 \cdot a_3^1 \cdots a_{n+1}^1 \] Here, \( a_1, a_2, \ldots, a_{n+1} \) are \( n+1 \) different prime numbers. 2. **Determine the Exponents**: The exponents of the prime factors in the expression are: - For \( a_1 \): \( m \) - For \( a_2, a_3, \ldots, a_{n+1} \): \( 1 \) each. 3. **Calculate the Number of Factors**: The formula to find the total number of factors of a number based on its prime factorization is: \[ (e_1 + 1)(e_2 + 1)(e_3 + 1) \cdots (e_k + 1) \] where \( e_i \) are the exponents of the prime factors. For our expression: - The exponent of \( a_1 \) contributes \( m + 1 \) to the factor count. - Each of the other \( n \) primes contributes \( 1 + 1 = 2 \). Therefore, the total number of factors is: \[ (m + 1) \cdot 2 \cdot 2 \cdots 2 \quad \text{(n times)} \] This can be simplified to: \[ (m + 1) \cdot 2^n \] 4. **Exclude the Factor 1**: Since we are asked for the number of different factors other than 1, we need to subtract 1 from the total count: \[ \text{Number of factors other than 1} = (m + 1) \cdot 2^n - 1 \] 5. **Conclusion**: The final expression gives us the number of different factors (other than 1) of \( a_1^m \cdot a_2^1 \cdots a_{n+1}^1 \). ### Final Answer: The number of different factors (other than 1) of \( a_1^m \cdot a_2^1 \cdots a_{n+1}^1 \) is: \[ (m + 1) \cdot 2^n - 1 \]
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