How many different factors are there for the number 48, excluding 1 and 48?
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12
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4
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8
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None of these
48 = 2^4 × 3. Total factors = (4+1)(1+1) = 10. The factors are: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48. Excluding 1 and 48, we have 8 factors: 2, 3, 4, 6, 8, 12, 16, 24.
To find the number of different factors of a given number, we can prime factorize the number and then use the exponents of the prime factors to calculate the number of factors.
Prime factorization of 48: [48 = 2^4 \times 3^1]
To find the number of factors, we consider all possible combinations of the exponents of the prime factors. For each prime factor, we can have exponents ranging from 0 to the highest exponent in the prime factorization.
For the prime factor 2, we have 4+1 = 5 possible exponents (0, 1, 2, 3, 4). For the prime factor 3, we have 1+1 = 2 possible exponents (0, 1).
Therefore, the total number of factors is the product of the number of possible exponents for each prime factor: (5 \times 2 = 10).
However, we need to exclude the factors 1 and 48. So, the total number of different factors is (10 - 2 = 8).
Hence, the correct answer is option C) 8.