To analyze whether \(2^{55} - 1\) is a Mersenne prime, we first note the definition of a Mersenne prime. A Mersenne prime is a prime number of the form \(2^p - 1\) where \(p\) is a prime number.
In this case, \(p = 55\), which is not a prime number (since it can be factored as \(5 \times 11\)). Therefore, \(2^{55} - 1\) cannot be a Mersenne prime.
Next, to find a factor of \(2^{55} - 1\), we can use the formula for factoring \(2^n - 1\). The factorization states that:
\[ 2^{n} - 1 = (2^{d} - 1)(2^{n/d} + 2^{(n/d)-1} + \ldots + 2 + 1) \]
for any divisor \(d\) of \(n\).
For \(n = 55\), we can look for its divisors. Notably, \(d = 5\) is a divisor. Using it, we can factor:
\[ 2^{55} - 1 = (2^{5} - 1)(2^{50} + 2^{45} + \ldots + 1) \]
Calculating \(2^5 - 1\):
\[ 2^5 - 1 = 32 - 1 = 31 \]
Now, let's evaluate this factorization:
- \(2^{5} - 1 = 31\) (which is prime), and
- \(2^{50} + 2^{45} + 2^{40} + 2^{35} + 2^{30} + 2^{25} + 2^{20} + 2^{15} + 2^{10} + 2^{5} + 1\) is a large number that adds together all powers of 2 from \(0\) to \(50\) (in groups).
Since \(2^{55} - 1\) clearly has a factor of \(31\), \(2^{55} - 1\) itself is not prime.
We have established that:
- No, \(2^{55} - 1\) is not a Mersenne prime.
- A factor of \(2^{55} - 1\) is \(2^{5} - 1 = 31\).
Now, among the options listed, the correct choice is:
No, \(2^{55} - 1\) is not a Mersenne prime. A factor of \(2^{55} - 1\) is \(2^{10} + 2^{9} + 2^{8} + 2^{7} + 2^{6} + 2^{5} + 2^{4} + 2^{3} + 2^{2} + 2 + 1 = 2,047.\)
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