Find GCD `(2^(120)-1,2^(100)-1)`.

Options

A. `2^(20)-1`
B. `2^(40)-1`
C. `2^(60)-1`
D. `2^(10)-1`

Answer

To find the GCD of \(2^{120} - 1\) and \(2^{100} - 1\), we can use a property of exponents and GCDs. The property states that:

\[ \text{GCD}(2^m - 1, 2^n - 1) = 2^{\text{GCD}(m, n)} - 1 \]

### Step 1: Identify the values of \(m\) and \(n\)
In this case, we have:
- \(m = 120\)
- \(n = 100\)

### Step 2: Calculate the GCD of \(m\) and \(n\)
Now, we need to find \(\text{GCD}(120, 100)\).

To find the GCD, we can use the prime factorization method:
- \(120 = 2^3 \times 3^1 \times 5^1\)
- \(100 = 2^2 \times 5^2\)

Now, we take the lowest power of each prime factor that appears in both factorizations:
- For \(2\): the minimum power is \(2^2\)
- For \(3\): it does not appear in \(100\), so we ignore it.
- For \(5\): the minimum power is \(5^1\)

Thus, the GCD is:
\[ \text{GCD}(120, 100) = 2^2 \times 5^1 = 4 \times 5 = 20 \]

### Step 3: Apply the GCD property
Now that we have \(\text{GCD}(120, 100) = 20\), we can substitute this back into our original property:
\[ \text{GCD}(2^{120} - 1, 2^{100} - 1) = 2^{\text{GCD}(120, 100)} - 1 = 2^{20} - 1 \]

### Final Answer
Thus, the GCD of \(2^{120} - 1\) and \(2^{100} - 1\) is:
\[ \boxed{2^{20} - 1} \]