Find the HCF of `2^(100)-1,2^(120)-1` ?

Options

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

Answer

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

\[ \text{HCF}(a^m - 1, a^n - 1) = a^{\text{HCF}(m, n)} - 1 \]

In our case, \(a = 2\), \(m = 100\), and \(n = 120\).

### Step 1: Find the HCF of the exponents \(m\) and \(n\)
We need to find \(\text{HCF}(100, 120)\).

**Finding HCF of 100 and 120:**
- The prime factorization of \(100\) is:
\[ 100 = 2^2 \times 5^2 \]
- The prime factorization of \(120\) is:
\[ 120 = 2^3 \times 3^1 \times 5^1 \]

**Finding common factors:**
- The common prime factors are \(2\) and \(5\).
- For \(2\), the minimum power is \(2^2\).
- For \(5\), the minimum power is \(5^1\).

Thus, we calculate:
\[ \text{HCF}(100, 120) = 2^2 \times 5^1 = 4 \times 5 = 20 \]

### Step 2: Apply the HCF to the formula
Now that we have \(\text{HCF}(100, 120) = 20\), we can substitute this back into our formula:

\[ \text{HCF}(2^{100} - 1, 2^{120} - 1) = 2^{\text{HCF}(100, 120)} - 1 = 2^{20} - 1 \]

### Step 3: Final Result
Thus, the HCF of \(2^{100} - 1\) and \(2^{120} - 1\) is:

\[ \text{HCF}(2^{100} - 1, 2^{120} - 1) = 2^{20} - 1 \]