Find the HCF of `(3^(125)-1)and(3^(35)-1)`.

Options

A. 5
B. 3
C. `(3^(5)-1)`
D. `(3^(35)-1)`

Answer

To find the HCF of \(3^{125} - 1\) and \(3^{35} - 1\), we can use a specific property of exponents. The property states that:

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

In our case, \(a = 3\), \(m = 125\), and \(n = 35\).

### Step 1: Identify the values
We have:
- \(a = 3\)
- \(m = 125\)
- \(n = 35\)

### Step 2: Calculate the HCF of \(m\) and \(n\)
We need to find \(\text{HCF}(125, 35)\).

To do this, we can use the prime factorization method or the Euclidean algorithm. Let's use the Euclidean algorithm:

1. Divide 125 by 35, which gives a quotient of 3 and a remainder of \(125 - 3 \times 35 = 125 - 105 = 20\).
2. Now, apply the algorithm again: Divide 35 by 20, which gives a quotient of 1 and a remainder of \(35 - 20 = 15\).
3. Next, divide 20 by 15, which gives a quotient of 1 and a remainder of \(20 - 15 = 5\).
4. Finally, divide 15 by 5, which gives a quotient of 3 and a remainder of \(15 - 3 \times 5 = 0\).

Since we reached a remainder of 0, the last non-zero remainder is the HCF. Thus, \(\text{HCF}(125, 35) = 5\).

### Step 3: Substitute back into the formula
Now, we substitute back into our formula:

\[ \text{HCF}(3^{125} - 1, 3^{35} - 1) = 3^{\text{HCF}(125, 35)} - 1 = 3^5 - 1 \]

### Step 4: Calculate \(3^5 - 1\)
Now we calculate \(3^5\):

\[ 3^5 = 243 \]

So,

\[ 3^5 - 1 = 243 - 1 = 242 \]

### Final Answer
Thus, the HCF of \(3^{125} - 1\) and \(3^{35} - 1\) is:

\[ \text{HCF}(3^{125} - 1, 3^{35} - 1) = 242 \]