`(97^10-1024)` is completely divisible by the number ?

Options

A. 12
B. 13
C. 11
D. 17

Answer

To determine if \( (97^{10} - 1024) \) is completely divisible by a certain number, we can follow these steps:

### Step 1: Rewrite \( 1024 \) as a power of \( 2 \)
We know that \( 1024 = 2^{10} \). Therefore, we can rewrite the expression as:
\[ 97^{10} - 2^{10} \]

### Step 2: Recognize the difference of powers
The expression \( a^n - b^n \) can be factored using the formula:
\[ a^n - b^n = (a - b)(a^{n-1} + a^{n-2}b + a^{n-3}b^2 + \ldots + b^{n-1}) \]
In our case, \( a = 97 \), \( b = 2 \), and \( n = 10 \). Thus, we can factor it as:
\[ (97 - 2)(97^9 + 97^8 \cdot 2 + 97^7 \cdot 2^2 + \ldots + 2^9) \]

### Step 3: Calculate \( 97 - 2 \)
Calculating \( 97 - 2 \):
\[ 97 - 2 = 95 \]

### Step 4: Identify the factors of \( 95 \)
Now we need to find the factors of \( 95 \):
\[ 95 = 5 \times 19 \]
The factors of \( 95 \) are \( 1, 5, 19, 95 \).

### Step 5: Check divisibility
Now we need to check if \( (97^{10} - 1024) \) is divisible by any of these factors. We can also check if \( 97^{10} - 1024 \) is divisible by \( 11 \) (since \( 11 \) is a prime number and a factor of \( 95 \)).

### Step 6: Check \( 97 \mod 11 \)
Calculating \( 97 \mod 11 \):
\[ 97 \div 11 = 8 \quad \text{(remainder 9)} \]
Thus, \( 97 \equiv 9 \mod 11 \).

### Step 7: Calculate \( 97^{10} \mod 11 \)
Now we need to calculate \( 9^{10} \mod 11 \). Using Fermat's Little Theorem, since \( 11 \) is prime:
\[ 9^{10} \equiv 1 \mod 11 \]
Thus,
\[ 97^{10} - 1024 \equiv 1 - 2^{10} \mod 11 \]
Calculating \( 2^{10} \mod 11 \):
\[ 2^{10} = 1024 \quad \text{and} \quad 1024 \mod 11 = 1 \]
Thus,
\[ 97^{10} - 1024 \equiv 1 - 1 \equiv 0 \mod 11 \]

### Conclusion
Since \( 97^{10} - 1024 \) is divisible by \( 11 \), we conclude that \( (97^{10} - 1024) \) is completely divisible by \( 11 \).

### Final Answer
The number by which \( (97^{10} - 1024) \) is completely divisible is \( 11 \).

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