Find the remainder when `(17)^(23) + (29)^(23)` is divided by 23.

Options

A. 0
B. 1
C. 2
D. 3

Answer

To find the remainder when \( (17)^{23} + (29)^{23} \) is divided by 23, we can use Fermat's Little Theorem, which states that if \( p \) is a prime number and \( a \) is an integer not divisible by \( p \), then:

\[ a^{p-1} \equiv 1 \mod p \]

### Step 1: Identify the values
Here, \( p = 23 \) and we have two bases: \( a = 17 \) and \( b = 29 \). Both 17 and 29 are not divisible by 23.

### Step 2: Apply Fermat's Little Theorem
According to Fermat's Little Theorem:

\[ 17^{22} \equiv 1 \mod 23 \]
\[ 29^{22} \equiv 1 \mod 23 \]

### Step 3: Calculate \( 17^{23} \) and \( 29^{23} \)
Now, we can express \( 17^{23} \) and \( 29^{23} \) as follows:

\[ 17^{23} = 17^{22} \cdot 17 \equiv 1 \cdot 17 \equiv 17 \mod 23 \]
\[ 29^{23} = 29^{22} \cdot 29 \equiv 1 \cdot 29 \equiv 29 \mod 23 \]

### Step 4: Reduce \( 29 \mod 23 \)
Next, we need to reduce \( 29 \) modulo \( 23 \):

\[ 29 \mod 23 = 29 - 23 = 6 \]

### Step 5: Combine the results
Now we can combine our results:

\[ (17^{23} + 29^{23}) \mod 23 = (17 + 6) \mod 23 = 23 \mod 23 = 0 \]

### Conclusion
Thus, the remainder when \( (17)^{23} + (29)^{23} \) is divided by 23 is:

\[ \boxed{0} \]