Find the remiander when `32^(32)` is divided 9.

Options

A. 4
B. 5
C. 3
D. 6

Answer

To find the remainder when \( 32^{32} \) is divided by \( 9 \), we can use the concept of modular arithmetic and properties of exponents. Here’s a step-by-step solution:

### Step 1: Simplify the base modulo 9
First, we simplify \( 32 \) modulo \( 9 \):
\[ 32 \div 9 = 3 \quad \text{(quotient)} \]
\[ 32 - (9 \times 3) = 32 - 27 = 5 \]
So, \( 32 \equiv 5 \mod 9 \).

### Step 2: Rewrite the expression
Now, we can rewrite \( 32^{32} \) as:
\[ 32^{32} \equiv 5^{32} \mod 9 \]

### Step 3: Use Euler's theorem
To simplify \( 5^{32} \mod 9 \), we can use Euler's theorem. First, we need to find \( \phi(9) \):
- The prime factorization of \( 9 \) is \( 3^2 \).
- Therefore, \( \phi(9) = 9 \left(1 - \frac{1}{3}\right) = 9 \times \frac{2}{3} = 6 \).

According to Euler's theorem, since \( 5 \) and \( 9 \) are coprime:
\[ 5^{\phi(9)} \equiv 1 \mod 9 \]
This means:
\[ 5^6 \equiv 1 \mod 9 \]

### Step 4: Reduce the exponent modulo \( \phi(9) \)
Now, we need to reduce the exponent \( 32 \) modulo \( 6 \):
\[ 32 \div 6 = 5 \quad \text{(quotient)} \]
\[ 32 - (6 \times 5) = 32 - 30 = 2 \]
So, \( 32 \equiv 2 \mod 6 \).

### Step 5: Calculate \( 5^{32} \mod 9 \)
Now we can reduce our expression:
\[ 5^{32} \equiv 5^2 \mod 9 \]
Calculating \( 5^2 \):
\[ 5^2 = 25 \]

### Step 6: Find \( 25 \mod 9 \)
Now, we find \( 25 \mod 9 \):
\[ 25 \div 9 = 2 \quad \text{(quotient)} \]
\[ 25 - (9 \times 2) = 25 - 18 = 7 \]
So, \( 25 \equiv 7 \mod 9 \).

### Conclusion
Thus, the remainder when \( 32^{32} \) is divided by \( 9 \) is:
\[ \boxed{7} \]