`(32^(32^32))/(9)` find R.

Options

A. 5
B. 7
C. 8
D. 4

Answer

To solve the problem \((32^{32^{32}}) \mod 9\) and find the remainder \(R\), we will follow these steps:

### Step 1: Simplify \(32 \mod 9\)
First, we need to find \(32 \mod 9\):
\[ 32 \div 9 = 3 \quad \text{(since } 9 \times 3 = 27\text{)} \]
\[ 32 - 27 = 5 \]
So, \(32 \equiv 5 \mod 9\).

### Step 2: Rewrite the expression
Now we can rewrite our original expression using this result:
\[ 32^{32^{32}} \mod 9 \equiv 5^{32^{32}} \mod 9 \]

### Step 3: Find the cyclicity of \(5^n \mod 9\)
Next, we need to find the pattern (cyclicity) of \(5^n \mod 9\):
- \(5^1 \mod 9 = 5\)
- \(5^2 \mod 9 = 25 \mod 9 = 7\)
- \(5^3 \mod 9 = 125 \mod 9 = 8\)
- \(5^4 \mod 9 = 625 \mod 9 = 4\)
- \(5^5 \mod 9 = 3125 \mod 9 = 2\)
- \(5^6 \mod 9 = 15625 \mod 9 = 1\)

The pattern repeats every 6 terms: \(5, 7, 8, 4, 2, 1\).

### Step 4: Find \(32^{32} \mod 6\)
Now we need to determine \(32^{32} \mod 6\) to find which power of 5 we will use:
\[ 32 \mod 6 = 2 \quad \text{(since } 6 \times 5 = 30\text{)} \]
Thus, we need to compute \(2^{32} \mod 6\).

### Step 5: Find the cyclicity of \(2^n \mod 6\)
Calculating \(2^n \mod 6\):
- \(2^1 \mod 6 = 2\)
- \(2^2 \mod 6 = 4\)
- \(2^3 \mod 6 = 8 \mod 6 = 2\)
- \(2^4 \mod 6 = 16 \mod 6 = 4\)

The pattern repeats every 2 terms: \(2, 4\). Since \(32\) is even, we have:
\[ 2^{32} \mod 6 \equiv 4 \]

### Step 6: Substitute back to find \(5^{32^{32}} \mod 9\)
Now we substitute back:
\[ 5^{32^{32}} \mod 9 \equiv 5^4 \mod 9 \]
From our earlier calculations:
\[ 5^4 \mod 9 = 4 \]

### Final Answer
Thus, the remainder \(R\) when \(32^{32^{32}}\) is divided by \(9\) is:
\[ \boxed{4} \]