Find the remainder when `32^(32^32)` is divided by 7

Options

A. 1
B. 2
C. 3
D. 4

Answer

To find the remainder when \( 32^{32^{32}} \) is divided by 7, we can follow these steps:

### Step 1: Simplify the Base
First, we simplify \( 32 \) modulo \( 7 \):
\[ 32 \mod 7 = 4 \]
Thus, we can rewrite the expression:
\[ 32^{32^{32}} \equiv 4^{32^{32}} \mod 7 \]

### Step 2: Find the Exponent Modulo 6
Next, we need to find \( 32^{32} \mod 6 \) because of Fermat's Little Theorem, which states that if \( p \) is a prime and \( a \) is not divisible by \( p \), then \( a^{p-1} \equiv 1 \mod p \). In our case, since \( 7 \) is prime, we use \( 6 \) (which is \( 7-1 \)) for the exponent.

First, we simplify \( 32 \mod 6 \):
\[ 32 \mod 6 = 2 \]
So, we need to compute:
\[ 2^{32} \mod 6 \]

### Step 3: Find \( 2^{32} \mod 6 \)
Now, we notice that \( 2^1 \equiv 2 \mod 6 \) and \( 2^2 \equiv 4 \mod 6 \), and \( 2^3 \equiv 8 \equiv 2 \mod 6 \). The powers of \( 2 \) modulo \( 6 \) repeat every \( 2 \) terms:
- \( 2^{2k} \equiv 4 \mod 6 \)
- \( 2^{2k+1} \equiv 2 \mod 6 \)

Since \( 32 \) is even, we have:
\[ 2^{32} \equiv 4 \mod 6 \]

### Step 4: Substitute Back into the Expression
Now we substitute back:
\[ 4^{32^{32}} \equiv 4^4 \mod 7 \]

### Step 5: Calculate \( 4^4 \mod 7 \)
Now we compute \( 4^4 \):
\[ 4^2 = 16 \equiv 2 \mod 7 \]
\[ 4^4 = (4^2)^2 \equiv 2^2 = 4 \mod 7 \]

### Final Result
Thus, the remainder when \( 32^{32^{32}} \) is divided by \( 7 \) is:
\[ \boxed{4} \]