What is the remainder when `2^(96)` is divided by 96 ?

Answer

To find the remainder when \( 2^{96} \) is divided by 96, we can use the Chinese Remainder Theorem (CRT) by breaking down 96 into its prime factors. The prime factorization of 96 is \( 2^5 \times 3 \).

### Step 1: Calculate \( 2^{96} \mod 32 \)
Since \( 32 = 2^5 \), we can find \( 2^{96} \mod 32 \).

\[ 2^{96} \text{ is clearly } 0 \text{ when divided by } 32. \]

**Hint:** Any power of 2 that is greater than or equal to \( 2^5 \) will be divisible by \( 32 \).

### Step 2: Calculate \( 2^{96} \mod 3 \)
Next, we need to find \( 2^{96} \mod 3 \). We can use Fermat's Little Theorem, which states that if \( p \) is a prime and \( a \) is not divisible by \( p \), then \( a^{p-1} \equiv 1 \mod p \).

Here, \( p = 3 \) and \( a = 2 \):

\[ 2^{2} \equiv 1 \mod 3 \]

Now, since \( 96 \mod 2 = 0 \):

\[ 2^{96} = (2^2)^{48} \equiv 1^{48} \equiv 1 \mod 3 \]

**Hint:** Use Fermat's Little Theorem to simplify calculations involving powers modulo a prime.

### Step 3: Solve the system of congruences
Now we have the following system of congruences:

1. \( x \equiv 0 \mod 32 \)
2. \( x \equiv 1 \mod 3 \)

Let \( x = 32k \) for some integer \( k \). Substituting into the second congruence:

\[ 32k \equiv 1 \mod 3 \]

Calculating \( 32 \mod 3 \):

\[ 32 \equiv 2 \mod 3 \]

Thus, we have:

\[ 2k \equiv 1 \mod 3 \]

To solve for \( k \), we can find the multiplicative inverse of \( 2 \mod 3 \). The inverse is \( 2 \) because:

\[ 2 \times 2 \equiv 1 \mod 3 \]

So, multiplying both sides of \( 2k \equiv 1 \) by \( 2 \):

\[ k \equiv 2 \mod 3 \]

This means:

\[ k = 3m + 2 \text{ for some integer } m \]

Substituting back into \( x \):

\[ x = 32(3m + 2) = 96m + 64 \]

Thus:

\[ x \equiv 64 \mod 96 \]

### Final Answer
The remainder when \( 2^{96} \) is divided by \( 96 \) is:

\[ \boxed{64} \]