The remainder obtained when `27^(50)` is divided by 12 is

Answer

To find the remainder when \( 27^{50} \) is divided by 12, we can follow these steps:

### Step 1: Simplify the Base
First, we can simplify \( 27 \) modulo \( 12 \):
\[ 27 \div 12 = 2 \quad \text{(remainder 3)} \]
Thus,
\[ 27 \equiv 3 \mod 12 \]

### Step 2: Rewrite the Expression
Now, we can rewrite \( 27^{50} \) in terms of \( 3 \):
\[ 27^{50} \equiv 3^{50} \mod 12 \]

### Step 3: Use Euler’s Theorem
To simplify \( 3^{50} \mod 12 \), we can use Euler's theorem. First, we need to find \( \phi(12) \):
- The prime factorization of \( 12 \) is \( 2^2 \times 3^1 \).
- Thus, \( \phi(12) = 12 \left(1 - \frac{1}{2}\right)\left(1 - \frac{1}{3}\right) = 12 \times \frac{1}{2} \times \frac{2}{3} = 4 \).

Since \( \gcd(3, 12) = 3 \neq 1 \), we cannot directly apply Euler’s theorem. Instead, we will compute \( 3^{50} \mod 12 \) directly.

### Step 4: Calculate Powers of 3 Modulo 12
Let’s compute the first few powers of \( 3 \) modulo \( 12 \):
- \( 3^1 \equiv 3 \mod 12 \)
- \( 3^2 \equiv 9 \mod 12 \)
- \( 3^3 \equiv 27 \equiv 3 \mod 12 \)
- \( 3^4 \equiv 81 \equiv 9 \mod 12 \)

We see a pattern:
- \( 3^{2n} \equiv 9 \mod 12 \)
- \( 3^{2n+1} \equiv 3 \mod 12 \)

### Step 5: Determine the Parity of the Exponent
Since \( 50 \) is even, we have:
\[ 3^{50} \equiv 9 \mod 12 \]

### Step 6: Conclusion
Thus, the remainder when \( 27^{50} \) is divided by \( 12 \) is:
\[ \boxed{9} \]

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