Question

When \(3^{333}\) is divided by 11, the remainder is:

• A. 1
• B. 10
• C. 33
• D. 11
• E. 5

Hint:

Look for patterns in pairs: (first and secon(d), (third and fourt(h), etc. Or consider a central number as a result of operations on surrounding numbers.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
This problem involves modular arithmetic. Since 11 is a prime number, we can use Fermat's Little Theorem to simplify the calculation of the remainder.
Step 2: Key Formula or Approach:
Fermat's Little Theorem: \( a^{p-1} \equiv 1 \pmod{p} \) where \( p \) is a prime and \( a \) is not divisible by \( p \).
Step 3: Detailed Explanation:
Here \( a = 3 \) and \( p = 11 \).
By the theorem: \( 3^{10} \equiv 1 \pmod{11} \).
We can rewrite the exponent: \( 333 = 10 \times 33 + 3 \).
Therefore:
\[ 3^{333} = (3^{10})^{33} \cdot 3^3 \] \[ 3^{333} \equiv (1)^{33} \cdot 27 \pmod{11} \] \[ 3^{333} \equiv 1 \cdot 27 \pmod{11} \] Now find the remainder when 27 is divided by 11:
\[ 27 = 11 \times 2 + 5 \] The remainder is 5.
Step 4: Final Answer:
The remainder is 5.