Question

What is the remainder obtained when the smallest prime number, when raised to the ninety second multiple of sixteen, is divided by two hundred and fifty seven?

• A. 1
• B. 2
• C. 3
• D. 5

Hint:

257 = 2^8 + 1, so 2^8 leaves remainder -1 (that is, 256) when divided by 257, and 2^16 leaves remainder 1. Reduce the exponent 1472 modulo 16 to find the remainder quickly.

Answer 1

The Correct Option is A

Step 1: Identify the numbers involved.
The smallest prime number is 2. The "ninety second multiple of sixteen" means 92 x 16 = 1472. So we need the remainder when \(2^{1472}\) is divided by 257.

Step 2: Spot a useful property of 257.
257 can be written as \(2^8 + 1\), so \(2^8 \equiv -1 \pmod{257}\).
Squaring both sides, \(2^{16} \equiv (-1)^2 = 1 \pmod{257}\). This tells us that powers of 2, taken modulo 257, repeat every 16 steps.

Step 3: Reduce the exponent 1472 using this repeating cycle.
Since \(2^{16} \equiv 1 \pmod{257}\), we only need the remainder when the exponent 1472 is divided by 16.
\[ 1472 \div 16 = 92, \text{ exactly, with no remainder.} \]
So \(1472 = 16 \times 92\), which means:
\[ 2^{1472} = (2^{16})^{92} \equiv 1^{92} = 1 \pmod{257} \]

Step 4: State the remainder.
So \(2^{1472}\) leaves a remainder of 1 when divided by 257.

Final Answer:
The remainder is 1. \[ \boxed{1} \]