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
• E. 6

Hint:

257 = \(2^8+1\), so \(2^8 \equiv -1\) and \(2^{16} \equiv 1 \pmod{257}\). Since 1472 is a multiple of 16, the remainder is 1.

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" is \(92 \times 16 = 1472\). So we need the remainder when \(2^{1472}\) is divided by 257.

Step 2: Use a property of 257.
257 is a prime number, and it can be written as \(2^8 + 1\). This means \(2^8 \equiv -1 \pmod{257}\).

Step 3: Find the order of 2 modulo 257.
Squaring both sides: \(2^{16} = (2^8)^2 \equiv (-1)^2 = 1 \pmod{257}\).

Step 4: Reduce the exponent.
Since \(1472 = 16 \times 92\), we can write:
\(2^{1472} = (2^{16})^{92} \equiv 1^{92} = 1 \pmod{257}\).

Step 5: Conclusion.
The remainder is 1.