When 2^256 is divided by 17, the remainder would be: (A) 1 (B) 16 (C) 14 (D) None of these
Question
When $2^{256}$ is divided by 17, the remainder would be: (A) 1 (B) 16 (C) 14 (D) None of these
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Answer
The correct option is (A) 1.
To determine the remainder when $2^{256}$ is divided by 17, let's rewrite $2^{256}$ in a different form as follows:
[ 2^{256} \text{ can be expressed as } \left(2^{4}\right)^{64} = (16)^{64} = (17 - 1)^{64}. ]
In the expansion of $(17 - 1)^{64}$ using the binomial theorem, every term in the expansion will be of the form:
[ \binom{64}{k} \cdot 17^k \cdot (-1)^{64 - k}. ]
For each term where $k \geq 1$, the factor $17^k$ ensures that it is divisible by 17. The only term that is not divisible by 17 is when $k = 0$, which simplifies to:
[ (-1)^{64} = 1. ]
Thus, every other term in the expansion is divisible by 17, and the only term that is not divisible gives a remainder of 1.
Alternatively, we can use Euler's Theorem which states that for any integer $a$ and a prime number $n$:
[ a^{\phi(n)} \equiv 1 \mod n \quad \text{if} \quad \gcd(a, n) = 1. ]
Here, $\phi(17) = 16$ since 17 is a prime number. Since $256 = 16 \times 16$, we have:
[ 2^{256} = (2^{16})^{16} \equiv 1^{16} \equiv 1 \mod 17. ]
Therefore, the remainder when $2^{256}$ is divided by 17 is 1.
Thus, the correct answer is (A) 1.
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