Question

Find remainder in 2021^3762 / 17

Answer 1

4

Answer 2

Solution:

1. Reduce the Base Modulo 17:

   $$2021 \equiv 2021 - 17 \times 118 = 2021 - 2006 = 15 \pmod{17}$$

   So, $2021^{3762} \equiv 15^{3762} \pmod{17}$.

2. Express in Simpler Form:

   Notice that $15 \equiv -2 \pmod{17}$. Therefore,

   $$15^{3762} \equiv (-2)^{3762} \pmod{17}$$

   Since the exponent is even,

   $$(-2)^{3762} = 2^{3762}$$

3. Apply Fermat's Little Theorem:

   For prime $17$, Fermat's Little Theorem gives

   $$2^{16} \equiv 1 \pmod{17}$$

   Reduce the exponent modulo 16:

   $$3762 \mod 16 = 3762 - 16 \times 235 = 3762 - 3760 = 2.$$

   Thus,

   $$2^{3762} \equiv 2^2 \equiv 4 \pmod{17}$$

Final Answer: The remainder when $2021^{3762}$ is divided by 17 is 4.

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Explanation (Minimal):

Reduce 2021 modulo 17 to get 15. Write 15 as $-2$ modulo 17. Since exponent is even, compute $2^{3762}$. Using Fermat’s theorem, reduce exponent modulo 16 to get $2^2 = 4$.