Question

The remainder when \( 4^{28^{2024}} \) is divided by 21 is __________.

Answer 1

Correct Answer: 1

We start with the given congruence:

\[ (428)^{2024} \equiv 8^{2024} \pmod{21} \]

Now, simplify \(8^2 \mod 21\):

\[ 8^2 = 64 \equiv 1 \pmod{21} \]

Hence,

\[ 8^{2024} = (8^2)^{1012} \equiv 1^{1012} \equiv 1 \pmod{21} \]

Therefore, the remainder is:

\[ \boxed{1} \]

Answer 2

Step 1: Simplify \(428^{2024} \mod 21\) Write 428 as:

$$428 = 420 + 8.$$

Thus:

$$428^{2024} = (420 + 8)^{2024}.$$

When divided by 21, 420 is a multiple of 21:

$$428^{2024} \equiv 8^{2024} \pmod{21}.$$

Step 2: Simplify \(8^{2024} \mod 21\) Write \(8^{2024}\) as:

$$8^{2024} = (8^2)^{1012}.$$

Calculate \(8^2\):

$$8^2 = 64.$$

Thus:

$$8^{2024} \equiv 64^{1012} \pmod{21}.$$

Step 3: Simplify \(64 \mod 21\) Since \(64 = 63 + 1 = 21 \times 3 + 1\), we have:

$$64 \equiv 1 \pmod{21}.$$

Thus:

$$64^{1012} \equiv 1^{1012} \pmod{21}.$$

Step 4: Final Result

$$8^{2024} \equiv 1 \pmod{21}.$$

Hence, the remainder when \(428^{2024}\) is divided by 21 is: 1.