The remainder when \( 4^{28^{2024}} \) is divided by 21 is __________.
Correct Answer: 1
Approach Solution - 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:
\[ \boxed{1} \]
Approach Solution -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.
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