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Question

Mathematics Question on Number Systems

The remainder when 42820244^{28^{2024}} is divided by 21 is __________.

Answer

Step 1: Simplify 4282024mod21428^{2024} \mod 21 Write 428 as:

428=420+8.428 = 420 + 8.

Thus:

4282024=(420+8)2024.428^{2024} = (420 + 8)^{2024}.

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

428202482024(mod21).428^{2024} \equiv 8^{2024} \pmod{21}.

Step 2: Simplify 82024mod218^{2024} \mod 21 Write 820248^{2024} as:

82024=(82)1012.8^{2024} = (8^2)^{1012}.

Calculate 828^2:

82=64.8^2 = 64.

Thus:

82024641012(mod21).8^{2024} \equiv 64^{1012} \pmod{21}.

Step 3: Simplify 64mod2164 \mod 21 Since 64=63+1=21×3+164 = 63 + 1 = 21 \times 3 + 1, we have:

641(mod21).64 \equiv 1 \pmod{21}.

Thus:

64101211012(mod21).64^{1012} \equiv 1^{1012} \pmod{21}.

Step 4: Final Result

820241(mod21).8^{2024} \equiv 1 \pmod{21}.

Hence, the remainder when 4282024428^{2024} is divided by 21 is: 1.

Explanation

Solution

Step 1: Simplify 4282024mod21428^{2024} \mod 21 Write 428 as:

428=420+8.428 = 420 + 8.

Thus:

4282024=(420+8)2024.428^{2024} = (420 + 8)^{2024}.

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

428202482024(mod21).428^{2024} \equiv 8^{2024} \pmod{21}.

Step 2: Simplify 82024mod218^{2024} \mod 21 Write 820248^{2024} as:

82024=(82)1012.8^{2024} = (8^2)^{1012}.

Calculate 828^2:

82=64.8^2 = 64.

Thus:

82024641012(mod21).8^{2024} \equiv 64^{1012} \pmod{21}.

Step 3: Simplify 64mod2164 \mod 21 Since 64=63+1=21×3+164 = 63 + 1 = 21 \times 3 + 1, we have:

641(mod21).64 \equiv 1 \pmod{21}.

Thus:

64101211012(mod21).64^{1012} \equiv 1^{1012} \pmod{21}.

Step 4: Final Result

820241(mod21).8^{2024} \equiv 1 \pmod{21}.

Hence, the remainder when 4282024428^{2024} is divided by 21 is: 1.