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JEE Main 20249 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
9 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Binomial Theorem

JEE Main 2024 · 9 April · Morning Shift · Mathematics · Binomial Theorem

The remainder when 428^ 2024 is divided by 21 is__________

The remainder when 4282024428^{2024} is divided by 21 is__________

Correct answer

0

Step-by-step solution

(428)2024=(420+8)2024=(21×20+8)2024=21 m+82024\begin{aligned} & (428)^{2024}=(420+8)^{2024} \\ & =(21 \times 20+8)^{2024} \\ & =21 \mathrm{~m}+8^{2024} \end{aligned}
Now 82024=(82)10128^{2024}=\left(8^2\right)^{1012}=(64)1012=(63+1)1012=(21×3+1)1012=2ln⁡+1\begin{aligned} & =(64)^{1012} \\ & =(63+1)^{1012} \\ & =(21 \times 3+1)^{1012} \\ & =2 \ln +1 \end{aligned}⇒\Rightarrow Remainder is 1.
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