Mathematics · Binomial Theorem

JEE Main 2024 — 9 April, Shift 1 — Question 25

The remainder when $428^{2024}$ is divided by $21$ is $\qquad$ .

Answer: 1

Numerical answer — enter this value.

Step-by-step solution

${{(428)}^{2024}}={{(420+8)}^{2024}}$ $={{(21\times 20+8)}^{2024}}$ $=21m+{{8}^{2024}}$ Now ${{8}^{2024}}={{\left( {{8}^{2}} \right)}^{1012}}$ $={{(64)}^{1012}}$ $={{(63+1)}^{1012}}$ $={{(21\times 3+1)}^{1012}}$ $=2\text{ln}+1$

=Remainder is 1

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Exam: JEE Main 2024
Paper: 9 April, Shift 1
Subject: Mathematics
Chapter: Binomial Theorem
Topic: Applications of Binomial Theorem

The coefficient of x^70 in x^2(1+x)^98+x^3(1+x)^97+ x^4(1+ x)^96+ . . + x^54(1+ x)^46 is ^99 C p- ^46 C q . Then a possible value to p+ q…

Answer: 1

Step-by-step solution

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