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Binomial Theorem question

2024 · 9 Apr · Shift 1 · Q54

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JEE Main
Mathematics
Binomial Theorem
2024 · 9 Apr · Shift 1 · Q54

Binomial Theorem question

2024 · 9 Apr · Shift 1 · Q54

JEE MainMathematicsBinomial TheoremNumerical+4 / −1

The remainder when

428^{2024}

is divided by 21 is

\underline{\hspace{2cm}}

.

Numerical answer

View written solutionFree

Correct answer: 1

We need to find the remainder when $428^{2024}$ is divided by $21$ .
First reduce $428$ modulo $21$ : $428 \div 21 = 20 \text{ remainder } 8$ So, $428 \equiv 8 \pmod{21}.$ Hence, $428^{2024} \equiv 8^{2024} \pmod{21}.$
Now compute powers of $8$ modulo $21$ : $8^2=64 \equiv 1 \pmod{21}$ because $64-63=1.$
Therefore, $8^{2024}=(8^2)^{1012} \equiv 1^{1012} \equiv 1 \pmod{21}.$
So the remainder when $428^{2024}$ is divided by $21$ is $\boxed{1}.$

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