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Binomial Theorem question

2024 · 29 Jan · Shift 2 · Q58

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Binomial Theorem
2024 · 29 Jan · Shift 2 · Q58

Binomial Theorem question

2024 · 29 Jan · Shift 2 · Q58

JEE MainMathematicsBinomial TheoremNumerical+4 / −1

Remainder when

64^{32^{32}}

is divided by 9 is equal to

\underline{\hspace{2cm}}

.

Numerical answer

View written solutionFree

Correct answer: 1

We need to find the remainder when $64^{32^{32}}$ is divided by $9$ .
First reduce the base modulo $9$ : $64 \equiv 1 \pmod{9}$ because $64 = 9\cdot 7 + 1.$
Therefore, $64^{32^{32}} \equiv 1^{32^{32}} \pmod{9}.$
Since any power of $1$ is $1$ , $1^{32^{32}} = 1.$
Hence the remainder when $64^{32^{32}}$ is divided by $9$ is $\boxed{1}.$
Comparison with stored correct answer:

Derived answer: $1$
Stored correct answer: $1$

They match.

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