Binomial Theorem Applications

Modulo Search: We need to find $2^{100} (mod 7)$ .
- Note that $2^{3} = 8 \equiv 1 (mod 7)$ .
Breakdown Power: Write $100$ in terms of $3$ ( $100 = 3 \times 33 + 1$ ).
- $2^{100} = (2^{3})^{33} \cdot 2^{1}$
Application of Congruence:
- $2^{100} \equiv (1)^{33} \cdot 2 (mod 7)$
- $2^{100} \equiv 1 \cdot 2 \equiv 2 (mod 7)$ .
Result: The remainder is $2$ .

This hard difficulty mathematics question is from the chapter binomial theorem and its simple applications, covering the topic of binomial theorem applications. It appeared in the 2025 exam.

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Binomial Theorem Applications

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