Binomial Theorem Applications
Hardmathematics
The remainder when 2¹⁰⁰ is divided by 7 is
Select the correct option:
Solution
Incorrect! Answer:
2
- Modulo Search: We need to find 2100(mod7).
- Note that 23=8≡1(mod7).
- Breakdown Power: Write 100 in terms of 3 (100=3×33+1).
- 2100=(23)33⋅21
- Application of Congruence:
- 2100≡(1)33⋅2(mod7)
- 2100≡1⋅2≡2(mod7).
- Result: The remainder is 2.
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About This Question
- Subject
- mathematics
- Chapter
- binomial theorem and its simple applications
- Topic
- binomial theorem applications
- Difficulty
- Hard
- Year
- 2025
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. Practice this and similar questions to strengthen your understanding of binomial theorem and its simple applications concepts.
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