Discussion
Find the remainder when \(2^{31}\) is divided by 5.
Answer: C
We look at the pattern of remainders of powers of 2 when divided by 5.
\(2^1 \div 5 \rightarrow rem=2\)
\(2^2 \div 5 \rightarrow rem=4\)
\(2^3 \div 5 \rightarrow rem=3\)
\(2^4 \div 5 \rightarrow rem=1\)
The cycle of remainders is (2, 4, 3, 1), which has a length of 4. To find the remainder of \(2^{31}\), we find the remainder of the power when divided by the cycle length: \(31 \div 4 \rightarrow rem=3\). The required remainder is the 3rd in the cycle, which is 3.
No comment is present. Be the first to comment.
Loading…
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development