Discussion
What is the remainder when \( 3^{40} \) is divided by 5?
Answer: C
Correct answer: 1
Look at the remainders of successive powers of 3 on division by 5: \[ 3^{1} = 3,\quad 3^{2} = 9 \equiv 4,\quad 3^{3} = 27 \equiv 2,\quad 3^{4} = 81 \equiv 1. \] The pattern 3, 4, 2, 1 then repeats, so the cycle length is 4. Since \( 40 \) is exactly divisible by 4, \( 3^{40} \) corresponds to the last term of the cycle, and the remainder is 1. A remainder of 0 in the exponent division always means the last term, not the first - which is the commonest slip in these questions. The result also follows from the unit digit: the unit digits of powers of 3 cycle 3, 9, 7, 1, so \( 3^{40} \) ends in 1, and a number ending in 1 leaves remainder 1 on division by 5. For a prime divisor Fermat’s little theorem gives the same conclusion directly: \( 3^{4} \equiv 1 \pmod 5 \).
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