Discussion
What is the smallest 5-digit number exactly divisible by 41?
Answer: A
The smallest 5-digit number is 10000.
Divide 10000 by 41.
\(10000 = 41 \times 243 + 37\).
The remainder is 37.
To find the number to be added to make it divisible by 41, we calculate (41 - 37) = 4.
The required number is 10000 + 4 = 10004.
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