Discussion
It is being given that \((2^{32} + 1)\) is completely divisible by a whole number. Which of the following numbers is completely divisible by this number?
Answer: D
Let \(2^{32} = x\).
Then, \((2^{32} + 1) = (x + 1)\).
Let \((x + 1)\) be completely divisible by the natural number N. Then,
\((2^{96} + 1) \\= [(2^{32})^{3} + 1]\\= (x^{3} + 1)\\= (x + 1)(x^{2} - x + 1)\)
which is completely divisible by N, since \((x + 1)\) is divisible by N.
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