What is the HCF of 8 (N 5 - N 3 + N ) and 28 (N 6 + 1)?

Aptitude Problems on H.C.F and L.C.M
Choose an option
  • A
    4 (N4 - N2 + 1)
  • B
    N3 - N + 4N2
  • C
    N3 - N + 3N2
  • D
    None of these

Answer

Correct Answer: 4 (N4 - N2 + 1)

Explanation

Let p(N) = 8 (N5 - N3 + N) = 4 x 2 x N (N4 - N2 + 1) and q(N) = 28 (N6 + 1) = 7 x 4 [( N2)3 + (1)3] = 4 x 7 (N2 + 1) ( N4 - N2 + 1) ∴ HCF of p(N) and q (N) = 4 (N4 - N2 + 1)

Discussion & Comments
No comments yet. Be the first to comment!
Join Discussion