LCM and HCF


  1. What is the HCF of 8 (N5 - N3 + N ) and 28 (N6 + 1)?









  1. View Hint View Answer Discuss in Forum

    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)

    Correct Option: A

    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)


  1. The HCF of (x4 - y4) and (x6 - y6) is









  1. View Hint View Answer Discuss in Forum

    Let f (x) = (x4 - y4)
    = (x2 - y2) (x2 + y2)
    = (x - y) (x + y) (x2 + y2)
    and g(x) = (x6 - y6)
    = (x3)2 - (y3)2
    = (x3 + y3) (x3 - y3)
    = (x + y) (x2 - xy + y2) (x - y)(x2 + xy + y2)
    = (x - y) (x + y) (x2 - xy + y2)
    (x2 + xy + y2)
    ∴ HCF of [f(x), g(x)] = (x - y) (x + y)
    = x2 - y2

    Correct Option: A

    Let f (x) = (x4 - y4)
    = (x2 - y2) (x2 + y2)
    = (x - y) (x + y) (x2 + y2)
    and g(x) = (x6 - y6)
    = (x3)2 - (y3)2
    = (x3 + y3) (x3 - y3)
    = (x + y) (x2 - xy + y2) (x - y)(x2 + xy + y2)
    = (x - y) (x + y) (x2 - xy + y2)
    (x2 + xy + y2)
    ∴ HCF of [f(x), g(x)] = (x - y) (x + y)
    = x2 - y2



  1. The HCF of (x3 - y2 - 2x) and (x3 + x2) is









  1. View Hint View Answer Discuss in Forum

    Let f(x) = x3 - y2 - 2x = x( x2 - x- 2)
    = x( x2 - 2x + x - 2)
    = x {x (x - 2) + 1 (x -2)}
    = x (x +1) (x - 2)
    and g(x) = x3 + x2
    = x2 (x +1) = x. x (x +1)
    ∴ LCM of [ f(x), g(x)] = x (x +1) . x . (x - 2)

    Correct Option: C

    Let f(x) = x3 - y2 - 2x = x( x2 - x- 2)
    = x( x2 - 2x + x - 2)
    = x {x (x - 2) + 1 (x -2)}
    = x (x +1) (x - 2)
    and g(x) = x3 + x2
    = x2 (x +1) = x. x (x +1)
    ∴ LCM of [ f(x), g(x)] = x (x +1) . x . (x - 2)
    = x2(x + 1)(x - 2)
    = x2(x2 - x - 2)
    = x4 - x3 - 2x2


  1. What is the HCF of a2 b4 + 2 a2 b2 and (ab)7 - 4a2 b2?









  1. View Hint View Answer Discuss in Forum

    a2 b4 + 2 a2 b2 = a2 b2 ( b2 +2).......................(i)
    and (ab)7 - 4 a2 b9 = a7 b7 - 4 a2 b9 = a2 b2(a5 b5 - 4b7)................(ii)

    from Eqs. (i) and (ii)
    HCF = a2 b2

    Correct Option: C

    a2 b4 + 2 a2 b2 = a2 b2 ( b2 +2).......................(i)
    and (ab)7 - 4 a2 b9 = a7 b7 - 4 a2 b9 = a2 b2(a5 b5 - 4b7)................(ii)

    from Eqs. (i) and (ii)
    HCF = a2 b2



  1. Find the largest number which divides 1305, 4665 and 6905 leaving same remainder in each case. Also, find the common Remainder.











  1. View Hint View Answer Discuss in Forum

    Given that
    x = 1305, y = 4665, z = 6905

    Then ,
    |x - y| = |1305 - 4665| = 3360
    |y - z] = |4665 - 6905| = 2240
    |z - x| =|6905 -1305| = 5600

    ∴ Required number = HCF of 3360, 2240 and 5600 = 1120

    On dividing 1305 by 1120, remainder is 185.
    On dividing 4665 by 1120, remainder is 185.
    On dividing 6905 by 1120, remainder is 185.
    ∴ Common remainder = 185

    Correct Option: C

    Given that
    x = 1305, y = 4665, z = 6905

    Then ,
    |x - y| = |1305 - 4665| = 3360
    |y - z] = |4665 - 6905| = 2240
    |z - x| =|6905 -1305| = 5600

    ∴ Required number = HCF of 3360, 2240 and 5600 = 1120

    On dividing 1305 by 1120, remainder is 185.
    On dividing 4665 by 1120, remainder is 185.
    On dividing 6905 by 1120, remainder is 185.
    ∴ Common remainder = 185