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Discussion :: Problems on H.C.F and L.C.M

  1. Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is:

  2. A.

    4

    B.

    5

    C.

    6

    D.

    8

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    Answer : Option A

    Explanation :

    N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305)

     

       = H.C.F. of 3360, 2240 and 5600 = 1120.

     

      Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4

     


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