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    The largest number of four digits which, when divided by 6, 12 and 18 leaves the same remainder 5 in each case is:
    Question

    The largest number of four digits which, when divided by 6, 12 and 18 leaves the same remainder 5 in each case is:

    A.

    9977

    B.

    9976

    C.

    9940

    D.

    9978

    Correct option is A

    Given:
    The number leaves a remainder of 5 when divided by 6, 12, and 18.
    Formula Used:
    N = LCM(6, 12, 18) × k + 5
    Solution:
    LCM(6, 12, 18) = 36
    N = 36k + 5
    Largest possible value of N is 9999.
    36k + 5 ≤ 9999
    36k ≤ 9994
    k ≤ 999436\frac {9994}{36} ​≈ 277.61
    N = 36 × 277 + 5 = 9972 + 5 = 9977
    Thus, the largest number of four digits which, when divided by 6, 12, and 18, leaves the same remainder of 5 is 9977.

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