Aptitude | Arithmetic Aptitude 6 | Question 4

X is a positive 4 digit number. Find the largest possible value of X, such that on dividing with 5, 6, 7 it gives the remainder zero.

(A)

9980

(B)

9870

(C)

9540

(D)

9640

Answer: (B)
Explanation:

The required number must be divisible by L.C.M. of 5,6 and 7.
L.C.M. of 5, 6 and 7 = 5 x 6 x 7 = 210

Let us divide 9999 by 210.

210) 9999 (47
      840
     ----
      1599
      1470
      ----
       129

Required number = 9999 – 129 = 9870

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