📜  资质|算术能力6 |问题4

📅  最后修改于: 2021-05-13 20:56:41             🧑  作者: Mango

可以被5、6和7整除的最大4位数字是:
(A) 9980
(B) 9870
(C) 9540
(D) 9640答案: (B)
解释:

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

这个问题的测验