从0到9十张卡片中选出三张,组成能被五整除的三位数共有几个

2025-04-04 16:02:03
推荐回答(4个)
回答1:

首先一共只有0-9,10个数字候选
末尾为0时,剩下9个数字能做首位的有9个,选出首位后,剩下只有8个,都能做第二位。这样的数共有9*8=72个。

末尾为5时,剩下的9个数字能作首位的只有8个(去掉0,因为只要三位数), 选出首位的后,剩下8个数字都可以做第二位。这样的数共有8*8=64个。

被5整除当且仅当末尾为0或5,所以这样的三位数共有64+72=136个。

回答2:

171个
末位为0时有9*10个
末位是5时,因为首位不为0所以有9*9个

回答3:

用组合数做
C(8,1)*C(9,1)+C(1,7)*C(1,9)=135

C(8,1)*C(9,1)末位为0时有9*8个
C(1,7)*C(1,9)末位是5时,因为首位不为0所以有7*9个
(因为说是从0到9十张卡片中选出三张,所以个位的数字和十位的数字和百位的数字一定不一样)

回答4:

我算出来是127个。。结尾是0或5.
当末尾是0的时候,有9×7,63种可能。
当末尾是5的时候,有8×8,64种可能。

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