A、B、C三人参加下面的游戏:有三张牌,每张上写着一个整数p、q、r并且0<P<q<r.洗牌后,分发给每人一

2025-03-16 06:50:56
推荐回答(1个)
回答1:

根据题意有:N(p+q+r)=39,
由于N≥2,
则N=3.
p+q+r=13.
由于A得20个球,因而r≥7.
如果r=7,
那么A三次所得球数只能是6+7+7=20,
这与p+q+r=13矛盾,
从而r>7.
由B三次得10个,且最后一次得了r个,
因p、q≥1,必有r≤8,
因此r=8,p+q=5,
由此p=1,q=4或p=2,q=3.
但由A三次走了20步,
只能得p=1,q=4.
现将已推算出各次每人走的步数列表:
A B C
8 1 4
8 1 4
4 8 1
观察此表知,第一次得q个球的是C.

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