一道澳大利亚的数学题,7月5日之前要哈Ten mathematics teachers composed 35 problems altogether for a mathematical contest.It is known that there were teachers who composed one,two and three problems.Prove that there was at least one teac

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一道澳大利亚的数学题,7月5日之前要哈Ten mathematics teachers composed 35 problems altogether for a mathematical contest.It is known that there were teachers who composed one,two and three problems.Prove that there was at least one teac
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一道澳大利亚的数学题,7月5日之前要哈Ten mathematics teachers composed 35 problems altogether for a mathematical contest.It is known that there were teachers who composed one,two and three problems.Prove that there was at least one teac
一道澳大利亚的数学题,7月5日之前要哈
Ten mathematics teachers composed 35 problems altogether for a mathematical contest.It is known that there were teachers who composed one,two and three problems.Prove that there was at least one teacher who composed at least five problems.
中文大意：有10个数学老师要给一个数学竞赛出35道题。已知有的老师出1道题，求证这些老师里面至少有1个老师至少出5道题。

一道澳大利亚的数学题,7月5日之前要哈Ten mathematics teachers composed 35 problems altogether for a mathematical contest.It is known that there were teachers who composed one,two and three problems.Prove that there was at least one teac
十个老师一共要出35个题目,现在知道有老师只出1个,2个和3个题目
那么至多还剩10-3=7个老师,最少还剩35-1-2-3=29个题目
反正法：
假设剩下的老师出的题目都小于5,那最多只能出7*4=28 个题目,28

你可以假设一种最少的情形，如果那样都有一个老师要出5道题，那么肯定至少有一个老师要出5道题
最少：1,2,3道题各有一位老师出，就还剩7个老师要出29道题平均起来每个老师至少要出4.1道题，由于 题目数为整数，所以至少有一个老师要出5道题

有的出1道,2道,3道,这是3个老师.
已经出了6道了吧.
35-6=29道.
假设其他7个老师全出4道,4乘7=28,还差了一道,所以肯定有一个老师要出5道题.

如果没有一个老师出5道题，那么10个老师最多出30道题。还剩余5道题。因为没有出4道题的，将5道题分开，所以至少要有一位老师出5道题。

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