一道gmat数学概率题1≤n≤99,what is the possibility of n(n+1) is a multiple of

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一道gmat数学概率题1≤n≤99,whatisthepossibilityofn(n+1)isamultipleof一道gmat数学概率题1≤n≤99,whatisthepossibilityofn

一道gmat数学概率题1≤n≤99,what is the possibility of n(n+1) is a multiple of
一道gmat数学概率题
1≤n≤99,what is the possibility of n(n+1) is a multiple of

一道gmat数学概率题1≤n≤99,what is the possibility of n(n+1) is a multiple of
是2/3.把从1到99分成33组,则每组都可写成如下形式:
(3k+1,3k+2,3k+3),0≤k≤32
例如(1,2,3,其中k=0)或者(4,5,6,k=1)
每个组中,当n=3k+2或者3k+3时,n(n+1)都可以被3整除,只有n=3k+1时不行.所以能被整除的数共有 33(33个组) x 2(每个组两个数能被整除) = 66个.
共有99个数,所以可能性是66/99 = 2/3