设任意三个连续自然数为n-1,n,n+1,
因为n-1,n,n+1中必有一个是3的倍数,至少有一个是偶数,
所以（n-1）*n*（n+1）既是3的倍数,也是2的倍数,
则积是6的倍数,即能被6整除.
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