作者ipas3 ()
看板GMAT
标题Re: [计量] PP1-DS-Q3
时间Wed Aug 26 14:22:49 2009
※ 引述《denizee3 (妮)》之铭言：
: How many different prime numbers are factors of the positive integer n ?
: (1) Four different prime numbers are factors of 2n.
: (2) Four different prime numbers are factors of n^2.
: Answer: B
: 我看了PP笔记里面的解释
: 但是我还是不太懂为什麽答案是B
: 是否可以请各位大大解释这一题的思路
: 谢谢!
(1) 若2(prime number)是n的prime factor,
则当2n有四个prime factors (2,x,y,z)时
n也有四个prime factors (2,x,y,z)
若2不是n的prime factor,
则当2n有四个prime factors (2,x,y,z)时
n只有三个prime factors (x,y,z) => insufficient
关键是2本身就可能成为一个prime factor
(2) 当n^2有四个prime factors时 = x^a * y^b * z^c * q^d
则n= x^(a/2) * y^(b/2) * z^(c/2) * q^(d/2)
其中 a/2 b/2 c/2 d/2 必为正整数 (否则n就不是positive integer)
n也有四个ptime factors => sufficient
结论是不管几次方prime number数目都一样
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※ 编辑: ipas3 来自: 76.172.54.4 (08/26 14:27)