GMAT数学题,GWD中的一道有关质因子数计算The positive integer k has exactly two positive prime factors,3 and 7.If k has a total of 6 positive factors,including 1 and k,what is the value of (1) is a factor of k.(2) is NOT a factor of k..
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![GMAT数学题,GWD中的一道有关质因子数计算The positive integer k has exactly two positive prime factors,3 and 7.If k has a total of 6 positive factors,including 1 and k,what is the value of (1) is a factor of k.(2) is NOT a factor of k..](/uploads/image/z/13177884-12-4.jpg?t=GMAT%E6%95%B0%E5%AD%A6%E9%A2%98%2CGWD%E4%B8%AD%E7%9A%84%E4%B8%80%E9%81%93%E6%9C%89%E5%85%B3%E8%B4%A8%E5%9B%A0%E5%AD%90%E6%95%B0%E8%AE%A1%E7%AE%97The+positive+integer+k+has+exactly+two+positive+prime+factors%2C3+and+7.If+k+has+a+total+of+6+positive+factors%2Cincluding+1+and+k%2Cwhat+is+the+value+of+%281%29+is+a+factor+of+k.%282%29+is+NOT+a+factor+of+k..)
GMAT数学题,GWD中的一道有关质因子数计算The positive integer k has exactly two positive prime factors,3 and 7.If k has a total of 6 positive factors,including 1 and k,what is the value of (1) is a factor of k.(2) is NOT a factor of k..
GMAT数学题,GWD中的一道有关质因子数计算
The positive integer k has exactly two positive prime factors,3 and 7.If k has a total of 6 positive factors,including 1 and k,what is the value of (1) is a factor of k.
(2) is NOT a factor of k..
背景知识
对于一个数n,假设他的质数因子是a,b,c,且n=(a^x)*(b^y)*(c^z)
则n的因子数为f=(x+1)(y+1)(z+1)
比如n=120,n=(2^3)*(3^1)*(5^1)
所以120的factors数为4*2*2=16个 为什么120有16个质因子啊?不是2*2*2*5*3就行了吗?
题目说质因子只有3和7
设k=(3^x)(7^y)
则(x+1)(y+1)=6 所以x、y=1或2
所以两个条件都可以得出
所以,只要告诉我为什么120有16个质因子就可以了
GMAT数学题,GWD中的一道有关质因子数计算The positive integer k has exactly two positive prime factors,3 and 7.If k has a total of 6 positive factors,including 1 and k,what is the value of (1) is a factor of k.(2) is NOT a factor of k..
不确定明白楼主的意思,先解释一下这些看看能不能帮上忙:
1.120有且只有3个质因子,而120共有16个因子:1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120.关键在于质因子和因子是不同的概念,楼主应该明白,相信是笔误.
2.关于公式(x+1)(y+1)(z+1).120有16个因子上面已经说明,没法再多作解释.一个数的因子可以从其质因子得到,就是选取其质因子进行组
合,所以是一个排列组合问题.问题可以叙述成:从x个a,y个b和z个c中挑选数字[0个数字(对应因子1)到x+y+z个数字(对应原数n)均可],共
有几种不同的选法?思路很直接,依次考虑a、b、c:
选a有x+1(选0个a到选所有x个a)种选法;
选b有x+1(选0个a到选所有x个b)种选法;
选c有x+1(选0个a到选所有x个c)种选法;
然后运用乘法原理,所以共有(x+1)(y+1)(z+1)个因子.
请楼主追问~