【PREP一题 数学求助】 p is the smallest prime factor of h(100) + 1，then p is

十口叶子

For every positive even integer n, the function h(n) is defined to be the product of all the even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) + 1, then p is
Between 2 and 10
Between 10 and 20
Between 20 and 30
Between 30and 40
Greater than 40

向各位大虾、NN们求助，谢谢~~~

子图

因为1不能被所有后面的数整除所以h（100）+1除以2到50分别会余1~

AmandaChen94

wxd 发表于 2010-10-28 21:54
h(100) = 2 x 4 x . . . x 100 = (2^50)x 50!.  Therefore, the integers 2, 3, 4, . . . , 50 are al ...

同意！               

ccp92

wxd 发表于 2010-10-28 21:54
h(100) = 2 x 4 x . . . x 100 = (2^50)x 50!.  Therefore, the integers 2, 3, 4, . . . , 50 are al ...

Mark一下！               

差点儿要事业

多谢解答！就是因为50!包含了所有小于50的整数，而+1之后显然这些数都不能成为h(100)+1的因子，所以前面四个选项都排除，只有最后一个最接近。

MichaelZhang322

子图 发表于 2013-1-24 10:10
因为1不能被所有后面的数整除所以h（100）+1除以2到50分别会余1~

感觉你解释的太好了！！

bapear

wxd 发表于 2010-10-28 21:54
h(100) = 2 x 4 x . . . x 100 = (2^50)x 50!.  Therefore, the integers 2, 3, 4, . . . , 50 are al ...

这题我也错了。你解释得好清楚，感谢～～

十口叶子

那个式子懂，但后面为什么2,3,4,……，50是h(100)的因数，就不能是h(100)+1的呀？后面解释还不是很懂。。。。

wxd

h(100) = 2 x 4 x . . . x 100 = (2^50)x 50!.  Therefore, the integers 2, 3, 4, . . . , 50 are all factors of h(100), which means that none of these integers can be a factor of h(100) + 1.  Thus the smallest prime factor of h(100) + 1 must be greater than 50.

wxd

同问  高手来解答啊