请教数学prep这题，答案是E，可是不会做

predy1

7.    905-!-item-!-187;#058&000575

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

 

(A) between 2 and 10

(B) between 10 and 20

(C) between 20 and 30

(D) between 30 and 40

(E) greater than 40

xianniu

 this is definitely a difficult number properties question. Let's first consider the prime factors of h(100). According to the given function,
h(100) = 2*4*6*8*...*100

By factoring a 2 from each term of our function, h(100) can be rewritten as
2^50*(1*2*3*...*50).

Thus, all integers up to 50 - including all prime numbers up to 50 - are factors of h(100).

Therefore, h(100) + 1
    cannot have any prime factors 50 or below, since dividing this value by any of these prime numbers will yield a remainder of 1.

Since the smallest prime number that can be a factor of h(100) + 1 has to be greater than 50, The correct answer is E.

samanthappa

h(100)=2*4*6*8*...100

=>(2*1)*(2*2)*(2*3)*(2*4)*.......*(2*50)

=>(2^50)*(1*2*3*4*...50)

=>h(100)+1=(2^50)*(1*2*3*4*...*50)+1

so when h(100)+1 is divided by any interger from 1 to 50, there will be a remaider 1

yoyo0327

谢谢啊~~~正想知道这道

predy1

太感谢 大N们了 厉害啊~

mao_gouhappy

2楼3楼都是NN啊!