PREP 一道数学题求解

pangpangshin

1，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
1) between 2 and 10
2)between 10 and 20
3)between 20 and 30
4)between 30 and 40
5)greater than 40

Answer is 5)greater than 40

abixiaoxinyue

This one has been explained before.

Basically,
1) These consecutive natural numbers h(100) and [h(100) + 1]  are co-prime, meaning that they do not share any prime numbers as their cofactors.  
2) h(100) contains all the prime numbers between 2 (from 2) and 47 (from 94).
3) Then prime numbers among factors of [h(100) + 1] would not include any prime numbers between 2 and 47.
4) Then the smallest prime facotr of [h(100) + 1] is bigger than 47.

avalon9527

h(n)=2*4*6*...*100=2的50次方*(1*2*3*4*5*....*50)
由此可以看出，从2到50均能被h(n)整除，而
h(n)+1=2的50次方*(1*2*3*...*50)+1
也就是说从2到50均被h(n)+1除 余数为1  
所以其质因子不可能是2到50之内的。

pangpangshin

Hey thank you so much! I got it right now. Really appreciate your answer.

xiexiang

2) h(100) contains all the prime numbers between 2 (from 2) and 47 (from 94).

请问，这一步有些不理解，原文不是说 h（100）是 product of 2 to n 吗？

谢谢

abixiaoxinyue

2, 4, 6, 8, 10, 12, 14, 16, 18, . . . 92, 94, 96, 98, 100
is the same as
2, 2*2, 2*3, 2*4, 2*5, 2*6, 2*7, 2*8, 2*9, . . . 2*46, 2*47, 2*48, 2*49, 2*50
So the product of the second series contains factors of 2 and 2, 3, 4 ,5, 6, 7, 8, 9, . . . 46, 47, 48, 49, 50.
47 is the biggest factor of h(100), which is a prime number.

pangpangshin

Thanks for the answer~~