求助 --- 两个PREP的数学题

pancy5858

请大家帮我看看这两道题,
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__.
答案: greater than 40

2. If each term in the sum a1+a2+......+an is either 7 or 77 and sum equals 350. which of the following could be equal to n?
答案: 40

多谢帮助提点!

大荣

1. h(100)+1 = 2^n*50!+1. Since all primes less than and equal to 50 are factors of 50!, any prime less than and equal to 50 isn't a factor of 2^n*50!+1. Thus, primes of this value must be greater than 50.

2. Suppose, the number of 7 is X and the number of 77 is Y.  Obviously, 7 *X + 77 *Y = 350. 7(X+Y) + 70 Y = 350.  There are several solutions of Y.

 Y =1,  X+Y = 40; Y=2, X+Y=30, Y=3, X+Y=20, Y=4, X+Y=10.

 Here n = X+Y, n = 10,20,30 or 40.

pancy5858

1. h(100)+1 = 2^n*50!+1. Since all primes less than and equal to 50 are factors of 50!, any prime less than and equal to 50 isn't a factor of 2^n*50!+1. Thus, primes of this value must be greater than 50.

-- by 会员 大荣 (2009/12/28 1:45:25)

回大荣, 第一题它的答案是选greater than 40, 不是50呢, 您再看看!?

gaojinye

因为〉50 所以〉40

下面是我引用的 解释得很好 作为补充吧：

Guest, 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.

running2k

好解。

对于第二题，其实很简单，因为和是350，如果是n是非10的倍数，尾数肯定不是0，所以就选择10的倍数选项。

因为〉50 所以〉40

下面是我引用的 解释得很好 作为补充吧：

Guest, 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.

-- by 会员 gaojinye (2009/12/28 2:15:10)

tjr

这一题应用这个原则：如果 k 是 n 的因数，同时 k 大于 1，那么 k 就不可能是 n + 1 的因数了。 这一题的 n 是偶数，那就把 n 写成 2m。那么 h(n) 就是
{2.1}{2.2}{2.3} . . . {2(m-1)}{2m} = (2^m)(m!)。
所以凡是小于或是等于 m 的质数就都是 h(n) 的因素，那就不可能是 h(n) + 1 的因数了。
h(100) = h(2?50)，所以 2 至 50 各整数都不能是 h(100) + 1 的因数，那么最少的质因数必须大于 50。题目那个大于 40 其实太不准确了