急急！快考试了~求教PREP数学题

carbonzju

楼上的 根号10>根号9=3,  所以x最大可以为3-6

carbonzju

1×2×3×4×5。。。×50 × 2^50 +1  

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.

DarthWalker

在曼哈顿看来的第一题解释
h(100)=2^50 * 50! = 2^50 * (1*2*3*...*50)

Here h(100) is divisible by any numer between 2 and 50. Right?
So h(100)+1 can't be divisile by any number between 2 and 50. h(100)+1 will always give a reminder of 1 when u divide the no by any no between 1 to 50.

So h(100)+1 doesnot have any factor between 2 and 50. Hence h(100)+1 doesnot have any prime factor between 2 and 50.
by:John

the idea is this: if a number is divisible by some prime p, then the next multiple of p will be p units bigger. for instance, 75 is divisible by 5. this means that the next greatest multiple of 5 is 80, which is 5 units away.
hopefully, this fact is clear. once you realize this, it follows that consecutive integers can't share ANY primes, because they're only 1 unit apart (too close together to work for any common factor except 1, which is trivially a factor of any integer at all, anywhere).
that's the basis for saying h(100) + 1 has no prime factors below 50. if it did, then you'd have two multiples of the same prime 1 unit apart, and that's impossible.

by:Ron


第二题如一楼所说x^4=100, 3<x<6,下午prep的时候刚做到，楼主什么时候考？

carbonzju

发贴前搜索下论坛不失为好办法。。。

DarthWalker

4 21...紧张啊...你呢？

-- by 会员 léa (2011/4/5 18:18:13)

这有什么好紧张的，看我头像就知道我几号考了，我都不敢紧张呢~