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.
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发表于 2011-4-5 17:22:09
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