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# hard math problem!

Mon 3 Nov, 2008 06:12 am
Find, with proof, the largest positive integer k with the following property:
There exists a positive number N such that N is divisible by all but three of the
integers 1, 2, 3, . . . , k, and furthermore those three integers (that donâ€™t divide N)
are consecutive
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markr

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Tue 4 Nov, 2008 02:32 am
@mobidic,
For N to be divisible by n, but not by any integer less than n, N must be prime or a power. Therefore, the three consecutive numbers must be a combination of primes and powers (x^y).

Given three consecutive integers, one is divisible by 3. Therefore, one of the consecutive numbers must be a power of the form (3x)^y.

We need to consider consecutive integers of the forms:
power, power, power
power, power, prime
power, prime, power
prime, power, power (7, 8, 9 is an example)
prime, power, prime

That's as far as I can go. My guess is k=13, and the consecutive numbers are 7, 8, and 9.
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