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Consecutive non-divisors (Posted on 2011-12-27) Difficulty: 4 of 5
A positive integer N is such that:

(i) N is divisible by all but three of the k positive integers 1, 2, 3, . . . , k; and:

(ii) Those three positive integers (that donít divide N) are consecutive.

(For example, if N=60 then the corresponding value of k is 10, since 7, 8, 9 doesn't divide 60.)

Determine (with proof) the maximum value of the positive integer k satisfying all the given conditions.

No Solution Yet Submitted by K Sengupta    
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Solution I looked it up (spoiler) | Comment 2 of 3 |
Well, since nobody else has posted yet, I looked up the last part of my proof.  

Here is a proof that consecutive integers greater than 8 cannot be a power of 2 and a power of 3.

And that implies that 7, 8, 9 are the largest set of three consecutive integers that do not divide the LCM of all positive integers less than them, since the middle one must be a power of 2 and one of the others must be a power of 3.  And thus, k = 13 (see my previous post).

In summary, the proof proceeds as follows:

Case a)  
  • If 2^m = 3^n + 1 then 2^m = 1 (mod 3) so m must be even
  • Let m = 2k
  • Then 3^n = 2^(2k) - 1 = (2^k - 1)(2^k + 1).  
  • These factors differ by 2, and must be powers of 3, so it can only be true if k = 1

Case b)
  • If  3^m = 2^n + 1, and n >= 3, then 3^m = 1 (mod 8)
  • Let m = 2k
  • Then 2^n = 3^(2k) - 1 = (3^k - 1)(3^k + 1).  
  • These factors differ by 2, and must be powers of 2, so it can only be true if k = 1

In fact, in a much more difficult proof, the following url says that "... the gap between powers of three and powers of two grows exponentially in the exponents p,q     " as a consequence of Hilbert's seventh problem.



Edited on December 27, 2011, 8:54 pm
  Posted by Steve Herman on 2011-12-27 20:38:39

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