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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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First Thoughts (spoiler?) | Comment 1 of 3
Hmm. 

1) I think that N is a little bit of a red herring.  What we are looking for are the largest three consecutive integers a<b<c such that they do not divide the Lowest Common Multiple of all of the numbers below them.  And then the maximal k is 2a - 1, because 2a is is a multiple of one of them, but nothing between (c+1) and (2a-1) is.  In other words, no N required to find k.  If (7,8,9) are the largest three, then k = 13.

2) a and b and c must each be primes raised to some power.  Otherwise, they will divide the LCM of all numbers less than a.

3) In fact, since one of any three consecutive numbers is divisible by 3, one of a, b or c must be a power of 3.  And since, at least one of them is divisible by 2, one of them must be a power of 2.  And it must be b, because at most one of them can be a power of 2.  So it is a or c that is the power of 3.

4) So, summarizing, our three consecutive numbers must be 
    3^x, 2^y, and p^z (or p^x, 2^y, 3^z) where p is prime and x, y and z are integers.

5) Unless there is a power of 2 great than 8 which equals 3^x +/- 1, then k = 13.  I'll explore later, unless somebody else proves it.  Gotta go back to work now. 



  Posted by Steve Herman on 2011-12-27 12:40:18
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