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Reciprocal Period Poser (Posted on 2016-10-29) Difficulty: 3 of 5
N is a duodecimal positive integer ≤ BB.
Find the value of N such that 1/N has the maximum period of its digits after the duodecimal point.

No Solution Yet Submitted by K Sengupta    
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solution | Comment 1 of 2
let N=2^a*3^b*k where k is not divisible by 2 or 3

Then the duodecimal representation of 1/N is given by a series of R digits and then a repeating series of L digits where
R=max(a,b)
and
L is the smallest positive integer such that 12^L = 1 mod k

now this L is maximized for prime k and we get the largest k for when N itself is prime.  Thus we want the largest prime less than BB (143) which is 139

thus the maximum period is 138 when N=B7

  Posted by Daniel on 2016-10-29 09:51:37
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