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