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 Floor Prime? (Posted on 2014-06-28)
Define F = Floor (10n/p), where p is a prime number greater than 5 and n is the cycle length of the repeating decimal 1/p.

Can F be a prime number?
If so, provide an example. If not, prove it.

 No Solution Yet Submitted by K Sengupta No Rating

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 Solution (spoiler) Comment 1 of 1
F(n) is non-prime for all p > 5.  This is because F is the repeating portion of the decimal for 1/p.

For instance, F(7) = 142857
Necessarily F(7) * 7 = 999999.
Because 7 is not the only prime factor of 999999 (as 3 is also a factor), 142857 is not prime.

In general, for any prime > 3,
F(p)*p = 999..999  (with n nines).

p is not divisible by 3, so F(p) is necessarily divisible by 9.

q.e.d.

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by the way, why do 2, 3, and 5 not have this property?
F(3) = 3, which is prime, because 3 is divisible by 3.
F(2) arguably = 5 and F(5) arguably = 2.  These do not follow the rule because 2 and 5 both divide 10, so F(2)*2 = F(5)*5 =10, not 9.

Edited on June 28, 2014, 9:35 pm
 Posted by Steve Herman on 2014-06-28 21:22:58

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