Find the smallest base ten positive integer with period 12.
Note: The period of an integer is the length of the repeating pattern of reciprocal. For example, repeating pattern of the reciprocal of 7 is 1/7 = 0.142857142857..... having a length of 6. So the period of 7 is 6.
(In reply to
This may be right by Jer)
When I factored
999999999999 = 3³*7*11*13*37*101*9901
I failed to consider that a product of two of these: 7 and 101 could give the desired result. I didn't check that there are not both factors of some shorted string of 9s.
When I looked up 9901 in OEIS I got
http://oeis.org/A007138
so I assumed I was right. I didn't read it carefully enough because I didn't notice this sequence is of primes. Of course 707 is not.
It is mysterious (to me anyway) that the product of 101 [period 4] and 7 [period 6] should have period 12.

Posted by Jer
on 20121201 00:35:54 