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.

.111111... has period 1 and is equal to 1/9
9 = 3² so 3 is the smallest integer with period 1.
.010101... has period 2 and is equal to 1/99
99 = 3²*11 and since 3 already has period 1, 11 is the smallest with period 2.

Jumping to .000000000001000000000001...
this is equal to 1/999999999999 (twelve 9's)
999999999999 = 3³*7*11*13*37*101*9901
All but the last of these is a factor of a shorter string of 9s
(and if you check each has a shorter period which is a factor of 12)
but checking the last: 1/9901 = 0.000100999899000100999899000100999899... with period 12.

So I believe the answer is 9901 but I feel I may be overlooking something.

Posted by Jer on 2012-11-30 11:11:06