SOLUTION: I didn't ask the type the question right before Find the smallest odd positive integer (m) so that the decimal expansion of 1/m has just 4 non recurring digits and has 2 repeating

Question 1046143: I didn't ask the type the question right before
Find the smallest odd positive integer (m) so that the decimal expansion of 1/m has just 4 non recurring digits and has 2 repeating digits.

Answer by KMST(5345) (Show Source): You can put this solution on YOUR website!
, ,
and for any integer that is not a multiple of 11, or 99,
the decimal part of will be a 2-digit repeating sequence.
So, we can represent all numbers whose decimal expansion is a 2-digit repeating sequence as
, where = an integer that is not a multiple of 11.
Then, , and
has pairs of repeating digits after the 4th decimal place.
--->--> .
For that to be an integer,
cannot have any factors not present in ,
and for to be an odd integer,
must have as a factor.
Choosing , we get with .
That is the smallest odd positive integer so that the decimal expansion of has pairs of repeating digits after the 4th decimal place.
However, may not be considered the right answer,
because the phrase "just 4 non recurring digits" probably means that the third decimal places must be different from the 5th one,
and the and fourth one must be different from the sixth one.
So, we need to put less factors into ,
leaving with more of the factors in to get
with (not right either),
with (not right either),
with , and so on.
We could keep trying to drop factors from ,
until we get to ,
when
has a 4th decimal digit that is different from the sixth one .
So, is my answer.

NOTE: there is an option that does not require trying so many factor combinations to get to ,
but it does require more algebra, so it may not get to the answer faster.
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