Repeated Decimal Expansion Corresponding to a Fraction Problem How do I solve the following question. 

The repeated decimal expansion $1.23\overline 6$ corresponds to
  which fraction?

a. $\frac{370}{300}$
b. $\frac{3710}{3001}$
c. $\frac{371}{301}$
d. $\frac{37100}{30001}$
e. $\frac{371}{300}$
Here is how I am trying to do but I get the wrong answer. 
$100α − α = 123.6 − 1.236 = 122.364$
Hence $99α = 122.364 \Rightarrow α = \frac{122.364}{99}$
Thank you.
 A: Working backwards, try some of those divisions.  
$371 / 300 = 1.23 \overline{6}$
A: It's multiple choice. In a test situation with a clock ticking down, sometimes you just have to pick the answer that looks the most sensible to you, hope for the best and move on.
But if this is homework, you can just put the five choices through a calculator at your leisure, plus other similar cases to help solidify your understanding.
Thus you would see that $$\frac{370}{300} \approx 1.23333333$$ The wrong digit repeats. Can the numerator be even? Maybe, but then the fraction wouldn't be in lowest terms... unless the denominator is odd, like in the next choice: $$\frac{3710}{3001} \approx 1.23625458$$ That doesn't seem to have any consecutive repeated digits.
Something tells me that in base 10, the only way to get an infinitely repeating 3 or 6 is for the denominator to be a nonzero multiple of 3, and for the numerator to not be a multiple of 3.
Clearly 3001 is not a multiple of 3, and in fact it's prime. 301 is not prime, but it's not a multiple of 3 either. 30001 is not prime either, but it is also quite obviously not a multiple of 3.
This leaves $$\frac{371}{300} \approx 1.23666667$$ Clearly we had a loss of machine precision there, but this has got to be the right answer.
A: How to solve the question:
multiply the number by two different powers of $10$ 
such that you can subtract them and obtain an integer.
$10^3a-10^2a=1236.\bar6 - 123.\bar6=1113$.
Therefore $900a=1113.$
Can you take it from here?
Note that $1113$ is divisible by $3$  (e.g., by the sum of digits test for divisibility by $3$) but not $9$ nor $100$.
A: $1.23\overline 6 = 1.23 + .00\overline 6= 1 + \frac{23}{100}+\frac{\frac{2}{3}}{100}= \frac{300}{300} + \frac{69}{300}+\frac{2}{300}= \frac{371}{300}$
