If each of the following fractions were written as a repeati

I have taken 15 seconds to resolve this problem! For this problem specifically, there are some very fast tricks that you can do remembering the properties of recurring decimals. You just need a little patience to understand a first time. Then you probably will never forget.

Here is a brief description useful in a some GMAT exercises:

First - We use the number that is repeating and the denominator has as many "9" digits as there are different digits in the block that repeats. e.g.
\(0.555555 = 5/9\)
\(0.13131313 = 13/99\)
\(0.432432432432 = 432/999\)

Second - If the sequence starts to repeat after some zeros, add the same number of zeros in the denominator. e.g.
\(0.005555 = 5/900\)
\(0.013131313 = 13/990\)
\(0.0004324324324... = 432/999000 = 54/124875 = 2/4625\)

Third - Terminating decimals + Repeating Decimals. e.g.
\(2.31555555 = 2.31 + 0.00555555 = 231/100 + 5/900\)
\(0.745454545 = 0.7 + 0.0454545 = 7/10 + 45/990\)

Last one - The reciprocal of a prime number "p", except 2 and 5, has a repeating sequence of p-1 digits, or a factor of p-1 digits. e.g.
\(1/7 = 0.142857 142857 142857...\) - As you can notice, the sequence has 6 digits = p-1 = 7-1 = 6 digits.
And if you multiply this fraction by a number you will only change the beginning of the sequence. e.g.
\(4/7 = 4*1/7 = 0.57 142857 142857 ...\)

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Once you have all these little rules in mind it is very fast to work in this problem.
A) 2/11 -> 11 is a factor of 99, so there are 2 repeating digits.
B) 1/3 -> 3 is a factor of 9, so there is 1 repeating digit.
C) 41/99 -> 99, so there are 2 repeating digits.
D) 2/3 -> 3 is a factor of 9, so there is 1 repeating digit..
E) 23/37 -> As I still don't have one item that is winning in repeating digits, this must be CORRECT

\(E\)

Good studies!

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