Appreciate any help on this question from OG 11th edition

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Hi everyone, I'm sure there's a trick to most questions if it seems too long to do it the obvious way, for example,

question 22 in the problem solving secion of og 11th edition goes as follows:

If each of the following fractions were written as a repeating decimal, which would have the longest sequence of different digits?
a) 2/11 b) 1/3 c) 41/99 d) 2/3 e) 23/37

Is there a quicker way to finding the answer to this than actually dividing each number to check?

Any help or tips for questions like this would be much appreciated

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by student22 » Wed Mar 31, 2010 6:45 pm
Yeah, convert the fractions so that the denominator is a series of 9s.

2/11 ---> 18/99
1/3 ---> 3/9
41/99 -> 41/99
2/3 ---> 6/9
23/37 --> 621/999

The numerator is the sequence of different digits. Answer is E.

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by ajith » Wed Mar 31, 2010 8:32 pm
student22 wrote:Yeah, convert the fractions so that the denominator is a series of 9s.

2/11 ---> 18/99
1/3 ---> 3/9
41/99 -> 41/99
2/3 ---> 6/9
23/37 --> 621/999

The numerator is the sequence of different digits. Answer is E.
2/11 ---> 18/99 = 0.1818....
1/3 ---> 3/9 = 0.333.....
41/99 -> 41/99 = 0.4141....
2/3 ---> 6/9 = 0.666....
23/37 --> 621/999 = 0.621621...
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by cyrusthegreat » Thu Apr 01, 2010 3:58 am
Thanks for the help folks

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by camilaross » Thu Apr 01, 2010 8:29 am
Thanks for the trick, that was very helpful.

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by arora007 » Mon Jan 03, 2011 4:21 am
student22 wrote:Yeah, convert the fractions so that the denominator is a series of 9s.

2/11 ---> 18/99
1/3 ---> 3/9
41/99 -> 41/99
2/3 ---> 6/9
23/37 --> 621/999

The numerator is the sequence of different digits. Answer is E.
nice trick... I took E as the answer by POE, but was never sure until i saw this... 999 =( 37 * 27) would be worth remembering!!
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