<PRE><B>Question 1901</B>
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Here is a variation of the process shown by tutor @alan3354.&lt;br&gt;
(1) n = 0.233333...
(2) 10n = 2.33333...&lt;br&gt;
Subtract (1) from (2):&lt;br&gt;
9n = 2.1
n = 2.1/9 = 21/90 = 7/30&lt;br&gt;
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Here is a different method, without using algebra.&lt;br&gt;
0.2333333... = 2.333333.../10 = (2 1/3)/10 = (7/3)/10 = 7/30.&lt;br&gt;
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And here is a shortcut that is based on the basic algebraic method.&lt;br&gt;
The numerator of the fraction is (a) all the digits up through the first cycle of the repeating digits, minus (b) just the non-repeating digits.  In this example, that is 23-2 = 21.&lt;br&gt;
The denominator of the fraction is one 9 for each repeating digit, followed by one 0 for each non-repeating digit.  In this example, with one non-repeating digit and one repeating digit, that is 90.&lt;br&gt;
The fraction is then 21/90 = 7/30.&lt;br&gt;
That shortcut is a great time saver if speed is important, as in a math competition.  Here are a couple of further examples of this method.&lt;br&gt;
2 non-repeating digits; one repeating:&lt;br&gt;
0.4166666... = (416-41)/900 = 375/900  ( = 5/12)&lt;br&gt;
3 repeating digits, 2 non-repeating:&lt;br&gt;
0.12345345345... = (12345-12)/99900 = 12333/99900  (which can also be simplified, if required)</PRE>