Which Fractions Repeat?

Date: 07/21/98 at 00:51:14
From: Melissa
Subject: Fractions

Is there any way you can determine whether a decimal is going to be 
terminating or repeating without doing the conversion? Also, is there 
any way you can tell how many decimal places it will have if it is 
indeed repeating? Finally, is there a way to determine what the length 
of the decimal places will be if a fraction in lowest terms is 
repeating? I don't have a clue as to how one can determine these things 
without doing the conversion. 

Thanks for your help.
Melissa

Date: 07/21/98 at 15:19:40
From: Doctor Rob
Subject: Re: Fractions

Thanks for asking, Melissa.

The way to answer the first question is to factor the denominator of 
the fraction (after it is reduced to lowest terms) into a product of 
powers of prime numbers. If the prime factors are all 2 or 5, then the 
decimal will terminate.

The answer to the second question is that if any other prime number 
appears as a factor, then the decimal will be nonterminating but 
repeating.

Example: If the denominator is 51200, then the decimal will terminate,
because 51200 = 2^11 * 5^2, and the only prime numbers dividing are 2 
and 5. If the denominator is 288, the decimal will repeat, since
288 = 2^5 * 3^2, and the prime number 3 appears in the factorization.

The answer to the third question is more difficult. It involves some
pretty complicated mathematics known as Number Theory. Probably the
simplest way to figure this out is the following. Write the denominator 
in the form N = 2^a * 5^b * M, where neither 2 nor 5 divides into M 
evenly, and a and b are some whole number exponents. Then find the 
smallest number all of whose digits are 9's which is a multiple of M. 
The number of 9's required is the length of the repeating part of the 
decimal expansion of your fraction.  

Example: N = 26 = 2*13, so a = 1, b = 0, and M = 13. Then divide:

      ----------
   13 ) 999....

13 doesn't go into 9 even once. It goes into 99 seven times, and 
7*13 = 91, so:

         7
      ----------
   13 ) 999....
        91
        ----
         89

Continue this, bringing down one nine at a time from the dividend, and
following the rules for long division, until a remainder of 0 is 
obtained:

         76923
      ---------
   13 ) 999999
        91
        ----
         89
         78
         ----
         119
         117
         -----
           29
           26
           ----
            39
            39
            ---
             0

This tells us that the repeating part of the decimal expansion of some
number of 13ths contains 6 digits. This implies that the repeating part 
of the decimal expansion of some number of 26ths also contains six 
digits.

You try one with denominator 85.

If a technically correct answer to the third question is needed, here 
it is. In terms of number theory, you need to find the order of 10 
modulo M, that is, the smallest power of 10 which leaves a remainder 
of 1 when divided by M. It is known to be a divisor of lambda(M), but 
which divisor can only be determined by testing. Lambda(M) can be found 
as the least common multiple of the numbers (p-1)*p^(e-1), where p is 
any prime dividing M, and p^e is the highest power of p dividing M. 
As an example, if M = 3^4*7 = 567, then:

   lamdba(567) = LCM([3-1]*3^3, [7-1]*7^0) = LCM(2*3^3, 2*3) = 54 

The actual repeating part has length 18, a divisor of 54, since 
3^4 is a divisor of 10^9 - 1, and 7 is a divisor of 10^6 - 1, and 
LCM(9, 6) = 18, so both 3^4 and 7 divide 10^18 - 1.

- Doctor Rob, The Math Forum
Check out our web site! http://mathforum.org/dr.math/