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Dividing 29/49

Date: 08/30/97 at 12:31:03
From: Fredericks
Subject: Division help

I need to divide 29/49 out until it repeats itself or terminates.  
I also need to do 30/49, 31/49, 32/49, 33/49, 34/49, 35/49.  
Is there a way to figure these problems out without doing them by 
long division?

Thank you!

Date: 09/04/97 at 15:30:28
From: Doctor Rob
Subject: Re: Division help

I don't think you can avoid division, but you can substitute a
multiplication and several additions for all but one of the 

Divide 1/49 out the long way. Then multiply it by 29. This gives 
you 29/49.  Then add 1/49 to get 30/49, add 1/49 again to get 31/49, 
and so on.

If all you want to know is how many places you have to go before 
the decimal equivalent of the fraction repeats, there are ways to 
discover that.  First you have to reduce the fraction to lowest 
terms. All the ones above are in lowest terms except 35/49.  
Now you have to factor the new denominator into prime number 
powers.  49 = 7^2.  In general, this will not be a single prime 
power, but a product of several.

First we will solve the problem for a prime power p^n.  
If the p = 2 or 5, the divisors of 10, the decimal will terminate 
after n places.  Then we say that the period k is 1.

If p is neither 2 nor 5, then you have to compute the smallest 
exponent k such that p^n divides 10^k - 1 evenly, with remainder 
zero.  It turns out to be p^(n-1) times the smallest exponent j 
such that p divides 10^j - 1 evenly, with remainder zero.  
Divide p into 999... until you get remainder zero. The number 
of 9's will be j. Then k = p^(n-1)*j, and this will be the number 
of places before you get a repeat. A sometimes useful fact is that 
j is a divisor of p-1. In your example case, j is a divisor of 6
(that is, 1, 2, 3, or 6), and k = 7*j.

If there are more than one prime power in the factorization, 
find the values of k for each prime power, and then take their 
Least Common Multiple (LCM). That will be the correct answer.

Proofs of all these facts depend on a branch of mathematics called 
Number Theory, which you may study at the college or university level.

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

Date: 07/31/01 at 13:50:29
From: Doctor Greenie
Subject: Re: Division help

Let's find a way to determine, with as fast and simple a method as 
possible, the decimal equivalents of the fractions n/49, for all 
values of n from 1 to 48.

For values of n that are multiples of 7, the fractions n/49 reduce to 
1/7, 2/7, ..., and 6/7.  These fractions have decimal equivalents 
(perhaps familiar to the reader) containing the following repeating 
groups of digits:

    1/7 =  .142857...
    2/7 =  .285714...
    3/7 =  .428571...
    4/7 =  .571428...
    5/7 =  .714285...
    6/7 =  .857142...

These decimal equivalents can be found using a single long division 
process - the one for finding the decimal equivalent of 1/7.  In this 
long division process, the repeating string of digits is found to be 
6 digits long, indicating that each of the 6 possible non-zero remainders 
was obtained at some step in the long division process. This means that 
the repeating decimal for 1/7,


must "contain" the repeating decimal forms of the fractions 
2/7, 3/7, ..., and 6/7.  The repeating fraction for 2/7 is the one that 
is, by inspection, twice as much as .142857.... - i.e., it is .285714.....

Also, note that the repeating decimal for 1/7,

contains the repeating decimal for 6/7, which is


Because 1/7 + 6/7 = 1, it must be true (and we see that it is), that the 
sum of the repeating decimal representations of 1/7 and 6/7 is the 
repeating decimal


which is equivalent to 1.  Another way of thinking of this is to view 
the decimal representation of 6/7 as the "9's complement" of the decimal 
representation of 1/7.

Let's use these ideas to shorten the amount of work required to find the 
decimal representations of all the fractions n/49, where n is between 
1 and 48 and is not a multiple of 7.

We start by performing a few steps of the long division process to find 
the decimal equivalent of 1/49:

        49 ) 1.000000000000000

If we are obervant here, we should see a pattern emerging. It appears that

    1/49 = (2/100) + (4/10000) + (8/1000000) + ...

which is an infinite geometric series with first term (2/100) and common 
ratio (2/100). The formula for the sum of an infinite geometric series 
tells us that this sum is equal to

      first term            (1/50)           (1/50)        1
   ------------------  =  ------------  =  ---------  =  ----
    1 - common ratio       1 - (1/50)       (49/50)       49

So this geometric series does indeed have the sum 1/49; this means we now 
have a way to find the repeating decimal representation of 1/49 by a method 
that is probably faster than long division - we simply "write" a string 
consisting of powers of 2, allowing two additional decimal places for each 
additional power of 2.

At the beginning, this is easy, because the powers of 2 are only 1 or 2 
digits each and so do not overlap...

    1/49 = .020408163264......

But then the digits begin overlapping, so we have to do some addition....

    1/49 = .020408163264

At some point we can stop this process before the decimal value starts 
repeating, because we know the decimal equivalent of 48/49 is going to 
show up somewhere in this string. With the decimal equivalent of 1/49 
being .02040816...., we know that the decimal equivalent of 48/49 will be 
the "9's complement" of 1/49, which is .97959183....  We see that, at this 
point in our calculation, we have reached the point where we have these 
digits; so we can complete the task of writing out the decimal representation 
of 1/49 by using the first several digits and then the 9's complement of 
those digits:

   1/49 = .020408163265306122448979591836734693877551020408.....

The digits have begun repeating now, so we now have the exact repeating 
decimal representation for 1/49:

   1/49 = .020408163265306122448979591836734693877551.....

Now, to find the decimal representations of 29/49, 30/49, and so on, we 
only need to find the strings of repeating digits within 1/49 that start 
with the correct few digits:

    29/49 = 29 * (.0204...) = (approximately) .59...., so

    29/49 = .591836734693877551020408163265306122448979.....

    30/49 = 30 * (.0204...) = (approximately) .61...., so

    30/49 = .612244897959183673469387755102040816326530.....

    31/49 = 31 * (.0204...) = (approximately) .63...., so

    31/49 = .632653061224489795918367346938775510204081.....

    32/49 = 32 * (.0204...) = (approximately) .65...., so

    32/49 = .653061224489795918367346938775510204081632.....

    33/49 = 33 * (.0204...) = (approximately) .67...., so

    33/49 = .673469387755102040816326530612244897959183.....

    34/49 = 34 * (.0204...) = (approximately) .69...., so

    34/49 = .693877551020408163265306122448979591836734.....

-Doctor Greenie,  The Math Forum
Associated Topics:
High School Number Theory

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