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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
divisions:

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,

0.142857142857142857...

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,

0.142857142857142857...

contains the repeating decimal for 6/7, which is

0.857142857142857142...

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

0.999999999999999999...

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:

.020408...
----------------------------------
49 ) 1.000000000000000
98
------
200
196
--------
400
392
--------
8

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

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
128
256
512
1024
2048
4096
8192
16384
32768
.......
-----------------------------------------------------------------
020408163265306122448979591......

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
http://mathforum.org/dr.math/
```
Associated Topics:
High School Number Theory

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