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Repeating Decimals


Date: 04/28/99 at 07:39:22
From: Grabowski
Subject: Long numbers

Hi,

I have been interested in working with fractions, and I would like to 
find patterns. When I divide one number by another, I sometimes get a 
simple decimal tail, like with 1/8 = 0.125, where the number finishes 
or turns into an endless line of zeros. Other times it repeats (as 
with 1/3). Then there are some more interesting repeaters, where a 
group of digitrs keeps repeating (eg. 1/7 =  0.142857...) I am 
interested in finding longer repeating groups in these number tails, 
but my calculator doesn't let me see more than a few digits. What can 
I do to have divisions like 1/13, and be able to read more than 
0.076923076, where I can't tell if the 076923 is really a repeating 
group? 

I want to try to see what fractions have longer tail repeats, (I know 
that pi is infinite), but I have to have a better type of calculator 
to work these out because ordinary long division has no joy in it when 
going to such a length. Does such a calculator exist? Also, I want to 
try to divide terms of the Fibonacci Series and other interesting 
sequences (prime numbers?) to see what the repeating situation is with 
them.

P. Grabowski


Date: 04/28/99 at 13:28:26
From: Doctor Rob
Subject: Re: Long numbers

Thanks for writing to Ask Dr. Math!

You have started on an odyssey which can lead you (eventually) to a 
fact known as Fermat's Little Theorem. It is a worthy journey, and you 
can learn much along the way.

The length of the repeating part is the smallest number of 9's such 
that the denominator divides 9999...999000...0000 evenly. For example, 
7 divides 999999 evenly, so the repeating part has length 6. 
Furthermore, the digits in the repeating part of 1/7 are the digits of 
the quotient 999999/7 = 142857.

 1/7 = 142857/999999 = 142857/10^6 + 142857/10^12 + 142857/10^18 + ...
     = 0.142857 + 0.000000142857 + 0.000000000000142857 + ...

You will recognize this as a geometric series.

Since 13 is a divisor of 999999, with quotient 76923, you can safely 
predict that 1/13 = 0.076923 076923 076923 076923 076923 076923 ... 
without needing a high-precision calculator!

Next you probably want to know in advance what number 
999...999000...000 is a multiple of some denominator N. This is done 
by factoring N as a product of prime powers, which can be done only 
one way. They fall into two classes: ones dividing 10 and ones not 
dividing 10 (the base of our system of numerals). The first consists 
of just {2,5}, and the second is all the other primes. The number of 
zeroes in the number in the first line of this paragraph is just the 
larger of the two exponents of 2 and 5. That's the easy part! And if 
there are no other primes, then that larger exponent E is such that N 
is a divisor of 10^E, and the decimal terminates after E decimal 
places. 1/N = Q/10^E, where Q = 10^E/N.

The next step is to find for the rest of the primes p dividing into N 
evenly what smallest power of 10 has the property that 10^k - 1 is a 
multiple of p. That number k is called the order of 10 modulo p. The 
most helpful fact about k is that it is a divisor of p - 1. Recall 
that for p = 7, we had k = 6, and, sure enough, 6 is a divisor of 7 - 
1 = 6. This limits you to just a few possibilities.  There is no easy 
way to tell which of these divisors of p - 1 is actually k, you just 
have to try them.

The next step is to find the highest power of p dividing 10^k - 1. We 
know that it is at least 1, but it may be more.  For example, if p = 
3, the divisors of 3-1 = 2 are 1 and 2.  It turns out that k = 1, 
because 3 is a divisor of 10^1 - 1 = 9, but, in fact, 3^2 is also a 
divisor of 9, too. Call that highest power p^e. Now if the power of p 
dividing N is p^n, and n <= e, then the period length of 1/p^n is k. 
If, on the other hand, n > e, then the period length of 1/p^n is 
k*p^(n-e). Now to get the length of the period of 1/N, take the least 
common multiple of the lengths of the periods of 1/p^n, for all primes 
other than 2 or 5 dividing N.

Example: Find the length of the period of 1/9450. Start by factoring 
9450 = 2 * 3^3 * 5^2 * 7. The number of zeroes will be the larger of 1 
and 2 (the exponents of 2 and 5 in this factorization), namely 2. Now 
first consider 3^3 = p^n. The order of 10 modulo 3 is 1, and 3^2 is a 
divisor of 10^1 - 1 = 9, so e = 2. Since n = 3 > 2, the period length 
of 1/3^3 is 1*3^(3-2) = 3. Next consider 7^1 = p^n. The order of 10 
modulo 7 is 6, as above, and 7^1 is the largest power of 7 dividing 
10^6 - 1 = 999999 = 3^3 * 7 * 11 * 13 * 37, so e = 1. Thus the period 
length of 1/7 is 6. Taking the LCM of 3 and 6 gives 6. That tells me 
that the number of 9's should be 6, and 9450 must divide evenly into 
99999900. Sure enough, it does, and the quotient is 10582. Thus

   1/9450 = 10582/99999900 = 0.00 010582 010582 010582 010582...

Notice the 2 digits to the right of the decimal before the repeating 
part begins, corresponding to the two zeroes at the end of the number 
99999900, corresponding to the larger of the exponents of 2 and 5 in 
the factorization of 9450. Similarly,

   4567/9450 = 48327994/99999900 = 0.48 328042 328042 328042 328042...

Example: Find the period length of 1/73. Since 73 is prime, we need to 
find the order of 10 modulo 73. It must be a divisor of 72, so it 
could be 72, 36, 24, 18, 12, 9, 8, 6, 4, 3, 2, or 1. It turns out to 
be 8, so k = 8, and 73 divides evenly into 99999999. The period length 
must be 8.  The quotient is 1369863, so the decimal expansion must be

   1/73 = 0.01369863 01369863 01369863 01369863...

Get the idea?

Proofs of the above facts are a part of a subject called Number 
Theory, which you can study as an advanced undergraduate at a college 
or university.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/   


Date: 05/21/99 at 15:41:35
From: S.J. Lean
Subject: Re: Long numbers

On April 28, 1999, Dr Rob gave me a good reply, except there is no 
information on the calculator (I mean, software calculator) I must 
find. 

For example, I have already look at sevenths, and find tail 142857 -- 
six digits long. But what is tail length of 1/7 times 1/7, or 1/49? 
When I work out it is very interesting:

   0.020408163265306122448979591836734693877551 ...

THEN digits repeat. I notice that this is 42 digits long! But 42 
happens to be six times seven and maybe it is only a coincidence, but 
maybe not. I have an idea that one over a number (n = 7) would give a 
tail length of such-and-such, (6 in this case) but that the tail 
length of one over the same number SQUARED (n^2 = 49) would turn out 
to be the first number TIMES the first tail length? This is MY "little 
theory", but how to check it out? I tried some examples and found it 
to be true.

But I cannot easily do trial and error without a better calculator, so 
how can I see more patterns for my clues? Do you know of any software 
for this specialization? I need it to work out some that are dozens of 
digits long (hundreds would be even better).

P. Grabowski


Date: 05/25/99 at 09:55:32
From: Doctor Rob
Subject: Re: Long numbers

Thanks for writing back to Ask Dr. Math!

The problem you are trying to deal with does not require high 
precision calculators to solve. I described the solution in a previous 
answer. If you insist on using such calculators, here is some 
information which may be helpful.

Here is a website with links to lots of on-line calculators:

  http://www-sci.lib.uci.edu/HSG/RefCalculators2.html   

It sounds to me as if you want a software symbolic computation program 
such as MAPLE, MATHEMATICA, MATLAB, GP/PARI, or the like, which 
includes high accuracy multiple precision arithmetic. Some of these 
are commercial products, which you have to buy. Others are freeware or 
shareware. Here is a webpage with links to software for math education 
which may be of some help:

  http://mathforum.org/mathed/math.software.archives.html   

You can use an ordinary 10-digit calculator to do the divisions you 
want to do, too. You want to find the decimal expansion of 1/N for 
some values of N less than, say, 10000000 = 10^7.  Enter 1000000000 
into your calculator. That is your first dividend, D1. Divide by N. 
The integer part of the quotient is the first part of the decimal 
expansion. Call that Q1. Then enter D1 and subtract Q1*N. That will 
give you the remainder R1. Now enter R1 followed by as many 0's as 
will fit in your calculator. That is your next dividend D2. Divide by 
N. The integer part of the quotient is the next part of the decimal 
expansion. Call that Q2. Then enter D2 and subtract Q2*N. That will 
give you the remainder R2. Continue this process as long as you like.

Example: Find the decimal expansion of 1/127.
   N = 127,
   D1 = 1000000000,
   D1/N = 1000000000/127 = 7874015.748...
so Q1 = 7874015, and then D1 - Q1*N = R1 = 95.

   D2 = 9500000000,
   D2/N = 9500000000/127 = 74803149.61...
so Q2 = 74803149, and then D2 - Q2*N = R2 = 77.

   D3 = 7700000000,
   D3/N = 7700000000/127 = 60629921.25...
so Q3 = 60629921, and then D3 - Q3*N = R3 = 33.

   D4 = 3300000000,
   D4/N = 3300000000/127 = 25984251.97...
so Q4 = 25984251, and then D4 - Q4*N = R4 = 123.

   D5 = 1230000000,
   D5/N = 1230000000/127 = 9685039.370...
so Q5 = 9685039, and then D5 - Q5*N = R5 = 47.

   D6 = 4700000000,
   D6/N = 4700000000/127 = 37007874.01...
so Q6 = 37007874.

The decimal expansion we have found so far is

   1/127 = 0.007874015748031496062992125984251968503937 007874...

and you can see the beginning of the second period. The length of the 
period is 42. This is because 10^42 - 1 is a multiple of 127, and no 
smaller power of 10 has that property. Note that 42 = (127-1)/3.

   999999999999999999999999999999999999999999/127 =
     7874015748031496062992125984251968503937 exactly.

If N is larger, you get fewer decimal places in the expansion at each 
step, so you have to take more steps, but the result is just as valid. 
This will not work if N is larger than D1.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Calculators, Computers
High School Number Theory
High School Sequences, Series
Middle School Fractions

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