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The Period of 61/97


Date: 01/12/2001 at 08:46:22
From: Celia
Subject: period of 61/97

I need to know the period of 61/97 (if there is one). I've calculated 
it to the 31st decimal place, but can't tell if it is a finite or 
infinite and periodic... Is there any way I can figure it out?

Thank you!
Celia


Date: 01/12/2001 at 15:52:02
From: Doctor Rick
Subject: Re: period of 61/97

Hi, Celia.

I can tell you that the period is finite; more particularly, that it 
is no greater than 97 digits. You've done at least 1/3 of the work.

How do I know that? If you've done the calculations by long division, 
then you've seen a sequence of remainders:

           0.628865979...
        -----------------
     97 ) 61.000000000...
          58 2
          ----
           2 80
           1 94
           ----
             860
             776
             ---
              840
              776
              ---
               640
               582
               ---
                580
                485
                ---
                 950
                 873
                 ---
                  770
                  679
                  ---
                   910
                   873
                   ---
                    37

The partial remainders are 28, 86, 84, 64, 58, 95, 77, 91, 37, ...

The first time you get a remainder that has shown up before, the next 
quotient digit will be the same as it was the first time, and the next 
remainder will be the same as it was the first time, and so on - in 
other words, the decimal begins repeating at that point.

Now we can apply the "pigeonhole principle." How many different 
remainders can there be? The remainder must be between 0 and 96. In 
fact, if it's 0 then the decimal will terminate, and we know that 
won't happen. (If you don't know why, I'd be glad to tell you.) There 
are thus only 96 different possible remainders. When we've gotten 97 
digits and 97 remainders, we can be sure that one of them has shown up 
before. You can imagine 96 "pigeonholes" labeled with the possible 
remainders. We have 97 numbers (the actual remainders) to put in them; 
at least two of the numbers must go into the same pigeonhole. 
Therefore, the decimal must begin repeating by the 97th digit.

If you're doing the calculation by a calculator (such as the 
calculator built into Windows) that handles up to 31 digits but no 
more, you can use this trick. Calculate 25 digits (so we have some 
digits to spare and we won't overtax the calculator in what follows):

     61/97 = 0.6288659793814432989690721

Then find the partial remainder after the last digit so far. Here's 
how to do it. Multiply the decimal by 97, subtract this from 61, and 
multiply by 1E25. I get a partial remainder of 63.

Next, divide the remainder by 97. You will get a continuation of the 
decimal expansion of 61/97. I get:

     63/97 = 0.64948453608247422680412371134021

When I calculated 61/97, I got 32 digits:

     61/97 = 0.62886597938144329896907216494845

Note that the last 7 digits, 6494845, match the first 7 digits of the 
continuation. This is a check on the accuracy of my method. The full 
decimal expansion so far is thus:

     61/97 = 0.62886597938144329896907216494845649484536082474226804
                                                           12371134021

You can continue in this way as far as you need in order to find where 
the decimal begins repeating. Have fun!

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
Middle School Division
Middle School Fractions

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