Lesson Finding digits of the decimal expansion to the number 1/977

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<H2>Finding digits of the decimal expansion to the number 1/977</H2>

<H3>Problem 1</H3>When expanded as a decimal, &nbsp;the fraction &nbsp;&nbsp;{{{1/977}}} &nbsp;has a repetend &nbsp;(the repeating part of the decimal) 
that begins right after the decimal point, &nbsp;and is &nbsp;976 &nbsp;digits long. 
If the last three digits of the repetend are &nbsp;ABC, &nbsp;compute the digits &nbsp;A, &nbsp;B, &nbsp;and &nbsp;C.

<B>Solution</B>

<pre>
Let {{{1/977}}} = x.  Since the repetend is 977 digits long, we see that  {{{(10^976)*x}}} has the same decimal
expansion as x.  Hence, the number  n = {{{(10^976)*x-x}}}  is an integer number and it is (it represents) the "repetend".

But, from the other side,  this number  n  is

9999...999999
    n = {{{(10^976-1)*x}}} = --------------- .
                              977

+---------------------------------------------------------------+
   |    The numerator is the integer formed by 976 digits of "9".  |
   +---------------------------------------------------------------+

So, we can write  9999...999999 = 977*n, where n is some integer number.    (1)

This number "n" is the "repetend part",  and our goal is to find three last digits of n.

+-------------------------------------------------------------------+
    | From this writing (1), it is clear that the last digit of n is 7, |
    |         providing 7*7 = 49 with the last digit 9;                 |
    |      so, we can write  n = 10m+7 for some integer m.              |
    +-------------------------------------------------------------------+

Substitute it into (1).  You will get

9999...999999 = 977*(10m+7) = 977*10m + 6839.

Hence,  9999...999999 = 977*10m + 6839.    (2)

Subtract 6839 from both sides of (2).  You will get

9999...9993160 = 977*10m           (3)

In (3),  divide both sides by 10.  You will get

9999...999316 = 977*m,             (4)

where the number in the left side has the last 3 digits 316 and all other preceding digits are "9",  and "m" is some integer.

+-------------------------------------------------------------------+
    | From this writing (4), it is clear that the last digit of m is 8, |
    |        providing 7*8 = 56 with the last digit 6;                  |
    |      so, we can write  m = 10k+8 for some integer k.              |
    +-------------------------------------------------------------------+

Substitute it into (4).  You will get

9999...999316 = 977*(10k+8) = 977*10k + 7816.

Hence,  9999...999316 = 977*10k + 7816.    (5)

Subtract 7816 from both sides of (5).  You will get

9999...991500 = 977*10k            (6)

In (6),  divide both sides by 10.  You will get

9999...999150 = 977*k,             (7)

where the number in the left side has the last 3 digits 134 and all other preceding digits are "9",  and "k" is some integer.

+-------------------------------------------------------------------+
    | From this writing (7), it is clear that the last digit of k is 0, |
    |        providing 7*0 = 0 with the last digit 0.                   |
    +-------------------------------------------------------------------+

Thus the last three digits in repetend part are  087.

<U>ANSWER</U>.  The last three digits in repetend part are  087.

So,  A = 0,  B = 8,  C = 7.
</pre>