Question 1208610: When expanded as a decimal, the fraction $\frac{1}{7}$ has a repetend (the repeating part of the decimal) of $142857$. The last three digits of the repetend are $857$.
When expanded as a decimal, the fraction $\frac{1}{13}$ has a repetend that is $6$ digits long. If the last three digits of the repetend are $ABC$, compute the digits $A$, $B$, and $C$.
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52781)   (Show Source):
You can put this solution on YOUR website! .
Since the method of solution is not predefined in the problem, choose the simplest way making long division of integer numbers.
Doing this way, find the 6-digit repetend, which is 076923.
So, the digits A, B and C are 9, 2, 3, respectively.
Solved (by the simplest way).
Answer by math_tutor2020(3816)  (Show Source):
You can put this solution on YOUR website!
You would use long division to divide 1 over 13.
13 goes out to the left, while 1 goes under the bar.
Here is a calculator that provides a step-by-step walkthrough
https://www.calculatorsoup.com/calculators/math/longdivisiondecimals.php
Adjust the drop-down menu for "decimal places" to 7 or larger.
The calculator will say that
1/13 = 0.076923076923076923... where "076923" repeats forever
The color coding is there to show when one block stops and the next begins.
The repetend is 076923
The last 3 digits are 9, 2, and 3.
A = 9
B = 2
C = 3
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