Difference between revisions of "2016 AMC 12B Problems/Problem 22"
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| − | + | Solution by e_power_pi_times_i | |
| + | If <math>\frac{1}{n} = abcdef</math> and <math>n < 1000</math>, <math>n</math> must be a factor of <math>999999</math>. Also, by the same procedure, <math>n+6</math> must be a factor of <math>9999</math>. Checking through all the factors of <math>999999</math> and <math>9999</math>, we see that <math>n =297</math> is a solution, so the answer is . | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2016|ab=B|num-b=21|num-a=23}} | {{AMC12 box|year=2016|ab=B|num-b=21|num-a=23}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 00:27, 22 February 2016
Problem
For a certain positive integer n less than 
, the decimal equivalent of 
is 
, a repeating decimal of period of 6, and the decimal equivalent of 
is 
, a repeating decimal of period 4. In which interval does 
lie?
![$\textbf{(A)}\ [1,200]\qquad\textbf{(B)}\ [201,400]\qquad\textbf{(C)}\ [401,600]\qquad\textbf{(D)}\ [601,800]\qquad\textbf{(E)}\ [801,999]$](http://latex.artofproblemsolving.com/b/a/0/ba0b05d779134dbe18999e12efb223c3b9bdd656.png)
Solution
Solution by e_power_pi_times_i
If 
and 
, must be a factor of 
. Also, by the same procedure, 
must be a factor of 
. Checking through all the factors of and , we see that 
is a solution, so the answer is .
See Also
| 2016 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 21 |
Followed by Problem 23 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
