Difference between revisions of "2016 AMC 12B Problems/Problem 22"
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− | For a certain positive integer n less than <math>1000</math>, the decimal equivalent of <math>\frac{1}{n}</math> is <math>0.\overline{abcdef}</math>, a repeating decimal of period of 6, and the decimal equivalent of <math>\frac{1}{n+6}</math> is <math>0.\overline{wxyz}</math>, a repeating decimal of period 4. In which interval does <math>n</math> lie? | + | For a certain positive integer <math>n</math> less than <math>1000</math>, the decimal equivalent of <math>\frac{1}{n}</math> is <math>0.\overline{abcdef}</math>, a repeating decimal of period of <math>6</math>, and the decimal equivalent of <math>\frac{1}{n+6}</math> is <math>0.\overline{wxyz}</math>, a repeating decimal of period <math>4</math>. In which interval does <math>n</math> lie? |
<math>\textbf{(A)}\ [1,200]\qquad\textbf{(B)}\ [201,400]\qquad\textbf{(C)}\ [401,600]\qquad\textbf{(D)}\ [601,800]\qquad\textbf{(E)}\ [801,999]</math> | <math>\textbf{(A)}\ [1,200]\qquad\textbf{(B)}\ [201,400]\qquad\textbf{(C)}\ [401,600]\qquad\textbf{(D)}\ [601,800]\qquad\textbf{(E)}\ [801,999]</math> |
Revision as of 00:37, 22 February 2016
Problem
For a certain positive integer less than , the decimal equivalent of is , a repeating decimal of period of , and the decimal equivalent of is , a repeating decimal of period . In which interval does lie?
Solution
Solution by e_power_pi_times_i
If , must be a factor of . Also, by the same procedure, must be a factor of . Checking through all the factors of and that are less than , 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 |
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