Difference between revisions of "2014 AIME II Problems/Problem 4"
Mathgeek2006 (talk | contribs) m (→Solution 2) |
|||
Line 42: | Line 42: | ||
==Solution 2== | ==Solution 2== | ||
Note that <math>\frac{33}{37}=\frac{891}{999} = 0.\overline{891}</math>. Also note that the period of <math>0.abab\overline{ab}+0.abcabc\overline{abc}</math> is at most <math>6</math>. Therefore, we only need to worry about the sum <math>0.ababab+ 0.abcabc</math>. Adding the two, we get | Note that <math>\frac{33}{37}=\frac{891}{999} = 0.\overline{891}</math>. Also note that the period of <math>0.abab\overline{ab}+0.abcabc\overline{abc}</math> is at most <math>6</math>. Therefore, we only need to worry about the sum <math>0.ababab+ 0.abcabc</math>. Adding the two, we get | ||
− | <cmath> \begin{ | + | <cmath> \begin{array}{ccccccc}&a&b&a&b&a&b\\ +&a&b&c&a&b&c\\ \hline &8&9&1&8&9&1\end{array} </cmath> |
− | From this, we can see that <math> | + | From this, we can see that <math>a=4</math>, <math>b=4</math>, and <math>c=7</math>, so our desired answer is <math>\boxed{447}</math> |
==Solution 3== | ==Solution 3== |
Revision as of 20:37, 13 March 2015
Problem
The repeating decimals and satisfy
where , , and are (not necessarily distinct) digits. Find the three digit number .
Solution 1
Notice repeating decimals can be written as the following:
where a,b,c are the digits. Now we plug this back into the original fraction:
Multiply both sides by This helps simplify the right side as well because :
Dividing both sides by and simplifying gives:
At this point, seeing the factor common to both a and b is crucial to simplify. This is because taking to both sides results in:
Notice that we arrived to the result by simply dividing by and seeing Okay, now it's pretty clear to divide both sides by in the modular equation but we have to worry about being multiple of Well, is a multiple of so clearly, couldn't be. Also, Now finally we simplify and get:
But we know is between and because it is a digit, so must be Now it is straightforward from here to find and :
and since a and b are both between and , we have . Finally we have the digit integer
Solution 2
Note that . Also note that the period of is at most . Therefore, we only need to worry about the sum . Adding the two, we get From this, we can see that , , and , so our desired answer is
Solution 3
Noting as above that and , let . Then
Solving for gives
Because must be integer, it follows that must be a multiple of (because clearly is not). Inspecting the equation, one finds that only yields a digit . Thus
See also
2014 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.