Difference between revisions of "2016 AMC 10B Problems/Problem 25"
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Since <math>x = \lfloor x \rfloor + \{ x \}</math>, we have | Since <math>x = \lfloor x \rfloor + \{ x \}</math>, we have | ||
| − | <cmath>f(x) = \sum_{k=2}^{10} (\lfloor k \lfloor x \rfloor + k \{ x \} \rfloor - k \lfloor x \rfloor)</cmath> | + | <cmath>f(x) = \sum_{k=2}^{10} (\lfloor k \lfloor x \rfloor +\rfloor k \{ x \} \rfloor - k \lfloor x \rfloor)</cmath> |
The function can then be simplified into | The function can then be simplified into | ||
Revision as of 22:23, 27 March 2017
Problem
Let 
, where 
denotes the greatest integer less than or equal to 
. How many distinct values does 
assume for 
?
Solution
Since 
, we have
![\[f(x) = \sum_{k=2}^{10} (\lfloor k \lfloor x \rfloor +\rfloor k \{ x \} \rfloor - k \lfloor x \rfloor)\]](http://latex.artofproblemsolving.com/d/4/4/d44458d2aba0380afa13ecc928ff916d2f2880b9.png)
The function can then be simplified into
![\[f(x) = \sum_{k=2}^{10} ( k \lfloor x \rfloor + \lfloor k \{ x \} \rfloor - k \lfloor x \rfloor)\]](http://latex.artofproblemsolving.com/f/c/e/fcef0323311a4ec76e2064194d95814155ed2e69.png)
which becomes
![\[f(x) = \sum_{k=2}^{10} \lfloor k \{ x \} \rfloor\]](http://latex.artofproblemsolving.com/0/8/c/08c34dd2e7a954f419f6c3192d31eb0e8ab01238.png)
We can see that for each value of k, 
can equal integers from 0 to k-1.
Clearly, the value of changes only when x is equal to any of the fractions 
.
So we want to count how many distinct fractions have the form 
where 
. We can find this easily by computing
![\[\sum_{k=2}^{10} \phi(k)\]](http://latex.artofproblemsolving.com/5/7/7/57764d9e498281fbf75592968356664165e205ff.png)
where 
is the Euler Totient Function. Basically counts the number of fractions with 
as its denominator (after simplification). This comes out to be 
.
Because the value of is at least 0 and can increase 31 times, there are a total of 
different possible values of .
See Also
| 2016 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
Followed by Last Problem | |
| 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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
