Difference between revisions of "2016 AMC 10B Problems/Problem 25"
Shivakannan (talk | contribs) (→See Also) |
Shivakannan (talk | contribs) (→Solution) |
||
Line 9: | Line 9: | ||
Since <math>x = \lfloor x \rfloor + \{ x \}</math>, we have | Since <math>x = \lfloor x \rfloor + \{ x \}</math>, we have | ||
− | <cmath>f(x) = \ | + | <cmath>f(x) = \product_{k=2}^{10} (\lfloor k \lfloor x \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 12:41, 31 March 2020
Problem
Let , where denotes the greatest integer less than or equal to . How many distinct values does assume for ?
Solution
Since , we have
\[f(x) = \product_{k=2}^{10} (\lfloor k \lfloor x \rfloor +k \{ x \} \rfloor - k \lfloor x \rfloor)\] (Error compiling LaTeX. )
The function can then be simplified into
which becomes
We can see that for each value of , can equal integers from to .
Clearly, the value of changes only when is equal to any of the fractions .
So we want to count how many distinct fractions less than have the form where . We can find this easily by computing
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 and can increase 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.