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 + | + | <cmath>f(x) = \sum_{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 23:24, 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
The function can then be simplified into
which becomes
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
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.