2016 AMC 10B Problems/Problem 25
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 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 0 and can increase 31 times, there are a total of
different possible values of
.
See Also
2016 AMC 10B (Problems • Answer Key • Resources) | ||
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