2016 AMC 10B Problems/Problem 25
Contents
Problem
Let , where denotes the greatest integer less than or equal to . How many distinct values does assume for ?
Solution 1
Since , we have
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 .
Solution 2
so we have Clearly, the value of changes only when is equal to any of the fractions . To get all the fractions, Graphing this function gives us different fractions but on an average, in each of the intervals don’t work. This means there are a total of different possible values of .
See Also
2016 AMC 10B (Problems • Answer Key • Resources) | ||
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