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 , 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
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Solution 2
Also seeing f(X) is periodic with a period of 1, you can list out 1+2+3…+9
+1 = 46 fractions. Of these, in each interval of 5, 3 fractions are repeated so 46 - 3(5) = 32 which is option A