Difference between revisions of "2016 AMC 10B Problems/Problem 25"
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Clearly, the value of <math>\lfloor k \{ x \} \rfloor</math> changes only when x is equal to any of the fractions <math>\frac{1}{k}, \frac{2}{k} \dots \frac{k-1}{k}</math>. | Clearly, the value of <math>\lfloor k \{ x \} \rfloor</math> changes only when x is equal to any of the fractions <math>\frac{1}{k}, \frac{2}{k} \dots \frac{k-1}{k}</math>. | ||
− | So we want to count how many distinct fractions have the form <math>\frac{m}{n}</math> where <math>n \le 10</math>. We can find this easily by computing | + | So we want to count how many distinct fractions less than <math>1</math> have the form <math>\frac{m}{n}</math> where <math>n \le 10</math>. We can find this easily by computing |
<cmath>\sum_{k=2}^{10} \phi(k)</cmath> | <cmath>\sum_{k=2}^{10} \phi(k)</cmath> |
Revision as of 11:08, 4 January 2019
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) | ||
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 |
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