Difference between revisions of "2016 AMC 10B Problems/Problem 25"
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We can see that for each value of <math>k</math>, <math>\lfloor k \{ x \} \rfloor</math> can equal integers from <math>0</math> to <math>k-1</math>. | We can see that for each value of <math>k</math>, <math>\lfloor k \{ x \} \rfloor</math> can equal integers from <math>0</math> to <math>k-1</math>. | ||
− | Clearly, the value of <math>\lfloor k \{ x \} \rfloor</math> changes only when <math>x</math> 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 <math>\{ x \}</math> 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 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 | 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 | ||
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Because the value of <math>f(x)</math> is at least <math>0</math> and can increase <math>31</math> times, there are a total of <math>\fbox{\textbf{(A)}\ 32}</math> different possible values of <math>f(x)</math>. | Because the value of <math>f(x)</math> is at least <math>0</math> and can increase <math>31</math> times, there are a total of <math>\fbox{\textbf{(A)}\ 32}</math> different possible values of <math>f(x)</math>. | ||
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==Solution 2== | ==Solution 2== |
Revision as of 01:54, 9 August 2020
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) | ||
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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