Difference between revisions of "2016 AMC 10B Problems/Problem 25"
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<math>\textbf{(A)}\ 32\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 45\qquad\textbf{(D)}\ 46\qquad\textbf{(E)}\ \text{infinitely many}</math> | <math>\textbf{(A)}\ 32\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 45\qquad\textbf{(D)}\ 46\qquad\textbf{(E)}\ \text{infinitely many}</math> | ||
− | + | ==Solution 1== | |
− | ==Solution== | + | |
Since <math>x = \lfloor x \rfloor + \{ x \}</math>, we have | Since <math>x = \lfloor x \rfloor + \{ x \}</math>, we have | ||
+ | |||
+ | <cmath>f(x) = \sum_{k=2}^{10} (\lfloor k \lfloor x \rfloor +k \{ x \} \rfloor - k \lfloor x \rfloor)</cmath> | ||
+ | |||
+ | The function can then be simplified into | ||
+ | |||
+ | <cmath>f(x) = \sum_{k=2}^{10} ( k \lfloor x \rfloor + \lfloor k \{ x \} \rfloor - k \lfloor x \rfloor)</cmath> | ||
+ | |||
+ | which becomes | ||
+ | |||
+ | <cmath>f(x) = \sum_{k=2}^{10} \lfloor k \{ x \} \rfloor</cmath> | ||
+ | |||
+ | 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>. |
+ | |||
+ | 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>. '''Explanation for this is provided below.''' We can find this easily by computing | ||
+ | |||
+ | <cmath>\sum_{k=2}^{10} \phi(k)</cmath> | ||
+ | |||
+ | where <math>\phi(k)</math> is the [[Euler Totient Function]]. Basically <math>\phi(k)</math> counts the number of fractions with <math>k</math> as its denominator (after simplification). This comes out to be <math>31</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>. | |
− | + | ===Explanation:=== | |
− | + | Arrange all such fractions in increasing order and take a current <math>\frac{m}{n}</math> to study. Let <math>p</math> denote the previous fraction in the list and <math>x_\text{old}</math> (<math>0 \le x_\text{old} < k</math> for each <math>k</math>) be the largest so that <math>\frac{x_\text{old}}{k} \le p</math>. Since <math>\text{ }\text{ }\frac{m}{n} > p</math>, we clearly have all <math>x_\text{new} \ge x_\text{old}</math>. Therefore, the change must be nonnegative. | |
− | < | + | But among all numerators coprime to <math>n</math> so far, <math>m</math> is the largest. Therefore, choosing <math>\frac{m}{n}</math> as <math>{x}</math> increases the value <math>\lfloor n \{ x \} \rfloor</math>. Since the overall change in <math>f(x)</math> is positive as fractions <math>m/n</math> increase, we deduce that all such fractions correspond to different values of the function. |
− | + | Minor Latex Edits made by MATHWIZARD2010. | |
− | + | ==Solution 2== | |
− | + | <math>x = \lfloor x \rfloor + \{ x \}</math> so we have <cmath>f(x) = \sum_{k=2}^{10} \lfloor k \{ x \} \rfloor.</cmath> 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>. To get all the fractions,graphing this function gives us <math>46</math> different fractions. But on average, <math>3</math> in each of the <math>5</math> intervals don’t work. This means there are a total of <math>\fbox{\textbf{(A)}\ 32}</math> different possible values of <math>f(x)</math>. | |
− | <cmath>\sum_{k=2}^{10} \ | ||
− | |||
− | + | ==Video Solution== | |
+ | https://www.youtube.com/watch?v=zXJrdDtZNbw | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2016|ab=B|num-b=24|after=Last Problem}} | {{AMC10 box|year=2016|ab=B|num-b=24|after=Last Problem}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 23:07, 22 January 2021
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 . Explanation for this is provided below. 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 .
Explanation:
Arrange all such fractions in increasing order and take a current to study. Let denote the previous fraction in the list and ( for each ) be the largest so that . Since , we clearly have all . Therefore, the change must be nonnegative.
But among all numerators coprime to so far, is the largest. Therefore, choosing as increases the value . Since the overall change in is positive as fractions increase, we deduce that all such fractions correspond to different values of the function.
Minor Latex Edits made by MATHWIZARD2010.
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 average, in each of the intervals don’t work. This means there are a total of different possible values of .
Video Solution
https://www.youtube.com/watch?v=zXJrdDtZNbw
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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