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AMC数学竞赛真题2016年10B 25

2018-12-17 重点归纳

AMC10数学竞赛是美国高中数学竞赛中的一项,是针对高中一年级及初中三年级学生的数学测试,该竞赛开始于2000年,分A赛和B赛,于每年的2月初和2月中举行,学生可任选参加一项即可。不管是对高校申请还是今后在数学领域的发展都极其有利!那么接下来跟随小编来看一下AMC10数学竞赛真题以及官方解答吧:

Problem 25

Let $f(x)=\sum_{k=2}^{10}(\lfloor kx \rfloor -k \lfloor x \rfloor)$, where $\lfloor r \rfloor$ denotes the greatest integer less than or equal to $r$. How many distinct values does $f(x)$ assume for $x \ge 0$?

$\textbf{(A)}\ 32\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 45\qquad\textbf{(D)}\ 46\qquad\textbf{(E)}\ \text{infinitely many}$

Solution

Since $x = \lfloor x \rfloor + \{ x \}$, we have

AMC数学竞赛

The function can then be simplified into

\[f(x) = \sum_{k=2}^{10} ( k \lfloor x \rfloor + \lfloor k \{ x \} \rfloor - k \lfloor x \rfloor)\]

which becomes

AMC8

We can see that for each value of k, $\lfloor k \{ x \} \rfloor$ can equal integers from 0 to k-1.

Clearly, the value of $\lfloor k \{ x \} \rfloor$ changes only when x is equal to any of the fractions $\frac{1}{k}, \frac{2}{k} \dots \frac{k-1}{k}$.

So we want to count how many distinct fractions have the form $\frac{m}{n}$ where $n \le 10$. We can find this easily by computing

美国数学竞赛AMC

where $\phi(k)$ is the Euler Totient Function. Basically $\phi(k)$ counts the number of fractions with $k$ as its denominator (after simplification). This comes out to be $31$.

Because the value of $f(x)$ is at least 0 and can increase 31 times, there are a total of $\fbox{\textbf{(A)}\ 32}$ different possible values of $f(x)$

以上就是小编对AMC10数学竞赛真题以及解析的介绍,希望对你有所帮助,如果想了解更多关于AMC数学竞赛报考点、南京AMC数学竞赛培训、美国数学竞赛AMC有用吗以及AMC学习资料等信息请持续关注AMC数学竞赛网。


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