2016 AMC 10A 真题与在线限时测验

考试时间还剩下：

1:15:00

1.

求 $\dfrac{11!-10!}{9!}$ 的值。

What is the value of $\dfrac{11!-10!}{9!}?$

$99$

$100$

$110$

$121$

$132$

答案：B

知识点：阶乘因式分解

难度评级：560

解答：

从分子中提取 $9!$ $\begin{align*} \dfrac{9!(11 \cdot 10 - 10)}{9!} &=11 \cdot 10 - 10 \\&= 110 - 10 \\&= 100. \end{align*}$ 所以正确答案是 B。

We can factor a $9!$ out the numerator and simplify. $\begin{align*} \dfrac{9!(11 \cdot 10 - 10)}{9!} &=11 \cdot 10 - 10 \\&= 110 - 10 \\&= 100. \end{align*}$ Thus, the correct answer is B .

2.

求使下式成立的 $x$ 值： $10^{x} \cdot 100^{2x}=1000^{5}$

For what value of $x$ does $10^{x} \cdot 100^{2x}=1000^{5}?$

$1$

$2$

$3$

$4$

$5$

答案：C

知识点：指数

难度评级：770

解答：

把所有底数写成 $10$ 的幂： $\small 10^x\cdot100^{2x}=10^x\cdot(10^2)^{2x}=10^{5x},$ 同时 $1000^5=(10^3)^5=10^{15}.$

因此 $5x=15$ ，所以 $x=3$ 。

所以正确答案是 C。

Rewrite all bases as powers of $10$ : $\small 10^x\cdot100^{2x}=10^x\cdot(10^2)^{2x}=10^{5x},$ and $1000^5=(10^3)^5=10^{15}.$

Therefore $5x=15$ , so $x=3$ .

Thus, the correct answer is C.

3.

Ben 每花一美元买贝果，David 就少花 $25$ 美分。Ben 比 David 多付 $\$12.50$ 。他们两人在贝果店一共花了多少钱？

For every dollar Ben spent on bagels, David spent $25$ cents less. Ben paid $\$12.50$ more than David. How much did they spend in the bagel store together?

$\$37.50$

$\$50.00$

$\$87.50$

$\$90.00$

$\$92.50$

答案：C

知识点：比与比例钱币

难度评级：870

解答：

Ben 每花一美元，David 花 $75$ 美分，所以两人的差额是 $25$ 美分。总差额为 $\$12.50$ ，也就是 Ben 花费的四分之一。

Ben 花了 $12.50\div0.25=50$ 美元，David 花了 $37.50$ 美元。两人共花 $87.50$ 美元。

所以正确答案是 C。

For each dollar Ben spent, David spent $75$ cents, so the difference was $25$ cents. The total difference $\$12.50$ is therefore one quarter of Ben's spending.

Ben spent $12.50\div0.25=50$ dollars, and David spent $37.50$ dollars. Together they spent $87.50$ dollars.

Thus, the correct answer is C.

4.

对所有实数 $x$ 和满足 $y \neq 0$ 的实数 $y$ ，余数定义为 $\text{rem} (x ,y)=x-y\left \lfloor \dfrac{x}{y} \right \rfloor$ 其中 $\left \lfloor \frac{x}{y} \right \rfloor$ 表示小于或等于 $\frac{x}{y}$ 的最大整数。求 $\text{rem} \left(\frac{3}{8}, -\frac{2}{5} \right)$ 的值。

The remainder can be defined for all real numbers $x$ and $y$ with $y \neq 0$ by $\text{rem} (x ,y)=x-y\left \lfloor \dfrac{x}{y} \right \rfloor$ where $\left \lfloor \frac{x}{y} \right \rfloor$ denotes the greatest integer less than or equal to $\frac{x}{y}.$ What is the value of $\text{rem} \left(\frac{3}{8}, -\frac{2}{5} \right)?$

$-\dfrac{3}{8}$

$-\dfrac{1}{40}$

$0$

$\dfrac{3}{8}$

$\dfrac{31}{40}$

答案：B

知识点：取整函数分数

难度评级：1140

解答：

$\begin{align*} \text{rem} \left(\dfrac{3}{8}, -\dfrac{2}{5}\right) &= \dfrac{3}{8} \\ &{}+ \dfrac{2}{5} \left \lfloor \dfrac{\frac{3}{8}}{-\frac{2}{5}}\right \rfloor \\ &= \dfrac{3}{8} \\ &{}+ \dfrac{2}{5} \left \lfloor -\dfrac{15}{16} \right \rfloor \\ &= \dfrac{3}{8} + \dfrac{2}{5} \cdot -1 \\ &= \dfrac{3}{8} - \dfrac{2}{5} \\ &= - \dfrac{1}{40} \end{align*}$ 所以正确答案是 B。

Using the formula, we get $\begin{align*} \text{rem} \left(\dfrac{3}{8}, -\dfrac{2}{5}\right) &= \dfrac{3}{8} \\ &{}+ \dfrac{2}{5} \left \lfloor \dfrac{\frac{3}{8}}{-\frac{2}{5}}\right \rfloor \\ &= \dfrac{3}{8} \\ &{}+ \dfrac{2}{5} \left \lfloor -\dfrac{15}{16} \right \rfloor \\ &= \dfrac{3}{8} + \dfrac{2}{5} \cdot -1 \\ &= \dfrac{3}{8} - \dfrac{2}{5} \\ &= - \dfrac{1}{40} \end{align*}$ Thus, the correct answer is B .

5.

一个长方体的整数边长之比为 $1:3:4$ 。下列哪个数可能是这个长方体的体积？

A rectangular box has integer side lengths in the ratio $1:3:4.$ Which of the following could be the volume of the box?

$48$

$56$

$64$

$96$

$144$

答案：D

知识点：体积比与比例

难度评级：900

解答：

设最短边长为 $s$ ，则另外两边为 $3s$ 和 $4s$ 。

体积为 $s \cdot 3s \cdot 4s = 12s^3.$ 测试 $s$ 的取值可知，当 $s = 2$ 时， $12s^3 = 96$ ，这是一个选项。

所以正确答案是 D。

Let $s$ be the side length of the smallest side. Then the other two sides are $3s$ and $4s.$

The volume is therefore $s \cdot 3s \cdot 4s = 12s^3.$ Testing out values of $s,$ we see that if $s = 2,$ then $12s^3 = 96,$ which is an answer choice.

Thus, the correct answer is D .

6.

Ximena 把整数 $1$ 到 $30$ 各列出一次。Emilio 抄写 Ximena 的数，但把每次出现的数字 $2$ 都替换成数字 $1$ 。Ximena 把她的数相加，Emilio 把他的数相加。Ximena 的和比 Emilio 的和大多少？

Ximena lists the whole numbers $1$ through $30$ once. Emilio copies Ximena's numbers, replacing each occurrence of the digit $2$ by the digit $1.$ Ximena adds her numbers and Emilio adds his numbers. How much larger is Ximena's sum than Emilio's?

$13$

$26$

$102$

$103$

$110$

答案：D

知识点：位值数字

难度评级：1020

解答：

个位数字 $2$ 被改成 $1$ 时，数少 $1$ 。这发生在 $2,12,22$ ，总共少 $3$ 。

十位数字 $2$ 被改成 $1$ 时，数少 $10$ 。这发生在 $20,21,\ldots,29$ ，总共少 $100$ 。

因此 Emilio 的和比 Ximena 的和少 $103$ ，所以 Ximena 的和大 $103$ 。

所以正确答案是 D。

When Emilio changes a units digit $2$ to $1$ , his copied number is smaller by $1$ . This happens in $2,12,22$ , for a loss of $3$ .

When he changes a tens digit $2$ to $1$ , his copied number is smaller by $10$ . This happens for $20,21,\ldots,29$ , for a loss of $100$ .

Thus Emilio's sum is $103$ less than Ximena's sum, so Ximena's sum is $103$ larger.

Thus, the correct answer is D.

7.

$7$ 个数据值 $60, 100, x, 40, 50, 200, 90$ 的平均数、中位数和众数都等于 $x$ 。求 $x$ 的值。

The mean, median, and mode of the $7$ data values $60, 100, x, 40, 50, 200, 90$ are all equal to $x.$ What is the value of $x?$

$50$

$60$

$75$

$90$

$100$

答案：D

知识点：平均数中位数（数据）众数

难度评级：960

解答：

七个数的和为 $540+x$ 。如果平均数等于 $x$ ，则 $\frac{540+x}{7}=x,$ 所以 $540=6x$ ， $x=90$ 。

当 $x=90$ 时，排序后的数据为 $40,50,60,90,90,100,200$ 。其中位数是 $90$ ，众数也是 $90$ ，所以符合条件。

所以正确答案是 D。

The sum of the seven data values is $540+x$ . If the mean equals $x$ , then $\frac{540+x}{7}=x,$ so $540=6x$ and $x=90$ .

With $x=90$ , the ordered data set is $40,50,60,90,90,100,200$ . Its median is $90$ , and its mode is also $90$ , so the value works.

Thus, the correct answer is D.

8.

Trickster Rabbit 同意每次 Foolish Fox 经过 Rabbit 家旁边的桥时，都把 Fox 的钱翻倍，但 Fox 每次过桥后必须向 Rabbit 支付 $40$ 枚硬币的过路费。付款在翻倍之后进行。Fox 起初很高兴，但他发现自己过桥三次后钱全没了。Fox 一开始有多少枚硬币？

Trickster Rabbit agrees with Foolish Fox to double Fox's money every time Fox crosses the bridge by Rabbit's house, as long as Fox pays $40$ coins in toll to Rabbit after each crossing. The payment is made after the doubling. Fox is excited about his good fortune until he discovers that all his money is gone after crossing the bridge three times. How many coins did Fox have at the beginning?

$20$

$30$

$35$

$40$

$45$

答案：C

知识点：逆推法

难度评级：1070

解答：

最后 Fox 有 $0$ 枚硬币，所以第三次付过路费之前有 $40$ 枚。

因此第三次翻倍前有 $40 \div 2 = 20$ 枚；第二次付过路费之前有 $20 + 40 = 60$ 枚。

第二次翻倍前有 $60 \div 2 = 30$ 枚；第一次付过路费之前有 $30 + 40 = 70$ 枚。

最后，第一次翻倍前 Fox 有 $70 \div 2 = 35$ 枚硬币。

所以正确答案是 C。

We know that Fox has $0$ coins at the end. Then before paying the final toll, Fox had $40$ coins.

Then he had $40 \div 2 = 20$ coins before the doubling. Then before paying the toll for the second crossing, he had $20 + 40 = 60$ coins.

Before the doubling on the second crossing, he had $60 \div 2 = 30$ coins. On the first crossing before the toll, Fox had $30 + 40 = 70$ coins.

Finally, before the first doubling, Fox had $70 \div 2 = 35$ coins.

Thus, the correct answer is C .

9.

一个由 $2016$ 枚硬币组成的三角形阵列，第一行有 $1$ 枚硬币，第二行有 $2$ 枚，第三行有 $3$ 枚，依此类推，第 $N$ 行有 $N$ 枚。求 $N$ 的各位数字之和。

A triangular array of $2016$ coins has $1$ coin in the first row, $2$ coins in the second row, $3$ coins in the third row, and so on up to $N$ coins in the $N$ th row. What is the sum of the digits of $N?$

$6$

$7$

$8$

$9$

$10$

答案：D

知识点：三角形数因式分解数字

难度评级：1070

解答：

回忆前 $N$ 个整数之和为 $\dfrac{N(N + 1)}{2}$ 。

我们要找 $N$ ，使得 $\dfrac{N(N + 1)}{2} = 2016.$ 交叉相乘并化简，得到 $N^2 + N - 4032 = 0.$ 因式分解得 $(N - 63)(N + 64) = 0$ 取正值， $N = 63$ 。各位数字和为 $9$ 。

所以正确答案是 D。

Recall that the sum of the first $N$ number is $\dfrac{N(N + 1)}{2}.$

We want to find $N$ such that $\dfrac{N(N + 1)}{2} = 2016.$ Cross-multiplying and simplifying gives us $N^2 + N - 4032 = 0.$ Factoring gives us $(N - 63)(N + 64) = 0$ We want the positive value so $N = 63.$ Adding together the digits gives us $9.$

Thus, the correct answer is D .

10.

一块地毯由如图所示的三种不同颜色组成。三种不同颜色区域的面积构成一个等差数列。最内层长方形宽为一英尺，外面的两个区域在四周各宽 $1$ 英尺。最内层长方形的长度是多少英尺？

A rug is made with three different colors as shown. The areas of the three differently colored regions form an arithmetic progression. The inner rectangle is one foot wide, and each of the two outer regions are $1$ foot wide on all four sides. What is the length in feet of the inner rectangle?

$1$

$2$

$4$

$6$

$8$

答案：B

知识点：面积等差数列

难度评级：1370

解答：

设内层长方形长度为 $l$ ，则内层面积为 $l$ 。

中间区域面积为 $(l + 2) \cdot 3 - l = 2l + 6.$ 外层区域面积为 $(l + 4) \cdot 5 - (l + 2) \cdot 3 = 2l + 14.$

这 $3$ 个面积构成等差数列，所以 $\begin{align*} l + 2l + 14 &= 2(2l + 6) \\ 3l + 14 &= 4l + 12 \\ l &= 2\end{align*}$

所以正确答案是 B。

Let $l$ be the length of the inner rectangle. Then the area of the inner rectangle is $l.$

The area of the middle region is going to be $(l + 2) \cdot 3 - l = 2l + 6.$ The area of the outer region is $(l + 4) \cdot 5 - (l + 2) \cdot 3 = 2l + 14.$

We know that these $3$ values form an arithmetic sequence. That means that $\begin{align*} l + 2l + 14 &= 2(2l + 6) \\ 3l + 14 &= 4l + 12 \\ l &= 2\end{align*}$

Thus, the correct answer is B .

11.

求阴影区域的面积。

Find the area of the shaded region.

$4\dfrac{3}{5}$

$5$

$5\dfrac{1}{4}$

$6\dfrac{1}{2}$

$8$

答案：D

知识点：面积分割三角形面积

难度评级：1070

解答：

可将阴影区域分成 $4$ 个底为 $1$ 的三角形。

其中两个三角形的高为 $8 \div 2 = 4$ ，另外两个三角形的高为 $5 \div 2 = \dfrac{5}{2}$ 。

面积总和为 $\begin{aligned} &2 \cdot \dfrac{1}{2} \left(1 \cdot \dfrac{5}{2}\right) \\ &\quad {}+ 2 \cdot \dfrac{1}{2} \left(1 \cdot 4\right) = 6\dfrac{1}{2} \end{aligned}$

所以正确答案是 D。

We can split the region into $4$ triangles with bases of $1.$

Two of the triangles have bases $8 \div 2 = 4$ and the other two have bases $5 \div 2 = \dfrac{5}{2}$

The sum of the areas of the triangles is $\begin{aligned} &2 \cdot \dfrac{1}{2} \left(1 \cdot \dfrac{5}{2}\right) \\ &\quad {}+ 2 \cdot \dfrac{1}{2} \left(1 \cdot 4\right) = 6\dfrac{1}{2} \end{aligned}$

Thus, the correct answer is D .

12.

从 $1$ 到 $2016$ 之间随机选取三个不同的整数。下列哪一句正确描述了这三个整数乘积为奇数的概率 $p$ ？

Three distinct integers are selected at random between $1$ and $2016,$ inclusive. Which of the following is a correct statement about the probability $p$ that the product of the three integers is odd?

$p \lt \dfrac{1}{8}$

$p = \dfrac{1}{8}$

$\dfrac{1}{8} \lt p \lt \dfrac{1}{3}$

$p = \dfrac{1}{3}$

$p \gt \dfrac{1}{3}$

答案：A

知识点：基本概率奇偶性无放回抽样

难度评级：1140

解答：

乘积为奇数，当且仅当选出的三个整数全是奇数。从 $1$ 到 $2016$ 中有 $1008$ 个奇数和 $1008$ 个偶数。

因为是不放回抽取， $p=\frac{1008}{2016}\cdot\frac{1007}{2015}\cdot\frac{1006}{2014}.$ 第一个因子为 $\frac12$ ，后两个因子都略小于 $\frac12$ ，所以 $p<\frac18$ 。

所以正确答案是 A。

The product is odd exactly when all three selected integers are odd. There are $1008$ odd and $1008$ even integers from $1$ to $2016$ .

Because the integers are selected without replacement, $p=\frac{1008}{2016}\cdot\frac{1007}{2015}\cdot\frac{1006}{2014}.$ The first factor is $\frac12$ , and each of the next two factors is slightly less than $\frac12$ . Therefore $p<\frac18$ .

Thus, the correct answer is A.

13.

五位朋友坐在电影院一排 $5$ 个座位中，从左到右编号为 $1$ 到 $5$ 。（“左”和“右”是从坐在座位上的人的视角看。）

电影期间 Ada 去大厅买爆米花。她回来时发现 Bea 向右移动了两个座位，Ceci 向左移动了一个座位，Dee 和 Edie 交换了座位，并且给 Ada 留下了一个端点座位。Ada 起身前坐在哪个座位？

Five friends sat in a movie theater in a row containing $5$ seats, numbered $1$ to $5$ from left to right. (The directions "left" and "right" are from the point of view of the people as they sit in the seats.)

During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had moved two seats to the right, Ceci had moved one seat to the left, and Dee and Edie had switched seats, leaving an end seat for Ada. In which seat had Ada been sitting before she got up?

$1$

$2$

$3$

$4$

$5$

答案：B

知识点：逻辑推理不变量

难度评级：1420

解答：

Dee 和 Edie 交换座位，不改变被占用座位的整体位置。

Bea 向右移动两个座位，Ceci 向左移动一个座位。

为了抵消 Bea 向右移动两个座位造成的总位移，Ada 也必须向左移动一个座位。

所有人的总位移必须为 $0.$

因此 Ada 原本坐在 $2$ 号座位，向左移动一位后坐到 $1$ 号端点座位。

所以正确答案是 B。

Note that Dee and Edie do not change the answers since their seats remain occupied throughout.

Since Bea moves to the right, this forces Ada and Ceci to move to offset this disruption.

Ceci only moves one seat, so Ada must also move one to cancel out the two seat shift that Bea did.

Bea moved to the right and Ceci moved to the left, so Ada must also move to the left to get a total displacement of $0.$

Therefore, Ada must have started off in seat $2$ to move one to the left to end up in seat $1.$

Thus, the correct answer is B .

14.

有多少种方法把 $2016$ 写成若干个二和若干个三的和，且不考虑顺序？例如， $1008\cdot 2 + 0\cdot 3$ 和 $402\cdot 2 + 404\cdot 3$ 是其中两种。

How many ways are there to write $2016$ as the sum of twos and threes, ignoring order? (For example, $1008\cdot 2 + 0\cdot 3$ and $402\cdot 2 + 404\cdot 3$ are two such ways.)

$236$

$336$

$337$

$403$

$672$

答案：C

知识点：丢番图方程奇偶性

难度评级：1280

解答：

题目可写成方程 $2x + 3y = 2016,$ 其中 $x$ 是二的个数， $y$ 是三的个数。

也就是数出从二〇一六中减去 $3$ 的倍数后，结果为偶数的情形数。

这对应于从 $(1008, 0)$ 到 $(0, 672)$ 的所有有序对，其中 $y$ 每次增加 $2$ 。

这样的 $y$ 和 $x$ 共有 $\dfrac{672}{2} + 1 = 337$ 组解。

所以正确答案是 C。

The problem can be rewritten as an equation $2x + 3y = 2016,$ where $x$ is the number of twos and $y$ is the number of threes.

The goal is to find the number of multiples of $3$ that can be subtracted from 2016 to result in an even number.

This can be achieved by the pairs of $(1008, 0)$ up to $(0, 672)$ with $y$ being incremented by $2.$

This gives us $\dfrac{672}{2} + 1 = 337$ solutions for $y$ and $x.$

Thus, the correct answer is C .

15.

如图，从一块圆形饼干面团中切出七块半径为 $1$ 英寸的饼干。相邻饼干相切，除中心饼干外，其他饼干都与面团边缘相切。剩下的边角料被重新塑造成另一块同样厚度的饼干。边角料饼干的半径是多少英寸？

Seven cookies of radius $1$ inch are cut from a circle of cookie dough, as shown. Neighboring cookies are tangent, and all except the center cookie are tangent to the edge of the dough. The leftover scrap is reshaped to form another cookie of the same thickness. What is the radius in inches of the scrap cookie?

$\sqrt{2}$

$1.5$

$\sqrt{\pi}$

$\sqrt{2\pi}$

$\pi$

答案：A

知识点：圆面积相切圆

难度评级：1140

解答：

圆形面团半径为 $3$ 英寸，因为从中心到边缘经过一个饼干直径再加一个饼干半径。

面团面积为 $3^2\pi = 9\pi$ ，七块饼干的总面积为 $7 \cdot 1^2\pi = 7\pi$ 。

剩余边角料的面积为 $9\pi - 7\pi = 2\pi$ ，所以它的半径为 $\sqrt{2}$ 。

所以正确答案是 A。

The circle of cookie dough has a radius of $3$ inches since it is the same as the diameter plus the radius of a cookie.

The area of the cookie dough is $3^2\pi = 9\pi,$ and the cookies have an area of $7 \cdot 1^2\pi = 7\pi.$

The area of the leftover scrap is therefore $9\pi - 7\pi = 2\pi.$ This means that its radius is $\sqrt{2}.$

Thus, the correct answer is A .

16.

顶点为 $A(0, 2), B(-3, 2)$ ，和 $C(-3, 0)$ 的三角形先关于 $x$ 轴反射，得到的像 $\triangle A'B'C'$ 再绕原点逆时针旋转 $90^{\circ}$ ，得到 $\triangle A''B''C''$ 。下列哪个变换能把 $\triangle A''B''C''$ 变回 $\triangle ABC$ ？

A triangle with vertices $A(0, 2), B(-3, 2),$ and $C(-3, 0)$ is reflected about the $x$ -axis, then the image $\triangle A'B'C'$ is rotated counterclockwise about the origin by $90^{\circ}$ to produce $\triangle A''B''C''.$ Which of the following transformations will return $\triangle A''B''C''$ to $\triangle ABC?$

绕原点逆时针旋转 $90^{\circ}$

counterclockwise rotation about the origin by $90^{\circ}$

绕原点顺时针旋转 $90^{\circ}$

clockwise rotation about the origin by $90^{\circ}$

关于 $x$ 轴反射

reflection about the $x$ -axis

关于直线 $y = x$ 反射

reflection about the line $y = x$

关于 $y$ 轴反射

reflection about the $y$ -axis

答案：D

知识点：变换坐标几何

难度评级：1220

解答：

为判断如何逆转这些变换，可以跟踪一个任意点。

设该点为 $(x, y)$ 。关于 $x$ 轴反射后，它变为 $(x, -y)$ 。

再逆时针旋转后得到 $(y, x)$ 。要把它变回 $(x, y)$ ，唯一的方法是关于 $y = x$ 反射。

所以正确答案是 D。

To figure out how to reverse the transformations, we can analyze a single point and see what happens to it.

Let $(x, y)$ be the point. After being reflected about the $x$ -axis, the point would go to $(x, -y).$

Rotating this counterclockwise would put it at $(y, x).$ The only transformation that puts this back at $(x, y)$ is reflection about $y = x.$

Thus, the correct answer is D .

17.

设 $N$ 是 $5$ 的正倍数。将一个红球和 $N$ 个绿球随机排成一行。令 $P(N)$ 为至少有 $\frac{3}{5}$ 的绿球在红球同一侧的概率。注意 $P(5)=1$ ，且当 $N$ 变大时 $P(N)$ 趋近于 $\frac{4}{5}$ 。使 $P(N) < \dfrac{321}{400}$ 的最小 $N$ 的各位数字之和是多少？

Let $N$ be a positive multiple of $5.$ One red ball and $N$ green balls are arranged in a line in random order. Let $P(N)$ be the probability that at least $\frac{3}{5}$ of the green balls are on the same side of the red ball. Observe that $P(5)=1$ and that $P(N)$ approaches $\frac{4}{5}$ as $N$ grows large. What is the sum of the digits of the least value of $N$ such that $P(N) < \dfrac{321}{400}?$

$12$

$14$

$16$

$18$

$20$

答案：A

知识点：基本概率不等式

难度评级：1790

解答：

先排好 $N$ 个绿球，再把红球放入 $N+1$ 个空隙之一。若红球左侧有 $k$ 个绿球，则右侧有 $N-k$ 个。

至少有 $\frac35N$ 个绿球在一侧，等价于 $k\le\frac25N$ 或 $k\ge\frac35N$ 。因此不满足条件的空隙为 $\frac25N+1,\frac25N+2,\ldots,\frac35N-1,$ 共 $\frac15N-1$ 个。

所以 $\begin{aligned} P(N) &= 1-\frac{N/5-1}{N+1} \\ &= \frac{4N+10}{5N+5}. \end{aligned}$ 解不等式 $\frac{4N+10}{5N+5}<\frac{321}{400}$ 得 $N>479$ 。最小的 $5$ 的正倍数是 $480$ ，其各位数字之和为 $12$ 。

所以正确答案是 A。

Think of first arranging the $N$ green balls, then placing the red ball in one of the $N+1$ gaps. If $k$ green balls are to the left of the red ball, then $N-k$ are to its right.

At least $\frac35N$ green balls are on one side exactly when $k\le\frac25N$ or $k\ge\frac35N$ . Thus the bad gaps are $\frac25N+1,\frac25N+2,\ldots,\frac35N-1,$ a total of $\frac15N-1$ gaps.

Therefore $\begin{aligned} P(N) &= 1-\frac{N/5-1}{N+1} \\ &= \frac{4N+10}{5N+5}. \end{aligned}$ Solving $\frac{4N+10}{5N+5}<\frac{321}{400}$ gives $N>479$ . The least positive multiple of $5$ is $480$ , whose digit sum is $12$ .

Thus, the correct answer is A.

18.

一个立方体的每个顶点要标上从 $1$ 到 $8$ 的一个整数，每个整数用一次，使得每个面的四个顶点上的数之和都相同。通过立方体旋转可以互相得到的标法视为相同。有多少种不同标法？

Each vertex of a cube is to be labeled with an integer $1$ through $8,$ with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?

$1$

$3$

$6$

$12$

$24$

答案：C

知识点：正方体分类讨论对称性

难度评级：2060

解答：

相对两个面包含全部八个标号，总和为 $36$ ，所以每个面的和必须为 $18$ 。

把标号 $1$ 放在一个顶点，设与它相邻的三个标号为 $a,b,c$ 。由每个面和为十八，另三个相邻顶点被迫为 $\begin{gathered} 17-a-b, \\ \quad 17-a-c, \\ \quad 17-b-c, \end{gathered}$ 而对顶点为 $a+b+c-16$ 。

假设三个相邻标号按递增顺序排列。检查从 $2,3,\ldots,8$ 中选三数的可能，合法三元组为 $(4,6,8)$ 、 $(4,7,8)$ 、 $(6,7,8)$ 。每个三元组给出两种不能通过旋转互相得到的标法，所以共有 $3\cdot2=6$ 种标法。

所以正确答案是 C。

Opposite faces together use all eight labels, whose sum is $36$ , so every face must have sum $18$ .

Put label $1$ at one vertex, and let the three adjacent labels be $a,b,c$ . The three vertices adjacent to those across faces are then forced to be $\begin{gathered} 17-a-b, \\ \quad 17-a-c, \\ \quad 17-b-c, \end{gathered}$ and the opposite vertex is $a+b+c-16$ .

Assume the three neighbor labels are listed in increasing order. Checking possible triples from $2,3,\ldots,8$ , the valid triples are $(4,6,8)$ , $(4,7,8)$ , and $(6,7,8)$ . Each gives two non-rotationally equivalent arrangements, so there are $3\cdot2=6$ arrangements.

Thus, the correct answer is C.

19.

在长方形 $ABCD$ 中， $AB=6$ ， $BC=3$ 。点 $E$ 在 $B$ 与 $C$ 之间，点 $F$ 在 $E$ 与 $C$ 之间，且 $BE=EF=FC$ 。线段 $\overline{AE}$ 和 $\overline{AF}$ 分别与 $\overline{BD}$ 相交于 $P$ 和 $Q$ 。

比值 $BP:PQ:QD$ 可写成 $r:s:t$ ，其中 $r,s,$ 和 $t$ 的最大公因数为 $1$ 。求 $r+s+t$ 。

In rectangle $ABCD,$ $AB=6$ and $BC=3.$ Point $E$ between $B$ and $C,$ and point $F$ between $E$ and $C$ are such that $BE=EF=FC.$ Segments $\overline{AE}$ and $\overline{AF}$ intersect $\overline{BD}$ at $P$ and $Q,$ respectively.

The ratio $BP:PQ:QD$ can be written as $r:s:t$ where the greatest common factor of $r,s,$ and $t$ is $1.$ What is $r+s+t?$

$7$

$9$

$12$

$15$

$20$

答案：E

知识点：相似矩形

难度评级：1720

解答：

因为 $BC=3$ ，且 $BE=EF=FC$ ，所以 $BE=1$ 、 $BF=2$ 。又 $AD=3$ 。

由 $\triangle APD\sim\triangle EPB$ ， $\frac{BP}{BD}=\frac{BE}{AD+BE}=\frac14.$ 由 $\triangle AQD\sim\triangle FQB$ ， $\frac{BQ}{BD}=\frac{BF}{AD+BF}=\frac25.$

因此 $BP=\frac14BD$ ， $PQ=\left(\frac25-\frac14\right)BD=\frac3{20}BD$ ，且 $QD=\frac35BD$ 。于是 $\begin{aligned} BP:PQ:QD &= \frac14:\frac3{20}:\frac35 \\ &= 5:3:12. \end{aligned}$ 和为 $5+3+12=20$ 。

所以正确答案是 E。

Since $BC=3$ and $BE=EF=FC$ , we have $BE=1$ and $BF=2$ . Also $AD=3$ .

From $\triangle APD\sim\triangle EPB$ , $\frac{BP}{BD}=\frac{BE}{AD+BE}=\frac14.$ From $\triangle AQD\sim\triangle FQB$ , $\frac{BQ}{BD}=\frac{BF}{AD+BF}=\frac25.$

Therefore $BP=\frac14BD$ , $PQ=\left(\frac25-\frac14\right)BD=\frac3{20}BD$ , and $QD=\frac35BD$ . Thus $\begin{aligned} BP:PQ:QD &= \frac14:\frac3{20}:\frac35 \\ &= 5:3:12. \end{aligned}$ The sum is $5+3+12=20$ .

Thus, the correct answer is E.

20.

对某个特定的 $N$ ，展开 $(a+b+c+d+1)^N$ 并合并同类项后，所得表达式中恰好有 $1001$ 项同时包含四个变量 $a$ 、 $b$ 、 $c$ 和 $d$ ，且每个变量的指数都为正。求 $N$ 。

For some particular value of $N,$ when $(a+b+c+d+1)^N$ is expanded and like terms are combined, the resulting expression contains exactly $1001$ terms that include all four variables $a,$ $b,$ $c,$ and $d,$ each to some positive power. What is $N?$

$9$

$14$

$16$

$17$

$19$

答案：B

知识点：隔板法组合

难度评级：1660

解答：

同时包含四个变量的项形如 $a^vb^wc^xd^y$ ，其中 $v,w,x,y\ge1$ ，还可乘以 $1$ 的某个幂 $z$ 。 $v+w+x+y+z=N.$

从 $v,w,x,y$ 各减去 $1$ 后，得到 $5$ 个非负整数，其和为 $N-4$ 。因此这样的项数为 $\binom{N}{4}.$

由于 $\binom{14}{4}=1001$ ，所以 $N=14$ 。

所以正确答案是 B。

A term that includes all four variables has form $a^vb^wc^xd^y$ , with $v,w,x,y\ge1$ , multiplied by some power of $1$ . If that power is $z$ , then $v+w+x+y+z=N.$

After subtracting $1$ from each of $v,w,x,y$ , this becomes a nonnegative stars-and-bars count with $5$ variables summing to $N-4$ . The number of such terms is $\binom{N}{4}.$

Since $\binom{14}{4}=1001$ , we get $N=14$ .

Thus, the correct answer is B.

21.

圆心为 $P, Q$ 和 $R$ 、半径分别为 $1, 2$ 和 $3$ 的三个圆位于直线 $l$ 的同侧，并分别在 $P', Q'$ 和 $R'$ 处与 $l$ 相切，其中 $Q'$ 在 $P'$ 和 $R'$ 之间。圆心为 $Q$ 的圆与另外两个圆都外切。求 $\triangle PQR$ 的面积。

Circles with centers $P, Q$ and $R,$ having radii $1, 2$ and $3,$ respectively, lie on the same side of line $l$ and are tangent to $l$ at $P', Q'$ and $R',$ respectively, with $Q'$ between $P'$ and $R'.$ The circle with center $Q$ is externally tangent to each of the other two circles. What is the area of triangle $\triangle PQR?$

$0$

$\sqrt{\dfrac{2}{3}}$

$1$

$\sqrt{6}-\sqrt{2}$

$\sqrt{\dfrac{3}{2}}$

答案：D

知识点：相切圆勾股定理梯形

难度评级：1970

解答：

由勾股定理，得到 $P'Q' = \sqrt{3^2 - 1^2} = 2\sqrt{2}$ 以及 $Q'R' = \sqrt{5^2 - 1^2} = 2\sqrt{6}.$

这是因为 $PQ = 1 + 2 = 3$ ， $QR = 2 + 3 = 5$ 。这些三角形的高也都是 $1$ 。

因此 $[Q'QPP'] = \dfrac{1}{2}(1 + 2)2\sqrt{2}$ $= 3\sqrt{2}.$

$[R'RQQ'] = \dfrac{1}{2}(2 + 3)2\sqrt{6}$ $= 5\sqrt{6}.$

$[R'RPP'] =$ $\dfrac{1}{2}(1 + 3)(2\sqrt{2} + 2\sqrt{6})$ $= 4\sqrt{2} + 4\sqrt{6}.$

现在可以把 $[PQR]$ 表示为 $[Q'QPP'] + [R'RQQ']$ $- [R'RPP'].$

$3\sqrt{2} + 5\sqrt{6} - 4\sqrt{2} - 4\sqrt{6}$ $= \sqrt{6} - \sqrt{2}.$

所以正确答案是 D。

Using the Pythagorean theorem, we get that $P'Q' = \sqrt{3^2 - 1^2} = 2\sqrt{2}$ and $Q'R' = \sqrt{5^2 - 1^2} = 2\sqrt{6}.$

This follows from $PQ = 1 + 2 = 3$ and $QR = 2 + 3 = 5.$ The heights of the triangles are also just $1.$

Then, we get that $[Q'QPP'] = \dfrac{1}{2}(1 + 2)2\sqrt{2}$ $= 3\sqrt{2}.$

We also get that $[R'RQQ'] = \dfrac{1}{2}(2 + 3)2\sqrt{6}$ $= 5\sqrt{6}.$

Finally, we have that $[R'RPP'] =$ $\dfrac{1}{2}(1 + 3)(2\sqrt{2} + 2\sqrt{6})$ $= 4\sqrt{2} + 4\sqrt{6}.$

Now, we can express $[PQR]$ as $[Q'QPP'] + [R'RQQ']$ $- [R'RPP'].$

This evaluates to $3\sqrt{2} + 5\sqrt{6} - 4\sqrt{2} - 4\sqrt{6}$ $= \sqrt{6} - \sqrt{2}.$

Thus, the correct answer is D .

22.

对某个正整数 $n$ ，数 $110n^3$ 有 $110$ 个正整数因数，包括 $1$ 和 $110n^3$ 本身。数 $81n^4$ 有多少个正整数因数？

For some positive integer $n,$ the number $110n^3$ has $110$ positive integer divisors, including $1$ and the number $110n^3.$ How many positive integer divisors does the number $81n^4$ have?

$110$

$191$

$261$

$325$

$425$

答案：D

知识点：因数个数质因数分解

难度评级：2110

解答：

质因数分解为 $110=2\cdot5\cdot11$ 。若 $110n^3$ 有 $110=2\cdot5\cdot11$ 个因数，则它恰有三个素因子，指数为 $1,4,10$ 的某种排列。

对于素数 $2,5,11$ ，它们在 $110n^3$ 中的指数都是 $1$ 加上 $3$ 的倍数。指数 $1,4,10$ 都满足这个形式，所以它们在 $n$ 中对应为 $0,1,3$ 的某种排列。

在 $81n^4=3^4n^4$ 中，来自 $n^4$ 的指数为 $0,4,12$ ，另有素数 $3$ 的指数 $4$ 。因数个数为 $\begin{aligned} &(0+1)\cdot(4+1)\cdot(12+1) \\ &\quad {}\cdot(4+1)=325. \end{aligned}$

所以正确答案是 D。

The prime factorization is $110=2\cdot5\cdot11$ . If $110n^3$ has $110=2\cdot5\cdot11$ divisors, then it has exactly three prime factors, with exponents $1,4,10$ in some order.

For each of the primes $2,5,11$ , its exponent in $110n^3$ is $1$ more than a multiple of $3$ . The exponents $1,4,10$ all have this form, so the corresponding exponents in $n$ are $0,1,3$ in some order.

In $81n^4=3^4n^4$ , the exponents from $n^4$ are therefore $0,4,12$ , in some order, along with the exponent $4$ on prime $3$ . Hence the divisor count is $\begin{aligned} &(0+1)\cdot(4+1)\cdot(12+1) \\ &\quad {}\cdot(4+1)=325. \end{aligned}$

Thus, the correct answer is D.

23.

二元运算 $\diamondsuit$ 满足 $a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c$ 以及 $a\,\diamondsuit \,a=1$ ，对所有非零实数 $a, b$ 和 $c$ 都成立。（这里 $\cdot$ 表示乘法。）方程 $2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100$ 的解可写成 $\frac{p}{q}$ ，其中 $p$ 和 $q$ 为互质正整数。求 $p+q$ 的值。

A binary operation $\diamondsuit$ has the properties that $a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c$ and that $a\,\diamondsuit \,a=1$ for all nonzero real numbers $a, b,$ and $c.$ (Here $\cdot$ represents multiplication). The solution to the equation $2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100$ can be written as $\frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. What is $p+q?$

$109$

$201$

$301$

$3049$

$33,601$

答案：A

知识点：自定义运算函数方程

难度评级：1820

解答：

因为 $a\diamondsuit a=1$ ，在 $a\diamondsuit(b\diamondsuit c)=(a\diamondsuit b)c$ 中令 $b=c$ ，得 $a\diamondsuit1=(a\diamondsuit b)b$ 。又由 $a\diamondsuit(a\diamondsuit a)=(a\diamondsuit a)a$ ，可得 $a\diamondsuit1=a$ ，因此 $a\diamondsuit b=\frac ab$ 。

因此代入方程得到 $\begin{aligned} 2016\diamondsuit(6\diamondsuit x) &= 2016\diamondsuit\frac6x \\ &= \frac{2016}{6/x} \\ &= 336x \\ &= 100. \end{aligned}$ 所以 $x=\frac{25}{84}$ ， $p+q=25+84=109$ 。

所以正确答案是 A。

Since $a\diamondsuit a=1$ , substituting $b=c$ in $a\diamondsuit(b\diamondsuit c)=(a\diamondsuit b)c$ gives $a\diamondsuit1=(a\diamondsuit b)b$ . Also, using $a\diamondsuit(a\diamondsuit a)=(a\diamondsuit a)a$ gives $a\diamondsuit1=a$ . Therefore $a\diamondsuit b=\frac ab$ .

The equation becomes $\begin{aligned} 2016\diamondsuit(6\diamondsuit x) &= 2016\diamondsuit\frac6x \\ &= \frac{2016}{6/x} \\ &= 336x \\ &= 100. \end{aligned}$ Thus $x=\frac{25}{84}$ , so $p+q=25+84=109$ .

Thus, the correct answer is A.

24.

一个四边形内接于半径为 $200\sqrt{2}$ 的圆。该四边形的三条边长为 $200$ 。第四条边长是多少？

A quadrilateral is inscribed in a circle of radius $200\sqrt{2}.$ Three of the sides of this quadrilateral have length $200.$ What is the length of the fourth side?

$200$

$200\sqrt{2}$

$200\sqrt{3}$

$300\sqrt{2}$

$500$

答案：E

知识点：圆内接四边形弦相似

难度评级：2300

解答：

设 $O$ 为圆心，令 $AD$ 与 $OB$ 、 $OC$ 分别交于 $E$ 、 $F$ 。等长弦 $AB,BC,CD$ 对应相等圆心角。

由相等角关系， $\triangle OAB\sim\triangle ABE$ 。因为 $OA=OB=200\sqrt{2}$ ，且 $AB=200$ ，相似关系给出 $AE=AB=200$ ， $BE=100\sqrt{2}$ 。同理 $FD=CD=200$ 。

在三角形 $OBC$ 中，同样的比例给出 $EF=\frac12BC=100$ 。因此 $\begin{aligned} AD &= AE+EF+FD \\ &= 200+100+200 \\ &= 500. \end{aligned}$

所以正确答案是 E。

Let $O$ be the circle's center, and let $AD$ meet $OB$ and $OC$ at $E$ and $F$ . Equal chords $AB,BC,CD$ give equal central angles.

From the equal-angle relationships, $\triangle OAB\sim\triangle ABE$ . Since $OA=OB=200\sqrt{2}$ and $AB=200$ , the similarity gives $AE=AB=200$ and $BE=100\sqrt{2}$ . Similarly, $FD=CD=200$ .

In triangle $OBC$ , the same scaling gives $EF=\frac12BC=100$ . Therefore $\begin{aligned} AD &= AE+EF+FD \\ &= 200+100+200 \\ &= 500. \end{aligned}$

Thus, the correct answer is E.

25.

有多少个正整数有序三元组 $(x,y,z)$ 同时满足 $\text{lcm}(x,y) = 72,$ $\text{lcm}(x,z) = 600,$ 以及 $\text{lcm}(y,z)=900$ ？

How many ordered triples $(x,y,z)$ of positive integers satisfy $\text{lcm}(x,y) = 72,$ $\text{lcm}(x,z) = 600,$ and $\text{lcm}(y,z)=900?$

$15$

$16$

$24$

$27$

$64$

答案：A

知识点：最小公倍数质因数分解分类讨论

难度评级：2390

解答：

先分解给定的最小公倍数： $\begin{gathered} 72=2^3\cdot3^2, \\ \quad 600=2^3\cdot3\cdot5^2, \\ \quad 900=2^2\cdot3^2\cdot5^2. \end{gathered}$ 因为 $\operatorname{lcm}(x,y)$ 没有因子 $5$ ，所以 $x$ 和 $y$ 都没有因子 $5$ ，而 $z$ 必须含 $5^2$ 。

写 $x=2^a3^b$ ， $y=2^c3^d$ ， $z=2^e3^f5^2$ 。 $\operatorname{lcm}(y,z)$ 中 $3$ 的指数迫使 $d=2$ ， $\operatorname{lcm}(x,z)$ 中 $2$ 的指数迫使 $a=3$ 。

剩余独立条件为 $\max(b,f)=1$ 和 $\max(c,e)=2$ 。前者有 $3$ 个有序选择，后者有 $5$ 个有序选择，所以共有 $3\cdot5=15$ 个三元组。

所以正确答案是 A。

Factor the given least common multiples: $\begin{gathered} 72=2^3\cdot3^2, \\ \quad 600=2^3\cdot3\cdot5^2, \\ \quad 900=2^2\cdot3^2\cdot5^2. \end{gathered}$ Since $\operatorname{lcm}(x,y)$ has no factor of $5$ , neither $x$ nor $y$ has a factor of $5$ , and $z$ must contain $5^2$ .

Write $x=2^a3^b$ , $y=2^c3^d$ , and $z=2^e3^f5^2$ . The power of $3$ in $\operatorname{lcm}(y,z)$ forces $d=2$ , and the power of $2$ in $\operatorname{lcm}(x,z)$ forces $a=3$ .

The remaining independent conditions are $\max(b,f)=1$ and $\max(c,e)=2$ . These have $3$ and $5$ ordered choices, respectively, so there are $3\cdot5=15$ triples.

Thus, the correct answer is A.