2011 AMC 10A 真题与在线限时测验

计时

1:15:00

1.

一个手机套餐每月费用为 $\$20$ ，外加每条短信 $5$ ¢，以及每分钟 $10$ ¢ 的超时通话费（超过 $30$ 小时的部分）。Michelle 一月份发送了 $100$ 条短信，并通话 $30.5$ 小时。她需要支付多少钱？

A cell phone plan costs $\$20$ each month, plus $5$ ¢ per text message sent, plus $10$ ¢ for each minute used over $30$ hours. In January Michelle sent $100$ text messages and talked for $30.5$ hours. How much did she have to pay?

$\$24.00$

$\$24.50$

$\$25.50$

$\$28.00$

$\$30.00$

答案：D

知识点：钱币单位换算

难度评级：720

小提示：

把短信费用和超时通话费用都换算成美元。

Convert the texting charge and excess-call charge into dollars

大提示：

只有多出的半小时按分钟收费。

Only the extra half hour is billed by the minute

解答：

月费为 $\$20$ 。短信费用为 $100\cdot 5=500$ 美分，即 $\$5$ 。她还多通话 $0.5$ 小时，即 $30$ 分钟，超出了套餐包含的 $30$ 小时。

因此超时通话费用为 $30\cdot10=300$ 美分，即 $\$3$ 。

总费用为 $\$20 + \$5 + \$3 = \$28\text{。}$

所以正确答案是 D。

Michelle has to pay $\$20$ for the monthly fee. Her text messages cost $100\cdot 5=500$ cents, or $\$5.$ Finally she talked for $0.5$ hours, $30$ minutes, over $30$ hours.

This means her extra charge is $30\cdot10=300$ cents, or $\$3.$

Her total cost is $\$20 + \$5 + \$3 = \$28.$

Thus, D is the correct answer.

2.

一个小洗发水瓶可装 $35$ 毫升洗发水，而一个大瓶可装 $500$ 毫升。Jasmine 想购买最少数量的小瓶，足以完全装满一个大瓶。她必须买多少瓶？

A small bottle of shampoo can hold $35$ milliliters of shampoo, whereas a large bottle can hold $500$ milliliters of shampoo. Jasmine wants to buy the minimum number of small bottles necessary to completely fill a large bottle. How many bottles must she buy?

$11$

$12$

$13$

$14$

$15$

答案：E

知识点：整除性估算

难度评级：660

小提示：

比较 $35$ 的倍数和 $500$ 。

Compare multiples of $35$ to $500$

大提示：

$35$ 的倍数中，第一个达到 $500$ 的值给出所需瓶数。

The first multiple of $35$ that reaches $500$ gives the number of bottles

解答：

所需瓶数为 $\dfrac{500}{35} = \dfrac{100}{7} = 14 \dfrac{2}{7}\text{。}$

因此最少必须购买 $15$ 个小瓶。

所以正确答案是 E。

The desired amount is $\dfrac{500}{35} = \dfrac{100}{7} = 14 \dfrac{2}{7}.$

This means that the smallest number of small bottles she must buy is $15.$

Thus, E is the correct answer.

3.

假设 $[a\ b]$ 表示 $a$ 和 $b$ 的平均数，而 $\{a\ b\ c\}$ 表示 $a$ 、 $b$ 和 $c$ 的平均数。求下式的值： $\{\{1 \ 1 \ 0\} \ [0 \ 1] \ 0\}$

Suppose $[a\ b]$ denotes the average of $a$ and $b,$ and $\{a\ b\ c\}$ denotes the average of $a,$ $b,$ and $c.$ What is the value of the following expression? $\{\{1 \ 1 \ 0\} \ [0 \ 1] \ 0\}$

$\dfrac{2}{9}$

$\dfrac{5}{18}$

$\dfrac{1}{3}$

$\dfrac{7}{18}$

$\dfrac{2}{3}$

答案：D

知识点：自定义运算平均数分数

难度评级：870

小提示：

先计算内部的平均数。

Evaluate the inner averages first

大提示：

使用 $\{1\ 1\ 0\}=\dfrac23$ 和 $[0\ 1]=\dfrac12$ 。

Use $\{1\ 1\ 0\}=\dfrac23$ and $[0\ 1]=\dfrac12$

解答：

先计算得 $\{1 \ 1 \ 0\} = \dfrac{1 + 1 + 0}{3} = \dfrac{2}{3}\text{。}$

另一个内层平均数为 $[0 \ 1] = \dfrac{1 + 0}{2} = \dfrac{1}{2}\text{。}$

最后得 $\left\{\dfrac{2}{3} \ \dfrac{1}{2} \ 0\right\} = \dfrac{\dfrac{2}{3} + \dfrac{1}{2} + 0}{3} = \dfrac{7}{18}\text{。}$

所以正确答案是 D。

We have that $\{1 \ 1 \ 0\} = \dfrac{1 + 1 + 0}{3} = \dfrac{2}{3}.$

We also get that $[0 \ 1] = \dfrac{1 + 0}{2} = \dfrac{1}{2}.$

Finally, $\left\{\dfrac{2}{3} \ \dfrac{1}{2} \ 0\right\} = \dfrac{\dfrac{2}{3} + \dfrac{1}{2} + 0}{3} = \dfrac{7}{18}.$

Thus, D is the correct answer.

4.

令 $X$ 和 $Y$ 为下列等差数列的和： $\begin{aligned} X &= 10+12+14+\cdots+100,\\ Y &= 12+14+16+\cdots+102 \end{aligned}\text{。}$ $Y - X$ 的值是多少？

Let $X$ and $Y$ be the following sums of arithmetic sequences: $\begin{aligned} X &= 10+12+14+\cdots+100,\\ Y &= 12+14+16+\cdots+102. \end{aligned}$ What is the value of $Y - X?$

$92$

$98$

$100$

$102$

$112$

答案：A

知识点：等差数列求和

难度评级：770

小提示：

两个和中大部分项会抵消。

Most terms in the two sums cancel

大提示：

只有 $X$ 的第一项和 $Y$ 的最后一项不匹配。

Only the first term of $X$ and the last term of $Y$ do not match

解答：

从 $12$ 到 $100$ 的所有偶数项在两个和中都出现，作差时会全部抵消。

因此 $Y - X = 102 - 10 = 92\text{。}$

所以正确答案是 A。

Note that the terms $12$ through $100$ are common to both sums. When we subtract, all these terms cancel out.

This means that $Y - X = 102 - 10 = 92.$

Thus, A is the correct answer.

5.

在一所小学，三年级、四年级、五年级学生每天平均分别跑 $12$ 、 $15$ 、 $10$ 分钟。三年级学生人数是四年级的两倍，四年级学生人数是五年级的两倍。这些学生每天平均跑多少分钟？

At an elementary school, the students in third grade, fourth grade, and fifth grade run an average of $12,$ $15,$ and $10$ minutes per day, respectively. There are twice as many third graders as fourth graders, and twice as many fourth graders as fifth graders. What is the average number of minutes run per day by these students?

$12$

$\dfrac{37}{3}$

$\dfrac{88}{7}$

$13$

$14$

答案：C

知识点：平均数比与比例

难度评级：1070

小提示：

三、四、五年级人数比为 $4:2:1$ 。

Use a $4:2:1$ ratio for the numbers of third, fourth, and fifth graders

大提示：

用权重 $4,2,1$ 计算加权平均。

Compute a weighted average with weights $4,2,1$

解答：

不妨设五年级有一人，则四年级有两人，三年级有四人。

这样做可以，因为平均数不受学生人数整体倍数的影响。

总跑步时间为 $12 \cdot 4 + 15 \cdot 2 + 10 = 88$ 分钟。总人数为 $1 + 2 + 4 = 7$ ，所以平均数为 $\dfrac{88}{7}$ 。

所以正确答案是 C。

WLOG, let there be one fifth grader. This then tells us that there are two fourth graders and four third graders.

We can do this, since we are only interested in the average, which is not impacted by the exact number of students.

The total number of minutes the students spend running is $12 \cdot 4 + 15 \cdot 2 + 10 = 88$ minutes. The total number of students is $1 + 2 + 4 = 7.$ The average is then $\dfrac{88}{7}.$

Thus, C is the correct answer.

6.

集合 $A$ 有 $20$ 个元素，集合 $B$ 有 $15$ 个元素。 $A \cup B$ ，即 $A$ 和 $B$ 的并集，最少可能有多少个元素？

Set $A$ has $20$ elements, and set $B$ has $15$ elements. What is the smallest possible number of elements in $A \cup B,$ the union of $A$ and $B?$

$5$

$15$

$20$

$35$

$300$

答案：C

知识点：韦恩图极限情形界定

难度评级：660

小提示：

并集必须包含较大集合的所有元素。

The union must contain all elements of the larger set

大提示：

让 $B$ 成为 $A$ 的子集可使并集最小。

Make $B$ a subset of $A$ to minimize the union

解答：

要使并集元素数最少，就要让两个集合重叠尽可能多。

可以让 $B$ 完全包含于 $A$ ，此时并集就是 $A$ ，共有 $20$ 个元素。

所以正确答案是 C。

To minimize the number of elements in the union, we want to maximize the overlap between the two sets.

We can then assume that $B$ is contained completely within $A,$ which means that the union is the same as $A,$ which has $20$ elements.

Thus, C is the correct answer.

7.

下列哪个方程没有解？

Which of the following equations does not have a solution?

$(x + 7)^2 = 0$

$|-3x| + 5 = 0$

$\sqrt{-x} - 2 = 0$

$\sqrt{x} - 8 = 0$

$|-3x| - 4 = 0$

答案：B

知识点：绝对值根式

难度评级：960

小提示：

找出哪个方程要求一个非负表达式等于负数。

Check which equation asks a nonnegative expression to be negative

大提示：

绝对值加 $5$ 不可能等于 $0$ 。

An absolute value plus $5$ cannot equal $0$

解答：

选项 A 化为 $x + 7 = 0$ $x = -7\text{，}$ 因此有解。

选项 B 化为 $|-3x| = -5\text{，}$ 绝对值不可能为负数，因此无解。

其他选项都有解，如下面计算所示。

选项 C 化为 $\sqrt{-x} = 2$ $-x = 4$ $x = -4\text{，}$ 所以有解。

选项 D 化为 $\sqrt{x} = 8$ $x = 64\text{，}$ 所以也有解。

最后，选项 E 化为 $|-3x| = 4$ $-3x = \pm 4$ $x = \pm \dfrac{4}{3}\text{，}$ 同样有解。

所以正确答案是 B。

A simplifies to $x + 7 = 0$ $x = -7,$ so it has a solution.

B simplifies to $|-3x| = -5,$ which has no solution since absolute value makes everything positive.

Let us make sure that all the other choices have solutions.

C simplifies to $\sqrt{-x} = 2$ $-x = 4$ $x = -4,$ which is fine.

D simplifies to $\sqrt{x} = 8$ $x = 64,$ which works.

Finally, E simplifies to $|-3x| = 4$ $-3x = \pm 4$ $x = \pm \dfrac{4}{3},$ which has a solution as well.

Thus, B is the correct answer.

8.

去年夏天，Town Lake 上的鸟中 $30 \%$ 是鹅， $25 \%$ 是天鹅， $10 \%$ 是鹭， $35 \%$ 是鸭。不是天鹅的鸟中，百分之多少是鹅？

Last summer $30 \%$ of the birds living on Town Lake were geese, $25 \%$ were swans, $10 \%$ were herons, and $35 \%$ were ducks. What percent of the birds that were not swans were geese?

$20$

$30$

$40$

$50$

$60$

答案：C

知识点：百分数比与比例

难度评级：870

小提示：

用不是天鹅的百分比作分母。

Use the percent that are not swans as the denominator

大提示：

所求百分比为 $\frac{30}{100-25}$ 。

The desired percent is $\frac{30}{100-25}$

解答：

不妨设共有 $100$ 只鸟，则不是天鹅的有 $75$ 只。所求百分比为 $\dfrac{30}{75} \cdot 100 = 40 \%\text{。}$

所以正确答案是 C。

WLOG, let there be $100$ birds. Then $75$ birds are not swans. The desired percentage is then $\dfrac{30}{75} \cdot 100 = 40 \%.$

Thus, C is the correct answer.

9.

一个长方形区域由直线 $y=a$ ， $y=-b$ ， $x=-c$ 和 $x=d$ 围成，其中 $a$ ， $b$ ， $c$ 和 $d$ 都是正数。下列哪一项表示该区域的面积？

A rectangular region is bounded by the graphs of the equations $y=a,$ $y=-b,$ $x=-c,$ and $x=d,$ where $a,$ $b,$ $c,$ and $d$ are all positive numbers. Which of the following represents the area of this region?

$ac+ad+bc+bd$

$ac-ad+bc-bd$

$ac+ad-bc-bd$

$-ac-ad+bc+bd$

$ac-ad-bc+bd$

答案：A

知识点：坐标几何矩形分配律

难度评级：900

小提示：

求长方形的竖直边长和水平边长。

Find the vertical and horizontal side lengths of the rectangle

大提示：

两条边长分别是 $a+b$ 和 $c+d$ 。

The side lengths are $a+b$ and $c+d$

解答：

该区域是长方形，两条边长分别为 $a - (-b) = a + b$ 和 $d - (-c) = c + d\text{。}$ 因此面积为 $(a + b)(c + d) =$ $ac + ad + bc + bd\text{。}$

所以正确答案是 A。

Note that the region is a rectangle with side lengths $a - (-b) = a + b$ and $d - (-c) = c + d.$ The area is then $(a + b)(c + d) =$ $ac + ad + bc + bd.$

Thus, A is the correct answer.

10.

Demeanor 老师班上 $30$ 名学生中的大多数人在学校书店买了铅笔。这些学生每人买了相同数量的铅笔，且这个数量大于 $1$ 。每支铅笔的价格（美分）大于每人购买的铅笔数。所有铅笔的总价为 $\$17.71$ 。每支铅笔多少美分？

A majority of the $30$ students in Ms. Demeanor’s class bought pencils at the school bookstore. Each of these students bought the same number of pencils, and this number was greater than $1.$ The cost of a pencil in cents was greater than the number of pencils each student bought, and the total cost of all the pencils was $\$17.71.$ What was the cost of a pencil in cents?

$7$

$11$

$17$

$23$

$77$

答案：B

知识点：质因数分解整除性

难度评级：1420

小提示：

分解 $1771$ 。

Factor $1771$

大提示：

买铅笔的人数是 $30$ 人中的多数，所以使用大于 $15$ 的因数。

The number of students is a majority of $30$ , so use the factor greater than $15$

解答：

设每人买 $p$ 支铅笔，买铅笔的学生有 $s$ 人，每支铅笔 $c$ 美分。

则 $psc = 1771 = 7 \cdot 11 \cdot 23\text{。}$

还必须满足 $30 \geq s \gt 15, p \gt 1, c \gt p\text{。}$

由质因数分解，唯一满足学生人数条件的是 $s = 23$ 。

剩下的因数是七和十一；结合价格大于购买数量的条件，得到 $p = 7$ 、 $c = 11$ 。

所以正确答案是 B。

Let $p$ be the number of pencils that each student bought, $s$ be the number of students that bought pencils, and $c$ be the cost of a pencil.

We have that $psc = 1771 = 7 \cdot 11 \cdot 23.$

We also have the following restrictions: $30 \geq s \gt 15, p \gt 1, c \gt p.$

From the above prime factorization, we have that $s = 23$ is the only value that satisfies the conditions.

Finally, we get that $p = 7$ and $c = 11$ are the only remaining values that satisfy the other conditions.

Thus, B is the correct answer.

11.

正方形 $EFGH$ 的每个顶点分别位于正方形 $ABCD$ 的一条边上。点 $E$ 在 $\overline{AB}$ 上，且 $AE=7\cdot EB$ 。 $EFGH$ 的面积与 $ABCD$ 的面积之比是多少？

Square $EFGH$ has one vertex on each side of square $ABCD.$ Point $E$ is on $\overline{AB}$ with $AE=7\cdot EB.$ What is the ratio of the area of $EFGH$ to the area of $ABCD?$

$\dfrac{49}{64}$

$\dfrac{25}{32}$

$\dfrac78$

$\dfrac{5\sqrt{2}}{8}$

$\dfrac{\sqrt{14}}{4}$

答案：B

知识点：正方形（几何）勾股定理面积比

难度评级：1420

小提示：

设 $EB=x$ ，于是 $AB=8x$ 。

Let $EB=x$ , so $AB=8x$

大提示：

内部正方形的一条边是一个 $7x$ 与 $x$ 为直角边的三角形的斜边。

A side of the inner square is the hypotenuse of a $7x$ -by- $x$ triangle

解答：

设 $x = EB$ ，则 $AB = 8x$ 。对 $EFGH$ 的一条边所在的直角三角形使用勾股定理，得该边长为 $\sqrt{(7x)^2 + x^2} = \sqrt{50x^2}$

因此所求面积比为 $\dfrac{\sqrt{50x^2}^2}{(8x)^2} = \dfrac{50}{64} = \dfrac{25}{32}\text{。}$

所以正确答案是 B。

Let $x = EB.$ Then $AB = 8x.$ Applying the Pythagorean Theorem to a side of $EFGH,$ we get $\sqrt{(7x)^2 + x^2} = \sqrt{50x^2}$

The desired ratio is then $\dfrac{\sqrt{50x^2}^2}{(8x)^2} = \dfrac{50}{64} = \dfrac{25}{32}.$

Thus, B is the correct answer.

12.

一支篮球队投进了一些三分球、两分球和一分罚球。他们通过两分球得到的分数与通过三分球得到的分数相同。他们投进的罚球数比投进的两分球数多一个。该队总得分为 $61$ 分。他们投进了多少个罚球？

The players on a basketball team made some three-point shots, some two-point shots, and some one-point free throws. They scored as many points with two-point shots as with three-point shots. Their number of successful free throws was one more than their number of successful two-point shots. The team’s total score was $61$ points. How many free throws did they make?

$13$

$14$

$15$

$16$

$17$

答案：A

知识点：一次方程

难度评级：960

小提示：

设投进的两分球数为 $x$ 。

Let $x$ be the number of two-point shots

大提示：

两分球得分等于三分球得分。

The points from two-point shots equal the points from three-point shots

解答：

设投进的两分球数为 $x$ ，则 $2x + 2x + (x + 1) = 61\text{，}$ 化简得 $5x + 1 = 61$ $x = 12\text{。}$

因此罚球数为 $12 + 1 = 13$ 。

所以正确答案是 A。

Let $x$ be the number of successful two-point shots. Then we have that $2x + 2x + (x + 1) = 61,$ which simplifies to $5x + 1 = 61$ $x = 12.$

The number of successful free throws is then $12 + 1 = 13.$

Thus, A is the correct answer.

13.

在 $200$ 到 $700$ 之间，有多少个偶整数的各位数字互不相同，且都取自集合 $\{1,2,5,7,8,9\}$ ？

How many even integers are there between $200$ and $700$ whose digits are all different and come from the set $\{1,2,5,7,8,9\}?$

$12$

$20$

$72$

$120$

$200$

答案：A

知识点：数字分类讨论乘法原理

难度评级：1280

小提示：

个位必须是 $2$ 或 $8$ 。

The units digit must be $2$ or $8$

大提示：

按百位是 $2$ 还是 $5$ 分情况。

Case on whether the hundreds digit is $2$ or $5$

解答：

百位只能是 $2$ 或 $5$ ，因此按这个数值分情况讨论。

情况 $1$ ：百位是 $2$ 。

此时个位只能是 $8$ ，十位还有 $4$ 种选择。

这一情况得到 $1 \cdot 4 = 4$ 个数。

情况 $2$ ：百位是 $5$ 。

同理，个位只能选 $2$ 或 $8$ ，十位还有 $4$ 种选择。

这一情况给出 $2 \cdot 4 = 8$ 个数。

因此整数总数为 $4 + 8 = 12$ 。

所以正确答案是 A。

Since the hundreds digit can only be a $2$ or $5,$ we can case on this value.

Case $1:$ hundreds digit is $2$

The only option for the units digit is $8,$ since the number must be even. This leaves $4$ options for the tens digit.

This gives us $1 \cdot 4 = 4$ numbers for this case.

Case $2:$ hundreds digit is $5$

Similarly to above, $2$ and $8$ are the only options for the units digit, leaving $4$ options for the tens digit.

This gives us $2 \cdot 4 = 8$ numbers for this case.

The total number of integers is then $4 + 8 = 12.$

Thus, A is the correct answer.

14.

一对标准的 $6$ -面均匀骰子掷一次。掷出的点数和决定一个圆的直径。圆的面积数值小于圆周长数值的概率是多少？

A pair of standard $6$ -sided fair dice is rolled once. The sum of the numbers rolled determines the diameter of a circle. What is the probability that the numerical value of the area of the circle is less than the numerical value of the circle’s circumference?

$\dfrac{1}{36}$

$\dfrac{1}{12}$

$\dfrac{1}{6}$

$\dfrac{1}{4}$

$\dfrac{5}{18}$

答案：B

知识点：骰子（概率）圆面积不等式

难度评级：1140

小提示：

比较 $\pi(\frac{d}{2})^2$ 和 $\pi d$ 。

Compare $\pi(\frac{d}{2})^2$ with $\pi d$

大提示：

不等式化为 $d<4$ 。

The inequality reduces to $d<4$

解答：

要使面积小于周长，必须有 $\pi r^2 \lt 2\pi r$ $r \lt 2\text{。}$

因此直径必须小于 $4$ 。满足条件的掷法有三种： $(1, 1), (1, 2), (2, 1)\text{。}$

因此概率为 $\dfrac{3}{36} = \dfrac{1}{12}\text{。}$

所以正确答案是 B。

For the area to be less than the circumference, we must have $\pi r^2 \lt 2\pi r$ $r \lt 2.$

This means the diameter must be less than $4.$ There are three possible rolls that satisfy this: $(1, 1), (1, 2), (2, 1).$

The probability is then $\dfrac{3}{36} = \dfrac{1}{12}.$

Thus, B is the correct answer.

15.

Roy 买了一辆新的电池汽油混合动力汽车。一次旅行中，汽车前 $40$ 英里只使用电池，之后的路程只使用汽油，每英里耗油 $0.02$ 加仑。整趟旅行平均每加仑行驶 $55$ 英里。这趟旅行总长多少英里？

Roy bought a new battery-gasoline hybrid car. On a trip the car ran exclusively on its battery for the first $40$ miles, then ran exclusively on gasoline for the rest of the trip, using gasoline at a rate of $0.02$ gallons per mile. On the whole trip he averaged $55$ miles per gallon. How long was the trip in miles?

$140$

$240$

$440$

$640$

$840$

答案：C

知识点：速率一次方程

难度评级：1370

小提示：

设只用汽油行驶的距离为 $x$ 。

Let $x$ be the gasoline-only distance

大提示：

每加仑英里数方程为 $\frac{x+40}{0.02x}=55$ 。

The miles per gallon equation is $\frac{x+40}{0.02x}=55$

解答：

设汽车只用汽油行驶的距离为 $x$ ，则 $\dfrac{40 + x}{0.02x} = 55\text{。}$ 交叉相乘并化简得 $40 = 0.1x$ $x = 400\text{。}$ 因此旅程总长为 $400 + 40 = 440$ 。

所以正确答案是 C。

Let $x$ be the distance the car drove solely on gasoline. We have that $\dfrac{40 + x}{0.02x} = 55.$ Cross-multiplying and simplifying gives $40 = 0.1x$ $x = 400.$ The total length of the trip is then $400 + 40 = 440.$

Thus, C is the correct answer.

16.

下列哪一项等于 $\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}$ ？

Which of the following is equal to $\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}?$

$3\sqrt2$

$2\sqrt6$

$\dfrac{7\sqrt2}{2}$

$3\sqrt3$

$6$

答案：B

知识点：根式代数变形

难度评级：1480

小提示：

将整个表达式平方。

Square the whole expression

大提示：

乘积 $(9-6\sqrt2)(9+6\sqrt2)$ 很小。

The product $(9-6\sqrt2)(9+6\sqrt2)$ is small

解答：

由于式子中含有平方根，可尝试把每个根号内的式子写成完全平方。

注意原式可改写为 $\begin{aligned} & \sqrt{6 - 6\sqrt2 + 3} \\ &{}+ \sqrt{6 + 6\sqrt2 + 3} \end{aligned}\text{。}$

因式分解并化简得 $\sqrt{(\sqrt6 - \sqrt3)^2} + \sqrt{(\sqrt6 + \sqrt3)^2}$ $= \sqrt6 - \sqrt3 + \sqrt6 + \sqrt3 = 2\sqrt6\text{。}$

所以正确答案是 B。

Since we have square roots, we can try to change the inside of each radical to be a perfect square.

Note that we can rewrite the expression as $\begin{aligned} & \sqrt{6 - 6\sqrt2 + 3} \\ &{}+ \sqrt{6 + 6\sqrt2 + 3}. \end{aligned}$

Factoring and simplifying gives us $\sqrt{(\sqrt6 - \sqrt3)^2} + \sqrt{(\sqrt6 + \sqrt3)^2}$ $= \sqrt6 - \sqrt3 + \sqrt6 + \sqrt3 = 2\sqrt6.$

Thus, B is the correct answer.

17.

在八项数列 $A$ 、 $B$ 、 $C$ 、 $D$ 、 $E$ 、 $F$ 、 $G$ 、 $H$ 中， $C$ 的值为 $5$ ，并且任意三个连续项之和为 $30$ 。 $A+H$ 是多少？

In the eight term sequence $A,$ $B,$ $C,$ $D,$ $E,$ $F,$ $G,$ $H,$ the value of $C$ is $5$ and the sum of any three consecutive terms is $30.$ What is $A+H?$

$17$

$18$

$25$

$26$

$43$

答案：C

知识点：递推找规律

难度评级：1370

小提示：

比较两个相邻的三项和。

Compare two consecutive three-term sums

大提示：

每隔三项会重复，所以 $A=D=G$ ，且 $C=F$ 。

The sequence repeats every three terms, so $A=D=G$ and $C=F$

解答：

由题设得到 $A + B + C = 30$ $B = 25 - A\text{。}$ 同理， $B + C + D = 30$ $D = A\text{。}$

继续对每三个连续项应用同一条件，得到 $E = 25 - A, F = 5, G = A\text{，}$ 最后 $H = 25 - A$ 。

因此所求和为 $A + 25 - A = 25\text{。}$

所以正确答案是 C。

From the condition about the sequence, we get that $A + B + C = 30$ $B = 25 - A.$ Similarly, we get $B + C + D = 30$ $D = A.$

Propagating these values through the sequence and repeating the condition for every consecutive triple, we get that $E = 25 - A, F = 5, G = A,$ and finally, $H = 25 - A.$

The desired sum is then $A + 25 - A = 25.$

Thus, C is the correct answer.

18.

圆 $A$ 、 $B$ 和 $C$ 的半径都为 $1$ 。圆 $A$ 和圆 $B$ 有一个切点。圆 $C$ 与 $\overline{AB}$ 的中点相切。圆 $C$ 内但圆 $A$ 和圆 $B$ 外的面积是多少？

Circles $A,$ $B,$ and $C$ each have radius $1.$ Circles $A$ and $B$ share one point of tangency. Circle $C$ has a point of tangency with the midpoint of $\overline{AB}.$ What is the area inside circle $C$ but outside circle $A$ and circle $B?$

$3 - \dfrac{\pi}{2}$

$\dfrac{\pi}{2}$

$2$

$\dfrac{3\pi}{4}$

$1+\dfrac{\pi}{2}$

答案：C

知识点：圆面积扇形面积分割

难度评级：1790

小提示：

使用圆心和交点形成的单位正方形。

Use the unit squares formed by the centers and intersection points

大提示：

两个四分之一圆的减法加上一个半圆，会留下一个简单面积。

Two quarter-circle subtractions plus a semicircle leave a simple area

解答：

所求区域面积等于圆 $C$ 的面积减去它与圆 $A$ 和圆 $B$ 的重叠区域面积。

由图可知，一个重叠区域的一半可由一个四分之一圆扇形减去一个直角三角形得到。

这个面积为 $\dfrac{1}{4} \pi \cdot 1^2 - \dfrac{1}{2} \cdot 1 \cdot 1 = \dfrac{\pi}{4} - \dfrac{1}{2}\text{。}$

共需减去四块这样的区域，所以最后的面积为 $\pi \cdot 1^2 - 4(\dfrac{\pi}{4} - \dfrac{1}{2}) = 2\text{。}$

所以正确答案是 C。

The area of this region is the area of circle $C$ minus the area of the overlapping regions with $A$ and $B.$

From the diagram, we can find the area of half of one of the overlapping regions by finding the area of the sector and subtracting the area of the triangle.

This area is then $\dfrac{1}{4} \pi \cdot 1^2 - \dfrac{1}{2} \cdot 1 \cdot 1 = \dfrac{\pi}{4} - \dfrac{1}{2}.$

There are four of these that we must subtract, which leaves us with a final answer of $\pi \cdot 1^2 - 4(\dfrac{\pi}{4} - \dfrac{1}{2}) = 2.$

Thus, C is the correct answer.

19.

$1991$ 年某镇人口是一个完全平方数。十年后，人口增加 $150$ 人，变成比一个完全平方数多 $9$ 。现在是 $2011$ 年，人口又增加 $150$ 人，再次成为完全平方数。下列哪一项最接近该镇这二十年人口增长的百分比？

In $1991$ the population of a town was a perfect square. Ten years later, after an increase of $150$ people, the population was $9$ more than a perfect square. Now, in $2011,$ with an increase of another $150$ people, the population is once again a perfect square. Which of the following is closest to the percent growth of the town’s population during this twenty-year period?

$42$

$47$

$52$

$57$

$62$

答案：E

知识点：平方差完全平方数百分数

难度评级：1750

小提示：

设 $1991$ 年人口为 $p^2$ 。

Let the $1991$ population be $p^2$

大提示：

使用 $(q-p)(q+p)=141$ ，并检验因数对。

Use $(q-p)(q+p)=141$ and test the factor pairs

解答：

设 $1991$ 年人口为 $p^2$ 。 $2001$ 年人口为 $q^2 + 9$ 。

于是 $p^2 + 150 = q^2 + 9$ $q^2 - p^2 = 141\text{。}$

因式分解得 $(q - p)(q + p) = 141\text{。}$

因为 $p$ 和 $q$ 都是整数，所以 $q - p$ 与 $q + p$ 的取值只可能是 $(1, 141)$ 和 $(3, 47)$ 。

检验第一对，得到 $q-p=1$ 和 $q+p=141\text{，}$ 两式相加后除以二，得到 $q = 71$ 、 $p = 70$ 。

但此时 $p^2 + 300$ 不是完全平方数，所以这一对不合要求。

检验另一对，得到 $q = 25$ 和 $p = 22$ 。

此时 $p^2 + 300 = 784$ 是完全平方数，人口增长率为 $\dfrac{300}{22^2} \cdot 100 \% \approx 62 \%\text{。}$

所以正确答案是 E。

Let the population in $1991$ be $p^2.$ Then let the population in $2001$ be $q^2 + 9.$

Using these values, we have $p^2 + 150 = q^2 + 9$ $q^2 - p^2 = 141.$

Factoring, we get $(q - p)(q + p) = 141.$

As $p$ and $q$ are integers, we have that the only possible values for $q - p$ and $q + p$ are $(1, 141)$ and $(3, 47).$

Trying the first pair, we have $q-p=1$ and $q+p=141,$ which adding together and dividing gives us $q = 71$ and $p = 70.$

We have that $p^2 + 300$ is not a square number, which means that this pair is the wrong one.

Trying the other pair and using the same strategy gives us $q = 25$ and $p = 22.$

Now, $p^2 + 300 = 784,$ which is a perfect square. The percent increase in population is then $\dfrac{300}{22^2} \cdot 100 \% \approx 62 \%.$

Thus, E is the correct answer.

20.

在一个半径为 $r$ 的圆周上，独立随机选取两个点。从每个点沿顺时针方向画一条长度为 $r$ 的弦。两条弦相交的概率是多少？

Two points on the circumference of a circle of radius $r$ are selected independently and at random. From each point a chord of length $r$ is drawn in a clockwise direction. What is the probability that the two chords intersect?

$\dfrac{1}{6}$

$\dfrac{1}{5}$

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

答案：D

知识点：几何概率弦弧

难度评级：1840

小提示：

长度为 $r$ 的弦对应 $60^\circ$ 的圆心角。

A chord of length $r$ subtends $60^\circ$

大提示：

固定一条弦，找出第二条弦起点所在的哪些弧会与它相交。

Fix one chord and look for the arc positions from which the second chord crosses it

解答：

半径为 $r$ 的圆中，长度为 $r$ 的弦所对的圆心角为 $60^\circ$ 。固定第一条弦，使其端点的角度为 $0^\circ$ 和 $60^\circ$ 。若第二条弦从角度 $\theta$ 开始，则另一端点在其顺时针方向 $60^\circ$ 处。

两条弦的端点恰好交替出现，当且仅当 $\theta$ 位于紧邻固定弦两个端点的两段 $60^\circ$ 圆弧之一。因此有利的起点共占圆周的 $120^\circ$ 。

所求概率为 $\dfrac{2}{6} = \dfrac{1}{3}\text{。}$

所以正确答案是 D。

A chord of length $r$ in a circle of radius $r$ subtends a $60^\circ$ arc. Fix the first chord, with endpoints at angles $0^\circ$ and $60^\circ.$ If the second chord starts at angle $\theta,$ its other endpoint is $60^\circ$ clockwise from there.

The endpoints of the two chords alternate exactly when $\theta$ lies in either of the two $60^\circ$ arcs immediately adjacent to the fixed chord’s endpoints. Thus the favorable starting positions occupy $120^\circ$ of the circle.

The desired probability is then $\dfrac{2}{6} = \dfrac{1}{3}.$

Thus, D is the correct answer.

21.

两枚重量相同的假币与 $8$ 枚相同的真币混在一起。每枚假币的重量不同于每枚真币。先从 $10$ 枚硬币中随机不放回选出一对硬币，再从剩下 $8$ 枚中随机不放回选出第二对硬币。已知第一对硬币的总重量等于第二对硬币的总重量。所选 $4$ 枚硬币全是真币的概率是多少？

Two counterfeit coins of equal weight are mixed with $8$ identical genuine coins. The weight of each of the counterfeit coins is different from the weight of each of the genuine coins. A pair of coins is selected at random without replacement from the $10$ coins. A second pair is selected at random without replacement from the remaining $8$ coins. The combined weight of the first pair is equal to the combined weight of the second pair. What is the probability that all $4$ selected coins are genuine?

$\dfrac{7}{11}$

$\dfrac{9}{13}$

$\dfrac{11}{15}$

$\dfrac{15}{19}$

$\dfrac{15}{16}$

答案：D

知识点：条件概率组合分类讨论

难度评级：1990

小提示：

两对重量相等要求每对中假币数量相同。

Equal pair weights require the same number of counterfeit coins in each pair

大提示：

统计全是真币的情况和每对各有一枚假币的情况。

Count the all-genuine case and the one-counterfeit-in-each-pair case

解答：

可能情况只有两类：两对都全是真币，或每一对都含一枚假币。

全是真币时，选第一对有 $\binom{8}{2} = 28$ 种，第二对有 $\binom{6}{2} = 15$ 种。

两对可交换，所以还要除以 $2$ ，得到 $28 \cdot 15 \div 2 = 210$ 种配置。

每对各有一枚假币时，选择两枚真币有 $\binom{8}{2} = 28$ 种，而假币的选择只有一种。

把两枚假币分别与两枚真币配对有两种方法。

因此这一类共有 $28 \cdot 2 = 56$ 种。

因此所求概率为 $\dfrac{210}{210 + 56} = \dfrac{210}{266} = \dfrac{15}{19}\text{。}$

所以正确答案是 D。

There are two cases: either both selected pairs contain only genuine coins or each selected pair has one counterfeit coin.

For the first case, there are $\binom{8}{2} = 28$ ways to choose the coins for the first pair and $\binom{6}{2} = 15$ choices for the second pair.

We also have to divide by $2$ since we can swap the pairs. This gives us $28 \cdot 15 \div 2 = 210$ configurations for this case.

For the second case, there are $\binom{8}{2} = 28$ ways to choose the non-counterfeit coins. There is only one choice for the counterfeit coins.

There are two ways to create the two pairs, two choices for which counterfeit coin goes with a genuine coin.

This means that there are $28 \cdot 2 = 56$ configurations for this case.

The desired probability is then $\dfrac{210}{210 + 56} = \dfrac{210}{266} = \dfrac{15}{19}.$

Thus, D is the correct answer.

22.

凸五边形 $ABCDE$ 的每个顶点都要指定一种颜色。有 $6$ 种颜色可选，并且每条对角线的两个端点必须颜色不同。有多少种不同的着色方法？

Each vertex of convex pentagon $ABCDE$ is to be assigned a color. There are $6$ colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?

$2520$

$2880$

$3120$

$3250$

$3750$

答案：C

知识点：图论分类讨论排列

难度评级：1840

小提示：

按使用 $5$ 、 $4$ 或 $3$ 种颜色分类。

Case on whether $5$ , $4$ , or $3$ colors are used

大提示：

重复颜色只能出现在相邻顶点上。

Repeated colors can occur only on adjacent vertices

解答：

只有 $3$ 种情况：所有顶点颜色都不同；恰有一对相邻顶点同色；或有 $2$ 对相邻顶点同色，且两对使用不同颜色。

情况 $1$ ：所有顶点颜色都不同。

此时有 $6! = 720$ 种着色方法。

情况 $2$ ：恰有一对相邻顶点同色。

选择各顶点颜色的方式有 $\dfrac{6!}{2} = 360$ 种，而同色的相邻顶点对有 $5$ 种选择。

因此这种情况共有 $360 \cdot 5 = 1800$ 种着色方法。

情况 $3$ ：有两对相邻顶点分别同色。

不在任何同色对中的顶点有 $5$ 种选择。再为两对和该单独顶点选择三种不同颜色，有 $6 \cdot 5 \cdot 4 = 120$ 种，所以这种情况共有 $120 \cdot 5 = 600$ 种着色方法。

所有情况合计 $720 + 1800 + 600 = 3120$ 种。

所以正确答案是 C。

Note that there are only $3$ cases: all the vertices are different, there is one pair of adjacent vertices with the same colors, or there are $2$ pairs (each pair has a different color).

Case $1:$ all vertices have different colors

This case just gives us $6! = 720$ different colorings.

Case $2:$ one pair of adjacent vertices has the same color

There are $\dfrac{6!}{2} = 360$ ways to choose the colors for this case. There are then $5$ options for the pair of vertices.

This gives us a total of $360 \cdot 5 = 1800$ colorings for this case.

Case $3:$ two pairs of adjacent vertices have the same color

There are $5$ choices for the vertex that is not in a pair. There are then $6 \cdot 5 \cdot 4 = 120$ choices for the colors. There are then a total of $120 \cdot 5 = 600$ colorings for this case.

There are a total of $720 + 1800 + 600 = 3120$ colorings for all the cases.

Thus, C is the correct answer.

23.

七名学生按如下方式从 $1$ 数到 $1000$ ：

• Alice 说出所有数字，但跳过每连续三个数字中的中间那个。也就是说，Alice 说 $1$ 、 $3$ 、 $4$ 、 $6$ 、 $7$ 、 $9$ 、 $\ldots$ 、 $997$ 、 $999$ 、 $1000$ 。

• Barbara 说出所有 Alice 没说的数字，但她也跳过自己连续三个数字中的中间那个。

• Candice 说出 Alice 和 Barbara 都没说的数字，但她也跳过自己连续三个数字中的中间那个。

• Debbie、Eliza 和 Fatima 也依次说出所有名字字母顺序在她们之前的学生都没说的数字，但各自也跳过自己连续三个数字中的中间那个。

• 最后，George 说出唯一没有别人说过的数字。

George 说的数字是多少？

Seven students count from $1$ to $1000$ as follows:

• Alice says all the numbers, except she skips the middle number in each consecutive group of three numbers. That is, Alice says $1,$ $3,$ $4,$ $6,$ $7,$ $9,$ $\ldots,$ $997,$ $999,$ $1000.$

• Barbara says all of the numbers that Alice doesn’t say, except she also skips the middle number in each consecutive group of three numbers.

• Candice says all of the numbers that neither Alice nor Barbara says, except she also skips the middle number in each consecutive group of three numbers.

• Debbie, Eliza, and Fatima say all of the numbers that none of the students with the first names beginning before theirs in the alphabet say, except each also skips the middle number in each of her consecutive groups of three numbers.

• Finally, George says the only number that no one else says.

What number does George say?

$37$

$242$

$365$

$728$

$998$

答案：C

知识点：等差数列过程模拟找规律

难度评级：2110

小提示：

追踪剩余数列的第一项和公差。

Track the first remaining term and the common difference

大提示：

每个学生都会把公差乘以三，并留下中间项。

Each student triples the common difference and shifts to the middle term

解答：

逐步追踪这些数列即可找出最后剩下的数。

Alice 没说的数为 $2, 5, 8, 11, 14, 17, \ldots, 998\text{。}$

Barbara 说完后剩下 $5, 14, 23, 32, 41, \ldots, 995\text{。}$

这些都是等差数列，而公差每次变为原来的 $3$ 倍。

Candice 说完后剩下 $14, 41, 68, 95, \ldots, 986\text{。}$

Debbie 说完后剩下 $41, 122, 203, \ldots, 932\text{，}$ Eliza 说完后剩下 $122, 365, 608, 851\text{。}$

最后，Fatima 说完后唯一剩下的数是 $365$ ，所以 George 说的就是这个数。

所以正确答案是 C。

We can walk through all the iterations to find what is left.

Alice does not say the numbers $2, 5, 8, 11, 14, 17, \ldots, 998.$

After Barbara says her numbers, the remaining ones are $5, 14, 23, 32, 41, \ldots, 995.$

Note that both of these are arithmetic sequences where the common difference is increased by a multiple of $3.$

This pattern continues as the numbers remaining after Candice says hers are $14, 41, 68, 95, \ldots, 986.$

Then after Debbie, they are $41, 122, 203, \ldots, 932$ and after Eliza, they are $122, 365, 608, 851.$

Finally, the only number left after Fatima goes is $365,$ which is the number that George will have to say.

Thus, C is the correct answer.

24.

两个不同的正四面体的所有顶点都在同一个单位立方体的顶点中。两个四面体交集形成区域的体积是多少？

Two distinct regular tetrahedra have all their vertices among the vertices of the same unit cube. What is the volume of the region formed by the intersection of the tetrahedra?

$\dfrac{1}{12}$

$\dfrac{\sqrt2}{12}$

$\dfrac{\sqrt3}{12}$

$\dfrac{1}{6}$

$\dfrac{\sqrt2}{6}$

答案：D

知识点：立体几何体积正方体相似

难度评级：2380

小提示：

两个四面体使用立方体两组交替顶点。

The two tetrahedra use the two alternating vertex sets of the cube

大提示：

每个面会从另一个四面体上切去一个半比例的角四面体。

Each face cuts off a half-scale corner tetrahedron from the other tetrahedron

解答：

两个正四面体使用立方体两组交替的四个顶点。每个四面体的边长都是 $\sqrt2$ ，也就是立方体的一条面对角线。

边长为 $s$ 的正四面体体积为 $\dfrac{\sqrt2}{12}s^3$ ，所以一个大四面体体积为 $\dfrac{\sqrt2}{12}(\sqrt2)^3=\dfrac13$ 。

一个四面体的每个面都会从另一个四面体切去一个相似比为 $\dfrac12$ 的角四面体，每个切去部分体积为原四面体的 $\dfrac18$ 。

共有四个这样的角部分被切去，所以交集体积是一个大四面体体积的 $1-4\cdot\dfrac18=\dfrac12$ 。故交集体积为 $\dfrac12\cdot\dfrac13=\dfrac16$ 。

所以正确答案是 D。

The two regular tetrahedra use the two alternating sets of four vertices of the cube. Each has edge length $\sqrt2$ , a face diagonal of the cube.

The volume of a regular tetrahedron with edge length $s$ is $\dfrac{\sqrt2}{12}s^3$ . Thus one large tetrahedron has volume $\dfrac{\sqrt2}{12}(\sqrt2)^3=\dfrac13$ .

Intersect one tetrahedron with the other. Each face of the first cuts from the second a corner tetrahedron similar to the original with scale factor $\dfrac12$ , so each cut-off piece has $\dfrac18$ of the large tetrahedron’s volume.

There are four such corner pieces, so the intersection has $1-4\cdot\dfrac18=\dfrac12$ of the volume of one large tetrahedron. Hence the intersection volume is $\dfrac12\cdot\dfrac13=\dfrac16$ .

Thus, D is the correct answer.

25.

令 $R$ 为一个正方形区域， $n \geq 4$ 为整数。称 $R$ 内部的一点 $X$ 为 $n$ 射线等分点，如果从 $X$ 发出的 $n$ 条射线能把 $R$ 分成 $n$ 个面积相等的三角形。有多少个点是 $100$ 射线等分点，但不是 $60$ 射线等分点？

Let $R$ be a square region and $n \geq 4$ an integer. A point $X$ in the interior of $R$ is called $n$ -ray partitional if there are $n$ rays emanating from $X$ that divide $R$ into $n$ triangles of equal area. How many points are $100$ -ray partitional but not $60$ -ray partitional?

$1500$

$1560$

$2320$

$2480$

$2500$

答案：C

知识点：面积分割格点容斥原理

难度评级：2490

小提示：

一个 $n$ 射线等分点位于一个 $(\frac{n}{2}-1)\times(\frac{n}{2}-1)$ 网格上。

An $n$ -ray point lies on an $(\frac{n}{2}-1)\times(\frac{n}{2}-1)$ grid

大提示：

从 $49\times49$ 网格中减去与 $29\times29$ 网格的重叠。

Subtract the overlap of the $49\times49$ and $29\times29$ grids

解答：

把正方形缩放为边长 $1$ ，并写成 $X=(u,v)$ ，其中 $u$ 和 $v$ 分别是它到左边和下边的距离。每个顶点都必须与 $X$ 相连；否则包含该顶点的某个区域就不是三角形。

这 $n$ 个三角形的面积都是 $\frac{1}{n}$ 。以正方形下边为底的三角形高为 $v$ ，所以底长为 $\frac{2}{nv}$ 。因此下边上的三角形数为 $\frac{nv}{2}$ ，它必须是正整数。对四条边作同样分析可知， $\begin{gathered} \dfrac{nu}{2},\quad\dfrac{n(1-u)}{2},\\ \dfrac{nv}{2},\quad\dfrac{n(1-v)}{2} \end{gathered}$ 都是正整数。反过来，只要这四个数都是整数，把各边分成相应数量的等长底边，再将分点与 $X$ 相连，就能得到所需的三角形。

当 $n=100$ 时，这说明 $u=\frac{i}{50}$ ，且 $v=\frac{j}{50}$ ，其中 $i,j\in\{1,2,\ldots,49\}$ 。因而所有 $100$ 射线等分点构成 $49\times49$ 网格。同理，所有 $60$ 射线等分点满足 $u=\frac{i}{30}$ 、 $v=\frac{j}{30}$ ，其中 $i,j\in\{1,2,\ldots,29\}$ 。

一个坐标同时属于两个网格，当且仅当 $\frac{i}{50}=\frac{j}{30}$ ，即 $3i=5j$ 。因此公共坐标为 $\frac{1}{10},\frac{2}{10},\ldots,\frac{9}{10}$ ，形成一个 $9\times9$ 的重叠网格。所求点数为 $49^2-9^2=2401-81=2320\text{。}$

所以正确答案是 C。

Scale the square to have side length $1,$ and write $X=(u,v),$ where $u$ and $v$ are its distances from the left and bottom sides. Every corner must be joined to $X$ ; otherwise one of the regions containing that corner would not be a triangle.

Each of the $n$ triangles has area $\frac{1}{n}.$ A triangle whose base lies on the bottom side has height $v,$ so its base has length $\frac{2}{nv}.$ Therefore the number of triangles along the bottom side is $\frac{nv}{2},$ which must be a positive integer. Applying the same argument to all four sides shows that $\begin{gathered} \dfrac{nu}{2},\quad\dfrac{n(1-u)}{2},\\ \dfrac{nv}{2},\quad\dfrac{n(1-v)}{2} \end{gathered}$ are positive integers. Conversely, whenever these four numbers are integers, subdividing each side into the indicated number of equal bases and joining the division points to $X$ produces the required triangles.

For $n=100,$ this says $u=\frac{i}{50}$ and $v=\frac{j}{50}$ for $i,j\in\{1,2,\ldots,49\}.$ Hence the $100$ -ray points form a $49\times49$ grid. Similarly, the $60$ -ray points have coordinates $u=\frac{i}{30}$ and $v=\frac{j}{30}$ with $i,j\in\{1,2,\ldots,29\}.$

A coordinate belongs to both grids exactly when $\frac{i}{50}=\frac{j}{30},$ or $3i=5j.$ Thus the common coordinates are $\frac{1}{10},\frac{2}{10},\ldots,\frac{9}{10},$ giving a $9\times9$ overlap. The requested number is $49^2-9^2=2401-81=2320.$

Thus, C is the correct answer.