2010 AMC 10B 真题与在线限时测验

计时

1:15:00

1.

求下式的值： $100(100-3)-(100 \cdot 100-3)$

What is the value of the following expression? $100(100-3)-(100 \cdot 100-3)$

$-20{,}000$

$-10{,}000$

$-297$

$-6$

$0$

答案：C

知识点：运算顺序整数运算

难度评级：560

小提示：

先展开两项，再相减。

Expand both terms before subtracting

大提示：

大部分 $100\cdot100$ 项会抵消。

Most of the $100\cdot100$ terms cancel

解答：

第一项为 $\begin{aligned}100(100 - 3) &= 100 \cdot 97 \\&= 9700\end{aligned}\text{，}$ 而第二项为 $\begin{aligned}100 \cdot 100 - 3 &= 10000 - 3\\ &= 9997\end{aligned}\text{。}$

因此差为 $9700 - 9997 = - 297\text{。}$

所以正确答案是 C。

We have that $\begin{aligned}100(100 - 3) &= 100 \cdot 97 \\&= 9700\end{aligned}$ and $\begin{aligned}100 \cdot 100 - 3 &= 10000 - 3\\ &= 9997.\end{aligned}$

As such, their difference is: $9700 - 9997 = - 297.$

Thus, C is the correct answer.

2.

Makayla 在 $9$ 小时工作日中参加了两场会议。第一场会议用了 $45$ 分钟，第二场会议用时是第一场的两倍。她工作日的百分之多少用于参加会议？

Makayla attended two meetings during her $9$ -hour work day. The first meeting took $45$ minutes and the second meeting took twice as long. What percent of her work day was spent attending meetings?

$15$

$20$

$25$

$30$

$35$

答案：C

知识点：百分数单位换算

难度评级：870

小提示：

把会议时间换成相同单位。

Convert the meeting times to the same units

大提示：

两场会议共 $45+90$ 分钟，而工作日为 $9$ 小时。

The meetings last $45+90$ minutes out of $9$ hours

解答：

$45$ 分钟等于 $\dfrac{45}{60} = \dfrac{3}{4}$ 小时。第二场会议用时为 $2 \cdot \dfrac{3}{4} = \dfrac{3}{2}$ 小时，所以两场会议共用 $\dfrac{3}{4} + \dfrac{3}{2} = \dfrac{9}{4}$ 小时。这占整个工作日的 $100 \% \cdot \dfrac{\frac{9}{4}}{9} = 100 \% \cdot \dfrac{1}{4} = 25 \%\text{。}$

所以正确答案是 C。

Note that $45$ minutes is $\dfrac{45}{60} = \dfrac{3}{4}$ hours. The second meeting is then $2 \cdot \dfrac{3}{4} = \dfrac{3}{2}$ hours. Both meetings take a total of $\dfrac{3}{4} + \dfrac{3}{2} = \dfrac{9}{4}$ hours then. The percent of Makayla’s work day spent attending meetings is $100 \% \cdot \dfrac{\frac{9}{4}}{9} = 100 \% \cdot \dfrac{1}{4} = 25 \%.$

Thus, C is the correct answer.

3.

一个抽屉里有红、绿、蓝、白四种颜色的袜子，每种颜色至少 $2$ 只。至少要从抽屉里取出多少只袜子，才能保证有一双同色袜子？

A drawer contains red, green, blue, and white socks with at least $2$ of each color. What is the minimum number of socks that must be pulled from the drawer to guarantee a matching pair?

$3$

$4$

$5$

$8$

$9$

答案：C

知识点：抽屉原理

难度评级：720

小提示：

考虑出现一双袜子之前的最坏情况。

Use the worst case before a pair appears

大提示：

你可能先各取出四种颜色的一只袜子。

You can draw one sock of each of the four colors first

解答：

为了尽量延迟出现同色一双，最多可以先取出每种颜色各一只袜子。

一共有 $4$ 种颜色，因此取出四只时仍可能没有同色一双。

再取一只就必然与其中一种颜色配成一双，所以需要 $4 + 1 = 5$ 只。

所以正确答案是 C。

To maximize the number of socks, we want to grab as many single socks as possible before getting a pair.

There are $4$ colors, which means that we can draw one sock of each color before drawing a pair.

This means that it takes at least $4 + 1 = 5$ socks to be drawn before a pair is guaranteed.

Thus, C is the correct answer.

4.

对实数 $x$ ，定义 $\heartsuit(x)$ 为 $x$ 和 $x^2$ 的平均数。求下式的值： $\heartsuit(1)+\heartsuit(2)+\heartsuit(3)$

For a real number $x,$ define $\heartsuit(x)$ to be the average of $x$ and $x^2.$ What is the value of the following expression? $\heartsuit(1)+\heartsuit(2)+\heartsuit(3)$

$3$

$6$

$10$

$12$

$20$

答案：C

知识点：自定义运算平均数

难度评级：870

小提示：

分别计算 $\heartsuit(1),\heartsuit(2),\heartsuit(3)$ 。

Evaluate $\heartsuit(1),\heartsuit(2),\heartsuit(3)$ separately

大提示：

$\heartsuit(x)=\dfrac{x+x^2}{2}$ 。

$\heartsuit(x)=\dfrac{x+x^2}{2}$

解答：

依次计算得 $\heartsuit(1) = \dfrac{1 + 1^2}{2} = 1\text{，}$ $\heartsuit(2) = \dfrac{2 + 2^2}{2} = 3\text{，}$ 以及 $\heartsuit(3) = \dfrac{3 + 3^2}{2} = 6\text{。}$

因此总和为 $1 + 3 + 6 = 10\text{。}$

所以正确答案是 C。

We have $\heartsuit(1) = \dfrac{1 + 1^2}{2} = 1,$ $\heartsuit(2) = \dfrac{2 + 2^2}{2} = 3,$ and $\heartsuit(3) = \dfrac{3 + 3^2}{2} = 6.$

Then $1 + 3 + 6 = 10.$

Thus, C is the correct answer.

5.

某个有 $31$ 天的月份中，星期一和星期三的天数相同。这个月的第一天可能是七天中的多少种？

A month with $31$ days has the same number of Mondays and Wednesdays. How many of the seven days of the week could be the first day of this month?

$2$

$3$

$4$

$5$

$6$

答案：B

知识点：日期与时间模运算

难度评级：960

小提示：

$31$ 天中，有三个星期几会多出现一次。

In $31$ days, three weekdays occur one extra time

大提示：

星期一和星期三必须同时是多出的日子，或同时不是。

Monday and Wednesday must either both be extra days or both not be extra days

解答：

注意 $31$ 除以 $7$ 时余数为 $3$ 。

因此，如果这个月从星期六、星期日、星期二或星期三开始，星期一和星期三的天数就不相等。

所以这个月只能从星期一、星期四或星期五开始。

所以正确答案是 B。

Note that $31$ days leaves a remainder of $3$ when divided by $7.$

This means that if the month starts on a Saturday, Sunday, Tuesday, or Wednesday, there will be an uneven number of Mondays and Wednesdays.

Then the month can only start on a Monday, Thursday, or Friday.

Thus, B is the correct answer.

6.

圆心为 $O$ ， $\overline{AB}$ 是直径，点 $C$ 在圆上，且 $\angle COB = 50^\circ$ 。 $\angle CAB$ 的度数是多少？

A circle is centered at $O,$ $\overline{AB}$ is a diameter and $C$ is a point on the circle with $\angle COB = 50^\circ.$ What is the degree measure of $\angle CAB?$

$20$

$25$

$45$

$50$

$65$

答案：B

知识点：圆等腰三角形导角

难度评级：960

小提示：

将 $\angle COB$ 与 $\angle AOC$ 联系起来。

Relate $\angle COB$ to $\angle AOC$

大提示：

$OA=OC$ ，所以 $\triangle AOC$ 是等腰三角形。

$OA=OC$ , so $\triangle AOC$ is isosceles

解答：

参考下图：

由于直径的两端与圆心在同一直线上， $\angle AOC = 180^{\circ} - 50^{\circ} = 130^{\circ}\text{。}$

$\triangle AOC$ 是等腰三角形，所以 $\angle CAO = \dfrac{180^{\circ} - 130^{\circ}}{2} = 25^{\circ}\text{。}$

因为 $\angle CAB = \angle CAO$ ，所以 $\angle CAB = 25^{\circ}$ 。

所以正确答案是 B。

Consider the following diagram:

We have that $\angle AOC = 180^{\circ} - 50^{\circ} = 130^{\circ}.$

Since $\triangle AOC$ is isosceles, we have that $\angle CAO = \dfrac{180^{\circ} - 130^{\circ}}{2} = 25^{\circ}.$

Since $\angle CAB = \angle CAO,$ we have that $\angle CAB = 25^{\circ}.$

Thus, B is the correct answer.

7.

一个三角形的边长为 $10$ 、 $10$ 和 $12$ 。一个长方形宽为 $4$ ，面积等于该三角形的面积。这个长方形的周长是多少？

A triangle has side lengths $10,$ $10,$ and $12.$ A rectangle has width $4$ and area equal to the area of the triangle. What is the perimeter of this rectangle?

$16$

$24$

$28$

$32$

$36$

答案：D

知识点：三角形面积勾股定理周长

难度评级：1220

小提示：

向长为 $12$ 的边作高。

Drop the altitude to the side of length $12$

大提示：

这条高把三角形分成两个 $6$ - $8$ - $10$ 直角三角形。

The altitude splits the triangle into two $6$ - $8$ - $10$ triangles

解答：

为求三角形的面积，可以向长为 $12$ 的边作高。

于是得到一个直角三角形，它的一条直角边为 $12 \div 2 = 6$ ，斜边为 $10$ 。

另一条直角边的长度为 $\sqrt{10^2 - 6^2} = \sqrt{64} = 8\text{。}$

三角形面积为 $\dfrac{8 \cdot 12}{2} = 48\text{。}$

于是长方形的长为 $48 \div 4 = 12$ 。它的周长为 $2(4 + 12) = 2 \cdot 16 = 32\text{。}$

所以正确答案是 D。

To find the area of the triangle, we can drop the altitude to the side of length $12.$

Then we have a right triangle with one leg $12 \div 2 = 6$ and hypotenuse $10.$

The other leg has length $\sqrt{10^2 - 6^2} = \sqrt{64} = 8.$

The area of the triangle is then $\dfrac{8 \cdot 12}{2} = 48.$

The length of the rectangle is then $48 \div 4 = 12.$ Its perimeter is $2(4 + 12) = 2 \cdot 16 = 32.$

Thus, D is the correct answer.

8.

一张学校戏剧票价为 $x$ 美元，其中 $x$ 是整数。一个 $9$ 年级小组买票共花 $\$48$ ，一个 $10$ 年级小组买票共花 $\$64$ 。 $x$ 可能有多少个值？

A ticket to a school play costs $x$ dollars, where $x$ is a whole number. A group of $9$ th graders buys tickets costing a total of $\$48,$ and a group of $10$ th graders buys tickets costing a total of $\$64.$ How many values for $x$ are possible?

$1$

$2$

$3$

$4$

$5$

答案：E

知识点：最大公约数因数个数

难度评级：1020

小提示：

票价必须同时整除两个总价。

The ticket price must divide both totals

大提示：

计算 $\gcd(48,64)$ 的正因数个数。

Count the positive divisors of $\gcd(48,64)$

解答：

$x$ 必须同时整除 $48$ 和 $64$ ，因此必须整除它们的最大公因数。

$48$ 和 $64$ 的最大公因数是 $16$ ，所以 $x$ 必须整除 $16$ 。

$16$ 有 $5$ 个正因数，即 $2$ 的不超过 $16$ 的各个幂。

所以正确答案是 E。

Note that $x$ must divide both $48$ and $64,$ which means that it must divide their greatest common divisor.

The greatest common divisor of $48$ and $64$ is $16,$ which means $x$ divides $16.$

$16$ has $5$ factors, namely all the powers of $2$ up to $16.$

Thus, E is the correct answer.

9.

Larry 的老师让他把数字代入 $a$ 、 $b$ 、 $c$ 、 $d$ 和 $e$ ，再计算下式： $a-(b-(c-(d+e)))$ Larry 忽略了括号，但加减号本身算对了，并且碰巧得到了正确结果。他给 $a$ 、 $b$ 、 $c$ 、 $d$ 代入的数分别是 $1$ 、 $2$ 、 $3$ 、 $4$ 。他给 $e$ 代入了什么数？

Lucky Larry’s teacher asked him to substitute numbers for $a,$ $b,$ $c,$ $d,$ and $e$ in the expression $a-(b-(c-(d+e)))$ and evaluate the result. Larry ignored the parentheses but added and subtracted correctly and obtained the correct result by coincidence. The numbers Larry substituted for $a,$ $b,$ $c,$ and $d$ were $1,$ $2,$ $3,$ and $4,$ respectively. What number did Larry substitute for $e?$

$-5$

$-3$

$0$

$3$

$5$

答案：D

知识点：运算顺序一次方程

难度评级：1280

小提示：

比较 Larry 忽略括号得到的值和正确值。

Compare Larry’s ignored-parentheses value to the correct value

大提示：

正确计算时，表达式为 $-2-e$ 。

Correctly evaluated, the expression is $-2-e$

解答：

忽略括号时，Larry 会得到 $1 - 2 - 3 - 4 + e = e - 8\text{。}$

按括号正确计算， $\begin{aligned} &1 - (2 -(3 - (4 + e)))\\ & = 1 - (2 - (-1 - e))\\ &= 1 - (3 + e) \\ &=-2 - e\end{aligned}\text{。}$

两者相等，所以 $e - 8 = -2 - e$ $e = 3\text{。}$

所以正确答案是 D。

Ignoring the parentheses, Larry would get $1 - 2 - 3 - 4 + e = e - 8.$

Evaluating with the parentheses, one would get $\begin{aligned} &1 - (2 -(3 - (4 + e)))\\ & = 1 - (2 - (-1 - e))\\ &= 1 - (3 + e) \\ &=-2 - e.\end{aligned}$

Both of these values are the same, so $e - 8 = -2 - e$ $e = 3.$

Thus, D is the correct answer.

10.

Shelby 在不下雨时以每小时 $30$ 英里的速度骑踏板车，在下雨时以每小时 $20$ 英里的速度骑行。今天上午她在晴天中骑行，傍晚在雨中骑行，共行驶 $16$ 英里，用时 $40$ 分钟。她在雨中骑了多少分钟？

Shelby drives her scooter at a speed of $30$ miles per hour if it is not raining, and $20$ miles per hour if it is raining. Today she drove in the sun in the morning and in the rain in the evening, for a total of $16$ miles in $40$ minutes. How many minutes did she drive in the rain?

$18$

$21$

$24$

$27$

$30$

答案：C

知识点：路程、速度与时间一次方程

难度评级：1370

小提示：

设雨中骑行时间为 $t$ 分钟。

Let $t$ be the rain time in minutes

大提示：

使用 $20\cdot\dfrac{t}{60}+30\cdot\dfrac{40-t}{60}=16$ 。

Use $20\cdot\dfrac{t}{60}+30\cdot\dfrac{40-t}{60}=16$

解答：

$40$ 分钟是 $\frac{2}{3}$ 小时。设 Shelby 在晴天中骑了 $h$ 小时。

则雨中骑了 $\frac{2}{3} - h$ 小时，所以总路程为 $30h + 20\left(\dfrac{2}{3} - h\right) = 10h + \dfrac{40}{3}$ 英里。这个值等于 $16$ ，因此 $10h + \dfrac{40}{3} = 16$ $h = \dfrac{4}{15}\text{。}$

因此雨中骑行时间为 $60\left(\dfrac{2}{3} - \dfrac{4}{15}\right) = 60 \cdot \dfrac{2}{5} = 24$ 分钟。

所以正确答案是 C。

Note that $40$ minutes is $\frac{2}{3}$ hours. Let Shelby drive $h$ hours in the sun.

Then she drives $\frac{2}{3} - h$ hours in the rain, which means she travels a total of $30h + 20\left(\dfrac{2}{3} - h\right) = 10h + \dfrac{40}{3}$ miles. We have this equals $16,$ so $10h + \dfrac{40}{3} = 16$ $h = \dfrac{4}{15}.$

Then Shelby drives $60\left(\dfrac{2}{3} - \dfrac{4}{15}\right) = 60 \cdot \dfrac{2}{5} = 24$ minutes in the rain.

Thus, C is the correct answer.

11.

一位购物者计划购买一件标价大于 $\$100$ 的商品，并且可以使用三张优惠券中的任意一张。优惠券 A 可按标价打 $15\%$ 折扣，优惠券 B 可直接减 $\$30$ ，优惠券 C 可打 $25\%$ 折扣，适用于标价超过 $\$100$ 的部分。

设 $x$ 和 $y$ 分别为使优惠券 A 节省的钱不少于优惠券 B 或 C 的最小和最大标价。求 $y - x$ 。

A shopper plans to purchase an item that has a listed price greater than $\$100$ and can use any one of the three coupons. Coupon A gives $15\%$ off the listed price, Coupon B gives $\$30$ off the listed price, and Coupon C gives $25\%$ off the amount by which the listed price exceeds $\$100.$

Let $x$ and $y$ be the smallest and largest prices, respectively, for which Coupon A saves at least as many dollars as Coupon B or C. What is $y - x?$

$50$

$60$

$75$

$80$

$100$

答案：A

知识点：百分数不等式

难度评级：1540

小提示：

用标价表示每张优惠券节省的钱。

Write each coupon’s savings in terms of the listed price

大提示：

优惠券 A 必须同时不少于 $\$30$ 和 $0.25(p-100)$ 。

Coupon A must beat both $\$30$ and $0.25(p-100)$

解答：

设商品标价为 $p$ 。优惠券 A 节省 $0.15p$ ，优惠券 B 节省 $\$ 30$ 。

优惠券 C 节省 $0.25(p - 100) = 0.25p - 25\text{。}$

因此必须满足 $0.15p \ge 30$ $p \geq 200$ 以及 $0.15p \ge 0.25p - 25$ $250 \geq p\text{。}$

这给出 $x = 200$ 、 $y = 250$ ，所以 $y - x = 50$ 。

所以正确答案是 A。

Let $p$ be the price of the item. Then coupon A saves $0.15p.$ Coupon B saves $\$ 30.$

Coupon C will save $0.25(p - 100) = 0.25p - 25.$

We must have that $0.15p \ge 30$ $p \geq 200$ and $0.15p \ge 0.25p - 25$ $250 \geq p.$

This shows that $x = 200$ and $y = 250.$ Therefore $y - x = 50.$

Thus, A is the correct answer.

12.

学年初，Wells 老师数学课上 $50\%$ 的学生对“你热爱数学吗？”回答“是”， $50\%$ 回答“否”。学年末， $70\%$ 回答“是”， $30\%$ 回答“否”。总共有 $x\%$ 的学生在学年初和学年末给出了不同回答。 $x$ 的最大可能值与最小可能值之差是多少？

At the beginning of the school year, $50\%$ of all students in Mr. Wells’ math class answered “Yes” to the question “Do you love math”, and $50\%$ answered “No.” At the end of the school year, $70\%$ answered “Yes” and $30\%$ answered “No.” Altogether, $x\%$ of the students gave a different answer at the beginning and end of the school year. What is the difference between the maximum and the minimum possible values of $x?$

$0$

$20$

$40$

$60$

$80$

答案：D

知识点：百分数极限情形界定

难度评级：1420

小提示：

想象共有 $100$ 名学生。

Think of $100$ students

大提示：

分别最小化和最大化改变答案的人数。

Minimize or maximize the number who switch answers

解答：

为了最小化 $x$ ，尽量让学生保持原答案。

由于回答“是”的比例从百分之五十增到百分之七十，至少有 $70 - 50 = 20$ 个百分点所对应的学生必须改变答案。

为了最大化 $x$ ，可以让所有原来回答“否”的学生都改成“是”，但最终仍需要百分之七十回答“是”。

这要求原来回答“是”的学生中有 $70 - 50 = 20$ 个百分点仍回答“是”，所以原来回答“是”的百分之五十中，最多有 $50 - 20 = 30$ 个百分点可以改变。

因此最多有 $50 + 30 = 80$ 个百分点的学生改变答案。最大值与最小值之差为 $80 - 20 = 60$ 。

所以正确答案是 D。

To minimize $x,$ we want to have as many kids as possible maintain their answer.

We then need at least $70 - 50 = 20$ percent of the students to change their answer.

To maximize $x,$ we can have everybody that answered no change their answer, but some who answered yes must stay yes.

We need $70 - 50 = 20$ percent of the people who said yes to stay yes, which means only $50 - 20 = 30$ percent can switch.

This makes the maximum percent of students that can switch $50 + 30 = 80$ percent. The difference is then $80 - 20 = 60.$

Thus, D is the correct answer.

13.

下列方程的所有解之和是多少？ $x = |2x-|60-2x||$

What is the sum of all the solutions of the equation below? $x = |2x-|60-2x||$

$32$

$60$

$92$

$120$

$124$

答案：C

知识点：绝对值分类讨论

难度评级：1660

小提示：

按 $60-2x$ 的符号分情况。

Split cases based on the sign of $60-2x$

大提示：

处理内层绝对值后，再解得到的绝对值方程。

After resolving the inner absolute value, solve the resulting absolute-value equations

解答：

先处理外层绝对值。有两种可能： $x = 2x - |60 - 2x|$ 或 $x = -2x + |60 - 2x|\text{。}$

这两个方程分别化为 $x = |60 - 2x|$ 和 $3x = |60 - 2x|\text{。}$

每个方程都有 $2$ 种情况。第一个方程给出 $x = 60 - 2x$ 和 $x = 2x - 60\text{。}$

分别求解得 $3x = 60$ $x = 20$ 以及 $-x = -60$ $x = 60\text{。}$

第二个方程给出 $3x = 60 - 2x$ 和 $3x = 2x - 60\text{。}$

再次分别求解，得 $5x = 60$ $x = 12$ 和 $x = -60$ 。由于原方程右边非负， $x$ 不能为负数。

所有解之和为 $20 + 60 + 12 = 92\text{。}$

所以正确答案是 C。

We first take care of the outer absolute value. We have either $x = 2x - |60 - 2x|$ or $x = -2x + |60 - 2x|.$

These simplify to $x = |60 - 2x|$ and $3x = |60 - 2x|.$

We have $2$ cases for each equation. For the first one, we have $x = 60 - 2x$ and $x = 2x - 60.$

Solving both gives us $3x = 60$ $x = 20$ and $-x = -60$ $x = 60.$

For the other equation, we have $3x = 60 - 2x$ and $3x = 2x - 60.$

Again solving both, we have $5x = 60$ $x = 12$ and $x = -60.$ Note that $x$ cannot be negative since the right side of the original equation is nonnegative.

Adding up all the solutions gives us $20 + 60 + 12 = 92.$

Thus, C is the correct answer.

14.

数 $1$ 、 $2$ 、 $3$ 、 $\cdots$ 、 $98$ 、 $99$ 和 $x$ 的平均数为 $100x$ 。求 $x$ 。

The average of the numbers $1,$ $2,$ $3,$ $\cdots,$ $98,$ $99,$ and $x$ is $100x.$ What is $x?$

$\dfrac{49}{101}$

$\dfrac{50}{101}$

$\dfrac{1}{2}$

$\dfrac{51}{101}$

$\dfrac{50}{99}$

答案：B

知识点：平均数等差数列一次方程

难度评级：1370

小提示：

使用和 $1+2+\cdots+99$ 。

Use the sum $1+2+\cdots+99$

大提示：

建立方程 $\dfrac{99\cdot50+x}{100}=100x$ 。

Set $\dfrac{99\cdot50+x}{100}=100x$

解答：

前 $n$ 个正整数的和为 $\dfrac{n(n + 1)}{2}$ 。

因此有 $\dfrac{\frac{99 \cdot 100}{2} + x}{100} = 100x\text{，}$ 化简得 $99 \cdot 50 = (100^2 - 1)x$ $= 101 \cdot 99x\text{，}$ 所以 $x = \dfrac{50}{101}$ 。

所以正确答案是 B。

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

Then, we have that $\dfrac{\frac{99 \cdot 100}{2} + x}{100} = 100x,$ which simplifies to $99 \cdot 50 = (100^2 - 1)x$ $= 101 \cdot 99x,$ by difference of squares. Dividing gives us $x = \dfrac{50}{101}.$

Thus, B is the correct answer.

15.

在一场 $50$ 题的选择题数学竞赛中，学生每答对一题得 $4$ 分，空题得 $0$ 分，答错一题得 $-1$ 分。Jesse 的总分为 $99$ 。Jesse 最多可能答对多少题？

On a $50$ -question multiple choice math contest, students receive $4$ points for a correct answer, $0$ points for an answer left blank, and $-1$ point for an incorrect answer. Jesse’s total score on the contest was $99.$ What is the maximum number of questions that Jesse could have answered correctly?

$25$

$27$

$29$

$31$

$33$

答案：C

知识点：一次方程不等式最优化

难度评级：1420

小提示：

设 $R$ 为答对题数， $W$ 为答错题数。

Let $R$ be correct answers and $W$ wrong answers

大提示：

使用 $4R-W=99$ 和 $R+W\le50$ 。

Use $4R-W=99$ and $R+W\le50$

解答：

设 Jesse 答对 $x$ 题，答错 $y$ 题。

因此必须同时满足 $4x-y=99$ 和 $x+y\leq50\text{。}$

我们有 $y = 4x - 99$ $5x - 99 \leq 50\text{。}$

整理得 $x \leq 29.8$ 。因为 $x$ 是整数，所以最大值为 $29$ 。

所以正确答案是 C。

Let $x$ be the number of questions Jesse answered correctly and $y$ be the number he answered incorrectly.

Then $4x-y=99$ and $x+y\leq50.$

We have that $y = 4x - 99$ $5x - 99 \leq 50.$

Rearranging and simplifying tells us that $x \leq 29.8.$ Since $x$ is an integer, its maximum value is $29.$

Thus, C is the correct answer.

16.

一个边长为 $1$ 的正方形和一个半径为 $\dfrac{\sqrt{3}}{3}$ 的圆有相同圆心。圆内但正方形外的面积是多少？

A square of side length $1$ and a circle of radius $\dfrac{\sqrt{3}}{3}$ share the same center. What is the area inside the circle, but outside the square?

$\dfrac{\pi}{3}-1$

$\dfrac{2\pi}{9}-\dfrac{\sqrt{3}}{3}$

$\dfrac{\pi}{18}$

$\dfrac{1}{4}$

$\dfrac{2\pi}{9}$

答案：B

知识点：圆面积扇形特殊直角三角形面积分割

难度评级：1790

小提示：

找出圆与正方形一边的交点。

Find where the circle cuts one side of the square

大提示：

每条边贡献一个位于正方形外的圆弓形区域。

One side contributes a circular segment outside the square

解答：

从 $O$ 向 $\overline{AB}$ 作高，垂足为 $X$ 。

由边长关系， $\triangle OBX$ 是 $30-60-90$ 特殊直角三角形。

扇形 $AOB$ 的面积为 $\dfrac{1}{6} \cdot \pi \cdot \left(\dfrac{\sqrt3}{3}\right)^2 = \dfrac{\pi}{18}\text{。}$

$\triangle AOB$ 的面积为 $\dfrac{1}{2} \cdot 2 \cdot \dfrac{\sqrt3}{6} \cdot \dfrac{1}{2} = \dfrac{\sqrt3}{12}\text{。}$

因此该边外、圆内的一个圆弓形面积为 $\dfrac{\pi}{18} - \dfrac{\sqrt3}{12}\text{。}$

这样的区域有 $4$ 个，总面积为 $4\left(\dfrac{\pi}{18} - \dfrac{\sqrt3}{12}\right) = \dfrac{2\pi}{9} - \dfrac{\sqrt3}{3}\text{。}$

所以正确答案是 B。

Drop the altitude from $O$ to $\overline{AB}$ at $X.$

Looking at the side lengths, we see that $\triangle OBX$ is a $30-60-90$ triangle.

This means that the area of sector $AOB$ is $\dfrac{1}{6} \cdot \pi \cdot \left(\dfrac{\sqrt3}{3}\right)^2 = \dfrac{\pi}{18}.$

We also have that the area of $\triangle AOB$ is $\dfrac{1}{2} \cdot 2 \cdot \dfrac{\sqrt3}{6} \cdot \dfrac{1}{2} = \dfrac{\sqrt3}{12}.$

The area of the sector outside the square and inside the circle is then $\dfrac{\pi}{18} - \dfrac{\sqrt3}{12}.$

There are $4$ of these regions, which gives us a total area of $4\left(\dfrac{\pi}{18} - \dfrac{\sqrt3}{12}\right) = \dfrac{2\pi}{9} - \dfrac{\sqrt3}{3}.$

Thus, B is the correct answer.

17.

Euclid 市的每所高中都派出一支 $3$ 人队伍参加数学竞赛。每位参赛者的分数都不同。Andrea 的分数是所有学生中的中位数，并且她是自己队伍中得分最高的。Andrea 的队友 Beth 和 Carla 分别排第 $37$ 和第 $64$ 。这个城市有多少所学校？

Every high school in the city of Euclid sent a team of $3$ students to a math contest. Each participant in the contest received a different score. Andrea’s score was the median among all students, and hers was the highest score on her team. Andrea’s teammates Beth and Carla placed $37$ th and $64$ th, respectively. How many schools are in the city?

$22$

$23$

$24$

$25$

$26$

答案：B

知识点：中位数（数据）不等式逻辑推理

难度评级：1600

小提示：

如果有 $n$ 所学校，则共有 $3n$ 名参赛者。

If there are $n$ schools, there are $3n$ contestants

大提示：

Andrea 的中位名次为 $\dfrac{3n+1}{2}$ 。

Andrea’s median place is $\dfrac{3n+1}{2}$

解答：

设有 $n$ 所学校，则比赛中共有 $3n$ 名参赛者。

因为中位数是某一名学生的分数， $3n$ 是奇数，所以 $n$ 是奇数。Carla 排第 $64$ 要求 $3n\ge64$ ，因此奇数 $n$ 至少为 $23$ 。

Andrea 的中位名次是 $\dfrac{3n+1}{2}$ 。因为她是队内最高分，所以名次在 Beth 之前，故 $\dfrac{3n+1}{2}\lt37$ 。这给出 $3n\lt73$ ，所以奇数 $n$ 至多为 $23$ 。

因此共有 $23$ 所学校。

所以正确答案是 B。

Let there be $n$ schools. Then there are $3n$ participants in the contest.

Because the median is one student’s score, $3n$ is odd, so $n$ is odd. Carla’s $64$ th place requires $3n\ge64,$ and hence the odd integer $n$ is at least $23.$

Andrea’s median place is $\dfrac{3n+1}{2}.$ Because she had the highest score on her team, she finished ahead of Beth, so $\dfrac{3n+1}{2}\lt37.$ This gives $3n\lt73,$ and therefore the odd integer $n$ is at most $23.$

Thus there are $23$ schools.

Thus, B is the correct answer.

18.

正整数 $a$ 、 $b$ 、 $c$ 从集合 $\{1, 2, 3,\dots, 2010\}$ 中随机且独立地有放回选取。

$abc + ab + a$ 能被 $3$ 整除的概率是多少？

Positive integers $a,$ $b,$ and $c$ are randomly and independently selected with replacement from the set $\{1, 2, 3,\dots, 2010\}.$

What is the probability that $abc + ab + a$ is divisible by $3?$

$\dfrac{1}{3}$

$\dfrac{29}{81}$

$\dfrac{31}{81}$

$\dfrac{11}{27}$

$\dfrac{13}{27}$

答案：E

知识点：模运算基本概率分类讨论

难度评级：1660

小提示：

将表达式分解为 $a(bc+b+1)$ 。

Factor the expression as $a(bc+b+1)$

大提示：

若 $a$ 不是 $3$ 的倍数，则在模 $3$ 下研究 $b(c+1)$ 。

If $a$ is not divisible by $3$ , work modulo $3$ with $b(c+1)$

解答：

注意 $abc + ab + a = a(bc + b + 1)\text{。}$ 因此若 $a$ 能被 $3$ 整除，整个表达式也能被它整除。

因为 $2010$ 能被 $3$ 整除， $a$ 能被 $3$ 整除的概率为 $\frac{1}{3}$ 。

现在考虑 $a$ 不能被 $3$ 整除的情况。要使表达式能被 $3$ 整除，必须有 $bc + b + 1$ 能被 $3$ 整除。

这意味着 $bc + b = b(c + 1) \equiv 2 \pmod{3}\text{。}$

唯一可能是一个因子为 $2$ （模 $3$ ），另一个因子为 $1$ （模 $3$ ）。

两种次序中的每一种，两个因子各有 $\frac{1}{3}$ 的概率得到所需余数，因此概率为 $\dfrac{1}{3} \cdot \dfrac{1}{3} = \dfrac{1}{9}\text{。}$

共有两种次序，所以这种情况的概率为 $\dfrac{2}{9}$ 。

总概率为 $\dfrac{1}{3} \cdot 1 + \dfrac{2}{3} \cdot \dfrac{2}{9} = \dfrac{13}{27}\text{。}$

所以正确答案是 E。

Note that $abc + ab + a = a(bc + b + 1).$ This means that if $a$ is divisible by $3,$ the whole expression is as well.

Since $2010$ is divisible by $3,$ we have that $a$ is divisible by $3$ with probability $\frac{1}{3}.$

Now consider $a$ not divisible by $3.$ For the expression to be divisible by $3,$ we must have that $bc + b + 1$ is divisible by $3.$

This means that $bc + b = b(c + 1) \equiv 2 \pmod{3}.$

The only possibility for this is that one of the factors is $2$ mod $3$ and the other is $1$ mod $3.$

For each of the two cases, there is a $\frac{1}{3}$ chance that each of the factors is the desired modulus, for a probability of $\dfrac{1}{3} \cdot \dfrac{1}{3} = \dfrac{1}{9}.$

There are two cases, which means that this happens with a $\dfrac{2}{9}$ probability.

The total probability is then $\dfrac{1}{3} \cdot 1 + \dfrac{2}{3} \cdot \dfrac{2}{9} = \dfrac{13}{27}.$

Thus, E is the correct answer.

19.

圆心为 $O$ 的圆面积为 $156\pi$ 。三角形 $ABC$ 是等边三角形， $\overline{BC}$ 是圆的一条弦， $OA = 4\sqrt{3}$ ，且点 $O$ 在 $\triangle ABC$ 外。求 $\triangle ABC$ 的边长。

A circle with center $O$ has area $156\pi.$ Triangle $ABC$ is equilateral, $\overline{BC}$ is a chord on the circle, $OA = 4\sqrt{3},$ and point $O$ is outside $\triangle ABC.$ What is the side length of $\triangle ABC?$

$2\sqrt{3}$

$6$

$4\sqrt{3}$

$12$

$18$

答案：B

知识点：弦等边三角形勾股定理

难度评级：1860

小提示：

设边长为 $s$ ，并从 $A$ 作高。

Let $s$ be the side length and drop the altitude from $A$

大提示：

使用圆半径 $\sqrt{156}$ 和等边三角形的高。

Use the circle radius $\sqrt{156}$ and the equilateral triangle altitude

解答：

参考下图：

由圆面积公式可知 $BO = \sqrt{156}$ ，因为它是半径。

延长 $\overline{AO}$ 与 $\overline{BC}$ 交于 $X$ 。设 $s$ 为 $\triangle ABC$ 的边长。

于是 $BX = \dfrac{s}{2}$ 且 $AX = \dfrac{s\sqrt3}{2}\text{。}$

$\triangle OXB$ 是直角三角形，所以 $(\sqrt{156})^2 = \left(\dfrac{s}{2}\right)^2$ $+ \left(\dfrac{s\sqrt3}{2} + 4\sqrt3\right)^2\text{。}$

化简得 $156 = s^2 + 12s + 48$ $s^2 + 12s - 108 = 0$ $(s - 6)(s + 18) = 0\text{。}$

由于 $s$ 为正， $s = 6$ 。

所以正确答案是 B。

Consider the following diagram:

Using the formula for the area of a circle, we have that $BO = \sqrt{156}$ since it is a radius.

Extend $\overline{AO}$ to intersect $\overline{BC}$ at $X.$ Let $s$ be the side length of $\triangle ABC.$

Then we have that $BX = \dfrac{s}{2}$ and $AX = \dfrac{s\sqrt3}{2}.$

We have that $\triangle OXB$ is right, which means that we can apply the Pythagorean Theorem. This gives us $(\sqrt{156})^2 = \left(\dfrac{s}{2}\right)^2$ $+ \left(\dfrac{s\sqrt3}{2} + 4\sqrt3\right)^2.$

Simplifying, we get $156 = s^2 + 12s + 48$ $s^2 + 12s - 108 = 0$ $(s - 6)(s + 18) = 0.$

Since $s$ is positive, we must have that $s = 6.$

Thus, B is the correct answer.

20.

两个圆位于正六边形 $ABCDEF$ 外。第一个圆与 $\overline{AB}$ 相切，第二个圆与 $\overline{DE}$ 相切。两个圆都与直线 $BC$ 和 $FA$ 相切。第二个圆面积与第一个圆面积的比是多少？

Two circles lie outside regular hexagon $ABCDEF.$ The first is tangent to $\overline{AB},$ and the second is tangent to $\overline{DE}.$ Both are tangent to lines $BC$ and $FA.$ What is the ratio of the area of the second circle to that of the first circle?

$18$

$27$

$36$

$81$

$108$

答案：D

知识点：正多边形切线特殊直角三角形面积比

难度评级：1960

小提示：

比较两个圆的半径。

Compare the two circle radii

大提示：

较大半径可由一个 $30$ - $60$ - $90$ 三角形得到。

The larger radius comes from a $30$ - $60$ - $90$ triangle

解答：

参考下图：

设正六边形边长为 $1$ 。较小圆内切于一个边长为 $1$ 的等边三角形。

这个等边三角形的内切圆半径为 $\dfrac{\sqrt3}{6}$ 。面积为 $\pi \cdot \left(\dfrac{\sqrt3}{6}\right)^2 = \dfrac{\pi}{12}\text{。}$

设较大圆圆心为 $O$ ，从 $O$ 向 $\overline{GH}$ 作垂线，垂足为 $J$ 。并连接 $\overline{OG}$ 。

$\triangle OJG$ 是直角三角形，且由 $\angle HGI = 60^{\circ}$ 可知， $\triangle OJG$ 是 $30-60-90$ 三角形。

设 $OJ = r$ ，则 $OG = 2r$ 。又 $OG$ 等于正六边形的高、等边三角形的高与圆半径之和。

于是 $OG = \dfrac{\sqrt3}{2} + \sqrt3 + r\text{。}$

代入 $OG$ ，得 $2r = \dfrac{\sqrt3}{2} + \sqrt3 + r\text{。}$ 化简得 $r = \dfrac{3\sqrt3}{2}\text{。}$

较大圆面积为 $\pi \cdot \left(\dfrac{3\sqrt3}{2}\right)^2 = \dfrac{27}{4}\pi\text{。}$

所求比值为 $\dfrac{\frac{27\pi}{4}}{\frac{\pi}{12}} = 81\text{。}$

所以正确答案是 D。

Consider the following diagram:

Assume the regular hexagon has side length $1.$ The smaller circle is inscribed in an equilateral triangle of side length $1.$

The inradius of this equilateral triangle is $\dfrac{\sqrt3}{6}.$ The area of the circle is then $\pi \cdot \left(\dfrac{\sqrt3}{6}\right)^2 = \dfrac{\pi}{12}.$

Let $O$ be the center of the larger circle. Drop the perpendicular from $O$ to $\overline{GH}$ at $J.$ Draw $\overline{OG}.$

We have that $\triangle OJG$ is right. Since $\angle HGI = 60^{\circ},$ we also have that $\triangle OJG$ is a $30-60-90$ triangle.

Let $OJ = r.$ Then $OG = 2r.$ We also have that $OG$ is the sum of the height of the hexagon, equilateral triangle, and radius of the circle.

Then $OG = \dfrac{\sqrt3}{2} + \sqrt3 + r.$

Substituting in $OG,$ we get $2r = \dfrac{\sqrt3}{2} + \sqrt3 + r.$ Simplifying gives us $r = \dfrac{3\sqrt3}{2}.$

The area of the larger circle is then $\pi \cdot \left(\dfrac{3\sqrt3}{2}\right)^2 = \dfrac{27}{4}\pi.$

The desired ratio is then $\dfrac{\frac{27\pi}{4}}{\frac{\pi}{12}} = 81.$

Thus, D is the correct answer.

21.

从 $1000$ 到 $10{,}000$ 之间随机选择一个回文数。它能被 $7$ 整除的概率是多少？

A palindrome between $1000$ and $10{,}000$ is chosen at random. What is the probability that it is divisible by $7?$

$\dfrac{1}{10}$

$\dfrac{1}{9}$

$\dfrac{1}{7}$

$\dfrac{1}{6}$

$\dfrac{1}{5}$

答案：E

知识点：回文数整除性基本概率

难度评级：1540

小提示：

将四位回文数写成 $abba$ 。

Write a four-digit palindrome as $abba$

大提示：

其值为 $1001a+110b$ ，而 $1001$ 能被 $7$ 整除。

The value is $1001a+110b$ , and $1001$ is divisible by $7$

解答：

任意一个 $4$ 位数都可写成 $abcd$ ，展开后为 $10^3a + 10^2b + 10c + d\text{。}$ 对回文数， $a = d$ 且 $b = c$ ，所以可化为 $1001a + 110b\text{。}$ 因为 $1001$ 能被 $7$ 整除，所以还需要 $110b$ 能被 $7$ 整除。

只有 $b$ 为 $0$ 或 $7$ 时才可能成立，因为 $110$ 不能被 $7$ 整除。

有 $9$ 种 $a$ 和 $2$ 种 $b$ ，符合条件的回文数共有 $9 \cdot 2 = 18$ 个。

四位回文数总数为 $9 \cdot 10$ ，因为千位有 $9$ 种选择、百位有 $10$ 种选择。

所求概率为 $\dfrac{18}{9 \cdot 10} = \dfrac{1}{5}\text{。}$

所以正确答案是 E。

Note that we can express any $4$ digit number as $abcd.$ This can be expressed in long form as $10^3a + 10^2b + 10c + d.$ Since in a palindrome, we have that $a = d$ and $b = c.$ We can simplify this to get $1001a + 110b.$ Note that $1001$ is divisible by $7.$ This means that $110b$ must also be divisible by $7.$

The only way for this to happen is if $b$ is $0$ or $7$ since $110$ is not divisible by $7.$

There are $9$ options for $a$ and $2$ options for $b,$ for a total of $9 \cdot 2 = 18$ palindromes.

The total number of palindromes is $9 \cdot 10$ since there are $9$ options for the thousands digit and $10$ options for the hundreds digit.

The desired probability is then $\dfrac{18}{9 \cdot 10} = \dfrac{1}{5}.$

Thus, E is the correct answer.

22.

七块不同的糖果要分到三个袋子中。红袋和蓝袋必须各至少得到一块糖，白袋可以为空。有多少种分法？

Seven distinct pieces of candy are to be distributed among three bags. The red bag and the blue bag must each receive at least one piece of candy; the white bag may remain empty. How many arrangements are possible?

$1930$

$1931$

$1932$

$1933$

$1934$

答案：C

知识点：补集计数容斥原理乘法原理

难度评级：1790

小提示：

先数所有 $3^7$ 种分法。

Count all $3^7$ distributions first

大提示：

减去红袋或蓝袋为空的情况，再加回重叠。

Subtract cases with the red or blue bag empty, then add back the overlap

解答：

可以用补集计数。无限制地分配糖果的方法总数为 $3^7 = 2187\text{。}$

要数无效安排，需要考虑红袋或蓝袋为空的情况。

若红袋为空，则每块糖只有 $2$ 个袋子可选，共 $2^7 = 128$ 种安排。蓝袋为空也同样有 $128$ 种。

这两类有一个重叠情况：红袋和蓝袋都为空。因此有效分法的数量为 $2187 - (128 + 128 - 1) = 1932\text{。}$

所以正确答案是 C。

We can count this with complementary counting. The total number of ways to distribute the candies with no restrictions is $3^7 = 2187.$

To find the number of invalid arrangements, we have to count the number of ways where either the red or blue bag is empty.

For the case where the red bag is empty, each candy has $2$ options for the bag that goes into. There are then $2^7 = 128$ arrangements for this case. Similarly, there are $128$ arrangements for the case where the blue bag is empty.

There is an overlap of one case where both bags are empty. The final answer is then $2187 - (128 + 128 - 1) = 1932.$

Thus, C is the correct answer.

23.

一个 $3 \times 3$ 数组中填入数字 $1$ 到 $9$ ，每个数字恰好使用一次，并且每一行和每一列中的数字都按递增顺序排列。这样的数组有多少个？

The entries in a $3 \times 3$ array include all the digits from $1$ through $9,$ arranged so that the entries in every row and column are in increasing order. How many such arrays are there?

$18$

$24$

$36$

$42$

$60$

答案：D

知识点：有限制的排列分类讨论

难度评级：2030

小提示：

中心格只能是 $4,5$ 或 $6$ 。

The center entry can only be $4,5,$ or $6$

大提示：

按中心格分类，并使用行列递增限制。

Case on the center and use the row-column increasing constraints

解答：

设 $a_{ij}$ 表示第 $i$ 行、第 $j$ 列的数。递增条件迫使 $a_{11}=1$ 、 $a_{33}=9$ ，而 $a_{22}$ 只能是 $4$ 、 $5$ 或 $6$ 。

若 $a_{22}=4$ ，则 $\{a_{12},a_{21}\}=\{2,3\}$ 。从 $5$ 、 $6$ 、 $7$ 和 $8$ 中选两个放入左下方的一对位置 $\{a_{31},a_{32}\}$ ；另外两个放入右上方的一对位置 $\{a_{13},a_{23}\}$ 。每一对都只有一种递增排列。因此共有 $2\binom42=12$ 个数阵：左上角周围的 $2$ 、 $3$ 有两种次序，左下方的一对有 $\binom42$ 种选择。把各数反向并旋转数阵，同样有 $12$ 个中心为 $6$ 的数阵。

若 $a_{22}=5$ ，选择位置 $a_{12}$ 、 $a_{13}$ 和 $a_{23}$ 中的三个数。它们可以是 $\{2,3,4,6,7,8\}$ 的任意三元子集，但不能是 $\{2,3,4\}$ 或 $\{6,7,8\}$ ；因为这两种选择会把三个小数或三个大数全放在中心的一侧，从而违反与中心数的必要大小关系。其余每一种选择都唯一确定剩余各数及其递增次序。因此共有 $\binom63-2=18$ 个数阵。

总数为 $12+18+12=42$ 。

所以正确答案是 D。

Let $a_{ij}$ be the entry in row $i$ and column $j.$ The increasing conditions force $a_{11}=1,$ $a_{33}=9,$ and $a_{22}$ to be $4,$ $5,$ or $6.$

If $a_{22}=4,$ then $\{a_{12},a_{21}\}=\{2,3\}.$ Choose which two of $5,$ $6,$ $7,$ and $8$ go in the bottom-left pair $\{a_{31},a_{32}\};$ the other two go in the top-right pair $\{a_{13},a_{23}\}.$ Each pair then has only one increasing order. There are $2\binom42=12$ arrays: two orders for $2,$ $3$ around the upper-left corner and $\binom42$ choices for the bottom-left pair. By reversing the digits and rotating the array, there are also $12$ arrays with center $6.$

If $a_{22}=5,$ choose the three entries in positions $a_{12},$ $a_{13},$ and $a_{23}.$ They can be any three of $\{2,3,4,6,7,8\}$ except $\{2,3,4\}$ or $\{6,7,8\};$ those two choices would put all three small or all three large entries on one side and violate a required comparison with the center. Every other choice uniquely determines the remaining entries and their increasing orders. This gives $\binom63-2=18$ arrays.

The total is $12+18+12=42.$

Thus, D is the correct answer.

24.

Raiders 和 Wildcats 的一场高中篮球赛在第一节结束时打平。Raiders 四节每节得分构成一个递增等比数列，Wildcats 四节每节得分构成一个递增等差数列。第四节结束时 Raiders 以一分获胜。两队得分都不超过 $100$ 。两队上半场总共得了多少分？

A high school basketball game between the Raiders and Wildcats was tied at the end of the first quarter. The number of points scored by the Raiders in each of the four quarters formed an increasing geometric sequence, and the number of points scored by the Wildcats in each of the four quarters formed an increasing arithmetic sequence. At the end of the fourth quarter, the Raiders had won by one point. Neither team scored more than $100$ points. What was the total number of points scored by the two teams in the first half?

$30$

$31$

$32$

$33$

$34$

答案：E

知识点：等比数列等差数列丢番图方程

难度评级：2180

小提示：

设第一节打平的分数为 $a$ 。

Let the tied first-quarter score be $a$

大提示：

使用等比和等差的总分，以及最终一分之差。

Use the geometric and arithmetic total scores and the one-point final margin

解答：

设两队第一节得分均为 $a$ 。设 Raiders 的公比为 $r$ ，Wildcats 的公差为 $d$ 。

将 $r=\frac{m}{n}$ 写成最简分数。由于 Raiders 四节得分都是整数， $a=n^3A$ ，其中 $A$ 为正整数，且总分为 $A(n^3+n^2m+nm^2+m^3)\le100$ ，所以只需检查 $\dfrac{3}{2},2,3,4$ 。

对 $r=\dfrac{3}{2}$ 或 $r=4$ ，Raiders 的可行总分无法使 Wildcats 总分成为 $4a+6d$ 的形式。对 $r=3$ ，方程 $40a=4a+6d+1$ 模 $6$ 不可能。

对 $r=2$ ，方程为 $15a=4a+6d+1$ ，所以 $11a=6d+1$ 。总分不超过 $100$ 的唯一正整数解是 $a=5,d=9$ 。

上半场总分为 $5+10+5+14=34$ 。

所以正确答案是 E。

Let the first-quarter score for each team be $a.$ Let the Raiders have common ratio $r$ and let the Wildcats have common difference $d.$

Write $r=\frac{m}{n}$ in lowest terms. Since the Raiders’ four quarter scores are integers, $a=n^3A$ for some positive integer $A$ , and their total is $A(n^3+n^2m+nm^2+m^3)\le100$ . Thus the only possible ratios are $\dfrac{3}{2},2,3,4$ .

For $r=\dfrac{3}{2}$ or $r=4$ , the only possible Raiders totals give Wildcats totals that are not of the form $4a+6d$ . For $r=3$ , the equation $40a=4a+6d+1$ is impossible modulo $6$ .

For $r=2$ , the equation is $15a=4a+6d+1$ , so $11a=6d+1$ . The only positive solution with total at most $100$ is $a=5,d=9$ .

The first-half total is $5+10+5+14=34.$

Thus, E is the correct answer.

25.

设 $a \gt 0$ ，且 $P(x)$ 是一个整系数多项式，满足 $\begin{aligned} P(1) &= P(3) \\ &= P(5) = P(7) = a \end{aligned}\text{，}$ 并且 $\begin{aligned} P(2)&=P(4)=P(6)\\ &=P(8)=-a \end{aligned}\text{。}$ $a$ 的最小可能值是多少？

Let $a \gt 0,$ and let $P(x)$ be a polynomial with integer coefficients such that $\begin{aligned} P(1) &= P(3) \\ &= P(5) = P(7) = a, \end{aligned}$ and $\begin{aligned} P(2)&=P(4)=P(6)\\ &=P(8)=-a. \end{aligned}$ What is the smallest possible value of $a?$

$105$

$315$

$945$

$7!$

$8!$

答案：B

知识点：多项式整除性最小公倍数

难度评级：2350

小提示：

使用 $P(x)-a$ 的根。

Use the roots of $P(x)-a$

大提示：

将 $2$ 、 $4$ 、 $6$ 和 $8$ 代入因式分解形式，迫使 $a$ 满足整除条件。

Plug $2,$ $4,$ $6,$ and $8$ into the factored form to force divisibility of $a$

解答：

因为 $1$ 、 $3$ 、 $5$ 和 $7$ 是 $P(x)-a$ 的根，可写成 $P(x)-a=(x-1)(x-3)$ $\cdot(x-5)(x-7)Q(x)$ ，其中 $Q(x)$ 也有整数系数。

将 $2$ 、 $4$ 、 $6$ 和 $8$ 代入 $x$ ，得到 $-2a=-15Q(2)$ $=9Q(4)$ $=-15Q(6)$ $=105Q(8)$ 。因此 $a$ 必须是 $\operatorname{lcm}(15,9,105)=315$ 的倍数。

这个下界可以达到：取 $Q(x)=42$ $+(x-2)(x-6)(60-8x)$ ，并定义 $P(x)=315$ $+(x-1)(x-3)$ $\cdot(x-5)(x-7)Q(x)$ 。该多项式有整数系数并满足所有要求。

所以正确答案是 B。

Because $1,$ $3,$ $5,$ and $7$ are roots of $P(x)-a$ , write $P(x)-a=(x-1)(x-3)$ $\cdot(x-5)(x-7)Q(x)$ , where $Q(x)$ has integer coefficients.

Substituting $2,$ $4,$ $6,$ and $8$ for $x$ gives $-2a=-15Q(2)$ $=9Q(4)$ $=-15Q(6)$ $=105Q(8)$ . Hence $a$ must be a multiple of $\operatorname{lcm}(15,9,105)=315$ .

This lower bound is attainable: take $Q(x)=42$ $+(x-2)(x-6)(60-8x)$ and define $P(x)=315$ $+(x-1)(x-3)$ $\cdot(x-5)(x-7)Q(x)$ . This polynomial has integer coefficients and satisfies the required values.

Thus, B is the correct answer.