2014 AMC 10B 真题与在线限时测验

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1.

Leah 有 $13$ 枚硬币，全部是一美分硬币和五美分硬币。如果她再多一枚五美分硬币，那么两种硬币的数量就相同。Leah 的硬币共值多少美分？

Leah has $13$ coins, all of which are pennies and nickels. If she had one more nickel than she has now, then she would have the same number of pennies and nickels. In cents, how much are Leah's coins worth?

$33$

$35$

$37$

$39$

$41$

答案：C

知识点：一次方程钱币

难度评级：560

解答：

设一美分硬币有 $p$ 枚，则五美分硬币有 $13-p$ 枚；按题意，五美分硬币数也等于 $p-1$ ，所以 $13-p=p-1$ 。因此有 $7$ 枚一美分硬币和 $6$ 枚五美分硬币。

硬币的总价值为 $7+6\cdot 5=37$ 美分。

所以正确答案是 C。

Let the number of pennies be $p.$ Then, the number of nickels is $13-p$ and $p-1,$ so $13-p=p-1.$ This means we have $7$ pennies and $6$ nickels.

Therefore, the number of cents is $7+6\cdot 5=37.$

Thus, the correct answer is C .

2.

求 $\dfrac{2^3 + 2^3}{2^{-3} + 2^{-3}}$ ？

What is $\dfrac{2^3 + 2^3}{2^{-3} + 2^{-3}}?$

$16$

$24$

$32$

$48$

$64$

答案：E

知识点：指数

难度评级：870

解答：

$\begin{align*} \dfrac{2^3 + 2^3}{2^{-3} + 2^{-3}} &= \dfrac{2\cdot 2^3}{2\cdot 2^{-3}} \\&=2^6 \\&= 64 \end{align*}$

所以正确答案是 E。

$\begin{align*} \dfrac{2^3 + 2^3}{2^{-3} + 2^{-3}} &= \dfrac{2\cdot 2^3}{2\cdot 2^{-3}} \\&=2^6 \\&= 64 \end{align*}$

Thus, the correct answer is E .

3.

Randy 旅程的前三分之一是在碎石路上行驶，接着在铺装路上行驶 $20$ 英里，剩下的五分之一在土路上行驶。Randy 的旅程总长多少英里？

Randy drove the first third of his trip on a gravel road, the next $20$ miles on pavement, and the remaining one-fifth on a dirt road. In miles how long was Randy's trip?

$30$

$\dfrac{400}{11}$

$\dfrac{75}{2}$

$40$

$\dfrac{300}{7}$

答案：E

知识点：分数一次方程

难度评级：900

解答：

设全程为 $t$ 英里，则 $\begin{align*}t &= \dfrac t5 + 20 + \dfrac t3 \\ t &= 20 + \dfrac {8t}{15} \\ \dfrac 7{15} t &= 20\\ t &= \dfrac{300}7.\end{align*}$

所以正确答案是 E。

Let the path distance be $t.$ Then, we get: $\begin{align*}t &= \dfrac t5 + 20 + \dfrac t3 \\ t &= 20 + \dfrac {8t}{15} \\ \dfrac 7{15} t &= 20\\ t &= \dfrac{300}7.\end{align*}$

Thus, the correct answer is E .

4.

Susie 买了 $4$ 个松饼和 $3$ 根香蕉。Calvin 买了 $2$ 个松饼和 $16$ 根香蕉，花费是 Susie 的两倍。一个松饼的价格是香蕉的多少倍？

Susie pays for $4$ muffins and $3$ bananas. Calvin spends twice as much paying for $2$ muffins and $16$ bananas. A muffin is how many times as expensive as a banana?

$\dfrac{3}{2}$

$\dfrac{5}{3}$

$\dfrac{7}{4}$

$2$

$\dfrac{13}{4}$

答案：B

知识点：方程组比与比例

难度评级：960

解答：

设一个松饼价格为 $m$ ，一根香蕉价格为 $b$ 。 $\begin{align*} 2(4m+3b) &= 16b + 2m\\ 8m+6b &= 16b + 2m \\ 10b &= 6m\\ m &= \dfrac 53 b.\end{align*}$

所以正确答案是 B。

Let the price for a muffin be $m$ and let the price for a banana be $b.$ Then, $\begin{align*} 2(4m+3b) &= 16b + 2m\\ 8m+6b &= 16b + 2m \\ 10b &= 6m\\ m &= \dfrac 53 b.\end{align*}$

Thus, the correct answer is B .

5.

Doug 用 $8$ 块大小相同的玻璃片制作一个正方形窗户，如图所示。每块玻璃片的高宽比为 $5 : 2$ ，玻璃片周围和之间的边框宽度都是 $2$ 英寸。这个正方形窗户的边长是多少英寸？

Doug constructs a square window using $8$ equal-size panes of glass, as shown. The ratio of the height to width for each pane is $5 : 2 ,$ and the borders around and between the panes are $2$ inches wide. In inches, what is the side length of the square window? \t\t

$\ 26$

$\ 28$

$\ 30$

$\ 32$

$\ 34$

答案：A

知识点：比与比例一次方程

难度评级：1280

解答：

设玻璃片较短边为 $x$ 。

由图中排列，一个方向的边长为 $5\cdot 2 + 4x$ ，另一个方向的玻璃片长度为 $2.5x$ 。

因此另一个方向的边长为 $3\cdot 2 + 2(2.5x)$ 。令两边相等并解 $x$ ： $\begin{align*}10+4x &= 6+5x \\ x&= 4\end{align*}$

边长为 $10+4\cdot 4 = 26.$

所以正确答案是 A。

Let the smaller side of a pane be a distance of $x.$

Then, the side length is $5\cdot 2 + 4x.$ Also, the other direction has pane lengths of $2.5x.$

This means the side length is $3\cdot 2 + 2(2.5x) .$ We solve for $x$ as follows: $\begin{align*}10+4x &= 6+5x \\ x&= 4\end{align*}$

Therefore, the side length is $10+4\cdot 4 = 26.$

Thus, the correct answer is A .

6.

Orvin 去商店时带的钱刚好够买 $30$ 个气球。到店后，他发现气球正在促销：按原价买 $1$ 个气球，第二个可减价 $\frac{1}{3}$ 。Orvin 最多可以买多少个气球？

Orvin went to the store with just enough money to buy $30$ balloons. When he arrived, he discovered that the store had a special sale on balloons: buy $1$ balloon at the regular price and get a second at $\frac{1}{3}$ off the regular price. What is the greatest number of balloons Orvin could buy?

$33$

$34$

$36$

$38$

$39$

答案：C

知识点：比与比例钱币配对与分组

难度评级：960

解答：

假设买 $6$ 个气球，其中 $3$ 个按原价购买，另 $3$ 个各按原价的 $\frac 23$ 购买。

因此用相当于 $5$ 个原价气球的钱可以买六个气球。带着买 $30$ 个原价气球的钱，最多可以买 $30 \cdot \frac 65 = 36$ 个气球。

所以正确答案是 C。

Suppose we buy $6$ balloons. Then, we can buy $3$ at full price and $3$ at a price of $\frac 23$ of a balloon.

Therefore, we can buy it at a price of $5$ balloons. Thus, with the money to buy $30$ balloons, we could buy $30 \cdot \frac 65 = 36$ balloons.

Thus, the correct answer is C .

7.

假设 $A > B > 0$ ，且 $A$ 比 $B$ 大 $x\%$ 。求 $x$ 。

Suppose $A > B > 0$ and $A$ is $x\%$ greater than $B.$ What is $x?$

$100\left(\frac{A-B}{B}\right)$

$100\left(\frac{A+B}{B}\right)$

$100\left(\frac{A+B}{A}\right)$

$100\left(\frac{A-B}{A}\right)$

$100\left(\frac{A}{B}\right)$

答案：A

知识点：百分数代数变形

难度评级：870

解答：

根据百分比增加的定义， $A = \dfrac {x+ 100}{100}B$ $= B + \dfrac x{100}B.$

$\begin{align*} (A-B) &= \dfrac x{100}B\\ 100\left(\dfrac{A-B} B\right) &=x.\end{align*}$

所以正确答案是 A。

By definition, we know $A = \dfrac {x+ 100}{100}B$ $= B + \dfrac x{100}B.$

This implies, $\begin{align*} (A-B) &= \dfrac x{100}B\\ 100\left(\dfrac{A-B} B\right) &=x.\end{align*}$

Thus, the correct answer is A .

8.

一辆卡车每 $t$ 秒行驶 $\dfrac{b}{6}$ 英尺。 $3$ 英尺等于一码。这辆卡车在 $3$ 分钟内行驶多少码？

A truck travels $\dfrac{b}{6}$ feet every $t$ seconds. There are $3$ feet in a yard. How many yards does the truck travel in $3$ minutes?

$\dfrac{b}{1080t}$

$\dfrac{30t}{b}$

$\dfrac{30b}{t}$

$\dfrac{10t}{b}$

$\dfrac{10b}{t}$

答案：E

知识点：单位换算速率

难度评级：960

解答：

卡车每 $t$ 秒行驶 $\dfrac{b}{18}$ 码，因为 $3$ 英尺等于一码；所以卡车每秒行驶 $\dfrac{b}{18t}$ 码。

$3$ 分钟为 $180$ 秒，因此行驶 $\dfrac{b}{18t}\cdot 180 = \dfrac {10b}t.$

所以正确答案是 E。

This means it travels $\dfrac{b}{18}$ yards in $t$ seconds since a yard is $3$ feet. Then, in one second, it travels $\dfrac{b}{18t}$ yards.

Therefore, in $3$ minutes which is $180$ seconds, it travels $\dfrac{b}{18t}\cdot 180 = \dfrac {10b}t.$

Thus, the correct answer is E .

9.

对实数 $w$ 和 $z$ ， $\cfrac{\frac{1}{w} + \frac{1}{z}}{\frac{1}{w} - \frac{1}{z}} = 2014.$ 求 $\frac{w+z}{w-z}$ 的值。

For real numbers $w$ and $z ,$ $\cfrac{\frac{1}{w} + \frac{1}{z}}{\frac{1}{w} - \frac{1}{z}} = 2014.$ What is $\frac{w+z}{w-z} ?$

$-2014$

$\dfrac{-1}{2014}$

$\dfrac{1}{2014}$

$1$

$2014$

答案：A

知识点：代数变形分数

难度评级：1020

解答：

分子和分母同乘后，得到 $\begin{align*} \dfrac{wz}{wz}\cdot \dfrac{\frac{1}{w} + \frac{1}{z}}{\frac{1}{w} - \frac{1}{z}} &= 2014\\ \dfrac{w+z}{z-w} &= 2014\\ \dfrac{w+z}{w-z} &= -1\cdot 2014\\ \dfrac{w+z}{w-z}&=-2014.\end{align*}$

所以正确答案是 A。

Observe that: $\begin{align*} \dfrac{wz}{wz}\cdot \dfrac{\frac{1}{w} + \frac{1}{z}}{\frac{1}{w} - \frac{1}{z}} &= 2014\\ \dfrac{w+z}{z-w} &= 2014\\ \dfrac{w+z}{w-z} &= -1\cdot 2014\\ \dfrac{w+z}{w-z}&=-2014.\end{align*}$

Thus, the correct answer is A .

10.

在下面的加法中， $A,$ $B,$ $C,$ 和 $D$ 是互不相同的数字。 $D$ 可以有多少个不同的值？ $\begin{array}{r} ABBCB \\ + \ BCADA \\ \hline DBDDD \end{array}$

In the addition shown below $A,$ $B,$ $C,$ and $D$ are distinct digits. How many different values are possible for $D?$ $\begin{array}{r} ABBCB \\ + \ BCADA \\ \hline DBDDD \end{array}$

$2$

$4$

$7$

$8$

$9$

答案：C

知识点：数字谜数字分类讨论

难度评级：1370

解答：

最左列没有产生第六位进位，所以 $A+B=D\le 9$ 。

个位列也是 $B+A=D$ ，所以个位列也没有进位。十位列给出 $C+D=D$ ，因此 $C=0$ 。

因为 $A$ 和 $B$ 是不同的非零数字， $D=A+B$ 可以是从 $3$ 到 $9$ 的任意数字。例如， $(A,B)=(1,2),$ $(1,3),$ $(2,3),$ $(2,4),$ $(2,5),$ $(2,6),$ $(2,7)$ 分别给出 $D=3,4,5,6,7,8,9$ 。

所以 $D$ 有 $7$ 个可能值，正确答案是 C。

From the leftmost column, there is no carry into a sixth digit, so $A+B=D\le 9$ .

The units column is $B+A=D$ , so it also has no carry. The tens column then gives $C+D=D$ , hence $C=0$ .

Since $A$ and $B$ are distinct nonzero digits, $D=A+B$ can be any digit from $3$ through $9$ . For example, $(A,B)=(1,2),$ $(1,3),$ $(2,3),$ $(2,4),$ $(2,5),$ $(2,6),$ $(2,7)$ give $D=3,4,5,6,7,8,9$ .

Thus there are $7$ possible values of $D$ , and the correct answer is C .

11.

对消费者来说，单次 $n\%$ 折扣比下面任何一种折扣都更划算：

（一）连续两次 $15\%$ 折扣；

（二）连续三次 $10\%$ 折扣；

（三）先 $25\%$ 折扣，再 $5\%$ 折扣。

$n$ 的最小正整数值是多少？

For the consumer, a single discount of $n\%$ is more advantageous than any of the following discounts:

(1) Two successive $15\%$ discounts.

(2) Three successive $10\%$ discounts.

(3) A $25\%$ discount followed by a $5\%$ discount.

What is the smallest possible positive integer value of $n?$

$\ \ 27$

$\ 28$

$\ 29$

$\ 31$

$\ 33$

答案：C

知识点：百分数不等式

难度评级：1540

解答：

需要求满足以下条件的最小 $n$ ： $1- \dfrac n{100} < (0.85)^2,$ $1- \dfrac n{100} < (0.9)^3,$ $1- \dfrac n{100} < (0.75)(0.95).$ 注意 $\begin{aligned} 0.75\cdot 0.95 &= 0.85^2-0.1^2 \\ &< 0.85^2,\end{aligned}$ 所以最后一个条件比第一个更严格。第二个条件给出 $1- \dfrac n{100} < 0.729.$ $n > 27.1.$

最后一个条件可写为 $1- \dfrac n{100} < 0.95\cdot 0.75$ $1- \dfrac n{100} < \dfrac 34 \cdot \dfrac{19}{20}$ $1- \dfrac n{100} < \dfrac{57}{80}$ $1- \dfrac n{100} < \dfrac{285}{400}.$

因此 $n > \dfrac{115}{4} = 28.75.$ 综合各条件，最小的 $n$ 为 $n=29$ 。

所以正确答案是 C。

We need to find the smallest possible $n$ such that $1- \dfrac n{100} < (0.85)^2,$ $1- \dfrac n{100} < (0.9)^3,$ $1- \dfrac n{100} < (0.75)(0.95).$ Note that $\begin{aligned} 0.75\cdot 0.95 &= 0.85^2-0.1^2 \\ &< 0.85^2,\end{aligned}$ so we don't need to worry about the first condintion since the last condition is true. Then, the second condition yields $1- \dfrac n{100} < 0.729.$ $n > 27.1.$

Also, we also can see that $1- \dfrac n{100} < 0.95\cdot 0.75$ $1- \dfrac n{100} < \dfrac 34 \cdot \dfrac{19}{20}$ $1- \dfrac n{100} < \dfrac{57}{80}$ $1- \dfrac n{100} < \dfrac{285}{400}.$

Therefore, $n > \dfrac{115}{4} = 28.75.$ Combining our conditions yields a smallest $n$ of $n=29.$

Thus, the correct answer is C .

12.

数 $2,014,000,000$ 的最大因数是它本身。它的第五大因数是多少？

The largest divisor of $2,014,000,000$ is itself. What is its fifth-largest divisor?

$125, 875, 000$

$201, 400, 000$

$251, 750, 000$

$402, 800, 000$

$503, 500, 000$

答案：C

知识点：因数质因数分解

难度评级：1280

解答：

第五大因数等于 $2,014,000,000$ 除以第五小因数。

$2,014,000,000$ 的素因数分解为 $2^7\cdot 5^6 \cdot 19\cdot 53.$ 最小的 $5$ 个因数是 $1,2,4,5,8$ 。所以第五小因数为 $8$ ，第五大因数为 $\dfrac{2014000000}8 =251750000.$

所以正确答案是 C。

The fifth-largest divisor is $2,014,000,000$ divided by the fifth smallest divisor.

The prime factorization of $2,014,000,000$ is: $2^7\cdot 5^6 \cdot 19\cdot 53.$ This makes the first $5$ smallest divisors $1,2,4,5,8$ Therefore, the fifth smallest divisor is $8,$ and the fifth largest divisor must be: $\dfrac{2014000000}8 =251750000.$

Thus, the correct answer is C .

13.

六个正六边形围绕一个边长为 $1$ 的正六边形，如图所示。求 $\triangle{ABC}$ 的面积。

Six regular hexagons surround a regular hexagon of side length $1$ as shown. What is the area of $\triangle{ABC}?$ \t\t

$2\sqrt{3}$

$3\sqrt{3}$

$1+3\sqrt{2}$

$2+2\sqrt{3}$

$3+2\sqrt{3}$

答案：B

知识点：正多边形等边三角形三角形面积

难度评级：1480

解答：

由旋转对称性， $AB = BC = AC$ ，所以这个三角形是等边三角形。

$AB$ 的四分之一是一个直角三角形中与 $60^{\circ}$ 相对的直角边，斜边为 $1$ ，所以 $\sin(60^\circ) = \dfrac{\sqrt 3}2.$

因此 $AB = 4\cdot \dfrac{\sqrt 3}2 = 2\sqrt 3.$

所以等边三角形的面积为 $\dfrac{s^2 \sqrt 3}4 = \dfrac{12\sqrt 3}4 = 3 \sqrt 3.$

所以正确答案是 B。

Since $AB = BC = AC$ by rotational symmetry, we know it is an equilateral triangle.

Then, one-fourth of $AB$ can be found as a the leg of a right triangle with hypotenuse $1$ and is opposite to the $60^{\circ}$ angle, making it $\sin(60^\circ) = \dfrac{\sqrt 3}2.$

As such, $AB = 4\cdot \dfrac{\sqrt 3}2 = 2\sqrt 3.$

Then, since it is an equilateral triangle, it has area $\dfrac{s^2 \sqrt 3}4 = \dfrac{12\sqrt 3}4 = 3 \sqrt 3.$

Thus, the correct answer is B .

14.

Danica 开着新车旅行了整数小时，平均速度为 $55$ 英里每小时。旅行开始时，里程表显示 $abc$ 英里，其中 $abc$ 是 $3$ 位数， $a\ge1$ ，且 $a+b+c\le7$ 。旅行结束时，里程表显示 $cba$ 英里。求 $a^2+b^2+c^2$ 的值。

Danica drove her new car on a trip for a whole number of hours, averaging $55$ miles per hour. At the beginning of the trip, $abc$ miles was displayed on the odometer, where $abc$ is a $3$ -digit number with $a\ge1$ and $a+b+c\le7.$ At the end of the trip, the odometer showed $cba$ miles. What is $a^2+b^2+c^2?$

$26$

$27$

$36$

$37$

$41$

答案：D

知识点：数字整除性

难度评级：1540

解答：

里程表读数 $cba$ 与 $abc$ 之差为 $100c + 10b+a - 100a - 10b-c$ $= 99(c-a)$ ，而这个差必须是 $55$ 的倍数。因为 $\gcd(55,99)$ 为 $11$ ，所以 $c-a$ 是 $5$ 的倍数，且 $c > a$ 。

在 $a+ b+c \leq 7$ 的限制下，唯一可能是 $a = 1, b = 0, c = 6$ ，其他组合都会有 $a+b+c > 7$ 。因此 $a^2+b^2+c^2 = 37$ 。

所以正确答案是 D。

We know that the difference of the numbers $cba$ and $abc$ is equal to: $100c + 10b+a - 100a - 10b-c$ $= 99(c-a)$ We know that this number also must be a multiple of $55.$ As $\gcd(55,99)$ is $11,$ we know that $c-a$ is a multiple of $5,$ and $c > a.$

This makes $a = 1, b = 0, c = 6$ the only possible value with $a+ b+c \leq 7$ as every other combination has $a+b+c > 7.$ As such, $a^2+b^2+c^2 = 37.$

Thus, the correct answer is D .

15.

在长方形 $ABCD$ 中， $DC = 2 \cdot CB$ ，点 $E$ 和 $F$ 在 $\overline{AB}$ 上，使得 $\overline{ED}$ 和 $\overline{FD}$ 如图三等分 $\angle ADC$ 。求 $\triangle DEF$ 的面积与长方形 $ABCD$ 面积之比。

In rectangle $ABCD,$ $DC = 2 \cdot CB$ and points $E$ and $F$ lie on $\overline{AB}$ so that $\overline{ED}$ and $\overline{FD}$ trisect $\angle ADC$ as shown. What is the ratio of the area of $\triangle DEF$ to the area of rectangle $ABCD?$ \t\t

$\ \ \dfrac{\sqrt{3}}{6}$

$\ \dfrac{\sqrt{6}}{8}$

$\ \dfrac{3\sqrt{3}}{16}$

$\ \dfrac{1}{3}$

$\ \dfrac{\sqrt{2}}{4}$

答案：A

知识点：三角学面积比矩形

难度评级：1660

解答：

设 $AD=h$ 。因为 $DC=AB=2h$ ，所以长方形 $ABCD$ 的面积为 $2h^2$ 。

射线 $DE$ 和 $DF$ 与 $DC$ 的夹角分别为 $60^\circ$ 和 $30^\circ$ 。

因此 $AE=h\tan30^\circ=\dfrac{h}{\sqrt3}$ 且 $AF=h\tan60^\circ=h\sqrt3.$

于是 $\begin{aligned}EF&=AF-AE\\&=h\sqrt3-\dfrac{h}{\sqrt3}\\&=\dfrac{2h\sqrt3}{3}.\end{aligned}$ $\triangle DEF$ 的面积为 $\dfrac12\cdot EF\cdot h=\dfrac{h^2\sqrt3}{3}.$

所求面积比为 $\dfrac{h^2\sqrt3/3}{2h^2}=\dfrac{\sqrt3}{6}.$ 所以正确答案是 A。

Let $AD=h.$ Since $DC=AB=2h,$ the area of rectangle $ABCD$ is $2h^2.$

The rays $DE$ and $DF$ make angles of $60^\circ$ and $30^\circ,$ respectively, with $DC.$

Therefore, $AE=h\tan30^\circ=\dfrac{h}{\sqrt3}$ and $AF=h\tan60^\circ=h\sqrt3.$

Hence $\begin{aligned}EF&=AF-AE\\&=h\sqrt3-\dfrac{h}{\sqrt3}\\&=\dfrac{2h\sqrt3}{3}.\end{aligned}$ The area of $\triangle DEF$ is $\dfrac12\cdot EF\cdot h=\dfrac{h^2\sqrt3}{3}.$

The desired ratio is $\dfrac{h^2\sqrt3/3}{2h^2}=\dfrac{\sqrt3}{6}.$ Thus, the correct answer is A.

16.

掷四枚公平的六面骰子。至少三枚骰子显示相同点数的概率是多少？

Four fair six-sided dice are rolled. What is the probability that at least three of the four dice show the same value?

$\dfrac{1}{36}$

$\dfrac{7}{72}$

$\dfrac{1}{9}$

$\dfrac{5}{36}$

$\dfrac{1}{6}$

答案：B

知识点：骰子（概率）分类讨论

难度评级：1420

解答：

共有 $6^4$ 个等可能有序结果。

恰好三枚相同时，重复点数有 $6$ 种，不同点数有 $5$ 种，不同骰子的位置有 $4$ 种，共 $6\cdot5\cdot4=120$ 种。

四枚全相同有 $6$ 种。

所求概率为 $\frac{120+6}{6^4}=\frac{126}{1296}=\frac7{72}$ 。

所以正确答案是 B。

There are $6^4$ equally likely ordered outcomes.

If exactly three dice show the same value, choose the repeated value in $6$ ways, the different value in $5$ ways, and the position of the different die in $4$ ways. This gives $6\cdot5\cdot4=120$ outcomes.

If all four dice match, there are $6$ outcomes.

The probability is $\frac{120+6}{6^4}=\frac{126}{1296}=\frac7{72}$ .

Thus, the correct answer is B .

17.

$10^{1002} - 4^{501}$ 的因数中，最大的 $2$ 的幂是多少？

What is the greatest power of $2$ that is a factor of $10^{1002} - 4^{501}?$

$2^{1002}$

$2^{1003}$

$2^{1004}$

$2^{1005}$

$2^{1006}$

答案：D

知识点：2的幂因式分解平方差

难度评级：1660

解答：

先提取明显的 $2$ 的幂： $10^{1002}-4^{501}=2^{1002}(5^{1002}-1)$ 。

因为 $5^{1002}-1=(5^{501}-1)(5^{501}+1)$ ，且 $501$ 为奇数，所以 $5^{501}-1$ 可被 $4$ 整除但不能被 $8$ 整除，而 $5^{501}+1$ 可被 $2$ 整除但不能被 $4$ 整除。

因此 $5^{1002}-1$ 正好贡献 $2^3$ ，整个表达式可被 $2^{1005}$ 整除，但不能被 $2^{1006}$ 整除。

所以正确答案是 D。

Factor out the obvious power of $2$ : $10^{1002}-4^{501}=2^{1002}(5^{1002}-1)$ .

Since $5^{1002}-1=(5^{501}-1)(5^{501}+1)$ , and $501$ is odd, $5^{501}-1$ is divisible by $4$ but not by $8$ , while $5^{501}+1$ is divisible by $2$ but not by $4$ .

Thus $5^{1002}-1$ contributes exactly $2^3$ , so the whole expression is divisible by $2^{1005}$ but not $2^{1006}$ .

Thus, the correct answer is D .

18.

一个由 $11$ 个正整数组成的列表平均数为 $10$ ，中位数为 $9$ ，且唯一众数为 $8$ 。列表中整数的最大可能值是多少？

A list of $11$ positive integers has a mean of $10,$ a median of $9,$ and a unique mode of $8.$ What is the largest possible value of an integer in the list?

$24$

$30$

$31$

$33$

$35$

答案：E

知识点：平均数中位数（数据）众数最优化

难度评级：1790

解答：

总和为 $11\cdot10=110$ 。

要最大化最大项，应最小化其余十项之和。按非递减顺序排列时，第六项为 $9$ ，且 $8$ 必须是唯一众数。若 $8$ 出现两次，前十项最小和为 $1+2+3+8+8+9$ $+10+11+12+13$ $=77$ ，最大项为 $33$ 。

若 $8$ 出现三次，前十项可取 $1,1,8,8,8,9,9,10,10,11$ ，其和为 $75$ ，最大项可为 $35$ 。

若 $8$ 出现四次或五次，前十项最小和至少为 $80$ ，最大项至多为 $30$ 。

因此最大可能值为 $35$ ，正确答案是 E。

The list has total sum $11\cdot10=110$ . To maximize the largest entry, minimize the sum of the other ten entries.

In nondecreasing order, the sixth entry is $9$ , and $8$ must be the unique mode. If $8$ appears twice, the least possible first ten entries sum to $1+2+3+8+8+9$ $+10+11+12+13$ $=77$ , giving largest entry $33$ .

If $8$ appears three times, the least possible first ten entries are $1,1,8,8,8,9,9,10,10,11$ , with sum $75$ , giving largest entry $35$ .

If $8$ appears four or five times, the least possible sum of the first ten entries is at least $80$ , so the largest entry is at most $30$ .

Therefore the largest possible entry is $35$ , and the correct answer is E .

19.

两个同心圆半径分别为 $1$ 和 $2$ 。在外圆上独立且均匀随机选取两点。连接这两点的弦与内圆相交的概率是多少？

Two concentric circles have radii $1$ and $2.$ Two points on the outer circle are chosen independently and uniformly at random. What is the probability that the chord joining the two points intersects the inner circle?

$\ \dfrac{1}{6}$

$\ \dfrac{1}{4}$

$\ \dfrac{2-\sqrt{2}}{2}$

$\ \dfrac{1}{3}$

$\ \dfrac{1}{2}$

答案：D

知识点：几何概率弦切线

难度评级：1600

解答：

不妨固定外圆上的一个端点。

第二个端点落在相应的一段圆弧内时，弦会与内圆相交。

临界情况是弦与内圆相切。观察圆心角，可得到两个直角三角形，其中邻边为 $1$ ，斜边为 $2,$ 所以 $\cos \left(\dfrac \theta 2\right) = \dfrac 12.$

因此 $\dfrac \theta 2 = 60^\circ,$ 从而 $\theta = 120 ^\circ .$

所以所求概率为 $\dfrac {120^\circ}{360^\circ} = \dfrac 13 .$

所以正确答案是 D。

First, without loss of generality, we could choose some point on the outer circle. Then, the second point can be chosen in a region on the other circle.

This region is such that it has a line that intersects the circle, so the edge of the region is such that the chord is perpendicular with the inner circle.

If we look at the angle at the center, we can see that it has 2 right triangles where the adjacent side is $1$ and the hypotenuse is $2,$ making $\cos \left(\dfrac \theta 2\right) = \dfrac 12.$

Thus, $\dfrac \theta 2 = 60^\circ,$ making $\theta = 120 ^\circ .$

Therefore, the probability is $\dfrac {120^\circ}{360^\circ} = \dfrac 13 .$

Thus, the correct answer is D .

20.

有多少个整数 $x$ 使 $x^4-51x^2+50$ 为负？

For how many integers $x$ is the number $x^4-51x^2+50$ negative?

$8$

$10$

$12$

$14$

$16$

答案：C

知识点：因式分解不等式区间内整数计数

难度评级：1280

解答：

先因式分解： $x^4-51x^2+50$ $= (x^2-50)(x^2-1).$

要使 $(x^2-50)(x^2-1) < 0$ ，两个因子必须异号。因为 $x^2-50 < x^2-1,$ 所以必须有 $x^2-50 < 0, x^2-1 > 0.$

这等价于 $1 < x^2 < 50,$ 即 $1< |x| \leq 7,$ 因此共有 $12$ 个整数解。

所以正确答案是 C。

First, note that $x^4-51x^2+50$ $= (x^2-50)(x^2-1).$

If $(x^2-50)(x^2-1) < 0$ means that one of the terms is negative.

Since $x^2-50 < x^2-1,$ it must be that $x^2-50 < 0, x^2-1 > 0.$ This means $1 < x^2 < 50,$ making $1< |x| \leq 7,$ resulting in $12$ solutions.

Thus, the correct answer is C .

21.

梯形 $ABCD$ 的平行边 $\overline{AB}$ 长 $33$ ，线段 $\overline {CD}$ 长 $21$ 。另外两边长为 $10$ 和 $14$ 。角 $A$ 和角 $B$ 都是锐角。求梯形 $ABCD$ 的较短对角线长度。

Trapezoid $ABCD$ has parallel sides $\overline{AB}$ of length $33$ and $\overline {CD}$ of length $21 .$ The other two sides are of lengths $10$ and $14 .$ The angles $A$ and $B$ are acute. What is the length of the shorter diagonal of $ABCD ?$

$10\sqrt{6}$

$25$

$8\sqrt{10}$

$18\sqrt{2}$

$26$

答案：B

知识点：梯形勾股定理

难度评级：1790

解答：

从 $C$ 和 $D$ 向 $AB$ 作垂线，垂足分别为 $E$ 和 $F$ 。不妨设 $BC=10$ ；交换两条腰只会把梯形翻转。

令 $FB = x$ ，高为 $h$ 。 $\begin{align*}AE &= 33-x-EF\\&=33-x-21 \\&= 12-x.\end{align*}$

由两个直角三角形得到 $14^2 = (12-x)^2 + h^2$ $10^2 = x^2 + h^2.$ 两式相减得 $96 = 144 - 24x$ $x = 2.$ 较短对角线长为 $\sqrt{ (21+x)^2+h^2}$ $= \sqrt{21^2 + 42x + (x^2 + h^2) }$ $= \sqrt{441+42\cdot 2 + 100}$ $= \sqrt{625}$ $=25.$

所以正确答案是 B。

Let the feet of the perpendiculars from $C$ and $D$ to $AB$ be $E$ and $F,$ respectively. Without loss of generality, let $BC=10$ ; interchanging the two legs only reflects the trapezoid. This yields the following diagram:

Then, let $FB = x$ and the altitude be $h.$ This means $\begin{align*}AE &= 33-x-EF\\&=33-x-21 \\&= 12-x.\end{align*}$

This suggests that: $14^2 = (12-x)^2 + h^2$ $10^2 = x^2 + h^2.$ Subtracting the equations, we get: $96 = 144 - 24x$ $x = 2.$ Then, we want to find $\sqrt{ (21+x)^2+h^2}$ $= \sqrt{21^2 + 42x + (x^2 + h^2) }$ $= \sqrt{441+42\cdot 2 + 100}$ $= \sqrt{625}$ $=25.$

Thus, the correct answer is B .

22.

八个半圆如图沿边长为 $2$ 的正方形内侧排列。与所有这些半圆相切的圆的半径是多少？

Eight semicircles line the inside of a square with side length $2$ as shown. What is the radius of the circle tangent to all of these semicircles? \t\t

$\dfrac{1+\sqrt2}4$

$\dfrac{\sqrt5-1}2$

$\dfrac{\sqrt3+1}4$

$\dfrac{2\sqrt3}5$

$\dfrac{\sqrt5}3$

答案：B

知识点：相切圆勾股定理

难度评级：1660

解答：

从正方形中心到一个半圆圆心的距离可由直角三角形的斜边求出。

一条直角边是从正方形中心到边中点的距离，长度为 $1$ 。

另一条直角边是从边中点到半圆圆心的距离，长度为 $\dfrac 12$ 。这也说明半圆半径为 $\dfrac 12$ 。

因此正方形中心到半圆圆心的距离为 $\sqrt{1^2 + \dfrac 12 ^2} = \dfrac {\sqrt 5}2.$ 再减去半圆半径 $\dfrac 12$ ，得到小圆半径 $\dfrac{\sqrt 5 -1}2 .$

所以正确答案是 B。

The distance from the center of the square to the center of the semicircles can be found as a hypotenuse of a right triangle.

One of the legs is from the center of the square to the center of one of the sides which is of distance $1.$

The other leg is from the center of the side to the center of one of the semicircles which is of distance $\dfrac 12.$ This also shows that the radius of the semicircles is $\dfrac 12.$

Therefore, the distance from the center of the square to the center of the semicircle is $\sqrt{1^2 + \dfrac 12 ^2} = \dfrac {\sqrt 5}2.$ Then , we subtract $\dfrac 12$ for the radius of the semicircle. This makes the radius of the circle $\dfrac{\sqrt 5 -1}2 .$

Thus, the correct answer is B .

23.

一个球内切于如图所示的截头正圆锥。截头圆锥的体积是球体积的两倍。截头圆锥下底半径与上底半径之比是多少？

A sphere is inscribed in a truncated right circular cone as shown. The volume of the truncated cone is twice that of the sphere. What is the ratio of the radius of the bottom base of the truncated cone to the radius of the top base of the truncated cone?

$\dfrac32$

$\dfrac{1+\sqrt5}2$

$\sqrt3$

$2$

$\dfrac{3+\sqrt5}2$

答案：E

知识点：圆锥球体积

难度评级：2300

解答：

设上底半径为 $1$ ，下底半径为 $R$ ，内切球半径为 $a$ 。

截面中球与上下底相切，所以截头圆锥高为 $2a$ 。由侧边、切点半径和两个底半径形成的直角三角形，可得 $R=a^2$ 。

截头圆锥的体积为 $\frac13\pi(R^2+R+1)(2a)$ $=\frac{2a\pi}{3}(a^4+a^2+1)$ 。

这个体积等于球体积的两倍，即 $\frac{8a^3\pi}{3}$ 。约去公因子，得到 $a^4-3a^2+1=0$ ，所以 $R^2-3R+1=0$ 。

取较大根， $R=\frac{3+\sqrt5}{2}$ ，所以正确答案是 E。

Let the top radius be $1$ , the bottom radius be $R$ , and the inscribed sphere radius be $a$ .

In the cross-section, the sphere is tangent to the two bases, so the frustum height is $2a$ . The right triangle formed by the side, a radius to the tangency point, and the base radii gives $R=a^2$ , as in the official diagram.

The frustum volume is $\frac13\pi(R^2+R+1)(2a)$ $=\frac{2a\pi}{3}(a^4+a^2+1)$ .

This is twice the sphere volume, $\frac{8a^3\pi}{3}$ . Cancelling gives $a^4-3a^2+1=0$ , so $R^2-3R+1=0$ .

Thus $R=\frac{3+\sqrt5}{2}$ , and the correct answer is E .

24.

数字 $1, 2, 3, 4, 5$ 要排成一个圆。若并非对每个 $1$ 到 $15$ 的 $n$ ，都能找到圆上一段连续出现的数字使其和为 $n$ ，则称这种排列为 $\textit{bad}$ 。只相差旋转或翻转的排列视为相同。有多少种不同的坏排列？

The numbers $1, 2, 3, 4, 5$ are to be arranged in a circle. An arrangement is $\textit{bad}$ if it is not true that for every $n$ from $1$ to $15$ one can find a subset of the numbers that appear consecutively on the circle that sum to $n.$ Arrangements that differ only by a rotation or a reflection are considered the same. How many different bad arrangements are there?

$1$

$2$

$3$

$4$

$5 .$

答案：B

知识点：环形排列分类讨论

难度评级：2390

解答：

单个数字给出 $1$ 到 $5$ ，它们的补集给出 $10$ 到 $14$ ，全部五个数给出 $15$ 。因此只需要检查能否得到和 $6$ 与 $7$ 。

若无法得到和 $6$ ，则 $1$ 不与 $5$ 相邻。通过旋转和翻转，可写成 $1bc5e$ 。相邻块 $bc$ 不能是 $\{2,3\}$ 或 $\{2,4\}$ ，因为 $1+2+3=6$ 且 $2+4=6$ 。于是 $e=2$ ，再避免连续块 $2,1,3$ ，得到坏排列 $14352$ 。

若无法得到 $7$ ，则 $2$ 不与 $5$ 相邻，可写成 $2bc5e$ 。此时 $bc$ 不能是 $\{3,4\}$ 或 $\{1,4\}$ ，所以 $e=4$ 。再避免连续块 $4,2,1$ ，得到 $b=3,\ c=1$ ，即坏排列 $23154$ 。

这两个排列确实都是坏排列，分别无法得到和 $6$ 与和 $7$ 。因此共有 $2$ 种坏排列。

所以正确答案是 B。

Single numbers give sums $1$ through $5$ , complements give sums $10$ through $14$ , and all five numbers give $15$ . So an arrangement is good exactly when consecutive blocks can make sums $6$ and $7$ .

If sum $6$ is impossible, then $1$ is not adjacent to $5$ . By rotating and reflecting, write the arrangement as $1bc5e$ . The adjacent pair $bc$ cannot be $\{2,3\}$ or $\{2,4\}$ , since $1+2+3=6$ and $2+4=6$ . Thus $e=2$ , and avoiding the consecutive block $2,1,3$ forces the bad arrangement $14352$ .

If sum $7$ is impossible, then $2$ is not adjacent to $5$ . Similarly write the arrangement as $2bc5e$ . Now $bc$ cannot be $\{3,4\}$ or $\{1,4\}$ , so $e=4$ . To avoid the consecutive block $4,2,1$ , the remaining order must be $b=3,\ c=1$ , giving $23154$ .

These two arrangements are indeed bad, one missing sum $6$ and the other missing sum $7$ . Hence there are $2$ bad arrangements.

Thus, the correct answer is B .

25.

一个小池塘中有十一片睡莲叶排成一行，标号为 $0$ 到 $10$ 。一只青蛙坐在 $1$ 号叶上。当青蛙在 $N$ 号叶上且 $0 < N < 10$ 时，它以概率 $\frac{N}{10}$ 跳到 $N-1$ 号叶，以概率 $1-\frac{N}{10}$ 跳到 $N+1$ 号叶。每次跳跃相互独立。

若青蛙到达 $0$ 号叶，它会被一条耐心等待的蛇吃掉。若青蛙到达 $10$ 号叶，它会离开池塘且不再回来。青蛙不被蛇吃掉而逃脱的概率是多少？

In a small pond there are eleven lily pads in a row labeled $0$ through $10.$ A frog is sitting on pad $1.$ When the frog is on pad $N,$ where $0 < N < 10,$ it will jump to pad $N-1$ with probability $\frac{N}{10}$ and to pad $N+1$ with probability $1-\frac{N}{10}.$ Each jump is independent of the previous jumps.

If the frog reaches pad $0$ it will be eaten by a patiently waiting snake. If the frog reaches pad $10$ it will exit the pond, never to return. What is the probability that the frog will escape without being eaten by the snake?

$\dfrac{32}{79}$

$\dfrac{161}{384}$

$\dfrac{63}{146}$

$\dfrac{7}{16}$

$\dfrac{1}{2}$

答案：C

知识点：随机游走递推概率对称性

难度评级：2440

解答：

设 $p_i$ 为从第 $i$ 片叶开始最终逃脱的概率。边界条件为 $p_0=0$ 、 $p_{10}=1$ ，由对称性 $p_5=\frac12$ 。

对 $1\le i\le 4$ ，有递推式 $p_i=\frac{i}{10}p_{i-1}+\frac{10-i}{10}p_{i+1}$ 。

从 $p_5=\frac12$ 向下推，得到 $p_4=\frac25p_3+\frac3{10}$ 和 $p_3=\frac3{10}p_2+\frac7{10}p_4=\frac5{12}p_2+\frac7{24}$ 。

接着可得 $p_2=\frac15p_1+\frac45p_3=\frac3{10}p_1+\frac7{20}$ ，而 $p_1=\frac9{10}p_2$ 。

代入 $p_2$ 的表达式，得到 $p_1=\frac9{10}\left(\frac3{10}p_1+\frac7{20}\right)$ ，所以 $p_1=\frac{63}{146}$ 。

所以正确答案是 C。

Let $p_i$ be the probability that the frog eventually escapes starting from pad $i$ . Then $p_0=0$ , $p_{10}=1$ , and by symmetry $p_5=\frac12$ .

For $1\le i\le 4$ , $p_i=\frac{i}{10}p_{i-1}+\frac{10-i}{10}p_{i+1}$ .

Working downward from $p_5=\frac12$ , we get $p_4=\frac25p_3+\frac3{10}$ , then $p_3=\frac3{10}p_2+\frac7{10}p_4=\frac5{12}p_2+\frac7{24}$ .

Next $p_2=\frac15p_1+\frac45p_3=\frac3{10}p_1+\frac7{20}$ . Finally $p_1=\frac9{10}p_2$ .

Substituting the expression for $p_2$ gives $p_1=\frac9{10}\left(\frac3{10}p_1+\frac7{20}\right)$ , so $p_1=\frac{63}{146}$ .

Thus, the correct answer is C .