2017 AMC 8 真题

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1.

下列哪个值最大?

Which of the following values is largest?

2+0+1+72+0+1+7

2×0+1+72 \times 0 +1+7

2+0×1+72+0 \times 1 + 7

2+0+1×72+0+1 \times 7

2×0×1×72 \times 0 \times 1 \times 7

答案:A
知识点:运算顺序
难度评级:370
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文字解答:

选项 (A) 的值为 1010

选项 (B) 的值为 0+1+7=80 + 1 + 7 = 8

选项 (C) 的值为 2+0+7=92 + 0 + 7 = 9

选项 (D) 的值为 2+0+7=92 + 0 + 7 = 9

选项 (E) 的值为 00

所以正确答案是 A

Option (A) evaluates to 10.10.

Option (B) evaluates to 0+1+7=8.0 + 1 + 7 = 8.

Option (C) evaluates to 2+0+7=9.2 + 0 + 7 = 9.

Option (D) evaluates to 2+0+7=9.2 + 0 + 7 = 9.

Option (E) evaluates to 0.0.

Thus, A is the correct answer.

2.

Alicia、Brenda 和 Colby 是最近学生会主席选举的候选人。下方饼图显示了三位候选人的得票分布。如果 Brenda 得到 3636 票,那么总共投出了多少票?

Alicia, Brenda, and Colby were the candidates in a recent election for student president. The pie chart below shows how the votes were distributed among the three candidates. If Brenda received 3636 votes, then how many votes were cast all together?

7070

8484

100100

106106

120120

答案:E
难度评级:450
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文字解答:

如果 3636 票是总票数的 30%30\%,那么总票数的 10%10\%1212 票。总票数就是 1012=12010 \cdot 12 = 120

所以正确答案是 E

If 3636 votes is 30%30\% of the total votes, then 10%10\% of the total votes is 1212 votes. The number of total votes would then be 1012=120.10 \cdot 12 = 120.

Thus, E is the correct answer.

3.

表达式 1684\sqrt{16\sqrt{8\sqrt{4}}} 的值是多少?

What is the value of the expression 1684?\sqrt{16\sqrt{8\sqrt{4}}}?

44

424\sqrt{2}

88

828\sqrt{2}

1616

答案:C
知识点:根式
难度评级:560
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文字解答:

这个表达式可化简为 1684=1616=64=8 \begin{align*} \sqrt{16\sqrt{8\sqrt{4}}} &= \sqrt{16\sqrt{16}} \\ &= \sqrt{64}\\ &= 8 \end{align*}

所以正确答案是 C

This expression can be reduced as follows: 1684=1616=64=8 \begin{align*} \sqrt{16\sqrt{8\sqrt{4}}} &= \sqrt{16\sqrt{16}} \\ &= \sqrt{64}\\ &= 8 \end{align*}

Thus, C is the correct answer.

4.

0.0003150.000315 乘以 7,928,5647,928,564 的乘积最接近下列哪一个数?

When 0.0003150.000315 is multiplied by 7,928,5647,928,564 the product is closest to which of the following?

210210

240240

2,1002,100

2,4002,400

24,00024,000

答案:D
知识点:估算
难度评级:720
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文字解答:

可以近似为 (3104)(8106)=24102=2400. \begin{align*} (3 \cdot 10^{-4})(8 \cdot 10^6) &= 24 \cdot 10^2 \\ &= 2400. \end{align*}

所以正确答案是 D

We can approximate the product as (3104)(8106)=24102=2400. \begin{align*} (3 \cdot 10^{-4})(8 \cdot 10^6) &= 24 \cdot 10^2 \\ &= 2400. \end{align*}

Thus, D is the correct answer.

5.

表达式 的值是多少? 123456781+2+3+4+5+6+7+8?\dfrac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1+2+3+4+5+6+7+8}?

What is the value of the expression 123456781+2+3+4+5+6+7+8?\dfrac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1+2+3+4+5+6+7+8}?

10201020

11201120

12201220

22402240

33603360

答案:B
知识点:阶乘
难度评级:770
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文字解答:

分母为 1+2++8=36=2361+2+\cdots+8=36=2\cdot3\cdot6。因此 4578=11204\cdot5\cdot7\cdot8=1120

所以正确答案是 B

The denominator is 1+2++8=36=236.1+2+\cdots+8=36=2\cdot3\cdot6. Canceling those three factors from the numerator leaves 4578=1120.4\cdot5\cdot7\cdot8=1120.

Thus, B is the correct answer.

6.

如果一个三角形的角度之比为 3:3:43:3:4,那么这个三角形最大角的度数是多少?

If the degree measures of the angles of a triangle are in the ratio 3:3:4,3:3:4, what is the degree measure of the largest angle of the triangle?

1818

3636

6060

7272

9090

答案:D
难度评级:770
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文字解答:

设三个角为 3x,3x3x, 3x4x4x。它们的和为 180180,所以 3x+3x+4x=1803x + 3x + 4x = 180,即 10x=18010x = 180,从而 x=18x = 18

最大角为 4x=724x = 72

所以正确答案是 D

We can let the three angles be equal to 3x,3x,3x, 3x, and 4x.4x. Then we know that their sum equals 180.180. From this we can set 3x+3x+4x=180,3x + 3x + 4x = 180, and solving this, we get 10x=18010x = 180 and x=18.x = 18.

The largest angle is 4x=72.4x = 72.

Thus, D is the correct answer.

7.

ZZ 是一个六位正整数,例如二十四万七千二百四十七,其前三位数字与后三位数字按相同顺序完全相同。下列哪个数一定也是 ZZ 的因数?

Let ZZ be a 6-digit positive integer, such as 247247, whose first three digits are the same as its last three digits taken in the same order. Which of the following numbers must also be a factor of Z?Z?

1111

1919

101101

111111

11111111

答案:A
难度评级:940
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文字解答:

ZZ 的形式为 nn。则 这说明 1111 一定是 的因数。 Z=1000n+n=1001n=71113n.\begin{aligned} Z&=1000n+n\\ &=1001n\\ &=7\cdot11\cdot13\cdot n. \end{aligned}

所以正确答案是 A

Let nn be the three-digit number formed by either repeated block. Then Z=1000n+n=1001n=71113n.\begin{aligned} Z&=1000n+n\\ &=1001n\\ &=7\cdot11\cdot13\cdot n. \end{aligned} Therefore, 1111 must be a factor of Z.Z.

Thus, A is the correct answer.

8.

Malcolm 今天放学后想去 Isabella 家,他知道她住在哪条街,但不知道门牌号。她告诉他:“我的门牌号是两位数,并且下列四个陈述中恰好有三个是真的。”

(一) 它是质数。

(二) 它是偶数。

(三) 它能被七整除。

(四) 它的一个数字是九。

这些信息使 Malcolm 能确定 Isabella 的门牌号。它的个位数字是多少?

Malcolm wants to visit Isabella after school today and knows the street where she lives but doesn't know her house number. She tells him, "My house number has two digits, and exactly three of the following four statements about it are true."

(1) It is prime.

(2) It is even.

(3) It is divisible by 7.

(4) One of its digits is 9.

This information allows Malcolm to determine Isabella's house number. What is its units digit?

44

66

77

88

99

答案:D
难度评级:1240
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文字解答:

陈述 (一) 和 (二) 不能同时为真,因为唯一的偶质数不是两位数。陈述 (一) 和 (三) 也不能同时为真,因为两位且能被 77 整除的数不是质数。由于只有一个陈述为假,陈述 (一) 必须为假,而 (二)、(三)、(四) 为真。

门牌号能被 2277 整除,所以能被 1414 整除。在两位的 1414 的倍数中,唯一含有数字 99 的是 9898。因此个位数字是 88

所以正确答案是 D

Statements (1) and (2) cannot both be true, because the only even prime is not two-digit. Statements (1) and (3) also cannot both be true, because a two-digit number divisible by 77 would not be prime. Since only one statement is false, statement (1) must be false, while statements (2), (3), and (4) are true.

The house number is divisible by 22 and 7,7, so it is divisible by 14.14. Among the two-digit multiples of 14,14, the only one with a digit of 99 is 98.98. Therefore, the units digit is 8.8.

Thus, D is the correct answer.

9.

Marcy 的所有弹珠都是蓝色、红色、绿色或黄色。她的弹珠中三分之一是蓝色,四分之一是红色,且有六颗绿色弹珠。Marcy 最少可能有多少颗黄色弹珠?

All of Marcy's marbles are blue, red, green, or yellow. One third of her marbles are blue, one fourth of them are red, and six of them are green. What is the smallest number of yellow marbles that Marcy could have?

11

22

33

44

55

答案:D
难度评级:1070
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文字解答:

如果弹珠总数同时能被 3344 整除,那么它必须能被 1212 整除。测试 1212 时,蓝色有 44 颗,红色有 33 颗,最多只剩 55 颗绿色,不可能。

若总数为 2424,则蓝色有 88 颗,红色有 66 颗。黄色弹珠数为 24866=424 - 8 - 6 - 6 = 4

所以正确答案是 D

If the number of marbles is divisible by both 33 and 4,4, then the number must be divisible by 12.12. If we test 12,12, we get that there are 44 blue marbles and 33 red marbles. This leaves a maximum of 55 green marbles, which is not possible.

If there are 2424 marbles, then there are 88 blue marbles and 66 red marbles. To find the number of yellow marbles, we get 24866=4.24 - 8 - 6 - 6 = 4.

Thus, D is the correct answer.

10.

一个盒子中有五张卡片,编号为一、二、三、四、五。从盒中随机不放回地选出三张卡片。所选卡片中的最大值为四的概率是多少?

A box contains five cards, numbered 1, 2, 3, 4, and 5. Three cards are selected randomly without replacement from the box. What is the probability that 4 is the largest value selected?

110\dfrac{1}{10}

15\dfrac{1}{5}

310\dfrac{3}{10}

25\dfrac{2}{5}

12\dfrac{1}{2}

答案:C
知识点:基本概率组合
难度评级:1020
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文字解答:

选择 33 张卡从 55 张卡中共有 (53)=10{5 \choose 3} = 10 种方式。如果 44 是最大值,则必须选中四,另外两张从 {1,2,3}\{1, 2, 3\} 中选,有 (32)=3{3 \choose 2} = 3 种方式。因此概率为 310\dfrac{3}{10}

所以正确答案是 C

The number of ways to choose 33 cards from 55 is (53)=10.{5 \choose 3} = 10. If 44 is the largest value selected, then the other two cards have to be chosen from {1,2,3}.\{1, 2, 3\}. There are (32)=3{3 \choose 2} = 3 ways to do this. The probability is then 310.\dfrac{3}{10}.

Thus, C is the correct answer.

11.

一个正方形地板由全等的正方形瓷砖铺成。如果位于两条对角线上的瓷砖总数是 3737,那么铺满地板共有多少块瓷砖?

A square-shaped floor is covered with congruent square tiles. If the total number of tiles that lie on the two diagonals is 37,37, how many tiles cover the floor?

148148

324324

361361

12961296

13691369

答案:C
难度评级:1140
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两条对角线共有 3737 块瓷砖,表示每条对角线上有 1919 块瓷砖,因为中间有一块重叠。 每行瓷砖数等于一条对角线上的瓷砖数,所以总瓷砖数为 192=36119^2 = 361

所以正确答案是 C

3737 tiles on both diagonals imply that there are 1919 tiles on each diagonal, since one tile overlaps in the middle. The total number of tiles would then be 192=36119^2 = 361 since the number of tiles in each row is equal to the number of tiles in one diagonal.

Thus, C is the correct answer.

12.

大于一的最小正整数在除以四、五、六时余数都是一。它位于下列哪一对数之间?

The smallest positive integer greater than 1 that leaves a remainder of 1 when divided by 4, 5, and 6 lies between which of the following pairs of numbers?

221919

22 and 1919

20203939

2020 and 3939

40405959

4040 and 5959

60607979

6060 and 7979

8080124124

8080 and 124124

答案:D
难度评级:1020
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如果一个数除以 445566 时余数都是 11,那么它比这些数的最小公倍数多一。最小公倍数是 6060,所以最小的这类正整数是 60+1=6160 + 1 = 61

所以正确答案是 D

If a number leaves a remainder of 11 when divided by 4,4, 5,5, and 6,6, then it is one more than the least common multiple of these numbers. The least common multiple is 60,60, so the smallest such positive integer is 60+1=61.60 + 1 = 61.

Thus, D is the correct answer.

13.

Peter、Emma 和 Kyler 互相下棋。Peter 赢了四局,输了二局。Emma 赢了三局,输了三局。如果 Kyler 输了三局,他赢了多少局?

Peter, Emma, and Kyler played chess with each other. Peter won 4 games and lost 2 games. Emma won 3 games and lost 3 games. If Kyler lost 3 games, how many games did he win?

00

11

22

33

44

答案:B
知识点:逻辑推理
难度评级:1020
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因为没有平局,总胜场数必须等于总负场数。总负场数为 2+3+3=82+3+3=8,所以 Kyler 赢了 843=18-4-3=1 局。

所以正确答案是 B

Across all players, every win is matched by one loss; drawn games, if any, contribute neither. The number of losses is 2+3+3=8,2+3+3=8, so the number of games Kyler won is 843=1.8-4-3=1.

Thus, B is the correct answer.

14.

Chloe 和 Zoe 都是 Demeanor 女士数学课上的学生。昨晚她们各自独立完成了家庭作业的一半题目,然后一起完成了另一半。Chloe 独立完成的题中只有 80%80\% 正确,但总体有 88%88\% 的答案正确。Zoe 独立完成的题中 90%90\% 正确。Zoe 的总体正确率是多少?

Chloe and Zoe are both students in Ms. Demeanor's math class. Last night they each solved half of the problems in their homework assignment alone and then solved the other half together. Chloe had correct answers to only 80%80\% of the problems she solved alone, but overall 88%88\% of her answers were correct. Zoe had correct answers to 90%90\% of the problems she solved alone. What was Zoe's overall percentage of correct answers?

8989

9292

9393

9696

9898

答案:C
知识点:百分数
难度评级:1370
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因为题目数量不影响答案,可以假设作业有 100100 道题。80%80\%50504040,所以 Chloe 独立答对 4040 道。88%88\%1001008888,所以 Chloe 总共答对 8888 道。因此共同完成部分她答对 8840=4888 - 40 = 48 道。

90%90\%50504545,所以 Zoe 独立答对 4545 道。共同完成部分她也答对 4848 道,所以总共答对 45+48=9345 + 48 = 93 道,整体正确率为 93%93\%

所以正确答案是 C

Since the answer is the same regardless of the number of problems, we can assume that there were 100100 problems on the assignment. 80%80\% of 5050 is 40,40, so Chloe answered 4040 questions correctly alone. 88%88\% of 100100 is 88,88, so Chloe answered 8888 questions correctly in total. This means that Chloe answered 8840=4888 - 40 = 48 together with Zoe.

90%90\% of 5050 is 45,45, so Zoe answered 4545 questions correctly by herself. We know that she answered 4848 questions correctly with Chloe, so she answered 45+48=9345 + 48 = 93 correctly in total. This means that her overall percentage is 93%.93\%.

Thus, C is the correct answer.

15.

在下方字母和数字的排列中,有多少条不同路径可以拼出 AMC8?从中间的 A 开始,路径每一步只能移动到上下左右相邻的字母或数字,不能斜着走。图中描出了一条这样的路径。

In the arrangement of letters and numerals below, by how many different paths can one spell AMC8? Beginning at the A in the middle, a path only allows moves from one letter to an adjacent (above, below, left, or right, but not diagonal) letter. One example of such a path is traced in the picture.

88

99

1212

2424

3636

答案:D
知识点:乘法原理
难度评级:1220
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AA 出发,有 44 种方式到达一个 MM。 从每个 MM 出发,有 33 种方式到达一个 CC。 从每个 CC 出发,有 22 种方式到达一个 88。 因此路径总数为 432=244 \cdot 3 \cdot 2 = 24

所以正确答案是 D

Starting from A,A, there are 44 ways to reach an M.M. From each M,M, there are 33 ways to reach a C.C. From each C,C, there are 22 ways to reach an 8.8. Multiplying all these possibilities, we get 432=24.4 \cdot 3 \cdot 2 = 24.

Thus, D is the correct answer.

16.

在下图中,在 BC\overline{BC} 上选择点 DD,使得 ACD\triangle ACDABD\triangle ABD 的周长相等。ABD\triangle ABD 的面积是多少?

In the figure below, choose point DD on BC\overline{BC} so that ACD\triangle ACD and ABD\triangle ABD have equal perimeters. What is the area of ABD?\triangle ABD?

34\dfrac{3}{4}

32\dfrac{3}{2}

22

125\dfrac{12}{5}

52\dfrac{5}{2}

答案:D
知识点:面积比周长
难度评级:1420
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要使两个三角形周长相等,BC\overline{BC} 必须分成 CD=3\overline{CD} = 3BD=2\overline{BD} = 2

ACD\triangle ACDABD\triangle ABD 有相同的高,所以面积与底边长度成比例。ABD\triangle ABD 的面积是 25\dfrac{2}{5}ABC\triangle ABC 的面积,因此 25342=125.\dfrac{2}{5} \cdot 3 \cdot \dfrac{4}{2} = \dfrac{12}{5}.

所以正确答案是 D

The only way to split BC\overline{BC} into two parts such that the two triangles have the same perimeter is if CD=3\overline{CD} = 3 and BD=2.\overline{BD} = 2.

ACD\triangle ACD and ABD\triangle ABD have the same altitudes, so their areas are proportional to their bases. This means that the area of ABD\triangle ABD is 25\dfrac{2}{5} the area of ABC,\triangle ABC, which is 25342=125.\dfrac{2}{5} \cdot 3 \cdot \dfrac{4}{2} = \dfrac{12}{5}.

Thus, D is the correct answer.

17.

我有一些金币和一些空宝箱。我试着每个宝箱放九枚金币,但这样会剩下二个宝箱空着。于是我改为每个宝箱放六枚金币,但这样会剩下三枚金币。我有多少枚金币?

Starting with some gold coins and some empty treasure chests, I tried to put 9 gold coins in each treasure chest, but that left 2 treasure chests empty. So instead I put 6 gold coins in each treasure chest, but then I had 3 gold coins left over. How many gold coins did I have?

99

2727

4545

6363

8181

答案:C
知识点:方程组
难度评级:1240
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设宝箱数为 nn,金币数为 gg。 则 且 解这个方程组得 n=7n = 7。 因此金币数为 67+3=456 \cdot 7 + 3 = 459(n2)=g9(n - 2) = g 6n+3=g.6n + 3 = g.

所以正确答案是 C

Let nn be the number of treasure chests and gg be the number of gold coins. Then 9(n2)=g9(n - 2) = g and 6n+3=g.6n + 3 = g. Solving this system yields n=7,n = 7, so the number of gold coins is 67+3=45.6 \cdot 7 + 3 = 45.

Thus, C is the correct answer.

18.

如下图所示,非凸四边形 ABCDABCD 中,BCD\angle BCD 是直角,AB=12AB=12BC=4BC=4CD=3CD=3AD=13AD=13。四边形 ABCDABCD 的面积是多少?

In the non-convex quadrilateral ABCDABCD shown below, BCD\angle BCD is a right angle, AB=12,AB=12, BC=4,BC=4, CD=3,CD=3, and AD=13.AD=13. What is the area of quadrilateral ABCD?ABCD?

1212

2424

2626

3030

3636

答案:B
难度评级:1430
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因为 BCD\angle BCD 是直角,对 BCD\triangle BCD 用勾股定理得 BD=5\overline{BD} = 5。又可得 DBA\angle DBA 是直角,因为 BDA\triangle BDA 的三边构成勾股三元组。

四边形 ABCDABCD 的面积等于 area(BDA)area(BCD)=121251243=306=24.\begin{align*} \text{area}(\triangle &BDA) - \text{area}(\triangle BCD) \\ &= \dfrac{1}{2} \cdot 12 \cdot 5 - \dfrac{1}{2} \cdot 4 \cdot 3 \\ &= 30 - 6 \\ &= 24. \end{align*}

所以正确答案是 B

Since BCD\angle BCD is a right angle, we can apply the Pythagorean theorem to BCD\triangle BCD to get that BD=5.\overline{BD} = 5. We also get that DBA\angle DBA is right since the sides of BDA\triangle BDA form a Pythagorean triple.

Then the area of ABCDABCD is equal to area(BDA)area(BCD)=121251243=306=24.\begin{align*} \text{area}(\triangle &BDA) - \text{area}(\triangle BCD) \\ &= \dfrac{1}{2} \cdot 12 \cdot 5 - \dfrac{1}{2} \cdot 4 \cdot 3 \\ &= 30 - 6 \\ &= 24. \end{align*}

Thus, B is the correct answer.

19.

对任意正整数 MM,记号 M!M! 表示从 11MM 的所有整数的乘积。使 5n5^n 成为 的因数的最大整数 nn 是多少? 98!+99!+100!98!+99!+100!

For any positive integer M,M, the notation M!M! denotes the product of the integers 11 through M.M. What is the largest integer nn for which 5n5^n is a factor of the sum: 98!+99!+100!98!+99!+100!

2323

2424

2525

2626

2727

答案:D
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该表达式可分解为 98!98!98!+99!+100!=98!(1+99+10099)=98!1002.\begin{aligned} 98!+99!+100! &=98!\bigl(1+99\\ &\qquad+100\cdot99\bigr)\\ &=98!\cdot100^2. \end{aligned}

每个 98/5+98/25\left\lfloor98/5\right\rfloor+\left\lfloor98/25\right\rfloor 含有两个因子 5598!98! 中因子 55 的个数为 19+3=2219+3=22 1002100^2 22+4=2622+4=26。这分别统计了能被 整除的数和能被 整除的数,从而计入第二个 因子。因此因子 的总数为 。

所以正确答案是 D

Factor out 98!98!: 98!+99!+100!=98!(1+99+10099)=98!1002.\begin{aligned} 98!+99!+100! &=98!\bigl(1+99\\ &\qquad+100\cdot99\bigr)\\ &=98!\cdot100^2. \end{aligned}

The exponent of 55 in 98!98! is 98/5+98/25\left\lfloor98/5\right\rfloor+\left\lfloor98/25\right\rfloor, which equals 19+3=22.19+3=22. The factor 1002100^2 contributes four more factors of 55, so the total exponent is 22+4=26.22+4=26.

Thus, D is the correct answer.

20.

1000100099999999(含端点)之间随机选一个整数。它是奇数且各位数字互不相同的概率是多少?

An integer between 10001000 and 9999,9999, inclusive, is chosen at random. What is the probability that it is an odd integer whose digits are all distinct?

1475\dfrac{14}{75}

56225\dfrac{56}{225}

107400\dfrac{107}{400}

725\dfrac{7}{25}

925\dfrac{9}{25}

答案:B
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这个数是奇数,所以个位数字有 55 种选择。千位数字不能为零,也不能等于个位数字,因此有 88 种选择。百位有 88 种选择,十位有 77 种选择。符合条件的数共有 5887=22405 \cdot 8 \cdot 8 \cdot 7 = 2240,所以概率为 22409000=56225\dfrac{2240}{9000} = \dfrac{56}{225}

所以正确答案是 B

Since the number is odd, the last digit is odd, giving 55 possibilities. The thousands digit cannot be zero or the number we already got, so that gives 88 possibilities. Similarly, the hundreds digit has 88 possibilities, and the tens digit has 77 possibilities. This gives a total of 5887=2240,5 \cdot 8 \cdot 8 \cdot 7 = 2240, making the probability 22409000=56225.\dfrac{2240}{9000} = \dfrac{56}{225}.

Thus, B is the correct answer.

21.

假设 aabbcc 是非零实数,且 a+b+c=0a+b+c=0。下列表达式可能的值是什么? aa+bb+cc+abcabc?\dfrac{a}{|a|}+\dfrac{b}{|b|}+\dfrac{c}{|c|}+\dfrac{abc}{|abc|}?

Suppose a,a, b,b, and cc are nonzero real numbers, and a+b+c=0.a+b+c=0. What are the possible value(s) for aa+bb+cc+abcabc?\dfrac{a}{|a|}+\dfrac{b}{|b|}+\dfrac{c}{|c|}+\dfrac{abc}{|abc|}?

00

111-1

11 and 1-1

222-2

22 and 2-2

00222-2

00, 22, and 2-2

00111-1

00, 11, and 1-1

答案:A
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因为 a+b+c=0a+b+c=0 且三个数都不为零,所以它们的符号只能是两个正数、一个负数,或两个负数、一个正数。第一种情况下,前三个符号分式的和为 11,而 abc/abc=1abc/|abc|=-1;第二种情况下,前三个分式的和为 1-1,而 abc/abc=1abc/|abc|=1。无论哪种情况,整个表达式都等于 00

所以正确答案是 A

Because a+b+c=0a+b+c=0 and none of the numbers is zero, their signs are either two positive and one negative or two negative and one positive. In the first case, the first three sign fractions sum to 11, while abc/abc=1abc/|abc|=-1. In the second case, the first three sum to 1-1, while abc/abc=1abc/|abc|=1. Either way, the entire expression equals 00.

Thus, A is the correct answer.

22.

在直角三角形 ABCABC 中,AC=12AC=12BC=5BC=5,且角 CC 是直角。如图,一个半圆内切于该三角形。这个半圆的半径是多少?

In the right triangle ABC,ABC, AC=12,AC=12, BC=5,BC=5, and angle CC is a right angle. A semicircle is inscribed in the triangle as shown. What is the radius of the semicircle?

76\dfrac{7}{6}

135\dfrac{13}{5}

5918\dfrac{59}{18}

103\dfrac{10}{3}

6013\dfrac{60}{13}

答案:D
知识点:切线相似
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OO 为内切半圆圆心,DD 为半圆与 AB\overline{AB} 的切点。于是 BD=5BD = 5,因为 BD\overline{BD}BC\overline{BC} 是从同一点作出的切线,所以 AD=8AD = 8。设 OD=rOD = r,且 OD\overline{OD} 垂直于 AB\overline{AB}。因此 ADOBCA\triangle ADO \sim \triangle BCA,有 r8=512\dfrac{r}{8} = \dfrac{5}{12},解得 r=103r = \dfrac{10}{3}

所以正确答案是 D

Let OO be the center of the inscribed semicircle and DD be the tangent point of the semicircle on AB.\overline{AB}. Then BD=5BD = 5 since BD\overline{BD} and BC\overline{BC} are tangents to the semicircle. Then AD=8AD = 8 and OD=r.OD = r. OD\overline{OD} is perpendicular to AB\overline{AB} so ADOBCA,\triangle ADO \sim \triangle BCA, so r8=512.\dfrac{r}{8} = \dfrac{5}{12}. Solving this, we get r=103.r = \dfrac{10}{3}.

Thus, D is the correct answer.

23.

连续四天,Linda 每天旅行一小时,并且她的速度使她每行进一英里所需分钟数都是整数。第一天之后,每天她的速度都降低,使得每行进一英里所需分钟数比前一天多 55 分钟。四天中每天她行进的距离也都是整数英里。这四次旅行一共多少英里?

Each day for four days, Linda traveled for one hour at a speed that resulted in her traveling one mile in an integer number of minutes. Each day after the first, her speed decreased so that the number of minutes to travel one mile increased by 55 minutes over the preceding day. Each of the four days, her distance traveled was also an integer number of miles. What was the total number of miles for the four trips?

1010

1515

2525

5050

8282

答案:C
知识点:因数等差数列
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Linda 每天旅行 6060 分钟。因为每天每英里所需分钟数是整数,且每天行进英里数也是整数,所以每天的每英里分钟数必须是 6060 的因数。6060 的因数为 1,2,3,4,5,6,10,12,15,20,301, 2, 3, 4, 5, 6, 10, 12, 15, 20, 306060 其中唯一一组四个相邻相差 55 的数是 5,10,155, 10, 152020。四天总路程为 605+6010+6015+6020=25\dfrac{60}{5} + \dfrac{60}{10} + \dfrac{60}{15} + \dfrac{60}{20} = 25 英里。

所以正确答案是 C

Linda traveled for 6060 minutes every day. Since one mile was traveled in an integer amount of minutes each day, her minutes per mile every day must be a factor of 60.60. The factors of 6060 are 1,2,3,4,5,6,10,12,15,20,30,1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60.60. The only sequence of four of these numbers that differ by 55 are 5,10,15,5, 10, 15, and 20.20. For the four days, she traveled 605+6010+6015+6020=25\dfrac{60}{5} + \dfrac{60}{10} + \dfrac{60}{15} + \dfrac{60}{20} = 25 miles.

Thus, C is the correct answer.

24.

Sanders 夫人有三个孙辈,他们定期给她打电话。一个每三天打一次,一个每四天打一次,一个每五天打一次。三人都在二千零一十六年十二月三十一日给她打了电话。接下来的一年中,有多少天她没有接到任何孙辈的电话?

Mrs. Sanders has three grandchildren, who call her regularly. One calls her every three days, one calls her every four days, and one calls her every five days. All three called her on December 31, 2016. On how many days during the next year did she not receive a phone call from any of her grandchildren?

7878

8080

144144

146146

152152

答案:D
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在一个 6060 天周期中,第一个孙辈打 2020 次,第二个打 1515 次,第三个打 1212 次,所以 20+15+12=4720 + 15 + 12 = 47。重复统计的天数为 60/12=560 / 12 = 560/15=460 / 15 = 460/20=360 / 20 = 3,因此先得 4747,再减去重复得到 47543=3547 - 5 - 4 - 3 = 35

6060 天三人都打电话,被加了三次又减了三次,所以要加回一次,得到每 6060 天有 3636 天收到电话,也就是 2424 天没有电话。二千零一十七年有 66 个完整的 6060 天周期,剩下第 361361 天和第 362362 天没有电话,因此没有电话的天数为 246+2=14624 \cdot 6 + 2 = 146

所以正确答案是 D

In a 6060-day period, the first child calls 2020 times, the second child calls 1515 times, and the third child calls 1212 times. 20+15+12=4720 + 15 + 12 = 47 overcounts, however. The first and second children call on the same day 60/12=560 / 12 = 5 times. The first and third children call on the same day 60/15=460 / 15 = 4 times. The second and third children call on the same day 60/20=360 / 20 = 3 times. Subtracting these from 4747 yields 47543=35.47 - 5 - 4 - 3 = 35.

The 6060th day is added in thrice and subtracted out thrice, so we need to add it back in. This means that for every 6060 days, Mrs. Sanders receives a call 3636 days, which means that she does not receive a call on 2424 days. There are 66 6060-day periods, and there are no calls on the 361361st or 362362nd day, which results in 246+2=14624 \cdot 6 + 2 = 146 total days with no phone calls.

Thus, D is the correct answer.

25.

如图,US\overline{US}UT\overline{UT} 都是长度为二的线段,且 mTUS=60m\angle TUS = 60^\circ

TR\overset{\large\frown}{TR}SR\overset{\large\frown}{SR} 都是半径为二的圆的六分之一。图中区域面积是多少?

In the figure shown, US\overline{US} and UT\overline{UT} are line segments each of length 2, and mTUS=60.m\angle TUS = 60^\circ.

Arcs TR\overset{\large\frown}{TR} and SR\overset{\large\frown}{SR} are each one-sixth of a circle with radius 2. What is the area of the region shown?

33π3\sqrt{3}-\pi

434π34\sqrt{3}-\dfrac{4\pi}{3}

232\sqrt{3}

432π34\sqrt{3}-\dfrac{2\pi}{3}

4+4π34+\dfrac{4\pi}{3}

答案:B
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可以延长 SU\overline{SU}TU\overline{TU},形成如下图形。

所求面积等于边长为 44 的等边三角形面积,减去两个半径为 22 的六分之一圆面积。边长为 ss 的等边三角形面积为 s234\dfrac{s^2\sqrt{3}}{4}。因此总面积为 423413π22=4343π\dfrac{4^2 \sqrt{3}}{4} - \dfrac{1}{3} \pi \cdot 2^2 = 4 \sqrt{3} - \dfrac{4}{3} \pi

所以正确答案是 B

We can extend SU\overline{SU} and TU\overline{TU} to form the following picture.

The area of this region is the area of an equilateral triangle with side length of 44 minus the area of two-sixths of a circle with radius 2.2. The area for an equilateral triangle with side length ss is s234.\dfrac{s^2\sqrt{3}}{4}. This means that the total area is 423413π22=4343π.\dfrac{4^2 \sqrt{3}}{4} - \dfrac{1}{3} \pi \cdot 2^2 = 4 \sqrt{3} - \dfrac{4}{3} \pi.

Thus, B is the correct answer.