2017 AMC 8 详解与视频讲解 | LIVE by Po-Shen Loh

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2017 AMC 8 详解

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所有题目均经美国数学协会（MAA）官方合法授权使用。

1.

下列哪个值最大？

Which of the following values is largest?

$2+0+1+7$

$2 \times 0 +1+7$

$2+0 \times 1 + 7$

$2+0+1 \times 7$

$2 \times 0 \times 1 \times 7$

知识点：运算顺序

难度评级：370

小提示：

按通常运算顺序计算每个表达式

Evaluate each expression using the usual order of operations.

大提示：

任何乘以 $0$ 的表达式都会损失部分或全部数值

Any expression that multiplies by $0$ will lose part or all of its value.

视频讲解：

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文字解答：

选项 $(A)$ 的值为 $10$ 。

选项 $(B)$ 的值为 $0 + 1 + 7 = 8$ 。

选项 $(C)$ 的值为 $2 + 0 + 7 = 9$ 。

选项 $(D)$ 的值为 $2 + 0 + 7 = 9$ 。

选项 $(E)$ 的值为 $0$ 。

所以正确答案是 A。

Option $(A)$ evaluates to $10.$

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

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

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

Option $(E)$ evaluates to $0.$

Thus, A is the correct answer.

2.

艾丽西亚、布伦达和科尔比是最近一次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。如果布伦达得到 $36$ 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 $36$ votes, then how many votes were cast all together?

$70$

$84$

$100$

$106$

$120$

知识点：百分数数据与图表解读

难度评级：450

小提示：

布伦达的扇形占饼图的 $30\%$

Brenda’s slice is $30\%$ of the pie chart.

大提示：

如果 $30\%$ 对应 $36$ 票，先求 $10\%$ 对应多少票

If $30\%$ corresponds to $36$ votes, find $10\%$ first.

视频讲解：

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文字解答：

如果 $36$ 票是总票数的 $30\%$ ，那么总票数的 $10\%$ 是 $12$ 票。总票数就是 $10 \cdot 12 = 120$ 。

所以正确答案是 E。

If $36$ votes is $30\%$ of the total votes, then $10\%$ of the total votes is $12$ votes. The number of total votes would then be $10 \cdot 12 = 120.$

Thus, E is the correct answer.

3.

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

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

$4$

$4\sqrt{2}$

$8$

$8\sqrt{2}$

$16$

知识点：根式

难度评级：560

小提示：

从最里面的平方根开始

Start with the innermost square root.

大提示：

化简 $\sqrt{4}$ 后，中间的根式会更容易

After simplifying $\sqrt{4}$ , the middle radical becomes easier.

视频讲解：

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文字解答：

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

所以正确答案是 C。

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

Thus, C is the correct answer.

4.

$0.000315$ 乘以 $7{,}928{,}564$ 的乘积最接近下列哪一个数？

When $0.000315$ is multiplied by $7{,}928{,}564$ the product is closest to which of the following?

$210$

$240$

$2{,}100$

$2{,}400$

$24{,}000$

知识点：估算

难度评级：720

小提示：

将每个因数四舍五入到一位有效数字

Round each factor to one significant digit.

大提示：

将 $0.000315$ 看作约 $3\cdot10^{-4}$ ，将 $7{,}928{,}564$ 看作约 $8\cdot10^6$

Think of $0.000315$ as about $3\cdot10^{-4}$ and $7{,}928{,}564$ as about $8\cdot10^6$ .

视频讲解：

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文字解答：

可以将乘积近似为 $\begin{align*} (3 \cdot 10^{-4})(8 \cdot 10^6) &= 24 \cdot 10^2 \\ &= 2{,}400 \end{align*}\text{。}$

所以正确答案是 D。

We can approximate the product as $\begin{align*} (3 \cdot 10^{-4})(8 \cdot 10^6) &= 24 \cdot 10^2 \\ &= 2{,}400. \end{align*}$

Thus, D is the correct answer.

5.

下面这个表达式的值是多少 $\dfrac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1+2+3+4+5+6+7+8}\text{？}$

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

$1020$

$1120$

$1220$

$2240$

$3360$

知识点：阶乘

难度评级：770

小提示：

先计算分母的和

First add the denominator.

大提示：

得到分母后，先约分而不是把分子全部乘出来

After the denominator is known, cancel factors from the numerator instead of multiplying everything out.

视频讲解：

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文字解答：

分母为 $1+2+\cdots+8=36=2\cdot3\cdot6$ 。从分子中约去这三个因数，剩下 $4\cdot5\cdot7\cdot8=1120$ 。

所以正确答案是 B。

The denominator is $1+2+\cdots+8=36=2\cdot3\cdot6.$ Canceling those three factors from the numerator leaves $4\cdot5\cdot7\cdot8=1120.$

Thus, B is the correct answer.

6.

如果一个三角形的角度之比为 $3:3:4$ ，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 $3:3:4,$ what is the degree measure of the largest angle of the triangle?

$18$

$36$

$60$

$72$

$90$

知识点：角度和比与比例

难度评级：770

小提示：

设三个角分别为 $3x$ 、 $3x$ 和 $4x$

Let the three angle measures be $3x,$ $3x,$ and $4x$ .

大提示：

三角形内角和为 $180^\circ$

The angles of a triangle add to $180^\circ$ .

视频讲解：

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文字解答：

设三个角为 $3x$ 、 $3x$ 和 $4x$ 。它们的和为 $180$ ，所以 $3x + 3x + 4x = 180$ ，即 $10x = 180$ ，从而 $x = 18$ 。

最大角为 $4x = 72$ 。

所以正确答案是 D。

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

The largest angle is $4x = 72.$

Thus, D is the correct answer.

7.

设 $Z$ 是一个 $6$ 位正整数，例如 $247247$ ，其前三位数字与后三位数字按相同顺序完全相同。下列哪个数一定是 $Z$ 的因数？

Let $Z$ 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 be a factor of $Z?$

$11$

$19$

$101$

$111$

$1111$

知识点：质因数分解位值

难度评级：940

小提示：

设 $n$ 为重复出现的三位数。用 $n$ 表示 $Z$

Let $n$ be the three-digit block that is repeated. Express $Z$ in terms of $n$ .

大提示：

分解 $1001$

Factor $1001$ .

视频讲解：

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文字解答：

设 $n$ 为任意一个重复数块所组成的三位数。那么 $\begin{aligned} Z&=1000n+n\\ &=1001n\\ &=7\cdot11\cdot13\cdot n \end{aligned}\text{。}$ 因此， $11$ 一定是 $Z$ 的因数。

所以正确答案是 A。

Let $n$ be the three-digit number formed by either repeated block. Then $\begin{aligned} Z&=1000n+n\\ &=1001n\\ &=7\cdot11\cdot13\cdot n. \end{aligned}$ Therefore, $11$ must be a factor of $Z.$

Thus, A is the correct answer.

8.

马尔科姆今天放学后想去伊莎贝拉家。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。”

$(1)$ 它是质数。

$(2)$ 它是偶数。

$(3)$ 它能被 $7$ 整除。

$(4)$ 它的一个数字是 $9$ 。

这些信息足以让马尔科姆确定伊莎贝拉的门牌号。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?

$4$

$6$

$7$

$8$

$9$

知识点：逻辑推理整除性

难度评级：1240

小提示：

恰好三个陈述为真，所以只有一个陈述为假

Since exactly three statements are true, only one statement can fail.

大提示：

一个两位数不可能既是质数又是大于 $2$ 的偶数，也不可能既是质数又能被 $7$ 整除。

A two-digit number cannot be both prime and an even multiple greater than $2$ , and it cannot be both prime and divisible by $7$ .

视频讲解：

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文字解答：

陈述 $(1)$ 和 $(2)$ 不能同时为真，因为唯一的偶质数不是两位数。陈述 $(1)$ 和 $(3)$ 也不能同时为真，因为能被 $7$ 整除的两位数不是质数。由于只有一个陈述为假，陈述 $(1)$ 必须为假，而陈述 $(2)$ 、 $(3)$ 和 $(4)$ 为真。

门牌号能被 $2$ 和 $7$ 整除，所以能被 $14$ 整除。在两位的 $14$ 的倍数中，唯一含有数字 $9$ 的是 $98$ 。因此个位数字是 $8$ 。

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 $7$ 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 $2$ and $7,$ so it is divisible by $14.$ Among the two-digit multiples of $14,$ the only one with a digit of $9$ is $98.$ Therefore, the units digit is $8.$

Thus, D is the correct answer.

9.

玛茜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。玛茜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?

$1$

$2$

$3$

$4$

$5$

知识点：分数最小公倍数

难度评级：1070

小提示：

弹珠总数必须同时能被 $3$ 和 $4$ 整除

The total number of marbles must be divisible by both $3$ and $4$ .

大提示：

测试最小的 $12$ 的倍数，直到能容纳 $6$ 颗绿色弹珠

Test the smallest multiples of $12$ until there is room for the $6$ green marbles.

视频讲解：

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文字解答：

如果弹珠总数同时能被 $3$ 和 $4$ 整除，那么它必须能被 $12$ 整除。测试 $12$ 时，蓝色有 $4$ 颗，红色有 $3$ 颗，最多只剩 $5$ 颗绿色，不可能。

若总数为 $24$ ，则蓝色有 $8$ 颗，红色有 $6$ 颗。黄色弹珠数为 $24 - 8 - 6 - 6 = 4$ 。

所以正确答案是 D。

If the number of marbles is divisible by both $3$ and $4,$ then the number must be divisible by $12.$ If we test $12,$ we get that there are $4$ blue marbles and $3$ red marbles. This leaves a maximum of $5$ green marbles, which is not possible.

If there are $24$ marbles, then there are $8$ blue marbles and $6$ red marbles. To find the number of yellow marbles, we get $24 - 8 - 6 - 6 = 4.$

Thus, D is the correct answer.

10.

一个盒子中有五张卡片，编号为 $1$ 、 $2$ 、 $3$ 、 $4$ 和 $5$ 。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 $4$ 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?

$\dfrac{1}{10}$

$\dfrac{1}{5}$

$\dfrac{3}{10}$

$\dfrac{2}{5}$

$\dfrac{1}{2}$

知识点：基本概率组合

难度评级：1020

小提示：

统计从 $5$ 张卡中选 $3$ 张的所有方式。

Count all ways to choose $3$ cards from the $5$ cards.

大提示：

如果 $4$ 是所选卡片中的最大值，另外两张必须从 $1$ 、 $2$ 和 $3$ 中选择

If $4$ is the largest selected card, the other two selected cards must come from $1,$ $2,$ and $3$ .

视频讲解：

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文字解答：

从 $5$ 张卡中选 $3$ 张共有 ${5 \choose 3} = 10$ 种方式。如果 $4$ 是所选卡片中的最大值，那么另外两张必须从 $\{1, 2, 3\}$ 中选，有 ${3 \choose 2} = 3$ 种方式。因此概率为 $\dfrac{3}{10}$ 。

所以正确答案是 C。

The number of ways to choose $3$ cards from $5$ is ${5 \choose 3} = 10.$ If $4$ is the largest value selected, then the other two cards have to be chosen from $\{1, 2, 3\}.$ There are ${3 \choose 2} = 3$ ways to do this. The probability is then $\dfrac{3}{10}.$

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 $37$ ，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 $37,$ how many tiles cover the floor?

$148$

$324$

$361$

$1296$

$1369$

知识点：对角线完全平方数

难度评级：1140

小提示：

在奇数乘奇数的正方形中，两条对角线只共享中心瓷砖

In an odd-by-odd square, the two diagonals share exactly the center tile.

大提示：

如果每条对角线上有 $n$ 块瓷砖，两条对角线合起来有 $2n-1$ 块

If each diagonal has $n$ tiles, the two diagonals together contain $2n-1$ tiles.

视频讲解：

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文字解答：

两条对角线共有 $37$ 块瓷砖，表示每条对角线上有 $19$ 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 $19^2 = 361$ 。

所以正确答案是 C。

$37$ tiles on both diagonals imply that there are $19$ tiles on each diagonal, since one tile overlaps in the middle. The total number of tiles would then be $19^2 = 361$ 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.

在大于 $1$ 的正整数中，除以 $4$ 、 $5$ 和 $6$ 时余数都是 $1$ 的最小的那个数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?

$2$ 和 $19$

$2$ and $19$

$20$ 和 $39$

$20$ and $39$

$40$ 和 $59$

$40$ and $59$

$60$ 和 $79$

$60$ and $79$

$80$ 和 $124$

$80$ and $124$

知识点：最小公倍数模运算

难度评级：1020

小提示：

除以 $4$ 、 $5$ 和 $6$ 都余 $1$ 的数，比这些数的最小公倍数的倍数多一

Numbers with remainder $1$ after division by $4,$ $5,$ and $6$ are one more than multiples of their least common multiple.

大提示：

求 $4$ 、 $5$ 和 $6$ 的最小公倍数，再加 $1$

Find the least common multiple of $4,$ $5,$ and $6$ , then add $1$ .

视频讲解：

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文字解答：

如果一个数除以 $4$ 、 $5$ 和 $6$ 时余数都是 $1$ ，那么它比这些数的最小公倍数多一。最小公倍数是 $60$ ，所以最小的这类正整数是 $60 + 1 = 61$ 。

所以正确答案是 D。

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

Thus, D is the correct answer.

13.

彼得、艾玛和凯勒互相下棋。彼得赢了 $4$ 局，输了 $2$ 局。艾玛赢了 $3$ 局，输了 $3$ 局。如果凯勒输了 $3$ 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?

$0$

$1$

$2$

$3$

$4$

知识点：逻辑推理

难度评级：1020

小提示：

每一场胜利都对应另一个人的一场失败

Every recorded win contributes one recorded loss to another player.

大提示：

比较三人总胜场数和总负场数

Compare total wins and total losses across all three players.

视频讲解：

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文字解答：

把三个人的记录合在一起，每一场胜利都对应一场失败；和棋若有，则对胜负总数都没有贡献。总负场数为 $2+3+3=8$ ，所以凯勒赢了 $8-4-3=1$ 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 $2+3+3=8,$ so the number of games Kyler won is $8-4-3=1.$

Thus, B is the correct answer.

14.

克洛伊和佐伊都是迪米纳老师数学课上的学生。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。克洛伊独立完成的题中只有 $80\%$ 正确，但总体有 $88\%$ 的答案正确。佐伊独立完成的题中 $90\%$ 正确。佐伊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 $80\%$ of the problems she solved alone, but overall $88\%$ of her answers were correct. Zoe had correct answers to $90\%$ of the problems she solved alone. What was Zoe’s overall percentage of correct answers?

$89$

$92$

$93$

$96$

$98$

知识点：百分数

难度评级：1370

小提示：

假设家庭作业有 $100$ 道题，使百分比具体化

Assume there are $100$ homework problems to make the percentages concrete.

大提示：

用克洛伊的总成绩求出共同完成部分答对了多少题

Use Chloe’s overall score to determine how many jointly solved answers were correct.

视频讲解：

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文字解答：

因为题目数量不影响答案，可以假设作业有 $100$ 道题。 $80\%$ 的 $50$ 是 $40$ ，所以克洛伊独立答对 $40$ 道。 $88\%$ 的 $100$ 是 $88$ ，所以克洛伊总共答对 $88$ 道。因此共同完成部分她答对 $88 - 40 = 48$ 道。

$90\%$ 的 $50$ 是 $45$ ，所以佐伊独立答对 $45$ 道。共同完成部分她也答对 $48$ 道，所以总共答对 $45 + 48 = 93$ 道，整体正确率为 $93\%$ 。

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 $100$ problems on the assignment. $80\%$ of $50$ is $40,$ so Chloe answered $40$ questions correctly alone. $88\%$ of $100$ is $88,$ so Chloe answered $88$ questions correctly in total. This means that Chloe answered $88 - 40 = 48$ together with Zoe.

$90\%$ of $50$ is $45,$ so Zoe answered $45$ questions correctly by herself. We know that she answered $48$ questions correctly with Chloe, so she answered $45 + 48 = 93$ correctly in total. This means that her overall percentage is $93\%.$

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 $\mathrm{AMC8}$ ？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 $\mathrm{AMC8}$ ? Beginning at the $A$ in the middle, a path allows only 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.

$8$

$9$

$12$

$24$

$36$

知识点：乘法原理

难度评级：1220

小提示：

从中心的 $A$ 出发，逐步统计选择数

Count choices one step at a time from the central $A$ .

大提示：

对每个 $M$ ，统计相邻的 $C$ ，再对每个 $C$ ，统计相邻的 $8$

From each $M$ , count adjacent $C$ ’s, then from each $C$ , count adjacent $8$ ’s.

视频讲解：

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文字解答：

从 $A$ 出发，有 $4$ 种方式到达一个 $M$ 。从每个 $M$ 出发，有 $3$ 种方式到达一个 $C$ 。从每个 $C$ 出发，有 $2$ 种方式到达一个 $8$ 。因此路径总数为 $4 \cdot 3 \cdot 2 = 24$ 。

所以正确答案是 D。

Starting from $A,$ there are $4$ ways to reach an $M.$ From each $M,$ there are $3$ ways to reach a $C.$ From each $C,$ there are $2$ ways to reach an $8.$ Multiplying all these possibilities, we get $4 \cdot 3 \cdot 2 = 24.$

Thus, D is the correct answer.

16.

在下图中，在边 $BC$ 上选择点 $D$ ，使得 $\triangle ACD$ 和 $\triangle ABD$ 的周长相等。 $\triangle ABD$ 的面积是多少？

In the figure shown below, choose point $D$ on side $BC$ so that $\triangle ACD$ and $\triangle ABD$ have equal perimeters. What is the area of $\triangle ABD?$

$\dfrac{3}{4}$

$\dfrac{3}{2}$

$2$

$\dfrac{12}{5}$

$\dfrac{5}{2}$

知识点：面积比周长

难度评级：1420

小提示：

用周长相等条件求 $BD$ 和 $CD$

Use the equal-perimeter condition to find $BD$ and $CD$ .

大提示：

一旦 $BC$ 被分割，两个三角形从 $A$ 引出的高相同，所以面积按底边长度比较

Once $BC$ is split, the two triangles have the same altitude from $A$ , so compare areas by base lengths.

视频讲解：

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文字解答：

要使两个三角形周长相等， $\overline{BC}$ 必须分成 $\overline{CD} = 3$ 和 $\overline{BD} = 2$ 。

$\triangle ACD$ 和 $\triangle ABD$ 有相同的高，所以面积与底边长度成比例。这说明 $\triangle ABD$ 的面积是 $\dfrac{2}{5}$ 倍的 $\triangle ABC$ 的面积，即 $\dfrac{2}{5} \cdot 3 \cdot \dfrac{4}{2} = \dfrac{12}{5}\text{。}$

所以正确答案是 D。

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

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

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 $9$ 枚金币，但这样会剩下 $2$ 个宝箱空着。于是我改为每个宝箱放 $6$ 枚金币，但这样会剩下 $3$ 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?

$9$

$27$

$45$

$63$

$81$

知识点：方程组

难度评级：1240

小提示：

设宝箱数为 $n$

Let $n$ be the number of treasure chests.

大提示：

为每箱 $9$ 枚的尝试和每箱 $6$ 枚的尝试分别写方程

Write one equation for the $9$ -coins attempt and another for the $6$ -coins attempt.

视频讲解：

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文字解答：

设宝箱数为 $n$ ，金币数为 $g$ 。那么 $9(n - 2) = g$ 且 $6n + 3 = g\text{。}$ 解这个方程组得 $n = 7$ ，所以金币数为 $6 \cdot 7 + 3 = 45$ 。

所以正确答案是 C。

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

Thus, C is the correct answer.

18.

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

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

$12$

$24$

$26$

$30$

$36$

知识点：勾股定理勾股数面积

难度评级：1430

小提示：

先用小直角三角形求 $BD$

First find $BD$ using the small right triangle.

大提示：

再把较大的三角形识别为 $5$ - $12$ - $13$ 直角三角形，并减去小三角形面积

Then recognize the larger triangle as a $5$ - $12$ - $13$ right triangle and subtract the small triangle’s area.

视频讲解：

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文字解答：

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

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

所以正确答案是 B。

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

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

Thus, B is the correct answer.

19.

对任意正整数 $M$ ，记号 $M!$ 表示从 $1$ 到 $M$ 的所有整数的乘积。使 $5^n$ 成为下列和式的因数的最大整数 $n$ 是多少？ $98!+99!+100!$

For any positive integer $M,$ the notation $M!$ denotes the product of the integers $1$ through $M.$ What is the largest integer $n$ for which $5^n$ is a factor of the sum: $98!+99!+100!$

$23$

$24$

$25$

$26$

$27$

知识点：阶乘勒让德公式

难度评级：1640

小提示：

从三项中提出 $98!$

Factor $98!$ out of all three terms.

大提示：

提出后，统计 $98!$ 和 $100^2$ 中因子 $5$ 的个数。

After factoring, count the powers of $5$ in $98!$ and in $100^2$ .

视频讲解：

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文字解答：

提出公因式 $98!$ ： $\begin{aligned} 98!+99!+100! &=98!\bigl(1+99\\ &\qquad+100\cdot99\bigr)\\ &=98!\cdot100^2 \end{aligned}\text{。}$

$98!$ 中因子 $5$ 的指数是 $\left\lfloor\frac{98}{5}\right\rfloor+\left\lfloor\frac{98}{25}\right\rfloor$ ，等于 $19+3=22$ 。因子 $100^2$ 还贡献四个因子 $5$ ，所以总指数是 $22+4=26$ 。

所以正确答案是 D。

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

The exponent of $5$ in $98!$ is $\left\lfloor\frac{98}{5}\right\rfloor+\left\lfloor\frac{98}{25}\right\rfloor$ , which equals $19+3=22.$ The factor $100^2$ contributes four more factors of $5$ , so the total exponent is $22+4=26.$

Thus, D is the correct answer.

20.

从 $1000$ 到 $9999$ （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 $1000$ and $9999,$ inclusive, is chosen at random. What is the probability that it is an odd integer whose digits are all distinct?

$\dfrac{14}{75}$

$\dfrac{56}{225}$

$\dfrac{107}{400}$

$\dfrac{7}{25}$

$\dfrac{9}{25}$

知识点：基本概率乘法原理

难度评级：1550

小提示：

直接统计符合条件的四位数，再除以四位整数总数

Count favorable four-digit numbers directly, then divide by the number of four-digit integers.

大提示：

先选奇数个位，再选非零千位，再选剩余两个数字

Choose the odd units digit first, then choose the nonzero thousands digit, then the remaining two digits.

视频讲解：

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文字解答：

这个数是奇数，所以个位数字有 $5$ 种选择。千位数字不能为零，也不能等于个位数字，因此有 $8$ 种选择。百位有 $8$ 种选择，十位有 $7$ 种选择。符合条件的数共有 $5 \cdot 8 \cdot 8 \cdot 7 = 2{,}240$ ，所以概率为 $\dfrac{2{,}240}{9{,}000} = \dfrac{56}{225}$ 。

所以正确答案是 B。

Since the number is odd, the last digit is odd, giving $5$ possibilities. The thousands digit cannot be zero or the number we already got, so that gives $8$ possibilities. Similarly, the hundreds digit has $8$ possibilities, and the tens digit has $7$ possibilities. This gives a total of $5 \cdot 8 \cdot 8 \cdot 7 = 2{,}240,$ making the probability $\dfrac{2{,}240}{9{,}000} = \dfrac{56}{225}.$

Thus, B is the correct answer.

21.

假设 $a$ 、 $b$ 和 $c$ 是非零实数，且 $a+b+c=0$ 。下列表达式可能的值是什么 $\dfrac{a}{|a|}+\dfrac{b}{|b|}+\dfrac{c}{|c|}+\dfrac{abc}{|abc|}\text{？}$

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

$0$

$1$ 和 $-1$

$1$ and $-1$

$2$ 和 $-2$

$2$ and $-2$

$0$ 、 $2$ 和 $-2$

$0$ , $2$ , and $-2$

$0$ 、 $1$ 和 $-1$

$0$ , $1$ , and $-1$

知识点：绝对值分类讨论

难度评级：1510

小提示：

因为 $a+b+c=0$ ，三个数不可能同号

Because $a+b+c=0$ , the three numbers cannot all have the same sign.

大提示：

每个分式 $\dfrac{x}{|x|}$ 只是 $x$ 的符号

Each fraction $\dfrac{x}{|x|}$ is just the sign of $x$ .

视频讲解：

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文字解答：

因为 $a+b+c=0$ 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 $1$ ，而 $\dfrac{abc}{|abc|}=-1$ ；第二种情况下，前三个分式的和为 $-1$ ，而 $\dfrac{abc}{|abc|}=1$ 。无论哪种情况，整个表达式都等于 $0$ 。

所以正确答案是 A。

Because $a+b+c=0$ 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 $1$ , while $\dfrac{abc}{|abc|}=-1$ . In the second case, the first three sum to $-1$ , while $\dfrac{abc}{|abc|}=1$ . Either way, the entire expression equals $0$ .

Thus, A is the correct answer.

22.

在直角三角形 $ABC$ 中， $AC=12$ 、 $BC=5$ ，且角 $C$ 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 $ABC,$ $AC=12,$ $BC=5,$ and angle $C$ is a right angle. A semicircle is inscribed in the triangle as shown. What is the radius of the semicircle?

$\dfrac{7}{6}$

$\dfrac{13}{5}$

$\dfrac{59}{18}$

$\dfrac{10}{3}$

$\dfrac{60}{13}$

知识点：切线相似

难度评级：1640

小提示：

画出到半圆与斜边相切点的半径

Draw the radius to the point where the semicircle touches the hypotenuse.

大提示：

使用同一点引出的切线段相等，再建立相似三角形

Use tangent lengths from the same external point, then set up similar triangles.

视频讲解：

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文字解答：

设 $O$ 为内切半圆圆心， $D$ 为半圆与 $\overline{AB}$ 的切点。于是 $BD = 5$ ，因为 $\overline{BD}$ 和 $\overline{BC}$ 都与半圆相切。由此得 $AD = 8$ ，并记 $OD = r$ 。因为 $\overline{OD}$ 垂直于 $\overline{AB}$ ，所以 $\triangle ADO \sim \triangle BCA$ ，有 $\dfrac{r}{8} = \dfrac{5}{12}$ ，解得 $r = \dfrac{10}{3}$ 。

所以正确答案是 D。

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

Thus, D is the correct answer.

23.

连续四天，琳达每天旅行一小时，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 $5$ 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 $5$ 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?

$10$

$15$

$25$

$50$

$82$

知识点：因数等差数列

难度评级：1610

小提示：

每天每英里所需分钟数必须整除 $60$

Each day’s minutes per mile must divide $60$ .

大提示：

列出 $60$ 的因数，并找出四个相差 $5$ 的因数

List the divisors of $60$ and find four of them spaced $5$ apart.

视频讲解：

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文字解答：

琳达每天旅行 $60$ 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 $60$ 的因数。 $60$ 的因数为 $1$ 、 $2$ 、 $3$ 、 $4$ 、 $5$ 、 $6$ 、 $10$ 、 $12$ 、 $15$ 、 $20$ 、 $30$ 和 $60$ 。其中唯一一组四个相邻相差 $5$ 的数是 $5$ 、 $10$ 、 $15$ 和 $20$ 。四天总路程为 $\dfrac{60}{5} + \dfrac{60}{10} + \dfrac{60}{15} + \dfrac{60}{20} = 25$ 英里。

所以正确答案是 C。

Linda traveled for $60$ 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 $60.$ The factors of $60$ are $1,$ $2,$ $3,$ $4,$ $5,$ $6,$ $10,$ $12,$ $15,$ $20,$ $30,$ and $60.$ The only sequence of four of these numbers that differ by $5$ are $5,$ $10,$ $15,$ and $20.$ For the four days, she traveled $\dfrac{60}{5} + \dfrac{60}{10} + \dfrac{60}{15} + \dfrac{60}{20} = 25$ miles in total.

Thus, C is the correct answer.

24.

桑德斯夫人有三个孙辈，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 $2016$ 年十二月 $31$ 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?

$78$

$80$

$144$

$146$

$152$

知识点：容斥原理最小公倍数

难度评级：1800

小提示：

电话模式每 $\operatorname{lcm}(3,4,5)$ 天重复

Calls repeat every $\operatorname{lcm}(3,4,5)$ days.

大提示：

统计每 $60$ 天周期中至少有一次电话的天数，再处理 $2017$ 年最后几天

Count days with at least one call in each $60$ -day cycle, then handle the last few days of $2017.$

视频讲解：

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文字解答：

在一个 $60$ 天周期中，第一个孙辈打 $20$ 次，第二个打 $15$ 次，第三个打 $12$ 次。但 $20 + 15 + 12 = 47$ 有重复计数：第一个和第二个孙辈在同一天打电话 $\frac{60}{12} = 5$ 次，第一个和第三个在同一天打电话 $\frac{60}{15} = 4$ 次，第二个和第三个在同一天打电话 $\frac{60}{20} = 3$ 次。从 $47$ 中减去这些，得 $47 - 5 - 4 - 3 = 35$ 。

第 $60$ 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 $60$ 天有 $36$ 天收到电话，也就是 $24$ 天没有电话。这一年有 $6$ 个完整的 $60$ 天周期，剩下第 $361$ 天和第 $362$ 天没有电话，因此没有电话的天数为 $24 \cdot 6 + 2 = 146$ 。

所以正确答案是 D。

In a $60$ -day period, the first child calls $20$ times, the second child calls $15$ times, and the third child calls $12$ times. $20 + 15 + 12 = 47$ overcounts, however. The first and second children call on the same day $\frac{60}{12} = 5$ times. The first and third children call on the same day $\frac{60}{15} = 4$ times. The second and third children call on the same day $\frac{60}{20} = 3$ times. Subtracting these from $47$ yields $47 - 5 - 4 - 3 = 35.$

The $60$ th day is added in thrice and subtracted out thrice, so we need to add it back in. This means that for every $60$ days, Mrs. Sanders receives a call $36$ days, which means that she does not receive a call on $24$ days. There are $6$ $60$ -day periods, and there are no calls on the $361$ st or $362$ nd day, which results in $24 \cdot 6 + 2 = 146$ total days with no phone calls.

Thus, D is the correct answer.

25.

如图， $\overline{US}$ 和 $\overline{UT}$ 都是长度为 $2$ 的线段，且 $m\angle TUS = 60^\circ$ 。

弧 $\overset{\large\frown}{TR}$ 和 $\overset{\large\frown}{SR}$ 都是半径为 $2$ 的圆的六分之一。图中区域面积是多少？

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

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

$3\sqrt{3}-\pi$

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

$2\sqrt{3}$

$4\sqrt{3}-\dfrac{2\pi}{3}$

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

知识点：等边三角形扇形面积分割

难度评级：1750

小提示：

延长两条直边，形成一个等边三角形

Extend the two straight sides to form an equilateral triangle.

大提示：

从该等边三角形中减去两个圆心角为 $60^\circ$ 、半径为 $2$ 的扇形。

Subtract two $60^\circ$ sectors of radius $2$ from that equilateral triangle.

视频讲解：

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文字解答：

可以延长 $\overline{SU}$ 和 $\overline{TU}$ ，形成如下图形。

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

所以正确答案是 B。

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

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

Thus, B is the correct answer.