2016 年 AMC 8 真题（P版）

1 答案 C

一小时有 \(60\) 分钟，所以总时间为 \(60 \cdot 11 + 5 = 665\) 分钟。

所以正确答案是 C。

English

There are \(60\) minutes in an hour, so the total time is \(60 \cdot 11 + 5 = 665\) minutes.

Thus, C is the correct answer.

2 答案 A

因为 \(M\) 是 \(\overline{AD}\) 的中点，所以底边 \(AM\) 长 \(4\)。从 \(C\) 到 \(\overline{AM}\) 的高为 \(AB=6\)。因此面积为 \(\dfrac{1}{2} \cdot 4 \cdot 6 = 12\)。

所以正确答案是 A。

English

Since \(M\) is the midpoint of \(\overline{AD},\) the base \(AM\) has length \(4.\) The altitude from \(C\) to \(\overline{AM}\) has length \(AB=6.\) The area is therefore \(\dfrac{1}{2} \cdot 4 \cdot 6 = 12.\)

Thus, A is the correct answer.

3 答案 A

由平均分可知四人总分为 \(4 \cdot 70 = 280\)。因此剩下的分数为 \[280 - 70 - 80 - 90 = 40\text{。}\]

所以正确答案是 A。

English

From the average, we can calculate the sum of the scores to be \(4 \cdot 70 = 280.\) This means that the remaining score is \[280 - 70 - 80 - 90 = 40.\]

Thus, A is the correct answer.

4 答案 B

为了更容易比较速度，将它们都换成每英里所需分钟数。

小时候他跑 \(15\) 英里用了 \(3 \cdot 60 + 30 = 210\) 分钟，所以速度为 \(\frac{210}{15} = 14\) 分钟每英里。

现在他走 \(10\) 英里用了 \(4 \cdot 60 = 240\) 分钟，所以速度为 \(\frac{240}{10} = 24\) 分钟每英里。

相减得，现在他走一英里多花 \(10\) 分钟。

所以正确答案是 B。

English

To better compare the rates, we can change his speed into minutes per mile.

As a boy he ran \(15\) miles in \(3 \cdot 60 + 30 = 210\) minutes, which means that he ran at a pace of \(\frac{210}{15} = 14\) minutes per mile.

As an adult, he can walk \(10\) miles in \(4 \cdot 60 = 240\) minutes, which means he walks at a pace of \(\frac{240}{10} = 24\) minutes per mile.

Subtracting the two, we get that he takes \(10\) more minutes to walk a mile as an adult.

Thus, B is the correct answer.

5 答案 E

第二个条件说明 \(N\) 的个位数字是 \(3\)，所以可能的数为 \(13\)、\(23\)、\(33\)、\(43\)、\(53\)、\(63\)、\(73\)、\(83\)、\(93\)。其中只有 \(73\) 除以 \(9\) 余 \(1\)，因为它的各位数字之和是 \(10\)。最后，\(73=6\cdot11+7\)，所以所求余数是 \(7\)。

所以正确答案是 E。

English

The second condition says that \(N\) ends in \(3\), so the possibilities are \(13,\) \(23,\) \(33,\) \(43,\) \(53,\) \(63,\) \(73,\) \(83,\) \(93.\) Of these, only \(73\) leaves remainder \(1\) when divided by \(9\), because its digit sum is \(10\). Finally, \(73=6\cdot11+7\), so the requested remainder is \(7\).

Thus, E is the correct answer.

6 答案 B

因为有 \(19\) 个人，每人对应一个名字长度，所以中间的长度是第十个值。从左侧开始数，数到第十个时得到 \(4\)。

所以正确答案是 B。

English

Since there are \(19\) people, each with one corresponding name length, the middle length will be the tenth one. Counting from the left side, the tenth value that we arrive upon is \(4.\)

Thus, B is the correct answer.

7 答案 B

任何指数为偶数的数都是完全平方数，所以可排除 A、C 和 E。另外，平方数的任意次幂仍然是平方数，所以也排除 D。

所以正确答案是 B。

English

Since any number with an even exponent is a perfect square, we can eliminate A, C, and E. Also, a square number to any power remains a square number, so that rules out D.

Thus, B is the correct answer.

8 答案 C

可以把这个和分组为 \[ \begin{gathered} (100 - 98) + (96 - 94) \\ {}+ \cdots + (4 - 2) \end{gathered} \] 每一对的值都是 \(2\)，共有 \(25\) 对。因此总和为 \(2 \cdot 25 = 50\)。

所以正确答案是 C。

English

We can group the sum as follows: \[ \begin{gathered} (100 - 98) + (96 - 94) \\ {}+ \cdots + (4 - 2). \end{gathered} \] Note that each pair evaluates to \(2\) and there are \(25\) pairs. Therefore, the total sum is \(2 \cdot 25 = 50.\)

Thus, C is the correct answer.

9 答案 B

将 \(2016\) 质因数分解为 \(2^5 \cdot 3^2 \cdot 7\)。因此 \(2016\) 的质因子是 \(2\)、\(3\) 和 \(7\)。它们的和为 \(12\)，所以正确答案是 B。

English

We can prime factorize \(2016\) as \(2^5 \cdot 3^2 \cdot 7.\) This shows that the prime divisors of \(2016\) are \(2,\) \(3,\) and \(7.\) The sum of these is \(12,\) so B is the correct answer.

10 答案 D

先算内层运算，得 \(5\ast x=15-x\)。再应用一次定义： \[ \begin{aligned} 2\ast(15-x)&=6-(15-x)\\ &=x-9\text{。} \end{aligned} \] 因此 \(x-9=1\)，所以 \(x=10\)。

所以正确答案是 D。

English

First, \(5\ast x=15-x\). Applying the operation again gives \[\begin{aligned} 2\ast(15-x)&=6-(15-x)\\ &=x-9. \end{aligned}\] Thus \(x-9=1\), so \(x=10\).

Thus, D is the correct answer.

11 答案 B

设十位数字为 \(a\)，个位数字为 \(b\)。原数与反序数相加得 \[ \begin{aligned} (10a+b)+(10b+a)&=132\\ 11(a+b)&=132\text{，} \end{aligned} \] 所以 \(a+b=12\)。可行的数字对为 \((3,9)\)、\((4,8)\)、\((5,7)\)、\((6,6)\)、\((7,5)\)、\((8,4)\)、\((9,3)\)，共得到 \(7\) 个两位数。

所以正确答案是 B。

English

Let \(a\) be the tens digit and \(b\) the units digit. Adding the number and its reversal gives \[\begin{aligned} (10a+b)+(10b+a)&=132\\ 11(a+b)&=132, \end{aligned}\] so \(a+b=12\). The valid digit pairs are \((3,9),\) \((4,8),\) \((5,7),\) \((6,6),\) \((7,5),\) \((8,4),\) \((9,3),\) giving \(7\) numbers.

Thus, B is the correct answer.

12 答案 B

假设有 \(12\) 名女生和 \(12\) 名男生，那么参加实地考察的有 \(9\) 名女生和 \(8\) 名男生，所以参加者中女生所占的比例是 \(\dfrac{9}{9+8}=\dfrac9{17}\)。

所以正确答案是 B。

English

Suppose there are \(12\) girls and \(12\) boys. Then \(9\) girls and \(8\) boys go on the field trip, so the fraction of field-trip students who are girls is \(\dfrac{9}{9+8}=\dfrac9{17}.\)

Thus, B is the correct answer.

13 答案 D

从六个数中选两个不同的数共有 \(\binom62=15\) 种方法。乘积为 \(0\) 当且仅当选出的一个数是 \(0\)，另一个数有 \(5\) 种选择。因此所求概率为 \(\dfrac5{15}=\dfrac13\)。

所以正确答案是 D。

English

There are \(\binom62=15\) ways to select two different numbers. A product is \(0\) exactly when one selected number is \(0\), and there are \(5\) choices for the other number. Thus the probability is \(\dfrac5{15}=\dfrac13.\)

Thus, D is the correct answer.

14 答案 A

卡尔开 \(350\) 英里用了 \(\frac{350}{35}\) 加仑汽油。

他又加了 \(8\) 加仑到原来剩下的 \(14 - 10 = 4\) 加仑中，一共有 \(12\) 加仑。

到达时油箱半满，所以后来又用了 \(12 - 7 = 5\) 加仑，对应 \(5 \cdot 35 = 175\) 英里。

因此他行驶的总路程为 \(350+175=525\) 英里。

所以正确答案是 A。

English

If Karl drove \(350\) miles, then he used \(\frac{350}{35}\) gallons of gas.

When he bought more gas, he added \(8\) gallons to \(14 - 10 = 4\) gallons, attaining a total of \(12\) gallons.

If his tank was half full when he arrived, he used \(12 - 7 = 5\) gallons, which equates to \(5 \cdot 35 = 175\) miles.

Therefore, he traveled a total distance of \(350+175=525\) miles.

Thus, A is the correct answer.

15 答案 C

使用平方差分解：

\[\begin{aligned} 13^4-11^4 &=(13^2+11^2)\\ &\qquad\cdot(13^2-11^2)\\ &=290\cdot48\\ &=32\cdot435 \end{aligned}\text{。}\]

因为 \(435\) 是奇数，所以 \(32\) 是能整除该表达式的最大的 \(2\) 的幂。

所以正确答案是 C。

English

We can factor this expression using difference of squares.

\[\begin{aligned} 13^4-11^4 &=(13^2+11^2)\\ &\qquad\cdot(13^2-11^2)\\ &=290\cdot48\\ &=32\cdot435. \end{aligned}\]

Because \(435\) is odd, \(32\) is the largest power of \(2\) that divides the expression.

Thus, C is the correct answer.

16 答案 D

因为安妮比邦妮快 \(25\%\)，所以邦妮每完成一圈时，安妮完成 \(1 \dfrac{1}{4}\) 圈。也就是说，邦妮每跑一圈，安妮领先四分之一圈。

安妮要整整领先一圈，需要邦妮完成 \(4\) 圈，此时安妮完成 \(5\) 圈。

所以正确答案是 D。

English

Since Annie is \(25\%\) faster than Bonnie, for every lap Bonnie finishes, Annie completes \(1 \dfrac{1}{4}\) laps. Therefore, Annie gains a quarter lap every time Bonnie finishes a lap.

With this in mind, for Annie to completely lap Bonnie, Bonnie must finish \(4\) laps, which means that Annie finished \(5\) laps.

Thus, D is the correct answer.

17 答案 D

没有限制时，密码总数为 \(10^4\)。这个条件排除了 \(10\) 个密码，因为前 \(3\) 位已经固定，最后一位可以任意。因此可接受密码数为 \[10{,}000 - 10 = 9990\text{。}\]

所以正确答案是 D。

English

The total number of passwords with no conditions is \(10^4.\) The condition removes \(10\) possible passwords since the first \(3\) are determined, and the last one can be anything. Therefore, the number of acceptable passwords is \[10{,}000 - 10 = 9990.\]

Thus, D is the correct answer.

18 答案 C

每场比赛淘汰 \(5\) 人。要决出冠军，需要淘汰 \(215\) 人。因此需要 \(\frac{215}{5} = 43\) 场比赛。

所以正确答案是 C。

English

Note that each race eliminates \(5\) people. For there to be a winner, \(215\) must be eliminated. Therefore, \(\frac{215}{5} = 43\) races are required to eliminate this number of people.

Thus, C is the correct answer.

19 答案 E

这些数的平均数为 \(\frac{10{,}000}{25} = 400\)。因为共有二十五个连续偶整数，平均数就是中间项。最大数比中间项大 \(12\) 个偶数间隔，所以等于 \(400 + 12 \cdot 2 = 424\)。

所以正确答案是 E。

English

The average of these numbers is \(\frac{10{,}000}{25} = 400.\) The largest number is \(12\) even numbers away, which means that it equals \(400 + 12 \cdot 2 = 424.\)

Thus, E is the correct answer.

20 答案 A

\(b\) 必须同时整除 \(12\) 和 \(15\)，所以它只能是 \(1\) 或 \(3\)。

如果 \(b = 1\)，则可取 \(a = 12\)、\(c = 15\)，它们的最小公倍数为 \(60\)。如果 \(b = 3\)，则可取 \(a = 4\)、\(c = 5\)，此时最小公倍数为 \(20\)。

所以正确答案是 A。

English

We know that \(b\) has to divide both \(12\) and \(15,\) so it must equal either \(1\) or \(3.\)

If \(b = 1,\) then \(a = 12\) and \(c = 15,\) making their least common multiple \(60.\) If \(b = 3,\) then we may take \(a = 4\) and \(c = 5.\) The least common multiple in this scenario is \(20.\)

Thus, A is the correct answer.

21 答案 B

\(3\) 枚红色筹码先于两枚绿色筹码都被抽出，当且仅当完整抽取顺序中的最后一枚筹码是绿色。选择两枚绿色筹码位置的等可能方式共有 \(10\) 种，其中最后一个位置为绿色的有 \(4\) 种。

因此所求概率为 \[\dfrac{4}{10} = \dfrac{2}{5}\text{。}\]

所以正确答案是 B。

English

The \(3\) reds are drawn before both green chips exactly when a green chip is the last chip in the full ordering. There are \(10\) equally likely ways to choose the two positions of the green chips, and \(4\) of them have a green chip in the last position.

Therefore, the desired probability is \[\dfrac{4}{10} = \dfrac{2}{5}.\]

Thus, B is the correct answer.

22 答案 C

设 \(I\) 为 \(\overline{AD}\) 的中点，\(G\) 为 \(\overline{EF}\) 的中点。再设 \(H\) 为 \(\overline{CF}\) 与 \(\overline{BE}\) 的交点。

\(\triangle BCE\) 的面积是 \(\dfrac12\cdot1\cdot4=2\)。由对称性可知，\(\triangle BCH\) 与 \(\triangle EFH\) 相似。它们的底边之比为 \(1 : 3\)，所以高之比也相同。因此 \(3IH = HG\)，从而 \(IH = 1\)。

所以 \([\triangle BCH]=\dfrac12\cdot1\cdot1=\dfrac12\)。于是 \([\triangle ECH]=2-\dfrac12=\dfrac32\)。图形关于中线对称，因此蝙蝠双翼的总面积为 \(2\cdot\dfrac32=3\)。

所以正确答案是 C。

English

Define \(I\) to be the midpoint of \(\overline{AD}\) and \(G\) to be the midpoint of \(\overline{EF}.\) Also define \(H\) to be the intersection of \(\overline{CF}\) and \(\overline{BE}.\)

The area of \(\triangle BCE\) is \(\dfrac12\cdot1\cdot4=2.\) By symmetry, we can see that \(\triangle BCH\) and \(\triangle EFH\) are similar. Since their bases are in a \(1 : 3\) ratio, so are their altitudes. This means that \(3IH = HG,\) which implies that \(IH = 1.\)

Therefore, \([\triangle BCH]=\dfrac12\cdot1\cdot1=\dfrac12.\) It follows that \([\triangle ECH]=2-\dfrac12=\dfrac32.\) Since the figure is symmetric, the total area of the bat wings is \(2\cdot\dfrac32=3.\)

Thus, C is the correct answer.

23 答案 C

因为 \(AE = EB = AB\)，它们都是全等圆的半径，所以三点构成等边三角形，\(\angle AEB = 60^{\circ}\)。

又因为 \(\overline{DB}\) 和 \(\overline{AC}\) 是直径，所以 \(\angle DEB=\angle AEC=90^\circ\)。因此 \[ \begin{aligned} \angle CED &= \angle DEB+\angle AEC\\ &\qquad-\angle AEB\\ &=90^\circ+90^\circ-60^\circ\\ &=120^\circ\text{。} \end{aligned} \]

所以正确答案是 C。

English

We know that \(AE = EB = AB\) since they are all radii of congruent circles, so they form an equilateral triangle, which means that \(\angle AEB = 60^{\circ}.\)

The segments \(\overline{DB}\) and \(\overline{AC}\) are diameters, so \(\angle DEB=\angle AEC=90^\circ.\) Therefore, \[\begin{aligned} \angle CED &= \angle DEB+\angle AEC\\ &\qquad-\angle AEB\\ &=90^\circ+90^\circ-60^\circ\\ &=120^\circ. \end{aligned}\]

Thus, C is the correct answer.

24 答案 A

因为 \(QRS\) 能被 \(5\) 整除，所以 \(S = 5\)。

因为 \(PQR\) 能被 \(4\) 整除，所以 \(QR\) 可以是 \(12\)、\(24\) 或 \(32\)。

如果 \(QR=12\)，则 \(R=2\)，那么 \(RST=25T\)，用剩余数字无法使其被 \(3\) 整除。如果 \(QR=32\)，则 \(R=2\)，也遇到同样问题。因此 \(QR=24\)，于是 \(RST=45T\)。在剩余数字中，只有 \(T=3\) 使 \(453\) 能被 \(3\) 整除。

因此 \(PQRST = 12453\)。

所以正确答案是 A。

English

Since \(QRS\) is divisible by \(5,\) we know that \(S = 5.\)

Since \(PQR\) is divisible by \(4,\) \(QR\) equals either \(12,\) \(24,\) or \(32.\)

If \(QR=12,\) then \(R=2,\) so \(RST=25T\) cannot be divisible by \(3\) using the remaining digits. If \(QR=32,\) then \(R=2\) again, giving the same obstacle. Thus \(QR=24,\) and then \(RST=45T.\) Among the remaining digits, only \(T=3\) makes \(453\) divisible by \(3\).

Therefore, \(PQRST = 12453.\)

Thus, A is the correct answer.

25 答案 B

设 \(O\) 是圆心，也就是 \(\overline{AB}\) 的中点。

由勾股定理可得 \(BC = 17\)。

三角形右半部分的面积为 \[ [\triangle BOC]=\dfrac12\cdot8\cdot15=60\text{。} \] 半径 \(OE\) 垂直于相切的边 \(BC\)，所以同一面积也等于 \(\dfrac12\cdot OE\cdot17\)。因此 \[ OE=\dfrac{120}{17}\text{。} \]

所以正确答案是 B。

English

Let \(O\) be the center of the circle, which is the midpoint of \(\overline{AB}.\)

We then get that \(BC = 17\) via the Pythagorean theorem.

The right half of the isosceles triangle has area \[[\triangle BOC]=\dfrac12\cdot8\cdot15=60.\] The radius \(OE\) is perpendicular to the tangent side \(BC\), so the same area is \(\dfrac12\cdot OE\cdot17.\) Hence \[OE=\dfrac{120}{17}.\]

Thus, B is the correct answer.