1991 年 AMC 8 真题（P版）

1 答案 B

竖式相减得 \[ \begin{aligned} &1{,}000{,}000{,}000{,}000 \\ &\quad {}- 777{,}777{,}777{,}777 \\ &= 222{,}222{,}222{,}223 \end{aligned}\text{。}\]

也可以想成：从 \(777{,}777{,}777{,}777\) 增加到 \(1{,}000{,}000{,}000{,}000\)，个位要加 \(3\)，其余十一位各加 \(2\)。

所以正确答案是 B。

English

Lining up the subtraction gives \[ \begin{aligned} &1{,}000{,}000{,}000{,}000 \\ &\quad {}- 777{,}777{,}777{,}777 \\ &= 222{,}222{,}222{,}223. \end{aligned} \]

Equivalently, to climb from \(777{,}777{,}777{,}777\) up to \(1{,}000{,}000{,}000{,}000\) you add \(3\) in the units place and \(2\) in each of the other eleven places.

Thus, the correct answer is B .

2 答案 C

\[\frac{16+8}{4-2} = \frac{24}{2} = 12\text{。}\]

所以正确答案是 C。

English

\[\frac{16+8}{4-2} = \frac{24}{2} = 12.\]

Thus, the correct answer is C .

3 答案 E

\(200{,}000 \times 200{,}000\) \(= 4 \times 10^{10}\) \(= 40{,}000{,}000{,}000\)，也就是四百亿。

所以正确答案是 E。

English

\(200{,}000 \times 200{,}000\) \(= 4 \times 10^{10}\) \(= 40{,}000{,}000{,}000,\) which is forty billion.

Thus, the correct answer is E .

4 答案 E

因为 \[ \begin{aligned} &991+993+995 \\ &\quad {}+997+999 \\ &= (1000-9)+(1000-7) \\ &\quad {}+(1000-5) \\ &\quad {}+(1000-3)+(1000-1) \\ &= 5000 - (9+7+5+3+1) \\ &= 5000 - 25 \end{aligned}\text{，}\]所以 \(N = 25\)。

所以正确答案是 E。

English

Since \[ \begin{aligned} &991+993+995 \\ &\quad {}+997+999 \\ &= (1000-9)+(1000-7) \\ &\quad {}+(1000-5) \\ &\quad {}+(1000-3)+(1000-1) \\ &= 5000 - (9+7+5+3+1) \\ &= 5000 - 25, \end{aligned} \] we get \(N = 25.\)

Thus, the correct answer is E .

5 答案 B

每个多米诺骨牌覆盖 \(2\) 个小正方形，所以能被不重叠多米诺完全覆盖的棋盘必须有偶数个小正方形。

各棋盘的格数为：\(3 \times 4 = 12\)，\(3 \times 5 = 15\)，\(4 \times 4 = 16\)，\(4 \times 5 = 20\)，\(6 \times 3 = 18\)。只有 \(3 \times 5 = 15\) 是奇数，所以不能被完全覆盖。（每个格数为偶数的棋盘都有一条边长是偶数，很容易用多米诺骨牌铺满。）

所以正确答案是 B。

English

Every domino covers exactly \(2\) squares, so any board that is completely covered by non-overlapping dominoes must contain an even number of small squares.

Counting squares: \(3 \times 4 = 12,\) \(3 \times 5 = 15,\) \(4 \times 4 = 16,\) \(4 \times 5 = 20,\) and \(6 \times 3 = 18.\) Only \(3 \times 5 = 15\) is odd, so that board cannot be covered. (Each of the even boards has a side of even length and is easily tiled with dominoes.)

Thus, the correct answer is B .

6 答案 C

五列的最大数依次为 \(13\)（第 \(1\) 列）、\(7\)（第 \(2\) 列）、\(15\)（第 \(3\) 列）、\(12\)（第 \(4\) 列）和 \(9\)（第 \(5\) 列）。

这些数中，只有 \(7\) 是它所在行中的最小数，因为第 \(2\) 行是 \(11\)、\(7\)、\(14\)、\(10\)、\(8\)。

所以正确答案是 C。

English

The largest entry in each column is \(13\) (column \(1\)), \(7\) (column \(2\)), \(15\) (column \(3\)), \(12\) (column \(4\)), and \(9\) (column \(5\)).

Of these, only \(7\) is the smallest number in its own row (row \(2\) is \(11,\) \(7,\) \(14,\) \(10,\) \(8\)).

Thus, the correct answer is C .

7 答案 D

分子可提取公因数：\(487{,}000\) \(\cdot\, (12{,}027{,}300 + 9{,}621{,}001)\)。

粗略估计为 \(500{,}000\) \(\times (10{,}000{,}000 + 10{,}000{,}000)\) \(= 500{,}000 \times 2\) \(\times 10^{7}\)。分母约为 \(20{,}000 \times 0.05 = 1000\)。

所以整个值约为 \(\dfrac{500{,}000 \times 2 \times 10^{7}}{1000}\) \(= 10^{10}\) \(= 10{,}000{,}000{,}000\)。

所以正确答案是 D。

English

Factor the numerator: \(487{,}000\) \(\cdot\, (12{,}027{,}300 + 9{,}621{,}001).\)

Rounding to leading digits, this is about \(500{,}000\) \(\times (10{,}000{,}000 + 10{,}000{,}000)\) \(= 500{,}000 \times 2\) \(\times 10^{7}.\) The denominator is about \(20{,}000 \times 0.05 = 1000.\)

So the value is roughly \(\dfrac{500{,}000 \times 2 \times 10^{7}}{1000}\) \(= 10^{10}\) \(= 10{,}000{,}000{,}000.\)

Thus, the correct answer is D .

8 答案 D

要得到较大的商，应让商为正，所以可以用两个正数或两个负数。

最好的正数组合是 \(\dfrac{8}{1} = 8\)。最好的负数组合是 \(\dfrac{-24}{-2} = 12\)。较大的是 \(12\)。

所以正确答案是 D。

English

For a large quotient it should be positive, so use two positive numbers or two negative numbers.

Best positive pair: \(\dfrac{8}{1} = 8.\) Best negative pair: \(\dfrac{-24}{-2} = 12.\) The larger is \(12.\)

Thus, the correct answer is D .

9 答案 B

不超过 \(46\) 的 \(3\) 的倍数有 \(15\) 个，\(5\) 的倍数有 \(9\) 个。\(15\) 的倍数有 \(3\) 个，即 \(15\)、\(30\)、\(45\)，它们被重复计算了。

用容斥原理，所求个数是 \(15 + 9 - 3 = 21\)。

所以正确答案是 B。

English

There are \(15\) multiples of \(3\) and \(9\) multiples of \(5\) up to \(46.\) The \(3\) multiples of \(15\) (namely \(15,\) \(30,\) \(45\)) were counted twice.

By inclusion-exclusion, the count is \(15 + 9 - 3 = 21.\)

Thus, the correct answer is B .

10 答案 B

边 \(BC\) 从 \((0,2)\) 到 \((4,2)\)，所以底长为 \(4\)。相对边 \(AD\) 在 \(x\) 轴上，所以高为 \(2\)。

面积是 \(4 \times 2 = 8\)。

所以正确答案是 B。

English

Side \(BC\) runs from \((0,2)\) to \((4,2),\) so the base is \(4.\) The opposite side \(AD\) lies on the \(x\)-axis, so the height is \(2.\)

The area is \(4 \times 2 = 8.\)

Thus, the correct answer is B .

11 答案 B

选定 \(5\) 后，另外两个不同的数必须和为 \(10\)。这样的数对是 \(1+9\)、\(2+8\)、\(3+7\)、\(4+6\)，共 \(4\) 组。

所以正确答案是 B。

English

With \(5\) chosen, the other two different numbers must sum to \(10.\) The pairs are \(1+9,\) \(2+8,\) \(3+7,\) \(4+6,\) giving \(4\) sets.

Thus, the correct answer is B .

12 答案 D

左边是 \(\dfrac{9}{3} = 3\)。右边是 \(\dfrac{5973}{N}\)，令它等于 \(3\)，得到 \(N = 1991\)。

也可以用规律：\((k-1) + k + (k+1) = 3k\)，所以除以 \(3\) 后得到中间项。这里中间项是 \(1991\)。

所以正确答案是 D。

English

The left side is \(\dfrac{9}{3} = 3.\) The right side is \(\dfrac{5973}{N},\) and setting it equal to \(3\) gives \(N = 1991.\)

Equivalently, \((k-1) + k + (k+1) = 3k,\) so dividing by \(3\) leaves the middle term. Here the middle term is \(1991.\)

Thus, the correct answer is D .

13 答案 C

因为 \(25 = 5^2\)，七个 \(25\) 提供 \(5^{14}\)。因为 \(8 = 2^3\)，三个 \(8\) 提供 \(2^9\)。

末尾零的个数是 \(\min(14, 9) = 9\)。

所以正确答案是 C。

English

Since \(25 = 5^2,\) the seven \(25\)’s give \(5^{14}.\) Since \(8 = 2^3,\) the three \(8\)’s give \(2^9.\)

The number of trailing zeros is \(\min(14, 9) = 9.\)

Thus, the correct answer is C .

14 答案 D

总分 \(11\) 并不能保证第一，例如 \(5+5+1\) 可能被另一个学生也得到 \(11\) 分。

但如果一个学生得到 \(5+5+3 = 13\) 分，那么剩下的名次使任何其他学生最多得到 \(3+3+5 = 11\) 分。因此 \(13\) 分能保证领先。

所以正确答案是 D。

English

A total of \(11\) (for example \(5+5+1\)) does not guarantee first place, since another student could also reach \(11.\)

But if one student scores \(5+5+3 = 13,\) the remaining places give every other student at most \(3+3+5 = 11.\) So \(13\) points guarantees the lead.

Thus, the correct answer is D .

15 答案 C

被切掉的小立方体有三个面原来在立体表面上：上面、前面、后面，所以移除了 \(3\) 平方英尺表面积。

切出凹口后，又露出三个新面：凹口底面和两个侧壁，共增加 \(3\) 平方英尺。因此表面积不变。

所以正确答案是 C。

English

The removed cube had three faces on the surface of the solid (top, front, and back), so \(3\) square feet of surface are removed.

Cutting it out exposes three new faces (the floor of the notch and its two side walls), adding \(3\) square feet. The surface area is unchanged.

Thus, the correct answer is C .

16 答案 B

第 \(1\) 次上半折到下半后，\(9\text{–}16\) 在底部。第 \(2\) 次下半折到上半后，\(9\text{–}12\) 在底部。第 \(3\) 次右半折到左半后，\(9\) 和 \(10\) 在底部。

第 \(4\) 次左半折到右半，会让 \(10\) 在底部，并把 \(9\) 带到最上面。

所以正确答案是 B。

English

Fold \(1\) (top over bottom) leaves squares \(9\text{–}16\) on the bottom. Fold \(2\) (bottom over top) leaves \(9\text{–}12\) on the bottom. Fold \(3\) (right over left) leaves \(9\) and \(10\) on the bottom.

Fold \(4\) (left over right) puts \(10\) on the bottom and brings \(9\) to the top.

Thus, the correct answer is B .

17 答案 C

第 \(k\) 排有 \(9+k\) 个座位，所以最多坐 \(\left\lceil \frac{9+k}{2} \right\rceil\) 人。第 \(1\) 到第 \(20\) 排的座位数从 \(10\) 到 \(29\)，对应最大人数为 \(5\)、\(6\)、\(6\)、\(7\)、\(7\)、\(8\)、\(8\)、\(\ldots\)、\(14\)、\(14\)、\(15\)。

这些数相加为 \(200\)。

所以正确答案是 C。

English

Row \(k\) has \(9+k\) seats, so it holds \(\left\lceil \frac{9+k}{2} \right\rceil\) students. For rows \(1\) through \(20\) (seats \(10\) through \(29\)) the maxima are \(5,\) \(6,\) \(6,\) \(7,\) \(7,\) \(8,\) \(8,\) \(\ldots,\) \(14,\) \(14,\) \(15.\)

These sum to \(200.\)

Thus, the correct answer is C .

18 答案 C

不论缺失的刻度是多少，每个 X 都代表相同数量的员工。共有 \(9\) 个 X 分布在工作年数 \(5\) 到 \(10\) 的各栏上，全图共有 \(30\) 个 X。

所以比例是 \(\dfrac{9}{30} = 30\%\)。

所以正确答案是 C。

English

No matter the missing scale, each X represents the same number of employees. There are \(9\) X’s over years \(5\) through \(10,\) and \(30\) X’s in all.

So the fraction is \(\dfrac{9}{30} = 30\%.\)

Thus, the correct answer is C .

19 答案 C

这十个数的总和是 \(100\)。要使其中一个数最大，另外九个不同正整数应尽可能小，即 \(1 + 2 + \cdots + 9 = 45\)。

最大的那个数就是 \(100 - 45 = 55\)。

所以正确答案是 C。

English

The ten numbers sum to \(100.\) To maximize one of them, the other nine (all different positive whole numbers) should be as small as possible: \(1 + 2 + \cdots + 9 = 45.\)

The largest number is then \(100 - 45 = 55.\)

Thus, the correct answer is C .

20 答案 A

三个数相加为 \(111A + 11B + C = 300\)。若 \(A = 1\)，数值太小；若 \(A \ge 3\)，数值太大，所以 \(A = 2\)。

此时 \(11B + C = 78\)，只能有 \(B = 7\) 且 \(C = 1\)。所以 \(C = 1\)。

所以正确答案是 A。

English

The three numbers add to \(111A + 11B + C = 300.\) Since \(A = 1\) is too small and \(A \ge 3\) is too large, \(A = 2.\)

Then \(11B + C = 78,\) which forces \(B = 7\) and \(C = 1.\) So \(C = 1.\)

Thus, the correct answer is A .

21 答案 A

从 \(32^\circ\) 到 \(20^\circ\) 是下降 \(12^\circ\)，也就是 \(4\) 个 \(3^\circ\)。

体积减少 \(4 \times 4 = 16\) 立方厘米，所以从 \(24\) 降到 \(24 - 16 = 8\)。

所以正确答案是 A。

English

From \(32^\circ\) to \(20^\circ\) is a \(12^\circ\) decrease, which is \(4\) steps of \(3^\circ.\)

The volume decreases by \(4 \times 4 = 16\) cubic centimeters, from \(24\) down to \(24 - 16 = 8.\)

Thus, the correct answer is A .

22 答案 D

乘积为奇数当且仅当两个数都是奇数。第一个转盘转到奇数 \(1\) 或 \(3\) 的概率是 \(\dfrac23\)，第二个转盘转到奇数 \(5\) 的概率是 \(\dfrac13\)。

所以乘积为奇数的概率是 \(\dfrac23 \times \dfrac13 = \dfrac29\)，乘积为偶数的概率是 \(1 - \dfrac29 = \dfrac79\)。

所以正确答案是 D。

English

The product is odd only when both numbers are odd. The first spinner is odd (\(1\) or \(3\)) with probability \(\dfrac23,\) and the second is odd (\(5\)) with probability \(\dfrac13.\)

So the product is odd with probability \(\dfrac23 \times \dfrac13 = \dfrac29,\) and even with probability \(1 - \dfrac29 = \dfrac79.\)

Thus, the correct answer is D .

23 答案 A

属于乐队或管弦乐队的女生人数为 \(100 + 80 - 60 = 120\)。所以属于至少一个团体的男生人数为 \(230 - 120 = 110\)。

乐队有 \(80\) 名男生，管弦乐队有 \(100\) 名男生，因此两个团体都有的男生人数是 \(80 + 100 - 110 = 70\)。属于乐队但不属于管弦乐队的男生有 \(80 - 70 = 10\) 人。

所以正确答案是 A。

English

Females in band or orchestra: \(100 + 80 - 60 = 120.\) So males in at least one group: \(230 - 120 = 110.\)

With \(80\) males in band and \(100\) in orchestra, the males in both are \(80 + 100 - 110 = 70.\) Hence males in band but not orchestra: \(80 - 70 = 10.\)

Thus, the correct answer is A .

24 答案 E

这个 \(3 \times 3 \times 3\) 正方体的体积是 \(27\)。由于小正方体更小且边长为整数，它们的边长只能是 \(1\) 或 \(2\)。原正方体内不能放下两个互不相交的边长为 \(2\) 的正方体，因为要将它们分开，某个方向上的总跨度至少为 \(4\)。因此最多只能放下一个边长为 \(2\) 的正方体。

剩余体积 \(27 - 8 = 19\)，由 \(19\) 个单位正方体填满。总数是 \(1 + 19 = 20\) 个正方体。

所以正确答案是 E。

English

The \(3 \times 3 \times 3\) cube has volume \(27.\) Since the cubes are smaller and have whole-number edge lengths, their edges are \(1\) or \(2\). Two edge-\(2\) cubes cannot be disjoint inside the original cube: separating them would require a total span of at least \(4\) in some direction. Thus at most one edge-\(2\) cube fits.

The remaining \(27 - 8 = 19\) of volume is filled by \(19\) unit cubes. That is \(1 + 19 = 20\) cubes.

Thus, the correct answer is E .

25 答案 C

每一次改变都会让当前黑色面积的 \(\dfrac34\) 保持黑色。五次改变后，黑色面积比例为 \[\left(\dfrac34\right)^5 = \dfrac{243}{1024}\text{。}\]

所以正确答案是 C。

English

Each change leaves \(\dfrac34\) of the current black area black. After five changes the black fraction is \[\left(\dfrac34\right)^5 = \dfrac{243}{1024}.\]

Thus, the correct answer is C .