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AMC10 2001 A

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AMC10 · 2001 (A)

Q1
The median of the list $n, n+3, n+4, n+5, n+6, n+8, n+10, n+12, n+15$ is 10. What is the mean?
列表 $n, n+3, n+4, n+5, n+6, n+8, n+10, n+12, n+15$ 的中位数是 10。平均数是多少？
Correct Answer: E
The middle number in the 9-number list is $n+6$, which is given as 10. Thus $n = 4$. Add the terms together to get $9n + 63 = 9 \cdot 4 + 63 = 99$. Thus the mean is $99/9 = 11$.
9 个数的列表中，中位数是中间的第 5 个数 $n+6$，给定为 10。因此 $n = 4$。各项之和为 $9n + 63 = 9 \cdot 4 + 63 = 99$。因此平均数是 $99/9 = 11$。
Q2
A number $x$ is 2 more than the product of its reciprocal and its additive inverse. In which interval does the number lie?
一个数 $x$ 比它的倒数与其加法逆元的乘积多 2。它位于哪个区间？
Correct Answer: C
The reciprocal of $x$ is $\frac{1}{x}$, and the additive inverse of $x$ is $-x$. The product of these is $\left(\frac{1}{x}\right) \cdot (-x) = -1$. So $x = -1 + 2 = 1$, which is in the interval $0 < x \le 2$.
$x$ 的倒数是 $\frac{1}{x}$，$x$ 的加法逆元是 $-x$。它们的乘积是 $\left(\frac{1}{x}\right) \cdot (-x) = -1$。所以 $x = -1 + 2 = 1$，位于区间 $0 < x \le 2$。
Q3
The sum of two numbers is $S$. Suppose 3 is added to each number and then each of the resulting numbers is doubled. What is the sum of the final two numbers?
两个数的和是 $S$。假设每个数加 3，然后将结果每个数乘以 2。最后两个数的和是多少？
Correct Answer: E
Suppose the two numbers are $a$ and $b$. Then the desired sum is $2(a+3) + 2(b+3) = 2(a+b) + 12 = 2S + 12$.
假设两个数是 $a$ 和 $b$。则所求和为 $2(a+3) + 2(b+3) = 2(a+b) + 12 = 2S + 12$。
Q4
What is the maximum number for the possible points of intersection of a circle and a triangle?
圆与三角形可能相交的最大点数是多少？
Correct Answer: E
The circle can intersect at most two points of each side of the triangle, so the number can be no greater than six. The figure shows that the number can indeed be six.
圆最多可与三角形每条边相交两点，因此点数不超过六。图示表明确实可以达到六点。
solution
Q5
How many of the twelve pentominoes pictured below have at least one line of symmetry?
下面图片中的十二个五连方中有多少个至少有一条对称轴？
stem
Correct Answer: D
Exactly six have at least one line of symmetry. They are:
恰好有六个至少有一条对称轴。它们是：
solution
Q6
Let $P(n)$ and $S(n)$ denote the product and the sum, respectively, of the digits of the integer $n$. For example, $P(23) = 6$ and $S(23) = 5$. Suppose $N$ is a two-digit number such that $N = P(N) + S(N)$. What is the units digit of $N$?
设 $P(n)$ 和 $S(n)$ 分别表示整数 $n$ 的各位数字的乘积和之和。例如，$P(23) = 6$ 和 $S(23) = 5$。假设 $N$ 是一个两位数，使得 $N = P(N) + S(N)$。$N$ 的个位数字是多少？
Correct Answer: E
Suppose $N = 10a + b$. Then $10a + b = ab + (a+b)$. It follows that $9a = ab$, which implies that $b = 9$, since $a \neq 0$.
设 $N = 10a + b$。则 $10a + b = ab + (a+b)$。由此得出 $9a = ab$，这意味着 $b = 9$，因为 $a \neq 0$。
Q7
When the decimal point of a certain positive decimal number is moved four places to the right, the new number is four times the reciprocal of the original number. What is the original number?
将某个正小数的十进制点向右移动四位后，新数字是原数字倒数的四倍。原数字是多少？
Correct Answer: C
If $x$ is the number, then moving the decimal point four places to the right is the same as multiplying $x$ by 10,000. That is, $10,000x = 4 \cdot (\frac{1}{x})$, which is equivalent to $x^2 = 4/10,000$. Since $x$ is positive, it follows that $x = 2/100 = 0.02$.
设原数字为 $x$，则向右移动小数点四位相当于将 $x$ 乘以 $10,000$。即 $10,000x = 4 \cdot (\frac{1}{x})$，这等价于 $x^2 = 4/10,000$。由于 $x$ 为正数，故 $x = 2/100 = 0.02$。
Q8
Wanda, Darren, Beatrice, and Chi are tutors in the school math lab. Their schedule is as follows: Darren works every third school day, Wanda works every fourth school day, Beatrice works every sixth school day, and Chi works every seventh school day. Today they are all working in the math lab. In how many school days from today will they next be together tutoring in the lab?
学校的数学实验室有四位助教：Wanda、Darren、Beatrice 和 Chi。他们的工作安排如下：Darren 每三个学校日工作一次，Wanda 每四个学校日工作一次，Beatrice 每六个学校日工作一次，Chi 每七个学校日工作一次。今天他们都在数学实验室工作。从今天起多少学校日后他们将再次一起在实验室辅导？
Correct Answer: B
The number of school days until they will next be together is the least common multiple of 3, 4, 6, and 7, which is 84.
他们下次一起工作的学校日数是最小公倍数 3、4、6 和 7 的最小公倍数，即 84。
Q9
The state income tax where Kristin lives is levied at the rate of $p\%$ of the first $28000$ of annual income plus $(p + 2)\%$ of any amount above $28000$. Kristin noticed that the state income tax she paid amounted to $(p + 0.25)\%$ of her annual income. What was her annual income?
Kristin 所在州的所得税税率为年收入前 $28000$ 美元的部分按 $p\%$ 征收，超过 $28000$ 美元的部分按 $(p + 2)\%$ 征收。Kristin 注意到她缴纳的州所得税相当于她年收入的 $(p + 0.25)\%$。她的年收入是多少？
Correct Answer: B
(B) If Kristin’s annual income is $x \ge 28{,}000$ dollars, then $\dfrac{p}{100}\cdot 28{,}000+\dfrac{p+2}{100}\cdot (x-28{,}000)=\dfrac{p+0.25}{100}\cdot x.$ Multiplying by 100 and expanding yields $28{,}000p+px+2x-28{,}000p-56{,}000=px+0.25x.$ So, $1.75x=\dfrac{7}{4}x=56{,}000$ and $x=32{,}000.$
（B）如果 Kristin 的年收入为 $x \ge 28{,}000$ 美元，则 $\dfrac{p}{100}\cdot 28{,}000+\dfrac{p+2}{100}\cdot (x-28{,}000)=\dfrac{p+0.25}{100}\cdot x.$ 两边同乘 100 并展开得到 $28{,}000p+px+2x-28{,}000p-56{,}000=px+0.25x.$ 所以，$1.75x=\dfrac{7}{4}x=56{,}000$，并且 $x=32{,}000.$
Q10
If $x, y, z$ are positive with $xy = 24, xz = 48, yz = 72$, then $x + y + z$ is
若 $x, y, z$ 为正数，且 $xy = 24$，$xz = 48$，$yz = 72$，则 $x + y + z$ 是
Correct Answer: D
(D) Since $x=\dfrac{24}{y}=48z$ we have $z=2y$. So $72=2y^2$, which implies that $y=6$, $x=4$, and $z=12$. Hence $x+y+z=22$.
（D）因为 $x=\dfrac{24}{y}=48z$ 所以有 $z=2y$。因此 $72=2y^2$，这意味着 $y=6$，$x=4$，且 $z=12$。因此 $x+y+z=22$。
Q11
Consider the dark square in an array of unit squares, part of which is shown. The first ring of squares around this center square contains 8 unit squares. The second ring contains 16 unit squares. If we continue this process, the number of unit squares in the $100^{th}$ ring is
考虑一个单位正方形阵列中的深色正方形，部分显示如下。围绕这个中心正方形的第一个环包含 8 个单位正方形。第二个环包含 16 个单位正方形。如果我们继续这个过程，第 $100^{th}$ 环中的单位正方形数量是
stem
Correct Answer: C
(C) The $n^{th}$ ring can be partitioned into four rectangles: two containing $2n+1$ unit squares and two containing $2n-1$ unit squares. So there are $$2(2n+1)+2(2n-1)=8n$$ unit squares in the $n^{th}$ ring. Thus, the $100^{th}$ ring has $8\cdot 100=800$ unit squares.
（C）第 $n^{th}$ 圈可以分成四个长方形：两个包含 $2n+1$ 个单位方格，两个包含 $2n-1$ 个单位方格。因此共有 $$2(2n+1)+2(2n-1)=8n$$ 个单位方格在第 $n^{th}$ 圈中。因此，第 $100^{th}$ 圈有 $8\cdot 100=800$ 个单位方格。
Q12
Suppose that n is the product of three consecutive integers and that n is divisible by 7. Which of the following is not necessarily a divisor of n?
假设 $n$ 是三个连续整数的乘积，并且 $n$ 能被 7 整除。以下哪一个不一定是 $n$ 的因数？
Correct Answer: D
(D) In any triple of consecutive integers, at least one is even and one is a multiple of 3. Therefore, the product of the three integers is both even and a multiple of 3. Since 7 is a divisor of the product, the numbers 6, 14, 21, and 42 must also be divisors of the product. However, 28 contains two factors of 2, and $n$ need not. For example, $5 \cdot 6 \cdot 7$ is divisible by 7, but not by 28.
（D）在任意三个连续整数中，至少有一个是偶数，且至少有一个是 3 的倍数。因此，这三个整数的乘积既是偶数又是 3 的倍数。由于 7 是该乘积的一个因数，那么 6、14、21 和 42 也必定都是该乘积的因数。然而，28 含有两个 2 的因子，而 $n$ 未必如此。例如，$5 \cdot 6 \cdot 7$ 能被 7 整除，但不能被 28 整除。
Q13
A telephone number has the form ABC −DEF −GHIJ, where each letter represents a different digit. The digits in each part of the number are in decreasing order; that is, A > B > C, D > E > F, and G > H > I > J. Furthermore, D, E, and F are consecutive even digits; G, H, I, and J are consecutive odd digits; and A + B + C = 9. Find A.
一个电话号码的形式是 ABC −DEF −GHIJ，其中每个字母代表不同的数字。号码每个部分的数字是递减的，即 A > B > C, D > E > F, 和 G > H > I > J。此外，D, E, 和 F 是连续的偶数数字；G, H, I, 和 J 是连续的奇数数字；并且 A + B + C = 9。求 A。
Correct Answer: E
(E) The last four digits (GHIJ) are either 9753 or 7531, and the remaining odd digit (either 1 or 9) is A, B, or C. Since $A + B + C = 9$, the odd digit among A, B, and C must be 1. Thus the sum of the two even digits in ABC is 8. The three digits in DEF are 864, 642, or 420, leaving the pairs 2 and 0, 8 and 0, or 8 and 6, respectively, as the two even digits in ABC. Of those, only the pair 8 and 0 has sum 8, so ABC is 810, and the required first digit is 8. The only such telephone number is 810-642-9753.
（E）最后四位数字（GHIJ）要么是 9753，要么是 7531，而剩下的奇数字（1 或 9）在 A、B、C 之中。由于 $A + B + C = 9$，因此 A、B、C 中的奇数字必须是 1。于是 ABC 中两个偶数字之和为 8。DEF 的三位数可能是 864、642 或 420，因此对应地，ABC 中两个偶数字分别是（2 和 0）、（8 和 0）或（8 和 6）。其中只有（8 和 0）的和为 8，所以 ABC 为 810，所求的首位数字是 8。唯一符合条件的电话号码是 810-642-9753。
Q14
A charity sells 140 benefit tickets for a total of $2001. Some tickets sell for full price (a whole dollar amount), and the rest sell for half price. How much money is raised by the full-price tickets?
一个慈善机构售出了 140 张总价为 $2001 的福利票。有些票按全价（整美元金额）出售，其余按半价出售。全价票筹集了多少钱？
Correct Answer: A
(A) Let $n$ be the number of full-price tickets and $p$ be the price of each in dollars. Then $np+(140-n)\cdot\frac{p}{2}=2001,$ so $p(n+140)=4002.$ Thus $n+140$ must be a factor of $4002=2\cdot3\cdot23\cdot29.$ Since $0\le n\le140,$ we have $140\le n+140\le280,$ and the only factor of $4002$ that is in the required range for $n+140$ is $174=2\cdot3\cdot29.$ Therefore, $n+140=174,$ so $n=34$ and $p=23.$ The money raised by the full-price tickets is $34\cdot23=782$ dollars.
（A）设 $n$ 为全价票的数量，$p$ 为每张票的价格（美元）。则 $np+(140-n)\cdot\frac{p}{2}=2001,$ 所以 $p(n+140)=4002.$ 因此 $n+140$ 必须是 $4002=2\cdot3\cdot23\cdot29$ 的一个因数。由于 $0\le n\le140,$ 有 $140\le n+140\le280,$ 而 $4002$ 在 $n+140$ 所需范围内的唯一因数是 $174=2\cdot3\cdot29.$ 因此 $n+140=174,$ 所以 $n=34,$ $p=23.$ 全价票筹集到的金额为 $34\cdot23=782$ 美元。
Q15
A street has parallel curbs 40 feet apart. A crosswalk bounded by two parallel stripes crosses the street at an angle. The length of the curb between the stripes is 15 feet and each stripe is 50 feet long. Find the distance, in feet, between the stripes.
一条街道有相距 40 英尺的平行路缘。一个由两条平行条纹界定的横道以一定角度穿过街道。条纹之间的路缘长度为 15 英尺，每条条纹长 50 英尺。求条纹之间的距离，单位英尺。
Correct Answer: C
(C) The crosswalk is in the shape of a parallelogram with base 15 feet and altitude 40 feet, so its area is $15\times 40=600\ \text{ft}^2$. But viewed another way, the parallelogram has base 50 feet and altitude equal to the distance between the stripes, so this distance must be $600/50=12$ feet.
（C）人行横道的形状是一个平行四边形，底为 15 英尺，高为 40 英尺，所以面积为 $15\times 40=600\ \text{ft}^2$。但换一种看法，这个平行四边形的底为 50 英尺，高等于条纹之间的距离，因此该距离必须是 $600/50=12$ 英尺。
solution
Q16
The mean of three numbers is 10 more than the least of the numbers and less than the greatest. The median of the three numbers is 5. What is their sum?
三个数的平均数比最小的数大10，并且小于最大的数。这三个数的中间数是5。它们的和是多少？
Correct Answer: D
(D) Since the median is 5, we can write the three numbers as $x$, 5, and $y$, where $\frac{1}{3}(x+5+y)=x+10$ and $\frac{1}{3}(x+5+y)+15=y$. If we add these equations, we get $\frac{2}{3}(x+5+y)+15=x+y+10$ and solving for $x+y$ gives $x+y=25$. Hence the sum of the numbers $x+y+5=30$.
（D）由于中位数是 5，我们可以把这三个数写成 $x$、5 和 $y$，其中 $\frac{1}{3}(x+5+y)=x+10$ 且 $\frac{1}{3}(x+5+y)+15=y$。 把这两个方程相加，得到 $\frac{2}{3}(x+5+y)+15=x+y+10$ 解出 $x+y$ 得 $x+y=25$。因此这三个数的和为 $x+y+5=30$。
Q17
Which of the cones below can be formed from a 252° sector of a circle of radius 10 by aligning the two straight sides?
下面哪个圆锥可以由半径为10的圆的252°扇形通过将两条直边对齐而成？
stem
Correct Answer: C
(C) The slant height of the cone is 10, the radius of the sector. The circumference of the base of the cone is the same as the length of the sector’s arc. This is $252/360 = 7/10$ of the circumference, $20\pi$, of the circle from which the sector is cut. The base circumference of the cone is $14\pi$, so its radius is 7.
（C）圆锥的斜高为 10，即扇形的半径。圆锥底面的周长等于扇形弧长。这段弧长是从中切出扇形的圆的周长 $20\pi$ 的 $252/360 = 7/10$。因此圆锥底面周长为 $14\pi$，其底面半径为 7。
Q18
The plane is tiled by congruent squares and congruent pentagons as indicated. The percent of the plane that is enclosed by the pentagons is closest to
平面被全等的正方形和全等的五边形如图所示铺满。五边形包围的平面百分比最接近
stem
Correct Answer: D
(D) The pattern shown at left is repeated in the plane. In fact, nine repetitions of it are shown in the statement of the problem. Note that four of the nine squares in the three-by-three square are not in the four pentagons that make up the three-by-three square. Therefore, the percentage of the plane that is enclosed by pentagons is $1-\dfrac{4}{9}=\dfrac{5}{9}=55\dfrac{5}{9}\%$
（D）左边所示的图案在平面上重复出现。事实上，题目陈述中展示了它的九次重复。注意，在这个 $3\times 3$ 的正方形中，九个小正方形里有四个不在构成该 $3\times 3$ 正方形的四个五边形内。因此，被五边形围成的平面所占百分比为 $1-\dfrac{4}{9}=\dfrac{5}{9}=55\dfrac{5}{9}\%$
solution
Q19
Pat wants to buy four donuts from an ample supply of three types of donuts: glazed, chocolate, and powdered. How many different selections are possible?
Pat想从充足的三种甜甜圈（釉面、巧克力和糖粉）中买四个甜甜圈。有多少种不同的选择？
Correct Answer: D
(D) The number of possible selections is the number of solutions of the equation $g+c+p=4$ where $g$, $c$, and $p$ represent, respectively, the number of glazed, chocolate, and powdered donuts. The 15 possible solutions to this equation are $(4,0,0)$, $(0,4,0)$, $(0,0,4)$, $(3,0,1)$, $(3,1,0)$, $(1,3,0)$, $(0,3,1)$, $(1,0,3)$, $(0,1,3)$, $(2,2,0)$, $(2,0,2)$, $(0,2,2)$, $(2,1,1)$, $(1,2,1)$, and $(1,1,2)$.
（D）可供选择的方式数等于方程的解的个数： $g+c+p=4$ 其中 $g$、$c$ 和 $p$ 分别表示糖霜甜甜圈、巧克力甜甜圈和糖粉甜甜圈的数量。该方程共有 15 组可能的解，分别为 $(4,0,0)$、$(0,4,0)$、$(0,0,4)$、$(3,0,1)$、$(3,1,0)$、$(1,3,0)$、$(0,3,1)$、$(1,0,3)$、$(0,1,3)$、$(2,2,0)$、$(2,0,2)$、$(0,2,2)$、$(2,1,1)$、$(1,2,1)$ 和 $(1,1,2)$。
Q20
A regular octagon is formed by cutting an isosceles right triangle from each of the corners of a square with sides of length 2000. What is the length of each side of the octagon?
通过从边长为2000的正方形的每个角切掉一个等腰直角三角形，形成一个正八边形。正八边形的每边长是多少？
Correct Answer: B
(B) Let $x$ represent the length of each side of the octagon, which is also the length of the hypotenuse of each of the right triangles. Each leg of the right triangles has length $\frac{x\sqrt{2}}{2}$, so $$2\cdot\frac{x\sqrt{2}}{2}+x=2000,\text{ and }x=\frac{2000}{\sqrt{2}+1}=2000(\sqrt{2}-1).$$
（B）设 $x$ 表示八边形每条边的长度，这也等于每个直角三角形斜边的长度。每个直角三角形的直角边长度为 $\frac{x\sqrt{2}}{2}$，所以 $$2\cdot\frac{x\sqrt{2}}{2}+x=2000,\text{ 且 }x=\frac{2000}{\sqrt{2}+1}=2000(\sqrt{2}-1).$$
Q21
A right circular cylinder with its diameter equal to its height is inscribed in a right circular cone. The cone has diameter 10 and altitude 12, and the axes of the cylinder and cone coincide. Find the radius of the cylinder.
一个直圆柱，其直径等于其高度，内接于一个直圆锥中。圆锥的直径为10，高为12，圆柱与圆锥的轴线重合。求圆柱的半径。
Correct Answer: B
(B) Let the cylinder have radius $r$ and height $2r$. Since $\triangle APQ$ is similar to $\triangle AOB$, we have $$\frac{12-2r}{r}=\frac{12}{5},\ \text{so } r=\frac{30}{11}.$$
（B）设圆柱的半径为 $r$，高为 $2r$。由于 $\triangle APQ$ 与 $\triangle AOB$ 相似，我们有 $$\frac{12-2r}{r}=\frac{12}{5}，\ \text{所以 } r=\frac{30}{11}。$$
solution
Q22
In the magic square shown, the sums of the numbers in each row, column, and diagonal are the same. Five of these numbers are represented by $v, w, x, y, z$. Find $y + z$.
在所示的幻方中，每行、每列和对角线上的数字之和相同。其中五个数字用 $v, w, x, y, z$ 表示。求 $y + z$。
stem
Correct Answer: D
(D) Since $v$ appears in the first row, first column, and on diagonal, the sum of the remaining two numbers in each of these lines must be the same. Thus, $25 + 18 = 24 + w = 21 + x,$ so $w = 19$ and $x = 22$. now 25,22, and 19 form a diagonal with a sum of 66, so we can find $v = 23$, $y = 26$, and $z = 20$. Hence $y + z = 46$.
（D）由于 $v$ 出现在第一行、第一列以及对角线上，因此这些线中其余两个数的和必须相同。因此， $25 + 18 = 24 + w = 21 + x,$ 所以 $w = 19$ 且 $x = 22$。现在 25、22 和 19 构成一条对角线，和为 66，因此可得 $v = 23$、$y = 26$、$z = 20$。因此 $y + z = 46$。
Q23
A box contains exactly five chips, three red and two white. Chips are randomly removed one at a time without replacement until all the red chips are drawn or all the white chips are drawn. What is the probability that the last chip drawn is white?
一个盒子中恰好有五张筹码，三张红色，两张白色。随机依次无放回取出筹码，直到取出所有红色筹码或所有白色筹码。最后一張取出的筹码为白色的概率是多少？
Correct Answer: E
(D) Think of continuing the drawing until all five chips are removed form the box. There are ten possible orderings of the colors: RRRWW, RRWRW, RWRRW, WRRRW, RRWWR, RWRWR, WRRWR, RWWRR, WRWRR, and WWRRR. The six orderings that end in R represent drawings that would have ended when the second white chip was drawn.
（D）设想继续抽取，直到盒子里的五个筹码都被取出。颜色共有十种可能的抽取顺序：RRRWW、RRWRW、RWRRW、WRRRW、RRWWR、RWRWR、WRRWR、RWWRR、WRWRR，以及 WWRRR。以 R 结尾的六种顺序表示：当第二个白色筹码被抽到时，抽取过程本应已经结束的那些情况。
Q24
In trapezoid $ABCD$, $\overline{AB}$ and $\overline{CD}$ are perpendicular to $\overline{AD}$, with $AB + CD = BC$, $AB < CD$, and $AD = 7$. What is $AB \cdot CD$?
在梯形 $ABCD$ 中， $\overline{AB}$ 和 $\overline{CD}$ 都垂直于 $\overline{AD}$，有 $AB + CD = BC$，$AB < CD$，且 $AD = 7$。求 $AB \cdot CD$。
Correct Answer: B
(B) Let $E$ be the foot of the perpendicular from $B$ to $CD$. Then $AB=DE$ and $BE=AD=7$. By the Pythagorean Theorem, $$ \begin{aligned} AD^2&=BE^2=BC^2-CE^2\\ &=(CD+AB)^2-(CD-AB)^2\\ &=(CD+AB+CD-AB)(CD+AB-CD+AB)\\ &=4\cdot CD\cdot AB. \end{aligned} $$ Hence, $AB\cdot CD=AD^2/4=7^2/4=49/4=12.25$.
（B）设 $E$ 为从 $B$ 到 $CD$ 的垂足。则 $AB=DE$ 且 $BE=AD=7$。由勾股定理， $$ \begin{aligned} AD^2&=BE^2=BC^2-CE^2\\ &=(CD+AB)^2-(CD-AB)^2\\ &=(CD+AB+CD-AB)(CD+AB-CD+AB)\\ &=4\cdot CD\cdot AB. \end{aligned} $$ 因此，$AB\cdot CD=AD^2/4=7^2/4=49/4=12.25$。
solution
Q25
How many positive integers not exceeding 2001 are multiples of 3 or 4 but not 5?
不超过 2001 的正整数中，能被 3 或 4 整除但不被 5 整除的有多少个？
Correct Answer: E
(B) For integers not exceeding 2001, there are $\lfloor 2001/3\rfloor = 667$ multiples of 3 and $\lfloor 2001/4\rfloor = 500$ multiples of 4. The total, 1167, counts the $\lfloor 2001/12\rfloor = 166$ multiples of 12 twice, so there are $1167-166=1001$ multiples of 3 or 4. From these we exclude the $\lfloor 2001/15\rfloor = 133$ multiples of 15 and the $\lfloor 2001/20\rfloor = 100$ multiples of 20, since these are multiples of 5. However, this excludes the $\lfloor 2001/60\rfloor = 33$ multiples of 60 twice, so we must re-include these. The number of integers satisfying the conditions is $1001-133-100+33=801$.
（B）对于不超过 2001 的整数，3 的倍数有 $\lfloor 2001/3\rfloor = 667$ 个，4 的倍数有 $\lfloor 2001/4\rfloor = 500$ 个。两者合计 1167，但其中 $\lfloor 2001/12\rfloor = 166$ 个 12 的倍数被重复计数了两次，所以 3 或 4 的倍数共有 $1167-166=1001$ 个。接着从中排除 $\lfloor 2001/15\rfloor = 133$ 个 15 的倍数和 $\lfloor 2001/20\rfloor = 100$ 个 20 的倍数，因为它们都是 5 的倍数。然而这样会把 $\lfloor 2001/60\rfloor = 33$ 个 60 的倍数重复排除两次，因此需要把它们加回来。满足条件的整数个数为 $1001-133-100+33=801$。