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AMC12每日一题(2001年真题#07)

2018-06-06
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　　AMC不但是美国顶尖数学人才的人才库，更为学校提供了解申请入学者在数学科目上的学习成就与表现评估。AMC成功地为许多学生因测验成绩优良而进入理想学校。藉由设计严谨的试题，达到激发应试者解决问题的能力，培养对数学的兴趣。试题由简至难兼具，使任何程度的学生都能感受到挑战，还可以筛选出特有天赋者。AMC12的主要目的是在刺激学生对数学的兴趣并且透过选择题的方式来开启学生对数学的才能。如果学生能预先练习必定能提高对数学的兴趣，最重要的是学生能集体参与对数学的练习远比一个人独自研读的效果来得好，特别在老师的指导之下，能够学习到如何分配时间解题。参予AMC12的学生应该不难发现测验的问题都很具挑战性，但测验的题型都不会超过学生的学习范围。这项测验希望每个考生能从竞赛中享受数学。今天课窝小编为大家整理了AMC12真题练习，希望考生们认真阅读，能够对你的考试有所帮助。

2001 AMC 12竞赛试题/第07题

Problem

How many positive integers not exceeding $2001$ are multiples of $3$ or $4$ but not $5$ ?

$\text{(A) }768 \qquad \text{(B) }801 \qquad \text{(C) }934 \qquad \text{(D) }1067 \qquad \text{(E) }1167$

Solution 1

Out of the numbers $1$ to $12$ four are divisible by $3$ and three by $4$ , counting $12$ twice. Hence $6$ out of these $12$ numbers are multiples of $3$ or $4$ .

The same is obviously true for the numbers $12k+1$ to $12k+12$ for any positive integer $k$ .

Hence out of the numbers $1$ to $60=5\cdot 12$ there are $5\cdot 6=30$ numbers that are divisible by $3$ or $4$ . Out of these $30$ , the numbers $15$ , $20$ , $30$ , $40$ , $45$ and $60$ are divisible by $5$ . Therefore in the set $\{1,\dots,60\}$ there are precisely $30-6=24$ numbers that satisfy all criteria from the problem statement.

Again, the same is obviously true for the set $\{60k+1,\dots,60k+60\}$ for any positive integer $k$ .

We have $1980/60 = 33$ , hence there are $24\cdot 33 = 792$ good numbers among the numbers $1$ to $1980$ . At this point we already know that the only answer that is still possible is $\boxed{\text{(B)}}$ , as we only have $20$ numbers left.

By examining the remaining $20$ by hand we can easily find out that exactly $9$ of them match all the criteria, giving us $792+9=\boxed{801}$ good numbers. This is correct.

Solution 2

We can solve this problem by finding the cases where the number is divisible by $3$ or $4$ , then subtract from the cases where none of those cases divide $5$ . To solve the ways the numbers divide $3$ or $4$ we find the cases where a number is divisible by $3$ and $4$ as separate cases. We apply the floor function to every case to get $\left\lfloor \frac{2001}{3} \right\rfloor$ , $\left\lfloor \frac{2001}{4} \right\rfloor$ , and $\left\lfloor \frac{2001}{12} \right\rfloor$ . The first two floor functions were for calculating the number of individual cases for $3$ and $4$ . The third case was to find any overlapping numbers. The numbers were $667$ , $500$ , and $166$ , respectively. We add the first two terms and subtract the third to get $1001$ . The first case is finished.

The second case is more or less the same, except we are applying $3$ and $4$ to $5$ . We must find the cases where the first case over counts multiples of five. Utilizing the floor function again on the fractions $\left\lfloor \frac{2001}{3*5} \right\rfloor$ , $\left\lfloor \frac{2001}{4*5} \right\rfloor$ , and $\left\lfloor \frac{2001}{3*4*5} \right\rfloor$ yields the numbers $133$ , $100$ , and $33$ . The first two numbers counted all the numbers that were multiples of either four with five or three with five less than $2001$ . The third counted the overlapping cases, which we must subtract from the sum of the first two. We do this to reach $200$ . Subtracting this number from the original $1001$ numbers procures $\boxed{\textbf{(B)}\ 801}$ .

Solution 3

First find the number of such numbers between 1 and 2000 (inclusive) and then add one to this result because 2001 is a multiple of 3.

There are $\frac45*2000=1600$ numbers that are not multiples of $5$ . $\frac23*\frac34*1600=800$ are not multiples of $3$ or $4$ , so $800$ numbers are. $800+1=\boxed{\textbf{(B)}\ 801}$

Solution 4

Take a good-sized sample of consecutive integers; for example, the first 25 positive integers. Determine that the numbers 3, 4, 6, 8, 12, 16, 18, 21, and 24 exhibit the properties given in the question. 25 is a divisor of 2000, so there are $\frac{9}{25}*2000=800$ numbers satisfying the given conditions between 1 and 2000. Since 2001 is a multiple of 3, add 1 to 800 to get $800+1=\boxed{\textbf{(B)}\ 801}$ .