OG中的新GRE数学题解析(13)
更新时间:2023-10-13 15:26:01 阅读量: 综合文库 文档下载
- gre数学题难度推荐度:
- 相关推荐
最权威的国际教育服务平台
OG中的新GRE数学题解析(13)
小编在此与大家分享新GRE OG中与新GRE数学的相关内容,希望各位在2015年能够搞定GRE数学真题。
17.Let S be the set of all positive integers n such that n2 is a multiple of both 24 and 108. Which of the following integers are divisors of every integer n in S ?
Indicate all such integers.
A 12
B 24
C 36
D 72
这个题有点小复杂,我先把OG上的解答贴上来。再写我自己的
To determine which of the integers in the answer choices is a divisor of every positive integer n in S, you must first understand the integers that are in S. Note that in this question you are given information about n2, not about n itself. Therefore, you must use the information about n2 to derive information about n. The fact that n2 is a multiple of both 24 and 108 implies that n2 is a multiple of the least common multiple of 24 and 108. To determine the least common multiple of 24 and 108, factor 24 and 108 into prime factors as (23)(3) and (22)(33), respectively. Because these are prime factorizations, you can conclude that the least common
资料来源:教育优选 http://www.jybest.cn/
最权威的国际教育服务平台
multiple of 24 and 108 is (23)(33). Knowing that n2 must be a multiple of (23)(33) does not mean that every multiple of (23)(33) is a possible value of n2, because n2 must be the square of an integer. The prime factorization of a square number must contain only even exponents. Thus, the least multiple of (23)(33) that is a square is (24)(34). This is the least possible value of n2, and so the least possible value of n is (22)(32), or 36. Furthermore, since every value of n2 is a multiple of (24)(34), the values of n are the positive multiples of 36; that is, S{36, 72, 108, 144, 180, . . .} .The question asks for integers that are divisors of every integer n in S, that is, divisors of every positive multiple of 36. Since Choice A, 12, is a divisor of 36, it is also a divisor of every multiple of 36. The same is true for Choice C, 36. Choices B and D, 24 and 72, are not divisors of 36, so they are not divisors of every integer in S. The correct answer consists of Choices A and C.
这个意思就是先求出24和108的最小公倍数,然后通过加倍使其成为一个整数的平方,这样就可以找出一系列的n了,这些n的公公因数应该有哪些?我找了前面的两个,36和72,所以AC可以选出来了,之后的所有数肯定包含了这两个选项,而BD因为不满足前面这两数,所以就排除了。
资料来源:教育优选 http://www.jybest.cn/
正在阅读:
OG中的新GRE数学题解析(13)10-13
小学组书法预赛 2(DOC)03-08
电气类硕士个人简历模板12-11
华为公司成功关键因素分析论文范文03-28
渗透结晶桩头处理 - 图文05-12
如果单位反馈控制系统的传递函数11-18
宏观经济分析及行业分析报告01-04
上海交大医学院预防医学教学大纲课稿03-18
童年的野味散文随笔11-20
部署万兆以太网的十个注意事项07-21
- 人教新课标必修4 Unit2 Working the land名师导航
- 毕业生“校漂族”大行其道 - 0
- 江苏各市中考作文题出炉 - 0
- 暑期精品班 - 三角形 - 图文
- 情人节送什么礼物好??超强礼物已抵达
- 工程项目管理制度1
- 第四次业务学习 2016
- 会计要素与会计科目
- 欠发达地区小企业会计准则运用问题研究
- 一级锅炉水G4题库
- BBD双进双出筒式磨煤机安装使用说明书 SM-1
- 初一数学有理数教案
- 渝北区房地产评估市场调研报告
- iWebMall 数据字典
- 2018年小学入学教育工作计划
- 计量专业实务与案例分析 - 模拟题三 - 2013年版
- 启示录讲义
- 路基灰土改良(方案)
- 人行反洗钱岗位准入培训测试题集
- 2015电大《学前儿童发展心理学》期末试题及答案
- 数学题
- 解析
- GRE
- 13
- 新课程理论讲座
- 信号基础知识培训 联锁设备 最全文档
- 购房老带新优惠政策实施方案
- 广播电视概论 复习精要(非常详细!考试、考研都用得着!)
- 苏教版六年级上册语文十二月份月考试卷
- 汽车标志大全(附说明)
- Oracle异常总结
- 伊朗工资酬金个人所得税(2016中英文对照)
- 康定斯基论点线面读书报告
- 卷烟社会存销比计算方法
- 剑桥国际少儿英语1第11单元文本
- 高中作文议论文记叙文范文 智慧话题作文(11)
- 潍坊市汽车配件零售公司名录2018版541家 - 图文
- 限高架安装交底
- 数字信号处理参考答案
- 操作系统实验报告 附思考题
- 第18课马克思主义诞生
- 百分数分数看线段图列算式44题专项练习(有答案)ok
- 《电源的电动势和内阻 闭合电路的欧姆定律》(DOC) - 图文
- 工厂监控设计方案