网站导航   4000-006-150  
小站教育
学生选择在小站备考:30天 524872名,今日申请2455人    备考咨询 >>

【备考每日练】GRE数学题之公倍数和公因数混合问题

2014年09月18日10:47 来源:小站整理
参与(0) 阅读(5552)
摘要:为更好地使得大家备考GRE考试,小站教育现推出GRE每日一题系列,为各位考生提供关于GRE考试的各种真题。希望考生们积极参加联系,同时做好备考工作,及时调整好考前状态。另外,大家在备考或者做题的时候一定要看清真题,千万不要马虎大意,紧紧围绕着真题进行答题。小站教育老师建议大家在备考GRE考试的时候一定要重视词汇的积累。最后小站教育祝大家在GRE考试中取得理想的成绩!

从2014年9月开始,小站教育将推出每日一题系列项目,借此帮助各位考生更好地备考GRE考试,我们也将在第一时间内提供题目的答案与解析。希望大家积极参加练习,努力提高自己的能力!小伙伴们一起来做题吧!来看看今天为大家带来的GRE数学练习题目。

【备考每日练】GRE数学题之公倍数和公因数混合问题图1

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 nin S ?

Indicate all such integers.

A 12

B 24

C 36

D 72

【备考每日练】GRE数学题之公倍数和公因数混合问题图2

正确答案:AC

解析: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 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。

往期练习题:

小站每日一练:GRE数学练习(汇总1)

小站每日一练:GRE数学练习(汇总2)

小站每日一练:GRE数学练习9

小站每日一练:GRE数学练习10

小站每日一练:GRE数学练习11

更多GRE考试讯息和GRE考试动态,请继续关注小站教育GRE频道


特别申明:本文内容来源网络,版权归原作者所有,如有侵权请立即与我们联系contactus@zhan.com,我们将及时处理。

GRE备考资料免费领取

免费领取
看完仍有疑问?想要更详细的答案?
备考问题一键咨询提分方案
获取专业解答

相关文章

【备考每日练】GRE数学题之倍数问题求解 考生不能不知的GRE数学中常用的10个概念最全汇总 平面几何题型应对有妙招 GRE数学三角函数专业术语总结 GRE考试时如何最短时间做完数学?高分学员为你分享 名师精准直击GRE数学考点 排列组合题型分别对待 GRE数学考试提分技巧有哪些?弃难投易是上策 GRE数学常见题型解题技巧讲解 数学高分并不难 【备考每日练】GRE数学练习题第3周大汇总

专题推荐

GRE关键词
版权申明| 隐私保护| 意见反馈| 联系我们| 关于我们| 网站地图| 最新资讯
© 2011-2024 ZHAN.com All Rights Reserved. 沪ICP备13042692号-23 举报电话:4000-006-150
沪公网安备 31010602002658号
增值电信业务经营许可证:沪B2-20180682