OG中的新GRE数学题解析(13)

最权威的国际教育服务平台资料来源：教育优选中的新GRE数学题解析（13）小编在此与大家分享新GREOG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.LetSbethesetofallpositiveintegersnsuchthatn2isamultipleofboth24and108.WhichofthefollowingintegersaredivisorsofeveryintegerninS?Indicateallsuchintegers.A12B24C36D72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TodeterminewhichoftheintegersintheanswerchoicesisadivisorofeverypositiveintegerninS,youmustfirstunderstandtheintegersthatareinS.Notethatinthisquestionyouaregiveninformationaboutn2,notaboutnitself.Therefore,youmustusetheinformationaboutn2toderiveinformationaboutn.Thefactthatn2isamultipleofboth24and108impliesthatn2isamultipleoftheleastcommonmultipleof24and108.Todeterminetheleastcommonmultipleof24and108,factor24and108intoprimefactorsas(23)(3)and(22)(33),respectively.Becausetheseareprimefactorizations,youcanconcludethattheleastcommon最权威的国际教育服务平台资料来源：教育优选(23)(33).Knowingthatn2mustbeamultipleof(23)(33)doesnotmeanthateverymultipleof(23)(33)isapossiblevalueofn2,becausen2mustbethesquareofaninteger.Theprimefactorizationofasquarenumbermustcontainonlyevenexponents.Thus,theleastmultipleof(23)(33)thatisasquareis(24)(34).Thisistheleastpossiblevalueofn2,andsotheleastpossiblevalueofnis(22)(32),or36.Furthermore,sinceeveryvalueofn2isamultipleof(24)(34),thevaluesofnarethepositivemultiplesof36;thatis,S{36,72,108,144,180,...}.ThequestionasksforintegersthataredivisorsofeveryintegerninS,thatis,divisorsofeverypositivemultipleof36.SinceChoiceA,12,isadivisorof36,itisalsoadivisorofeverymultipleof36.ThesameistrueforChoiceC,36.ChoicesBandD,24and72,arenotdivisorsof36,sotheyarenotdivisorsofeveryintegerinS.ThecorrectanswerconsistsofChoicesAandC.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。