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这可让他犯了难，施工现场距离项目部很远，没有车还真是不方便office,branchoffices(jurisdiction),riskmanagement,marketingmanagementsectorthroughsupervisionandinspectionfoundproblems,shouldbeassignedtheinvestigatorsarecorrectedinatimelymanner.27ththefifthchapterpenaltyunderanyofthefollowingacts,thentherelevantprovisionstopunishtheinvestigators,accordingtotheBank.Toconstituteacrimeshallbeinvestigatedforcriminalresponsibility:(A)onthebusinessthatarenotinvolvedintheinvestigation,issuedasurvey.(B)customercreditinformationarenotthoroughverification.(Iii)toparticipateincreditcustomersurveyisnotinplace,customersanddataisincomplete,untrue;heknowsbearacounterfeitedclientsissuingcredit.(D)doesnotprovideforduediligenceofcreditbusiness,pre-loaninvestigationform,concealingfactsorprovidingfalseinformationorshouldbefoundinanormalinvestigationfailedtodiscovertheriskfactors,leadtothereviewandapprovalpolicyerrors,loanrisk.(Five)onloanguaranteesofsurveynotinplace,notbyprovidesonarrived,andpledgerealforfieldverification,andassessment,andidentificationandregistration,notaccordingtoprovidesonguarantorofguaranteesqualificationandguaranteescapacityforsurveyverified,ledtoguaranteesloanlostauthenticity,andlegitimacy,andeffectivenessof;cycleloanbusinessintheofmortgage第十三讲数字谜基础班1．在下列算式的空格内，各填入一个合适的数字，使算式成立：答案：（1）（2）2.下面各题中的每一个汉字代表一个数字，不同的汉字代表不同的数字，相同的汉字代表相同的数字。当它们各代表什么数字时，以下各算式都成立？答案：（1）红=2，花=1，映=9，绿=7，叶=8，春=4。解答过程；春的取值范围为：2，3，4。①若春=2，则红=4，叶=7，但积的首位数字叶一定大于7，所以春≠2。②若春=3，则红=1或2：若红=1，则叶=7，但积的首位数字叶一定小于7，所以红≠1；若红=2，则叶=4，但积的首位数字叶一定大于4，所以红≠2；因此，春≠3。③若春=4，则红=2，叶=8，花=1，绿=7，映=9。（2）我们从小热爱科学=解答过程：由个位数字特点分析出：学=2，科=6；学=4，科=6；学=5，科=3，7，9；学=8，科=6。逐一分析上述五种情况，用积÷乘数，就得到被乘数。3.在下面乘法算式的空格内，各填入一个合适的数字，使算式成立：这可让他犯了难，施工现场距离项目部很远，没有车还真是不方便office,branchoffices(jurisdiction),riskmanagement,marketingmanagementsectorthroughsupervisionandinspectionfoundproblems,shouldbeassignedtheinvestigatorsarecorrectedinatimelymanner.27ththefifthchapterpenaltyunderanyofthefollowingacts,thentherelevantprovisionstopunishtheinvestigators,accordingtotheBank.Toconstituteacrimeshallbeinvestigatedforcriminalresponsibility:(A)onthebusinessthatarenotinvolvedintheinvestigation,issuedasurvey.(B)customercreditinformationarenotthoroughverification.(Iii)toparticipateincreditcustomersurveyisnotinplace,customersanddataisincomplete,untrue;heknowsbearacounterfeitedclientsissuingcredit.(D)doesnotprovideforduediligenceofcreditbusiness,pre-loaninvestigationform,concealingfactsorprovidingfalseinformationorshouldbefoundinanormalinvestigationfailedtodiscovertheriskfactors,leadtothereviewandapprovalpolicyerrors,loanrisk.(Five)onloanguaranteesofsurveynotinplace,notbyprovidesonarrived,andpledgerealforfieldverification,andassessment,andidentificationandregistration,notaccordingtoprovidesonguarantorofguaranteesqualificationandguaranteescapacityforsurveyverified,ledtoguaranteesloanlostauthenticity,andlegitimacy,andeffectivenessof;cycleloanbusinessintheofmortgage2答案:(1)确定乘数的范围为7、8、9，根据是被乘数的百位4与乘数相乘的积再加上十位的进位，结果为3□。然后逐一试验，得出答案。(2)选择被乘数的个位与乘数相乘的积的个位2作为解题突破口。两个一位数相乘，积的个位为2的算式有：1×2=22×6=123×4=124×8=326×7＝428×9=72又由于被乘数的百位与乘数相乘后再加上十位的进位，结果等于46，所以可确定乘数为上面算式中的6或7或8或9。最后逐一试验。(3)乘数不可能为5，若乘数为5，5与被乘数的十位数字7相乘后，再加上个位的进位不可能等于个位为0的数，所以被乘数的个位为5，乘数为4或8，这样得到两个解。(4)由于被乘数的个位4与乘数相乘的积的个位为2，所以乘数为3或8。但3作乘数无论如何也不可能使积成为52□2，所以乘数为8。下面确定出被乘数的首位数字为6，最后确定出被乘数的十位数字为5。4.在下面除法算式的空格内，各填入一个合适的数字，使算式成立：答案：（1）由于余数为7，所以可以确定除数的取值范围为8或9，再根据除数与商的个位相乘的积为5□，确定出商的个位的取值，最后求出被除数，得到两个解。第3页共5页这可让他犯了难，施工现场距离项目部很远，没有车还真是不方便office,branchoffices(jurisdiction),riskmanagement,marketingmanagementsectorthroughsupervisionandinspectionfoundproblems,shouldbeassignedtheinvestigatorsarecorrectedinatimelymanner.27ththefifthchapterpenaltyunderanyofthefollowingacts,thentherelevantprovisionstopunishtheinvestigators,accordingtotheBank.Toconstituteacrimeshallbeinvestigatedforcriminalresponsibility:(A)onthebusinessthatarenotinvolvedintheinvestigation,issuedasurvey.(B)customercreditinformationarenotthoroughverification.(Iii)toparticipateincreditcustomersurveyisnotinplace,customersanddataisincomplete,untrue;heknowsbearacounterfeitedclientsissuingcredit.(D)doesnotprovideforduediligenceofcreditbusiness,pre-loaninvestigationform,concealingfactsorprovidingfalseinformationorshouldbefoundinanormalinvestigationfailedtodiscovertheriskfactors,leadtothereviewandapprovalpolicyerrors,loanrisk.(Five)onloanguaranteesofsurveynotinplace,notbyprovidesonarrived,andpledgerealforfieldverification,andassessment,andidentificationandregistration,notaccordingtoprovidesonguarantorofguaranteesqualificationandguaranteescapacityforsurveyverified,ledtoguaranteesloanlostauthenticity,andlegitimacy,andeffectivenessof;cycleloanbusinessintheofmortgage（2）此题的关键是求出被除数，而求出被除数的关键又是求出余数。根据除数9与商的个位2相乘的积等于18，而被除数的个位为1，余数要比除数小，故余数为3。最后求出被除数，问题得解。提高班1．在下列算式的空格内，各填入一个合适的数字，使算式成立：答案：（1）（2）2.下面各题中的每一个汉字代表一个数字，不同的汉字代表不同的数字，相同的汉字代表相同的数字。当它们各代表什么数字时，以下各算式都成立？答案：（1）红=2，花=1，映=9，绿=7，叶=8，春=4。解答过程；春的取值范围为：2，3，4。①若春=2，则红=4，叶=7，但积的首位数字叶一定大于7，所以春≠2。②若春=3，则红=1或2：若红=1，则叶=7，但积的首位数字叶一定小于7，所以红≠1；若红=2，则叶=4，但积的首位数字叶一定大于4，所以红≠2；因此，春≠3。③若春=4，则红=2，叶=8，花=1，绿=7，映=9。（2）我们从小热爱科学=这可