Datasets:

math-eval

/

TAL-SCQ5K

like 47

[ "2018年湖北武汉创新杯小学高年级五年级竞赛初赛数学思维能力等级测试第1题4分" ]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$3$$ " } ], [ { "aoVal": "C", "content": "$$4$$ " } ], [ { "aoVal": "D", "content": "$$5$$ " } ] ]

[ "知识标签->拓展思维->计算模块->小数->小数乘除->小数乘法运算" ]

[ "$$0.025\\times 0.04=0.001$$ " ]

[ "2020年希望杯五年级竞赛模拟第33题", "2020年新希望杯五年级竞赛第33题" ]

[ [ { "aoVal": "A", "content": "黄 " } ], [ { "aoVal": "B", "content": "黑 " } ], [ { "aoVal": "C", "content": "红 " } ], [ { "aoVal": "D", "content": "橙 " } ], [ { "aoVal": "E", "content": "蓝 " } ], [ { "aoVal": "F", "content": "紫 " } ], [ { "aoVal": "G", "content": "绿 " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "共猜对$$5$$次，由题意知：$$1$$号箱被猜$$2$$次橙，$$2$$号箱被猜$$2$$次黑，则$$1$$号箱为橙或$$2$$号箱为黑． 假设$$1$$号为橙色，则拉斐尔和多纳泰罗一个猜中$$1$$次，一人猜中两次． 假设拉斐尔猜中$$1$$次，则$$2$$号不为黑，$$3$$号不为绿，则李奥纳多猜中$$3$$号为黄，米开朗琪罗全猜错． 假设有问题，则拉斐尔猜中$$2$$次，$$1$$号为橙，$$2$$号不为黑，$$3$$号为绿，则李奥纳多全猜错，则$$1$$号为橙，$$2$$号为黑，$$3$$号不为绿，拉斐尔猜中$$2$$次，李奥纳多猜中$$2$$号为黑，多纳泰罗猜中$$1$$号为橙，米开朗琪罗猜中$$3$$号为紫． " ]

[ "2021年新希望杯二年级竞赛初赛第30题5分" ]

[ [ { "aoVal": "A", "content": "星期二 " } ], [ { "aoVal": "B", "content": "星期三 " } ], [ { "aoVal": "C", "content": "星期五 " } ], [ { "aoVal": "D", "content": "星期六 " } ], [ { "aoVal": "E", "content": "星期日 " } ] ]

[ "拓展思维->思想->逐步调整思想" ]

[ "前一半$$7$$天的情况：真、真、假、真、假、假、真， 后一半$$7$$天的情况：真、假、真、假、真、真、假， $$\\text{A}$$选项：若这一天是星期二，则前一半人在今天说的是真话，那么明天应该说真话才合理，但星期三他们说的是假话，相互矛盾．后一半人在今天说的是假话，那么明天应该说假话才合理，但是星期三他们说的是真话，也相互矛盾，故$$\\text{A}$$错误； $$\\text{B}$$选项：若这一天是星期三，则前一半人在今天说的是假话，那么明天应该说假话才合理，但星期四他们说的是真话，相互矛盾．后一半人在今天说的是真话，那么明天应该说真话才合理，但是星期四他们说的是假话，也相互矛盾，故$$\\text{B}$$错误； $$\\text{C}$$选项：若这一天是星期五，则前一半人在今天说的是假话，那么明天应该说假话才合理，星期六他们说的正好是假话，符合．后一半人在今天说的是真话，那么明天应该说真话才合理，星期六他们说的正好是真话，符合，故$$\\text{C}$$正确； $$\\text{D}$$选项：若这一天是星期六，则前一半人在今天说的是假话，那么明天应该说假话才合理，但星期日他们说的是真话，相互矛盾．后一半人在今天说的是真话，那么明天应该说真话才合理，但是星期日他们说的是假话，也相互矛盾，故$$\\text{D}$$错误； E选项：若这一天是星期日，则前一半人在今天说的是真话，那么明天应该说真话才合理，星期一他们说的是真话，符合．后一半人在今天说的是假话，那么明天应该说假话才合理，但是星期一他们说的是真话，相互矛盾，故$$\\text{E}$$错误． 故选$$\\text{C}$$． " ]

[ "2009年第7届创新杯六年级竞赛初赛第5题4分" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$3$$ " } ], [ { "aoVal": "D", "content": "$$4$$ " } ] ]

[ "拓展思维->能力->构造模型->模型思想" ]

[ "$$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde4\\times (1-99 \\%)\\div (1-98 \\%)$$ $$=4\\times 1 \\%\\div 2 \\%$$ $$=0.04\\div 2 \\%$$ $$=2$$（吨）， 答：运抵武昌后，葡萄还剩$$2$$吨． 故选$$\\text{B}$$． " ]

[ "2011年第9届全国创新杯小学高年级六年级竞赛第1题" ]

[ [ { "aoVal": "A", "content": "$$20$$ " } ], [ { "aoVal": "B", "content": "$$25$$ " } ], [ { "aoVal": "C", "content": "$$30$$ " } ], [ { "aoVal": "D", "content": "$$35$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->分百应用题->求分率" ]

[ "$$1\\div \\left( 1-20 \\% \\right)-1=25 \\%$$． " ]

[ "2017年全国小学生数学学习能力测评四年级竞赛复赛第7题3分" ]

[ [ { "aoVal": "A", "content": "$$50$$ " } ], [ { "aoVal": "B", "content": "$$52$$ " } ], [ { "aoVal": "C", "content": "$$55$$ " } ] ]

[ "拓展思维->能力->数据处理" ]

[ "先用错误的被除数$$1205$$减去余数$$5$$，再除以商$$48$$求出除数；再用原题的被除数$$1250$$除以除数求出商． $$(1205-5)\\div 48$$， $$=1200\\div 48$$， $$=25$$， $$1250\\div 25=50$$． 答：正确的商是$$50$$． 故选$$\\text{A}$$． " ]

[ "2009年亚太杯竞赛初赛第19题5分" ]

[ [ { "aoVal": "A", "content": "无法造 " } ], [ { "aoVal": "B", "content": "一个 " } ], [ { "aoVal": "C", "content": "两个 " } ], [ { "aoVal": "D", "content": "三个 " } ], [ { "aoVal": "E", "content": "以上各项皆否 " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "长方形的长为$$a$$，宽为$$b$$，那么长方形的周长为$$2a+2b$$，一定是$$2$$的倍数，但是该题中要求用尽所有的枝条，所有枝条的长度和为$$9\\times 1+6\\times 2+3\\times 4=33$$（厘米），是奇数，所以矛盾，无法造． " ]

[ "2016年第12届全国新希望杯小学高年级六年级竞赛复赛第3题4分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{9}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{1}{5}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{1}{6}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{1}{3}$$ " } ] ]

[ "拓展思维->拓展思维->计数模块->统计与概率->概率->典型问题->摸小球" ]

[ "$$6$$个球中选$$2$$个，一共有$$C_{6}^{2}=15$$种情况，$$3$$个红球中选$$2$$个，一共有$$C_{3}^{2}=3$$种．则都是红球的概率为$$P=\\frac{C_{3}^{2}}{C_{6}^{2}}=\\frac{3}{15}=\\frac{1}{5}$$． " ]

[ "2015年华杯赛四年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$6$$ " } ] ]

[ "拓展思维->拓展思维->组合模块->抽屉原理->构造型抽屉原理" ]

[ "因为每人先给$$6$$分，所以得分情况有以下$$25$$种： $$0$$、$$1$$、$$2$$、$$3$$、$$4$$、$$5$$、$$6$$、$$7$$、$$8$$、$$9$$、$$10$$、$$11$$、$$12$$、$$13$$、$$14$$、$$15$$、$$16$$、$$17$$、$$18$$、$$20$$、$$21$$、$$22$$、$$25$$、$$26$$、$$30$$，因此根据抽屉原理，$$51\\div 25=2\\cdots \\cdots 1$$，所以至少有$$3$$人得分相同。 " ]

[ "2019年第23届广东世界少年奥林匹克数学竞赛五年级竞赛决赛第7题5分" ]

[ [ { "aoVal": "A", "content": "$$110$$ " } ], [ { "aoVal": "B", "content": "$$115$$ " } ], [ { "aoVal": "C", "content": "$$117$$ " } ], [ { "aoVal": "D", "content": "$$120$$ " } ] ]

[ "拓展思维->思想->对应思想", "Overseas Competition->知识点->数论模块->因数与倍数->因数个数定理" ]

[ "$$990$$的因数有：$$1$$；$$990$$；$$2$$；$$495$$；$$3$$；$$330$$；$$5$$；$$198$$；$$6$$；$$165$$；$$9$$；$$110$$；$$10$$；$$99$$；$$11$$；$$90$$；$$15$$；$$66$$；$$18$$；$$55$$；$$22$$；$$45$$；$$30$$；$$33$$，一共$$24$$个， 平均数为：$$(1+990+2+495+3+330+5+198+6+165+9+110+10+99+11+90+15+66+18+55+22+45+30+33)\\div 24$$ $$=2808\\div 24$$ $$=117$$， 故选答案$$\\text{C}$$． " ]

[ "2014年全国迎春杯六年级竞赛复赛第2题" ]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$6$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->定义新运算->观察规律型->找规律型定义新运算" ]

[ "由新定义：$$n!=1\\times 2\\times 3\\times \\ldots\\times n$$得： $$2014!=1\\times 2\\times 3\\times 4\\times 5\\times \\ldots\\times 2021\\times 2022$$ $$=1\\times 3\\times 4\\times 6\\times 7\\times 8\\times \\ldots\\times 2021\\times 2022\\times 10$$ 所以$$1\\times 3\\times 4\\times 6\\times 7\\times 8\\times \\ldots\\times 2021\\times 2022\\times 10$$是$$10$$的倍数， 所以$$2022!$$的个位数为$$0$$； $$3!=1\\times 2\\times 3=6$$ 所以$$2022!-3!$$的个位数也就为：$$10-6=4$$． " ]

[ "2013年华杯赛四年级竞赛初赛", "2013年华杯赛四年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "小东 " } ], [ { "aoVal": "B", "content": "小西 " } ], [ { "aoVal": "C", "content": "小南 " } ], [ { "aoVal": "D", "content": "小北 " } ] ]

[ "拓展思维->拓展思维->组合模块->逻辑推理->假设型逻辑推理" ]

[ "解：根据题干分析可得，小南与小北说的话是相互矛盾的，所以两人中一定有一个人说的是正确的。 假设小北说的是正确的，则小南说``小东说的不对''是错，可得：小东说的对，这样与已知只有一个人说对了相矛盾，所以此假设不成立，故小南说的是正确的。 故选：$$\\text{C}$$。 " ]

[ "2014年全国迎春杯三年级竞赛复赛第2题" ]

[ [ { "aoVal": "A", "content": "$$3$$和$$4$$ " } ], [ { "aoVal": "B", "content": "$$3$$和$$6$$ " } ], [ { "aoVal": "C", "content": "$$4$$和$$6$$ " } ], [ { "aoVal": "D", "content": "$$6$$和$$3$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "由``$$\\square $$+$$\\square $$+$$\\square $$$$+$$△$$+$$△$$+$$△$$=27$$''可得``$$\\square $$+△$$=9$$''，又由于``$$△$$+$$△$$+$$\\square $$$$=12$$''可得△$$=3$$，$$\\square $$=$$6$$．题目问的是$$\\square $$和△分别代表什么，所以选$$\\text{D}$$． " ]

[ "2017年全国华杯赛小学中年级竞赛初赛模拟第1题", "小学中年级三年级上学期其它" ]

[ [ { "aoVal": "A", "content": "$$145$$ " } ], [ { "aoVal": "B", "content": "$$140$$ " } ], [ { "aoVal": "C", "content": "$$125$$ " } ], [ { "aoVal": "D", "content": "$$120$$ " } ] ]

[ "拓展思维->知识点->应用题模块->和差倍问题->和差问题->二量和差问题->明和暗差" ]

[ "因为甲给乙$$40$$元，还比乙多$$10$$元，所以甲比乙多$$40\\times 2+10=90$$元， 所以甲：$$(200+90)\\div 2=145$$（元） " ]

[ "2020年第23届世界少年奥林匹克数学竞赛四年级竞赛决赛（中国区）第9题3分" ]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$6$$ " } ] ]

[ "拓展思维->拓展思维->组合模块->抽屉原理->最不利原则" ]

[ "从$$1$$至$$9$$这$$9$$个整数中， $$1+9=2+8=3+7=4+6$$，都等于$$10$$； 根据最不利原则，取的数都是$$(1,9)$$、$$(2,8)$$、$$(3,7)$$、$$(4,6)$$组合中的$$1$$个， 再取一个$$5$$，即取第$$6$$个数时，能保证其中有两个数的和等于$$10$$． 故选$$\\text{D}$$． " ]

[ "2010年迎春杯六年级竞赛复赛", "六年级其它", "2010年迎春杯五年级竞赛复赛" ]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$8$$ " } ], [ { "aoVal": "D", "content": "$$12$$ " } ], [ { "aoVal": "E", "content": "以上都不正确 " } ] ]

[ "Overseas Competition->知识点->组合模块->构造与论证", "拓展思维->拓展思维->组合模块->操作与策略->操作问题->数字操作" ]

[ "每次操作，每个数字个数的奇偶性都会变化．$$1$$、$$3$$、$$5$$原来都是偶数个，它们的个数奇偶性永远保持一致；$$2$$、$$4$$原来都是奇数个，它们的个数奇偶性也永远保持一致，而且和$$1$$、$$3$$、$$5$$的个数奇偶性不同$$.$$ 最后剩下$$2$$个数字，只能是$$2$$和$$4$$，乘积为$$8$$，答案选$$\\text{C}$$. " ]

[ "2015年北京华杯赛小学高年级竞赛初赛A卷第1题" ]

[ [ { "aoVal": "A", "content": "甲、乙 " } ], [ { "aoVal": "B", "content": "乙、丙 " } ], [ { "aoVal": "C", "content": "甲、丙 " } ], [ { "aoVal": "D", "content": "乙、丁 " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "题目要求有两个人去，可以使用假设法，若甲去，则乙去，乙去则丙也去．三个人去，矛盾，所以甲不去．若丙不去则乙不去，那么只有丁去，矛盾，所以丙去．丙去则丁不去，由两个人去得到结论，乙要去． 所以答案是$$\\text{B}$$，丙和乙去． " ]

[ "2020年新希望杯五年级竞赛决赛（8月）第17题" ]

[ [ { "aoVal": "A", "content": "星期一 " } ], [ { "aoVal": "B", "content": "星期二 " } ], [ { "aoVal": "C", "content": "星期三 " } ], [ { "aoVal": "D", "content": "星期四 " } ] ]

[ "拓展思维->拓展思维->应用题模块->周期问题->时间中的周期问题->日期中的周期" ]

[ "$$2020\\div 4=505$$，$$2020$$年是闰年． 从$$2019$$年国庆节经过$$366$$天是$$2020$$年国庆节． $$366\\div 7=52$$（周）$$\\cdots \\cdots 2$$（天）， 故$$2020$$年的国庆节是星期四． 故选$$\\text{D}$$． " ]

[ "2008年陈省身杯小学高年级六年级竞赛" ]

[ [ { "aoVal": "A", "content": "$$\\frac{6}{25}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{22}{25}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{44}{25}$$ " } ], [ { "aoVal": "D", "content": "$$1\\frac{22}{25}$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->分百应用题->量率对应已知单位1" ]

[ "根据题意得，一堆煤$$2$$吨，每天用去它的$$\\frac{1}{25}$$，$$3$$天后还剩（ ）吨，可列式$$2-3\\times \\frac{1}{25}\\times 2=\\frac{44}{25}$$，故答案为$$\\text{C}$$． " ]

[ "2017年全国华杯赛小学高年级竞赛" ]

[ [ { "aoVal": "A", "content": "$$\\frac{ab}{a+b}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{a+b}{2}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{2ab}{a+b}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{2a+2b}{ab}$$ " } ] ]

[ "拓展思维->思想->转化与化归的思想" ]

[ "设甲、乙两地之间的距离为$$s$$千米．去时所用的时间为$$\\frac{s}{a}$$，返回时所用的时间为$$\\frac{s}{b}$$，往返的平均速度为 $$\\frac{2s}{\\frac{s}{a}+\\frac{s}{b}}=\\frac{2s}{\\frac{bs+as}{ab}}=\\frac{2s\\times ab}{as+bs}=\\frac{2ab}{a+b}$$． " ]

[ "2017年全国美国数学大联盟杯小学中年级三年级竞赛初赛第35题" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$1$$ " } ], [ { "aoVal": "C", "content": "$$2$$ " } ], [ { "aoVal": "D", "content": "$$3$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "原式$$=120$$，各位是$$0$$． " ]

[ "2020年广东广州羊排赛四年级竞赛第9题4分" ]

[ [ { "aoVal": "A", "content": "$$20$$ " } ], [ { "aoVal": "B", "content": "$$50$$ " } ], [ { "aoVal": "C", "content": "$$40$$ " } ], [ { "aoVal": "D", "content": "$$30$$ " } ], [ { "aoVal": "E", "content": "$$34$$ " } ] ]

[ "拓展思维->能力->逻辑分析->代数逻辑推理" ]

[ "在一场比赛中，如果一胜一负， 则胜得$$2$$分，负得$$0$$分， 总分为$$2+0=2$$分， 如果赛平，则总分为$$1+1=2$$分， 即在一场比赛中，无论结果如何，比赛总分是不变的，都是$$2$$分， $$6$$支队伍进行单循环赛，共有比赛：$$5+4+3+2+1=15$$场， 所以六支队伍的总得分是：$$15\\times 2=30$$分， 故选：$$\\text{D}$$． " ]

[ "2018年湖北武汉新希望杯六年级竞赛训练题（五）第14题" ]

[ [ { "aoVal": "A", "content": "$$66$$ " } ], [ { "aoVal": "B", "content": "$$67$$ " } ], [ { "aoVal": "C", "content": "$$68$$ " } ], [ { "aoVal": "D", "content": "$$69$$ " } ], [ { "aoVal": "E", "content": "$$70$$ " } ] ]

[ "海外竞赛体系->知识点->计算模块->比较与估算", "拓展思维->拓展思维->计算模块->比较与估算->放缩法->首尾放缩法" ]

[ "$$A=(1+2+3+\\cdots +11)+\\left( \\frac{11}{111}+\\frac{11}{112}+\\frac{11}{113}+\\cdots +\\frac{11}{121} \\right)\\textgreater66+\\frac{11}{121}\\times 11=67$$，$$A\\textless{}66+\\frac{11}{111}\\times 11\\textless{}68$$，所以$$67\\textless{}A\\textless{}68$$，$$A$$的整数部分是$$67$$． " ]

[ "2013年希望杯三年级竞赛复赛", "2013年希望杯二年级竞赛复赛", "2013年希望杯四年级竞赛复赛", "2013年希望杯五年级竞赛复赛", "2013年希望杯六年级竞赛复赛", "2013年希望杯一年级竞赛复赛" ]

[ [ { "aoVal": "A", "content": "$$7$$ " } ], [ { "aoVal": "B", "content": "$$8$$ " } ], [ { "aoVal": "C", "content": "$$9$$ " } ], [ { "aoVal": "D", "content": "$$10$$ " } ] ]

[ "拓展思维->拓展思维->数论模块->因数与倍数->因数个数定理" ]

[ "$$42=2\\times 3\\times 7$$，因数个数等于指数加$$1$$再连乘，$$\\left( 1+1 \\right)\\times \\left( 1+1 \\right)\\times \\left( 1+1 \\right)=8$$（个）。所以选B。 " ]

[ "2019年第24届YMO二年级竞赛决赛第4题3分" ]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$6$$ " } ], [ { "aoVal": "D", "content": "$$7$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "根据题意可得： $$15=1+2+3+9$$；$$15=1+2+4+8$$； $$15=1+2+5+7$$；$$15=1+3+4+7$$； $$15=1+3+5+6$$；$$15=2+3+4+6$$； 一共有$$6$$种． 答：共有$$6$$种不同的分法． 故选：$$\\text{C}$$． " ]

[ "2018年IMAS小学中年级竞赛（第二轮）第1题4分" ]

[ [ { "aoVal": "A", "content": "$$45$$ " } ], [ { "aoVal": "B", "content": "$$48$$ " } ], [ { "aoVal": "C", "content": "$$51$$ " } ], [ { "aoVal": "D", "content": "$$54$$ " } ], [ { "aoVal": "E", "content": "$$57$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->数列与数表->数列规律" ]

[ "$$100-97+94-91+88-85+\\cdots +4-1$$ $$=\\left( 100-97 \\right)+\\left( 94-91 \\right)+\\left( 88-85 \\right)+\\cdots +\\left( 4-1 \\right)$$ $$=\\underbrace{3+3+3+\\cdots +3}_{17项}$$ $$=3\\times 17$$ $$=51$$． 故选$$\\text{C}$$． ", "<p>$$100-97+94-91+88-85+\\cdots +4-1$$</p>\n<p>$$=\\left( 100+94+88+\\cdots +4 \\right)-\\left( 97+91+85+\\cdots +1 \\right)$$</p>\n<p>$$=\\frac{\\left( 100+4 \\right)\\times 17}{2}-\\frac{\\left( 97+1 \\right)\\times 17}{2}$$</p>\n<p>$$=\\left( 104-98 \\right)\\times \\frac{17}{2}$$</p>\n<p>$$=6\\times \\frac{17}{2}$$</p>\n<p>$$=51$$．</p>\n<p>故选$$\\text{C}$$．</p>" ]

[ "2005年第3届创新杯六年级竞赛初赛第10题", "2005年六年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "$$2$$克和$$9$$克 " } ], [ { "aoVal": "B", "content": "$$2$$克和$$11$$克 " } ], [ { "aoVal": "C", "content": "$$2$$克和$$13$$克 " } ], [ { "aoVal": "D", "content": "$$2$$克和$$15$$克 " } ] ]

[ "拓展思维->拓展思维->计数模块->枚举法综合->枚举法->有序枚举" ]

[ "我们只用分析所给四个选项中的重量能否称出，而$$9=5+3+1\\text{，}11=5+5+1\\text{，}13=5+5+3$$都可称出，选$$D$$。 " ]

[ "2020年希望杯二年级竞赛模拟第4题" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$12$$ " } ], [ { "aoVal": "C", "content": "$$14$$ " } ], [ { "aoVal": "D", "content": "$$15$$ " } ] ]

[ "拓展思维->思想->数形结合思想" ]

[ "后一个数等于前一个数$$+$$ 3 " ]

[ "2019年第23届广东世界少年奥林匹克数学竞赛二年级竞赛初赛第5题5分" ]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$6$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "从题干中可以知道，今年哥哥$$16$$岁，弟弟$$8$$岁，二者相差，$$16-8=8$$（岁）， 那么题目问几年前，哥哥的年龄是弟弟的$$3$$倍， 也就是二者年龄差应该是弟弟的$$2$$倍，$$8\\div 2=4$$（岁），$$8-4=4$$（年）， 即$$4$$年前，哥哥的年龄是弟弟的$$3$$倍． 我们可以验证一下，四年前，哥哥的年龄是：$$16-4=12$$（岁），弟弟的年龄是：$$8-4=4$$（岁）． $$12$$岁正好是$$4$$岁的$$3$$倍． " ]

[ "2019年浙江杭州滨江区杭州江南实验学校五年级竞赛模拟（江南杯）第12题1分" ]

[ [ { "aoVal": "A", "content": "$$340-400$$ " } ], [ { "aoVal": "B", "content": "$$200-230$$ " } ], [ { "aoVal": "C", "content": "$$230-350$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "假设都按照最低价$$10.5$$一个，$$22$$个总价是$$231$$元，假设都按照最高价$$15.5$$元一个，总价是$$341$$元，所以这些玩具的总价不少于$$230$$元，也不高于$$340$$元．据此解答． $$10.5\\times22=231$$（元）， $$15.5\\times22=341$$（元）， 所以这些玩具的总价不少于$$230$$元，也不高于$$340$$元． 故答案为∶$$\\text{C}$$． " ]

[ "2015年全国世奥赛竞赛A卷第6题" ]

[ [ { "aoVal": "A", "content": "$$2:9$$ " } ], [ { "aoVal": "B", "content": "$$1:9$$ " } ], [ { "aoVal": "C", "content": "$$1:8$$ " } ], [ { "aoVal": "D", "content": "$$1:7$$ " } ] ]

[ "知识标签->拓展思维->应用题模块->比例应用题->复杂的比例问题" ]

[ "第一次倒完后，$$1$$号杯中剩下$$50$$克糖，$$2$$号杯中有$$50$$克糖和$$25$$克盐．此时把$$2$$号杯的$$\\frac{2}{7}$$倒入$$1$$号杯，那么$$2$$号杯的糖和盐都倒了$$\\frac{2}{7}$$，即糖倒了$$50\\times \\frac{2}{7}=\\frac{100}{7}$$（克），盐倒了$$25\\times \\frac{2}{7}=\\frac{50}{7}$$（克）．那么$$1$$号杯的糖就变成了$$50+\\frac{100}{7}=\\frac{450}{7}$$（克），盐$$\\frac{50}{7}$$克． 所以含盐量和含糖量之比为$$\\frac{50}{7}:\\frac{450}{7}=1:9$$． " ]

[ "2014年全国迎春杯四年级竞赛复赛第5题" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$1$$ " } ], [ { "aoVal": "C", "content": "$$3$$ " } ], [ { "aoVal": "D", "content": "$$9$$ " } ] ]

[ "拓展思维->拓展思维->数论模块->尾数特征->末一位数" ]

[ "因为从$$5!=1\\times 2\\times 3\\times 4\\times 5$$ 开始，都含有$$2$$和$$5$$，那么个位必然是$$0$$，那么我们只要计算$$-4!+3!-2!+1!$$的个位数字即可，得出个位数字为$$1$$． " ]

[ "2019年全国小学生数学学习能力测评四年级竞赛复赛第8题3分" ]

[ [ { "aoVal": "A", "content": "$$6$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$3$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "除去第$$1$$小时以外的费用：$$62-15\\times2=32$$（元）； 每人除了第一小时以外所花的时间：$$32\\div 2\\div 8=2$$（小时）； 共时长：$$2+1=3$$（小时）． 故选$$\\text{C}$$． " ]

[ "2021年新希望杯二年级竞赛初赛第30题5分" ]

[ [ { "aoVal": "A", "content": "星期二 " } ], [ { "aoVal": "B", "content": "星期三 " } ], [ { "aoVal": "C", "content": "星期五 " } ], [ { "aoVal": "D", "content": "星期六 " } ], [ { "aoVal": "E", "content": "星期日 " } ] ]

[ "拓展思维->思想->分类讨论思想" ]

[ "前一半$$7$$天的情况：真、真、假、真、假、假、真， 后一半$$7$$天的情况：真、假、真、假、真、真、假， $$\\text{A}$$选项：若这一天是星期二，则前一半人在今天说的是真话，那么明天应该说真话才合理，但星期三他们说的是假话，相互矛盾．后一半人在今天说的是假话，那么明天应该说假话才合理，但是星期三他们说的是真话，也相互矛盾，故$$\\text{A}$$错误； $$\\text{B}$$选项：若这一天是星期三，则前一半人在今天说的是假话，那么明天应该说假话才合理，但星期四他们说的是真话，相互矛盾．后一半人在今天说的是真话，那么明天应该说真话才合理，但是星期四他们说的是假话，也相互矛盾，故$$\\text{B}$$错误； $$\\text{C}$$选项：若这一天是星期五，则前一半人在今天说的是假话，那么明天应该说假话才合理，星期六他们说的正好是假话，符合．后一半人在今天说的是真话，那么明天应该说真话才合理，星期六他们说的正好是真话，符合，故$$\\text{C}$$正确； $$\\text{D}$$选项：若这一天是星期六，则前一半人在今天说的是假话，那么明天应该说假话才合理，但星期日他们说的是真话，相互矛盾．后一半人在今天说的是真话，那么明天应该说真话才合理，但是星期日他们说的是假话，也相互矛盾，故$$\\text{D}$$错误； E选项：若这一天是星期日，则前一半人在今天说的是真话，那么明天应该说真话才合理，星期一他们说的是真话，符合．后一半人在今天说的是假话，那么明天应该说假话才合理，但是星期一他们说的是真话，相互矛盾，故$$\\text{E}$$错误． 故选$$\\text{C}$$． " ]

[ "2014年第15届上海中环杯小学低年级二年级竞赛初赛第16题" ]

[ [ { "aoVal": "A", "content": "甲 " } ], [ { "aoVal": "B", "content": "乙 " } ], [ { "aoVal": "C", "content": "丙 " } ], [ { "aoVal": "D", "content": "丁 " } ] ]

[ "知识标签->拓展思维->组合模块->逻辑推理->假设型逻辑推理->真假话->找矛盾" ]

[ "单双数加减的性质是：单数个单数和为单，双数个单数和为双． 甲：$$9+9+9+1+7=35$$，正确． 乙：$$9+9+9+1+1+7=36$$，正确． 丙：三个单数的和为单数，结果$$24$$为双数，错误． 丁：$$7+7+7+0=21$$，正确． " ]

[ "2007年第5届创新杯六年级竞赛第6题5分" ]

[ [ { "aoVal": "A", "content": "$$1004$$ " } ], [ { "aoVal": "B", "content": "$$1002$$ " } ], [ { "aoVal": "C", "content": "$$1000$$ " } ], [ { "aoVal": "D", "content": "$$998$$ " } ] ]

[ "拓展思维->思想->整体思想" ]

[ "∵有$$2007$$盏亮着的电灯，现按其顺序编号为$$1$$，$$2$$，$$\\cdots$$，$$2007$$， ∴编号为$$2$$的倍数的灯有$$(2007-1)\\div2=1003$$（只）， 编号为$$3$$的倍数的灯有$$2007\\div3=669$$（只）， 编号为$$5$$的倍数的灯的有$$(2007-2)\\div5=401$$（只）， 其中既是$$3$$的倍数也是$$5$$的倍数有$$(2007-12)\\div15=133$$（只）， 既是$$2$$的倍数也是$$3$$的倍数有$$(2007-3)\\div6=334$$（只）， 既是$$2$$的倍数也是$$5$$的倍数有$$(2007-7)\\div10=200$$（只）， 既是$$2$$的倍数也是$$5$$的倍数，还是$$3$$的倍数有$$(2007-27)\\div30=66$$（只）， 只拉$$1$$次的：$$1003-334-200+66=535$$， $$669-334-133+66=268$$， $$401-200-133+66=134$$， 拉$$3$$次的$$66$$， 所以亮的就是$$2007-535-268-134-66=1004$$（只）． 故选$$\\text{A}$$． " ]

[ "2017年全国华杯赛小学高年级竞赛" ]

[ [ { "aoVal": "A", "content": "是$$5 \\%$$ " } ], [ { "aoVal": "B", "content": "是$$10 \\%$$ " } ], [ { "aoVal": "C", "content": "是$$7.5 \\%$$ " } ], [ { "aoVal": "D", "content": "不能确定 " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "如果甲瓶盐水为$$a$$升，则含盐$$5a \\%$$，如果乙瓶盐水为$$b$$升，则含盐$$10 \\%b$$，依照浓度的计算公式，混合后盐水的浓度是：$$\\frac{5 \\%a+10 \\%b}{a+b}$$，因为不知$$a$$和$$b$$的数值，所以两瓶混合后盐水的浓度不能确定． " ]

[ "2014年全国迎春杯四年级竞赛复赛第8题" ]

[ [ { "aoVal": "A", "content": "$$5000$$ " } ], [ { "aoVal": "B", "content": "$$7320$$ " } ], [ { "aoVal": "C", "content": "$$8000$$ " } ], [ { "aoVal": "D", "content": "$$8640$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->工程问题->合作工程问题->已知工时反推->两人合作" ]

[ "我们可以理解为小元每打$$10$$份的字数就会打错$$1$$份，小芳每打$$10$$份的字数就会打错$$2$$份，即小芳打$$5$$份的字数只能正确$$4$$份． 假设小元打的总字数为$$80$$份，那么他正确的为$$8\\times 9=72$$份，根据题意可知小芳正确的为$$72\\div 2=36$$份， 那么小芳打字的总分数为$$36\\div 4\\times 5=45$$份．小元和小芳的总字数份为$$80+45=125$$份，共$$10000$$字，即每份字数为$$10000\\div125=80$$． 小元和小芳共正确的字数为$$(72+36)\\times 80=8640$$． " ]

[ "2016年陈省身杯六年级竞赛国际青少年数学解题能力展示活动第11题" ]

[ [ { "aoVal": "A", "content": "$$16$$ " } ], [ { "aoVal": "B", "content": "$$32$$ " } ], [ { "aoVal": "C", "content": "$$48$$ " } ], [ { "aoVal": "D", "content": "$$64$$ " } ] ]

[ "拓展思维->思想->对应思想", "课内体系->能力->运算求解" ]

[ "成本$$:$$利润$$:$$售价$$=8:12:20=2:3:5$$，卖出$$\\frac{3}{4}$$后，售价为$$\\frac{3}{4}\\times 5=\\frac{15}{4}$$(份)，此时获利份数为$$\\frac{15}{4}-2=\\frac{7}{4}$$(份)，全成本为$$224\\div \\frac{7}{4}\\times 2=256$$（元），钢笔共有$$256\\div 8=32$$(支)． " ]

[ "2016年第14届全国创新杯小学高年级五年级竞赛复赛第1题" ]

[ [ { "aoVal": "A", "content": "$$67$$ " } ], [ { "aoVal": "B", "content": "$$73$$ " } ], [ { "aoVal": "C", "content": "$$158$$ " } ], [ { "aoVal": "D", "content": "$$22$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "设这个数为$$a$$，则$$\\left { \\begin{matrix}a\\div 3\\cdot \\cdot \\cdot \\cdot \\cdot \\cdot 2 a\\div 5\\cdot \\cdot \\cdot \\cdot \\cdot \\cdot 3 a\\div 7\\cdot \\cdot \\cdot \\cdot \\cdot \\cdot 4 \\end{matrix} \\right.$$，整理得$$\\left { \\begin{matrix}a\\div 3\\cdot \\cdot \\cdot \\cdot \\cdot \\cdot 11 a\\div 5\\cdot \\cdot \\cdot \\cdot \\cdot \\cdot 3 a\\div 7\\cdot \\cdot \\cdot \\cdot \\cdot \\cdot 11 \\end{matrix} \\right.$$，所以$$a=21k+11$$ 所以当$$k=7$$时，$$a=158$$． " ]

[ "2019年第23届广东世界少年奥林匹克数学竞赛一年级竞赛初赛第9题5分" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$15$$ " } ], [ { "aoVal": "C", "content": "$$16$$ " } ], [ { "aoVal": "D", "content": "$$20$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "除了涛涛前面的人和涛涛本身，剩下的就是排在涛涛后面的人， 所以共有：$$26-10-1=15$$（人）． 故选：$$\\text{B}$$． " ]

[ "2019年第23届广东世界少年奥林匹克数学竞赛一年级竞赛决赛第8题5分" ]

[ [ { "aoVal": "A", "content": "$$5$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$7$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->年龄问题->年龄问题基本关系->年龄差" ]

[ "从题干中可以知道，小明今年$$14$$岁，小强今年$$8$$岁， 那么今年二者的年龄差是$$14-8=6$$（岁）； 无论经过几年，小明和小强的年龄差还是不变的； 所以$$2$$年后，小明比小强大$$6$$岁； 可以换另一种方法检验一下：$$2$$年后小明$$16$$岁，小强$$10$$岁；$$16-10=6$$（岁）． 所以选择$$\\text{B}$$选项． " ]

[ "2013年第11届全国创新杯五年级竞赛第4题5分" ]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$7$$ " } ] ]

[ "拓展思维->七大能力->逻辑分析" ]

[ "对于赛况分析试题，尤其对于与分数相关的试题，最重要的是思维方式，本题如果从整体上来考虑比赛所产生的总分值，问题将迎刃而解，依题意可知比赛总场次为$$24$$场比赛之中，若平局则将会让所有队伍的总分增加$$2$$分（比赛双方均得$$1$$分），若出现了胜败，则所有队伍的总分增加$$3$$分，而现在所有队伍获得的总分值为：$$22+19+14+12=67$$（分），$$24$$场比赛，有$$3$$分，有$$2$$分，总分为$$67$$分，可当做鸡兔同笼问题解答，易得平局有$$5$$场． " ]

[ "2008年二年级竞赛学而思杯" ]

[ [ { "aoVal": "A", "content": "你是真城的人吗? " } ], [ { "aoVal": "B", "content": "你是假城的人吗? " } ], [ { "aoVal": "C", "content": "你是说真话的人吗? " } ], [ { "aoVal": "D", "content": "你是说假话的人吗? " } ], [ { "aoVal": "E", "content": "你是这座城的人吗? " } ] ]

[ "拓展思维->拓展思维->组合模块->逻辑推理->假设型逻辑推理" ]

[ "对于A，B，C，D，不论问哪个城里的人，很明显得到的是相同的答案，无法判断真假，只有问E时，真城里的人会回答``是''，而假城里的人的答案是``不是''. " ]

[ "2012年第10届创新杯四年级竞赛初赛第1题6分" ]

[ [ { "aoVal": "A", "content": "$$80.5$$ " } ], [ { "aoVal": "B", "content": "$$81.5$$ " } ], [ { "aoVal": "C", "content": "$$82.5$$ " } ], [ { "aoVal": "D", "content": "$$83.5$$ " } ] ]

[ "拓展思维->能力->实践应用" ]

[ "排除法： $$80.5\\times 3-(75+87.5)=241.5-162.5=79$$； $$81.5\\times 3-(75+87.5)=244.5-162.5=82$$； $$82.5\\times 3-(75+87.5)=247.5-162.5=85$$； $$83.5\\times 3-(75+87.5)=250.5-162.5=88\\textgreater87.5$$． 故选$$\\text{D}$$． " ]

[ "2010年五年级竞赛创新杯", "2010年六年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$6$$ " } ], [ { "aoVal": "D", "content": "$$7$$ " } ] ]

[ "拓展思维->思想->枚举思想" ]

[ "若只有一个砝码，那么这个砝码有放左、放右和不放$$3$$种不同选择，其中左、右都不放的情况不计，剩下的情况会出现左右重复，只能称出$$\\frac{3-1}{2}=1$$（种）不同的质量；取$$1$$克，可以称出$$1$$克； 若有两个砝码，共有$$3\\times 3=9$$（种）放法，最多称出$$\\frac{{{3}^{2}}-1}{2}=4$$（种）不同的质量；取$$1$$克和$$3$$克，即可称出$$1$ $4$$克； 若有三个砝码，共有$${{3}^{3}}=27$$（种）放法，最多称出$$\\frac{{{3}^{3}}-1}{2}=13$$（种）不同的质量；取$$1$$克、$$3$$克和$$9$$克，即可称出$$1$ $13$$克； 若有四个砝码，共有$${{3}^{4}}=81$$（种）放法，最多称出$$\\frac{{{3}^{4}}-1}{2}=40$$（种）不同的质量．取$$1$$克、$$3$$克、$$9$$克和$$27$$克，即可称出$$1$ $40$$克． 由此我们只需$$1$$克、$$3$$克、$$9$$克、$$27$$克这样的$$4$$个砝码，即可称出$$1$ $40$$克所有整数克的质量． " ]

[ "2015年华杯赛六年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$12$$ " } ], [ { "aoVal": "B", "content": "$$14$$ " } ], [ { "aoVal": "C", "content": "$$16$$ " } ], [ { "aoVal": "D", "content": "$$18$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->归一归总问题->乘除法应用->除法应用" ]

[ "解：设另一张卡号是$$x$$。则： $$2x+2\\leqslant 20$$ $$2x+2-2\\leqslant 20-2$$ $$2x\\leqslant 18$$ $$2x\\div 2\\leqslant 18\\div 2$$ $$x\\leqslant 9$$ 又因为$$x\\geqslant 1$$ $$9\\times 2=18$$（张） $$\\because 4$$、$$6$$、$$8$$即可以做为一倍量的数，也可以作为$$2$$倍量多$$2$$的数， $$\\therefore $$即总共可以取出：$$18-3\\times 2=12$$（张）。 答：小明最多可以取出$$12$$张卡片。 故选：$$A$$。 " ]

[ "2014年全国迎春杯三年级竞赛初赛第6题" ]

[ [ { "aoVal": "A", "content": "$$2$$时$$45$$分 " } ], [ { "aoVal": "B", "content": "$$2$$时$$49$$分 " } ], [ { "aoVal": "C", "content": "$$2$$时$$50$$分 " } ], [ { "aoVal": "D", "content": "$$2$$时$$53$$分 " } ] ]

[ "拓展思维->拓展思维->应用题模块->周期问题->基本排列的周期问题" ]

[ "$$3\\times 5+1\\times 4=19$$，折好第$$5$$个时已经到了$$2$$时$$49$$分． " ]

[ "2019年华夏杯一年级竞赛复赛（华南赛区）第16题6分" ]

[ [ { "aoVal": "A", "content": "小奥 " } ], [ { "aoVal": "B", "content": "小林 " } ], [ { "aoVal": "C", "content": "小匹 " } ], [ { "aoVal": "D", "content": "小克 " } ] ]

[ "知识标签->课内知识点->数与运算->数的运算的实际应用（应用题）->整数的简单实际问题->加法的实际应用" ]

[ "小克$$20$$颗糖发放给$$4$$个小朋友，每发完一次就少$$4$$个糖，$$20=4+4+4+4+4$$． $$20$$颗糖刚好可以发$$5$$个轮回，最后一颗糖刚好发到小克，所以答案是小克． " ]

[ "2014年全国迎春杯五年级竞赛初赛第3题" ]

[ [ { "aoVal": "A", "content": "$$18$$ " } ], [ { "aoVal": "B", "content": "$$19$$ " } ], [ { "aoVal": "C", "content": "$$20$$ " } ], [ { "aoVal": "D", "content": "$$21$$ " } ] ]

[ "知识标签->学习能力->七大能力->逻辑分析" ]

[ "$$48\\div 2.5=19\\cdots 0.5$$，$$19+1=20$$（辆）． " ]

[ "2011年全国希望杯四年级竞赛初赛第16题" ]

[ [ { "aoVal": "A", "content": "$$60$$ " } ], [ { "aoVal": "B", "content": "$$40$$ " } ], [ { "aoVal": "C", "content": "$$30$$ " } ], [ { "aoVal": "D", "content": "$$20$$ " } ] ]

[ "知识标签->数学思想->对应思想" ]

[ "一个半小时是$$90$$分钟，$$2$$个半小时是$$2\\times 60+30=150$$分钟．易知王强步行单程需要$$150\\div 2=75$$分钟，则坐车单程要$$90-75=15$$分钟，所以来回都坐车要$$15\\times 2=30$$分钟． " ]

[ "2018年第8届北京学而思综合能力诊断二年级竞赛年度教学质量第10题" ]

[ [ { "aoVal": "A", "content": "$$200$$ " } ], [ { "aoVal": "B", "content": "$$160$$ " } ], [ { "aoVal": "C", "content": "$$180$$ " } ], [ { "aoVal": "D", "content": "$$170$$ " } ] ]

[ "课内体系->能力->运算求解", "拓展思维->拓展思维->计数模块->加乘原理->乘法原理->物品搭配" ]

[ "$$2\\times 2\\times 3=12$$（种）． " ]

[ "2017年河南郑州联合杯竞赛第2题4分" ]

[ [ { "aoVal": "A", "content": "$$3$$次 " } ], [ { "aoVal": "B", "content": "$$4$$次 " } ], [ { "aoVal": "C", "content": "$$5$$次 " } ], [ { "aoVal": "D", "content": "$$6$$次 " } ] ]

[ "拓展思维->拓展思维->应用题模块->周期问题->蜗牛爬井问题->蜗牛爬井求天数（爬出去类）" ]

[ "第一次跳时，向上跳$$6$$米，又下滑$$3$$米，此时距离井底$$3$$米，第二次跳完距离井底$$6$$米，第三次跳完距离井底$$9$$米，第四次先向上跳$$6$$米，此时青蛙跳出了井． " ]

[ "2020年希望杯六年级竞赛模拟第25题" ]

[ [ { "aoVal": "A", "content": "$$\\textgreater$$ " } ], [ { "aoVal": "B", "content": "$$\\textless{}$$ " } ], [ { "aoVal": "C", "content": "$$=$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "因为$$\\frac{1}{{{2}^{2}}}\\textless{}\\frac{1}{1\\times 2}$$，$$\\frac{1}{{{3}^{2}}}\\textless{}\\frac{1}{2\\times 3}$$，$$\\frac{1}{{{4}^{2}}}\\textless{}\\frac{1}{3\\times 4}$$， 所以$$\\frac{1}{{{2}^{2}}}+\\frac{1}{{{3}^{2}}}+\\frac{1}{{{4}^{2}}}+\\cdots +\\frac{1}{{{2020}^{2}}}\\textless{}\\frac{1}{1\\times 2}+\\frac{1}{2\\times 3}+\\cdots +\\frac{1}{2019\\times 2020}$$， 因为$$\\frac{1}{1\\times 2}+\\frac{1}{2\\times 3}+\\frac{1}{3\\times 4}\\cdots +\\frac{1}{2019\\times 2020}$$ $$=1-\\frac{1}{2}+\\frac{1}{2}-\\frac{1}{3}+\\frac{1}{3}-\\frac{1}{4}+\\cdots +\\frac{1}{2019}-\\frac{1}{2020}$$ $$=1-\\frac{1}{2020}$$ $$=\\frac{2019}{2020}$$， 所以$$\\frac{1}{{{2}^{2}}}+\\frac{1}{{{3}^{2}}}+\\cdots +\\frac{1}{{{2020}^{2}}}\\textless{}\\frac{2019}{2020}$$， 所以$$1+\\frac{1}{{{2}^{2}}}+\\frac{1}{{{3}^{2}}}+\\frac{1}{{{4}^{2}}}+\\cdots +\\frac{1}{{{2020}^{2}}}\\textless{}2$$． 故选$$\\text{B}$$． " ]

[ "2020年广东广州羊排赛五年级竞赛第7题3分" ]

[ [ { "aoVal": "A", "content": "$$5$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$7$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ], [ { "aoVal": "E", "content": "$$2$$ " } ] ]

[ "拓展思维->拓展思维->数论模块->因数与倍数->因数与倍数基础", "Overseas Competition->知识点->数论模块->质数与合数->质数与合数判定" ]

[ "最小质数是$$2$$，所以千位是$$2$$； 最小合数是$$4$$，所以百位是$$4$$；最小的自然数是$$0$$，所以十位是$$0$$，这个四位数是$$3$$的倍数，那么这个四位数各个位数之和是$$3$$的倍数，$$2+4+0=6$$． $$\\text{A}$$选项：$$6+5=11$$，不是$$3$$的整数倍，故$$\\text{A}$$错误； $$\\text{B}$$选项：$$6+6=12$$，是$$3$$的整数倍，故$$\\text{B}$$正确； $$\\text{C}$$选项：$$6+7=13$$，不是$$3$$的整数倍，故$$\\text{C}$$错误； $$\\text{D}$$选项：$$6+8=14$$，不是$$3$$的整数倍，故$$\\text{D}$$错误； $$\\text{E}$$选项：$$6+2=8$$，不是$$3$$的整数倍，故$$\\text{E}$$错误． 故选$$\\text{B}$$． " ]

[ "2020年新希望杯二年级竞赛决赛（8月）第4题", "2020年新希望杯二年级竞赛初赛（个人战）第4题" ]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$10$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->归一归总问题->乘除法应用->除法应用" ]

[ "根据$$1$$斤菠萝可以做$$2$$个菠萝蛋糕，那么要做$$10$$个菠萝蛋糕，需要菠萝$$10\\div2=5$$（斤）．故选择$$\\text{B}$$． " ]

[ "2017年IMAS小学高年级竞赛（第一轮）第15题4分" ]

[ [ { "aoVal": "A", "content": "$$30$$ " } ], [ { "aoVal": "B", "content": "$$34$$ " } ], [ { "aoVal": "C", "content": "$$36$$ " } ], [ { "aoVal": "D", "content": "$$39$$ " } ], [ { "aoVal": "E", "content": "$$41$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "因$$a+c\\geqslant 1+2=3$$．故$$e\\geqslant 3$$，因此取$$e=3$$，此时可判断出$$a$$、$$c$$分别为$$1$$与$$2$$，且知$$b$$、$$d$$均不能为$$0$$，否则数码$$f$$会与$$b$$或$$d$$相等，故$$b+d\\geqslant 4+5=9$$，即$$f=9$$，例如$$14+25=39$$、$$15+24=39$$． 故选$$\\text{D}$$． " ]

[ "2019年浙江杭州滨江区杭州江南实验学校五年级竞赛模拟（江南杯）第6题2分" ]

[ [ { "aoVal": "A", "content": "$$8$$ " } ], [ { "aoVal": "B", "content": "$$1$$ " } ], [ { "aoVal": "C", "content": "$$9$$ " } ] ]

[ "拓展思维->思想->对应思想", "课内体系->能力->数据处理" ]

[ "列竖式 故$$9\\div 11=0.\\dot{8}\\dot{1}$$． 周期为$$2$$故$$50\\div 2=25$$， 故第$$50$$位为$$1$$． 故答案为：$$\\text{B}$$选项． " ]

[ "2019年第23届广东世界少年奥林匹克数学竞赛三年级竞赛决赛第2题5分" ]

[ [ { "aoVal": "A", "content": "$$120$$ " } ], [ { "aoVal": "B", "content": "$$124$$ " } ], [ { "aoVal": "C", "content": "$$126$$ " } ], [ { "aoVal": "D", "content": "$$130$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "两次分糖剩余之差：$$24+16=40$$（颗）， 每次分糖数之差：$$7-5=2$$（颗）． 小朋友人数：$$40\\div2=20$$（人）， 糖果总数：$$20\\times5+24=124$$（颗）． 所以选择$$\\text{B}$$． " ]

[ "2017年全国美国数学大联盟杯小学高年级五年级竞赛初赛第36题" ]

[ [ { "aoVal": "A", "content": "$$14$$ " } ], [ { "aoVal": "B", "content": "$$16$$ " } ], [ { "aoVal": "C", "content": "$$28$$ " } ], [ { "aoVal": "D", "content": "$$30$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "这是一个等比数列问题，$${{a}_{n}}={{a}_{1}}{{q}^{n-1}}={{2}^{n}}$$，$${{S}_{n}}=\\frac{{{a}_{1}}\\left( 1-{{q}^{n}} \\right)}{1-q}=\\frac{2\\left( 1-{{2}^{n}} \\right)}{1-2}=2\\left( {{2}^{n}}-1 \\right)=30$$，所以选$$D$$． " ]

[ "2017年河南郑州豫才杯四年级竞赛初赛第6题" ]

[ [ { "aoVal": "A", "content": "商是$$15$$，余数是$$20$$ " } ], [ { "aoVal": "B", "content": "商是$$13$$，余数是$$6$$ " } ], [ { "aoVal": "C", "content": "商是$$24$$，余数是$$11$$ " } ], [ { "aoVal": "D", "content": "商是$$15$$，余数是$$2$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "$$56\\times 15+11=851$$，$$851\\div 65=13\\cdots \\cdots 6$$，所以$$\\text{B}$$选项正确． " ]

[ "2013年全国华杯赛小学高年级竞赛初赛B卷第4题" ]

[ [ { "aoVal": "A", "content": "$$20$$ " } ], [ { "aoVal": "B", "content": "$$51.36$$ " } ], [ { "aoVal": "C", "content": "$$31.36$$ " } ], [ { "aoVal": "D", "content": "$$10.36$$ " } ] ]

[ "拓展思维->能力->实践应用" ]

[ "甲学校花钱$$56\\times 8.06=451.36$$元；乙学校要买糖$$56\\div (1+5 \\%)=\\frac{160}{3}$$kg，单价$$8.06-0.56=7.5$$元，甲学校花钱$$\\frac{160}{3}\\times7.5=400$$元；乙学校将比甲学校少花$$51.36$$元． " ]

[ "2022年第九届鹏程杯四年级竞赛初赛第3题5分" ]

[ [ { "aoVal": "A", "content": "$$2022$$ " } ], [ { "aoVal": "B", "content": "$$2021$$ " } ], [ { "aoVal": "C", "content": "$$2020$$ " } ], [ { "aoVal": "D", "content": "$$2019$$ " } ], [ { "aoVal": "E", "content": "$$2018$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->整数->整数乘除->整数乘除法巧算" ]

[ "无 " ]

[ "2013年第25届广东广州五羊杯六年级竞赛第8题5分" ]

[ [ { "aoVal": "A", "content": "$$38$$ " } ], [ { "aoVal": "B", "content": "$$43$$ " } ], [ { "aoVal": "C", "content": "$$48$$ " } ], [ { "aoVal": "D", "content": "$$53$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "显然两个素数中至少有一个是偶数，而素数中只有$$2$$是偶数． 将$$2$$代入，可得另一素数为$$41$$，故和为$$43$$． 故选$$\\text{B}$$． " ]

[ "2016年河南郑州联合杯小学高年级六年级竞赛复赛第17题2分" ]

[ [ { "aoVal": "A", "content": "甲 " } ], [ { "aoVal": "B", "content": "乙 " } ], [ { "aoVal": "C", "content": "甲、乙均可 " } ], [ { "aoVal": "D", "content": "不知道 " } ] ]

[ "拓展思维->能力->实践应用" ]

[ "经济问题；甲：$$1000$$元可以买到$$1000\\div 50 \\% =2000$$元，乙：$$1000\\div 200=5$$，$$100\\times 5=500$$元，$$1000$$元可以买到：$$1000+500=1500$$元，$$2000\\textgreater1500$$，所以去甲商场合算． " ]

[ "2019年第24届YMO六年级竞赛决赛第7题3分" ]

[ [ { "aoVal": "A", "content": "$$149$$ " } ], [ { "aoVal": "B", "content": "$$244$$ " } ], [ { "aoVal": "C", "content": "$$264$$ " } ], [ { "aoVal": "D", "content": "$$274$$ " } ] ]

[ "拓展思维->拓展思维->组合模块->操作与策略->归纳递推->斐波那契数列递推" ]

[ "设第$$n$$级有$${{a}_{n}}$$种登的方式， 则第$$n-3$$，$$n-2$$，$$n-1$$级再分别登$$3$$，$$2$$，$$1$$级可到第$$n$$级， 则$${{a}_{n}}={{a}_{n-3}}+{{a}_{n-2}}+{{a}_{n-1}}$$， 而$${{a}_{1}}=1$$，$${{a}_{2}}=1+1=2$$，$${{a}_{3}}=1+1+2=4$$， 故$${{a}_{4}}=1+2+4=7$$， $${{a}_{5}}=2+4+7=13$$， $${{a}_{6}}=4+7+13=24$$， $${{a}_{7}}=7+13+24=44$$， $${{a}_{8}}=13+24+44=81$$， $${{a}_{9}}=24+44+81=149$$， $${{a}_{10}}=44+81+149=274$$． 故选：$$\\text{D}$$． " ]

[ "2014年第25届亚太杯四年级竞赛决赛第18题5分" ]

[ [ { "aoVal": "A", "content": "小董懂英文但不懂法语． " } ], [ { "aoVal": "B", "content": "小董懂法语但不懂英文． " } ], [ { "aoVal": "C", "content": "小董既不懂英文也不懂法语． " } ], [ { "aoVal": "D", "content": "如果小董懂英文，小董一定不懂法语． " } ], [ { "aoVal": "E", "content": "如果小董不懂法语，那么他一定懂英文． " } ] ]

[ "拓展思维->拓展思维->组合模块->逻辑推理->条件型逻辑推理->神推理" ]

[ "逻辑推理． 小董并非既懂英文又懂法语． 则有以下$$3$$种情况： $$(1)$$小董懂英文但不懂法语． $$(2)$$小董懂法语但不懂英文． $$(3)$$小董既不懂英文也不懂法语． 故选$$\\text{D}$$． " ]

[ "2014年第10届全国新希望杯小学高年级六年级竞赛复赛第2题4分" ]

[ [ { "aoVal": "A", "content": "$$11$$ " } ], [ { "aoVal": "B", "content": "$$14$$ " } ], [ { "aoVal": "C", "content": "$$21$$ " } ], [ { "aoVal": "D", "content": "$$28$$ " } ] ]

[ "知识标签->学习能力->七大能力->运算求解" ]

[ "154按照操作依次为$$77$$，$$84$$，$$42$$，$$21$$，$$28$$，$$14$$，$$7$$，$$14$$，7\\ldots\\ldots，不可能得到$$11$$． ", "<p>154和$$7$$都是$$7$$点倍数，每次操作后最终结果都是$$7$$的倍数，不可能得到11.</p>" ]

[ "2019年第7届湖北长江杯六年级竞赛复赛B卷第5题3分" ]

[ [ { "aoVal": "A", "content": "$$15$$ " } ], [ { "aoVal": "B", "content": "$$26$$ " } ], [ { "aoVal": "C", "content": "$$27$$ " } ], [ { "aoVal": "D", "content": "$$28$$ " } ] ]

[ "拓展思维->思想->方程思想" ]

[ "根据年龄差不会变这一特征，从年龄差入手，年龄差$$+$$老师现在的年龄$$=39$$， 所以老师$$+$$学生$$=42$$，设老师今年岁数为$$x$$， 则学生的岁数是$$42-x$$岁，再根据年龄差$$+$$老师现在的年龄$$=39$$， 列出方程解决问题， $$x-(42---x)+x=39$$， $$3x---42=39$$， $$3x=39+42$$， $$3x=81$$， $$x=27$$， 也就是老师今年$$27$$岁， 所以选择$$\\text{C}$$选项． " ]

[ "2017年河南郑州豫才杯五年级竞赛初赛第2题" ]

[ [ { "aoVal": "A", "content": "$$5$$ " } ], [ { "aoVal": "B", "content": "$$40$$ " } ], [ { "aoVal": "C", "content": "$$50$$ " } ], [ { "aoVal": "D", "content": "$$500$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "$$200000\\div 500\\div 8=50$$（分钟），所以她需要平均每天坚持阅读$$50$$分钟才能完成阅读计划． " ]

[ "2019年第24届YMO五年级竞赛决赛第6题3分", "2020年第24届YMO五年级竞赛决赛第6题3分" ]

[ [ { "aoVal": "A", "content": "小$$Y$$ " } ], [ { "aoVal": "B", "content": "小$$M$$ " } ], [ { "aoVal": "C", "content": "小$$O$$ " } ], [ { "aoVal": "D", "content": "小$$T$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "对比发现，小$$O$$与小$$Y$$说的矛盾，相互对立， 则小$$O$$与小$$Y$$必一对一错， 则小$$M$$说假话，则小$$M$$会开车，选$$\\text{B}$$． " ]

[ "2008年四年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "68 " } ], [ { "aoVal": "B", "content": "70 " } ], [ { "aoVal": "C", "content": "71 " } ], [ { "aoVal": "D", "content": "72 " } ] ]

[ "拓展思维->拓展思维->应用题模块->间隔问题" ]

[ "24根电线杆之间有23个间隔，每个间隔的长是$$115\\div 23=5$$(米)，又第2根和第16根电线杆之间有14个间隔，所以，它们相隔$$5\\times 14=70$$(米) " ]

[ "2020年第24届YMO三年级竞赛决赛第8题3分", "2017年华杯赛四年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$6$$ " } ], [ { "aoVal": "D", "content": "$$7$$ " } ] ]

[ "拓展思维->拓展思维->组合模块->抽屉原理->简单抽屉原理" ]

[ "先找到和为$$10$$的组合：$$(1$$，$$9)$$、$$(2$$，$$8)$$、$$(3$$，$$7)$$、$$(4$$，$$6)$$，剩下$$5$$、$$10$$，若前面四组数中各取一个，后面$$5$$、$$10$$全取，这样还不能满足题目要求的两数和为$$10$$，如果再取剩下四个中的任意一个，必定会凑出和为$$10$$的组合，故最少要取$$6+1=7$$（个）。 " ]

[ "2012年第8届全国新希望杯五年级竞赛复赛第2题" ]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{12}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{1}{4}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{2}{3}$$ " } ], [ { "aoVal": "D", "content": "$$9$$ " } ] ]

[ "拓展思维->拓展思维->计数模块->统计与概率->概率->计数求概率" ]

[ "亮绿灯的可能性为：$$9\\div (3+9+24)=\\frac{1}{4}$$，故选$$\\text{B}$$． ~ " ]

[ "2012年第10届创新杯四年级竞赛初赛第2题6分" ]

[ [ { "aoVal": "A", "content": "$$9$$ " } ], [ { "aoVal": "B", "content": "$$10$$ " } ], [ { "aoVal": "C", "content": "$$11$$ " } ], [ { "aoVal": "D", "content": "$$12$$ " } ] ]

[ "拓展思维->能力->公式记忆->符号化数学原理" ]

[ "计数------分类枚举． 一位数：$$0$$个， 两位数：$$32$$，共$$1$$个， 三位数：$$104$$、$$140$$、$$212$$、$$320$$、$$500$$，共$$5$$个， 四位数：$$1004$$、$$1040$$、$$1112$$、$$1220$$、$$1400$$、$$2012\\cdots $$ 故$$2012$$为第$$12$$个． 故选$$\\text{D}$$． " ]

[ "2018年湖北武汉新希望杯五年级竞赛训练题（五）第2题" ]

[ [ { "aoVal": "A", "content": "$$400$$ " } ], [ { "aoVal": "B", "content": "$$420$$ " } ], [ { "aoVal": "C", "content": "$$440$$ " } ], [ { "aoVal": "D", "content": "$$460$$ " } ] ]

[ "拓展思维->能力->实践应用" ]

[ "甲仓库比乙仓库多$$100+80=180$$(吨)，乙仓库：$$180\\div (2.5-1)=120$$（吨），甲仓库：$$120+180=300$$（吨），$$120+300=420$$（吨）． " ]

[ "2014年第10届全国新希望杯五年级竞赛复赛第4题" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$13$$ " } ], [ { "aoVal": "C", "content": "$$20$$ " } ], [ { "aoVal": "D", "content": "$$23$$ " } ] ]

[ "拓展思维->思想->方程思想" ]

[ "欣+希+望+贝=$$49$$ 欣=希+望=$$2$$望+$$3$$ 希=望+贝=望+$$3$$ 则$$4$$望+$$9=49$$，所以望=$$10$$， 那么欣欣捐了$$2\\times 10+3=23$$． " ]

[ "2020年新希望杯六年级竞赛（2月）第17题" ]

[ [ { "aoVal": "A", "content": "$$40$$ " } ], [ { "aoVal": "B", "content": "$$180$$ " } ], [ { "aoVal": "C", "content": "$$80$$ " } ], [ { "aoVal": "D", "content": "$$220$$ " } ] ]

[ "拓展思维->能力->实践应用" ]

[ "$$80 \\%$$的猫认为自己是猫，$$20 \\%$$的狗认为自己是猫，猫狗混合后有$$36 \\%$$的认为自己是猫，所以狗：猫$$=\\left(80 \\%-36 \\%\\right):\\left(36 \\%-20 \\%\\right)=11:4$$， 狗：$$140\\div \\left( 11-4 \\right)\\times 11=220$$（只）． " ]

[ "2021年第8届鹏程杯五年级竞赛初赛第22题4分" ]

[ [ { "aoVal": "A", "content": "$$2020$$ " } ], [ { "aoVal": "B", "content": "$$2021$$ " } ], [ { "aoVal": "C", "content": "$$4040$$ " } ], [ { "aoVal": "D", "content": "$$4042$$ " } ], [ { "aoVal": "E", "content": "以上都不对 " } ] ]

[ "拓展思维->拓展思维->数论模块->整除->整除特征->整除特征综合" ]

[ "$$\\underbrace{99\\cdots99}_{2021个}\\times\\underbrace{99\\cdots99}_{2021个}+1\\underbrace{99\\cdots99}_{2021个}$$ $$=\\underbrace{99\\cdots99}_{2021个}\\times\\underbrace{99\\cdots99}_{2021个}+1\\underbrace{00\\cdots00}_{2021个}+\\underbrace{99\\cdots99}_{2021个}$$ $$=\\underbrace{99\\cdots99}_{2021个}\\times(\\underbrace{99\\cdots99}_{2021个}+1)+1\\underbrace{00\\cdots00}_{2021个}$$ $$=\\underbrace{99\\cdots99}_{2021个}\\times 1\\underbrace{00\\cdots00}_{2021个}+1\\underbrace{00\\cdots00}_{2021个}$$ $$=(\\underbrace{99\\cdots99}_{2021个}+1)\\times 1\\underbrace{00\\cdots00}_{2021个}$$ $$=1\\underbrace{00\\cdots00}_{2021个}+\\times 1\\underbrace{00\\cdots00}_{2021个}$$ $$=1\\underbrace{00\\cdots00}_{4042个}$$ $$n$$最大取$$4042$$． 故选：$$\\text{D}$$． " ]

[ "2009年全国迎春杯小学中年级四年级竞赛初赛第2题" ]

[ [ { "aoVal": "A", "content": "$$48$$ " } ], [ { "aoVal": "B", "content": "$$50$$ " } ], [ { "aoVal": "C", "content": "$$54$$ " } ], [ { "aoVal": "D", "content": "$$56$$ " } ] ]

[ "拓展思维->能力->实践应用" ]

[ "同倍数变化问题．把$$1$$个田格本和$$5$$个练习本捆绑成一组，那么每发$$3$$个横格本，就发一组田格本和练习本，田格本和练习本的总量是横线本的$$2$$倍，每次发的数量也是$$2$$倍，所以剩下的也是$$2$$倍，即$$48$$本． " ]

[ "2013年第9届全国新希望杯小学高年级五年级竞赛复赛第1题" ]

[ [ { "aoVal": "A", "content": "$$95$$分 " } ], [ { "aoVal": "B", "content": "$$96$$分 " } ], [ { "aoVal": "C", "content": "$$97$$分 " } ], [ { "aoVal": "D", "content": "$$98$$分 " } ] ]

[ "拓展思维->拓展思维->应用题模块->平均数问题->综合题->综合题普通型" ]

[ "$$445-87.25\\times 4=96$$．~ " ]

[ "2019年第23届广东世界少年奥林匹克数学竞赛四年级竞赛决赛第7题5分" ]

[ [ { "aoVal": "A", "content": "$$50$$ " } ], [ { "aoVal": "B", "content": "$$100$$ " } ], [ { "aoVal": "C", "content": "$$500$$ " } ], [ { "aoVal": "D", "content": "$$1000$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "$$1$$到$$1000$$中，一共有$$500$$个奇数，$$500$$个偶数， 奇数之和$$=1+3+5+\\cdots \\cdots +997+999$$， 偶数之和$$=2+4+6+\\cdots \\cdots +998+1000$$， 可以用比每个奇数大$$1$$的偶数去减该奇数，正好所有的奇数都可以找到唯一比它大$$1$$的偶数， 所有偶数之和比所有奇数之和多$$500$$， 故选$$\\text{C}$$． " ]

[ "2013年小机灵杯三年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$5$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$7$$ " } ] ]

[ "拓展思维->拓展思维->计数模块->加乘原理->加法原理" ]

[ "解：罗马数字是由罗马人发明的，它一共由$$7$$个数字组成； 故选:C。 " ]

[ "六年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "第一根钢管剩下的部分长些 " } ], [ { "aoVal": "B", "content": "第二根钢管剩下的部分长些 " } ], [ { "aoVal": "C", "content": "两根钢管剩下的部分同样长 " } ], [ { "aoVal": "D", "content": "以上三种结果都有可能 " } ] ]

[ "拓展思维->拓展思维->应用题模块->分百应用题->认识单位1" ]

[ "当两根钢管都是$$1$$米时，则剩下的钢管相等；当两根钢管都大于$$1$$米时，则用去$$\\frac{3}{10}$$米的那一根剩下的多；当两根钢管都小于$$1$$米时，用去$$\\frac{3}{10}$$的那一根剩下的多。 " ]

[ "2017年全国华杯赛小学高年级竞赛初赛模拟" ]

[ [ { "aoVal": "A", "content": "$$96$$ " } ], [ { "aoVal": "B", "content": "$$120$$ " } ], [ { "aoVal": "C", "content": "$$144$$ " } ], [ { "aoVal": "D", "content": "$$240$$ " } ] ]

[ "拓展思维->能力->抽象概括" ]

[ "$$\\text{P}_{4}^{4}\\times \\text{P}_{3}^{2}=4\\times 3\\times 2\\times 3\\times 2=144$$． 另一种方法：先捆绑，再插空，$$2!\\times 4!\\times C_{4}^{3}=144$$． " ]

[ "2017年全国美国数学大联盟杯小学中年级四年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "＄$$0.90$$ " } ], [ { "aoVal": "B", "content": "＄$$0.99$$ " } ], [ { "aoVal": "C", "content": "＄$$1.19$$ " } ], [ { "aoVal": "D", "content": "＄$$1.29$$ " } ] ]

[ "拓展思维->能力->实践应用" ]

[ "$$3$$个quarter（二十五美分）$$+4$$个dime（十美分）$$+4$$个penny（一美分）． " ]

[ "2018~2019学年浙江杭州西湖区杭州市行知小学五年级上学期期中期中竞赛第19题1分" ]

[ [ { "aoVal": "A", "content": "甲数$$=$$乙数 " } ], [ { "aoVal": "B", "content": "甲数$$\\textgreater$$乙数 " } ], [ { "aoVal": "C", "content": "甲数$$\\textless{}$$乙数 " } ] ]

[ "知识标签->拓展思维->计算模块->小数->小数乘除->小数除法运算" ]

[ "$$1.2=\\frac{6}{5}$$，即甲数$$\\times \\frac{6}{5}=$$乙数$$\\times \\frac{5}{6}$$（且甲、乙$$\\ne 0$$），根据积的特性，当积一定时，一个因数越大，则另一个因数越小； 那么$$\\frac{6}{5}\\textgreater\\frac{5}{6}$$，则甲$$\\textless{}$$乙． " ]

[ "2014年迎春杯四年级竞赛初赛", "2014年迎春杯三年级竞赛初赛", "2014年迎春杯三年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$20$$ " } ], [ { "aoVal": "B", "content": "$$10$$ " } ], [ { "aoVal": "C", "content": "$$11$$ " } ], [ { "aoVal": "D", "content": "$$0$$ " } ] ]

[ "拓展思维->拓展思维->组合模块->逻辑推理->条件型逻辑推理->神推理" ]

[ "解：因为最开始开灯和关灯的房间各是$$10$$间，由于第一间的灯是开着的， 所以，第一间人看到的，开灯的$$9$$间，关灯的$$10$$间， 之后，他就关灯， 以后无论开灯的出来看，还是关灯的出来看，始终关灯的多， 即：一轮结束，灯全部会关闭。 故选：A。 " ]

[ "2018年湖北武汉新希望杯五年级竞赛训练题（三）第1题" ]

[ [ { "aoVal": "A", "content": "一个分数的分母越小，它的分数单位就越小 " } ], [ { "aoVal": "B", "content": "分数都比整数小 " } ], [ { "aoVal": "C", "content": "假分数都大于$$1$$ " } ], [ { "aoVal": "D", "content": "$$2$$米的$$\\frac{1}{3}$$和$$1$$米的$$\\frac{2}{3}$$一样长 " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "$$2$$米的$$\\frac{1}{3}$$和$$1$$米的$$\\frac{2}{3}$$一样长． " ]

[ "2016年全国学而思杯小学高年级六年级竞赛" ]

[ [ { "aoVal": "A", "content": "$$\\dfrac{{{a}^{11}}}{10}$$ " } ], [ { "aoVal": "B", "content": "$$-\\dfrac{{{a}^{11}}}{10}$$ " } ], [ { "aoVal": "C", "content": "$$-\\dfrac{{{a}^{10}}}{11}$$ " } ], [ { "aoVal": "D", "content": "$$-\\dfrac{{{a}^{11}}}{11}$$ " } ] ]

[ "拓展思维->思想->逐步调整思想" ]

[ "分别观察分子、分母的规律，以及符号的规律．从左到右数的第$$10$$个式子是$$-\\dfrac{{{a}^{11}}}{10}$$． " ]

[ "2018年第6届湖北长江杯五年级竞赛初赛A卷第7题3分" ]

[ [ { "aoVal": "A", "content": "$$21$$ " } ], [ { "aoVal": "B", "content": "$$28$$ " } ], [ { "aoVal": "C", "content": "$$14$$ " } ], [ { "aoVal": "D", "content": "$$36$$ " } ] ]

[ "拓展思维->拓展思维->几何模块->立体图形->长方体与正方体->长方体与正方体基本概念运用->正方体的基本概念" ]

[ "正方体的棱长是$$18$$、$$12$$、$$6$$的倍数， 其中$$36$$是它们的最小公倍数， $$36\\div 18=2$$（层）， $$36\\div 12=3$$（层）， $$36\\div 6=6$$（层）， 需要砖块为：$$2\\times 3\\times 6=36$$（块）． 故答案选：$$\\text{D}$$． " ]

[ "2019年浙江杭州滨江区杭州江南实验学校五年级竞赛模拟（江南杯）第6题2分" ]

[ [ { "aoVal": "A", "content": "$$8$$ " } ], [ { "aoVal": "B", "content": "$$1$$ " } ], [ { "aoVal": "C", "content": "$$9$$ " } ] ]

[ "课内体系->知识点->数的认识->认、读、写数->小数->循环小数", "拓展思维->思想->对应思想" ]

[ "列竖式 故$$9\\div 11=0.\\dot{8}\\dot{1}$$． 周期为$$2$$故$$50\\div 2=25$$， 故第$$50$$位为$$1$$． 故答案为：$$\\text{B}$$选项． " ]

[ "2017年河南郑州豫才杯四年级竞赛初赛第14题" ]

[ [ { "aoVal": "A", "content": "$$10$$分钟~~~ " } ], [ { "aoVal": "B", "content": "$$50$$分钟~~~ " } ], [ { "aoVal": "C", "content": "$$1$$小时$$30$$分钟 " } ], [ { "aoVal": "D", "content": "$$100$$分钟 " } ] ]

[ "拓展思维->拓展思维->应用题模块->归一归总问题->双归一问题" ]

[ "$$200000\\div 300\\div 7\\approx 100$$分钟． " ]

[ "2014年全国迎春杯五年级竞赛初赛第8题", "2014年全国迎春杯六年级竞赛初赛第9题" ]

[ [ { "aoVal": "A", "content": "$$46$$ " } ], [ { "aoVal": "B", "content": "$$47$$ " } ], [ { "aoVal": "C", "content": "$$48$$ " } ], [ { "aoVal": "D", "content": "没有符合条件的数 " } ] ]

[ "拓展思维->思想->逐步调整思想" ]

[ "由已知条件，$$4$$个质数中一定有$$11$$，那么则满足$$a\\times b\\times c=a+b+c+11$$，其中$$a$$、$$b$$、$$c$$都是质数．若$$a$$、$$b$$、$$c$$都是奇数，那么等式左边是奇数，右边为偶数，矛盾．若$$a$$、$$b$$、$$c$$中有$$1 $$个偶数，那么一定是$$2$$．即$$a\\times b\\times 2=a+b+2+11$$，此时，根据奇偶性，$$a$$、$$b$$中也必有一个偶数为$$2$$，解得$$a$$、$$b$$、$$c$$、$$d$$为$$2$$、$$2$$、$$5$$、$$11$$和为$$20$$．选项中ABC均不符合条件，故选D． " ]

[ "2015年第13届全国创新杯六年级竞赛第6题", "小学高年级六年级其它2015年数学思维能力等级测试初试第6题4分" ]

[ [ { "aoVal": "A", "content": "$$21$$ " } ], [ { "aoVal": "B", "content": "$$22$$ " } ], [ { "aoVal": "C", "content": "$$24$$ " } ], [ { "aoVal": "D", "content": "$$28$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "第一次走了$$\\frac{2}{7}$$，第二次一共走了$$\\frac{5}{7}$$，则全程是$$9\\div \\left( \\frac{5}{7}-\\frac{2}{7} \\right)=21$$千米． " ]

[ "2018年IMAS小学中年级竞赛（第一轮）第12题4分", "2018年IMAS小学高年级竞赛（第一轮）第3题3分" ]

[ [ { "aoVal": "A", "content": "$$94$$ " } ], [ { "aoVal": "B", "content": "$$94.5$$ " } ], [ { "aoVal": "C", "content": "$$95$$ " } ], [ { "aoVal": "D", "content": "$$95.5$$ " } ], [ { "aoVal": "E", "content": "$$96$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "可知\\uline{小杨}在期末考试中，语文与数学两科的总分为$$97\\times 2=194$$分， 所以三科的总分为$$194+94=288$$分， 即三科的平均分为$$288\\div 3=96$$分． 故选$$\\text{E}$$． ", "<p>可知<u>小杨</u>在期末考试中，语文与数学两科的平均分数为$$97$$分，而英语考了$$94$$分，</p>\n<p>所以语文与数学两科总共比英语多了$$\\left( 97-94 \\right)\\times 2=6$$分，</p>\n<p>所以三科的总平均分数比英语多了$$6\\div 3=2$$分，</p>\n<p>即三科的平均分为$$94+2=96$$分．</p>\n<p>故选$$\\text{E}$$．</p>" ]

[ "2013年第25届广东广州五羊杯六年级竞赛第10题5分" ]

[ [ { "aoVal": "A", "content": "$$82$$ " } ], [ { "aoVal": "B", "content": "$$83$$ " } ], [ { "aoVal": "C", "content": "$$84$$ " } ], [ { "aoVal": "D", "content": "$$85$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "三科总分为$$85\\times 3=255$$（分）． 设最低分的科目英语为$$x$$分， 则有$$x+4\\textgreater255-2x-4\\textgreater x$$， 解得$$\\begin{cases}x\\textless{}84 x\\textgreater82 \\end{cases}$$， 所以$$x=83$$（分）． " ]

[ "2017年第15届全国希望杯六年级竞赛" ]

[ [ { "aoVal": "A", "content": "$$\\frac{9}{10}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{9}{20}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{11}{45}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{22}{45}$$ " } ] ]

[ "拓展思维->能力->归纳总结->归纳推理" ]

[ "因为$$\\frac{1}{n\\left( n+1 \\right)\\left( n+2 \\right)}=\\frac{1}{2}\\times \\frac{\\left( n+2 \\right)-n}{n\\left( n+1 \\right)\\left( n+2 \\right)}=\\frac{1}{2}\\left[ \\frac{1}{n\\left( n+1 \\right)}-\\frac{1}{\\left( n+1 \\right)\\left( n+2 \\right)} \\right]$$ 所以原式$$=\\frac{1}{2}\\times \\left( \\frac{1}{1\\times 2}-\\frac{1}{2\\times 3}+\\frac{1}{2\\times 3}-\\frac{1}{3\\times 4}+\\frac{1}{3\\times 4}-\\frac{1}{4\\times 5}...+\\frac{1}{8\\times 9}-\\frac{1}{9\\times 10} \\right)$$ $$=\\frac{1}{2}\\times \\left[ \\frac{1}{2}-\\frac{1}{90} \\right]=\\frac{11}{45}$$． " ]

[ "2019年第23届广东世界少年奥林匹克数学竞赛五年级竞赛决赛第5题5分" ]

[ [ { "aoVal": "A", "content": "$$6$$ " } ], [ { "aoVal": "B", "content": "$$9$$ " } ], [ { "aoVal": "C", "content": "$$12$$ " } ], [ { "aoVal": "D", "content": "$$15$$ " } ] ]

[ "拓展思维->拓展思维->数论模块->因数与倍数->因数与倍数基础" ]

[ "$$600$$的因数有$$1$$，$$2$$，$$3$$，$$4$$，$$5$$，$$6$$，$$8$$，$$10$$，$$12$$，$$15$$，$$20$$，$$24$$，$$25$$，$$30$$，$$40$$，$$50$$，$$60$$，$$75$$，$$100$$，$$120$$，$$150$$，$$200$$，$$300$$，$$600$$，其中是$$6$$的倍数的有$$6$$，$$12$$，$$24$$，$$30$$，$$60$$，$$120$$，$$150$$，$$300$$，$$600$$这$$9$$个． 故选$$\\text{B}$$． " ]

[ "2019年广东深圳全国小学生数学学习能力测评六年级竞赛初赛第7题3分" ]

[ [ { "aoVal": "A", "content": "$$9:11$$ " } ], [ { "aoVal": "B", "content": "$$1:10$$ " } ], [ { "aoVal": "C", "content": "$$19:81$$ " } ], [ { "aoVal": "D", "content": "$$19:100$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "依题意得，糖水质量是$$900$$克，糖与水的质量比是$$1:9$$，那么糖的质量是$$90$$克，水的质量是$$900-90=810$$（克），又加入了$$100$$克糖，现在糖与水的质量比是：$$(90+100):810=190:810=19:81$$． 故选$$\\text{C}$$． " ]