Datasets:

---

math-eval

/

TAL-SCQ5K

like 57

[ "2011年第7届全国新希望杯小学高年级六年级竞赛第6题4分" ]

[ [ { "aoVal": "A", "content": "$$108$$ " } ], [ { "aoVal": "B", "content": "$$96$$ " } ], [ { "aoVal": "C", "content": "$$88$$ " } ], [ { "aoVal": "D", "content": "$$81$$ " } ] ]

[ "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" ]

[ "先算卡片的总值． 最小值$$100\\div 10=10$$次，共减$$10\\times 27$$； 最多$$14$$次，共减去$$14\\times 27$$， 因此差为$$(14-10)\\times 27=108$$． " ]

[ "2015年全国迎春杯四年级竞赛初赛第5题" ]

[ [ { "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->鸡兔同笼问题->分组法解鸡兔同笼->腿倍型" ]

[ "白天比晚上多了一个鸡头，还多了一只鹤脚；无论晚上还是白天，足数和头数的差都一样，所以鹤的数量和鸡的数量是一样的．将鸡和鹤打一个包，则在白天这个包和兔子腿数一样为$$4$$，在晚上这个包和兔子头数一样为$$1$$；则可以得出晚上的头数为$$56\\div 4=14$$（个）． " ]

[ "2016年全国小学生数学学习能力测评四年级竞赛初赛第6题3分" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$7$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ], [ { "aoVal": "E", "content": "$$14$$ " } ] ]

[ "拓展思维->能力->运算求解", "Overseas Competition->知识点->组合模块->操作与策略->统筹规划" ]

[ "$36\\div5=7R1$. Thus, for the first $35$ boxes, there is a total of $$14\\times7=98$$ balls. Therefore, there are $100-98=2$ balls. " ]

[ "六年级其它导引", "2015年世界少年奥林匹克数学竞赛六年级竞赛复赛A卷第10题10分", "2018年陕西西安碑林区西安市铁一中学小升初（二十六）第20题5分" ]

[ [ { "aoVal": "A", "content": "$$38\\frac{19}{20}$$ " } ], [ { "aoVal": "B", "content": "$$38\\frac{17}{20}$$ " } ], [ { "aoVal": "C", "content": "$$38\\frac{13}{20}$$ " } ], [ { "aoVal": "D", "content": "$$38\\frac{1}{20}$$ " } ] ]

[ "拓展思维->思想->转化与化归的思想" ]

[ "算式中的分母是裂项计算的最基本形式，但分子比较复杂，我们可以从前几项找找规律： $$\\frac{{{1}^{2}}+{{2}^{2}}}{1\\times 2}=\\frac{5}{2}=2\\frac{1}{2}$$，$$\\frac{{{2}^{2}}+{{3}^{2}}}{2\\times 3}=\\frac{13}{6}=2\\frac{1}{6}$$，$$\\frac{{{3}^{2}}+{{4}^{2}}}{3\\times 4}=\\frac{25}{12}=2\\frac{1}{12}$$． 我们发现一规律：每一项减去$$2$$后，分子就变成了$$1$$． 再来试试最后一项：$$\\frac{{{19}^{2}}+{{20}^{2}}}{19\\times 20}=\\frac{761}{380}=2\\frac{1}{380}$$， 也满足这个规律，这是为什么呢？ 观察每一项的分子和分母，我们发现分子的每个加数都与分母大小接近，可以做如下变形： $$\\frac{{{19}^{2}}+{{20}^{2}}}{19\\times 20}=\\frac{19\\times \\left( 20-1 \\right)+20\\times \\left( 19+1 \\right)}{19\\times 20}$$ $$=\\frac{19\\times 20\\times 2+\\left( 20-19 \\right)}{19\\times 20}$$ $$=2+\\frac{1}{19\\times 20}$$． 算式中的每一项都能像上面一样进行变形，所以： 原式$$=2\\frac{1}{1\\times 2}+2\\frac{1}{2\\times 3}+\\cdots +2\\frac{1}{19\\times 20}$$ $$=2\\times 19+\\frac{1}{1\\times 2}+\\frac{1}{2\\times 3}+\\cdots +\\frac{1}{19\\times 20}$$ $$=38+1-\\frac{1}{20}$$ $$=38\\frac{19}{20}$$． " ]

[ "2019年第23届广东世界少年奥林匹克数学竞赛六年级竞赛初赛第8题5分" ]

[ [ { "aoVal": "A", "content": "$$24$$ " } ], [ { "aoVal": "B", "content": "$$25$$ " } ], [ { "aoVal": "C", "content": "$$26$$ " } ], [ { "aoVal": "D", "content": "$$27$$ " } ] ]

[ "拓展思维->能力->构造模型->模型思想" ]

[ "$$1\\times 8\\times 9=72$$， ∴各位数字乘积为$$72$$的有 $$189$$，$$198$$，$$819$$，$$891$$，$$918$$，$$981$$，$$6$$个． ∵$$2\\times 4\\times 9=72$$， ∴三位数各位数字乘积为$$72$$的有 $$249$$，$$294$$，$$429$$，$$492$$，$$924$$，$$942$$，$$6$$个． ∵$$3\\times 3\\times 8=72$$， ∴三位数各位数字乘积为$$72$$的有 $$338$$，$$383$$，$$833$$，$$3$$个． ∵$$3\\times 4\\times 6=72$$， ∴三位数各位数字乘积为$$72$$的有 $$346$$，$$364$$，$$463$$，$$436$$，$$643$$，$$634$$，共$$6$$个． ∵$$2\\times 6\\times 6=72$$， ∴三位数各位数字乘积为$$72$$的有 $$266$$，$$626$$，$$662$$，$$3$$个． 综上所述共有$$6+6+3+6+3=24$$个． 故选$$\\text{A}$$． " ]

[ "2020年广东广州羊排赛五年级竞赛第8题3分" ]

[ [ { "aoVal": "A", "content": "$$15$$ " } ], [ { "aoVal": "B", "content": "$$105$$ " } ], [ { "aoVal": "C", "content": "$$205$$ " } ], [ { "aoVal": "D", "content": "$$366$$ " } ] ]

[ "拓展思维->能力->实践应用" ]

[ "根据周期问题的公式：总数$$\\div $$周期$$=$$周期数$$\\cdots \\cdots $$余数，余数是几，就是周期中的第几个．本题最后这一天是周三，刚好是周期中的第一天，说明余数就是$$1$$，又因为一周有$$7$$天，依次判断选项中的数除以$$7$$的余数． $$\\text{A}$$.$$15\\div 7=2$$（周）$$\\cdots \\cdots 1$$（天）；$$\\text{B}$$.$$105\\div 7=15$$（周）； $$\\text{C}$$.$$205\\div 7=29$$（周）$$\\cdots \\cdots 2$$（天）；$$\\text{D}$$.$$366\\div 7=52$$（周）$$\\cdots \\cdots 2$$（天）． 故选$$\\text{A}$$． " ]

[ "2006年第4届创新杯六年级竞赛复赛第6题" ]

[ [ { "aoVal": "A", "content": "$$19$$ " } ], [ { "aoVal": "B", "content": "$$17$$ " } ], [ { "aoVal": "C", "content": "$$13$$ " } ], [ { "aoVal": "D", "content": "非上述答案 " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "原算式中有约数$$2$$，$$3$$，$$5$$，$$7$$，$$11$$，$$13$$，没有约数$$17$$，那么不是$$n$$的约数的最小质数是$$17$$． " ]

[ "2014年全国迎春杯六年级竞赛复赛第8题" ]

[ [ { "aoVal": "A", "content": "$$12$$ " } ], [ { "aoVal": "B", "content": "$$36$$ " } ], [ { "aoVal": "C", "content": "$$48$$ " } ], [ { "aoVal": "D", "content": "$$60$$ " } ] ]

[ "拓展思维->拓展思维->计数模块->加乘原理->组数问题->有特殊要求的组数问题" ]

[ "设这个数为$$\\overline{ABCBA}$$，$$A$$位可以填$$11$$，$$ 88$$，$$ 69$$，$$96$$，$$4$$种情况，$$B$$位可以填$00$$，$$11$$，$$88$$，$$69$$，$$96$，$$5$$种情况，$$C$$位可以填$$0$$，$$1$$，$$8$$，$$3$$种情况，$$4\\times 5\\times 3=60$$（个）． " ]

[ "2017年四川六年级竞赛排位赛第20题2分" ]

[ [ { "aoVal": "A", "content": "$$18$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{174}{7}$$ " } ], [ { "aoVal": "C", "content": "$$144$$ " } ], [ { "aoVal": "D", "content": "$$3$$ " } ] ]

[ "拓展思维->能力->数据处理" ]

[ "甲$$:$$丙$$=4:3=20:15$$，丙$$:$$乙$$=5:3=15:9$$， 则甲$$:$$乙$$:$$丙$$=20:9:15$$，每份为$$87\\div (20+9)=3$$， 则丙数$$-$$乙数$$=(15-9)\\times 3=18$$． 故选$$\\text{A}$$． " ]

[ "2017年第15届全国希望杯五年级竞赛第1试试题第18题", "其它改编题" ]

[ [ { "aoVal": "A", "content": "$$A$$ " } ], [ { "aoVal": "B", "content": "$$B$$ " } ], [ { "aoVal": "C", "content": "$$C$$ " } ], [ { "aoVal": "D", "content": "$$D$$ " } ], [ { "aoVal": "E", "content": "$$E$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "$$A$$和$$E$$说的话对立，$$C$$和$$D$$说的话对立，必有两对两错，故$$B$$说的是错的，则是$$E$$射中的． " ]

[ "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8题5分" ]

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

[ "拓展思维->拓展思维->应用题模块->鸡兔同笼问题->假设法解鸡兔同笼->基本型->原型题" ]

[ "枚举法： 当有$$1$$只兔，$$3$$只鸡时，共$$1\\times4+3\\times2=10$$（条）腿． 当有$$2$$只兔，$$2$$只鸡时，共$$2\\times4+2\\times2=12$$（条）腿． 所以有$$2$$只兔． " ]

[ "2016年全国美国数学大联盟杯小学中年级三年级竞赛初赛" ]

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

[ "知识标签->拓展思维->计算模块->整数->四则混合运算" ]

[ "$$35$$个题目需要$$30$$分钟，$$315$$里面有$$9$$个$$35$$，所以需要的时间是$$30\\times 9=270$$分钟，等于$$4.5$$小时． " ]

[ "2016年新希望杯六年级竞赛训练题（五）第6题" ]

[ [ { "aoVal": "A", "content": "$$100$$ " } ], [ { "aoVal": "B", "content": "$$150$$ " } ], [ { "aoVal": "C", "content": "$$200$$ " } ], [ { "aoVal": "D", "content": "$$250$$ " } ] ]

[ "拓展思维->拓展思维->行程模块->直线型行程问题->路程速度时间->单人变速问题" ]

[ "甲跑完全程共花了$$3000\\div 8=375$$秒，乙在这段时间跑了$$3000-125=2875$$米． 设乙用$$10$$米/秒速度跑了$$x$$秒，则他用$$5$$米/秒的速度跑了$$\\left( 375-x \\right)$$秒．$$10x+\\left( 375-x \\right)\\times 5=2875$$，解得$$x=200$$，用$$5$$米/秒的速度所跑的路程为$$3000-200\\times 10=1000$$米． 所用时间为$$1000\\div 5=200$$秒． " ]

[ "2018年第22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6分" ]

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

[ "拓展思维->思想->整体思想" ]

[ "小文的左边有人：$$12-1=11$$（人）， 所以从右边数小文排在：$$20-11=9$$（个）． 故选择$$\\text{C}$$． " ]

[ "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4题5分" ]

[ [ { "aoVal": "A", "content": "$$6$$ " } ], [ { "aoVal": "B", "content": "$$7$$ " } ], [ { "aoVal": "C", "content": "$$8$$ " } ], [ { "aoVal": "D", "content": "$$9$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->间隔问题->直线型两端都没有->锯木头类型问题" ]

[ "锯成$$5$$段，需要锯$$4$$次，所以每次$$1$$分钟．锯成$$10$$段，需要锯$$9$$次，所以需要$$9$$分钟，故选择$$\\text{D}$$． " ]

[ "2014年全国迎春杯三年级竞赛初赛第8题" ]

[ [ { "aoVal": "A", "content": "红珠 " } ], [ { "aoVal": "B", "content": "黄珠 " } ], [ { "aoVal": "C", "content": "绿珠 " } ], [ { "aoVal": "D", "content": "白珠 " } ] ]

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

[ "$$2000\\div (4+3+2+1)=200$$． " ]

[ "2019年第22届世界少年奥林匹克数学竞赛四年级竞赛复赛（中国区）第4题3分" ]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$8$$ " } ], [ { "aoVal": "D", "content": "$$12$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "第一把钥匙最坏的情况要试$$3$$次，把这把钥匙和这把锁拿出；剩下的$$3$$把锁和$$3$$把钥匙，最坏的情况要试$$2$$次，把这把钥匙和这把锁拿出；剩下的$$2$$把锁和$$2$$把钥匙，最坏的情况要试$$1$$次，把这把钥匙和这把锁拿出；剩下的$$1$$把锁和$$1$$把钥匙就不用试了；$$3+2+1=6$$（次）． 故选$$\\text{B}$$． " ]

[ "2016年IMAS小学高年级竞赛第一轮检测试题第13题4分" ]

[ [ { "aoVal": "A", "content": "$$100$$ " } ], [ { "aoVal": "B", "content": "$$105$$ " } ], [ { "aoVal": "C", "content": "$$108$$ " } ], [ { "aoVal": "D", "content": "$$109$$ " } ], [ { "aoVal": "E", "content": "$$215$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "每两个数相加，和最小为$$1+2=3$$，最大为$$n+(n-1)=2n-1$$，现在共有$$215$$种不同的数值，即最大的数值为$$215+3-1=217$$，$$217=2n-1$$，$$n=109$$． ", "<p>最大值和最小值之间有多少数字就是多少种可能性．$$2n -1-3=215$$．</p>" ]

[ "2017年第15届湖北武汉创新杯小学高年级五年级竞赛决赛第5题" ]

[ [ { "aoVal": "A", "content": "$$600$$亩 " } ], [ { "aoVal": "B", "content": "$$630$$亩 " } ], [ { "aoVal": "C", "content": "$$660$$亩 " } ], [ { "aoVal": "D", "content": "$$690$$亩 " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "$$440000\\div \\frac{2000}{3}=660$$（亩）． " ]

[ "2018年第6届湖北长江杯五年级竞赛初赛A卷第6题3分", "2018年上海小学高年级五年级竞赛A卷" ]

[ [ { "aoVal": "A", "content": "$$8:00$$ " } ], [ { "aoVal": "B", "content": "$$18:40$$ " } ], [ { "aoVal": "C", "content": "$$19:20$$ " } ], [ { "aoVal": "D", "content": "$$20:00$$ " } ] ]

[ "拓展思维->拓展思维->数论模块->因数与倍数->公因数与公倍数->公倍数与最小公倍数->两数的最小公倍数" ]

[ "$$10$$和$$8$$的最小公倍数是$$40$$， $$40\\times20=800$$（分钟）， $$6$$时$$20$$分再过$$800$$分钟为$$20$$时， 所以第$$20$$次同时发车为$$20$$时． 故选$$\\text{D}$$． " ]

[ "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": "$$7$$ " } ] ]

[ "知识标签->数学思想->分类讨论思想" ]

[ "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，总分最多$$24\\times 3=72$$分，平$$1$$场总分少$$1$$分，$$72-67=5$$分，所以平局$$5$$场. " ]

[ "2017年第22届全国华杯赛小学中年级竞赛初赛第4题10分" ]

[ [ { "aoVal": "A", "content": "$$90$$ " } ], [ { "aoVal": "B", "content": "$$105$$ " } ], [ { "aoVal": "C", "content": "$$120$$ " } ], [ { "aoVal": "D", "content": "$$135$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "在同样时间内，猎豹跑$$2$$步可以跑$$4$$米，孤狸跑$$3$$步可以跑$$3$$米，那么$${{V}_{豹}}:{{V}_{狐}}=4:3$$，猎豹跑距离气$${{S}_{豹}}=30\\times \\frac{4}{4-3}=120$$米． " ]

[ "2017年第16届湖北武汉创新杯六年级竞赛邀请赛训练题（三）" ]

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

[ "拓展思维->思想->对应思想" ]

[ "$$\\frac{a}{7}+\\frac{b}{11}=\\frac{11a+7b}{77}=1.03$$ $$11a+7b\\approx 79.31~ ， 11a+7b=79$$ 由于$$a$$和$$b$$都是正整数，$$a=4$$，$$b=5$$，$$a+b=9$$. " ]

[ "2014年全国华杯赛小学高年级竞赛初赛A卷第2题" ]

[ [ { "aoVal": "A", "content": "$$40$$ " } ], [ { "aoVal": "B", "content": "$$42$$ " } ], [ { "aoVal": "C", "content": "$$48$$ " } ], [ { "aoVal": "D", "content": "$$50$$ " } ] ]

[ "知识标签->数学思想->方程思想" ]

[ "假设小龙所有的题目都做对了，则小龙的得分为：$$50\\times 3=150$$（分）；实际上小龙的得分为$$118$$分，假设与实际相差：$$150-118=32$$（分）；小龙一道题由对变错会损失：$$3+1=4$$（分），所以小龙错了：$$32\\div 4=8$$（道）题，则小龙答对了：$$50-8=42$$（道）题． " ]

[ "2017年第15届湖北武汉创新杯六年级竞赛初赛第5题" ]

[ [ { "aoVal": "A", "content": "$$3.4$$ " } ], [ { "aoVal": "B", "content": "$$4.8$$ " } ], [ { "aoVal": "C", "content": "$$6.0$$ " } ], [ { "aoVal": "D", "content": "$$6.6$$ " } ] ]

[ "拓展思维->思想->方程思想" ]

[ "设每支铅笔$$x$$元，每支圆珠笔$$y$$元，则$$3x+5y=5.4$$，$$5x+7y=7.8$$，得$$2x+2y=2.4$$，所以$$4x+4y=4.8$$． " ]

[ "2013年第12届全国小机灵杯小学中年级三年级竞赛初赛第6题8分" ]

[ [ { "aoVal": "A", "content": "$$42$$ " } ], [ { "aoVal": "B", "content": "$$47$$ " } ], [ { "aoVal": "C", "content": "$$51$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->定义新运算->直接运算型->普通型" ]

[ "$$3\\nabla 4=2\\times 3+3\\times 4=18$$，$$2\\Delta \\left( 3\\nabla 4 \\right)=2\\Delta 18=3\\times 2+2\\times 18=42$$． " ]

[ "2017年第16届湖北武汉创新杯六年级竞赛邀请赛训练题（一）" ]

[ [ { "aoVal": "A", "content": "$$45$$ " } ], [ { "aoVal": "B", "content": "$$46$$ " } ], [ { "aoVal": "C", "content": "$$47$$ " } ], [ { "aoVal": "D", "content": "$$48$$ " } ] ]

[ "拓展思维->能力->归纳总结->归纳推理" ]

[ "原式$$=\\left( 1+2+3+\\cdots \\cdots +9 \\right)+\\left( \\frac{1}{6}+\\frac{1}{12}+\\cdots \\cdots +\\frac{1}{110} \\right)$$ $$=45+\\left( \\frac{1}{2\\times 3}+\\frac{1}{3\\times 4}+\\cdots \\cdots +\\frac{1}{10\\times 11} \\right)$$ $$=45+\\left( \\frac{1}{2}\\times \\frac{1}{11} \\right)$$ $$=45\\frac{9}{22}$$ 所以符合题意的最小值为$$46$$． " ]

[ "2018年湖北武汉新希望杯五年级竞赛训练题（五）第1题" ]

[ [ { "aoVal": "A", "content": "$$32$$ " } ], [ { "aoVal": "B", "content": "$$34$$ " } ], [ { "aoVal": "C", "content": "$$36$$ " } ], [ { "aoVal": "D", "content": "$$38$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->方阵问题->实心方阵->实心方阵的增减", "课内体系->思想->数形结合思想" ]

[ "$$9\\times 9-7\\times 7=32$$（棵）． " ]

[ "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4题5分" ]

[ [ { "aoVal": "A", "content": "$$17$$ " } ], [ { "aoVal": "B", "content": "$$16$$ " } ], [ { "aoVal": "C", "content": "$$15$$ " } ], [ { "aoVal": "D", "content": "$$14$$ " } ] ]

[ "拓展思维->能力->构造模型->模型思想" ]

[ "$$25-10+1=16$$． 故选：$$\\text{B}$$． " ]

[ "2012年IMAS小学中年级竞赛第一轮检测试题第2题3分" ]

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

[ "拓展思维->拓展思维->计算模块->整数->整数加减->整数减法运算->减法横式" ]

[ "13-2=11；11-5=6 " ]

[ "2012年IMAS小学高年级竞赛第一轮检测试题第18题4分" ]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$6$$ " } ], [ { "aoVal": "D", "content": "$$7$$ " } ], [ { "aoVal": "E", "content": "$$8$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "老王接下来的休息时间为：周二和周三；周五和周六；周一和周二；周四和周五；周日和周一；$$\\cdots\\cdots$$． 所以他再次在星期天休息时已经过了$$10\\times 4+9=49$$天，刚好$$7$$个星期． " ]

[ "2016年第14届全国创新杯五年级竞赛初赛第8题" ]

[ [ { "aoVal": "A", "content": "$$8$$ " } ], [ { "aoVal": "B", "content": "$$9$$ " } ], [ { "aoVal": "C", "content": "$$13$$ " } ], [ { "aoVal": "D", "content": "$$20$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "斐波那契数列从第三项开始，等于前两项的和，即$$1$$、$$2$$、$$3$$、$$5$$、$$8$$、$$13$$、$$21$$． 即第二长的为$$13$$． " ]

[ "2020年希望杯二年级竞赛模拟第17题" ]

[ [ { "aoVal": "A", "content": "班长，学习委员，体育委员 " } ], [ { "aoVal": "B", "content": "学习委员，体育委员，班长 " } ], [ { "aoVal": "C", "content": "学习委员，班长，体育委员 " } ], [ { "aoVal": "D", "content": "班长，体育委员，学习委员 " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "由②小舒比学习委员的年龄大；和③小贝和学习委员年龄不同．可知小舒和小贝都不是学习委员，所以学习委员是小皮； 已知小舒比小皮大，并且小皮的年龄比体育委员的年龄大，所以小舒不是体育委员，所以小舒是班长；那么小贝就是体育委员． 故选$$\\text{C}$$． " ]

[ "2011年第9届全国创新杯小学高年级六年级竞赛第7题" ]

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

[ "拓展思维->思想->对应思想" ]

[ "$$6\\oplus x=6+x-1=5+x$$；$$3\\otimes 5=3\\times 5-1=14$$；$$\\left( 5+x \\right)\\oplus 14=5+x+14-1=x+18$$； $$4\\otimes \\left( 18+x \\right)=4\\times \\left( 18+x \\right)-1=4x+71$$；$$4x+71=79$$，得$$x=2$$． " ]

[ "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9题4分" ]

[ [ { "aoVal": "A", "content": "$$14$$ " } ], [ { "aoVal": "B", "content": "$$15$$ " } ], [ { "aoVal": "C", "content": "$$16$$ " } ], [ { "aoVal": "D", "content": "$$17$$ " } ] ]

[ "拓展思维->能力->实践应用" ]

[ "设$$10$$元的有$$x$$本，$$15$$元的有$$y$$本，则$$10x+15y=90$$，解得$$\\begin{cases} x=6 y=2 \\end{cases}$$，$$\\begin{cases} x=3 y=4 \\end{cases}$$，共$$6+2+3+4=15$$本． " ]

[ "2011年北京五年级竞赛", "2011年北京六年级期中" ]

[ [ { "aoVal": "A", "content": "对 " } ], [ { "aoVal": "B", "content": "不对 " } ] ]

[ "知识标签->数学思想->枚举思想" ]

[ "将这四个人用$$4$$个点表示，如果两个人之间送过礼，就在两点之间连一条线． 由于每人送出$$2$$件礼物，图中共有$$4\\times 2=8$$条线，由于每人礼品都分赠给$$2$$个人，所以每两点之间至多有$$1+1=2$$条线．四点间，每两点连一条线，一共$$6$$条线，现在有$$8$$条线，说明必有两点之间连了$$2$$条线，还有另外两点（有一点可以与前面的点相同）之间也连了$$2$$条线． 即为所证结论． " ]

[ "2014年全国迎春杯三年级竞赛初赛第12题" ]

[ [ { "aoVal": "A", "content": "一 " } ], [ { "aoVal": "B", "content": "四 " } ], [ { "aoVal": "C", "content": "五 " } ], [ { "aoVal": "D", "content": "六 " } ] ]

[ "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" ]

[ "星期六有：$$21\\to 28\\to 4(35)\\to 11\\to 18\\to 25$$，所以 $$31$$日是星期五． $$10+31=41$$（天），$$41\\div7=5\\cdots 6$$ ，差一天是星期六，所以$$31$$日是星期五． " ]

[ "2017年第15届湖北武汉创新杯五年级竞赛初赛第11题" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$27$$ " } ], [ { "aoVal": "C", "content": "$$37$$ " } ], [ { "aoVal": "D", "content": "无法计算 " } ] ]

[ "知识标签->学习能力->七大能力->运算求解" ]

[ "原式$$=\\frac{37\\times 1001001001\\times 73}{2701\\times 1001001001}$$ $$=\\frac{2701\\times 1001001001}{2701\\times 1001001001}$$ $$=1$$． " ]

[ "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": "$$274$$ " } ], [ { "aoVal": "C", "content": "$$504$$ " } ], [ { "aoVal": "D", "content": "$$927$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "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$${{a}_{4}}=1+2+4=7$$， $${{a}_{5}}=2+4+7=13$$， $${{a}_{6}}=4+7+13=24$$， $${{a}_{7}}=7+13+24=44$$， $${{a}_{8}}=13+24+44=81$$， $${{a}_{9}}=24+44+81=149$$， $${{a}_{10}}=44+81+149=274$$， $${{a}_{11}}=81+149+274=504$$， $${{a}_{12}}=149+274+504=927$$． 故选：$$\\text{D}$$． " ]

[ "2019年第7届湖北长江杯五年级竞赛复赛B卷第8题3分" ]

[ [ { "aoVal": "A", "content": "赚了 " } ], [ { "aoVal": "B", "content": "亏了 " } ], [ { "aoVal": "C", "content": "不亏不赚 " } ], [ { "aoVal": "D", "content": "不确定 " } ] ]

[ "拓展思维->拓展思维->应用题模块->经济问题->基本经济概念->已知售价利润求成本" ]

[ "售价为$$200$$元，$$A$$赚了十分之一，$$A$$的进价为$$200\\div \\left( 1+10 \\% \\right)=\\frac{2000}{11}$$（元）， $$B$$亏了十分之一，$$B$$的进价为$$200\\div \\left( 1-10 \\% \\right)=\\frac{2000}{9}$$（元）， $$\\frac{2000}{11}+\\frac{2000}{9}=\\frac{40000}{99}$$（元）$$\\textgreater400$$元， 进价$$\\textgreater$$售价，亏了． 故选：$$\\text{B}$$． " ]

[ "2016年全国华杯赛小学高年级竞赛初赛第1题" ]

[ [ { "aoVal": "A", "content": "$$2017$$ " } ], [ { "aoVal": "B", "content": "$$2016$$ " } ], [ { "aoVal": "C", "content": "$$2015$$ " } ], [ { "aoVal": "D", "content": "$$2014$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "$${{\\left( {{10}^{2016}}-1 \\right)}^{2}}=\\left( {{10}^{2016}}-2 \\right)\\times {{10}^{2016}}+1=\\underbrace{999\\cdots 99}_{2015}8\\underbrace{000\\cdots 00}_{2015}1$$ " ]

[ "2012年全国华杯赛小学高年级竞赛初赛网络版第3题" ]

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

[ "拓展思维->能力->逻辑分析" ]

[ "$$（1）$$$$（3）$$$$（4）$$正确． " ]

[ "2013年小机灵杯五年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "出入相补原理 " } ], [ { "aoVal": "B", "content": "等差数列求和 " } ], [ { "aoVal": "C", "content": "十进制计数法 " } ] ]

[ "拓展思维->拓展思维->组合模块->智巧趣题" ]

[ "解：古时候的原始人捕猎，捕到一只野兽对应一根手指。等到$$10$$根手指都用完，就在绳子上打一个结，这就是运用现在的数学中的十进制计数法； 故选：$$C$$。 " ]

[ "2012年IMAS小学中年级竞赛第一轮检测试题第2题3分" ]

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

[ "拓展思维->拓展思维->计算模块->整数->整数加减->整数减法运算->减法横式" ]

[ "13-2=11；11-5=6 " ]

[ "2017年IMAS小学高年级竞赛（第一轮）第11题4分" ]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$7$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ], [ { "aoVal": "E", "content": "$$9$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "小李记录下的三个差之和等于六个数码的其中三个数码减去另外三个数码， 所以最大可能值为$$6+5+4-3-2-1=9$$． 故选$$\\text{E}$$． " ]

[ "2018年湖北武汉新希望杯五年级竞赛训练题（四）第4题" ]

[ [ { "aoVal": "A", "content": "$$84$$ " } ], [ { "aoVal": "B", "content": "$$89$$ " } ], [ { "aoVal": "C", "content": "$$90$$ " } ], [ { "aoVal": "D", "content": "$$96$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->经济问题->分段计价问题" ]

[ "$$4\\times 10+3.5\\times (24-10)=89$$（元）． " ]

[ "2006年华杯赛五年级竞赛初赛", "2006年华杯赛六年级竞赛初赛" ]

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

[ "拓展思维->拓展思维->数论模块->分解质因数" ]

[ "因为$$2007005=2005\\times 1000+2005=2005\\times 1001=\\left( 5\\times 401\\times )( 7\\times 11\\times 13 \\right)$$ " ]

[ "2012年全国希望杯五年级竞赛初赛第2题" ]

[ [ { "aoVal": "A", "content": "$$41$$ " } ], [ { "aoVal": "B", "content": "$$51$$ " } ], [ { "aoVal": "C", "content": "$$61$$ " } ], [ { "aoVal": "D", "content": "$$71$$ " } ] ]

[ "知识标签->学习能力->七大能力->运算求解" ]

[ "$$21.49+52.37-0.4+5.51-11.37-6.6$$$$=(21.49+5.51)+(52.37-11.37)-(0.4+6.6)$$$$=27+41-7$$$$=61$$． " ]

[ "2016年第14届全国创新杯五年级竞赛初赛第9题" ]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{45}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{2}{75}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{7}{225}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{8}{225}$$ " } ] ]

[ "拓展思维->思想->枚举思想" ]

[ "三位数一共有$$900$$个满足条件的魔力数有$$\\text{C}_{7+2-1}^{2}=28$$个，则$$\\frac{28}{900}=\\frac{7}{225}$$． " ]

[ "2007年四年级竞赛创新杯" ]

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

[ "拓展思维->拓展思维->应用题模块->间隔问题->直线型两端都没有->剪绳子" ]

[ "将绳折成$$3$$段再对折，相当于折成$$6$$段，一刀与这$$6$$段共有$$6$$个交点，所以将绳剪成$$7$$段。 " ]

[ "2018年全国小学生数学学习能力测评四年级竞赛复赛第9题3分" ]

[ [ { "aoVal": "A", "content": "$$9$$ " } ], [ { "aoVal": "B", "content": "$$10$$ " } ], [ { "aoVal": "C", "content": "$$16$$ " } ] ]

[ "拓展思维->能力->构造模型->模型思想" ]

[ "利用假设的方法来解答， 假设$$26$$副都是跳棋，那么$$26$$副跳棋可供$$6\\times 26=156$$（人）进行活动， 与$$120$$人相差$$156-120=36$$（人）， 每副象棋和跳棋的人数相差$$6-2=4$$（人）， 用一共相差的人数$$36$$除以$$4$$就求出象棋有多少副，$$36\\div 4=9$$（副）． 故选$$\\text{A}$$． " ]

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

[ [ { "aoVal": "A", "content": "$$3\\times 3\\div 3+3$$ " } ], [ { "aoVal": "B", "content": "$$3\\div 3+3\\times 3$$ " } ], [ { "aoVal": "C", "content": "$$3\\times 3-3+3$$ " } ], [ { "aoVal": "D", "content": "$$3\\div 3+3\\div 3$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->整数->四则混合运算" ]

[ "解：A、$$3\\times 3\\div 3+3$$ $$=3+3$$ $$=6$$ B、$$3\\div 3+3\\times 3$$ $$=1+9$$ $$=10$$ C、$$3\\times 3-3+3$$ $$=9-3+3$$ $$=9$$ D、$$3\\div 3+3\\div 3$$ $$=1+1$$ $$=2$$ 故选：C。 " ]

[ "2021年新希望杯一年级竞赛初赛第14题5分" ]

[ [ { "aoVal": "A", "content": "鼠和鸡 " } ], [ { "aoVal": "B", "content": "鸡和兔 " } ], [ { "aoVal": "C", "content": "兔和狗 " } ], [ { "aoVal": "D", "content": "兔和鼠 " } ], [ { "aoVal": "E", "content": "鼠和狗 " } ] ]

[ "拓展思维->能力->实践应用" ]

[ "根据题意分析可知，狗和猫都与鸡相邻，即鸡的左右两边是猫和狗，又因为猫与老鼠都与兔相邻，所以猫的旁边是兔子和老鼠，兔子和老鼠在猫的同一侧，由此可知，与猫相邻的是鸡和兔． 故选$$\\text{B}$$． " ]

[ "2017年第16届湖北武汉创新杯五年级竞赛邀请赛训练题（三）" ]

[ [ { "aoVal": "A", "content": "$$16$$ " } ], [ { "aoVal": "B", "content": "$$17$$ " } ], [ { "aoVal": "C", "content": "$$18$$ " } ], [ { "aoVal": "D", "content": "$$19$$ " } ] ]

[ "拓展思维->拓展思维->数论模块->整除->整除特征->整除初识" ]

[ "在$$20100$$以内不能被$$2$$，$$3$$，$$5$$，$$7$$整除的数其实就是比$$20$$大的质数，只要理解了这一点，就可以直接得出答案．$$100$$以内有$$25$$质数，$$20$$以内有$$8$$个质数，$$258=17$$（个）． " ]

[ "2019年第24届YMO二年级竞赛决赛第6题3分" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$15$$ " } ], [ { "aoVal": "C", "content": "$$18$$ " } ], [ { "aoVal": "D", "content": "$$20$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "根据题意分析可知，用工作总量除以工作效率就能得到工作时间，小$$\\text{Y}$$的工作效率为：$$1\\div 2=\\frac{1}{2}$$，小$$\\text{M}$$的工作效率为：$$1\\div 3=\\frac{1}{3}$$，所以两人同时组装，完成的时间为： $$15 \\div \\left( \\frac{1}{2}+ \\frac{1}{3}\\right)$$ $$=15 \\div \\frac{5}{6}$$ $$=18$$（小时） 故答案为：$$\\text{C}$$． " ]

[ "2011年北京五年级竞赛", "2011年其它", "2019年陕西西安新城区爱知中学小升初入学真卷2第24题7分" ]

[ [ { "aoVal": "A", "content": "$$300$$ " } ], [ { "aoVal": "B", "content": "$$350$$ " } ], [ { "aoVal": "C", "content": "$$400$$ " } ], [ { "aoVal": "D", "content": "$$450$$ " } ], [ { "aoVal": "E", "content": "$$480$$ " } ] ]

[ "海外竞赛体系->知识点->数论模块->质数与合数->特殊质数运用->特殊质数2", "拓展思维->思想->整体思想" ]

[ "两车相遇时甲走了全程的$$\\frac{5}{9}$$，乙走了全程的$$\\frac{4}{9}$$， 之后甲的速度减少$$20 \\%$$，乙的速度增加$$20 \\%$$， 此时甲、乙的速度比为$$5 \\times (1 - 20 \\% )：$$4$$ \\times (1 + 20 \\% ) = 5：6$$ ， 所以甲到达$$B$$地时，乙又走了$$\\frac{4}{9} \\times \\frac{6}{5} = \\frac{8}{{15}}$$，距离 $$A$$地$$\\frac{5}{9} - \\frac{8}{{15}} = \\frac{1}{{45}}$$， 所以$$A$$、$$B$$两地的距离为$$10 \\div \\frac{1}{{45}} = 450$$ $$($$千米$$)$$． " ]

[ "2013年第11届全国创新杯五年级竞赛第3题5分" ]

[ [ { "aoVal": "A", "content": "$$2$$厘米 " } ], [ { "aoVal": "B", "content": "$$3$$厘米 " } ], [ { "aoVal": "C", "content": "$$4$$厘米 " } ], [ { "aoVal": "D", "content": "$$7$$厘米 " } ] ]

[ "拓展思维->能力->图形认知" ]

[ "根据题意我们必须将$$42$$拆分成$$3$$个自然数的乘积，且恰好其中长与高之和应为$$18\\div 2=9cm$$，易得：$$42=7\\times 2\\times 3$$，故长为$$7$$厘米，宽为$$2$$厘米，高为$$3$$厘米． " ]

[ "2019年美国数学大联盟杯五年级竞赛初赛第5题5分" ]

[ [ { "aoVal": "A", "content": "$$30$$ " } ], [ { "aoVal": "B", "content": "$$70$$ " } ], [ { "aoVal": "C", "content": "$$105$$ " } ], [ { "aoVal": "D", "content": "$$175$$ " } ] ]

[ "拓展思维->拓展思维->数论模块->分解质因数->分解质因数（式）", "Overseas Competition->知识点->数论模块->分解质因数" ]

[ "题目翻译：下面哪一项不是三个完全不同的质数的乘积？ 把每个选项的数进行分解质因数． $$30=2\\times3\\times5$$，是三个不同的质数的乘积． $$70=2\\times5\\times7$$，是三个不同的质数的乘积． $$105=3\\times5\\times7$$，是三个不同的质数的乘积． $$175=5\\times5\\times7$$，不是三个不同的质数的乘积． " ]

[ "2014年全国华杯赛小学高年级竞赛初赛B卷第6题", "2016年吉林长春二道区吉大附中力旺实验小学小升初第9题2分" ]

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

[ "拓展思维->思想->对应思想" ]

[ "可以直接判断乙必定说的是真话，可以直接判断乙必定是说真话的，乙中的数字不管怎么变换都不可能是$$9$$的倍数． 因为七位数的数字之和为$$2+0+1+4+3+1+5=16$$，不是$$9$$的倍数； 如果丙说真话，那么他手中的数字是$$0$$和$$5$$，可以实现对调位置后被$$10$$整除； 如果甲说真话，那么他手中的数字只能是$$5$$和$$2$$，可以实现对调位置后被$$8$$整除； 如果丁说真话，那么他手中的数字只能是$$0$$和$$3$$，这样才能使得奇数位数字之和减去偶数位数字之和的差是$$11$$的倍数$$(2314015$$，$$(5+0+1+2)-(1+4+3)=0$$)． 综上，如果丙说真话，那甲和丁都是说谎话的人，两个人说谎话，不符合题意，所以说谎话的人是丙，选$$\\text{C}$$． " ]

[ "2017年IMAS小学高年级竞赛（第二轮）第5题4分" ]

[ [ { "aoVal": "A", "content": "$$4.42$$ " } ], [ { "aoVal": "B", "content": "$$0.42$$ " } ], [ { "aoVal": "C", "content": "$$4.41$$ " } ], [ { "aoVal": "D", "content": "$$4$$ " } ], [ { "aoVal": "E", "content": "$$2.42$$ " } ] ]

[ "拓展思维->思想->逆向思想" ]

[ "可知原来的正数与其整数部分相加所得之和为$$22.1\\div 5=4.42$$， 因此原来的正数之小数部分为$$0.42$$，且整数部分为$$4\\div 2=2$$， 所以原来这个正数是$$2.42$$． 故选$$\\text{E}$$． " ]

[ "2018年第6届湖北长江杯六年级竞赛初赛A卷第7题3分" ]

[ [ { "aoVal": "A", "content": "$$8$$ " } ], [ { "aoVal": "B", "content": "$$9$$ " } ], [ { "aoVal": "C", "content": "$$10$$ " } ], [ { "aoVal": "D", "content": "$$12$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->工程问题->合作工程问题->休息工程问题" ]

[ "把这条公路的工程量看成单位``$$1$$''， 那么甲的工作效率就是$$\\frac{1}{24}$$，乙的工作效率就是$$\\frac{1}{30}$$， 合作的工作效率就是二者的和， 用总工作量减去甲独干的工作量就是合干的工作量， 用合干的工作量除以合作的工作效率就是合干的时间，也就是乙队干的时间． $$\\left( 1-\\frac{1}{24}\\times 6 \\right)\\div \\left( \\frac{1}{24}+\\frac{1}{30} \\right)$$ $$=\\left( 1-\\frac{1}{4} \\right)\\div \\frac{3}{40}$$ $$=\\frac{3}{4}\\div \\frac{3}{40}$$ $$=10$$（天）． 答：乙队修了$$10$$天． 故选$$\\text{C}$$． " ]

[ "2016年全国美国数学大联盟杯小学高年级六年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "星期三 " } ], [ { "aoVal": "B", "content": "星期四 " } ], [ { "aoVal": "C", "content": "星期五 " } ], [ { "aoVal": "D", "content": "星期六 " } ] ]

[ "知识标签->课内知识点->数的认识->数的特征->倍数->两个数的公倍数" ]

[ "$$8\\times 11=88$$，$$88$$除以$$7$$余$$4$$，所以正好是周四. " ]

[ "2017年第15届湖北武汉创新杯五年级竞赛决赛第6题" ]

[ [ { "aoVal": "A", "content": "$$11$$米 " } ], [ { "aoVal": "B", "content": "$$22$$米 " } ], [ { "aoVal": "C", "content": "$$33$$米 " } ], [ { "aoVal": "D", "content": "$$44$$米 " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "$$176-660\\div 5=44\\text{m/min}$$． " ]

[ "2019年第24届YMO二年级竞赛决赛第3题3分" ]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$8$$ " } ], [ { "aoVal": "D", "content": "$$10$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->间隔问题->直线型两端都没有->剪绳子" ]

[ "根据题意分析可知，一根细绳对折两次后被分成了$$4$$段，从中间剪开后，分为两个一样长的短段和一个长段，长段的距离是短段的$$2$$倍， 所以细绳原来长：$$2+1+1=4$$（米）． 故选$$\\text{A}$$． " ]

[ "2021年第24届世界少年奥林匹克数学竞赛四年级竞赛复赛（中国区）第15题3分" ]

[ [ { "aoVal": "A", "content": "$$18$$ " } ], [ { "aoVal": "B", "content": "$$24$$ " } ], [ { "aoVal": "C", "content": "$$36$$ " } ], [ { "aoVal": "D", "content": "$$42$$ " } ] ]

[ "拓展思维->拓展思维->组合模块->组合模块最值问题->最值原理在几何中的应用" ]

[ "$$26\\div2=13$$（厘米），要使长方形的面积最大，它的长和宽最接近，$$7+6=13$$（厘米），$$7\\times6=42$$（平方厘米），所以长方形的面积最大是$$42$$平方厘米． 故答案为：$$\\text{D}$$． " ]

[ "2017年河南郑州联合杯竞赛第10题4分" ]

[ [ { "aoVal": "A", "content": "$$a\\textgreater b$$~ " } ], [ { "aoVal": "B", "content": "$$a\\textless{}b$$~~~~~~~ " } ], [ { "aoVal": "C", "content": "$$a=b$$~ " } ], [ { "aoVal": "D", "content": "无法判断 " } ] ]

[ "拓展思维->拓展思维->计算模块->比较与估算->比较大小综合->糖水原理法->加糖" ]

[ "比较大小；$$\\frac{1234567890}{2345678901}\\textgreater\\frac{1234567890-2016}{2345678901-2016}\\textgreater\\frac{1234567890-2017}{2345678901-2016}$$，则$$a\\textgreater b$$． " ]

[ "2003年六年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "正比例关系 " } ], [ { "aoVal": "B", "content": "反比例关系 " } ], [ { "aoVal": "C", "content": "不成比例关系 " } ], [ { "aoVal": "D", "content": "不能确定 " } ] ]

[ "拓展思维->拓展思维->计算模块->比和比例->比例->正比例与反比例" ]

[ "$$s=a{{l}^{2}}$$，其中a为块数，$${l}$$为边长，故不成比例 " ]

[ "2019年第23届广东世界少年奥林匹克数学竞赛五年级竞赛初赛第3题5分" ]

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

[ "拓展思维->能力->运算求解" ]

[ "$$32=2\\times2\\times2\\times2\\times2={{4}^{2}}\\times2$$，不是平方数． $$324={{18}^{2}}$$是平方数． $$32444={{2}^{2}}\\times8111$$不是平方数． $$324444={{2}^{2}}\\times3\\times19\\times1423$$不是平方数． $$3244444={{2}^{2}}\\times7\\times115873$$，而$$115873$$不含质因数$$7$$，故不是． $$32444444={{2}^{2}}\\times23\\times352657$$，而$$352657$$不含质因数$$23$$，故不是． $$324444444={{2}^{2}}\\times3\\times27037037$$，而$$27037037$$不含质因数$$37$$，故不是． 所以选择$$\\text{A}$$． " ]

[ "2017年第15届湖北武汉创新杯五年级竞赛决赛第8题" ]

[ [ { "aoVal": "A", "content": "$$2064$$ " } ], [ { "aoVal": "B", "content": "$$2068$$ " } ], [ { "aoVal": "C", "content": "$$7099$$ " } ], [ { "aoVal": "D", "content": "$$516$$ " } ] ]

[ "拓展思维->拓展思维->数论模块->位值原理与进制->位值原理运用->位值原理的综合应用" ]

[ "五个数两两相加再求和，每个数计算$$4$$次，则五个数的和为：$$2064\\div 4=516$$． " ]

[ "2020年广东广州羊排赛三年级竞赛第8题5分" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$20$$ " } ], [ { "aoVal": "C", "content": "$$15$$ " } ], [ { "aoVal": "D", "content": "$$5$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "一个工人$$3$$小时磨了$$60$$千克面粉， 所以$$1$$小时磨面粉$$60\\div3=20$$（千克）， 这个工人磨完$$200$$千克面粉，一共要$$200\\div20=10$$（时）， 所以答案为$$10$$小时． " ]

[ "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34题" ]

[ [ { "aoVal": "A", "content": "$$8$$ " } ], [ { "aoVal": "B", "content": "$$9$$ " } ], [ { "aoVal": "C", "content": "$$10$$ " } ], [ { "aoVal": "D", "content": "$$11$$ " } ] ]

[ "Overseas Competition->知识点->几何模块->直线型->图形认知->三角形->三角形三边关系", "拓展思维->能力->逻辑分析" ]

[ "To be the lengths of a triangle, the sum of two sides must be longer than the third side. Therefore, we just need to make the third number exactly the sum of the two former numbers, which is also the Fibonacci sequence. $$1$$，$$2$$，$$3$$，$$5$$，$$8$$，$$13$$，$$21$$，$$34$$，$$55$$，$$89$$. The last number to be chosen is $$89$$. A total of $$10$$ numbers. Choose $$C$$． " ]

[ "2013年华杯赛六年级竞赛初赛" ]

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

[ "拓展思维->拓展思维->应用题模块->工程问题->合作工程问题" ]

[ "解：三人一共搬了： $$\\left( 1+1 \\right)\\div \\left( \\frac{1}{10}+\\frac{1}{12}+\\frac{1}{15} \\right)$$ $$=2\\div \\frac{1}{4}$$ $$=8$$（天） 阳阳帮妈妈运的天数： $$\\left( 1-\\frac{1}{12}\\times 8 \\right)\\div \\frac{1}{15}$$ $$=\\frac{1}{3}\\times 15$$ $$=5$$（天） 答：阳阳帮妈妈运了$$5$$天。 故选：C。 " ]

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

[ "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": "$$150$$ " } ], [ { "aoVal": "B", "content": "$$140$$ " } ], [ { "aoVal": "C", "content": "$$125$$ " } ], [ { "aoVal": "D", "content": "$$120$$ " } ] ]

[ "拓展思维->拓展思维->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：$$(210+90)\\div 2=150$$ " ]

[ "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->能力->逻辑分析" ]

[ "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$$\\text{C}$$． " ]

[ "2006年第4届创新杯五年级竞赛复赛第10题" ]

[ [ { "aoVal": "A", "content": "$$26$$ " } ], [ { "aoVal": "B", "content": "$$28$$ " } ], [ { "aoVal": "C", "content": "$$30$$ " } ], [ { "aoVal": "D", "content": "$$24$$ " } ] ]

[ "拓展思维->能力->构造模型->模型思想" ]

[ "甲每到$$4$$的倍数就休息，而乙每到$$7$$的倍数和比$$7$$的倍数少一天都休息． 因为$$4$$和$$7$$的最小公倍数是$$28$$， 因为今年是平年， 所以在$$28$$的倍数体息的日子时：$$365\\div28\\approx 15$$（天）， 而在比$$7$$的倍数少一天休息时，甲乙第一次重逢的日子是第二十天，以后每隔$$28$$天就共同休息一天，$$365-20=345$$（天），$$345\\div28\\approx 12$$（天）， 所以甲乙两人共同休息的天数是$$15+12+1=28$$（天）． " ]

[ "2016年新希望杯六年级竞赛训练题（二）第4题" ]

[ [ { "aoVal": "A", "content": "$$11$$ " } ], [ { "aoVal": "B", "content": "$$12$$ " } ], [ { "aoVal": "C", "content": "$$13$$ " } ], [ { "aoVal": "D", "content": "$$14$$ " } ] ]

[ "拓展思维->拓展思维->行程模块->流水行船问题->基本流水行船问题->四个速度->基本行程" ]

[ "该船顺流行$$120$$千米，再逆流行$$72$$千米，共用时$$24$$小时；若逆流行$$40$$千米，再顺流行$$120$$千米，共用时$$16$$小时． 所以，逆流$$72-40=32$$千米，用时$$24-16=8$$小时，逆流船速是$$\\frac{32}{8}=4$$千米/时，则顺流船速$$40\\div \\left( 8-\\frac{24}{4} \\right)=20$$千米/时，静水船速是$$\\left( 4+20 \\right)\\div 2=12$$千米/时． " ]

[ "2017年全国小学生数学学习能力测评五年级竞赛复赛第10题3分" ]

[ [ { "aoVal": "A", "content": "$$5096303$$ " } ], [ { "aoVal": "B", "content": "$$5096304$$ " } ], [ { "aoVal": "C", "content": "$$5096305$$ " } ], [ { "aoVal": "D", "content": "$$5096306$$ " } ] ]

[ "拓展思维->拓展思维->数论模块->尾数特征->末一位数" ]

[ "因为两个相邻整数相乘，积的个位数字只可能是$$0$$，$$2$$，$$6$$之中的一个，可以直接排除$$\\text{A}$$、$$\\text{B}$$、$$\\text{C}$$，只留下$$\\text{D}$$． ", "<p>分解质因数后重新搭配：$$5096306=\\left( 2\\times 1129 \\right)\\times \\left( 37\\times 61 \\right)=2258\\times 2257$$．注意：遇到像这种数比较大的时候，一般都会想到尾数规律，进而排除不成立的选项，而不是第二种直接求出具体的两个数．</p>" ]

[ "2006年四年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "2 " } ], [ { "aoVal": "C", "content": "6 " } ], [ { "aoVal": "D", "content": "8 " } ] ]

[ "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" ]

[ "每4个3相乘，尾数是1.因$$2006=4\\times 501+2$$，所以2006个3相乘的尾数为$$3\\times 3=9$$;每4个7相乘，尾数为1，因$$100=4\\times 25$$，所以100个7相乘的尾数为1;故两个积的个位数字相减，结果为$$9-1=8$$，选D " ]

[ "2015年上海走美杯五年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$18$$ " } ], [ { "aoVal": "B", "content": "$$16$$ " } ], [ { "aoVal": "C", "content": "$$10$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->还原问题->两量还原问题" ]

[ "第一次取后还剩下$$\\left( 4-1 \\right)\\times 2=6$$（个），原来有$$\\left( 6+2 \\right)\\times 2=16$$（个）。 " ]

[ "2010年六年级竞赛创新杯", "2010年第8届创新杯六年级竞赛初赛第8题4分" ]

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

[ "拓展思维->拓展思维->应用题模块->列方程解应用题->列方程解应用题等量代换->等量代换加减型" ]

[ "甲和乙一天可以完成$$\\frac{5}{12}$$，乙和丙一天可以完成$$\\frac{4}{15}$$，甲和丙一天可以完成$$\\frac{7}{20}$$，则三个工程队一天一共可以完成$$\\frac{1}{2}\\times \\left( \\frac{5}{12}+\\frac{4}{15}+\\frac{7}{20} \\right)=\\frac{31}{60}$$，则甲工程队的效率为：$$\\frac{31}{60}-\\frac{4}{15}=\\frac{1}{4}$$，乙工程队的效率为：$$\\frac{5}{12}-\\frac{1}{4}=\\frac{1}{6}$$，丙工程队的效率为：$$\\frac{4}{15}-\\frac{1}{6}=\\frac{1}{10}$$．按要求，只有甲、乙工程队能在一星期内完成．甲、乙、丙$$3$$队一天的工程款是：$$\\left( 1800\\div 2\\frac{2}{5}+1500\\div 3\\frac{3}{4}+1600\\div 2\\frac{6}{7} \\right)\\div 2=855$$（元）. 甲工程队一天的工程款是:$$855-1500\\div 3\\frac{3}{4}=455$$（元），甲工程队总的工程款是:$$455\\times 4=1820$$（元）； 乙工程队一天的工程款是：$$855-1600\\div 2\\frac{6}{7}=295$$（元），乙工程队总的工程款是:$$295\\times 6=1770$$（元）．所以乙工程队所需费用最省，乙工程队被招标. ", "<p>甲乙每天$$1800\\div 2\\frac{2}{5}=1800\\div \\frac{12}{5}=750$$（元）；</p>\n<p>乙丙每天$$1500\\div 3\\frac{3}{4}=1500\\div \\frac{15}{4}=400$$（元）；</p>\n<p>甲丙每天$$1600\\div 2\\frac{6}{7}=1600\\div \\frac{20}{7}=560$$（元），</p>\n<p>甲乙丙每天：$$\\left( 750+400+560 \\right)\\div 2=1710\\div 2=855$$（元），</p>\n<p>则甲每天：$$855-400=455$$（元），</p>\n<p>乙每天：$$855-560=295$$（元），</p>\n<p>丙每天：$$855-750=105$$（元）．</p>\n<p>因为一星期完成任务，所以乙队最省．</p>\n<p>答：被招标的是乙工程队．</p>\n<p>故选$$\\text{B}$$．</p>\n" ]

[ "2017年新希望杯六年级竞赛训练题（一）第1题" ]

[ [ { "aoVal": "A", "content": "女生人数与全班人数的比是$$4:9$$，男生人数与女生人数的比是$$4:5$$． " } ], [ { "aoVal": "B", "content": "比的前项和后项同时乘一个相同的数，比值不变． " } ], [ { "aoVal": "C", "content": "最简单的整数比，就是比的前项和后项都是质数的比． " } ], [ { "aoVal": "D", "content": "如果$$A:B=C$$，那么是比的前项，$$A$$是比的后项． " } ] ]

[ "拓展思维->拓展思维->计算模块->比和比例->比->求比值" ]

[ "$$\\text{A}$$项，男生人数与女生人数的比是$$5:4$$．$$\\text{B}$$项，同时乘的数不能是$$0$$．$$\\text{C}$$项，最简比的前后项是互质数，不一定都是质数． " ]

[ "2008年第6届创新杯六年级竞赛初赛A卷第1题5分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{7}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{2}{15}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{1}{8}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{15}{112}$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->比较与估算->比较大小综合->分数通分法->通分子" ]

[ "由于$$\\frac{1}{7}=\\frac{2}{14}\\textgreater\\frac{2}{15}\\textgreater\\frac{2}{16}=\\frac{1}{8}$$，$$\\frac{15}{112}\\textgreater\\frac{14}{112}=\\frac{1}{8}$$，所以$$\\frac{1}{8} \\textless{} \\frac{1}{7}$$，$$\\frac{1}{8} \\textless{} \\frac{2}{15}$$，$$\\frac{1}{8} \\textless{} \\frac{15}{112}$$因此，这些分数中最小的是$$\\frac{1}{8}$$． " ]

[ "2019年广东广州羊排赛六年级竞赛第10题3分" ]

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

[ "拓展思维->拓展思维->数论模块->余数问题->余数的性质->余数找规律" ]

[ "根据余数的性质，和的余数等于余数的和， $$1$$，$$1$$，$$2$$，$$3$$，$$5$$，$$8$$，$$13$$，$$21$$，$$\\cdots $$，$$\\div 4$$余$$1$$，$$1$$，$$2$$，$$3$$，$$1$$，$$0$$，$$1$$，$$1$$，$$2$$，$$3$$，$$1$$，$$0$$， $$1$$，$$1$$，$$\\cdots $$，$$6$$个为一周期，$$2019\\div 6=336$$（组）$$\\cdots \\cdots $$$$3$$（个）， 为周期的第$$3$$个，余数为$$2$$． " ]

[ "2017年河南郑州豫才杯四年级竞赛初赛第3题" ]

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

[ "拓展思维->拓展思维->计算模块->单位换算->面积单位换算" ]

[ "（$$1$$）我国东北三省总面积是$$78$$万平方千米，所以错了；（$$2$$）$$100$$万次$$1$$心跳是 一个正常人$$9.9$$天心跳的次数；（$$3$$）$$1$$层楼大概有$$3$$米，$$3$$层楼的高度大约有$$9$$米，$$1$$亿张$$A4$$纸摞起来的高度相当于$$9$$米；（$$4$$）光的速度大约是每秒$$3$$亿米每秒，所以错了；综上所述，（$$1$$）和（$$4$$）错了．~~~~ " ]

[ "2011年全国希望杯四年级竞赛初赛第20题" ]

[ [ { "aoVal": "A", "content": "$$DEBAC$$ " } ], [ { "aoVal": "B", "content": "$$CABDE$$ " } ], [ { "aoVal": "C", "content": "$$DCBAE$$ " } ], [ { "aoVal": "D", "content": "$$CADBE$$ " } ] ]

[ "拓展思维->拓展思维->组合模块->逻辑推理->假设型逻辑推理->真假话->半真半假", "课内体系->知识点->数学广角->推理->假设法判断真假" ]

[ "假设法． 第一步：假设甲说的前半句是真的，那么$$D$$是第$$1$$名， 那么此时丙说的前半句错，后半句对．则$$A$$是第$$4$$名． 同理乙的后半句对，$$C$$是第$$4$$名．矛盾． 由此至甲的后半句对． 第二步：已知$$E$$是第$$5$$名，$$D$$不是第$$1$$名． 和第一名有关的话只剩下丁说的，设$$C$$是第$$1$$名． 则戊：``第$$2$$名是$$C$$，第$$4$$名是$$B$$''．可知前错后对，$$B$$是第$$4$$名． 且有乙：``第二名是$$A$$，第四名是$$C$$''．可知，$$A$$是第$$2$$名． $$D$$是第$$3$$名． " ]

[ "2006年第4届创新杯五年级竞赛复赛第8题", "2006年五年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "15 " } ], [ { "aoVal": "B", "content": "18 " } ], [ { "aoVal": "C", "content": "20 " } ], [ { "aoVal": "D", "content": "24 " } ] ]

[ "拓展思维->拓展思维->计数模块->枚举法综合->字典排序法->非数字排列" ]

[ "$$7=1+1+1+4$$有$$4$$种放法. $$7=1+1+2+3$$有$$12$$种放法. $$7=2+2+2+1$$有$$4$$种放法. 共有$$4+12+4=20$$种不同的放法. " ]

[ "2015年第14届春蕾杯一年级竞赛初赛第5题1分" ]

[ [ { "aoVal": "A", "content": "小王 " } ], [ { "aoVal": "B", "content": "小胖 " } ], [ { "aoVal": "C", "content": "小朱 " } ], [ { "aoVal": "D", "content": "园园 " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "根据题意分析可知，小胖比小王高，所以小胖的身高高于小王，小胖说比园园矮，所以园园的身高高于小胖，小朱没有小王高，所以小王的身高高于小朱，由此可知，四个人的身高从高到低位：园园$$\\textgreater$$小胖$$\\textgreater$$小王$$\\textgreater$$小朱． 故选$$\\text{D}$$． " ]

[ "2017年全国华杯赛竞赛初赛模拟题1第4题" ]

[ [ { "aoVal": "A", "content": "$$6$$ " } ], [ { "aoVal": "B", "content": "$$8$$ " } ], [ { "aoVal": "C", "content": "$$9$$ " } ], [ { "aoVal": "D", "content": "$$10$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "分类计数 用$$1$$种颜色染色，则有$$2$$种不同的染色方式； 用$$2$$种颜色染色，再分类如下计数： （$$1$$）仅$$1$$个面是红色，有$$1$$种不同的染色方式； （$$2$$）仅$$2$$个面是红色，相对和相邻，有$$2$$种不同的染色方式； （$$3$$）仅$$3$$个面是红色，其中有$$2$$个相对，经适当转动，固定为红色上底面和红色下底面，仅有$$1$$种不同的染色方式：$$3$$个红色面两两相邻，仅有$$1$$种不同的染色方式； （$$4$$）有$$4$$个红色面，即有$$2$$个黄色面，不同染色方式的个数同（$$2$$）； （$$5$$）有$$5$$个红色面，即有$$1$$个黄色面，不同染色方式的个数同（$$1$$）． 共有$$10$$种不同的染色方式． " ]

[ "2018年湖北武汉创新杯小学高年级六年级竞赛初赛数学思维能力等级测试第1题4分" ]

[ [ { "aoVal": "A", "content": "甲获利润多 " } ], [ { "aoVal": "B", "content": "乙获利润多 " } ], [ { "aoVal": "C", "content": "两人利润相同 " } ], [ { "aoVal": "D", "content": "无法比较谁多 " } ] ]

[ "拓展思维->思想->赋值思想" ]

[ "假设成本为$$100$$元，则甲的利润$$ =100\\times 20 \\%\\times 15 =300$$（元），乙的利润$$=100\\times 15 \\%\\times 24=360$$（元）． " ]

[ "2012年IMAS小学中年级竞赛第一轮检测试题第10题3分" ]

[ [ { "aoVal": "A", "content": "$$12$$ " } ], [ { "aoVal": "B", "content": "$$20$$ " } ], [ { "aoVal": "C", "content": "$$23$$ " } ], [ { "aoVal": "D", "content": "$$25$$ " } ], [ { "aoVal": "E", "content": "$$27$$ " } ] ]

[ "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->年龄和" ]

[ "姐姐与妹妹的年龄差是$$4$$岁，当两人年龄和为$$50$$岁时，妹妹的年龄等于$$(50-4)\\div 2=23$$（岁）． " ]

[ "2014年迎春杯四年级竞赛复赛" ]

[ [ { "aoVal": "A", "content": "$$540$$ " } ], [ { "aoVal": "B", "content": "$$600$$ " } ], [ { "aoVal": "C", "content": "$$630$$ " } ], [ { "aoVal": "D", "content": "$$650$$ " } ] ]

[ "拓展思维->拓展思维->组合模块->数字谜->巧填算符" ]

[ "解：由于$$25+9+7+5+3+1=50$$，所以我们猜测$$0\\sim 50$$之间的所有偶数都有可能得到， $$0\\sim 50$$所有偶数的总和是$$(0+50)\\times 26\\div 2=650$$； 当把$$1$$前面的$$+$$号变成$$-$$号，可得$$25+9+7+5+3-1=48$$，比$$50$$小$$1\\times 2$$， 当把$$3$$前面的$$+$$号变成$$-$$号，可得$$25+9+7+5-3+1=44$$，比$$50$$小$$3\\times 2$$， 当把$$3$$和$$1$$前面的$$+$$号变成$$-$$号，可得$$25+9+7+5-3-1=42$$，比$$50$$小$$4\\times 2$$， 当把$$5$$前面的$$+$$号变成$$-$$号，可得$$25+9+7-5+3+1=40$$，比$$50$$小$$5\\times 2$$， $$\\cdots \\cdots $$ $$22=1+5+7+9$$，因此当把$$1$$，$$5$$，$$7$$，$$9$$前面的$$+$$号变成$$-$$号，可得$$25-9-7-5+3-1=6$$， $$24=3+5+7+9$$，因此当把$$3$$，$$5$$，$$7$$，$$9$$前面的$$+$$号变成$$-$$号，可得$$25-9-7-5-3+1=2$$， $$25=1+3+5+7+9$$，因此当把$$1$$，$$3$$，$$5$$，$$7$$，$$9$$前面的$$+$$号变成$$-$$号，可得$$25-9-7-5-3-1=0$$， 根据上述规律可得，但是数字$$2$$和$$23$$无法凑出来，那么偶数$$4$$和$$46$$无法取到， 所以答案是：$$650-4-46=600$$。 故选：B。 " ]

[ "2011年第9届全国创新杯小学高年级六年级竞赛第18题" ]

[ [ { "aoVal": "A", "content": "$$16.4$$ " } ], [ { "aoVal": "B", "content": "$$16.5$$ " } ], [ { "aoVal": "C", "content": "$$16.6$$ " } ], [ { "aoVal": "D", "content": "$$17.5$$ " } ] ]

[ "拓展思维->七大能力->逻辑分析" ]

[ "可用方程解决，假设原来一等奖的平均分为$$a$$，二等奖的平均分为$$b$$，根据题意可以得到：$$10a+20b=6\\left( a+3 \\right)+24\\left( b+2 \\right)$$，得$$a-b=16.5$$． " ]

[ "2016年新希望杯小学高年级六年级竞赛训练题（六）第6题" ]

[ [ { "aoVal": "A", "content": "$$1000.0$$ " } ], [ { "aoVal": "B", "content": "$$1063.8$$ " } ], [ { "aoVal": "C", "content": "$$1276.6$$ " } ], [ { "aoVal": "D", "content": "$$1333.3$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->分百应用题->认识单位1" ]

[ "全年打印纸总共花了$$20000\\times \\frac{3}{4}=15000$$元，设第一个月打印纸花了$$x$$元，则全年$$12x+\\frac{\\left( 0.1+0.6 \\right)\\times 6}{2}x=15000$$，解得：$$x=\\frac{150000}{141}$$． 那么$$8$$月份用了$$x\\times 120 \\% =\\frac{180000}{141}=1276.6$$元． " ]

[ "小学高年级六年级其它2014年数学思维能力等级测试第4题", "2014年第12届全国创新杯六年级竞赛第4题" ]

[ [ { "aoVal": "A", "content": "$$70$$ " } ], [ { "aoVal": "B", "content": "$$71$$ " } ], [ { "aoVal": "C", "content": "$$85$$ " } ], [ { "aoVal": "D", "content": "$$87$$ " } ] ]

[ "拓展思维->思想->方程思想" ]

[ "设每册$$n$$道题，则有 $$\\left { \\begin{matrix}351\\textgreater4n 689\\textless{}8n \\end{matrix} \\right.\\Rightarrow \\left { \\begin{matrix}n\\textless{}87\\frac{3}{4} n\\textgreater86\\frac{1}{8} \\end{matrix} \\right.$$ 故$$n$$只能为$$87$$． " ]

[ "2016年全国中环杯四年级竞赛初赛第11题" ]

[ [ { "aoVal": "A", "content": "$$S\\textgreater R\\textgreater T\\textgreater P\\textgreater Q$$ " } ], [ { "aoVal": "B", "content": "$$S\\textgreater T\\textgreater R\\textgreater Q\\textgreater P$$ " } ], [ { "aoVal": "C", "content": "$$S\\textgreater R\\textgreater T\\textgreater Q\\textgreater P$$ " } ], [ { "aoVal": "D", "content": "$$S\\textgreater T\\textgreater R\\textgreater P\\textgreater Q$$ " } ] ]

[ "知识标签->拓展思维->计算模块->方程基础->不等式->不等式组求解" ]

[ "$$Q+T=800$$①； $$Q+R=900$$②；$$P+T=700$$③；$$Q+S\\geqslant 1000$$④；$$R+T\\geqslant 1000$$⑤；由①②得：$$R\\textgreater T$$；由①③得：$$Q\\textgreater P$$；由②④得：$$S\\textgreater R$$；由②⑤得：$$T\\textgreater Q$$；所以：$$S\\textgreater R\\textgreater T\\textgreater Q\\textgreater P$$． " ]

[ "2008年第6届创新杯六年级竞赛复赛第10题4分", "2008年六年级竞赛创新杯" ]

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

[ "拓展思维->拓展思维->组合模块->操作与策略->操作问题" ]

[ "用符号$$a\\to b$$表示``擦去原数$$a$$，改写为$$b$$''的一次操作. 如:$$5\\to 2$$(擦1次就可得到2); $${{2}^{3}}=8\\to 3\\to 2$$(擦2次就可得到2); $${2\\times 2\\times 3\\times 5\\times 7}=420\\to 8={{2}^{3}}\\to 3\\to 2$$(擦3次就可得到2). 再试几次，没有出现需要擦4次或更多次才出现2的情况，因此初步断定选B. 严格证明如下: （1）如果黑板上最初写的是奇数，只要擦1次就可出现2. （2）如果黑板上最初写的是偶数，擦过1次，改写的数如果是奇数，显然再擦1次就可出现2(共擦2次即可). （3）如果黑板上最初写的是偶数，擦过1次，改写的数如果还是偶数，那么可以证明这个偶数不可能有奇质数因子，即这个改写的数为$${{2}^{k}}$$($$k\\geqslant 1$$)，从而第2次改写得3，第3次改写得2. 事实上，设最初写的偶数为$$a$$，改写的偶数为$$b$$，则1，2，3，\\ldots，$$b-1$$都是$$a$$的因数，而$$b$$不能整除$$a$$，假设$$b$$有奇质数因子，则$$b={{2}^{k}}{{p}_{1}}^{{{a}_{1}}}\\cdots {{p}_{s}}^{{{a}_{s}}}$$为质因数分解式，其中$${{p}_{1}} \\textless{} {{p}_{2}}\\cdots \\textless{} {{p}_{s}}$$，均为奇质数，$$s\\geqslant 1$$，$$k\\geqslant 1$$.显然，$$1 \\textless{} {{2}^{k}} \\textless{} b$$，$$1 \\textless{} {{p}_{i}}^{{{a}_{i}}} \\textless{} b$$($$i=1$$，2\\ldots，$$s$$)，所以$${{2}^{k}}\\textbar a$$，$${{p}_{i}}^{{{a}_{i}}}\\textbar a$$($$i=1$$，2，\\ldots，$$s$$)，从而$${{2}^{k}}p_{1}^{{{a}_{1}}}\\cdots {{p}_{s}}^{{{a}_{s}}}\\textbar a$$，即$$b\\textbar a$$，与$$b$$不能整除$$a$$矛盾，故$$b$$没有奇质数因子. " ]

[ "2009年五年级竞赛创新杯", "2009年四年级竞赛创新杯", "2009年三年级竞赛创新杯", "2009年六年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "24种 " } ], [ { "aoVal": "B", "content": "36种 " } ], [ { "aoVal": "C", "content": "48种 " } ], [ { "aoVal": "D", "content": "60种 " } ] ]

[ "拓展思维->拓展思维->计数模块->排列组合->排列组合综合" ]

[ "因为翻译可以从除甲外的4人中选1个担当，有4种方法:又导游、保洁可从余下的4人中选2个担当，有$$4\\times 3=12$$种选法，所以有$$4\\times 12=48$$种选派方案. " ]

[ "2017年河南郑州东风杯竞赛初赛" ]

[ [ { "aoVal": "A", "content": "■~~~~~ " } ], [ { "aoVal": "B", "content": "◇~~~~~~~~~~~~ " } ], [ { "aoVal": "C", "content": "●~~~~~ " } ], [ { "aoVal": "D", "content": "无法确定 " } ] ]

[ "拓展思维->能力->图形认知" ]

[ "观察可知，■◇◇●●●，$$6$$个图形一个循环周期，$$40\\div 6=6\\cdots \\cdots4$$，所以第$$40$$个图形是一个周期中的第$$4$$个，是●． " ]

[ "2017年第15届全国希望杯六年级竞赛" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{10}{11}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{10}{22}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{10}{111}$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->小数->循环小数->循环小数混合运算" ]

[ "原式$$=2\\frac{16}{999}\\times \\frac{1}{20\\frac{16-1}{90}+2\\frac{19}{90}}$$ $$=\\frac{2\\times 999+16}{999}\\times \\frac{1}{22\\frac{34}{90}}$$ $$=\\frac{2000+16-2}{999}\\times \\frac{90}{2014}$$ $$=\\frac{10}{111}$$ " ]