Datasets:

---

math-eval

/

TAL-SCQ5K

like 57

[ "2017年第20届世界少年奥林匹克数学竞赛四年级竞赛决赛（中国区）第2题3分" ]

[ [ { "aoVal": "A", "content": "减少$$14$$ " } ], [ { "aoVal": "B", "content": "减少$$3$$ " } ], [ { "aoVal": "C", "content": "增加$$3$$ " } ], [ { "aoVal": "D", "content": "增加$$14$$ " } ] ]

[ "拓展思维->思想->转化与化归的思想" ]

[ "两数相减，当减数增加$$7$$，被减数不变则差减少了$$7$$． 又因为要使得差增加$$7$$， 所以被减数应增加：$$7+7=14$$， 故选$$\\text{D}$$． " ]

[ "2020年亚洲国际数学奥林匹克公开赛（AIMO）二年级竞赛决赛第23题8分" ]

[ [ { "aoVal": "A", "content": "$$198$$ " } ], [ { "aoVal": "B", "content": "$$0$$ " } ], [ { "aoVal": "C", "content": "$$8$$ " } ], [ { "aoVal": "D", "content": "$$1990$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->整数->整数提取公因数->整数乘法巧算之提取公因数（普通型）" ]

[ "原式$$\\begin{eqnarray}\\&=\\&199\\times \\left( 11-2-8-1 \\right)\\end{eqnarray}$$ $$\\begin{eqnarray} \\&=\\&199\\times 0 \\&=\\&0\\end{eqnarray}$$ " ]

[ "其它改编自2015年全国希望杯六年级竞赛初赛第1题" ]

[ [ { "aoVal": "A", "content": "$$\\frac{31}{32}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{15}{16}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{7}{8}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{3}{4}$$ " } ] ]

[ "拓展思维->能力->数据处理" ]

[ "原式$$=\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{8}+\\frac{1}{16}+\\left( \\frac{1}{32}+\\frac{1}{32} \\right)-\\frac{1}{32}$$$$=1-\\frac{1}{32}=\\frac{31}{32}$$． " ]

[ "2004年六年级竞赛创新杯", "2004年第2届创新杯六年级竞赛复赛第9题" ]

[ [ { "aoVal": "A", "content": "$$2:5:8$$ " } ], [ { "aoVal": "B", "content": "$$4:5:6$$ " } ], [ { "aoVal": "C", "content": "$$3:5:7$$ " } ], [ { "aoVal": "D", "content": "$$5:6:7$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->浓度问题->浓度基本题型" ]

[ "设取Ⅰ号溶液$$6x$$升，那么一号溶液中有柠檬汁$$x$$升，油$$2x$$升，醋$$3x$$升.同样设取Ⅱ号溶液$$12y$$升，那么Ⅱ号溶液中有柠檬汁$$3y$$升，油$$4y$$升，醋$$5y$$升。现在可以用排除法对四个选项进行分析，为了便于描述，现设混合后三种液体的比为$$a:b:c$$.。 我们先观察选项$$A$$，由$$2(x+3y)=2x+6y\\textgreater2x+4y$$得$$2a\\textgreater b$$，而选项$$A$$中的$$2\\times 2 \\textless{} 5$$，所以选项$$A$$不能被调出； 再来观察选项$$B$$，$$(2x+4y)-(x+3y)=x+y$$，由$$4(x+y)=4x+4y\\textgreater2x+4y$$得$$4(b-a)\\textgreater b$$即$$3b\\textgreater4a$$，而选项$$B$$中的$$3\\times 5 \\textless{} 4\\times 4$$，所以选项$$B$$不能被调出；同理选项$$D$$中的$$3\\times 6 \\textless{} 4\\times 5$$，选项$$D$$也不能被调出； 那么能被调出的只有选项$$C$$。实际上，当$$x=3$$，$$y=1$$时，混合后三种液体的比正好为$$3:5:7$$，选$$C$$.。 " ]

[ "2018年河南郑州K6联赛竞赛初赛第23题3分" ]

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

[ "知识标签->拓展思维->计数模块->统计与概率->概率->基本概率->可能的情况" ]

[ "可能性；要想红色朝上次数最少，则$$1$$面涂红色，$$2$$面染黄色． " ]

[ "2014年迎春杯三年级竞赛复赛" ]

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

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

[ "解：枚举法，$$10$$能变成哪三个数相加，$$10=1+1+8=1+2+7=1+3+6=1+4+5=2+2+6=2+3+5=2+4+4=3+3+4$$，共$$8$$种。 故选:C。 " ]

[ "2017年第20届世界少年奥林匹克数学竞赛四年级竞赛初赛（中国区）第5题5分" ]

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

[ "拓展思维->能力->逻辑分析->代数逻辑推理" ]

[ "$2$枚$$5$$角；$$1$$枚$$5$$角$$+2$$枚$$2$$角$$+1$$枚$$1$$角；$$1$$枚$$5$$角$$+1$$枚$$2$$角$$+3$$枚$$1$$角；$$1$$枚$$5$$角$$+5$$枚$$1$$角；$$5$$枚$$2$$角；$$4$$枚$$2$$角$$+2$$枚$$1$$角；$$3$$枚$$2$$角$$+4$$枚$$1$$角；一共是$$7$$种情况． " ]

[ "2005年第3届创新杯五年级竞赛初赛第2题" ]

[ [ { "aoVal": "A", "content": "$$8$$分 " } ], [ { "aoVal": "B", "content": "$$7.5$$分 " } ], [ { "aoVal": "C", "content": "$$7$$分 " } ], [ { "aoVal": "D", "content": "$$6.5$$分 " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "数学、语文、英语三科的总成绩是$$92\\times 3=276$$分， 语文、英语两科的总成绩是$$90\\times 2=180$$分， 那么数学的得分为$$276-180=96$$分， 又英语比语文高$$3$$分， 那么语文的成绩为$$\\left( 180-3 \\right)\\div 2=88.5$$分， 所以数学比语文高$$96-88.5=7.5$$分． 故选$$\\text{B}$$． " ]

[ "2020年新希望杯二年级竞赛初赛（团战）第12题" ]

[ [ { "aoVal": "A", "content": "第一名 " } ], [ { "aoVal": "B", "content": "第二名 " } ], [ { "aoVal": "C", "content": "第三名 " } ], [ { "aoVal": "D", "content": "第四名 " } ] ]

[ "拓展思维->能力->构造模型->模型思想" ]

[ "小明后面有$$5$$人，前面有$$4$$人，则他此时排在第$$5$$，之后，没有人超过他，小明又超过了$$3$$人后，他此时就是第二名． " ]

[ "2018年湖北武汉新希望杯六年级竞赛训练题（五）第4题" ]

[ [ { "aoVal": "A", "content": "$$11101$$ " } ], [ { "aoVal": "B", "content": "$$1011$$ " } ], [ { "aoVal": "C", "content": "$$10101$$ " } ], [ { "aoVal": "D", "content": "$$11001$$ " } ] ]

[ "拓展思维->拓展思维->数论模块->位值原理与进制->进制的性质与应用->进制间的互化" ]

[ "短除法，倒取余数 " ]

[ "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6题3分" ]

[ [ { "aoVal": "A", "content": "$$11$$ " } ], [ { "aoVal": "B", "content": "$$18$$ " } ], [ { "aoVal": "C", "content": "$$19$$ " } ], [ { "aoVal": "D", "content": "$$21$$ " } ] ]

[ "拓展思维->拓展思维->组合模块->操作与策略->统筹规划->简单时间统筹问题->同时进行问题" ]

[ "从题干可以知道，小明早晨起床，要完成这几件事：起床穿衣$$3$$分钟，刷牙洗脸$$5$$分钟，在火炉上烧水煮面要$$10$$分钟，吃面要$$8$$分钟，整理房间$$5$$分钟． 为了最优化时间，先起床穿衣，再在火炉上烧水煮面，煮面的过程中刷牙洗脸、整理房间；最后吃面； $$3+10+8=21$$（分钟）； 即为了尽快做完这些事，最少用$$21$$分钟． 故选：$$\\text{D}$$． " ]

[ "2013年第11届全国创新杯五年级竞赛第6题5分" ]

[ [ { "aoVal": "A", "content": "$$1$$与$$101$$之间的$$2$$倍数 " } ], [ { "aoVal": "B", "content": "$$1$$与$$101$$之间的$$3$$倍数 " } ], [ { "aoVal": "C", "content": "$$1$$与$$101$$之间的$$4$$倍数 " } ], [ { "aoVal": "D", "content": "$$1$$与$$101$$之间的$$6$$倍数 " } ] ]

[ "知识标签->学习能力->七大能力->逻辑分析" ]

[ "一个等差数串的平均数，是这个数串首项与末项的平均数，简化计算后易知$$1$$与$$101$$之间$$4$$的倍数的平均数最大，其值为$$(4+100)\\div 2=52$$． " ]

[ "2018年第22届广东世界少年奥林匹克数学竞赛三年级竞赛决赛第5题5分" ]

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

[ "拓展思维->能力->运算求解" ]

[ "已知一条毛毛虫由幼虫长成成虫，每天长大一倍，$$30$$天能长到$$20$$厘米，可根据最后结果出发，逐步向前一步一步推理； 按从后往前推理可知最后一天长的是前一天的$$2$$倍，最后一天是$$20$$厘米，长到$$10$$厘米的时候是$$30$$天$$-1$$天； 根据以上分析可得$$29$$天时是$$10$$厘米，由此可推出长到$$5$$厘米时是几天，进而得出答案． $$30-1-1=28$$（天）， 答：长到$$5$$厘米时要用$$28$$天． " ]

[ "2019年广东深圳全国小学生数学学习能力测评四年级竞赛初赛第9题3分" ]

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

[ "拓展思维->能力->运算求解" ]

[ "每天长一倍，第$$3$$天是$$32$$平方米，那么第$$4$$天，就应该是$$32\\times 2=64$$平方米． 故选$$\\text{A}$$． " ]

[ "2016年IMAS小学高年级竞赛第二轮检测试题第2题4分" ]

[ [ { "aoVal": "A", "content": "$$b\\textless{}d\\textless{}a\\textless{}c$$ " } ], [ { "aoVal": "B", "content": "$$d\\textless{}b\\textless{}a\\textless{}c$$ " } ], [ { "aoVal": "C", "content": "$$d\\textless{}a\\textless{}b\\textless{}c$$ " } ], [ { "aoVal": "D", "content": "$$d\\textless{}b\\textless{}c\\textless{}a$$ " } ], [ { "aoVal": "E", "content": "$$b\\textless{}d\\textless{}c\\textless{}a$$ " } ] ]

[ "知识标签->学习能力->七大能力->逻辑分析" ]

[ "因为$$\\frac{1}{a-2013}=\\frac{1}{b+2014}=\\frac{1}{c-2015}+\\frac{1}{d+2016}$$， 所以$$a-2013=b+2014=c-2015=d+2016$$， 所以$$d\\textless{}b\\textless{}a\\textless{}c$$． " ]

[ "2018年湖北武汉新希望杯五年级竞赛训练题（一）第6题" ]

[ [ { "aoVal": "A", "content": "广西人 " } ], [ { "aoVal": "B", "content": "湖南人 " } ], [ { "aoVal": "C", "content": "四川人 " } ], [ { "aoVal": "D", "content": "江苏人 " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "从层数看，丁\\textgreater 甲\\textgreater 丙\\textgreater 乙，江苏\\textgreater 广西\\textgreater 四川\\textgreater 湖南，所以甲是广西人． " ]

[ "2012年第8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": "$$B\\textgreater A\\textgreater C\\textgreater D$$ " } ], [ { "aoVal": "B", "content": "$$B\\textgreater C\\textgreater A\\textgreater D$$ " } ], [ { "aoVal": "C", "content": "$$C\\textgreater B\\textgreater A\\textgreater D$$ " } ], [ { "aoVal": "D", "content": "$$A\\textgreater B\\textgreater D\\textgreater C$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "根据题意有$$B$$和$$C$$最重，第二重的是$$A$$和$$B$$，因此$$B\\textgreater C\\textgreater A\\textgreater D$$． " ]

[ "2017年全国亚太杯五年级竞赛初赛第24题" ]

[ [ { "aoVal": "A", "content": "$$1440$$ " } ], [ { "aoVal": "B", "content": "$$1340$$ " } ], [ { "aoVal": "C", "content": "$$1250$$ " } ], [ { "aoVal": "D", "content": "$$1025$$ " } ] ]

[ "知识标签->数学思想->转化与化归的思想" ]

[ "第一次相遇甲和乙共走了半圈，此时乙走了$$250$$ 米． 而第二次相遇，甲和乙一共走了$$3$$个半圈，所以此时乙走了$$250\\times 3=750$$米． 因为第二次相遇在甲走完一周前$$80$$米处，所以半圈的路线长为$$750-80=670$$米． 所以此圆形场地的周长为$$670\\times 2=1340$$米． " ]

[ "2017年河南郑州豫才杯六年级竞赛初赛第5题" ]

[ [ { "aoVal": "A", "content": "$$40$$ " } ], [ { "aoVal": "B", "content": "$$45$$ " } ], [ { "aoVal": "C", "content": "$$50$$ " } ], [ { "aoVal": "D", "content": "$$55$$ " } ] ]

[ "拓展思维->思想->逐步调整思想" ]

[ "比例解行程；相等的时间内，甲跑了$$200$$米时，乙跑了$$160$$米，甲乙的速度比是$$200:160=5:4$$，要使两人同时到达终点，乙跑了$$200$$米的时间内甲要跑$$200\\div 4\\times 5=250$$米，所以同学甲的起跑线应后退$$250-200=50$$米． " ]

[ "2015年IMAS小学中年级竞赛第二轮检测试题第2题4分" ]

[ [ { "aoVal": "A", "content": "$$3735$$ " } ], [ { "aoVal": "B", "content": "$$605$$ " } ], [ { "aoVal": "C", "content": "$$615$$ " } ], [ { "aoVal": "D", "content": "$$625$$ " } ] ]

[ "知识标签->学习能力->七大能力->运算求解" ]

[ "由题意，发现规律$$a\\otimes b=a+aa+aaa+\\cdots +\\underbrace{aa\\cdots a}_{b}$$，则$$5\\otimes 3=5+55+555=615$$，故选$$\\text{C}$$． " ]

[ "2018年美国数学大联盟杯五年级竞赛初赛第13题5分" ]

[ [ { "aoVal": "A", "content": "$$111111$$ " } ], [ { "aoVal": "B", "content": "$$222222$$ " } ], [ { "aoVal": "C", "content": "$$444444$$ " } ], [ { "aoVal": "D", "content": "$$888888$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "在两个连续整数中，其中一个是双数，一个是单数．一个双数和一个单数的和一定是一个单数，所以两个连续整数的和也定是单数． 在$$\\text{A}$$、$$\\text{B}$$、$$\\text{C}$$、$$\\text{D}$$，$$4$$个选项中，只有$$\\text{A}$$是单数，其余$$3$$个都是双数． 故答案选：$$\\text{A}$$ " ]

[ "2017年第4届广东深圳鹏程杯小学高年级竞赛第3题6分" ]

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

[ "知识标签->课内知识点->数与运算->除法->整数除法->带余除法" ]

[ "根据余数的可减性，这个数可以同时整除$$90-69=21$$和$$125-90=35$$，所以这个数能整除$$（21,35）=7$$，即这个数是$$7$$的因数，而$$7$$的因数有$$1$$和$$7$$，又因为这个数大于$$1$$，所以经检验，这个整数$$m$$为$$7$$． " ]

[ "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->思想->对应思想" ]

[ "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->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" ]

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

[ "2009年第7届创新杯六年级竞赛初赛第3题4分" ]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$8$$ " } ], [ { "aoVal": "C", "content": "$$12$$ " } ], [ { "aoVal": "D", "content": "$$6$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "$$P$$和$$Q$$一天可以完成$$\\frac{1}{2}$$，$$Q$$和$$R$$一天可以完成$$\\frac{1}{4}$$，$$P$$和$$R$$一天可以完成$$\\frac{1}{2.4}=\\frac{5}{12}$$，则三人一天一共可以完成$$\\frac{1}{2}\\times \\left( \\frac{1}{2}+\\frac{1}{4}+\\frac{5}{12} \\right)=\\frac{7}{12}$$，$$P$$的工作效率为$$\\frac{7}{12}-\\frac{1}{4}=\\frac{1}{3}$$，因此$$P$$单独做需要$$3$$天完成. " ]

[ "2005年六年级竞赛创新杯", "2005年第3届创新杯六年级竞赛初赛第1题" ]

[ [ { "aoVal": "A", "content": "一定是偶数 " } ], [ { "aoVal": "B", "content": "一定是质数 " } ], [ { "aoVal": "C", "content": "一定是合数 " } ], [ { "aoVal": "D", "content": "可能是偶数，也可能是质数，也可能是合数 " } ] ]

[ "课内体系->知识模块->数与代数", "拓展思维->拓展思维->数论模块->质数与合数" ]

[ "$$3+5=8$$，$$8$$为偶数也为合数；$$2+3=5$$，$$5$$为质数，故选D。 " ]

[ "2010年五年级竞赛明心奥数挑战赛", "2010年六年级竞赛明心奥数挑战赛" ]

[ [ { "aoVal": "A", "content": "$$9.2$$ " } ], [ { "aoVal": "B", "content": "$$8.17$$ " } ], [ { "aoVal": "C", "content": "$$8.7$$ " } ], [ { "aoVal": "D", "content": "$$9.21$$ " } ], [ { "aoVal": "E", "content": "$$8.71$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->比较与估算->比较大小综合->小数比较大小" ]

[ "$$9.21\\textgreater9.2\\textgreater9\\textgreater8.71\\textgreater8.7\\textgreater8.17$$，$$9.2-9=0.2 \\textless{} 9-8.71=0.29$$。 " ]

[ "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": "$$2$$；$$100$$ " } ], [ { "aoVal": "B", "content": "$$4$$；$$100$$ " } ], [ { "aoVal": "C", "content": "$$8$$；$$100$$ " } ], [ { "aoVal": "D", "content": "$$8$$；$$200$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "若通过$$860$$米隧道时速度不变则需要$$120\\times 2=240$$（秒），火车过桥速度：$$\\left( 860-320 \\right)\\div \\left( 240-105 \\right)=4$$（米/秒）；火车车身长：$$105\\times 4-320=100$$（米）． " ]

[ "2019年全国小学生数学学习能力测评四年级竞赛复赛第10题3分" ]

[ [ { "aoVal": "A", "content": "像山一样高 " } ], [ { "aoVal": "B", "content": "像一栋房子高 " } ], [ { "aoVal": "C", "content": "像一个人一样高 " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "$$2$$的十次方为$$1024$$，$$2$$的五次方为$$32$$， 所以$$0.1\\times1024\\times1024\\times32=3355443.2$$毫米$$=3355.4432$$米． 故像山一样高． 故选：$$\\text{A}$$． " ]

[ "2018年IMAS小学高年级竞赛（第一轮）第4题3分" ]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$6$$ " } ], [ { "aoVal": "E", "content": "$$7$$ " } ] ]

[ "拓展思维->能力->数据处理" ]

[ "$$\\frac{2}{3}=\\frac{6}{9}=\\frac{2+4}{3+6}$$． 故选$$\\text{B}$$． ", "<p>可知原分数的分母为分子的$$1.5$$倍．</p>\n<p>当分母增加$$6$$之后，为了使这个比例不变，分子应该加上$$6\\div 1.5=4$$．</p>\n<p>故选$$\\text{B}$$．</p>" ]

[ "2014年全国迎春杯四年级竞赛初赛第7题" ]

[ [ { "aoVal": "A", "content": "$$32$$ " } ], [ { "aoVal": "B", "content": "$$24$$ " } ], [ { "aoVal": "C", "content": "$$35$$ " } ], [ { "aoVal": "D", "content": "$$36$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->盈亏问题->盈亏基本类型->盈亏基本类型盈亏问题" ]

[ "如果每天吃$$3$$个，十多天吃完，最后一天只吃了$$2$$个，说明糖果至少有$$3\\times 10+2=32$$（个），且糖果数应除以$$3$$ 余$$2$$；如果每天吃$$4$$个，不到$$10$$天就吃完了，最后一天吃了$$3$$个，说明糖果至多有$$4\\times 8+3=35$$（个），且除以$$4$$余$$3$$． 综上，糖果有$$35$$个． " ]

[ "2015年第13届全国创新杯五年级竞赛复赛第4题" ]

[ [ { "aoVal": "A", "content": "$$12$$ " } ], [ { "aoVal": "B", "content": "$$15$$ " } ], [ { "aoVal": "C", "content": "$$18$$ " } ], [ { "aoVal": "D", "content": "$$21$$ " } ] ]

[ "拓展思维->能力->运算求解->程序性计算" ]

[ "$$151+197+238-31=555$$是除数的倍数，$$555=3\\times 5\\times 37$$，由于余数和是$$31$$，检验$$37$$成立，商分别是$$4$$、$$5$$、$$6$$，所以和是$$15$$． " ]

[ "其它改编自2012年全国希望杯六年级竞赛初赛第2题" ]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{2009}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{1}{2010}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{1}{8040}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{1}{4020}$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->裂项与通项归纳->分数裂项->分数裂差->两分数间接裂差" ]

[ "$$\\begin{align}\\& ~ \\frac{251}{2008\\times 2009}+\\frac{251}{2009\\times 2010} \\& =251\\times(\\frac{1}{2008}-\\frac{1}{2009}+\\frac{1}{2009}-\\frac{1}{2010}) \\&=251\\times (\\frac{1}{2008}-\\frac{1}{2010}) \\& =\\frac{1}{8040} \\end{align}$$． " ]

[ "2014年全国迎春杯四年级竞赛初赛第8题" ]

[ [ { "aoVal": "A", "content": "$$44$$ " } ], [ { "aoVal": "B", "content": "$$43$$ " } ], [ { "aoVal": "C", "content": "$$42$$ " } ], [ { "aoVal": "D", "content": "$$41$$ " } ] ]

[ "拓展思维->七大能力->运算求解" ]

[ "倒推．$$44$$ 对应的是$$44-2=42$$，颠倒后是$$24$$，除以$$2$$ 为$$12$$．符合条件．其他的均不符合条件． " ]

[ "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" ]

[ "$$2000\\div (4+3+2+1)=200$$． " ]

[ "2018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": "$$5$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$7$$ " } ] ]

[ "拓展思维->思想->转化与化归的思想" ]

[ "用$$8$$个$$3$$和$$1$$个$$0$$组成的九位数，其中$$1$$个$$3$$必须放在最高位上， 所以这个$$9$$位数最后两位只有三种情况：即$$03$$，$$30$$，$$33$$； $$4$$的倍数的特点是末尾两位数是$$4$$的倍数，整个数就是$$4$$的倍数， 要使除以$$4$$余数是$$1$$，那么末尾两位数是比$$4$$的倍数多$$1$$的数， 只有$$33$$符合要求，所以前$$7$$位是由$$1$$个$$0$$和$$6$$个$$3$$组成，可以是： $$303333333$$， $$330333333$$， $$333033333$$， $$333303333$$， $$333330333$$， $$333333033$$， 一共有$$6$$个． 故选$$\\text{B}$$． " ]

[ "2013年IMAS小学高年级竞赛第一轮检测试题第18题4分" ]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$6$$ " } ], [ { "aoVal": "E", "content": "$$7$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "设这三个数从小到大分别为$$a$$、$$b$$、$$c$$．显然可知$$a$$、$$b$$、$$c$$不可以都等于$$1$$，否则$$abc=1\\ne 1+1+1=3$$，所以$$c\\geqslant 2$$．由题意可知，$$a$$只能是$$1$$，否则三个数的乘积将大于它们之和（因为$$abc\\geqslant 2bc=bc+bc\\textgreater2b+c\\geqslant a+b+c$$），所以$$1+b+c=bc$$． 同样，如果$$b$$大于或等于$$3$$，则$$bc$$大于或等于$$c$$的$$3$$倍，而$$1+b+c$$小于$$c$$的$$3$$倍，所以$$b$$只能是$$2$$． 因此，$$1+2+c=2c$$，得$$c=3$$． 故这个三个正整数之总和为$$1+2+3=6$$，故选$$\\text{D}$$． " ]

[ "2018年第6届湖北长江杯六年级竞赛初赛B卷第6题3分" ]

[ [ { "aoVal": "A", "content": "$$15$$ " } ], [ { "aoVal": "B", "content": "$$24$$ " } ], [ { "aoVal": "C", "content": "$$40$$ " } ], [ { "aoVal": "D", "content": "$$120$$ " } ] ]

[ "拓展思维->能力->数据处理" ]

[ "分数的分子和分母同时乘或者除以一个相同的数（$$0$$除外），分数的大小不变．这叫做分数的基本性质．一个分数的分母是分子的$$5$$倍，比如分子是$$A$$，分母是$$5A$$；扩大$$3$$倍这个分数变为：$$\\frac{A}{5A}=\\frac{3A}{15A}$$；根据题意我们道，相当于分子增加$$8$$，也就是$$3A-A=8$$，所以$$A=4$$； $$15A-5A=15\\times 4-5\\times 4=40$$， 故选$$\\text{C}$$． " ]

[ "2017年第15届湖北武汉创新杯六年级竞赛初赛第2题" ]

[ [ { "aoVal": "A", "content": "$$44$$ " } ], [ { "aoVal": "B", "content": "$$47$$ " } ], [ { "aoVal": "C", "content": "$$52$$ " } ], [ { "aoVal": "D", "content": "$$62$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "$$37\\times 4+67\\times 7-57\\times 10=47$$． 故选$$\\text{B}$$． " ]

[ "竞赛" ]

[ [ { "aoVal": "A", "content": "300 " } ], [ { "aoVal": "B", "content": "360 " } ], [ { "aoVal": "C", "content": "400 " } ], [ { "aoVal": "D", "content": "480 " } ] ]

[ "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->长方体表面积" ]

[ "相乘分解，因为$$96\\times 40\\times 60={{\\left( {{2}^{5}}\\times 3\\times 5 \\right)}^{2}}$$，所以长方体体积为$${{2}^{5}}\\times 3\\times 5=480$$（立方厘米） " ]

[ "2011年全国华杯赛竞赛初赛第4题" ]

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

[ "知识标签->课内题型->综合与实践->智巧趣题->逻辑推理->排除矛盾法" ]

[ "$$5$$名学生说的话相互矛盾，只能有$$1$$人说的是真话，则只有李复习了，说的是真话． 故选：$$\\text{B}$$． " ]

[ "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", "Overseas Competition->知识点->组合模块->操作与策略->归纳递推" ]

[ "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：$$\\text{D}$$． " ]

[ "2016年创新杯六年级竞赛训练题（四）第5题" ]

[ [ { "aoVal": "A", "content": "$$24$$ " } ], [ { "aoVal": "B", "content": "$$15$$ " } ], [ { "aoVal": "C", "content": "$$30$$ " } ], [ { "aoVal": "D", "content": "$$10$$ " } ] ]

[ "知识标签->课内知识点->数学广角->排列组合->排列数" ]

[ "$$24$$ " ]

[ "2013年第25届广东广州五羊杯六年级竞赛第2题5分" ]

[ [ { "aoVal": "A", "content": "$$0.7\\text{m/s}$$ " } ], [ { "aoVal": "B", "content": "$$0.8\\text{m/s}$$ " } ], [ { "aoVal": "C", "content": "$$1\\text{m/s}$$ " } ], [ { "aoVal": "D", "content": "$$1.25\\text{m/s}$$ " } ] ]

[ "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->平均速度->设数法" ]

[ "平均速度$$=$$总路程$$\\div $$总时间，将上山的路程看作单位``$$1$$''， 则有$$\\dfrac{2}{\\dfrac{1}{0.5}+ \\dfrac{1}{2}}= \\dfrac{4}{5}$$． 故选$$\\text{B}$$． " ]

[ "2013年全国希望杯六年级竞赛初赛第18题" ]

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

[ "知识标签->数学思想->转化与化归的思想" ]

[ "甲和丙的话相互矛盾，一定是一真一假，所以乙说的是假话，那么乙获奖了． " ]

[ "2014年全国迎春杯竞赛" ]

[ [ { "aoVal": "A", "content": "$$144$$ " } ], [ { "aoVal": "B", "content": "$$100$$ " } ], [ { "aoVal": "C", "content": "$$81$$ " } ], [ { "aoVal": "D", "content": "$$64$$ " } ] ]

[ "拓展思维->思想->转化与化归的思想" ]

[ "从$$2$$月$$6$$日到$$2$$月$$17$$日为止，一共有$$17-6+1=12$$（天）其中有$$2$$个星期六，星期日．工作了$$12-4=8$$（天），共完成$$1+3+5+7+9+11+13+15=64$$（题）． " ]

[ "2021年第4届山东青岛市南区京山杯六年级竞赛决赛A卷第1题4分" ]

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

[ "拓展思维->思想->方程思想" ]

[ "设成本为$$x$$元，由题意可得等量关系：$$\\left( 1+50 \\% \\right)\\times $$成本价$$=$$售价，进而得到方程，可算出成本价，再利用售价$$-$$成本价$$=$$利润．此题主要考查了一元一次方程的应用，关键是正确理解题意，找出题目中的等量关系，列出方程． 设成本为$$x$$元，由题意得： $$\\left( 1+50 \\% \\right)x=105$$， 解得：$$x=70$$， $$105-70=35$$（元）， 故选$$\\text{D}$$． " ]

[ "2003年第1届创新杯六年级竞赛初赛第8题", "2003年六年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "1万年 " } ], [ { "aoVal": "B", "content": "2万年 " } ], [ { "aoVal": "C", "content": "3万年 " } ], [ { "aoVal": "D", "content": "4万年 " } ] ]

[ "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" ]

[ "这个人1秒钟读2个数字，那么1天可以读$$2\\times 60\\times 60\\times 24=172800$$(个)数字，一年（平年）可以读$$172800\\times 365=6307200$$个数字，那么1万年大约可以读6307亿个数字，现要读12411亿个数字，大约要读$$12411\\div 6307\\approx 2$$(万年). " ]

[ "2017年第20届世界少年奥林匹克数学竞赛四年级竞赛初赛（中国区）第4题5分" ]

[ [ { "aoVal": "A", "content": "$$12$$、$$18$$ " } ], [ { "aoVal": "B", "content": "$$9$$、$$21$$ " } ], [ { "aoVal": "C", "content": "$$17$$、$$13$$ " } ], [ { "aoVal": "D", "content": "$$19$$、$$11$$ " } ] ]

[ "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" ]

[ "假设都是鸭子，共有脚：$$2\\times 30=60$$（只）， 所以有兔子$$(86-60)\\div (4-2)=13$$（只）， 鸭子：$$30-13=17$$（只）． 故选$$\\text{C}$$． " ]

[ "2015年上海走美杯五年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$104$$ " } ], [ { "aoVal": "B", "content": "$$88$$ " } ], [ { "aoVal": "C", "content": "$$46$$ " } ], [ { "aoVal": "D", "content": "$$96$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->还原问题->两量还原问题" ]

[ "方法一：第一次取后还剩下$$\\left( 24-3 \\right)\\times 2=42$$（个），原来有$$\\left( 42+2 \\right)\\times 2=88$$（个）； 方法二：设原有苹果$$x$$个，$$(\\frac{x}{2}-2)\\times \\frac{1}{2}+3=24$$，解得$$x=88$$。 " ]

[ "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->思想->转化与化归的思想" ]

[ "由新定义：$$n!=1×2×3×\\ldots×n$$得： $$2014!=1×2×3×4×5×\\ldots×2013×2014$$ $$=1×3×4×6×7×8×\\ldots×2013×2014×10$$ 所以$$1×3×4×6×7×8×\\ldots×2013×2014×10$$是$$10$$的倍数， 所以$$2014!$$的个位数为$$0$$； $$3!=1×2×3=6$$ 所以$$2014!-3!$$的个位数也就为：$$10-6=4$$． " ]

[ "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1题5分" ]

[ [ { "aoVal": "A", "content": "一 " } ], [ { "aoVal": "B", "content": "二 " } ], [ { "aoVal": "C", "content": "三 " } ], [ { "aoVal": "D", "content": "四 " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "从$$2019$$年的$$1$$月$$1$$日至$$2019$$年的$$10$$月$$1$$日共：$$31+28+31+30+31+30+31+31+30+1=274$$（天）， 根据``二、三、四、五、六、日、一''的周期规律：$$274\\div 7=39$$（周）$$\\cdots \\cdots 1$$（天）． 所以$$10$$月$$1$$日是这个周期的第$$1$$天，即星期二． 故选$$\\text{B}$$． " ]

[ "2015年第13届全国创新杯五年级竞赛复赛第1题" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$2014$$ " } ], [ { "aoVal": "C", "content": "$$2015$$ " } ], [ { "aoVal": "D", "content": "$$2016$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "$${{2015}^{2}}-(2015-1)\\times (2015+1)={{2015}^{2}}-{{2015}^{2}}+1=1$$． " ]

[ "2020年第24届YMO三年级竞赛决赛第4题3分", "2019年第24届YMO三年级竞赛决赛第4题3分" ]

[ [ { "aoVal": "A", "content": "$$80-27$$ " } ], [ { "aoVal": "B", "content": "$$82-29$$ " } ], [ { "aoVal": "C", "content": "$$82+29$$ " } ], [ { "aoVal": "D", "content": "$$82-27$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->数列与数表->数列规律->数列找规律填数", "Overseas Competition->知识点->组合模块->操作与策略->归纳递推" ]

[ "规律：每个算式第一个数是从$$100$$开始，依次少$$2$$， 第二个数是从$$2$$开始，依次多$$3$$， 算式中间的符号是``$$+-$$''循环， 则第$$10$$个算式，中间符号为``$$-$$''， 第$$1$$个数为$$100-9\\times 2=82$$， 第$$2$$个数为$$2+3\\times 9=29$$， 则第$$10$$个算式为``$$82-29$$''． 故选$$\\text{B}$$． " ]

[ "2015年世界少年奥林匹克数学竞赛三年级竞赛复赛A卷第7题8分" ]

[ [ { "aoVal": "A", "content": "$$12$$ " } ], [ { "aoVal": "B", "content": "$$15$$ " } ], [ { "aoVal": "C", "content": "$$18$$ " } ], [ { "aoVal": "D", "content": "$$21$$ " } ] ]

[ "拓展思维->能力->运算求解", "课内体系->能力->实践应用" ]

[ "因为蜗牛白天爬$$5$$米，晚上下滑$$3$$米，所以相当于蜗牛一天往上爬$$5-3=2$$米，这只蜗牛前五天一共爬了$$5\\times 2=10$$米．因为这只蜗牛是第六天爬上柱子顶端的，所以这根柱子最长有$$10+5=15$$米． 故答案为：$$15$$． " ]

[ "2019年第24届YMO四年级竞赛决赛第9题3分", "2020年第24届YMO四年级竞赛决赛第9题3分", "2021年第24届世界少年奥林匹克数学竞赛四年级竞赛决赛第9题3分" ]

[ [ { "aoVal": "A", "content": "$$2017$$ " } ], [ { "aoVal": "B", "content": "$$2018$$ " } ], [ { "aoVal": "C", "content": "$$2019$$ " } ], [ { "aoVal": "D", "content": "$$2020$$ " } ] ]

[ "拓展思维->思想->转化与化归的思想" ]

[ "$$5$$，$$a$$，$$b$$，$$c$$，$$4035$$是等差数列， 则$$5$$，$$b$$，$$4035$$也是等差数列， 则$$b=\\left( 4035+5 \\right)\\div 2$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde=4040\\div 2$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde=2020$$． 选$$\\text{D}$$． " ]

[ "2018年全国小学生数学学习能力测评五年级竞赛初赛第9题3分" ]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$3$$ " } ], [ { "aoVal": "C", "content": "$$2.5$$ " } ], [ { "aoVal": "D", "content": "$$3.5$$ " } ] ]

[ "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" ]

[ "相遇问题中，路程和$$=$$速度和$$\\times $$相遇时间， 所以：相遇时间$$=$$路程和$$\\div $$速度和． 甲船逆水而上速度是甲船静水速度减去水速，即：$$36-$$水速； 乙船顺水而下速度是乙船静水速度加上水速，即：$$28+$$水速； 甲乙速度和为：$$36-$$水速$$+28+$$水速$$=64$$（千米/小时）． 甲乙相遇时间为：$$192\\div64=3$$（小时）． 故选：$$\\text{B}$$． " ]

[ "2018年四川成都锦江区四川师范大学附属第一实验中学小升初模拟11第9题3分", "2015年湖北武汉世奥赛五年级竞赛模拟训练题（四）第4题" ]

[ [ { "aoVal": "A", "content": "$$36$$ " } ], [ { "aoVal": "B", "content": "$$45$$ " } ], [ { "aoVal": "C", "content": "$$55$$ " } ], [ { "aoVal": "D", "content": "$$65$$ " } ] ]

[ "拓展思维->拓展思维->组合模块->抽屉原理->最不利原则" ]

[ "从最不利的情况考虑，第$$1$$把锁需要试$$11$$次；第$$2$$把锁需要试$$10$$次$$\\ldots \\ldots $$；一共需要试$$2+3+\\cdots \\cdots +11=65$$（次）． " ]

[ "2019年第24届YMO三年级竞赛决赛第3题3分", "2020年第24届YMO三年级竞赛决赛第3题3分" ]

[ [ { "aoVal": "A", "content": "$$20$$ " } ], [ { "aoVal": "B", "content": "$$28$$ " } ], [ { "aoVal": "C", "content": "$$36$$ " } ], [ { "aoVal": "D", "content": "$$45$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "根据题意，$$10$$个相同的小球分给$$3$$个人，每人至少$$1$$个，就是将$$10$$个球分成$$3$$组， 一个人最多分：$$10-1-1=8$$（支）， $$8+7+6+5+4+3+2+1=36$$（种）， 故选$$\\text{C}$$． " ]

[ "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5题5分" ]

[ [ { "aoVal": "A", "content": "$$24$$ " } ], [ { "aoVal": "B", "content": "$$96$$ " } ], [ { "aoVal": "C", "content": "$$100$$ " } ], [ { "aoVal": "D", "content": "$$120$$ " } ] ]

[ "拓展思维->能力->数据处理" ]

[ "由于首位不能为$$0$$，故有$$4$$种选择，其它三位，任意选$$3$$个数即可，根据分步计数原理可得一共有$$4\\times 4\\times 3\\times 2=96$$个． " ]

[ "2016年全国中环杯五年级竞赛中小学生思维能力训练活动决赛" ]

[ [ { "aoVal": "A", "content": "$$123456$$ " } ], [ { "aoVal": "B", "content": "$$132465$$ " } ], [ { "aoVal": "C", "content": "$$145632$$ " } ], [ { "aoVal": "D", "content": "$$325416$$ " } ], [ { "aoVal": "E", "content": "$$146325$$ " } ] ]

[ "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" ]

[ "（$$1$$）因为$$E+U+L= 6$$； 而$$1+2+3=6$$； 所以$${E， U ，L}=~ {1，$$2$$，$$3$$ }$$； （$$2$$）因为$$S+R+U+T = 18$$； 而$$6+5+4+3=18$$； 所以$${S， R ，$$U$$ ，$$T$$ }= {6，$$5$$，$$4$$，$$3$$ }$$； （$$3$$）因为$$U\\times T=15$$； 而$$15=1\\times15=3\\times5$$； 所以$${U，T}={3，$$5$$ }$$； （$$4$$）因为$$S\\times L = 8 $$； 而$$8=1\\times8=2\\times4$$； 所以$${S，L}= {2，$$4$$ }$$． 由（$$1$$）和（$$3$$），得$$U= 3$$，则$$T= 5$$； 由（$$1$$）和（$$4$$），得$$L= 2$$，则$$S= 4$$； 最后分别结合（$$1$$）和（$$2$$），得$$E =1$$，$$R = 6$$； 故六位数$$\\overline{EULSRT}=132465$$． " ]

[ "2019年广东深圳全国小学生数学学习能力测评五年级竞赛初赛第9题3分" ]

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

[ "拓展思维->能力->运算求解" ]

[ "假设她全做对，可得：$$15\\times 10=150$$（分）， 现在得了$$88$$分，少了：$$150-88=62$$（分）， 做错一题，不但得不到$$10$$分，还扣$$4$$分， 说明错一题，少得$$10+4=14$$（分）， 不答得$$0$$分，说明不答少得$$10$$分， ∵$$62\\div 14=4$$（题）$$\\cdots 6$$（分），$$6$$不是$$10$$的倍数， 不合题意； $$62\\div 14=3$$（题）$$\\cdots 20$$（分），$$20$$是$$10$$的倍数， 符合题意； ∴未答的题有$$20\\div 10=2$$（题）． 选$$\\text{A}$$． " ]

[ "2019年第7届湖北长江杯五年级竞赛复赛A卷第9题3分" ]

[ [ { "aoVal": "A", "content": "$$12$$ " } ], [ { "aoVal": "B", "content": "$$13$$ " } ], [ { "aoVal": "C", "content": "$$36$$ " } ], [ { "aoVal": "D", "content": "$$38$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "设椅子的单价是$$x$$元，则桌子的价格是$$3x$$元， $$3x\\times4+x\\times9=252$$ $$21x=252$$ $$x=12$$（元）， 桌子的单价：$$3\\times12=36$$（元）． 答：桌子、椅子的单价分别是$$12$$元、$$36$$元． 故答案选：$$\\text{C}$$． " ]

[ "其它改编题", "2017年全国华杯赛小学中年级竞赛初赛模拟第5题" ]

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

[ "拓展思维->思想->转化与化归的思想" ]

[ "把和为$$12$$的两个数分成一组，这样就把这$$11$$个数分成$$6$$组：$$(1,11)$$，$$\\left( 2,10 \\right)$$，$$\\left( 3,9 \\right)$$，$$(4,8)$$，$$(5,7)$$，$$(6)$$．要保证一定有两个数的和为$$12$$，就要保证至少有两个数属于同一组． 由抽屉原理可知，从这$$12$$个数中选出$$7$$个数，就一定有两个数属于同一组．此时这两个数的和就是$$12$$． 如果我们从$$6$$组中各取一个数，则取出的这$$6$$个数中，没有两个数的和是$$12$$，因此本题的答案就是至少选出$$7$$个不同的数． " ]

[ "2015年第4届广东广州羊排赛六年级竞赛第8题1分" ]

[ [ { "aoVal": "A", "content": "锐角三角形 " } ], [ { "aoVal": "B", "content": "直角三角形 " } ], [ { "aoVal": "C", "content": "钝角三角形 " } ], [ { "aoVal": "D", "content": "等腰三角形 " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "最大的角为$$180\\div (2+3+5)\\times 5=90$$（度），是直角三角形． 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初赛第6题5分" ]

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

[ "拓展思维->能力->推理推导->言语逻辑推理", "Overseas Competition->知识点->组合模块->逻辑推理" ]

[ "根据（$$1$$）可知，$$1$$号比$$2$$号快．根据（$$2$$）可知，$$2$$号比$$3$$号快．根据（$$3$$）可知，$$4$$号比$$3$$号快．根据（$$4$$）可知，$$4$$号比$$1$$号快．所以$$4$$号快于$$1$$号，$$1$$号快于$$2$$号，$$2$$号快于$$3$$号．故最快的是$$4$$号．故$$\\text{ABC}$$错误，$$\\text{D}$$正确． 故选$$\\text{D}$$． " ]

[ "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7题5分" ]

[ [ { "aoVal": "A", "content": "$$14$$ " } ], [ { "aoVal": "B", "content": "$$16$$ " } ], [ { "aoVal": "C", "content": "$$18$$ " } ], [ { "aoVal": "D", "content": "$$19$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "假设这个自然数为$$n$$，$$n$$除$$62$$、$$90$$、$$130$$的余数分别为$$a$$、$$b$$、$$c$$， 则$$62-a$$，$$90-b$$，$$130-c$$都是$$n$$的倍数， 可得$$\\left( 62-a \\right)+\\left( 90-b \\right)+\\left( 130-c \\right)=62+90+130-\\left( a+b+c \\right)$$， 得$$282-24=258$$，$$258=2\\times 3\\times 43$$， 则$$n$$可能为$$2$$、$$3$$、$$6$$、$$43$$， 又由于三个余数的和为$$24$$， 则$$abc$$中至少有一个要大于$$8$$， 由于除数大于余数， 因此$$n=43$$， 所以$$a=19$$，$$b=4$$，$$c=1$$， 因此三个余数中最大的是$$19$$， 故选$$\\text{D}$$． " ]

[ "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": "$$16\\frac{4}{11}$$ " } ], [ { "aoVal": "B", "content": "$$13\\frac{11}{13}$$ " } ], [ { "aoVal": "C", "content": "$$18\\frac{6}{13}$$ " } ], [ { "aoVal": "D", "content": "$$21\\frac{9}{11}$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "根据题意我们可以得到如果以$$2$$点整为初始时刻，那么从初始时刻开始，时针和分针共走了$$120 {}^{}\\circ $$，已知时针、分针速度分别为$$6 {}^{}\\circ $$每分，$$0.5 {}^{}\\circ $$每分，那么此时为下午$$2$$点过$$120\\left( 6 {}^{}\\circ +0.5 {}^{}\\circ ~\\right)=18\\frac{6}{13}$$分． " ]

[ "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": "质数 " } ], [ { "aoVal": "B", "content": "合数 " } ], [ { "aoVal": "C", "content": "奇数 " } ], [ { "aoVal": "D", "content": "偶数 " } ] ]

[ "拓展思维->拓展思维->数论模块->奇数与偶数->奇数与偶数的乘法规律" ]

[ "奇数$$\\times 3$$只能是奇数．最容易做错的是选合数，反例：$$1\\times 3=3$$，$$3$$不是合数． " ]

[ "2012年美国数学大联盟杯六年级竞赛初赛第33题5分(每题5分)" ]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$66$$ " } ], [ { "aoVal": "C", "content": "$$528$$ " } ], [ { "aoVal": "D", "content": "$$660$$ " } ] ]

[ "Overseas Competition->知识点->数论模块->因数与倍数->公因数与公倍数", "拓展思维->能力->运算求解" ]

[ "公约数，亦称``公因数''．它是指能同时整除几个整数的数．如果一个整数同时是几个整数的约数，称这个整数为它们的``公约数''；公因数中最大的称为最大公因数． $$264$$和$$528$$的最大公因数为$$264$$； $$264$$和$$660$$的最大公因数为$$132$$． 故选$$\\text{D}$$． " ]

[ "2017年全国华杯赛小学高年级竞赛" ]

[ [ { "aoVal": "A", "content": "小宏长跑的总路程多于小海长跑的总路程 " } ], [ { "aoVal": "B", "content": "小宏长跑的总路程少于小海长跑的总路程 " } ], [ { "aoVal": "C", "content": "小宏长跑的总路程等于小海长跑的总路程 " } ], [ { "aoVal": "D", "content": "小宏长跑的次数少于小海长跑的次数 " } ] ]

[ "拓展思维->能力->数据处理" ]

[ "小海跑$$1$$次的路程$$=\\frac{5}{3}\\times $$小宏跑$$1$$次的路程，故小海长跑$$11$$次$$=11\\times \\frac{5}{3}\\times $$小宏跑$$1$$次的路程$$=$$小宏长跑$$18\\frac{1}{3}$$次，所以选$$B$$． " ]

[ "2021年鹏程杯六年级竞赛初赛第23题" ]

[ [ { "aoVal": "A", "content": "$$6125$$ " } ], [ { "aoVal": "B", "content": "$$6135$$ " } ], [ { "aoVal": "C", "content": "$$6145$$ " } ], [ { "aoVal": "D", "content": "$$6155$$ " } ], [ { "aoVal": "E", "content": "以上都不对 " } ] ]

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

[ "原来甲乙合作工效$$\\left( \\frac{1}{100}+\\frac{1}{150} \\right)\\times (1+20 \\%)=\\frac{1}{50}$$， 完成前面$$\\frac{2}{5}$$的工作量需工时$$\\frac{2}{5}\\div \\frac{1}{50}=20$$（天）， 完成后面$$\\frac{3}{5}$$的工作量需工时$$90-20=70$$（天）， 工作效率是$$\\frac{3}{5}\\div 70=\\frac{3}{350}$$， 工效之差为$$\\frac{1}{50}-\\frac{3}{350}=\\frac{4}{350}=\\frac{2}{175}$$， 总工作量为$$70\\div \\frac{2}{175}=6125$$． 故选$$\\text{A}$$． " ]

[ "2020年新希望杯二年级竞赛初赛（团战）第52题" ]

[ [ { "aoVal": "A", "content": "$$90$$ " } ], [ { "aoVal": "B", "content": "$$82$$ " } ], [ { "aoVal": "C", "content": "$$81$$ " } ], [ { "aoVal": "D", "content": "$$80$$ " } ], [ { "aoVal": "E", "content": "$$79$$ " } ] ]

[ "拓展思维->拓展思维->计数模块->枚举法综合->字典排序法->组数->数字组数（规定数位大小）" ]

[ "根据题意分析可知，个数数字与十位数字相同的两位数有$$10$$个，两位数一共有$$91$$个， 由此可知，个位数字与十位数字不同的两位数有$$91-10=81$$（个）． 故选$$\\text{C}$$． " ]

[ "2011年北京五年级竞赛" ]

[ [ { "aoVal": "A", "content": "$$35$$千克 " } ], [ { "aoVal": "B", "content": "$$45$$千克 " } ], [ { "aoVal": "C", "content": "$$55$$千克 " } ] ]

[ "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）->分数的简单实际问题" ]

[ "相当于把石灰水分成了$$101$$份，石灰占$$1$$份．所以石灰有$$4545\\times \\frac{1}{{1{ + }100}}{ = }45$$（千克）． " ]

[ "2018年湖北武汉新希望杯小学高年级六年级竞赛训练题（二）第2题", "2017年新希望杯小学高年级六年级竞赛训练题（二）第2题" ]

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

[ "拓展思维->拓展思维->计算模块->数列与数表->数列规律" ]

[ "$$123-76=47$$，$$76-47=29$$，$$47-29=18$$，$$29-18=11$$，$$18-11=7$$,$$11-7=4$$，$$7-4=3$$． " ]

[ "2020年四川绵阳涪城区绵阳东辰国际学校超越杯四年级竞赛第2题2分" ]

[ [ { "aoVal": "A", "content": "$$32$$ " } ], [ { "aoVal": "B", "content": "$$33$$ " } ], [ { "aoVal": "C", "content": "$$34$$ " } ], [ { "aoVal": "D", "content": "$$35$$ " } ], [ { "aoVal": "E", "content": "$$36$$ " } ] ]

[ "拓展思维->能力->实践应用", "Overseas Competition->知识点->计算模块->整数->整数乘除->整数除法运算" ]

[ "根据题意分析可知，利用关系式：数量$$=$$总价$$\\div $$单价，用爸爸带的钱除以每个笔记本的单价即可得到最多能买多少个笔记本，列式为：$$269\\div 8=33$$（个）$$\\cdots \\cdots 5$$（元），剩下的$$5$$元不足以再买一个笔记本， 所以最多能买$$33$$个， 故选$$\\text{B}$$． " ]

[ "2014年北京六年级竞赛", "2016年陕西西安小升初交大附中入学真卷9第11题", "2014年全国迎春杯六年级竞赛初赛第6题", "2014年全国迎春杯五年级竞赛初赛第6题" ]

[ [ { "aoVal": "A", "content": "$$6$$ " } ], [ { "aoVal": "B", "content": "$$28$$ " } ], [ { "aoVal": "C", "content": "$$56$$ " } ], [ { "aoVal": "D", "content": "$$70$$ " } ], [ { "aoVal": "E", "content": "$$75$$ " } ] ]

[ "拓展思维->思想->方程思想" ]

[ "设丁拿了$$a$$件礼物，则四人花同样的钱，每人可以拿到$$a+\\frac{3+7+14}{4}=a+6$$件礼物， 实际情况：丁少拿了$$6$$件，乙多拿了$$1$$件，给丁$$14$$元，则货物单价$$14$$元， 丙多拿了$$14-6=8$$件，$$3$$件给甲，$$5$$件给丁，$$5\\times14=70$$元． " ]

[ "2018年全国小学生数学学习能力测评六年级竞赛初赛第9题3分" ]

[ [ { "aoVal": "A", "content": "$$90$$ " } ], [ { "aoVal": "B", "content": "$$99$$ " } ], [ { "aoVal": "C", "content": "$$110$$ " } ], [ { "aoVal": "D", "content": "$$180$$ " } ] ]

[ "拓展思维->拓展思维->数论模块->因数与倍数->公因数与公倍数->因倍应用题->倍数应用题" ]

[ "如果按照$$2:3:4$$发书，相当于将书分成$$2+3+4=9$$份；如果按照$$2:4:5$$发书，相当于将书分成$$2+4+5=11$$份，因此这批书是$$9$$和$$11$$的公倍数，只有$$\\text{B}$$选项符合． " ]

[ "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": "$$120$$ " } ] ]

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

[ "$$990$$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$$\\text{C}$$． " ]

[ "2013年第11届创新杯三年级竞赛初赛第5题6分" ]

[ [ { "aoVal": "A", "content": "$$51$$ " } ], [ { "aoVal": "B", "content": "$$50$$ " } ], [ { "aoVal": "C", "content": "$$53$$ " } ], [ { "aoVal": "D", "content": "$$57$$ " } ] ]

[ "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" ]

[ "也就是说，有这样一个两位数，首先他是一个十位是$$5$$的两位数，并且除以$$7$$的余数是$$4$$，那么最终只可能是$$53$$． " ]

[ "2017~2018学年浙江杭州五年级期中", "2017年第22届湖北武汉华杯赛三年级竞赛复赛（华罗庚金杯）第2题10分" ]

[ [ { "aoVal": "A", "content": "$$1986$$ " } ], [ { "aoVal": "B", "content": "$$2034$$ " } ], [ { "aoVal": "C", "content": "$$2176$$ " } ], [ { "aoVal": "D", "content": "$$2458$$ " } ], [ { "aoVal": "E", "content": "$$2526$$ " } ] ]

[ "拓展思维->拓展思维->组合模块->数字谜", "Overseas Competition->知识点->组合模块->数字谜->竖式数字谜" ]

[ "因为字母代表$$2\\sim 9$$八个数字，故$$N$$只能为$$2$$． $$\\frac{\\begin{matrix} \\begin{matrix} + \\end{matrix} \\begin{matrix}O T S \\end{matrix} \\begin{matrix}N W I \\end{matrix} \\begin{matrix}E O X \\end{matrix} \\end{matrix}}{\\begin{matrix} N I N E \\end{matrix}}$$， $$O+E+X=E$$，故$$O+X=10$$，可为$$3+7$$，$$4+6$$；$$W+I=9$$，当$$O=3$$，$$I=4$$或$$5$$， 当$$I=4$$，$$W=5$$，或$$I=5$$，$$W=4$$， $$T+S$$需要为$$20$$或$$21$$，不可能； 当$$O=7$$，$$I=4$$或$$5$$，$$T+S=16$$或$$17$$，此时$$E=6$$． $$\\frac{\\begin{matrix} \\begin{matrix} + \\end{matrix} \\begin{matrix}7 8 9 \\end{matrix} \\begin{matrix}2 4 5 \\end{matrix} \\begin{matrix}6 7 3 \\end{matrix} \\end{matrix}}{\\begin{matrix} 2 5 2 6 \\end{matrix}}$$或$$\\frac{\\begin{matrix} \\begin{matrix} + \\end{matrix} \\begin{matrix}7 9 8 \\end{matrix} \\begin{matrix}2 4 5 \\end{matrix} \\begin{matrix}6 7 3 \\end{matrix} \\end{matrix}}{\\begin{matrix} 2 5 2 6 \\end{matrix}}$$ " ]

[ "2020年广东广州羊排赛六年级竞赛第6题3分" ]

[ [ { "aoVal": "A", "content": "$$1:300$$ " } ], [ { "aoVal": "B", "content": "$$300:1$$ " } ], [ { "aoVal": "C", "content": "$$1:30000000$$ " } ], [ { "aoVal": "D", "content": "$$30000000:1$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "$$300$$千米$$=30000000$$厘米； 比例尺$$=$$图上距离$$:$$实际距离$$=1:300000000$$． 故选$$\\text{C}$$． " ]

[ "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2题5分" ]

[ [ { "aoVal": "A", "content": "比原价低 " } ], [ { "aoVal": "B", "content": "比原价高 " } ], [ { "aoVal": "C", "content": "与原价相同 " } ], [ { "aoVal": "D", "content": "无法判断 " } ] ]

[ "拓展思维->拓展思维->应用题模块->经济问题->基本经济概念->折扣问题" ]

[ "先把原价看作单位``$$1$$''； 涨价后的价钱是原价的$$\\left( 1+15 \\% \\right)$$； 后又降价$$15 \\%$$，是降低涨价后的价格的$$15 \\%$$， 据此可知现在的价格是原价的百分之几，进而得出结论． $$\\left( 1+15 \\% \\right)\\times \\left( 1-15 \\% \\right)$$ $$=1.15\\times 0.85$$ $$=97.8 \\%$$； $$97.8 \\% ~\\textless{} ~1$$，即现价比原价低． 故选$$\\text{A}$$． " ]

[ "2020年新希望杯三年级竞赛初赛（团战）第35题" ]

[ [ { "aoVal": "A", "content": "$$\\left( 784-455 \\right)\\times 39+44\\times 11$$ " } ], [ { "aoVal": "B", "content": "$$11\\times 1+22\\times 2+33\\times 3+44\\times 4+55\\times 5+66\\times 6+77\\times 7+88\\times 8+99\\times 9$$ " } ], [ { "aoVal": "C", "content": "$$11\\times 1+22\\times 2+33\\times 3+44\\times 4+55\\times 5+66\\times 6+77\\times 7+88\\times 8+99\\times 9$$ " } ], [ { "aoVal": "D", "content": "$$123\\times 456+789$$ " } ], [ { "aoVal": "E", "content": "$$2\\times 4\\times 6\\times \\cdots \\times 2018\\times 2020-1\\times 3\\times 5\\times \\cdots \\times 2017\\times 2019$$ " } ] ]

[ "拓展思维->拓展思维->数论模块->奇数与偶数->奇数与偶数的混合计算" ]

[ "暂无 " ]

[ "2019年第23届广东世界少年奥林匹克数学竞赛二年级竞赛决赛第4题5分" ]

[ [ { "aoVal": "A", "content": "$$40$$ " } ], [ { "aoVal": "B", "content": "$$80$$ " } ], [ { "aoVal": "C", "content": "$$100$$ " } ], [ { "aoVal": "D", "content": "$$120$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "根据题意分析可知， 一张桌子比一把椅子多$$20$$元， 那么一把椅子就比一张桌子少$$20$$元， 所以一把椅子的价钱是：$$60-20=40$$（元）， 那么一套桌椅的价钱是：$$60+40=100$$（元）． 故选：$$\\text{C}$$． " ]

[ "2018年第22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6分" ]

[ [ { "aoVal": "A", "content": "$$100$$ " } ], [ { "aoVal": "B", "content": "$$80$$ " } ], [ { "aoVal": "C", "content": "$$60$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->间隔问题->直线型两端都有->爬楼梯问题" ]

[ "小红每天回家要走：$$20\\times \\left(5-1\\right)=80$$（级）． 故选：$$\\text{B}$$． " ]

[ "2019年广东深圳全国小学生数学学习能力测评四年级竞赛初赛第7题3分" ]

[ [ { "aoVal": "A", "content": "$$9:05$$ " } ], [ { "aoVal": "B", "content": "$$9:35$$ " } ], [ { "aoVal": "C", "content": "$$9:55$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "根据选项$$\\text{C}$$，现在是$$9:55$$，那么$$5$$分钟前分针的位置是指向$$10$$，$$5$$分钟后时针也是指向$$10$$．所以$$\\text{C}$$选项答案是满足题目要求的． " ]

[ "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->课内知识点->数的认识->数的特征->倍数->倍的认识" ]

[ "154按照操作依次为$$77$$，$$84$$，$$42$$，$$21$$，$$28$$，$$14$$，$$7$$，$$14$$，7\\ldots\\ldots，不可能得到$$11$$． 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. " ]

[ "2014年迎春杯四年级竞赛复赛" ]

[ [ { "aoVal": "A", "content": "$$7$$ " } ], [ { "aoVal": "B", "content": "$$9$$ " } ], [ { "aoVal": "C", "content": "$$11$$ " } ], [ { "aoVal": "D", "content": "$$13$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->数列与数表->等差数列->等差数列求通项" ]

[ "根据题意得$$({{a}_{1}}+{{a}_{8}})\\times 8\\div 2=({{a}_{9}}+{{a}_{12}})\\times 4\\div 2$$，因为$${{a}_{8}}={{a}_{1}}+14$$，$${{a}_{9}}={{a}_{1}}+16$$，$${{a}_{12}}={{a}_{1}}+22$$， 所以$$({{a}_{1}}+{{a}_{1}}+14)\\times 8\\div 2=({{a}_{1}}+16+{{a}_{1}}+22)\\times 4\\div 2$$，解得$${{a}_{1}}=5$$，因此$${{a}_{2}}=5+2=7$$。 " ]

[ "2006年四年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "$$9$$ " } ], [ { "aoVal": "B", "content": "$$11.$$ " } ], [ { "aoVal": "C", "content": "$$14$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->小数->小数基础->小数点的移动规律" ]

[ "小数点向右移动$$1$$位，则新数比原数扩大$$10$$倍，即增加$$9$$倍，所以原数为$$59.94\\div \\left( 10-1 \\right)=6.66$$，选$$\\text{A}$$. " ]

[ "2022年第9届广东深圳鹏程杯四年级竞赛初赛第13题5分" ]

[ [ { "aoVal": "A", "content": "$$25$$ " } ], [ { "aoVal": "B", "content": "$$22$$ " } ], [ { "aoVal": "C", "content": "$$21$$ " } ], [ { "aoVal": "D", "content": "$$20$$ " } ], [ { "aoVal": "E", "content": "$$17$$ " } ] ]

[ "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" ]

[ "无 " ]

[ "2011年全国华杯赛竞赛初赛第4题" ]

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

[ "知识标签->数学思想->转化与化归的思想" ]

[ "$$5$$名学生说的话相互矛盾，只能有$$1$$人说的是真话，则只有李复习了，说的是真话． 故选：$$\\text{B}$$． " ]

[ "2008年第6届创新杯四年级竞赛初赛B卷第9题5分", "2008年四年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "50 " } ], [ { "aoVal": "B", "content": "49 " } ], [ { "aoVal": "C", "content": "34 " } ], [ { "aoVal": "D", "content": "33 " } ] ]

[ "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" ]

[ "这串数的规律是``奇奇偶''循环的，在前99个数中，有33个偶数，第100个数是奇数，所以，这串数中前100个数有33个偶数. " ]

[ "2008年四年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "$$39$$ " } ], [ { "aoVal": "B", "content": "$$40$$ " } ], [ { "aoVal": "C", "content": "$$41$$ " } ], [ { "aoVal": "D", "content": "$$42$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->间隔问题" ]

[ "车队$$145$$秒行的路程为$$5\\times 145=725$$（米） 车队的长度为$$725-200=525$$（米） 车队共有车$$\\left( 525-5 \\right) \\div\\left( 5+8 \\right)+1=41$$（辆） " ]

[ "2017年河南郑州联合杯竞赛附加赛第一场第3题2分" ]

[ [ { "aoVal": "A", "content": "$$20$$ " } ], [ { "aoVal": "B", "content": "$$40$$ " } ], [ { "aoVal": "C", "content": "$$60$$ " } ], [ { "aoVal": "D", "content": "$$80$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "被减数$$+$$减数$$+$$差$$=160$$，减数$$=3$$差，又被减数$$=$$减数$$+$$差； 则有被减数$$=$$减数$$+$$差$$=160\\div 2=80$$，即$$3$$差$$+$$差$$=80$$，所以差$$=80\\div 4=20$$． 故选A． " ]

[ "2018年第6届湖北长江杯六年级竞赛初赛B卷第8题3分" ]

[ [ { "aoVal": "A", "content": "$$19$$ " } ], [ { "aoVal": "B", "content": "$$20$$ " } ], [ { "aoVal": "C", "content": "$$21$$ " } ], [ { "aoVal": "D", "content": "$$22$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "第一杯的糖：$$200\\times 30 \\%=60$$（克）， 第二杯的糖：$$300\\times 15 \\%=45$$（克）， 两杯混到一起后浓度为 $$(45+60)\\div (200+300)$$ $$=105\\div 500$$ $$=0.21$$ $$=21 \\%$$， 甜度为$$21$$． 故选$$\\text{C}$$． " ]

[ "2021年鹏程杯六年级竞赛初赛第3题" ]

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

[ "拓展思维->思想->对应思想" ]

[ "如果这三个数中有$$2$$个奇数和$$1$$个偶数，那么等式左边必为偶数，等式右边必为奇数，不可能．如果这三个数均为奇数，那么等式左边必为奇数，而等式右边必为偶数，不可能．因此，这三个数中最多有$$1$$个奇数，例如$$2+4+1=2\\times 4-1$$． 故选$$\\text{A}$$． " ]

[ "2017年河南郑州豫才杯六年级竞赛" ]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{6}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{1}{12}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{1}{14}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{1}{15}$$ " } ] ]

[ "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" ]

[ "由题意可知，红$$:$$白$$=1:6$$，黑$$:$$红$$=1:2$$，则黑$$:$$红$$:$$白$$=1:2:12$$．摸出黑色小球的可能性为$$\\frac{1}{1+2+12}=\\frac{1}{15}$$． " ]

[ "2006年第4届创新杯五年级竞赛复赛第5题", "2006年五年级竞赛创新杯" ]

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

[ "拓展思维->拓展思维->数论模块->因数与倍数->公因数与公倍数->因倍应用题->倍数应用题" ]

[ "不妨将100厘米长的木棍视为有刻度的米尺，那么从左到右每隔6厘米染一个红点，这些红点在米尺上的量数为(单位:厘米): $$6$$，$$12$$，$$18$$，$$24$$，$$30$$，$$36$$，$$\\cdots$$，$$60$$;$$66$$，$$\\cdots$$，$$90$$;$$96$$① 从右到左每隔5厘米染一个红点，这些红点在米尺上的量数为(单位:厘米):$$5$$，$$10$$，$$15$$，$$20$$，$$25$$，$$30$$;$$35$$，$$\\cdots$$ ，$$60$$;$$65$$，$$\\cdots$$ ，$$90$$;$$95$$② 比较上述两个数列①和②，知 在$$0\\sim30$$内的红点，有两组$$\\left( 5,6 \\right)\\left( 24,25 \\right)$$，两个相邻红点相距1厘米; 在$$30\\sim60$$内的红点，有两组$$\\left( 35,36 \\right)\\left( 54,55 \\right)$$，两个相邻红点相距1厘米; 在$$60\\sim90$$内的红点，有两组$$\\left( 65,66 \\right)\\left( 84,85 \\right)$$，两个相邻红点相距1厘米; 在$$90\\sim100$$内的红点，有一组$$\\left( 95,96 \\right)$$，两个相邻红点相距$$1$$厘米; 因此，沿红点锯开，长1厘米得短木棍有7条 " ]

[ "2006年四年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "2.565 " } ], [ { "aoVal": "B", "content": "2.56 " } ], [ { "aoVal": "C", "content": "2.855 " } ], [ { "aoVal": "D", "content": "2.85 " } ] ]

[ "拓展思维->拓展思维->计算模块->小数->小数基础->小数点的移动规律" ]

[ "小数点向右移动一位，则扩大10倍，增加9倍.因此原数为$$25.65\\div \\left( 10-1 \\right)=2.85$$，故选D. " ]