Datasets:

math-eval

/

TAL-SCQ5K

like 34

"2023-07-07T00:00:00"

[ "2005年五年级竞赛创新杯" ]

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

[ "拓展思维->拓展思维->数论模块->完全平方数->平方数的尾数特征" ]

[ "一个完全平方数的末尾只能是$$0$$、$$1$$、$$4$$、$$5$$、$$6$$、$$9$$，所以在$$1$$、$$2$$、$$3$$、$$5$$、$$7$$、$$8$$中，$$2$$、$$3$$、$$7$$、$$8$$这$$4$$个数不是一个整数的平方的个位数。 故选：D。 " ]

"2023-07-07T00:00:00"

[ "2013年全国学而思杯二年级竞赛第1题" ]

[ [ { "aoVal": "A", "content": "$$60$$ " } ], [ { "aoVal": "B", "content": "$$70$$ " } ], [ { "aoVal": "C", "content": "$$80$$ " } ], [ { "aoVal": "D", "content": "$$90$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->整数->整数加减->整数加减巧算之分组法" ]

[ "此题分为两种思路：一种可以直接使用凑整的方法，共有$$4$$对，$$20\\times ~4=80$$；另一种可以使用等差数列求和的方法：$$\\left( 3+17 \\right)\\times ~8\\div2=80$$． " ]

"2023-07-07T00:00:00"

[ "2018年第22届广东世界少年奥林匹克数学竞赛六年级竞赛决赛第2题5分" ]

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

[ "拓展思维->思想->对应思想" ]

[ "甲、乙路程比$$=6:5$$， 甲、乙时间比$$=11:10$$， 故甲、乙的速度比为$$\\frac{6}{11}:\\frac{5}{10}=12:11$$． 故选：$$\\text{B}$$． " ]

"2023-07-07T00:00:00"

[ "2017年第16届湖北武汉创新杯五年级竞赛邀请赛训练题（一）" ]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$1$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->小数->循环小数->循环小数的周期问题" ]

[ "$$\\frac{1}{7}=0.\\dot{1}4285\\dot{7}$$，循环节有$$6$$个数字，$$2017\\div 6=336\\cdots \\cdots 1$$，所以小数点后第$$2017$$位数字是$$1$$． " ]

"2023-07-07T00:00:00"

[ "2016年全国AMC六年级竞赛8第5题" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$4$$ " } ], [ { "aoVal": "D", "content": "$$5$$ " } ], [ { "aoVal": "E", "content": "$$7$$ " } ] ]

[ "拓展思维->能力->运算求解", "Overseas Competition->知识点->数论模块->余数问题->中国剩余定理" ]

[ "翻译：$$N$$是一个两位数，$$N$$除以$$9$$余$$1$$，除以$$10$$ 余$$3$$，那么$$N$$除以$$11$$余． $$N\\div 9=\\cdots \\cdots 1$$，且个位是$$3$$，所以$$N=9\\times 8+1=73$$（$$N$$是一个两位数），$$73\\div 11=6\\cdots \\cdots 7$$． " ]

"2023-07-07T00:00:00"

[ "2020年新希望杯六年级竞赛（2月）第13题", "2020年希望杯六年级竞赛模拟第13题" ]

[ [ { "aoVal": "A", "content": "$$6.5$$ " } ], [ { "aoVal": "B", "content": "$$7$$ " } ], [ { "aoVal": "C", "content": "$$7.5$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ], [ { "aoVal": "E", "content": "$$8.5$$ " } ] ]

[ "拓展思维->能力->构造模型->模型思想" ]

[ "哥哥工效：$$\\frac{1}{6}$$，弟弟工效：$$\\frac{1}{9}$$， 则$$1\\div \\left( \\frac{1}{6}+\\frac{1}{9} \\right)=\\frac{18}{5}=3$$（周期）$$\\cdots \\cdots \\frac{3}{5}$$， $$1-3\\times \\left( \\frac{1}{6}+\\frac{1}{9} \\right)=\\frac{1}{6}$$， $$\\frac{1}{6}\\div\\frac{1}{6}=1$$（小时）， 共$$3\\times 2+1=7$$（小时）． " ]

"2023-07-07T00:00:00"

[ "2016年第12届全国新希望杯五年级竞赛决赛第1题" ]

[ [ { "aoVal": "A", "content": "$$\\frac{64}{69}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{42}{102}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{60}{135}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{48}{108}$$ " } ] ]

[ "拓展思维->能力->数感认知->分数数字加工" ]

[ "分子、分母缩小相同倍数，分子分母之间差也缩小相同的倍数，所以$$60\\div 5=12$$．则$$\\frac{4}{9}=\\frac{4\\times 12}{9\\times 12}=\\frac{48}{108}$$． " ]

"2023-07-07T00:00:00"

[ "2013年全国美国数学大联盟杯小学高年级竞赛初赛第15题" ]

[ [ { "aoVal": "A", "content": "$${{4}^{13}}$$ " } ], [ { "aoVal": "B", "content": "$${{2}^{26}}$$ " } ], [ { "aoVal": "C", "content": "$${2}^{4}$$ " } ], [ { "aoVal": "D", "content": "$${{16}^{1155}}$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->乘方->乘方的运算->乘方的幂运算" ]

[ "$$2^{3}\\times2^{5}\\times2^{7}\\div2^{11}=2^{3+5+7-11}=2^{4}$$． " ]

"2023-07-07T00:00:00"

[ "2014年全国学而思杯二年级竞赛第3题" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$20$$ " } ], [ { "aoVal": "C", "content": "$$30$$ " } ], [ { "aoVal": "D", "content": "$$40$$ " } ], [ { "aoVal": "E", "content": "$$50$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->和差倍问题->倍的基本计算->两量倍", "海外竞赛体系->知识点->应用题模块->分百应用题->认识单位1" ]

[ "$$10\\times 4=40$$（元）． " ]

"2023-07-07T00:00:00"

[ "2009年全国迎春杯六年级竞赛初赛第3题" ]

[ [ { "aoVal": "A", "content": "答案请写在答题卡上 " } ] ]

[ "知识标签->拓展思维->应用题模块->盈亏问题->盈亏基本类型->亏亏问题" ]

[ "答案请写在答题卡上 " ]

"2023-07-07T00:00:00"

[ "2012年IMAS小学中年级竞赛第一轮检测试题第17题4分" ]

[ [ { "aoVal": "A", "content": "$$6$$ " } ], [ { "aoVal": "B", "content": "$$12$$ " } ], [ { "aoVal": "C", "content": "$$444$$ " } ], [ { "aoVal": "D", "content": "$$888$$ " } ], [ { "aoVal": "E", "content": "$$6012$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "$$444888=444000+888=74\\times 6000+74\\times 12=74\\times 6012$$． " ]

"2023-07-07T00:00:00"

[ "2009年全国迎春杯小学中年级竞赛复赛第7题10分" ]

[ [ { "aoVal": "A", "content": "$$180$$ " } ], [ { "aoVal": "B", "content": "$$160$$ " } ], [ { "aoVal": "C", "content": "$$120$$ " } ], [ { "aoVal": "D", "content": "$$80$$ " } ] ]

[ "知识标签->拓展思维->计数模块->加乘原理->乘法原理->物品搭配" ]

[ "假如给小强的是智力拼图，则有$$2\\times 5\\times 4\\times 3=120$$种方法． 假如给小强的是遥控汽车，则有$$1\\times 5\\times 4\\times 3=60$$种方法． 总共有$$120+60=180$$种方法． 若将遥控汽车给小强，则学习机要给小玉，此时另外$$3$$个孩子在剩余$$5$$件礼物中任选$$3$$件，有$$5\\times 4\\times 3=60$$种方法;若将遥控汽车给小玉，则智力拼图要给小强，此时也有$$60$$种方法;若遥控汽车既不给小强、也不给小玉，则智力拼图要给小强，学习机要给小玉，此时仍然有$$60$$种方法. 所以共有$$60+60+60=180$$种方法. " ]

"2023-07-07T00:00:00"

[ "2021年鹏程杯六年级竞赛初赛第14题" ]

[ [ { "aoVal": "A", "content": "$$2694$$ " } ], [ { "aoVal": "B", "content": "$$2695$$ " } ], [ { "aoVal": "C", "content": "$$2696$$ " } ], [ { "aoVal": "D", "content": "$$2697$$ " } ], [ { "aoVal": "E", "content": "以上都不对 " } ] ]

[ "拓展思维->能力->公式记忆->符号化数学原理" ]

[ "完全平方数除以$$4$$只能余$$0$$或$$1$$，两个完全平方数相减，其差除以$$4$$余数只能是$$0$$或$$1$$或$$3$$， 即除以$$4$$余$$2$$的数不能写成两个完全平方数的差． 由于$$(n+1)^{2}-(n-1)^{2}=4n$$ $$(2n+1)^{2}-(2n)^{2}=4n+1$$， $$(2n+2)^{2}-(2n+1)^{2}=4n+3$$， 所以，除$$4n+2$$外，均可表示成两个数平方差， 即每$$4$$个连续的自然数中，有一个不是``鹏程数''， 所以第$$2021$$个``鹏程数''为：$$2695$$． 故选$$\\text{B}$$． " ]

"2023-07-07T00:00:00"

[ "2017年第15届全国希望杯小学中年级四年级竞赛第1试试题第19题" ]

[ [ { "aoVal": "A", "content": "$$120$$ " } ], [ { "aoVal": "B", "content": "$$119$$ " } ], [ { "aoVal": "C", "content": "$$118$$ " } ], [ { "aoVal": "D", "content": "$$117$$ " } ] ]

[ "知识标签->拓展思维->应用题模块->列方程解应用题->一元一次方程解应用题" ]

[ "黑子个数$$=$$白子个数$$\\times 2$$；黑子个数$$=$$次数$$\\times 3+31$$，白子个数$$=$$次数$$\\times 2+1$$； 设取的次数为$$x$$次，则白子有($$2x+1$$)个，黑子有($$3x+31$$)个， 由题意得$$2\\times (2x+1)=3x+31$$， 解得$$x=29$$， 则袋子中原有黑子$$3\\times 29+31=118$$（个）． 假设黑子每次取的个数也是白子的$$2$$倍，即黑子每次取$$2\\times2=4$$（个），白子每次取$$2$$个，则： $$(31-1\\times2)\\div(2\\times2-3)$$ $$=29\\div1$$ $$=29$$（次） $$3\\times29+31$$ $$=87+31$$ $$=118$$（个）． 答：袋中原有黑子 $$118$$个． " ]

"2023-07-07T00:00:00"

[ "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40题" ]

[ [ { "aoVal": "A", "content": "$$20$$ " } ], [ { "aoVal": "B", "content": "$$30$$ " } ], [ { "aoVal": "C", "content": "$$10$$ " } ], [ { "aoVal": "D", "content": "$$40$$ " } ] ]

[ "拓展思维->思想->整体思想" ]

[ "$$(40+20)\\div 2=30$$． " ]

"2023-07-07T00:00:00"

[ "2011年第9届全国创新杯小学高年级六年级竞赛第2题" ]

[ [ { "aoVal": "A", "content": "$$30$$ " } ], [ { "aoVal": "B", "content": "$$42$$ " } ], [ { "aoVal": "C", "content": "$$56$$ " } ], [ { "aoVal": "D", "content": "$$60$$ " } ] ]

[ "拓展思维->能力->公式记忆->符号化数学原理" ]

[ "加上首位两站共$$8$$个站，共有$C_{8}^{2}=28$种组合，考虑车票的次序共有：$$28×2=56$$种． " ]

"2023-07-07T00:00:00"

[ "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->思想->对应思想" ]

[ "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． " ]

"2023-07-07T00:00:00"

[ "2013年第9届全国新希望杯小学高年级五年级竞赛复赛第2题" ]

[ [ { "aoVal": "A", "content": "$$3.5$$元 " } ], [ { "aoVal": "B", "content": "$$5$$元 " } ], [ { "aoVal": "C", "content": "$$6$$元 " } ], [ { "aoVal": "D", "content": "$$7$$元 " } ] ]

[ "拓展思维->拓展思维->应用题模块->列方程解应用题->多元一次方程解应用题->小数系数方程（组）解题" ]

[ "设买$$x$$千克梨，$$y$$千克苹果 $$\\begin{cases}4x+7y=84 4.5x+7y=87.5 \\end{cases}\\Rightarrow 0.5x=3.5\\Rightarrow x=7$$． " ]

"2023-07-07T00:00:00"

[ "2015年全国美国数学大联盟杯五年级竞赛初赛第35题" ]

[ [ { "aoVal": "A", "content": "＄$$16$$ " } ], [ { "aoVal": "B", "content": "＄$$32$$ " } ], [ { "aoVal": "C", "content": "＄$$48$$ " } ], [ { "aoVal": "D", "content": "＄$$80$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "$$64\\times(1+25 \\%)=80$$，最后答案是$$\\text{D}$$． " ]

"2023-07-07T00:00:00"

[ "2017年全国华杯赛小学高年级竞赛" ]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{3}{2}$$ " } ], [ { "aoVal": "D", "content": "$$7$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->方程基础->一元一次方程->整数系数方程" ]

[ "如果$$a=\\frac{3}{2}$$，则有$$2=3x+6y=2\\times \\left( \\frac{3}{2}x+3y \\right)=14$$． " ]

"2023-07-07T00:00:00"

[ "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5题" ]

[ [ { "aoVal": "A", "content": "10米 " } ], [ { "aoVal": "B", "content": "9.75米 " } ], [ { "aoVal": "C", "content": "9.25米 " } ], [ { "aoVal": "D", "content": "10.25米 " } ] ]

[ "拓展思维->拓展思维->应用题模块->比例应用题" ]

[ "当甲到达终点时，乙离终点还有5米，那么甲、乙的速度比是甲:乙$$=100:\\left( 100-5 \\right)=100:95=20:19$$，当乙到达终点时，丙还有5米，那么乙、丙的速度比是乙:丙$$=100:\\left( 100-5 \\right)=20:19$$，现把这两个比例中乙的份数变为一样的，那么甲:乙$$=20:19=400:380$$，乙:丙$$=20:19=380:361$$，从而甲:乙:丙$$=400:380:361$$，即甲:丙$$=400:361$$，所以当甲到达终点跑了100米时，丙跑了$$100\\times \\frac{361}{400}=90.25$$(米)，离终点还有$$100-90.25=9.75$$(米)，选$$B$$ " ]

"2023-07-07T00:00:00"

[ "2018年第22届广东世界少年奥林匹克数学竞赛一年级竞赛决赛第3题5分" ]

[ [ { "aoVal": "A", "content": "亮 " } ], [ { "aoVal": "B", "content": "不亮 " } ], [ { "aoVal": "C", "content": "不能确定 " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "开单数次数，改变原来灯的状态，开双数次，不改变原来的状态．总共开的次数为：$$4+5=9$$（次）， 所以改变了原来的状态，原来是开着的， 所以现在不亮． 故选择$$\\text{B}$$． " ]

"2023-07-07T00:00:00"

[ "2011年全国走美杯三年级竞赛初赛第5题" ]

[ [ { "aoVal": "A", "content": "$$2916$$ " } ], [ { "aoVal": "B", "content": "$$856$$ " } ], [ { "aoVal": "C", "content": "$$795$$ " } ], [ { "aoVal": "D", "content": "$$660$$ " } ] ]

[ "知识标签->数学思想->分类讨论思想" ]

[ "把页码进行分类： ①页码是一位数的是$$1\\tilde{ }9$$，共$$9$$个数字； ②页码是两位数的是$$10\\tilde{ }99$$，共$$90\\times 2=180$$个数字； ③页码是三位数的是$$100\\tilde{ }301$$，共$$202\\times 3=606$$个数字；所以，总共有``数字''个数为$$9+180+606=795$$个． " ]

"2023-07-07T00:00:00"

[ "2016年第12届全国新希望杯小学高年级六年级竞赛复赛第1题4分" ]

[ [ { "aoVal": "A", "content": "$$65$$元 " } ], [ { "aoVal": "B", "content": "$$66$$元 " } ], [ { "aoVal": "C", "content": "$$67$$元 " } ], [ { "aoVal": "D", "content": "$$68$$元 " } ] ]

[ "拓展思维->拓展思维->应用题模块->列方程解应用题->多元一次方程解应用题->小数系数方程（组）解题" ]

[ "假设原价为$$x$$元，进价为$$y$$元，则$$\\left { \\begin{matrix}0.9x-y=60 0.8x-y=46 \\end{matrix} \\right.$$，解得$$\\left { \\begin{matrix}x=140 y=66 \\end{matrix} \\right.$$，即成本为$$66$$元． " ]

"2023-07-07T00:00:00"

[ "2015年第13届全国创新杯小学高年级五年级竞赛初赛第10题" ]

[ [ { "aoVal": "A", "content": "$$S$$ " } ], [ { "aoVal": "B", "content": "$$Q$$ " } ], [ { "aoVal": "C", "content": "$$O$$ " } ], [ { "aoVal": "D", "content": "$$T$$ " } ] ]

[ "拓展思维->拓展思维->组合模块->字母规律->数字与字母结合" ]

[ "$$2015\\div 6\\cdot \\cdot \\cdot \\cdot \\cdot \\cdot 5$$，则$$2015$$个数应为$$S$$． " ]

"2023-07-07T00:00:00"

[ "2020年第1届广东深圳超常思维竞赛五年级竞赛初赛第15题4分" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$3$$ " } ], [ { "aoVal": "D", "content": "$$5$$ " } ], [ { "aoVal": "E", "content": "$$10$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->列方程解应用题->多元一次方程解应用题->整数系数多元一次方程组解应用题" ]

[ "设甲池原有水量为$$v$$，每天降于甲池的雨量为$$x$$，那么$$\\frac{v+2x}{3\\times 2}$$及$$\\frac{v+4x}{2\\times 4}$$都是表示一辆水车一天抽出甲池的水量，即$$\\frac{v+2x}{6}=\\frac{v+4x}{8}$$，得$$v=4x$$， 所以$$x=\\frac{v}{4}$$． 由此可知一辆水车一天抽甲池的水量为$$\\frac{v+2x}{6}=\\frac{v+\\frac{v}{2}}{6}=\\frac{v}{4}$$， 从上面两式，得知降于甲池一天的雨水，恰被一辆水车抽尽． 因乙池的面积是甲池的$$3$$倍，故要当天抽尽乙池当天的雨水，恰需要$$3$$辆水车．又因甲池原有水量为$$v$$，那么乙池原有水量为$$3v$$．已知一辆水车抽甲池的水量为$$\\frac{v}{4}$$，那么一辆水车抽尽乙池原有水量，必须$$3v\\div \\frac{v}{4}=12$$天．今要$$6$$天抽尽乙池原有水量，就需要$$2$$辆水车．故要$$6$$天抽尽乙池的雨量和原有水量，共需要用$$3+2=5$$辆水车． 故选：$$\\text{D}$$． （算数解法）（$$1$$）因甲池的水（包括下的雨水），用$$3$$辆水车，$$2$$天抽尽；若一天抽完，要用$$6$$辆水车．同样，用$$2$$辆水车，$$4$$天抽尽；若一天抽尽，要用$$8$$辆水车． 因为多经过两天，多下了两天雨，所以多用$$8-6=2$$辆水车， 所以，甲池每天的雨量要用一辆水车才能抽尽． 若只抽甲池原存水量，用$$3-1=2$$辆水车，$$2$$天抽尽；用一辆水车，$$4$$天抽尽， 所以每辆水车每天抽甲池原存水量的$$\\frac{1}{4}$$. （$$2$$）因乙池是甲池的$$3$$倍，所以抽当天下的雨水，需要$$3$$辆水车． 又每辆水车是抽乙池原存水量的$$\\frac{1}{4}\\div 3=\\frac{1}{12}$$．今要$$6$$天抽尽乙池的水，每天应抽出$$\\frac{1}{6}$$． 所以只抽乙池原存水量，每天需$$\\frac{1}{6}\\div \\frac{1}{12}=2$$辆水车， 所以抽乙池雨水和原存水量共用的水车是$$3+2=5$$辆这就是说，$$5$$辆水车在$$6$$天抽尽乙池的水． 故选：$$\\text{D}$$． " ]

"2023-07-07T00:00:00"

[ "2015年全国学而思杯小学低年级竞赛学前组第2题", "2015年第5届全国学而思综合能力诊断学前班竞赛第2题" ]

[ [ { "aoVal": "A", "content": "下午$$2$$时 " } ], [ { "aoVal": "B", "content": "下午$$3$$时 " } ], [ { "aoVal": "C", "content": "下午$$4$$时 " } ], [ { "aoVal": "D", "content": "我不知道 " } ] ]

[ "知识标签->拓展思维->组合模块->时间问题->认识钟表" ]

[ "现在是下午$$2$$时，$$1$$小时后是下午$$3$$时． " ]

"2023-07-07T00:00:00"

[ "2014年全国迎春杯四年级竞赛复赛第4题" ]

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

[ "拓展思维->思想->方程思想" ]

[ "根据题意得$$({{a}_{1}}+{{a}_{8}})\\times 8\\div 2=({{a}_{9}}+{{a}_{12}})\\times4\\div 2$$，因为$${{a}_{8}}={{a}_{1}}+14$$，$${{a}_{9}}={{a}_{1}}+16$$，$${{a}_{12}}={{a}_{1}}+22$$， 所以$$({{a}_{1}}+{{a}_{1}}+14)\\times 8\\div2=({{a}_{1}}+16+{{a}_{1}}+22)\\times 4\\div 2$$，解得$${{a}_{1}}=5$$，因此$${{a}_{2}}=5+2=7$$． " ]

"2023-07-07T00:00:00"

[ "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": "$$8$$ " } ], [ { "aoVal": "B", "content": "$$12$$ " } ], [ { "aoVal": "C", "content": "$$32$$ " } ], [ { "aoVal": "D", "content": "$$48$$ " } ] ]

[ "Overseas Competition->知识点->计算模块->小数", "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" ]

[ "$$123456789$$除以$$10000$$，小数点朝前移动四位，答案是$$12345.6789$$，此时，十位是$$4$$，千分位是$$8$$，积为$$32$$． " ]

"2023-07-07T00:00:00"

[ "2014年第26届广东广州五羊杯六年级竞赛第1题4分" ]

[ [ { "aoVal": "A", "content": "$$1 \\%$$ " } ], [ { "aoVal": "B", "content": "$$2 \\%$$ " } ], [ { "aoVal": "C", "content": "$$2.4 \\%$$ " } ], [ { "aoVal": "D", "content": "$$2.8 \\%$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "原浓度为 $$4 \\%=\\frac{4}{100}=\\frac{12}{300}\\overrightarrow{加水}$$ 后来浓度为$$3 \\%=\\frac{3}{100}=\\frac{12}{400}$$，可理解为原有$$12$$克糖（只加水，前后糖不变），原来有糖水$$300$$克，加$$100$$克水，变成$$400$$克糖水．最后再加$$100$$克水，浓度变成$$\\frac{12}{500}=\\frac{24}{1000}=2.4 \\%$$． 故选$$\\text{C}$$． " ]

"2023-07-07T00:00:00"

[ "2014年全国迎春杯四年级竞赛初赛第13题" ]

[ [ { "aoVal": "A", "content": "$$154$$ " } ], [ { "aoVal": "B", "content": "$$156$$ " } ], [ { "aoVal": "C", "content": "$$158$$ " } ], [ { "aoVal": "D", "content": "$$160$$ " } ] ]

[ "拓展思维->拓展思维->数论模块->奇数与偶数->奇数与偶数的加减规律" ]

[ "从$$1$$开始连续奇数相加应该等于项数的平方．因为$$961={{31}^{2}}$$，所以擦去的第一个奇数为$$31\\times2-1+2=63$$． 设第二段有$$n$$个数，$$[65+65+2(n-1)]\\times n\\div 2=1001=7\\times 11\\times 13$$，经尝试，$$n=13$$．所以擦去的第二个数为$$65+2\\times(13-1)+2=91$$． 两个数的和为$$63+91=154$$． " ]

"2023-07-07T00:00:00"

[ "2012年全国美国数学大联盟杯小学高年级竞赛初赛第34题" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$12$$ " } ], [ { "aoVal": "C", "content": "$$13$$ " } ], [ { "aoVal": "D", "content": "$$15$$ " } ] ]

[ "拓展思维->拓展思维->计数模块->枚举法综合->整数分拆->简单拆分->加法拆数（指定个数）" ]

[ "$$2+3+5=10$$ $$2+3+7=12$$ $$3+5+7=15$$ 只有$$13$$不行． " ]

"2023-07-07T00:00:00"

[ "2019年希望杯五年级竞赛模拟第33题" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$3$$ " } ], [ { "aoVal": "D", "content": "$$4$$ " } ], [ { "aoVal": "E", "content": "$$5$$ " } ] ]

[ "拓展思维->思想->分类讨论思想" ]

[ "由题意可知每人只猜对一半，并且每个名次都有人说对，因此可以通过假设法来解决．假设甲说的第一名是勇士是对的，第三名是掘金是错的．由此推出乙说的第三名是勇士是错的，第五名是爵士是对的．丙说的第二名是勇士是错的，第四名是火箭是对的．戊说的第$$5$$名是火箭是错的，第三名是开拓者是对的．丁说的第五名开拓者是错的，第二名是掘金是对的．因此可以得出队伍的名次依次为：勇士、掘金、开拓者、火箭、爵士，开拓者获得第三名． " ]

"2023-07-07T00:00:00"

[ "2004年第3届全国希望杯四年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$28$$ " } ], [ { "aoVal": "C", "content": "$$32$$ " } ], [ { "aoVal": "D", "content": "$$38$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "箱子里有红球$$13$$个、黄球$$10$$个、蓝球$$15$$个，最差的情况是：取出的球为$$15$$个蓝球、$$13$$个红球、$$3$$个黄球这$$31$$个球，此时只要再取出一个必为黄色；所以至少要从中取出$$31 + 1 = 32$$（个）球，才能保证三种颜色的球都至少有$$4$$个． " ]

"2023-07-07T00:00:00"

[ "2015年全国美国数学大联盟杯小学高年级五年级竞赛初赛第29题" ]

[ [ { "aoVal": "A", "content": "$$153$$ " } ], [ { "aoVal": "B", "content": "$$351$$ " } ], [ { "aoVal": "C", "content": "$$370$$ " } ], [ { "aoVal": "D", "content": "$$371$$ " } ] ]

[ "拓展思维->能力->运算求解", "Overseas Competition->知识点->计算模块->数列与数表->数列规律->多重数列" ]

[ "Calculate following the rule. $$1458-702-351-153-153-153-\\cdots $$ All of the terms since the $$10$$th term are $$153$$. " ]

"2023-07-07T00:00:00"

[ "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6题3分" ]

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

[ "拓展思维->思想->对应思想" ]

[ "一共赛了$$4\\times (4-1)\\div2=6$$场，每人各有$$3$$场比赛， 因为甲，乙，丙三人胜的场数相同，若甲，乙，丙各胜$$1$$场， 则丁胜$$6-1\\times3=3$$场，即丁全胜，不合题意（甲胜了丁）． 若甲，乙，丙各胜$$2$$场，则丁胜$$6-2\\times3=0$$场，即丁全输，符合题意． 所以，丁胜了$$0$$场． 故选$$\\text{A}$$． " ]

"2023-07-07T00:00:00"

[ "2014年走美杯二年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$24$$ " } ], [ { "aoVal": "B", "content": "$$23$$ " } ], [ { "aoVal": "C", "content": "$$22$$ " } ], [ { "aoVal": "D", "content": "$$21$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->应用题模块排队问题" ]

[ "解：$$10-1+13-1=21$$（人） 答：这队少先队员共有$$21$$人。 故答案选：D。 " ]

"2023-07-07T00:00:00"

[ "2014年全国迎春杯五年级竞赛复赛第15题" ]

[ [ { "aoVal": "A", "content": "$$141$$ " } ], [ { "aoVal": "B", "content": "$$152$$ " } ], [ { "aoVal": "C", "content": "$$171$$ " } ], [ { "aoVal": "D", "content": "$$175$$ " } ] ]

[ "拓展思维->拓展思维->组合模块->逻辑推理->假设型逻辑推理->答案（数字）正误问题" ]

[ "AB看到自己手里的数，就说我知道了。这里必须理解是什么情况。绝不单单是，其中一个数很大，不能乘以$$2$$，乘以$$2$$就超出$$100$$，而是利用了一个条件$$-\\/-\\/-\\/-\\/-\\/-\\/-$$六个不同的因数 AB拿的数必须具有这样的特点，他手中的数要么乘$$2$$就会超出两位数，且这个数至少有$$6$$个因数，要么乘以一个数后，至少有$$6$$个因数，他手中的数不能有$$6$$个或者$$6$$个以上的因数，否则他不能确定这个数是自己手中的数还是要乘以一个数得到的数。（一个是最大的，一个是次大的） 这样我们根据枚举就能枚举出所有可以满足的情况 枚举如下， $2\\times 3\\times 13=78, 3\\times 13=39$ 可能 $2\\times 3\\times 17=102 $排除 $2\\times 2\\times 13=52, 2\\times 13=26$ 因为$$26\\times 3=78$$ 所以排除 $2\\times 2\\times 17=68, 2\\times 17=34$ 可能 $2\\times 2\\times 19=76, 2\\times 19=38$ 可能 $2\\times 2\\times 23=92, 2\\times 23=46$ 可能 $2\\times 2\\times 29$ 排除 $2\\times 3\\times 3\\times 3=54, 3\\times 3\\times 3=27$ 可能（这一点容易丢） $3\\times 3\\times 5=45$ 因为$$45\\times 2$$ 排除 $3\\times 3\\times 11=99, 3\\times 11=33$，因为$$33\\times 2=66$$ 排除 $2\\times 5\\times 7=70, 5\\times 7=35$ 可能 $2\\times 3\\times 3\\times 5=90, 5\\times 9=45$ 排除因为$$45$$就已经$$6$$个因数了，不能确定是$$45$$还是90 $2\\times 7\\times 7=98, 7\\times 7=49$ 可能 这样就剩下⑦个数了。同时我将他们的因数都写出来。 78 39 26 13 6 3 2 1 一共$$8$$个因数 76 38 19 4 2 1~~一共$$6$$个因数 68 34 17 4 2 1~~一共$$6$$个因数 92 46 23 4 2 1~~一共$$6$$个因数 54 27 18 9 6 3 2 1 一共$$8$$个因数 70 35 14 10 7 5 2 1 一共$$8$$个因数 98 49 14 7 2 1~~~一共$$6$$个因数 $$C$$和$$D$$能从$$AB$$知道，可以很``容易''判断出这$$7$$个数，但此时$$C$$和$$D$$也说自己知道了。很显然，他们手里不可能是上面重复的数，如7,6，$$4$$ ， 3， 2， 1。否则无法确定老师手中的数。 既然能确定，那我们就能排除掉98,因为除了98, 49，没有$$2$$个数能给$$C$$和$$D$$了。 同理，也可以排除76~~68 和92。 如果是$$78$$，$$E$$ 就是$$13$$后的第一个数$$6$$，如果是$$70$$，由于还有一个$$7$$，所以$$E$$拿的一定是$$14$$，这样才能确保$$E$$一定比$$F$$的大。如果是$$54$$，$$E$$只能是6. 所以可以知道$$C$$和$$D$$之间必须有一个数存在才行，这样的话，这个数就能判断只能是$$70$$了，$$E$$是$$14$$，那$$C$$和$$D$$就是$$10$$和$$5$$，$$F$$就是7 答案就是70+35+14+10+7+5=141 " ]

"2023-07-07T00:00:00"

[ "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 " } ], [ { "aoVal": "E", "content": "体育委员，班长，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$$\\text{C}$$． " ]

"2023-07-07T00:00:00"

[ "2018年湖北武汉新希望杯小学高年级五年级竞赛训练题（三）第2题" ]

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

[ "拓展思维->能力->运算求解", "Overseas Competition->知识点->应用题模块->分百应用题->认识单位1" ]

[ "$$42=2\\times 3\\times 7$$，$$(1+1)\\times (1+1)\\times (1+1)=8$$（个）． " ]

"2023-07-07T00:00:00"

[ "2020年希望杯一年级竞赛模拟第2题" ]

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

[ "拓展思维->拓展思维->计数模块->加乘原理->加法原理->握手碰杯问题" ]

[ "因为由题干可知，一共有$$4$$个人，每两人之间需要握一次手，我们可以将这$$4$$个人分别用$$A$$、$$B$$、$$C$$和$$D$$来指代，则所有的握手情况为：$$AB$$，$$AC$$，$$AD$$，$$BC$$，$$BD$$和$$CD$$，所以一共要握$$6$$次，故本题答案为$$\\text{C}$$． " ]

"2023-07-07T00:00:00"

[ "2019年第7届湖北长江杯五年级竞赛复赛A卷第8题3分" ]

[ [ { "aoVal": "A", "content": "$$5$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$9$$ " } ], [ { "aoVal": "D", "content": "无数 " } ] ]

[ "拓展思维->拓展思维->计算模块->分数->分数基础->分数的性质" ]

[ "依据分数单位的意义，即把单位``$$1$$''平均分成若干份取一份的数，叫做分数单位，以及最简分数的概念，即可进行解答． 因为$$\\frac{9}{10}$$、 $$\\frac{7}{10}$$、$$\\frac{3}{10}$$、$$\\frac{11}{10}\\cdots $$的分数单位都是$$\\frac{1}{10}$$，也就是说只要分母是$$10$$的最简分数，它的分数单位都是$$\\frac{1}{10}$$，这样的分数有无数个． 故选：$$\\text{D}$$． " ]

"2023-07-07T00:00:00"

[ "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38题" ]

[ [ { "aoVal": "A", "content": "$$25$$ " } ], [ { "aoVal": "B", "content": "$$26$$ " } ], [ { "aoVal": "C", "content": "$$27$$ " } ], [ { "aoVal": "D", "content": "$$28$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "$$40-(40\\times 4-95)\\div (4+1)=27$$． " ]

"2023-07-07T00:00:00"

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

[ [ { "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）． " ]

"2023-07-07T00:00:00"

[ "2016年全国小学生数学学习能力测评五年级竞赛初赛第8题3分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{8}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{1}{2}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{3}{8}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{5}{8}$$ " } ] ]

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

[ "全部情况有：$$2\\times 2=4$$（种）， 最终得到一个正面一个反面可有以下$$2$$种情况： 正反、反正， 故所求概率为$$\\frac{2}{4}$$． 故答案为：$$\\text{B}$$． " ]

"2023-07-07T00:00:00"

[ "2011年第7届全国新希望杯六年级竞赛第3题4分" ]

[ [ { "aoVal": "A", "content": "$$3:27$$ " } ], [ { "aoVal": "B", "content": "$$4:17$$ " } ], [ { "aoVal": "C", "content": "$$5:14$$ " } ], [ { "aoVal": "D", "content": "$$6:22$$ " } ] ]

[ "拓展思维->拓展思维->行程模块->时钟问题->时钟上的追及问题->已知时间求角度" ]

[ "略． " ]

"2023-07-07T00:00:00"

[ "2017年全国小学生数学学习能力测评六年级竞赛复赛第8题3分" ]

[ [ { "aoVal": "A", "content": "赚了 " } ], [ { "aoVal": "B", "content": "亏了 " } ], [ { "aoVal": "C", "content": "不赚不亏 " } ], [ { "aoVal": "D", "content": "无法比较 " } ] ]

[ "拓展思维->能力->实践应用" ]

[ "可把衣服的成本价看作是单位``$$1$$''，根据分数除法的意义，列式求出衣服的成本价，再分别求出赚的钱数和亏的钱数，再进行比较．据此解答． 本题的关键是求出每件衣服的成本价，再求出每件衣服赚的钱数和亏的钱数，再进行比较． $$80\\div \\left( 1+\\frac{1}{4} \\right)$$ $$=80\\div \\frac{5}{4}$$ $$=64$$（元）， $$80\\div \\left( 1-\\frac{1}{4} \\right)$$ $$=80\\div \\frac{3}{4}$$ $$=106\\frac{2}{3}$$（元）， $$106\\frac{2}{3}-80-\\left( 80-64 \\right)$$ $$=26\\frac{2}{3}-16$$ $$=10\\frac{2}{3}$$（元）， 所以两件衣服合算在一起亏了$$10\\frac{2}{3}$$元． 故选$$\\text{B}$$． " ]

"2023-07-07T00:00:00"

[ "2011年四年级竞赛创新杯", "2011年第9届创新杯四年级竞赛初赛第2题6分" ]

[ [ { "aoVal": "A", "content": "10 " } ], [ { "aoVal": "B", "content": "15 " } ], [ { "aoVal": "C", "content": "20 " } ], [ { "aoVal": "D", "content": "21 " } ] ]

[ "拓展思维->拓展思维->计数模块->加乘原理->加乘原理综合" ]

[ "枚举法:$$4=4+0+0+0=3+1+0+0=2+1+1+0=2+2+0+0=1+1+1+1$$，对应4，0，0，0的四位数有1个，对应3，1，0，0的四位数有6个，对应2，1，1，0的四位数有9个，对应2，2，0，0的四位数有3个，对应1，1，1，1的四位数有1个，故$$1+6+9+3+1=20$$(个)或者也可用隔板法 " ]

"2023-07-07T00:00:00"

[ "2017年希望杯六年级竞赛复赛" ]

[ [ { "aoVal": "A", "content": "$$115$$ " } ], [ { "aoVal": "B", "content": "$$2015$$ " } ], [ { "aoVal": "C", "content": "$$2017$$ " } ], [ { "aoVal": "D", "content": "$$2019$$ " } ] ]

[ "拓展思维->拓展思维->数论模块->质数与合数->质数与合数判定" ]

[ "解：依题意可知： 两数和为奇数，那么一定有一个偶数。偶质数是$$2$$。 当$$b=2$$时，$$5a+2=2027,a=405$$不符合题意。 当$$a=2$$时，$$10+b=2027,b=2017$$符合题意， $$a+b=2+2017=2019$$。 故答案为：D。 " ]

"2023-07-07T00:00:00"

[ "2018年湖北武汉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$$\\text{A:B=C}$$，那么$$\\text{A}$$是比的前项，$$\\text{B}$$是比的后项，$$\\text{C}$$是比的比值 " } ] ]

[ "拓展思维->拓展思维->计算模块->比和比例->比->比的认识" ]

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

"2023-07-07T00:00:00"

[ "2003年第1届创新杯五年级竞赛复赛第3题", "2003年五年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "$$0.11564$$ " } ], [ { "aoVal": "B", "content": "$$1.1564$$ " } ], [ { "aoVal": "C", "content": "$$11.564$$ " } ], [ { "aoVal": "D", "content": "$$115.64$$ " } ] ]

[ "拓展思维->能力->数据处理" ]

[ "除数$$4.13$$变成$$41.3$$扩大到原来的$$10$$倍，被除数没有变，那么商缩小到原来的$$10$$倍，又现在的商比正确的少$$2.52$$，那么现在的商为:$$2.52\\div \\left( 10-1 \\right)=0.28$$，所以被除数为$$0.28\\times 41.3=11.564$$ " ]

"2023-07-07T00:00:00"

[ "2018年湖北武汉锦奥杯小学高年级五年级竞赛初赛湖北赛区第16题10分" ]

[ [ { "aoVal": "A", "content": "$$80$$米/分 " } ], [ { "aoVal": "B", "content": "$$100$$米/分 " } ], [ { "aoVal": "C", "content": "$$150$$米/分 " } ], [ { "aoVal": "D", "content": "$$180$$米/分 " } ] ]

[ "知识标签->拓展思维->行程模块->流水行船问题->基本流水行船问题->水中坠物" ]

[ "从丢桶到找到桶，桶一共顺流而下的时间：$$4050\\div 45=90$$（分钟）， 渔民从丢桶到发现桶丢了的时间等于渔民从返回找桶到找到桶的时间$$90\\div 2=45$$（分钟） 小船顺流而下的速度：$$6525\\div 45=145$$（米/分） 小船在静水中的速度：$$145-45=100$$（米/分） " ]

"2023-07-07T00:00:00"

[ "2014年全国学而思杯二年级竞赛第6题" ]

[ [ { "aoVal": "A", "content": "一 " } ], [ { "aoVal": "B", "content": "二 " } ], [ { "aoVal": "C", "content": "三 " } ], [ { "aoVal": "D", "content": "四 " } ], [ { "aoVal": "E", "content": "五 " } ] ]

[ "海外竞赛体系->知识点->应用题模块->分百应用题->认识单位1", "拓展思维->能力->逻辑分析->代数逻辑推理" ]

[ "$$2$$、$$9$$、$$16$$、$$23 $$星期三；或者$$(23-2+1)\\div7=3$$（周）$$\\cdots 1$$（天），周三那天． " ]

"2023-07-07T00:00:00"

[ "2005年五年级竞赛创新杯", "2005年第3届创新杯五年级竞赛初赛第7题" ]

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

[ "拓展思维->拓展思维->行程模块->直线型行程问题->两人相遇与追及问题->相遇问题->同时出发相向而行" ]

[ "爸爸从家出发去追小明时，小明已经离家$$70\\times 12=840$$(米)，那么爸爸追上小明需要的时间为$$840\\div \\left( 280-70 \\right)=4$$(分钟)，选B． " ]

"2023-07-07T00:00:00"

[ "2017年IMAS小学高年级竞赛（第二轮）第1题4分" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$16$$ " } ], [ { "aoVal": "C", "content": "$$18$$ " } ], [ { "aoVal": "D", "content": "$$20$$ " } ], [ { "aoVal": "E", "content": "$$32$$ " } ] ]

[ "拓展思维->思想->整体思想" ]

[ "由图可知，从第一个三角形开始，以每五个三角形为一个周期，每个周期内有$$3$$个黑色三角形与$$2$$个白色三角形，即黑色三角形比白色三角形多$$1$$个．可知$$80$$个三角形共有$$16$$个周期， 所以黑色三角形比白色三角形总共多$$16$$个． 故选$$\\text{B}$$． " ]

"2023-07-07T00:00:00"

[ "小学中年级三年级其它第九讲最小公倍数与最大公约数", "2016年创新杯六年级竞赛训练题（二）第3题" ]

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

[ "拓展思维->能力->运算求解" ]

[ "$$\\left( 甲，乙\\right)=1$$，则甲、乙互质；$$\\left(乙，丙 \\right)=$$质数；$$\\left(甲，丙 \\right)=$$质数的平方，最小的质数的平方为$$4$$，$$\\therefore $$丙最小：$$3\\times ~4=12$$，甲：$$4$$，乙$$3$$．$$12+4+3=19$$． " ]

"2023-07-07T00:00:00"

[ "2004年五年级竞赛创新杯", "2011年五年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "$$A$$，黄 " } ], [ { "aoVal": "B", "content": "$$B$$，蓝 " } ], [ { "aoVal": "C", "content": "$$C$$，红 " } ], [ { "aoVal": "D", "content": "$$B$$，黄 " } ] ]

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

[ "根据题意，因为乙没有得到$$B$$盒，而$$B$$盒中装有蓝球，所以乙没有得到蓝球；又乙没有得到黄球，所以乙得到的是红球，又$$A$$盒中装的不是红球，乙没有得到$$B$$盒，所以乙得到的$$C$$盒红球，又甲没有得到$$A$$盒，所以甲得到的是$$B$$盒蓝球，所以丙得到的为$$A$$盒黄球，选$$A$$。 " ]

"2023-07-07T00:00:00"

[ "2019年第24届YMO三年级竞赛决赛第8题3分", "2017年第22届全国华杯赛小学中年级竞赛初赛第2题10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->组合模块->抽屉原理->最不利原则" ]

[ "将和为$$10$$的两个数作为一组，有$$\\left( 1,9 \\right)$$，$$\\left( 2,8 \\right)$$，$$\\left( 3,7 \\right)$$，$$\\left( 4,6 \\right)$$、$$\\left( 5,10 \\right)$$，考虑极端情况，若在前$$4$$组中各取一个数字，再取$$5$$和$$10$$，不会有两数和为$$10$$，若再在剩下的数中取任何一个，都会使两数凑成$$10$$，所以至少需要取$$7$$个数． " ]

"2023-07-07T00:00:00"

[ "2015年IMAS小学高年级竞赛第一轮检测试题第1题4分" ]

[ [ { "aoVal": "A", "content": "$$88075$$ " } ], [ { "aoVal": "B", "content": "$$88800$$ " } ], [ { "aoVal": "C", "content": "$$88200$$ " } ], [ { "aoVal": "D", "content": "$$74000$$ " } ], [ { "aoVal": "E", "content": "$$80800$$ " } ] ]

[ "拓展思维->思想->转化与化归的思想", "Overseas Competition->知识点->计算模块->整数->整数乘除->整数乘除法巧算" ]

[ "$$32\\times 37\\times 75$$ $$=4\\times 8\\times 37\\times 3\\times 25$$ $$=(4\\times 25)\\times (37\\times 3)\\times 8$$ $$=100\\times 111\\times 8$$ $$=88800$$． " ]

"2023-07-07T00:00:00"

[ "2009年第7届创新杯六年级竞赛初赛第2题4分" ]

[ [ { "aoVal": "A", "content": "$$2.5\\text{cm}$$ " } ], [ { "aoVal": "B", "content": "$$5.1\\text{cm}$$ " } ], [ { "aoVal": "C", "content": "$$7.5\\text{cm}$$ " } ], [ { "aoVal": "D", "content": "$$8.2\\text{cm}$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->比例应用题->方程法解比例问题" ]

[ "根据已知条件得下半身长是$$160\\times 0.6=96\\text{cm}$$， 设选的高跟鞋的高度是$$x\\text{cm}$$，则根据黄金分割的定义得： $$\\frac{96+x}{160+x}=0.618$$， 解得：$$x\\approx 7.5\\text{cm}$$． 故选$$\\text{C}$$． " ]

"2023-07-07T00:00:00"

[ "2018年IMAS小学高年级竞赛（第二轮）第2题4分" ]

[ [ { "aoVal": "A", "content": "$$12.5$$ " } ], [ { "aoVal": "B", "content": "$$13.75$$ " } ], [ { "aoVal": "C", "content": "$$17.5$$ " } ], [ { "aoVal": "D", "content": "$$30$$ " } ], [ { "aoVal": "E", "content": "$$55$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "四个人总共花费了$$50+5\\times 4=70$$元， 故平均每人花费$$70\\div 4=17.5$$元． 故选$$\\text{C}$$． ", "<p>每人坐车的平均花费为$$50\\div 4=12.5$$元，</p>\n<p>平均每人各总共花费$$12.5+5=17.5$$元．</p>\n<p>故选$$\\text{C}$$．</p>" ]

"2023-07-07T00:00:00"

[ "2014年全国迎春杯四年级竞赛初赛第5题", "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（五）第5~0题3分", "2019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（八）第6题3分" ]

[ [ { "aoVal": "A", "content": "$$216$$ " } ], [ { "aoVal": "B", "content": "$$324$$ " } ], [ { "aoVal": "C", "content": "$$273$$ " } ], [ { "aoVal": "D", "content": "$$301$$ " } ] ]

[ "拓展思维->能力->实践应用" ]

[ "每只猴子多分了$$3$$个，分了$$5\\times 9+(9-3)+57=108$$ （个），那么共$$108\\div 3=36$$（只）猴子．共$$36\\times6+57=273$$（个）桃子． " ]

"2023-07-07T00:00:00"

[ "2006年六年级竞赛创新杯", "2006年五年级竞赛创新杯", "2006年第4届创新杯四年级竞赛复赛第6题", "2006年四年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "$$6$$分钟 " } ], [ { "aoVal": "B", "content": "$$24$$分钟 " } ], [ { "aoVal": "C", "content": "$$30$$分钟 " } ], [ { "aoVal": "D", "content": "$$36$$分钟 " } ] ]

[ "拓展思维->拓展思维->应用题模块->平均数问题->公式类" ]

[ "$$48\\times 5\\div 8=30$$（分钟） " ]

"2023-07-07T00:00:00"

[ "2017年河南郑州联合杯竞赛第6题4分" ]

[ [ { "aoVal": "A", "content": "$$96$$级 " } ], [ { "aoVal": "B", "content": "$$88$$级 " } ], [ { "aoVal": "C", "content": "$$80$$级 " } ], [ { "aoVal": "D", "content": "$$90$$级 " } ] ]

[ "拓展思维->拓展思维->行程模块" ]

[ "$$16\\times \\left( 6-1 \\right)=80$$级． " ]

"2023-07-07T00:00:00"

[ "2013年全国学而思杯三年级竞赛第6题" ]

[ [ { "aoVal": "A", "content": "$$123$$ " } ], [ { "aoVal": "B", "content": "$$132$$ " } ], [ { "aoVal": "C", "content": "$$213$$ " } ], [ { "aoVal": "D", "content": "$$231$$ " } ], [ { "aoVal": "E", "content": "$$312$$ " } ] ]

[ "Overseas Competition->知识点->组合模块->逻辑推理", "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->找矛盾" ]

[ "方法一：由于$$3$$人中有且仅有$$1$$人全预测正确，故可枚举全正确的人的情况： 若小何全正确，则小琳只有第$$3$$句正确，小俊只有第$$2$$句正确，符合要求； 若小琳全正确，则小俊全错，不符合要求； 若小俊全正确，则小琳全错，不符合要求； 方法二：注意到小琳和小俊的预测全都不同，故知全正确的人不可能在这两人之中（否则另一个人就全错，不符合要求） 综上，小何全正确，答案为$$123$$. " ]

"2023-07-07T00:00:00"

[ "2019年第24届YMO二年级竞赛决赛第7题3分", "五年级其它" ]

[ [ { "aoVal": "A", "content": "故事书 " } ], [ { "aoVal": "B", "content": "科技书 " } ], [ { "aoVal": "C", "content": "画册 " } ], [ { "aoVal": "D", "content": "字典 " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "故事书和科技书共有$$17$$本， 则画册和字典共有$$35-17=18$$本， 则画册与字典都不可能是$$9$$本，否则画册与字典都是$$9$$本， 则只能是故事书和科技书中的一类为$$9$$本，故故事书为$$9$$本， 则科技书为$$17-9=8$$本， 则画册为$$16-8=8$$本，矛盾， 故科技书为$$9$$本， 故选$$\\text{B}$$． " ]

"2023-07-07T00:00:00"

[ "2021年新希望杯六年级竞赛初赛第11题5分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{99}{100}\\textgreater{} \\frac{100}{101}\\textgreater{} \\frac{199}{201}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{199}{201}\\textgreater{} \\frac{100}{101}\\textgreater{} \\frac{99}{100}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{99}{100}\\textgreater{} \\frac{199}{201}\\textgreater{} \\frac{100}{101}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{100}{101}\\textgreater{} \\frac{199}{201}\\textgreater\\frac{99}{100}$$ " } ], [ { "aoVal": "E", "content": "$$\\frac{100}{101}\\textgreater{} \\frac{99}{100}\\textgreater{} \\frac{199}{201}$$ " } ] ]

[ "拓展思维->能力->数感认知->分数数字加工" ]

[ "$$\\frac{99}{100}=1- \\frac{1}{100}=1- \\frac{2}{200}$$， $$\\frac{100}{101}=1- \\frac{1}{101}=1- \\frac{2}{202}$$， $$\\frac{199}{201}=1- \\frac{2}{201}$$， 因为$$\\frac{2}{202}\\textless{} \\frac{2}{201}\\textless{} \\frac{2}{200}$$， 所以$$1- \\frac{2}{202}\\textgreater1- \\frac{2}{201}\\textgreater1- \\frac{2}{200}$$， 即$$\\frac{100}{101}\\textgreater{} \\frac{199}{201}\\textgreater{} \\frac{99}{100}$$． 故选$$\\text{D}$$． " ]

"2023-07-07T00:00:00"

[ "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->拓展思维->计算模块->公式类运算->完全平方公式" ]

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

"2023-07-07T00:00:00"

[ "2016年第20届四川成都华杯赛小学中年级竞赛B卷第1~6题60分", "2015年第20届全国华杯赛小学中年级四年级竞赛初赛A卷" ]

[ [ { "aoVal": "A", "content": "狮子、老虎 " } ], [ { "aoVal": "B", "content": "老虎、豹子 " } ], [ { "aoVal": "C", "content": "狮子、豹子 " } ], [ { "aoVal": "D", "content": "老虎、大象 " } ], [ { "aoVal": "E", "content": "狮子、大象 " } ] ]

[ "知识标签->拓展思维->组合模块->逻辑推理->条件型逻辑推理->连线法" ]

[ "在逻辑推理中，原命题成立，则逆否命题也成立. 从题意出发： （1）狮子去则老虎去，逆否命题；老虎不去则狮子也不去 （2）不派豹子则不派老虎，逆否命题：派老虎则要派豹子 （3）派豹子则大象不愿意去，逆否命题；大象去则不能派豹子 从（2）出发可以看出答案为$$B$$ 题目要求有两个动物去，可以使用假设法，若狮子去，则老虎去，老虎去则豹子也去，三个动物去，矛盾，所以狮子不去，若豹子不去则老虎不去，那么只有大象去，矛盾，所以豹子去，豹子去则大象不去，由两骄气去得到结论，老虎要去，所以答案是$$B$$，豹子和老虎去. " ]

"2023-07-07T00:00:00"

[ "2014年全国迎春杯三年级竞赛复赛第7题" ]

[ [ { "aoVal": "A", "content": "$$152$$ " } ], [ { "aoVal": "B", "content": "$$145$$ " } ], [ { "aoVal": "C", "content": "$$140$$ " } ], [ { "aoVal": "D", "content": "$$154$$ " } ] ]

[ "拓展思维->拓展思维->组合模块->数字谜->横式数字谜->横式数字谜的最值" ]

[ "结果最大应该是$$7\\times 8+96=152$$． " ]

"2023-07-07T00:00:00"

[ "2012年IMAS小学高年级竞赛第二轮测试真题第4题" ]

[ [ { "aoVal": "A", "content": "一 " } ], [ { "aoVal": "B", "content": "二 " } ], [ { "aoVal": "C", "content": "三 " } ], [ { "aoVal": "D", "content": "四 " } ], [ { "aoVal": "E", "content": "五 " } ] ]

[ "拓展思维->思想->整体思想" ]

[ "$$500\\div 24=20$$（天）$$\\cdots$$（$$20$$小时） $$20$$（天）$$\\div 7=2$$（个）$$\\cdots$$$$6$$（天） 星期二$$+2$$个星期$$6$$天$$=$$星期一 $$9$$点$$+20$$个小时$$=29$$个小时$$=$$一天$$5$$个小时． ∴星期二凌晨$$5$$点． " ]

"2023-07-07T00:00:00"

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

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

[ "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" ]

[ "将这一串数写成除以$$3$$的余数，则为：$$2$$，$$0$$，$$2$$，$$2$$，$$1$$，$$0$$，$$1$$，$$1$$，$$2$$，$$0$$，$$2\\cdots \\cdots $$ 所以重复的为：``$$2$$，$$0$$，$$2$$，$$2$$，$$1$$，$$0$$，$$1$$，$$1$$''， 故周期为$$8$$．$$2019\\div 8=252$$（组）$$\\cdots \\cdots 3$$（个），则答案为$$2$$，故选择$$\\text{B}$$． " ]

"2023-07-07T00:00:00"

[ "2014年迎春杯三年级竞赛初赛", "2014年迎春杯五年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$15$$ " } ], [ { "aoVal": "B", "content": "$$24$$ " } ], [ { "aoVal": "C", "content": "$$30$$ " } ], [ { "aoVal": "D", "content": "$$35$$ " } ] ]

[ "拓展思维->拓展思维->组合模块->数字谜" ]

[ "$$2014=2\\times 19\\times 53$$，五个一位数之和最大，则两位数应最小 $$2\\times \\left( a+\\overline{1b} \\right)\\times \\left( c+d+e+\\overline{3f} \\right)=2014$$可得$$a+b=9=9+0$$，$$c+d+e+f=23=8+6+5+4$$，则五个一位数的和的最大值为$$2+a+c+d+e=2+9+8+6+5=30$$。 " ]

"2023-07-07T00:00:00"

[ "2017年新希望杯小学高年级六年级竞赛训练题（三）第1题", "2018年湖北武汉新希望杯小学高年级六年级竞赛训练题（三）第1题" ]

[ [ { "aoVal": "A", "content": "$$400$$ " } ], [ { "aoVal": "B", "content": "$$500$$ " } ], [ { "aoVal": "C", "content": "$$600$$ " } ], [ { "aoVal": "D", "content": "$$700$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->浓度问题->抓不变量" ]

[ "$$1000\\times (1-96 \\%)\\div (1-92 \\%)=500$$（千克）． " ]

"2023-07-07T00:00:00"

[ "2018年IMAS小学高年级竞赛（第一轮）第15题4分" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$12$$ " } ], [ { "aoVal": "C", "content": "$$15$$ " } ], [ { "aoVal": "D", "content": "$$20$$ " } ], [ { "aoVal": "E", "content": "$$24$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "设能被$$9$$整除的回文数为 $$\\overline{aba}$$ ﹐其中$$ 1\\leqslant a\\leqslant 9$$ ，$$ 0\\leqslant b\\leqslant 9$$能被$$9$$整除的数的各位数码之和也能被$$9$$整除﹐反之亦然．因此$$a+b+a=2a+b$$能被$$9$$整除． 当$$2a+b=27$$时，只有$$999$$一个数， 当$$2a+b=18$$时，有$$585$$、$$666$$、$$747$$、$$828$$、$$909$$等五个数， 当$$2a+b=9$$时，有$$171$$、$$252$$、$$333$$、$$414$$等四个数， 故符合要求的回文数共有$$1+5+4=10$$个，故选（$$\\text{A}$$）． ", "<p>设能被$$9$$整除的回文数为 $$\\overline{aba}$$ ﹐其中$$ 1\\leqslant a\\leqslant 9$$ ，$$ 0\\leqslant b\\leqslant 9$$．能被$$9$$整除的数的各位数码之和也能被$$9$$整除﹐反之亦然．</p>\n<p>当$$a=1$$时，只有$$171$$一个数，</p>\n<p>当$$a=2$$时，只有$$252$$一个数，</p>\n<p>当$$a=3$$时，只有$$333$$一个数，</p>\n<p>当$$a=4$$时，只有$$414$$一个数，</p>\n<p>当$$a=5$$时﹐只有$$585$$一个数，</p>\n<p>当$$a=6$$时，只有$$666$$一个数，</p>\n<p>当$$a=7$$时，只有$$747$$一个数，</p>\n<p>当$$a=8$$时，只有$$828$$一个数，</p>\n<p>当$$a=9$$时，只有$$909$$、$$999$$二个数，</p>\n<p>故符合要求的回文数共有$$10$$个．故选（$$\\text{A}$$）．</p>" ]

"2023-07-07T00:00:00"

[ "2016年创新杯五年级竞赛训练题（三）第7题", "小学高年级六年级其它2015年数学思维能力等级测试复试第6题" ]

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

[ "拓展思维->能力->运算求解" ]

[ "分组，（$$1$$、$$2$$、$$4$$、$$8$$），（$$3$$、$$6$$、$$12$$），（$$5$$、$$10$$），$$7$$、$$9$$、$$11$$，最少$$9$$个． " ]

"2023-07-07T00:00:00"

[ "2017年河南郑州小升初豫才杯第13题3分", "2017年安徽合肥庐江县小升初第18题1分", "2017年河南郑州豫才杯竞赛第13题" ]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{3}$$ " } ], [ { "aoVal": "B", "content": "$$50 \\%$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{1}{4}$$ " } ], [ { "aoVal": "D", "content": "无法确定 " } ] ]

[ "拓展思维->拓展思维->计数模块->统计与概率->概率->概率基本概念" ]

[ "玩``石头、剪刀、布''游戏，只能出现胜、负、平三种情况，所以两人获胜的可能性都是$$1\\div 3=\\frac{1}{3}$$． " ]

"2023-07-07T00:00:00"

[ "2014年全国迎春杯三年级竞赛初赛第13题", "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（四）第5题3分" ]

[ [ { "aoVal": "A", "content": "$$8$$ " } ], [ { "aoVal": "B", "content": "$$9$$ " } ], [ { "aoVal": "C", "content": "$$46$$ " } ], [ { "aoVal": "D", "content": "$$52$$ " } ] ]

[ "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" ]

[ "盈盈类问题：共有$$(10-1)\\div (5-4)=9$$（只）船． " ]

"2023-07-07T00:00:00"

[ "2015年湖北武汉世奥赛五年级竞赛模拟训练题（三）第1题" ]

[ [ { "aoVal": "A", "content": "开 " } ], [ { "aoVal": "B", "content": "关 " } ], [ { "aoVal": "C", "content": "无法确定 " } ] ]

[ "拓展思维->七大能力->逻辑分析" ]

[ "第$$10$$个人进来后，一共拉了开关$$\\left( 1+2+3+\\ldots \\ldots +10 \\right)$$下，这$$10$$个加数中有$$5$$个偶数，任意数量的偶数之和一定是偶数；有$$5$$个奇数，奇数个奇数的和一定是奇数．因为偶数$$+$$奇数$$=$$奇数，而这盏灯开始是亮着的，则拉奇数下开关，灯一定是关着的． " ]

"2023-07-07T00:00:00"

[ "2017年新希望杯六年级竞赛训练题（三）第6题", "2018年湖北武汉新希望杯六年级竞赛训练题（三）第6题" ]

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

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

[ "先选择$$1$$、$$3$$、$$5$$、$$7$$、$$9$$、$$11$$、$$13$$、$$15$$，再选择$$4$$、$$12$$、$$16$$，共$$11$$个数． " ]

"2023-07-07T00:00:00"

[ "2019年第7届湖北长江杯五年级竞赛复赛B卷第3题3分" ]

[ [ { "aoVal": "A", "content": "$$40392$$ " } ], [ { "aoVal": "B", "content": "$$14382$$ " } ], [ { "aoVal": "C", "content": "$$40293$$ " } ], [ { "aoVal": "D", "content": "$$28438$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "被$$88$$整除即同时被$$11$$和$$8$$整除，若一个整数的未尾三位数能被$$8$$整除，则这个数能被$$8$$整除，$$34B$$能被$$8$$整除，得最后一位$$B$$为$$4$$，若一个整数的奇位数字之和与偶位数字之和的差能被$$11$$整除，则这个数能被$$11$$整除. 奇数位数字之和为$$2+3+4=9$$，偶数位数字之和为$$3+A+4=7+A$$，得第三位$$A$$为$$2$$，所以这个数是$$322344$$，这个数除以$$8$$所得的商是$$322344\\div8=40293$$，因此，答案选：$$\\text{C}$$． " ]

"2023-07-07T00:00:00"

[ "2018年第22届广东世界少年奥林匹克数学竞赛五年级竞赛初赛第1题6分" ]

[ [ { "aoVal": "A", "content": "$$24$$ " } ], [ { "aoVal": "B", "content": "$$114$$ " } ], [ { "aoVal": "C", "content": "$$132$$ " } ] ]

[ "拓展思维->思想->枚举思想" ]

[ "$$25=1+24=2+23=3+22=4+21=5+20=6+19$$ $$=7+18=8+17=9+16=10+15=11+14=12+13$$． 所以：$$1\\times 24=24$$，$$2\\times 23=46$$，$$3\\times 22=66$$，$$4\\times 21=84$$，$$5\\times 20=100$$，$$6\\times 19=114$$，$$7\\times 18=126$$，$$8\\times 17=136$$，$$9\\times 16=144$$，$$10\\times 15=150$$，$$11\\times 14=154$$，$$12\\times 13=156$$， 故$$132$$是不可能的． 故选择$$\\text{C}$$． " ]

"2023-07-07T00:00:00"

[ "2003年第1届创新杯六年级竞赛复赛第22题" ]

[ [ { "aoVal": "A", "content": "$$24000$$ " } ], [ { "aoVal": "B", "content": "$$10000$$ " } ], [ { "aoVal": "C", "content": "$$70$$ " } ], [ { "aoVal": "D", "content": "$$80$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->牛吃草问题->牛吃草转化型->生活中的牛吃草->其他问题" ]

[ "设每亿人每年消耗资源量为$$1$$份， 每年新生资源量：$$(80\\times 300-100\\times 100)\\div (300-100)=70$$（份） 即为保证不断发展，地球上最多养活$$70$$亿人． " ]

"2023-07-07T00:00:00"

[ "2020年新希望杯二年级竞赛初赛（个人战）第6题", "2020年新希望杯二年级竞赛决赛（8月）第6题" ]

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

[ "拓展思维->思想->对应思想" ]

[ "加上李叔叔自己，一共：$$5+6-1=10$$（人）， 故选择：$$\\text{C}$$． " ]

"2023-07-07T00:00:00"

[ "2004年第2届创新杯五年级竞赛初赛第4题" ]

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

[ "拓展思维->能力->逻辑分析" ]

[ "此题应采取求最值的方法解决，看看取最小数与最大数时应取几个，因为$$5\\div 0.6\\approx 8.3$$，$$5\\div \\underbrace{0.666\\cdots 666}_{10{个}6}\\approx 7.5$$，故至少是$$8$$个． 故选$$\\text{C}$$． " ]

"2023-07-07T00:00:00"

[ "2019年第24届YMO三年级竞赛决赛第7题3分", "2020年第24届YMO三年级竞赛决赛第7题3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->还原问题->单一变量还原问题" ]

[ "第$$1$$次井外$$12$$米， 第$$2$$次井外共有$$2\\times 2=4$$（米）， 故井深$$12-4=8$$（米）． 故选$$\\text{C}$$． " ]

"2023-07-07T00:00:00"

[ "2004年第2届创新杯六年级竞赛复赛第5题" ]

[ [ { "aoVal": "A", "content": "$$\\left( \\frac{1}{17}+\\frac{1}{19} \\right)\\times 20$$ " } ], [ { "aoVal": "B", "content": "$$\\left( \\frac{1}{24}+\\frac{1}{29} \\right)\\times30$$ " } ], [ { "aoVal": "C", "content": "$$\\left( \\frac{1}{31}+\\frac{1}{17} \\right)\\times40$$ " } ], [ { "aoVal": "D", "content": "$$\\left( \\frac{1}{41}+\\frac{1}{47} \\right)\\times50$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "$$\\left( \\frac{1}{17}-\\frac{1}{19} \\right)\\div 20=\\frac{2}{17\\times 19}\\div 20=\\frac{1}{17\\times 19}\\times \\frac{1}{10}$$;$$\\left( \\frac{1}{15}-\\frac{1}{21} \\right)\\div 60=\\frac{6}{15\\times 21}\\div 60=\\frac{1}{15\\times 21}\\times \\frac{1}{10}$$; $$\\left( \\frac{1}{13}-\\frac{1}{23} \\right)\\div 100=\\frac{10}{13\\times 23}\\div 100=\\frac{1}{13\\times 23}\\times \\frac{1}{10};$$$$\\left( \\frac{1}{11}-\\frac{1}{25} \\right)\\div 140=\\frac{14}{11\\times 25}\\div 140=\\frac{1}{11\\times 25}\\times \\frac{1}{10};$$ 只需比较$$\\frac{1}{17\\times 19}$$，$$\\frac{1}{15\\times 21}$$，$$\\frac{1}{13\\times 23}$$，$$\\frac{1}{11\\times 25}$$的大小，根据和一定，两数越接近乘 积越大，则$$11\\times 25 \\textless{} 13\\times 23 \\textless{} 15\\times 21 \\textless{} 17\\times 19$$，那么 $$\\frac{1}{11\\times 25}\\textgreater\\frac{1}{13\\times 23}\\textgreater\\frac{1}{15\\times 21}\\textgreater\\frac{1}{17\\times 19}$$，所以答案为$$D$$ " ]

"2023-07-07T00:00:00"

[ "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": "$$10$$ " } ], [ { "aoVal": "E", "content": "$$12$$ " } ] ]

[ "Overseas Competition->知识点->组合模块->抽屉原理->最不利原则", "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$$\\text{B}$$． " ]

"2023-07-07T00:00:00"

[ "2017年河南郑州联合杯竞赛第9题4分" ]

[ [ { "aoVal": "A", "content": "$$2$$个 " } ], [ { "aoVal": "B", "content": "$$5$$个 " } ], [ { "aoVal": "C", "content": "$$4$$个 " } ], [ { "aoVal": "D", "content": "$$11$$个 " } ] ]

[ "拓展思维->思想->逆向思想" ]

[ "最不利原则；$$4+1=5$$个，则最少需要摸出$$5$$个珠子． " ]

"2023-07-07T00:00:00"

[ "2020年广东广州羊排赛四年级竞赛第17题5分" ]

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

[ "课内体系->知识点->式与方程->简易方程->解简易方程->整数简易方程->解方程：ax＋b＝c这种类型", "拓展思维->拓展思维->计算模块->方程基础->一元一次方程" ]

[ "$$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde5x+12=52-3x$$ 解：$$5x+3x=52-12$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde8x=40$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde x=40\\div8$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde x=5$$． " ]

"2023-07-07T00:00:00"

[ "2018年第22届广东世界少年奥林匹克数学竞赛四年级竞赛初赛第1题6分" ]

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

[ "拓展思维->能力->抽象概括" ]

[ "这组数是平方数列， 所以第$$6$$个空为$$6^{2}=36$$， 故选择$$\\text{C}$$． " ]

"2023-07-07T00:00:00"

[ "2008年第6届创新杯六年级竞赛初赛B卷第9题5分", "2008年六年级竞赛创新杯" ]

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

[ "拓展思维->拓展思维->组合模块->组合模块最值问题->和一定型最值问题->多数之积的最值->拆分数的数目不确定" ]

[ "由于希望播出的天数要尽可能的多，所以，在每天播出的集数互不相等的条件下，每天播放的集数应可能的少，又$$1+2+3+4+5+6+7=28$$，如果各天播出的集数分别为1、2、3、4、5、6、7时，那么七天共可播出28集，还剩2集未播出，由于已有过一天播出2集的情况，因此，这余下2集不能再单独一天播出，而只好把它们分到以前的日子里播出，例如，各天播出的集数安排为1、2、3、4、5、7、8或1、2、3、4、5、6、9均可，所以最多可以播7天. " ]

"2023-07-07T00:00:00"

[ "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5题4分" ]

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

[ "拓展思维->拓展思维->组合模块->数字谜->横式数字谜->与数论的结合" ]

[ "由个位分析，$$x=3$$，$$x= 8$$，但是$$7788$$不能被$$38$$整除，则$$\\overline{yz6}=7788\\div 33=236$$．则$$x+y+z=8$$． " ]

"2023-07-07T00:00:00"

[ "2012年第13届上海中环杯小学中年级三年级竞赛初赛第9题" ]

[ [ { "aoVal": "A", "content": "$$9$$ " } ], [ { "aoVal": "B", "content": "$$36$$ " } ], [ { "aoVal": "C", "content": "$$45$$ " } ], [ { "aoVal": "D", "content": "$$27$$ " } ] ]

[ "拓展思维->能力->实践应用" ]

[ "设一张桌子$$x$$元，一把椅子$$y$$元． 则可得出方程：$$\\begin{cases}2x+3y=99 x=4y \\end{cases}$$，代入消元法，解得方程：$$x=36$$，$$y=9$$． 所以一张桌子$$36$$元，一把椅子$$9$$元． " ]

"2023-07-07T00:00:00"

[ "2016年第3届广东深圳鹏程杯六年级竞赛集训材料第十八章综合训练题（八）第13题" ]

[ [ { "aoVal": "A", "content": "$$150$$ " } ], [ { "aoVal": "B", "content": "$$151$$ " } ], [ { "aoVal": "C", "content": "$$152$$ " } ], [ { "aoVal": "D", "content": "$$153$$ " } ] ]

[ "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" ]

[ "人数必须是整数，总人数是$$16$$，$$15$$，$$5$$，$$6$$的公倍数。求公倍数可以由最小公倍数来求，所以可以求出这次竞赛共有$$240$$（$$16$$、$$15$$、$$5$$、$$6$$的最小公倍数）人参加。 $$80$$～$$90$$分的人数占$$\\left( 1-20 \\% -\\frac{1}{6} \\right)=\\frac{19}{30}$$． 由此可求出$$80$$～$$90$$分的人数． $$240\\times \\frac{19}{30}=152$$（人） 答：成绩在$$80$$～$$90$$分的有$$152$$人． 拓展：$$3$$个数短除法，百分数（六年级上）也叫百分比、百分率 " ]

"2023-07-07T00:00:00"

[ "2017年第16届湖北武汉创新杯六年级竞赛邀请赛训练题（二）" ]

[ [ { "aoVal": "A", "content": "$$100$$ " } ], [ { "aoVal": "B", "content": "$$140$$ " } ], [ { "aoVal": "C", "content": "$$160$$ " } ], [ { "aoVal": "D", "content": "$$120$$ " } ] ]

[ "拓展思维->思想->转化与化归的思想" ]

[ "设这$$10$$名同学的平均得分是$$x$$分，那么$$10$$名同学的总得分是$$10x$$分，则： $$10x6\\times \\left( x20 \\right)=4\\times 150$$，$$10x-6x=480$$，$$\\begin{matrix}4x=480 \\end{matrix}$$，$$x=120$$， 答 ：这$$10$$名同学的平均得分是$$120$$分. " ]

"2023-07-07T00:00:00"

[ "竞赛" ]

[ [ { "aoVal": "A", "content": "$$3510$$ " } ], [ { "aoVal": "B", "content": "$$3750$$ " } ], [ { "aoVal": "C", "content": "$$2510$$ " } ], [ { "aoVal": "D", "content": "$$5210$$ " } ] ]

[ "拓展思维->拓展思维->行程模块->直线型行程问题->多人相遇与追及问题->多人相遇追及问题" ]

[ "那$$1$$分钟是甲和丙相遇，所以距离是$$\\left( 60+70 \\right)\\times 1=130$$（米），这距离是乙、丙相遇时间里甲、乙的路程差，所以乙、丙相遇时间$$=130\\div (65-60)=26$$（分），所以路程$$=26\\times (65+70)=3510$$（米）． " ]

"2023-07-07T00:00:00"

[ "2021年第8届鹏程杯五年级竞赛初赛第9题4分" ]

[ [ { "aoVal": "A", "content": "$$35$$ " } ], [ { "aoVal": "B", "content": "$$40$$ " } ], [ { "aoVal": "C", "content": "$$45$$ " } ], [ { "aoVal": "D", "content": "$$50$$ " } ] ]

[ "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->公式法" ]

[ "深圳和广州的路程为$$60$$千米，则从深圳到广州用时$$60\\div30=2$$小时，从广州到深圳用时$$60\\div60=1$$小时，总共用时$$1+2=3$$小时，平均速度$$=60\\times2\\div3=40$$千米$$/$$小时． 答案：$$B$$． " ]

"2023-07-07T00:00:00"

[ "2019年第22届世界少年奥林匹克数学竞赛四年级竞赛决赛（中国区）第4题2分" ]

[ [ { "aoVal": "A", "content": "$$94$$ " } ], [ { "aoVal": "B", "content": "$$95$$ " } ], [ { "aoVal": "C", "content": "$$96$$ " } ], [ { "aoVal": "D", "content": "$$97$$ " } ] ]

[ "拓展思维->思想->逐步调整思想" ]

[ "$$A$$，$$B$$，$$C$$的总分是：$$95\\times 3=285$$（分）， $$B$$，$$C$$，$$D$$的总分是：$$94\\times 3=282$$（分）， $$A$$比$$D$$就多考了︰$$285-282=3$$（分）， 因为$$E$$是第三名考了$$96$$分，所以，$$D$$有两种可能： 一是$$D$$比$$E$$考得少，鉴于$$A$$是第一名，又比$$D$$多三分，$$A$$只能是$$98$$分，而$$D$$是$$95$$分，$$B$$，$$C$$中有一人考$$97$$分，这样的话，$$B$$，$$C$$中的另一个考得分数就是：$$285-98-97=90$$，这与所有人得分都大于$$91$$是矛盾的，所以，$$D$$的名次一定在$$E$$的前面；即$$D$$是第二名；$$D$$是第二名，得分就要多于$$96$$分，结合$$A$$比$$D$$多$$3$$分，可知$$D$$的得分是$$97$$分，$$A$$的得分是：$$100$$分． 故选$$\\text{D}$$． " ]

"2023-07-07T00:00:00"

[ "2021年新希望杯三年级竞赛初赛第30题5分" ]

[ [ { "aoVal": "A", "content": "星期一 " } ], [ { "aoVal": "B", "content": "星期二 " } ], [ { "aoVal": "C", "content": "星期三 " } ], [ { "aoVal": "D", "content": "星期四 " } ], [ { "aoVal": "E", "content": "星期五 " } ], [ { "aoVal": "F", "content": "星期六 " } ], [ { "aoVal": "G", "content": "星期日 " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "周期问题： 元旦$$1$$月$$1$$日是星期日，则周期：星期日、星期一、星期二、星期三、星期四、星期五、星期六$$\\cdots\\cdots$$ 从$$1$$月$$1$$日到$$2$$月$$1$$日一共有$$31+1=32$$（天）， $$32\\div7=4$$（周）$$\\cdots\\cdots4$$（天）， 所以$$2$$月$$1$$日是星期三，则$$2$$月份的周期：星期三、星期四、星期五、星期六、星期日、星期一、星期二$$\\cdots\\cdots$$ 闰年的$$2$$月有$$29$$天，$$29\\div7=4$$（周）$$\\cdots\\cdots1$$（天）， 所以$$2$$月$$29$$日是星期三， 因此这一年的$$2$$月有$$5$$个星期三． " ]