Datasets:

math-eval

/

TAL-SCQ5K

like 41

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

[ "2015年湖北武汉世奥赛五年级竞赛模拟训练题（二）第2题" ]

[ [ { "aoVal": "A", "content": "$$15$$ " } ], [ { "aoVal": "B", "content": "$$18$$ " } ], [ { "aoVal": "C", "content": "$$21$$ " } ], [ { "aoVal": "D", "content": "$$24$$ " } ] ]

[ "知识标签->拓展思维->应用题模块->列方程解应用题->列方程解应用题等量代换->等量代换传递型" ]

[ "$$12$$辆大卡车装的重量等于$$12\\div 2\\times 5=30$$辆小卡车，则原计划运$$1$$次的货物现在要运$$\\left( 18+30 \\right)\\div 8=6$$（次），所以一共需运$$6\\times 4=24$$次． " ]

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

[ "2004年五年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "$$17$$束 " } ], [ { "aoVal": "B", "content": "$$18$$束 " } ], [ { "aoVal": "C", "content": "$$19$$束 " } ], [ { "aoVal": "D", "content": "$$20$$束 " } ] ]

[ "拓展思维->拓展思维->计数模块->容斥原理->三量容斥" ]

[ "由容斥原理：$$50-\\left( 16+15+21-7-8-10+5 \\right)=18$$（束），选$$B $$。 " ]

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

[ "2012年第10届全国创新杯小学高年级六年级竞赛第2题4分" ]

[ [ { "aoVal": "A", "content": "$$2012$$ " } ], [ { "aoVal": "B", "content": "$$4024$$ " } ], [ { "aoVal": "C", "content": "$$3700$$ " } ], [ { "aoVal": "D", "content": "$$3800$$ " } ] ]

[ "拓展思维->思想->整体思想" ]

[ "根据题意可以得到：$$2012+412\\left( n-1 \\right)=572n$$，$$n=10$$，十个数总和为$$5720$$，要使最大的数尽量大，则其他的尽量小，所以最大数最大值为 $$5720-2012-1\\times 8=3700$$． " ]

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

[ "2014年全国学而思杯一年级竞赛第10题" ]

[ [ { "aoVal": "A", "content": "美国 " } ], [ { "aoVal": "B", "content": "俄国 " } ], [ { "aoVal": "C", "content": "中国 " } ], [ { "aoVal": "D", "content": "日本 " } ] ]

[ "知识标签->课内知识点->数学广角->推理->解决简单逻辑推理问题" ]

[ "中国比日本、美国快，美国比俄国快，所以日本、美国、俄国都不是第一，而韩国也不是第一个到的，所以中国是第一． " ]

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

[ "2003年六年级竞赛创新杯", "2003年第1届创新杯六年级竞赛初赛第2题" ]

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

[ "拓展思维->拓展思维->应用题模块->列方程解应用题" ]

[ "现在只知道第一根钢管用去$$\\frac{3}{10}$$米，第二根钢管用去$$\\frac{3}{10}$$，并不知道这两只钢管的长度，它们可能很长，也可能很短，这样就可以任意由我们来假设这两根钢管的长度。我们先来看看选项$$A$$是否成立，$$A$$要求第一根钢管剩下的部分长些，那我们可以假设第一根钢管的长为$$10$$米，第二根钢管的长为$$1$$米，显然这样就可以满足选项$$A$$。再来看看选项$$B$$，$$B$$要求第二根钢管剩下的部分长些，那我们可以假设第一根钢管的长为$$1$$米，第二根钢管剩下的部分同样长，我们假设第一根钢管的长为$$1$$米，那么用去$$\\frac{3}{10}$$米后还剩下$$\\frac{7}{10}$$米，若要第二根钢管剩下的长度也为$$\\frac{7}{10}$$米，只要第二根钢管原来的长为$$\\frac{7}{10}\\div \\left( 1-\\frac{3}{10} \\right)=1$$（米）即可，选项$$C$$也可以满足。 " ]

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

[ "2017年第13届湖北武汉新希望杯小学高年级五年级竞赛决赛第6题" ]

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

[ "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", "Overseas Competition->知识点->组合模块->逻辑推理->条件型逻辑推理->表格法" ]

[ "由②④⑤推断出乙为第二名，由于甲丁一起打球，甲丙一起上班，再由③推断出丙丁一个是第一名，另一个为第三名，则第四名为甲． " ]

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

[ "2009年第7届创新杯四年级竞赛初赛第8题4分" ]

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

[ "拓展思维->思想->逐步调整思想" ]

[ "全体同学走后，黑板上的数是$$\\left(30+5\\right)\\div5=7$$；最后一名学生走过之前，黑板上的数是$$\\left(7+7\\right)\\div2=7$$，或$$\\left(7+14\\right)\\div3=7$$，总之，最后一名学生（即第$$40$$名学生）走过之前，黑板上的数还是$$7$$． 同理，第$$39$$名学生来到之时，黑板上的数还是$$7\\cdots \\cdots $$由此可知，第$$1$$名学生到来之时，黑板上的数还是$$7$$，即黑板上最初的数是$$7$$． " ]

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

[ "2018年第22届广东世界少年奥林匹克数学竞赛三年级竞赛初赛第4题6分" ]

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

[ "拓展思维->思想->对应思想" ]

[ "这列数按照``$$3$$，$$1$$，$$2$$''的周期排列，则，$$53\\div 3=17$$（组）$$\\cdots \\cdots 2$$（个）， 所以第$$53$$个数位这个周期的第$$2$$个数，即为$$1$$． 故选择$$\\text{B}$$． " ]

"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复赛第3题" ]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$9$$ " } ], [ { "aoVal": "D", "content": "$$12$$ " } ] ]

[ "拓展思维->能力->构造模型->模型思想" ]

[ "一台功效$$\\frac{1}{45}$$，五台需要$$9$$小时． " ]

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

[ "2003年六年级竞赛创新杯", "2003年第1届创新杯六年级竞赛初赛第7题" ]

[ [ { "aoVal": "A", "content": "40\\% " } ], [ { "aoVal": "B", "content": "45\\% " } ], [ { "aoVal": "C", "content": "55\\% " } ], [ { "aoVal": "D", "content": "60\\% " } ] ]

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

[ "正方形的棱长最大为3厘米，正方形的体积最大为$$3\\times 3\\times 3=27$$(立方厘米)，原来长方形体积为$$5\\times 4\\times 3=60$$(立方厘米)，那么正方形的体积比原来的体积减少$$\\left( 60-27 \\right)\\div 60\\times 100 \\%=55 \\%$$. " ]

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

[ "2017年第17届世奥赛六年级竞赛决赛第2题" ]

[ [ { "aoVal": "A", "content": "$$100n\\left( n+1 \\right)+25$$ " } ], [ { "aoVal": "B", "content": "$$100\\left( n+1 \\right)+25$$ " } ], [ { "aoVal": "C", "content": "$$100{{n}^{2}}+25$$ " } ], [ { "aoVal": "D", "content": "$$100{{n}^{2}}+100n+5$$ " } ] ]

[ "知识标签->学习能力->七大能力->逻辑分析" ]

[ "$${{\\left( 10n+5 \\right)}^{2}}=100n\\left( n+1 \\right)+25=100{{n}^{2}}+100n+25$$． " ]

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

[ "2017年第15届全国希望杯六年级竞赛" ]

[ [ { "aoVal": "A", "content": "$$50$$ " } ], [ { "aoVal": "B", "content": "$$48\\frac{6}{25}$$ " } ], [ { "aoVal": "C", "content": "$$46\\frac{6}{25}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{6}{25}$$ " } ] ]

[ "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" ]

[ "原式$$=\\frac{{{5}^{2}}-1+2}{{{5}^{2}}-1}+\\frac{{{7}^{2}}-1+2}{{{7}^{2}}-1}+\\frac{{{9}^{2}}-1+2}{{{9}^{2}}-1}+...+\\frac{{{99}^{2}}-1+2}{{{99}^{2}}-1}$$ $$=1+\\frac{2}{{{5}^{2}}-1}+1+\\frac{2}{{{7}^{2}}-1}+1+\\frac{2}{{{9}^{2}}-1}+...+1+\\frac{2}{{{99}^{2}}-1}$$ $$=48+\\frac{2}{4\\times 6}+\\frac{2}{6\\times 8}+\\frac{2}{8\\times 10}+...+\\frac{2}{98\\times 100}$$ $$=48+\\left( \\frac{1}{4}-\\frac{1}{6} \\right)+\\left( \\frac{1}{6}-\\frac{1}{8} \\right)+\\left( \\frac{1}{8}-\\frac{1}{10} \\right)+...+\\left( \\frac{1}{98}-\\frac{1}{100} \\right)$$ $$=48+\\frac{1}{4}-\\frac{1}{100}$$ $$=48\\frac{6}{25}$$ " ]

"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3题5分" ]

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

[ "拓展思维->能力->运算求解" ]

[ "不变量为水，则水：$$600\\times (1-7 \\%)=558$$（克）， 后总：$$558\\div (1-10 \\%)=620$$（克）， 加糖：$$620-600=20$$（克）． 故选$$\\text{B}$$． " ]

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

[ "2019年第7届湖北长江杯六年级竞赛复赛B卷第10题3分" ]

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

[ "拓展思维->思想->对应思想" ]

[ "我们把乙的时间看做单位``$$1$$''，则甲的工作时间就是$$1-\\frac{2}{3}$$，然后分别求出他们的工作效率，进一步求出答案． $$1\\div \\left( 1-\\frac{1}{3} \\right)\\div (1\\div 1)$$$$=\\frac{3}{2}\\div 1$$$$=3:2$$． 故选$$\\text{C}$$． " ]

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

[ "2014年陕西西安小升初西工大附中入学真卷（六）第1题3分", "2017年河南郑州东风杯竞赛初赛" ]

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

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

[ "根据分数的基本性质可知，$$\\frac{a\\times 2}{b\\times 2}$$和$$\\frac{a\\div 2}{b\\div 2}$$与$$\\frac{a}{b}$$的大小相等．$$\\frac{a}{b}$$的分子和分母同时加$$2$$得到的分数比$$\\frac{a}{b}$$大．$$\\frac{a}{b}$$的分子和分母同时减$$2$$得到的分数比$$\\frac{a}{b}$$小．所以$$\\frac{a+2}{b+2}$$最大． " ]

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

[ "2017年IMAS小学高年级竞赛（第一轮）第16题4分" ]

[ [ { "aoVal": "A", "content": "$$120$$ " } ], [ { "aoVal": "B", "content": "$$1200$$ " } ], [ { "aoVal": "C", "content": "$$1320$$ " } ], [ { "aoVal": "D", "content": "$$2640$$ " } ], [ { "aoVal": "E", "content": "$$2880$$ " } ] ]

[ "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" ]

[ "小华上班时间为$$8$$小时， 此期间分针转了$$8$$圈， 即$$8\\times 360=2880$$度， 而时针每小时转$$30$$度， $$8$$小时共转了$$8\\times 30=240$$度， 分针转过的度数比时针转过的度数多了$$2880-240=2640$$度， 故选$$\\text{D}$$． " ]

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

[ "2018年全国小学生数学学习能力测评六年级竞赛初赛第8题3分", "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8第3题3分", "2018年湖南长沙小升初数学入学试卷第4题4分" ]

[ [ { "aoVal": "A", "content": "$$1\\frac{1}{3}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{3}{10}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{3}{4}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{5}{6}$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->工程问题->简单工程问题->基本工程问题" ]

[ "首先根据王师傅加工一批零件，$$\\frac{1}{2}$$小时加工了这批零件的$$\\frac{3}{8}$$，工作效率$$=$$工作量$$\\div $$工作时间，求出每小时加工这批零件的几分之几；求出剩下的工作量，然后根据工作时间$$=$$工作量$$\\div $$工作效率，求出全部加工完还需要多少小时即可． 解；$$\\frac{3}{8}\\div \\frac{1}{2}=\\frac{3}{4}$$ $$(1-\\frac{3}{8})\\div \\frac{3}{4}$$ $$=\\frac{5}{8}\\div \\frac{3}{4}$$ $$=\\frac{5}{6}$$（小时） 答:全部加工完还需要$$\\frac{5}{6}$$小时． 故选$$\\rm D$$． " ]

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

[ "2019年亚洲国际数学奥林匹克公开赛（AIMO）四年级竞赛决赛第5题3分" ]

[ [ { "aoVal": "A", "content": "$$45825\\div 33$$ " } ], [ { "aoVal": "B", "content": "$$29484\\div 36$$ " } ], [ { "aoVal": "C", "content": "$$41552\\div 56$$ " } ], [ { "aoVal": "D", "content": "$$39715\\div 65$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->整数->整数乘除->整数除法运算->表外除法计算" ]

[ "$$\\text{B}$$答案为$$819$$，$$\\text{C}$$答案为$$742$$，$$\\text{D}$$答案为$$611$$，$$BCD$$都能被整除． 故选$$\\text{A}$$. " ]

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

[ "2016年新希望杯六年级竞赛训练题（三）第5题" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$15$$ " } ], [ { "aoVal": "C", "content": "$$6$$ " } ], [ { "aoVal": "D", "content": "$$12$$ " } ] ]

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

[ "设$$x$$年后$$3$$个孙子的年龄之和等于祖父的年龄． $$67+x=15+13+9+3x$$，$$x=15$$． " ]

"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->拓展思维->数论模块->奇数与偶数->奇数与偶数的加减规律" ]

[ "第$$2014$$个人进来后，一共拉了开关$$\\left( 1+2+3+\\ldots \\ldots +2014 \\right)$$下，这$$2014$$个加数中有$$2014\\div 2=1007$$个偶数，任意数量的偶数之和一定是偶数；有$$1007$$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年河南郑州联合杯六年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{10}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{1}{11}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{1}{9}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{1}{8}$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->分百应用题->求分率" ]

[ "盐水共$$10+100=110$$克，则盐占盐水的$$\\frac{10}{110}=\\frac{1}{11}$$． " ]

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

[ "2019年第7届湖北长江杯六年级竞赛复赛B卷第3题3分" ]

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

[ "拓展思维->思想->分类讨论思想" ]

[ "完成本题要根据每个题目的内容分别进行分析，才能确定正确的选项有几个． （$$1$$）相邻两个偶数相差$$2$$，由此可设这$$9$$个连续的偶数中的中间的那个为$$x$$，则这$$9$$个连续偶数的和为： $$x-2\\times 4+x-2\\times 3+\\cdots +x+x+2+\\cdots +x+2\\times 4$$，则其平均数为$$9x\\div 9=x$$，即$$90$$则最小的为$$90-2\\times 4=82$$，最大的为$$90+2\\times 4=98$$．所以，$$9$$个连续偶数的平均数是$$90$$，这些数中最小的一个是$$2$$，最大的是$$18$$是错误的； （$$2$$）暗室中共有$$3$$种不同颜色的袜子，最差情况是取出三只后，每种颜色各一只，此时只要再取出一只即能确保取出一双相同颜色的袜子，即最少取出$$3+1=4$$只；正确． （$$3$$）由题意可知，游泳或骑自行车会其中至少一项的有$$48-12=36$$人，这$$36$$人中，不会游泳的有$$36-27=9$$人，不会骑自行车的有$$36-25=11$$人，则班既会游泳又会骑自行车的有$$36-(9+11)=16$$人．正确； （$$4$$）如果有一级，则有一种走法：$$1$$；如有两级，则有$$2$$级走法即：$$1$$，$$1$$，如果有$$3$$级有$$3$$种：$$111$$，$$12$$，$$21$$；四级有$$5$$种：$$1111$$，$$22$$，$$121$$，$$211$$，$$112$$；$$\\cdots $$，由此可以发现，从第$$3$$个数开始，每一个数都等于它前面的$$2$$个数之和．共有$$10$$级，$$1$$，$$2$$，$$3$$，$$5$$，$$8$$，$$13$$，$$21$$，$$34$$，$$55$$，$$89$$．则到第十级共有$$89$$种不同走法．正确． 所以，（$$1$$）（$$3$$）（$$4$$）的描述都是正确的，共$$3$$句． 故选$$\\text{C}$$ ． " ]

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

[ "2006年第4届创新杯四年级竞赛初赛A卷第9题", "2006年四年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "28元 " } ], [ { "aoVal": "B", "content": "56元 " } ], [ { "aoVal": "C", "content": "70元 " } ], [ { "aoVal": "D", "content": "112元 " } ] ]

[ "拓展思维->拓展思维->应用题模块->平均数问题->移多补少" ]

[ "出同样多的钱应拿同样多的货，多拿货就要多出钱，以丁为标准，甲、乙、丙比丁共多$$3+7+14=24$$件，这24件本应平均分配，即每人要拿$$24\\div 4=6$$件(每人所得件数本应是丁现有件数加6)，而乙比丁实际多拿7件，也就是说乙比原计划多拿$$7-6=1$$件，乙要多拿出14元，则每件14元.丙比计划多拿$$14-6=8$$件，这样丙应再拿出8件的钱，即$$14\\times 8=112$$元.又因为丁比计划少拿6件，应取回$$6\\times 14=84$$元，但丁已经收了乙的14元，所以丙还应付给丁$$84-14=70$$元，丙另外的钱，3件共$$112-70=42$$(元)应给甲，故选C " ]

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

[ "2017年河南郑州豫才杯四年级竞赛初赛第1题" ]

[ [ { "aoVal": "A", "content": "$$100$$个月 " } ], [ { "aoVal": "B", "content": "$$700$$个星期 " } ], [ { "aoVal": "C", "content": "$$36500$$天~~~~~~~~~ " } ], [ { "aoVal": "D", "content": "$$87600$$小时 " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "$$10$$岁相当于$$120$$个月，$$480$$个星期，$$3650$$天，$$87600$$小时． " ]

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

[ "2017年四川成都六年级竞赛“全能明星”选拔赛第1题2分", "2017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（五）第1题2分" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$0.5$$ " } ], [ { "aoVal": "D", "content": "$$1.5$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "$$\\frac{8}{9}\\times \\left[ \\frac{3}{4}-\\left( \\frac{7}{16}-\\frac{1}{4} \\right) \\right]=0.5$$． ", "<p>$$\\frac{8}{9}\\times \\left[ \\frac{3}{4}-\\left( \\frac{7}{16}-\\frac{1}{4} \\right) \\right]$$</p>\n<p>$$=\\frac{8}{9}\\times \\left( \\frac{3}{4}-\\frac{7}{16}+\\frac{1}{4} \\right)$$</p>\n<p>$$=\\frac{8}{9}\\times \\frac{9}{16}$$</p>\n<p>$$=\\frac{1}{2}$$</p>\n<p>$$=0.5$$</p>\n\n" ]

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

[ "2006年第4届创新杯四年级竞赛复赛第1题" ]

[ [ { "aoVal": "A", "content": "$$8$$ " } ], [ { "aoVal": "B", "content": "$$16$$ " } ], [ { "aoVal": "C", "content": "$$20$$ " } ], [ { "aoVal": "D", "content": "$$24$$ " } ] ]

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

[ "实际每天加工的个数为$$80\\times 5\\div 4=100$$(个)；每天比计划增加的个数是$$100-80=20$$ (个)，故选：$$\\text{C}$$． " ]

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

[ "2008年六年级竞赛创新杯", "2008年第6届创新杯六年级竞赛初赛B卷第1题5分" ]

[ [ { "aoVal": "A", "content": "53 " } ], [ { "aoVal": "B", "content": "40 " } ], [ { "aoVal": "C", "content": "39 " } ], [ { "aoVal": "D", "content": "15 " } ] ]

[ "拓展思维->拓展思维->计算模块->分数->分数基础->分数的分类" ]

[ "$$x$$可取7，8，11，13，14.其和为53. " ]

"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->思想->转化与化归的思想" ]

[ "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"

[ "2008年第6届创新杯四年级竞赛初赛A卷第1题5分" ]

[ [ { "aoVal": "A", "content": "$$2007$$ " } ], [ { "aoVal": "B", "content": "$$2008$$ " } ], [ { "aoVal": "C", "content": "$$0$$ " } ], [ { "aoVal": "D", "content": "$$20072008$$ " } ] ]

[ "拓展思维->思想->转化与化归的思想" ]

[ "$$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde2008\\times 20072007-2007\\times 20082008$$ $$=2008\\times 2007\\times 10001-2007\\times 2008\\times 10001$$ $$=0$$． 故选$$\\text{C}$$． " ]

"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6题3分" ]

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

[ "拓展思维->思想->方程思想" ]

[ "根据题意分析可知，总分数$$\\div $$考的次数$$=$$平均分数；可设他至少要考$$x$$次才能尽快达到$$110$$分以上，那么 淘气的总分数为：$$120x+105\\times 4$$，淘气考试的次数为：$$x+4$$，淘气平均分应为$$\\geqslant $$$$100$$，根据公式列方程解答即可． 设淘气至少要考$$x$$次才能尽快达到$$110$$分以上， $$\\left( 120x+105\\times 4 \\right)$$$$\\div \\left( x+4 \\right)\\geqslant 110$$， $$120x+420\\geqslant 110x+440$$， $$120x-110x\\geqslant 440-420$$， $$10x\\geqslant 20$$， $$x\\geqslant 2$$． 即再考$$2$$次满分平均分可达到$$110$$分以上． 故选答案$$\\text{B}$$． " ]

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

[ "2020年新希望杯四年级竞赛初赛第26题", "2020年希望杯四年级竞赛模拟第26题" ]

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

[ "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" ]

[ "根据题目可知，打水分别按照小和尚，高和尚和胖和尚的顺序进行的，也就是说以每$$3$$天为一个周期进行循环的，所以小和尚第$$50$$次打水应该经过了$$49$$个周期多$$1$$天． 也就是$$3\\times 49+1=148$$（天），小和尚第一次打水是星期一， 那么经过$$148$$天应该是$$148\\div 7=21$$（周）$$\\cdots \\cdots 1$$（天），所以小和尚第$$50$$次打水仍然是星期一． 故选：$$\\text{A}$$． " ]

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

[ "其它改编自2015年全国希望杯六年级竞赛初赛第11题" ]

[ [ { "aoVal": "A", "content": "$$\\frac{2}{5}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{1}{3}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{7}{19}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{10}{19}$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->分百应用题->求分率" ]

[ "设女生人数为$$10$$份，男生人数为$$9$$份，则参加演出的人数为$$\\frac{2}{5}\\times 10+\\frac{1}{3}\\times 9=7$$，占全班人数的$$\\frac{7}{10+9}=\\frac{7}{19}$$． " ]

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

[ "2004年第2届创新杯六年级竞赛复赛第3题" ]

[ [ { "aoVal": "A", "content": "一个 " } ], [ { "aoVal": "B", "content": "两个 " } ], [ { "aoVal": "C", "content": "三个 " } ], [ { "aoVal": "D", "content": "四个 " } ] ]

[ "拓展思维->拓展思维->数论模块->位值原理与进制->位值原理运用->计算中的位值原理" ]

[ "设这个两位数为$$\\overline{ab}$$，依题意有$$\\overline{ab}=a+7b$$，即$$10a+b=a+7b$$，整理得$$a=\\frac{2}{3}b$$，又$$a$$，$$b$$均为一位数且$$a$$不为$$0$$，那么$$b$$只能为$$3$$，$$6$$，$$9$$对应的$$a$$分别为$$2$$，$$4$$，$$6$$，所以这样的两位数有$$23$$，$$46$$，$$69$$三个，故选$$\\text{C}$$． ", "<p>设十位数是$$x$$，个位数字是$$y$$，</p>\n<p>根据题意，得$$x+7y=10x+y$$，即$$3x=2y$$，</p>\n<p>而$$1\\leqslant x\\leqslant 9$$，$$0\\leqslant y\\leqslant 9$$，且$$x$$，$$y$$都是整数，</p>\n<p>根据条件同时满足的$$x$$，$$y$$的值有：$$2$$，$$3$$；$$4$$，$$6$$；$$6$$，$$9$$共$$3$$组．</p>\n<p>即这样的两位数是$$23$$，$$46$$，$$69$$共$$3$$个．</p>\n<p>故选：$$\\text{C}$$．</p>" ]

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

[ "2021年第8届鹏程杯四年级竞赛初赛第5题4分", "2021年第8届鹏程杯五年级竞赛初赛第5题4分" ]

[ [ { "aoVal": "A", "content": "$$1010$$ " } ], [ { "aoVal": "B", "content": "$$1011$$ " } ], [ { "aoVal": "C", "content": "$$2020$$ " } ], [ { "aoVal": "D", "content": "$$2021$$ " } ], [ { "aoVal": "E", "content": "以上都不对 " } ] ]

[ "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" ]

[ "分组运算，$$2$$和$$3$$一组，$$4$$和$$5$$一组，$$\\cdots \\cdots 2020$$和$$2021$$一组，每组结果都是$$1$$，一共有$$2020\\div2=1010$$组，所以和是$$1010$$，前面还剩下一个$$1$$，$$1+1010=1011$$． 故选$$\\text{B}$$． " ]

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

[ "2012年IMAS小学高年级竞赛第一轮检测试题第7题3分" ]

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

[ "拓展思维->能力->实践应用" ]

[ "甲早到$$5-(-10)=15$$分钟，丙早到$$5-3=2$$分钟，丁早到$$(-5)-(-10)=5$$分钟，乙迟到$$10-(-5)=15$$分钟． " ]

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

[ "2017年北京学而思杯四年级竞赛年度教学质量测评第20题3分" ]

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

[ "拓展思维->拓展思维->组合模块->操作与策略->归纳递推" ]

[ "略 " ]

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

[ "2018年全国小学生数学学习能力测评四年级竞赛初赛第10题3分" ]

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

[ "拓展思维->能力->逻辑分析->代数逻辑推理" ]

[ "$$4\\times\\left(1+2+3\\right)=24$$，$$5\\times\\left(2+3\\right)-1=24$$， $$6\\times 2\\times\\left(3-1\\right)=24$$，$$7\\times 3+2+1=24$$， $$8\\times 3\\div\\left(2-1\\right)=24$$，$$9\\times 3-2-1=24$$， $$10\\times 2+3+1=24$$，$$11\\times 2+3-1=24$$， $$12\\times\\left(3+1\\right)\\div 2=24$$． " ]

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

[ "2018年第17届春蕾杯一年级竞赛初赛第14题6分" ]

[ [ { "aoVal": "A", "content": "金 " } ], [ { "aoVal": "B", "content": "银 " } ], [ { "aoVal": "C", "content": "铜 " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "推理题目 " ]

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

[ "2021年新希望杯六年级竞赛初赛第28题5分" ]

[ [ { "aoVal": "A", "content": "$$145504$$ " } ], [ { "aoVal": "B", "content": "$$135184$$ " } ], [ { "aoVal": "C", "content": "$$145164$$ " } ], [ { "aoVal": "D", "content": "$$135524$$ " } ], [ { "aoVal": "E", "content": "$$145584$$ " } ] ]

[ "拓展思维->拓展思维->数论模块->高斯记号->高斯记号的复杂应用" ]

[ "列出一些分数找规律： （$$1$$）$$\\left[ \\frac{1}{7} \\right]$$、$$\\left[ \\frac{3}{7} \\right]$$、$$\\left[ \\frac{5}{7} \\right]$$都为$$0$$； （$$2$$）$$\\left[ \\frac{7}{7} \\right]$$、$$\\left[ \\frac{9}{7} \\right]$$、$$\\left[ \\frac{11}{7} \\right]$$、$$\\left[ \\frac{13}{7} \\right]$$都为$$1$$； （$$3$$）$$\\left[ \\frac{15}{7} \\right]$$、$$\\left[ \\frac{17}{7} \\right]$$、$$\\left[ \\frac{19}{7} \\right]$$都为$$2$$； （$$4$$）$$\\left[ \\frac{21}{7} \\right]$$、$$\\left[ \\frac{23}{7} \\right]$$、$$\\left[ \\frac{25}{7} \\right]$$、$$\\left[ \\frac{27}{7} \\right]$$都为$$3$$； （$$5$$）$$\\left[ \\frac{29}{7} \\right]$$、$$\\left[ \\frac{31}{7} \\right]$$、$$\\left[ \\frac{33}{7} \\right]$$都为$$4$$； $$\\cdots $$； （$$287$$）$$\\left[ \\frac{2009}{7} \\right]$$、$$\\left[ \\frac{2011}{7} \\right]$$、$$\\left[ \\frac{2013}{7} \\right]$$、$$\\left[ \\frac{2015}{7} \\right]$$都为$$287$$； （$$288$$）$$\\left[ \\frac{2017}{7} \\right]$$、$$\\left[ \\frac{2019}{7} \\right]$$、$$\\left[ \\frac{2021}{7} \\right]$$都为$$288$$， 所以题目式$$=4\\times \\left( 1+3+5+\\cdots +287 \\right)+3\\times \\left( 2+4+\\cdots +288 \\right)$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde=4\\times \\left( 2+4+6+\\cdots +288-144 \\right)+3\\times \\left( 2+4+\\cdots +288 \\right)$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde=7\\times \\left( 2+4+6+\\cdots +288 \\right)-4\\times 144$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde=7\\times 2\\times \\left( 1+2+\\cdots +144 \\right)-576$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde=7\\times 2\\times \\frac{\\left( 1+144 \\right)\\times 144}{2}-576$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde=145584$$． " ]

"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1题3分" ]

[ [ { "aoVal": "A", "content": "$$30$$ " } ], [ { "aoVal": "B", "content": "$$36$$ " } ], [ { "aoVal": "C", "content": "$$34$$ " } ], [ { "aoVal": "D", "content": "$$32$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "已知$$f\\left( m,n \\right)=a{{m}^{3}}+b{{n}^{3}}$$，$$f\\left( 1,2 \\right)=a\\times {{1}^{3}}+b\\times {{2}^{3}}=a+8b=4$$， 那么$$f\\left( 2,4 \\right)=a\\times {{2}^{3}}+b\\times {{4}^{3}}=8a+64b=8\\times \\left( a+8b \\right)=8\\times 4=32$$． 故选$$\\text{D}$$． " ]

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

[ "2022年小学高年级六年级竞赛（小学奥数5-2-5整除与分类计数综合专项训练）第6题", "2022年小学高年级六年级竞赛（小学奥数5-2-5整除与分类计数综合专项训练）" ]

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

[ "拓展思维->拓展思维->数论模块->因数与倍数" ]

[ "【详解】 要使得$$2008+a$$能被$$2007-a$$整除，我们可以将条件等价的转化为只要让$$\\frac{2008+a}{2007-a}$$是一个整数即可。下面是一个比较难的技巧，我们知道若$$a$$可以使得$$\\frac{2008+a}{2007-a}$$是一个整数，那么$$a$$也同样可以使得$$\\frac{2008+a}{2007-a}+1=\\frac{2008+a+2007-a}{2007-a}=\\frac{4015}{2007-a}$$是一个整数，这样只要$$2007-a$$是$$4015$$的约数即可，将$$4015$$分解可知其共有$$8$$个因数，其中$$4015$$是最大的一个，但是显然没有可以让$$2007-a$$等于$$4015$$的$$a$$的值，其余的$$7$$个均可以有对应的$$a$$的值，所以满足条件的$$a$$的取值共有$$7$$个。 " ]

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

[ "2010年第8届创新杯六年级竞赛初赛第3题4分", "2018年四川成都武侯区成都西川中学小升初面试真卷（四）第8题3分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{3}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{1}{4}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{2}{5}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{1}{6}$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "数字和是$$3$$的倍数有：$$（1，2），（1，5），（2，4），（3，6），（3，3），（4，5）$$，则$$\\frac{6\\times 2}{6\\times 6}=\\frac{1}{3}$$． 故选$$\\text{A}$$． " ]

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

[ "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->能力->实践应用" ]

[ "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$$\\text{B}$$． " ]

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

[ "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": "$$11\\times 2+22\\times 2+33\\times 3+44\\times 4+55\\times 5+66\\times 6+77\\times 7+88\\times 8+99\\times 9$$ " } ], [ { "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->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）" ]

[ "暂无 " ]

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

[ "2011年全国华杯赛竞赛初赛第4题" ]

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

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

[ "任何两人说的话都不能同时为真，所以最多有一个人说的是真话，如果有一个人复习了，那么李说的是真话，符合题意；如果没有人复习了，那么张说的是真话，矛盾． " ]

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

[ "2016年新希望杯六年级竞赛训练题（一）第6题" ]

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

[ "拓展思维->能力->逻辑分析" ]

[ "$$8$$、$$5$$、$$6$$的最小公倍数是$$120$$，这三人相会的周期是$$120$$天，$$120\\div 7=17$$（周）$$\\cdots\\cdots1$$（天），下次碰面将在星期三． " ]

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

[ "2016年创新杯六年级竞赛训练题（二）第6题" ]

[ [ { "aoVal": "A", "content": "$$10 \\%$$ " } ], [ { "aoVal": "B", "content": "$$11 \\%$$ " } ], [ { "aoVal": "C", "content": "$$12 \\%$$ " } ], [ { "aoVal": "D", "content": "$$15 \\%$$ " } ] ]

[ "拓展思维->思想->方程思想" ]

[ "设普通型``生死水''的浓度为$$x \\%$$，初始重量为$$100$$，连续两次加入的艾草浸液和水仙根粉末重量都是$$a$$，那么$$\\left { \\begin{matrix} \\dfrac{x}{100+a}=9 \\% \\dfrac{x+a}{100+2a}=23 \\% \\end{matrix}\\Rightarrow \\left { \\begin{matrix} 100x-9a=900 100x+54a=2300 \\end{matrix}\\Rightarrow x=11 \\right. \\right.$$，故普通型``生死水''的浓度为$$11 \\%$$． " ]

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

[ "2012年全国华杯赛小学高年级竞赛初赛第1题", "2017年全国小升初八中入学备考课程" ]

[ [ { "aoVal": "A", "content": "$$30$$ " } ], [ { "aoVal": "B", "content": "$$40$$ " } ], [ { "aoVal": "C", "content": "$$50$$ " } ], [ { "aoVal": "D", "content": "$$60$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->分数->分数运算->分小四则混合运算" ]

[ "$$\\left[ \\left( 0.8+\\frac{1}{5} \\right)\\times 24+6.6 \\right]\\div \\frac{9}{14}-7.6$$ $$=\\left[ (0.8+0.2)\\times 24+6.6 \\right]\\div \\frac{9}{14}-7.6$$ $$=\\left[ 1\\times 24+6.6 \\right]\\div \\frac{9}{14}-7.6$$ $$=30.6\\div \\frac{9}{14}-7.6$$ $$=30.6\\times \\frac{14}{9}-7.6$$ $$=47.6-7.6$$ $$=40$$． " ]

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

[ "2014年华杯赛六年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$11$$点$$40$$分 " } ], [ { "aoVal": "B", "content": "$$11$$点$$50$$分 " } ], [ { "aoVal": "C", "content": "$$12$$点 " } ], [ { "aoVal": "D", "content": "$$12$$点$$10$$分 " } ] ]

[ "拓展思维->拓展思维->行程模块->直线型行程问题->路程速度时间->单人简单行程问题" ]

[ "根据题意，实际走比计划走全程省了$$5+4=9$$（分钟）。将全程平均分成$$9$$等份，则在每一等份中实际走比计划走省$$1$$分钟。所以当走完$$5$$等份时，到达的地点时间实际和计划的时间相同，此时是丙地即在全程的$$\\frac{5}{9}$$处。计划走全程用$$3$$小时即$$180$$分钟，则走到丙地用$$180\\times \\frac{5}{9}=100$$（分钟）。$$10$$点$$10$$分出发走$$100$$分钟是$$11$$点$$50$$分。 " ]

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

[ "2008年五年级竞赛创新杯", "2008年第6届创新杯五年级竞赛初赛A卷第9题5分" ]

[ [ { "aoVal": "A", "content": "6 " } ], [ { "aoVal": "B", "content": "7 " } ], [ { "aoVal": "C", "content": "8 " } ], [ { "aoVal": "D", "content": "10 " } ] ]

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

[ "记快车、慢车的速度为$${{v}_{\\text{快}}}$$米/秒、$${{v}_{\\text{慢}}}$$米/秒，且$${{v}_{\\text{快}}}={{v}_{\\text{慢}}}$$，快、慢两列火车相向而行，则相对速度为$${{v}_{\\text{快}}}+{{v}_{\\text{慢}}}=3{{v}_{\\text{慢}}}$$（米/秒）.设坐在快火车上的人见慢车驶过的时间是$$x$$秒.则$$\\begin{cases}\\frac{50}{3{{v}_{\\text{慢}}}}=5 \\frac{80}{3{{v}_{\\text{慢}}}}=x \\end{cases}$$ 即$$\\begin{cases}80=x\\times 3{{v}_{\\text{慢}}}\\left( 1 \\right) 50=5\\times {{v}_{\\text{慢}}}\\left( 2 \\right) \\end{cases}$$$$\\left( 2 \\right)\\div \\left( 1 \\right)$$ 得$$\\frac{50}{80}=\\frac{5}{x}$$.解得$$x=8$$ " ]

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

[ "2020年希望杯二年级竞赛模拟第15题" ]

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

[ "拓展思维->思想->对应思想" ]

[ "根据题意分析可知，用黄小鸭前面的数量加上后面的数量再加上自己所以一共有： $$2+3+1=6$$（只）小鸭， 故答案为：$$\\text{C}$$． " ]

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

[ "2017年河南郑州豫才杯竞赛第10题", "2017年河南郑州小升初豫才杯第二场第10题" ]

[ [ { "aoVal": "A", "content": "$$99$$ " } ], [ { "aoVal": "B", "content": "$$197$$ " } ], [ { "aoVal": "C", "content": "$$97$$ " } ], [ { "aoVal": "D", "content": "$$196$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "分析可知，三角形数的规律为第$$n$$个数比第$$n-1$$个数大$$n$$（$$n$$大于$$1$$），则第$$99$$个三角形数比第$$98$$个三角形数大$$99$$，第$$98$$个三角形数比第$$97$$个三角形数大$$98$$，所以所求的差为$$99+98=197$$． " ]

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

[ "2014年迎春杯三年级竞赛复赛" ]

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

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

[ "解：根据分析，由丁说的可得``丁$$\\textgreater$$乙''，根据丙说的可得``戊$$\\textgreater$$丙''， 根据甲说的可得``丙$$\\textgreater$$甲$$\\textgreater$$丁''，综合可得``戊$$\\textgreater$$丙$$\\textgreater$$甲$$\\textgreater$$丁$$\\textgreater$$乙''。 故选：D。 " ]

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

[ "2019年第24届YMO二年级竞赛决赛第4题3分" ]

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

[ "拓展思维->拓展思维->计数模块->枚举法综合->整数分拆->整数拆分应用->加法拆数（应用）" ]

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

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

[ "2015年第13届全国创新杯五年级竞赛复赛第7题" ]

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

[ "拓展思维->能力->逻辑分析->代数逻辑推理" ]

[ "$$23\\times 2023+(1+2+\\cdot \\cdot \\cdot +2023)m$$除以$$14$$的余数， $$23\\div 14$$余$$9$$， $$23\\div 14$$余$$7$$， $$1+2+\\cdot \\cdot \\cdot +2023=2023\\times 1012$$刚好是$$14$$的倍数， 那么只用看$$23\\times 2023$$除以$$14$$的余数是$$7$$． " ]

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

[ "2014年IMAS小学高年级竞赛第二轮检测试题第2题4分" ]

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

[ "拓展思维->能力->运算求解" ]

[ "因为$$2+3+4+5+6+7$$ $$=5+9+13$$ $$=14+13$$ $$=27$$（个） 所以最多有$$6$$位小朋友． " ]

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

[ "2017年华杯赛小学高年级竞赛（深圳营一）第4题10分" ]

[ [ { "aoVal": "A", "content": "$$n\\textgreater11$$ " } ], [ { "aoVal": "B", "content": "$$n\\textless{}11$$ " } ], [ { "aoVal": "C", "content": "$$n=11$$ " } ], [ { "aoVal": "D", "content": "$$n$$有多个值 " } ] ]

[ "拓展思维->思想->分类讨论思想" ]

[ "若$$n\\textgreater11$$，则$$11$$整除$$\\varphi \\left( n \\right)$$，但$$11$$不能整除$$22n-32$$． 因此，$$n\\textgreater11$$不符合要求．故，$$n\\leqslant 11$$． 若$$7\\textless{}n\\leqslant 11$$，则$$\\varphi \\left( n \\right)=2\\times 3\\times 5\\times 7=210$$， 由$$210=22n-32$$， 得$$n=11$$． 若$$5\\textless{}6\\leqslant 7$$，则$$\\varphi \\left( n \\right)=2\\times 3\\times 5=30$$， 由$$30=22n-32$$，得正整数$$n$$不存在． 若$$3\\textless{}n\\leqslant 5$$，则$$\\varphi \\left( n \\right)=2\\times 3=6$$， 由$$6=22n-32$$，得正整数$$n$$不存在． 若$$n=3$$，则$$\\varphi \\left( n \\right)=2$$，由$$2=22n-32$$， 得正整数$$n$$不存在． ∴满足条件的正整数$$n$$只有$$1$$个，$$n=11$$． 解法二：由$$\\varphi \\left( n \\right)=22n-32$$，得$$\\varphi \\left( n \\right)-1024=22\\left( n-48 \\right)$$． 由于$$\\varphi \\left( n \\right)$$是偶数，但不是$$4$$的倍数，因此，$$n-48$$是奇数． 若$$n-48\\geqslant 3$$，则$$n-48$$含有奇数的素数因子$$p$$， 即$$p$$为奇素数，且$$p$$整除$$n-48$$． 由$$n-48\\textless{}n$$知，$$p$$整除$$\\varphi \\left( n \\right)$$．由此$$p$$整除$$1024$$矛盾． 故，$$n-48\\textless{}3$$，即$$n\\leqslant 49$$，且$$n$$为奇数． ∵$$n\\leqslant 49$$时，$$22n-32\\leqslant 22\\times 49-32=1046$$， ∴$$\\varphi \\left( n \\right)\\leqslant 1046$$． 又$$2\\times 3\\times 5\\times 7=210$$，$$2\\times 3\\times 5\\times 7\\times 11=210\\times 11\\textgreater1046$$． ∴$$n\\leqslant 11$$，即$$n=3，$$5$$，$$7$$，$$9$$，11$$． 将$$n=3，$$5$$，$$7$$，$$9$$，11$$分别代入$$\\varphi \\left( n \\right)=22n-32$$验证， $$n=3$$时，$$\\varphi \\left( 3 \\right)=2$$，$$22n-32=34$$，不符合要求． $$n=$$5时，$$\\varphi \\left( 5 \\right)=2\\times 3=6$$，$$22n-32=78$$，不符合要求． $$n=7$$时，$$\\varphi \\left( 7 \\right)=2\\times 3\\times 5=30$$，$$22n-32=122$$，不符合要求． $$n=9$$时，$$\\varphi \\left( 9 \\right)=2\\times 3\\times 5\\times 7=210$$，$$22n-32=166$$，不符合要求． $$n=11$$时，$$\\varphi \\left( 11 \\right)=2\\times 3\\times 5\\times 7=210$$，$$22n-32=210$$，符合要求． ∴满足条件的正整数$$n$$只有$$1$$个，$$n=11$$． 故选$$\\text{C}$$． " ]

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

[ "2016年第21届全国华杯赛小学中年级四年级竞赛初赛B卷", "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5题3分" ]

[ [ { "aoVal": "A", "content": "$$35$$ " } ], [ { "aoVal": "B", "content": "$$40$$ " } ], [ { "aoVal": "C", "content": "$$41$$ " } ], [ { "aoVal": "D", "content": "$$47$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "由前$$11$$场的总分减去前$$10$$场的总分即为第$$11$$场的得分：$$34\\times 11-333=374-333=41$$分． " ]

"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39题" ]

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

[ "拓展思维->能力->运算求解" ]

[ "十位数是$$0$$． " ]

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

[ "2017年河南郑州小升初豫才杯第二场第13题", "2017年河南郑州豫才杯竞赛第13题" ]

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

[ "拓展思维->能力->逻辑分析" ]

[ "第一题做对的$$25$$人中，有$$10$$人全对，则有$$25-10=15$$人是只做对第一题，也是做错第二题的；已知第二题总共有$$18$$人做错，那么多的$$3$$人就是两题都错的．故选$$C$$． " ]

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

[ "2016年创新杯六年级竞赛训练题（二）第5题" ]

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

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

[ "满足这样的尾数有（$$0$$，$$0$$，$$0$$），（$$0$$，$$5$$，$$5$$），故末三位可能为$$000$$，$$250$$，$$500$$，$$750$$共$$4$$种． " ]

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

[ "2017年河南郑州联合杯竞赛第2题4分" ]

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

[ "拓展思维->思想->逐步调整思想" ]

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

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

[ "2017年全国迎春杯六年级竞赛初赛第3题" ]

[ [ { "aoVal": "A", "content": "$$320$$ " } ], [ { "aoVal": "B", "content": "$$410$$ " } ], [ { "aoVal": "C", "content": "$$504$$ " } ], [ { "aoVal": "D", "content": "$$610$$ " } ], [ { "aoVal": "E", "content": "$$728$$ " } ] ]

[ "海外竞赛体系->知识点->应用题模块->分百应用题", "拓展思维->拓展思维->应用题模块->分百应用题->转化单位1->统一单位1" ]

[ "分数应用题，既考查了量率对应的思想，也考查了单位$$1$$变化问题的处理，非常全面． 由于卧底的$$20 \\% $$被驱离出岛，相当于原来的$$\\frac{2}{3}\\times 20 \\% =\\frac{2}{15}$$被驱离，那么$$728$$人相当于原来人数的$$1-\\frac{2}{15}=\\frac{13}{15}$$，则原来人数为$$728\\div \\frac{13}{15}=840$$人，那么现在还剩下的卧底有$$840\\times \\frac{2}{3}\\times \\left( 1-20 \\% ~\\right)+840\\times \\left( 1-\\frac{2}{3} \\right)\\times \\frac{1}{5}=504$$人． " ]

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

[ "2017年第13届湖北武汉新希望杯五年级竞赛决赛第1题" ]

[ [ { "aoVal": "A", "content": "互质的两个数没有公因数 " } ], [ { "aoVal": "B", "content": "$$12$$和$$18$$的最小公倍数是$$72$$ " } ], [ { "aoVal": "C", "content": "$$24$$和$$30$$的最大公因数是$$6$$ " } ], [ { "aoVal": "D", "content": "两个质数的和一定是偶数 " } ] ]

[ "知识标签->数学思想->转化与化归的思想" ]

[ "$$\\text{A}$$选项：互质的两个数公因数为$$1$$，$$\\text{B}$$选项：$$12$$和$$18$$的最小公倍数是$$36$$，$$\\text{D}$$选项：$$2+3=5$$为奇数． " ]

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

[ "2017年河南郑州豫才杯小学高年级五年级竞赛初赛第15题" ]

[ [ { "aoVal": "A", "content": "$$4$$小时 ~~~ " } ], [ { "aoVal": "B", "content": "$$4.5$$小时~~ " } ], [ { "aoVal": "C", "content": "$$5$$小时 " } ], [ { "aoVal": "D", "content": "$$6$$小时 " } ] ]

[ "知识标签->拓展思维->应用题模块->经济问题->分段计价问题" ]

[ "设最多在停车场停了$$x$$小时，$$3+\\left( x-1 \\right)\\times 2\\times 1.5=13.5$$，解得$$x=4.5$$，所以他们的车在停车场最多停$$4.5$$小时． " ]

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

[ "2020年广东广州羊排赛四年级竞赛第9题4分" ]

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

[ "拓展思维->能力->逻辑分析->代数逻辑推理" ]

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

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

[ "2016年河南郑州联合杯小学高年级六年级竞赛复赛第14题2分", "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12第8题3分", "2014年四川成都小升初七中嘉祥外国语学校第30题", "小学高年级六年级上学期其它北师大版53天天练" ]

[ [ { "aoVal": "A", "content": "$$480$$ " } ], [ { "aoVal": "B", "content": "$$300$$~ " } ], [ { "aoVal": "C", "content": "$$360$$~ " } ], [ { "aoVal": "D", "content": "$$440$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->浓度问题->浓度基本题型->已知溶液浓度求溶质" ]

[ "浓度问题；盐的重量占盐水重量的$$\\frac{1}{9}$$，则盐占水的$$\\frac{1}{8}$$， $$15\\div \\left( \\frac{1}{8}-\\frac{1}{11} \\right)$$ $$=15\\div\\frac{3}{88}$$ $$=440$$． 故选$$\\text{D}$$. " ]

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

[ "2016年创新杯五年级竞赛训练题（一）第8题" ]

[ [ { "aoVal": "A", "content": "$$15$$ " } ], [ { "aoVal": "B", "content": "$$20$$ " } ], [ { "aoVal": "C", "content": "$$25$$ " } ], [ { "aoVal": "D", "content": "$$30$$ " } ] ]

[ "拓展思维->能力->实践应用" ]

[ "平时逆水航行所用时间是顺水航行所用时间的$$2$$倍，所以顺水航行速度是逆水航行的$$2$$倍，即$${{V}_{水}}+8=2\\times \\left( 8-{{V}_{水}} \\right)$$，解得：$${{V}_{}}=\\frac{8}{3}$$，所以水速为$$\\frac{8}{3}$$千米/小时，变为原来的$$2$$倍后变为$$\\frac{16}{3}$$千米/小时．设甲、乙两地相距$$S$$千米，则：$$\\frac{S}{\\frac{16}{3}+8}+\\frac{S}{8-\\frac{16}{3}}=9$$ 解得：$$S=20$$，所以甲、乙两地相距$$20$$千米． " ]

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

[ "2006年第4届创新杯五年级竞赛初赛A卷第2题", "2006年五年级竞赛创新杯" ]

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

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

[ "整点的分钟数$$60n$$应为9的倍数，即$$n$$最小为3，所以12点后最早为3点时既亮灯又响铃. " ]

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

[ "2014年全港小学数学挑战赛四年级竞赛初赛A卷第13题5分" ]

[ [ { "aoVal": "A", "content": "$$31$$ " } ], [ { "aoVal": "B", "content": "$$33$$ " } ], [ { "aoVal": "C", "content": "$$35$$ " } ], [ { "aoVal": "D", "content": "$$37$$ " } ] ]

[ "拓展思维->思想->逆向思想" ]

[ "因为这$$5$$个奇数是连续的，所以每相邻两个奇数之间相差$$2$$，用$$5$$个连续奇数的和除以$$5$$，可得到最中间的奇数，最大的数与最中间的奇数相差$$4$$，所以用最中间的奇数加上$$4$$即可得到当中最大的奇数，故列式为：$$145\\div 5+4=29+4=33$$，即最大的奇数是$$33$$．故答案为：$$33$$． " ]

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

[ "2016年全国学而思杯一年级竞赛样卷第7题" ]

[ [ { "aoVal": "A", "content": "甲姓张，乙姓黄，丙姓李． " } ], [ { "aoVal": "B", "content": "甲姓黄，乙姓李，丙姓张． " } ], [ { "aoVal": "C", "content": "甲姓黄，乙姓张，丙姓李． " } ], [ { "aoVal": "D", "content": "甲姓李，乙姓张，丙姓黄． " } ] ]

[ "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" ]

[ "甲姓黄，乙姓张，丙姓李． " ]

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

[ "其它改编自2012年全国希望杯六年级竞赛初赛第3题" ]

[ [ { "aoVal": "A", "content": "$$3.\\dot{1}41592\\dot{6}$$ " } ], [ { "aoVal": "B", "content": "$$3.1\\dot{4}1592\\dot{6}$$ " } ], [ { "aoVal": "C", "content": "$$3.14\\dot{1}592\\dot{6}$$ " } ], [ { "aoVal": "D", "content": "$$3.14159\\dot{2}\\dot{6}$$ " } ] ]

[ "拓展思维->能力->数感认知->小数数字加工" ]

[ "要使小数最小，则循环节开始的几位尽量小，因此从$$1$$开始循环，下一位为$$4$$，依次往下，最小的为$$3.\\dot{1}41592\\dot{6}$$． " ]

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

[ "2016年第16届世奥赛六年级竞赛决赛第7题" ]

[ [ { "aoVal": "A", "content": "$$6$$ " } ], [ { "aoVal": "B", "content": "$$8$$ " } ], [ { "aoVal": "C", "content": "$$10$$ " } ], [ { "aoVal": "D", "content": "$$16$$ " } ] ]

[ "拓展思维->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->方程法解其他问题" ]

[ "找到功能做效率的等量关系列出方程，设这份文件总共有$$x$$页，$$\\frac{x}{10}+\\frac{x}{8}-0.2=\\frac{x}{5}$$，解得$$x=8$$． " ]

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

[ "2020年第24届YMO四年级竞赛决赛第9题3分", "2019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->能力->公式记忆->言语化数学原理" ]

[ "$$5$$，$$a$$，$$b$$，$$c$$，$$4035$$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$$\\text{D}$$． " ]

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

[ "2019年第24届YMO二年级竞赛决赛第1题3分" ]

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

[ "拓展思维->能力->运算求解", "Overseas Competition->知识点->计数模块->加乘原理->排队问题" ]

[ "根据题意分析可知，小$$Y$$后面有$$42-22=20$$（人），$$22-20-1=1$$（人）．那么小$$M$$应该在小$$Y$$的正前方，所以小$$Y$$和小$$M$$之间有$$0$$个人． 故选：$$\\text{A}$$． " ]

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

[ "2016年全国小学生数学学习能力测评四年级竞赛初赛第8题3分" ]

[ [ { "aoVal": "A", "content": "$$432$$ " } ], [ { "aoVal": "B", "content": "$$434$$ " } ], [ { "aoVal": "C", "content": "$$436$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->整数->整数乘除->整数乘法运算->表外乘法计算" ]

[ "一个因数为$$54\\div 3=18$$，另一个因数为$$96\\div 4=24$$， 所以原来两数的积为$$18\\times 24=432$$． 故选：$$\\text{A}$$． " ]

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

[ "2016年创新杯六年级竞赛训练题（三）第1题" ]

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

[ "拓展思维->能力->抽象概括" ]

[ "按照互质的定义枚举判断即可． " ]

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

[ "2017年第22届全国华杯赛小学中年级竞赛初赛第3题10分", "2021年陕西西安六年级下学期小升初模拟分类专题五十五（计数原理 排列组合 容斥原理 抽屉原理）第2题", "2019年陕西西安碑林区西北工业大学附属中学小升初入学真卷六第4题3分" ]

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

[ "拓展思维->能力->运算求解" ]

[ "由题知，密码由三位数构成，且只包括$$8$$与$$5$$两个数字． 以$$5$$为百位数字，有$$558$$、$$585$$、$$588$$三种情况； 以$$8$$为百位数字，有$$855$$、$$885$$、$$858$$三种情况． 所以最少要试$$3+3=6$$（次），才能确保打开箱子． " ]

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

[ "2016年新希望杯六年级竞赛训练题（五）第1题" ]

[ [ { "aoVal": "A", "content": "$$44$$ " } ], [ { "aoVal": "B", "content": "$$100$$ " } ], [ { "aoVal": "C", "content": "$$125$$ " } ], [ { "aoVal": "D", "content": "$$150$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->分百应用题->量率对应求单位1" ]

[ "$$20\\div \\left( 125 \\% -1 \\right)=80$$（米），$$80\\div \\left( 1-20 \\% \\right)=100$$（米）． " ]

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

[ "2015年第13届全国创新杯五年级竞赛复赛第7题" ]

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

[ "拓展思维->拓展思维->数论模块->余数问题->余数的性质->余数性质综合" ]

[ "$$23\\times 2015+(1+2+\\cdot \\cdot \\cdot +2015)m$$除以$$14$$的余数，由于$$1+2+\\cdot \\cdot \\cdot +2015=2015\\times 1008$$是$$14$$的倍数，那么只用看$$23\\times 2015$$除以$$14$$的余数是$$5$$． " ]

"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13题" ]

[ [ { "aoVal": "A", "content": "$$2:1$$ " } ], [ { "aoVal": "B", "content": "$$12:25$$ " } ], [ { "aoVal": "C", "content": "$$13:25$$ " } ], [ { "aoVal": "D", "content": "$$1:2$$ " } ] ]

[ "拓展思维->能力->运算求解", "Overseas Competition->知识点->计算模块->比和比例" ]

[ "$4$的倍数有$$25$$个 $8$的倍数有$$12$$个 " ]

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

[ "走美杯三年级竞赛", "走美杯六年级竞赛" ]

[ [ { "aoVal": "A", "content": "$$50$$ " } ], [ { "aoVal": "B", "content": "$$100$$ " } ], [ { "aoVal": "C", "content": "$$2500$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->数列与数表->数列规律->金字塔数列" ]

[ "原式$$=50\\times50=2500$$ " ]

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

[ "2015年第11届全国新希望杯五年级竞赛复赛第4题" ]

[ [ { "aoVal": "A", "content": "星星 " } ], [ { "aoVal": "B", "content": "希希 " } ], [ { "aoVal": "C", "content": "望望 " } ], [ { "aoVal": "D", "content": "贝贝 " } ] ]

[ "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" ]

[ "星星和希希比贝贝高，希希比望望高，希希不是最高，那么只能是星星． " ]

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

[ "拓展思维->能力->数据处理" ]

[ "从数列的第一个数开始余数分别是$$1$$,$$1$$,$$2$$,$$3$$,$$0$$,$$3$$,$$3$$,$$1$$,$$4$$,$$0$$,$$4$$,$$4$$,$$3$$,$$2$$,$$0$$,$$2$$,$$2$$,$$4$$,$$1$$,$$0$$.周期为$$20$$ $$2017\\div 5=100$$（组）$$\\ldots \\ldots 17$$（个），所以余数是此数列第$$17$$个余数，是$$2$$． " ]

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

[ "2016年创新杯六年级竞赛训练题（二）第2题" ]

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

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

[ "$$\\left( 7+7.5+8+8.5+9 \\right)\\div 5=8$$． " ]

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

[ "2015年第11届全国新希望杯五年级竞赛复赛第6题", "2016年创新杯五年级竞赛训练题（四）第6题" ]

[ [ { "aoVal": "A", "content": "$$0.7$$千米/小时 " } ], [ { "aoVal": "B", "content": "$$1.4$$干米/小时 " } ], [ { "aoVal": "C", "content": "$$0.9$$千米/小时 " } ], [ { "aoVal": "D", "content": "$$1.8$$千米/小时 " } ] ]

[ "知识标签->数学思想->转化与化归的思想" ]

[ "原来的速度和是船速和$$60\\div 2=30$$，那么船速是$$30\\div 2=15$$千米/时，第二次时间为$$60\\div 40=1.5$$小时．两船两次相向运动的速度差都是$$2$$个水速，路程差为$$0.45\\times 2=0.9千米，那么$$0.9$$除以（2-1.5）等于$$1.8$$，$$1.8$$除以$$2$$等于$$0.9$$，故水速为$$0.9$$千米/时． " ]

"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->思想->枚举思想" ]

[ "（1）这个数的因数个数肯定不低于$$6$$个（假定这个数为$$N$$，且拿到的$$6$$个数从大到小分别是$$ABCDEF$$~） （2）有两个人同时第一时间知道结果，这说明以下几个问题： 第一种情况：有一个人知道了最后的结果，这个结果是怎么知道的呢？很简单，他拿到的因数在$$50\\sim 99$$之间（也就是说$$A$$ 的$$2$$倍是$$3$$位数，所以$$A$$其实就是$$N$$） 第二种情况：有一个人拿到的不是最后结果，但是具备以下条件： 这个数的约数少于$$6$$个，比如：有人拿到$$36$$，但他不能判断$$N$$究竟是$$36$$还是$$72$$． 这个数小于$$50$$，不然这个数就只能也是$$N$$了． 这个数大于$$33$$，比如：有人拿到$$29$$，那么他不能断定$$N$$ 是$$58$$还是$$87$$；(这里有个特例是$$27$$，因为$$27\\times 2=54$$，因数个数不少于$$6$$个；$$27\\times 3=81$$，因数个数少于$$6$$个，所以如果拿到$$27$$可以判断$$N$$只能为$$54$$） 这个数还不能是是质数，不然不存在含有这个因数的两位数． 最关键的是，这两人的数是$$2$$倍关系 但是上述内容并不完全正确，需要注意还有一些``奇葩''数：$$17$$、$$19$$、$$23$$也能顺利通过第一轮． 因此，这两个人拿到的数有如下可能： （$$54$$，$$27$$）（$$68$$，$$34$$）（$$70$$，$$35$$）（$$76$$，$$38$$）（$$78$$，$$39$$）（$$92$$，$$46$$）（$$98$$，$$49$$） （3）为了对比清晰，再来把上面所有的情况的因数都列举出来： （$$54$$，$$27$$，$$18$$，$$9$$，$$6$$，$$3$$，$$2$$，$$1$$） （$$68$$，$$34$$，$$17$$，$$4$$，$$2$$，$$1$$）（$$\\times$$） （$$70$$，$$35$$，$$14$$，$$10$$，$$7$$，$$5$$，$$2$$，$$1$$） （$$76$$，$$38$$，$$19$$，$$4$$，$$2$$，$$1$$）（$$\\times$$） （$$78$$，$$39$$，$$26$$，$$13$$，$$6$$，$$3$$，$$2$$，$$1$$） （$$92$$，$$46$$，$$23$$，$$4$$，$$2$$，$$1$$）（$$\\times$$） （$$98$$，$$49$$，$$14$$，$$7$$，$$2$$，$$1$$） 对于第一轮通过的数，用红色标注，所以$$N$$不能是$$68$$、$$76$$、$$92$$中的任意一个． 之后在考虑第二轮需要通过的两个数． 用紫色标注的$$6$$、$$3$$、$$2$$、$$1$$，因为重复使用，如果出现了也不能判断$$N$$是多少，所以不能作为第二轮通过的数． 用绿色标注的$$14$$和$$7$$也不能作为第二轮通过的数，这样$$N$$也不是$$98$$． 那么通过第二轮的数只有黑色的数． 所以$$N$$ 只能是$$54$$、$$70$$、$$78$$中的一个． 再来观察可能满足$$E$$和$$F$$所说的内容： （$$54$$，$$27$$，$$18$$，$$9$$，$$6$$，$$3$$，$$2$$，$$1$$） （$$70$$，$$35$$，$$14$$，$$10$$，$$7$$，$$5$$，$$2$$，$$1$$） （$$78$$，$$39$$，$$26$$，$$13$$，$$6$$，$$3$$，$$2$$，$$1$$） 因为$$F$$说他的数在$$C$$和$$D$$之间，发现上面的数据只有当$$N=70$$的时候，$$F=7$$，在$$CD$$$$(10$$和$$5)$$之间，是唯一满足条件的一种情况． 又因为$$E$$ 确定自己比$$F$$的大，那么他拿到的数一定是该组中剩余数里最大的．所以$$E$$拿到的是$$14$$（$$N=70$$~）． 所以$$N=70$$，六个人拿的数之和为：$$70+35+14+10+7+5=141$$． " ]

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

[ "2016年IMAS小学高年级竞赛第一轮检测试题第13题4分" ]

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

[ "拓展思维->拓展思维->组合模块->操作与策略->归纳递推->汉诺塔递推" ]

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

"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决赛第3题5分" ]

[ [ { "aoVal": "A", "content": "$$50$$ " } ], [ { "aoVal": "B", "content": "$$52$$ " } ], [ { "aoVal": "C", "content": "$$54$$ " } ], [ { "aoVal": "D", "content": "$$56$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "解：设小李吃$$x$$个葡萄，则大李吃（$$68-x$$）个葡萄， $$68-x=4x-2$$， $$68+2=4x+x$$， $$70=5x$$， $$\\textasciitilde\\textasciitilde x=14$$， ∴$$68-x=68-14=54$$（个）． 故选$$\\text{C}$$． " ]

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

[ "2018年第22届广东世界少年奥林匹克数学竞赛一年级竞赛初赛第5题6分" ]

[ [ { "aoVal": "A", "content": "$$15$$ " } ], [ { "aoVal": "B", "content": "$$20$$ " } ], [ { "aoVal": "C", "content": "$$25$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->平均数问题->移多补少->不等变相等（无图）" ]

[ "小灰兔给了小白兔$$5$$个萝卜后， 此时小白兔和小灰兔都有$$15+5=20$$（个）萝卜， 则小灰兔原来有萝卜：$$20+5=25$$（个）． 故选：$$\\text{C}$$． " ]

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

[ "2016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": "$$8$$ " } ], [ { "aoVal": "B", "content": "$$9$$ " } ], [ { "aoVal": "C", "content": "$$23$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "$$(12-1)\\div 8\\times (8+8)+1$$ $$=11\\times 2+1$$ $$=23$$（根）． 答：他会走到第$$23$$根电线杆． 故选：$$\\text{C}$$． " ]

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

[ "2014年全国迎春杯四年级竞赛初赛第11题" ]

[ [ { "aoVal": "A", "content": "$$25526250$$ " } ], [ { "aoVal": "B", "content": "$$26650350$$ " } ], [ { "aoVal": "C", "content": "$$27775250$$ " } ], [ { "aoVal": "D", "content": "$$28870350$$ " } ] ]

[ "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" ]

[ "将A、B、C、D 逐一代入检验．只有B 满足$$（1）$$、$$（2）$$、$$（3）$$、$$（4）$$、$$（5）$$． " ]

"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决赛第4题5分" ]

[ [ { "aoVal": "A", "content": "$$96$$ " } ], [ { "aoVal": "B", "content": "$$108$$ " } ], [ { "aoVal": "C", "content": "$$120$$ " } ], [ { "aoVal": "D", "content": "$$132$$ " } ] ]

[ "拓展思维->能力->图形认知" ]

[ "由于相邻两层数量相差$$8$$，已知最内层有$$28$$个棋子 则 第二层棋子有：$$28+8=36$$（个）， 第三层棋子有：$$36+8=44$$（个）， 所以三层一共有$$28+36+44=108$$（个）． 故选：$$\\text{B}$$． " ]

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

[ "2006年五年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "$$20$$段 " } ], [ { "aoVal": "B", "content": "$$24$$段 " } ], [ { "aoVal": "C", "content": "$$28$$段 " } ], [ { "aoVal": "D", "content": "$$30$$段 " } ] ]

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

[ "在没有重合点时，内点数为$$\\left( 10-1 \\right)+\\left( 12-1 \\right)+\\left( 15-1 \\right)=34$$，$$\\left[ 10,12,15 \\right]=60$$，不妨设木棍长$$60$$厘米； 则第一种刻度线分割的长度为$$6$$厘米； 则第二种刻度线分割的长度为$$5$$厘米； 则第三种刻度线分割的长度为$$4$$厘米； 第一种与第二种刻度线重合处为$$\\left[ 6,5 \\right]=30$$厘米，全棍重合$$1$$次； 第二种与第三种刻度线重合处为$$\\left[ 5,4 \\right]=20$$厘米，全棍重合$$2$$次； 第三种与第一种刻度线重合处为$$\\left[ 4,6 \\right]=12$$厘米，全棍重合$$4$$次； 则总内点数为$$34-7=27$$，于是总段数为$$27+1=28$$（段）。 " ]

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

[ "2011年北京五年级竞赛" ]

[ [ { "aoVal": "A", "content": "6，15 " } ], [ { "aoVal": "B", "content": "6，21 " } ], [ { "aoVal": "C", "content": "15，6 " } ], [ { "aoVal": "D", "content": "24，8 " } ] ]

[ "拓展思维->思想->逆向思想" ]

[ "解包含与排除题，画图是一种很直观、简捷的方法，可以帮助解决问题，画图时注意把不同的对象与不同的区域对应清楚．建议教师帮助学生画图分析，清楚的分析每一部分的含义． 如图，$$A$$圆表示学画画的人，$$B$$圆表示学钢琴的人，$$C$$表示既学钢琴又学画画的人，图中$$A$$圆不含阴影的部分表示只学画画的人，有：$$43-37=6$$人，图中$$B$$圆不含阴影的部分表示只学钢琴的人，有：$$58-37=21$$人． " ]

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

[ "2017年第15届全国希望杯六年级竞赛" ]

[ [ { "aoVal": "A", "content": "$$108$$；$$27$$ " } ], [ { "aoVal": "B", "content": "$$108$$；$$18$$ " } ], [ { "aoVal": "C", "content": "$$120$$；$$12$$ " } ], [ { "aoVal": "D", "content": "$$120$$；$$18$$ " } ] ]

[ "拓展思维->能力->构造模型->模型思想" ]

[ "三个不同的自然数$$a$$，$$b$$，$$c$$的乘积是$$210$$，当三个数越接近，它们的和越小；反之，它们的和越大．因为$$210=1\\times 2\\times 3\\times 5\\times 7$$，所以当$$a$$，$$b$$，$$c$$分别取$$1$$，$$2$$，$$105$$时，它们的和最大，为$$1+2+105=108$$；当$$a$$，$$b$$，$$c$$分别取$$5$$，$$6$$，$$7$$时，它们的和最小，为$$5+6+7=18$$． 故选$$\\text{B}$$． " ]

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

[ "2011年北京五年级竞赛", "2019年广东广州海珠区中山大学附属中学小升初第42题10分" ]

[ [ { "aoVal": "A", "content": "2.8小时 " } ], [ { "aoVal": "B", "content": "5.6小时 " } ], [ { "aoVal": "C", "content": "3.6小时 " } ], [ { "aoVal": "D", "content": "1.5小时 " } ] ]

[ "知识标签->学习能力->七大能力->实践应用" ]

[ "在距西村$$30$$公里处和乙相聚，则甲比乙多走$$60$$公里，而甲骑自行车每小时比乙快$$12$$公里，所以，甲乙相聚时所用时间是$$60 \\div 12 = 5$$（小时），所以甲从西村到和乙相聚用了$$5 - 3.5 = 1.5$$（小时），所以，甲速是$$30 \\div 1.5 = 20$$（公里/小时），所以，丙速是$$20 - 15 = 5$$（公里/小时），东村到西村的距离是：$$20 \\times 3.5 = 70$$（公里），所以，甲丙相遇时间是：$$\\left( {2 \\times 70} \\right) \\div \\left( {20 + 5} \\right) = 5.6$$$$($$小时$$)$$． " ]

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

[ "2017年第17届世奥赛六年级竞赛决赛第13题" ]

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

[ "拓展思维->拓展思维->组合模块->时间问题->时间计算" ]

[ "到达纽约时间，韩国时间是$$7$$月$$15$$日$$21:00$$，纽约时间是$$7$$月$$15$$日$$8：00$$；纽约出发时间，韩国时间是$$8$$月$$1$$日$$20:00$$，纽约时间是$$8$$月$$1$$日$$7:00$$．依次可知思思在纽约一共给父母打了$$8$$个电话． " ]

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

[ "五年级竞赛创新杯", "六年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "$$(\\frac{1}{17}-\\frac{1}{19})\\div 20$$ " } ], [ { "aoVal": "B", "content": "$$(\\frac{1}{15}-\\frac{1}{21})\\div 60$$ " } ], [ { "aoVal": "C", "content": "$$(\\frac{1}{13}-\\frac{1}{23})\\div 100$$ " } ], [ { "aoVal": "D", "content": "$$(\\frac{1}{11}-\\frac{1}{25})\\div 140$$ " } ] ]

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

[ "A，B，C，D依次为$$\\frac{1}{17\\times 19\\times 10}，\\frac{1}{15\\times 21\\times 10}，\\frac{1}{13\\times 23\\times 10}，\\frac{1}{11\\times 25\\times 10}$$，和同近积大，$$11\\times 25\\times 10$$最小，所以$$\\frac{1}{11\\times 25\\times 10}$$最大. " ]

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

[ "2013年美国数学大联盟杯小学高年级竞赛初赛第4题5分" ]

[ [ { "aoVal": "A", "content": "$$111$$ " } ], [ { "aoVal": "B", "content": "$$1101$$ " } ], [ { "aoVal": "C", "content": "$$1110$$ " } ], [ { "aoVal": "D", "content": "$$11000$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "$$\\sqrt{4\\times 9\\times 16}$$，可以看成$$\\sqrt{{{2}^{2}}\\times {{3}^{2}}\\times {{4}^{2}}}$$，也就是$$2\\times 3\\times 4=24$$， 所以答案应该是$$24$$． 但是答案都是$$2$$进制表示，$$24$$的$$2$$进制为$$11000$$． " ]