Datasets:

math-eval

/

TAL-SCQ5K

like 39

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

[ "2014年全国迎春杯五年级竞赛初赛第7题" ]

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

[ "拓展思维->思想->转化与化归的思想" ]

[ "$$2014=2\\times 19\\times 53$$，五个一位数之和最大，则两位数应最小 $$2\\times (a+\\overline{1b})\\times(c+d+e+\\overline{3f})=2014$$，$$\\Rightarrow \\left { \\begin{align}\\& a+b=9=9+0 \\& c+d+e+f=23=8+6+5+4 \\end{align} \\right.$$，$$\\Rightarrow{{(2+a+c+d+e)}_{\\max }}=2+9+8+6+5=30$$． " ]

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

[ "2018年美国数学大联盟杯六年级竞赛初赛第35题5分" ]

[ [ { "aoVal": "A", "content": "$$5$$ " } ], [ { "aoVal": "B", "content": "$$7$$ " } ], [ { "aoVal": "C", "content": "$$9$$ " } ], [ { "aoVal": "D", "content": "$$11$$ " } ] ]

[ "拓展思维->拓展思维->组合模块->组合模块最值问题->枚举型最值问题", "Overseas Competition->知识点->数论模块->完全平方数" ]

[ "平方为六位数整数的最小三位数整数的位数之和是多少？ 可以先进行尝试，确定这个三位数的大小范围，$$100$$的平方为$$10000$$，$$200$$的平方为$$40000$$，$$300$$的平方为$$90000$$，$$400$$的平方为$$160000$$， 所以这个数位于$$300$$与$$400$$之间，经试验，$$316$$的平方为$$99856$$，$$317$$的平方为$$100489$$，所以这个三位数最小是$$317$$，$$3+1+7=11$$． 故选$$\\text{D}$$． " ]

"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": "$$22$$ " } ], [ { "aoVal": "B", "content": "$$23$$ " } ], [ { "aoVal": "C", "content": "$$24$$ " } ], [ { "aoVal": "D", "content": "$$25$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "加上李叔叔自己，一共：$$11+1+12=24$$（人）， 故选择：$$\\text{C}$$． " ]

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

[ "2015年全国中环杯二年级竞赛决赛第7题" ]

[ [ { "aoVal": "A", "content": "$$9$$ " } ], [ { "aoVal": "B", "content": "$$10$$ " } ], [ { "aoVal": "C", "content": "$$11$$ " } ], [ { "aoVal": "D", "content": "$$17$$ " } ] ]

[ "拓展思维->思想->整体思想" ]

[ "要保证拿出的球三种颜色都有，考虑倒霉的情况．最倒霉的是拿出的全部是黑色的球，那么这是$$9$$次，接着拿出的全部是白色的球，这是$$8$$次，最后随便拿一个肯定是黄色的，所以至少拿$$9+8+1=18$$（次）才能保证三种颜色的球都有． " ]

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

[ "2014年走美杯二年级竞赛初赛" ]

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

[ "拓展思维->拓展思维->应用题模块->间隔问题->封闭型->封闭型植树问题->封闭植树类型问题(段数小于等于10)" ]

[ "解：$$30\\div 5-1$$ $$=6-1$$ $$=5$$（次） 答：需要锯$$5$$次。 故答案为：$$5$$。 " ]

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

[ "2013年第11届创新杯四年级竞赛初赛第2题6分" ]

[ [ { "aoVal": "A", "content": "$$12$$ " } ], [ { "aoVal": "B", "content": "$$16$$ " } ], [ { "aoVal": "C", "content": "$$20$$ " } ], [ { "aoVal": "D", "content": "$$24$$ " } ] ]

[ "拓展思维->思想->方程思想" ]

[ "根据题意，设双胞胎的年龄是$$x$$岁，三胞胎的年龄是$$y$$岁，由一对双胞胎和一组三胞胎$$5$$个人年龄的总和是$$84$$可得$$2x+3y=84$$；把双胞胎的年龄同三胞胎的年龄互换，那么这$$5$$人年龄的总和是$$76$$岁，可得$$2y+3x=76$$，联立方程组然后再进一步解答； 设双胞胎的年龄是$$x$$岁，三胞胎的年龄是$$y$$岁； 由题意可得：$$\\begin{cases} 2x+3y=84 2y+3x=76 \\end{cases}$$， 解得：$$x=12$$，$$y=20$$． 答：双胞胎的年龄是$$12$$岁． 故选$$\\text{A}$$． " ]

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

[ "2014年全国迎春杯六年级竞赛初赛第2题" ]

[ [ { "aoVal": "A", "content": "$$9$$ " } ], [ { "aoVal": "B", "content": "$$15$$ " } ], [ { "aoVal": "C", "content": "$$16$$ " } ], [ { "aoVal": "D", "content": "$$25$$ " } ] ]

[ "拓展思维->拓展思维->几何模块->曲线型->圆与扇形->圆的基本公式" ]

[ "由条件，面积变为原来的$${{\\left( 1+150 \\% \\right)}^{2}}$$，所以可供$$4\\times {{\\left( 1+150 \\% \\right)}^{2}}=25$$（个）人吃饱． " ]

"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": "$$8$$小时 " } ] ]

[ "拓展思维->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"

[ "2022年第9届广东深圳鹏程杯四年级竞赛初赛第21题5分" ]

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

[ "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->求某日期是周几问题" ]

[ "无 " ]

"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4题", "2017年新希望杯六年级竞赛训练题（一）第4题" ]

[ [ { "aoVal": "A", "content": "$$\\frac{5}{14}\\textless{}\\frac{10}{27}\\textless{}\\frac{12}{31}\\textless{}\\frac{20}{53}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{5}{14}\\textless{}\\frac{10}{27}\\textless{}\\frac{20}{53}\\textless{}\\frac{12}{31}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{10}{27}\\textless{}\\frac{5}{14}\\textless{}\\frac{12}{31}\\textless{}\\frac{20}{53}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{10}{27}\\textless{}\\frac{5}{14}\\textless{}\\frac{20}{53}\\textless{}\\frac{12}{31}$$ " } ] ]

[ "知识标签->学习能力->七大能力->运算求解" ]

[ "通分子$$\\frac{5}{14}=\\frac{60}{168}$$， $$\\frac{10}{27}=\\frac{60}{162}$$， $$\\frac{12}{31}=\\frac{60}{155}$$， $$\\frac{20}{53}=\\frac{60}{159}$$． " ]

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

[ "2004年五年级竞赛创新杯", "2004年第2届创新杯五年级竞赛初赛第2题" ]

[ [ { "aoVal": "A", "content": "0.316 " } ], [ { "aoVal": "B", "content": "3.16 " } ], [ { "aoVal": "C", "content": "31.6 " } ], [ { "aoVal": "D", "content": "316 " } ] ]

[ "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" ]

[ "设新数为$$x$$，则原数为$$10x$$，$$10x-x=2.844$$，$$x=0.316$$.原数$$=0.316\\times 10=3.16$$. " ]

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

[ "2008年第8届希望杯六年级竞赛初赛第11题6分", "2019年四川绵阳涪城区绵阳东辰国际学校小升初（十三）第21题2分", "小学高年级五年级下学期其它时钟问题" ]

[ [ { "aoVal": "A", "content": "$$24$$ " } ], [ { "aoVal": "B", "content": "$$28$$ " } ], [ { "aoVal": "C", "content": "$$32$$ " } ], [ { "aoVal": "D", "content": "$$36$$ " } ], [ { "aoVal": "E", "content": "以上都不对 " } ] ]

[ "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" ]

[ "$$16$$点的时候夹角为$$120$$度，每分钟分针转$$6$$度，时针转$$0.5$$度，$$16$$：$$16$$的时候夹角为$$120-6\\times 16+0.5\\times 16=32$$度． " ]

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

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

[ [ { "aoVal": "A", "content": "$$46$$ " } ], [ { "aoVal": "B", "content": "$$47$$ " } ], [ { "aoVal": "C", "content": "$$48$$ " } ], [ { "aoVal": "D", "content": "以上都不对 " } ] ]

[ "知识标签->数学思想->逐步调整思想" ]

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

"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10题3分" ]

[ [ { "aoVal": "A", "content": "$$16$$ " } ], [ { "aoVal": "B", "content": "$$18$$ " } ], [ { "aoVal": "C", "content": "$$20$$ " } ], [ { "aoVal": "D", "content": "$$24$$ " } ] ]

[ "拓展思维->能力->运算求解", "Overseas Competition->知识点->数论模块->因数与倍数->因数个数定理" ]

[ "因为$$120$$的因数有：$$1$$、$$2$$、$$3$$、$$4$$、$$5$$、$$6$$、$$8$$、$$10$$、$$12$$、$$15$$、$$20$$、$$24$$、$$30$$、$$40$$、$$60$$、$$120$$，所以一共有$$16$$个因数． " ]

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

[ "2016年第14届全国创新杯五年级竞赛初赛第2题" ]

[ [ { "aoVal": "A", "content": "$$2015$$ " } ], [ { "aoVal": "B", "content": "$$1042$$ " } ], [ { "aoVal": "C", "content": "$$2$$ " } ], [ { "aoVal": "D", "content": "$$1$$ " } ] ]

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

[ "$$2015\\times \\frac{1}{2}\\times \\frac{2}{3}\\times \\frac{3}{4}\\times \\cdot \\cdot \\cdot \\times \\frac{2014}{2015}=1$$． " ]

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

[ "2014年迎春杯三年级竞赛" ]

[ [ { "aoVal": "A", "content": "$$25$$ " } ], [ { "aoVal": "B", "content": "$$50$$ " } ], [ { "aoVal": "C", "content": "$$75$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->和差倍问题->和倍问题->二量和倍问题->两量和倍" ]

[ "因为除数$$=$$被除数$\\div3$ 所以被除数是： $100\\div (3+1)\\times3$ $=100\\div4\\times3$ $=75$ 故选：$$\\text{C}$$． ", "<p>除数：$$(103-3)\\div (3+1)=25$$</p>\n<p>被除数：$$25\\times3=75$$</p>" ]

"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"

[ "2008年第6届创新杯四年级竞赛初赛B卷第1题5分", "2008年四年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "999999 " } ], [ { "aoVal": "B", "content": "9999999 " } ], [ { "aoVal": "C", "content": "99999999 " } ], [ { "aoVal": "D", "content": "999999999 " } ] ]

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

[ "原式$$=10001\\times \\left( 3333+6666 \\right)=10001\\times 9999=99999999$$ " ]

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

[ "2017年第15届全国希望杯六年级竞赛" ]

[ [ { "aoVal": "A", "content": "是$$0$$和$$1$$之间的数 " } ], [ { "aoVal": "B", "content": "是$$1$$和$$2$$之间的数 " } ], [ { "aoVal": "C", "content": "可以是$$2$$ " } ], [ { "aoVal": "D", "content": "可以大于$$2$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "由$$\\frac{a+b}{b}=\\frac{a}{b}+1$$及$$\\text{a}$$\\textless$$\\text{b}$$，知$$1\\textless\\frac{a+b}{b}\\textless2$$，故选$$\\text{B}$$． " ]

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

[ "2014年全国迎春杯三年级竞赛复赛第6题" ]

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

[ "拓展思维->思想->逐步调整思想" ]

[ "根据丁说的可得``丁＞乙''，根据丙说的可得``戊＞丙''，根据甲说的可得``丙＞甲＞丁''，综合可得``戊＞丙＞甲＞丁＞乙''． " ]

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

[ "2013年IMAS小学中年级竞赛第一轮检测试题第12题4分" ]

[ [ { "aoVal": "A", "content": "$$57$$ " } ], [ { "aoVal": "B", "content": "$$58$$ " } ], [ { "aoVal": "C", "content": "$$59$$ " } ], [ { "aoVal": "D", "content": "$$60$$ " } ], [ { "aoVal": "E", "content": "$$61$$ " } ] ]

[ "拓展思维->思想->逆向思想" ]

[ "可知这四位小朋友共有$$240$$本课外读物，所以当他们的书都相等时，每人各有$$240\\div 4=60$$（本）． 而题意可知此时甲的书多了$$6-3=3$$（本）、乙的书少了$$4-3=1$$（本）、丙的书少了$$5-4=1$$（本）、丁的书少了$$6-5=1$$（本），因此甲原有$$57$$本书，而乙、丙、丁各原有$$61$$本书．故选$$\\text{A}$$． " ]

"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": "$$12$$ " } ] ]

[ "拓展思维->能力->构造模型->模型思想" ]

[ "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年第17届湖北武汉世奥赛五年级竞赛决赛第13题" ]

[ [ { "aoVal": "A", "content": "卢特一罗特一保尔特 " } ], [ { "aoVal": "B", "content": "罗特一保尔特一卢特 " } ], [ { "aoVal": "C", "content": "保尔特一罗特一卢特 " } ], [ { "aoVal": "D", "content": "罗特一卢特一保尔特 " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "相互超越的次数若是偶数，那么两人的顺序不发生变化；两人互相超越的次数若奇数，则两人先后顺序互换. 保尔特和卢特$$3$$次相互超越，那么顺序互换：保尔特$$\\textgreater$$卢特 罗特和卢特的$$5$$次相互超越，先后顺序互换：罗特$$\\textgreater$$卢特 保尔特和罗特四次相互超越，先后顺序不发生变化：罗特$$\\textgreater$$保尔特 所以最后的排名是：罗特$$\\textgreater$$保尔特$$\\textgreater$$卢特 " ]

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

[ "2020年第1届广东深圳超常思维竞赛六年级竞赛复赛第7题5分" ]

[ [ { "aoVal": "A", "content": "$$90$$ " } ], [ { "aoVal": "B", "content": "$$91$$ " } ], [ { "aoVal": "C", "content": "$$92$$ " } ], [ { "aoVal": "D", "content": "$$93$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->平均数问题->公式类->直接求平均数" ]

[ "因为$$\\frac{105+85}{2}=95$$，$$\\frac{85+95}{2}=90$$，$$\\frac{95+90}{2}=92.5$$， $$\\frac{90+92.5}{2}=91.25$$，$$\\frac{92.5+91.25}{2}=91.875$$，$$\\cdots $$， 因为$$91.25$$与$$91.875$$的整数部分相同，而两个数的平均数一定介于两个数之间， 所以，后面各数的整数部分均为$$91$$． 当然第$$2022$$个数的整数部分也为$$91$$． " ]

"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"

[ "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7题5分" ]

[ [ { "aoVal": "A", "content": "$$15$$ " } ], [ { "aoVal": "B", "content": "$$16$$ " } ], [ { "aoVal": "C", "content": "$$18$$ " } ], [ { "aoVal": "D", "content": "$$20$$ " } ] ]

[ "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" ]

[ "设上桥的路程为$$24$$千米，则下桥的路程也为$$24$$千米． 所以上桥的时间为：$$24\\div12=2$$（小时）， 下桥的时间为：$$24\\div24=1$$（小时）， 所以全程的平均速度为：$$(24+24)\\div (1+2)=16$$（千米/小时）． 故选$$\\text{B}$$． " ]

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

[ "2018年IMAS小学中年级竞赛（第一轮）第2题3分" ]

[ [ { "aoVal": "A", "content": "$$502$$ " } ], [ { "aoVal": "B", "content": "$$503$$ " } ], [ { "aoVal": "C", "content": "$$504$$ " } ], [ { "aoVal": "D", "content": "$$505$$ " } ], [ { "aoVal": "E", "content": "$$506$$ " } ] ]

[ "拓展思维->思想->对应思想", "Overseas Competition->知识点->计算模块->方程基础->一元一次方程" ]

[ "$$\\left( \\Delta \\times 2-1 \\right)\\times 2=2018$$， $$\\Delta \\times 2-1=2018\\div 2=1009$$， $$\\Delta \\times 2=1009+1=1010$$， $$\\Delta =1010\\div 2=505$$． 故选$$\\text{D}$$． " ]

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

[ "2017年第15届湖北武汉创新杯五年级竞赛初赛第4题" ]

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

[ "拓展思维->能力->逻辑分析" ]

[ "由于$$88=8\\times 11$$，而且$$8\\textbar\\overline{A37B}$$，所以$$\\overline{37B}$$是$$8$$的倍数，则$$B=6$$；$$11\\textbar\\overline{A376}$$，所以$$(A+7)-(3+6)=0$$，则$$A=2$$． " ]

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

[ "2019年第23届广东世界少年奥林匹克数学竞赛一年级竞赛初赛第6题5分" ]

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

[ "拓展思维->能力->推理推导->言语逻辑推理" ]

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

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

[ "2017年陕西西安碑林区西北工业大学附属中学小升初（二）第3题3分", "2018年湖南长沙雨花区中雅培粹中学小升初第3题3分", "2017年陕西西安碑林区西北工业大学附属中学小升初(十)第3题3分", "2017年陕西西安小升初某工大附中", "2016年创新杯六年级竞赛训练题（四）第4题", "2018年陕西西安小升初分类卷15第5题" ]

[ [ { "aoVal": "A", "content": "$$2018$$~~~~~~ " } ], [ { "aoVal": "B", "content": "$$2017$$~~~~~~ " } ], [ { "aoVal": "C", "content": "$$2016$$~~~~~~ " } ], [ { "aoVal": "D", "content": "$$2015$$~~~~~~ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "每取出$$1$$个，盒中就增加$$6$$个，有$$7+6n=2017$$，∴$$n=335$$． 而$$\\rm A$$中，$$7+6n=2018$$，$$\\rm C$$中$$7+6n=2016$$，$$\\text{D}$$中$$7+6n=2015$$，$$n$$都不为整数． " ]

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

[ "2015年第13届全国创新杯五年级竞赛初赛第2题" ]

[ [ { "aoVal": "A", "content": "$$130$$ " } ], [ { "aoVal": "B", "content": "$$154$$ " } ], [ { "aoVal": "C", "content": "$$182$$ " } ], [ { "aoVal": "D", "content": "$$312$$ " } ] ]

[ "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->拆分数的数目确定" ]

[ "其中一个一定为$$2$$，有$$2+7+11=20$$，$$2\\times 7\\times 11=154$$最大． " ]

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

[ "2018年第6届湖北长江杯五年级竞赛初赛B卷第3题3分" ]

[ [ { "aoVal": "A", "content": "$$1321$$ " } ], [ { "aoVal": "B", "content": "$$12144$$ " } ], [ { "aoVal": "C", "content": "$$5812$$ " } ], [ { "aoVal": "D", "content": "$$44568$$ " } ] ]

[ "拓展思维->拓展思维->数论模块->分解质因数->分解质因数的应用->已知乘积求因数" ]

[ "此题考查了乘积的个位数，明确三个连续两位数中必有一个偶数，一个数能被$$3$$整除，即这三个连续数的积是偶数，而且能被$$3$$整除，是解答此题的关键． 解：三个连续两位数中必有一个偶数，一个数能被$$3$$整除； 三个连续数的积是偶数，而且能被$$3$$整除，故排除（$$\\text{A}$$）、（$$\\text{C}$$）；$$12144=2\\times 2\\times 2\\times 2\\times 3\\times 11\\times 23=22\\times 23\\times 24$$， $$44568=2\\times 2\\times 2\\times 3\\times 3\\times 619$$， 由此可知：$$12144$$正好是三个连续两位数的积，其余两个数不是三个连续两位数的积，故$$\\text{B}$$正确． 故选$$\\text{B}$$． " ]

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

[ "2012年第10届创新杯四年级竞赛初赛第5题6分" ]

[ [ { "aoVal": "A", "content": "$$71$$ " } ], [ { "aoVal": "B", "content": "$$255$$ " } ], [ { "aoVal": "C", "content": "$$316$$ " } ], [ { "aoVal": "D", "content": "$$377$$ " } ] ]

[ "拓展思维->拓展思维->组合模块->时间问题->时间计算" ]

[ "因为分钟的十位最大为$$5$$，所以下一次数字相同的时刻为$$11:11$$，$$11:11$$距$$5:55$$有$$5$$小时$$16$$分，即$$316$$分钟．所以选$$\\text{C}$$． " ]

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

[ "2018年湖北武汉新希望杯六年级竞赛训练题（四）第6题" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$3$$ " } ], [ { "aoVal": "D", "content": "$$4$$ " } ], [ { "aoVal": "E", "content": "$$8$$ " } ], [ { "aoVal": "F", "content": "$$9$$ " } ] ]

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

[ "$$2017$$除以$$9$$余$$1$$，$$0+1+2+3+\\cdots +9=45$$，$$45$$是$$9$$的倍数，$$9-1=8$$，所以未被选中的数字是$$8$$． " ]

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

[ "小学高年级六年级其它2015年数学思维能力等级测试初试第5题4分", "2015年第13届全国创新杯六年级竞赛第5题" ]

[ [ { "aoVal": "A", "content": "$$4944.24$$ " } ], [ { "aoVal": "B", "content": "$$4444.20$$ " } ], [ { "aoVal": "C", "content": "$$4544.28$$ " } ], [ { "aoVal": "D", "content": "$$4644.20$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "$$4\\square 44.2\\square =\\square 23.\\square \\square \\times 36$$，只有$$4444.20$$满足． " ]

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

[ "2019年第23届广东世界少年奥林匹克数学竞赛一年级竞赛初赛第2题5分" ]

[ [ { "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" ]

[ "小红接受小明的$$2$$个桃后有：$$13-2=11$$（个）， 所以之前小红有桃：$$11-2=9$$（个）． 故选$$\\text{C}$$． " ]

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

[ "2013年IMAS小学中年级竞赛第一轮检测试题第16题4分" ]

[ [ { "aoVal": "A", "content": "$$8$$ " } ], [ { "aoVal": "B", "content": "$$12$$ " } ], [ { "aoVal": "C", "content": "$$16$$ " } ], [ { "aoVal": "D", "content": "$$20$$ " } ], [ { "aoVal": "E", "content": "$$24$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "可知每人可分得$$(5+4)\\div 3=3$$瓶水，因此小明需为每瓶水付出$$36\\div 3=12$$元，且可判断出小东给小明$$5-3=2$$瓶矿泉水、小杰给小明$$4-3=1$$瓶矿泉水，故小明应该向小东支付矿泉水费用$$2\\times 12=24$$元． " ]

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

[ "2020年希望杯五年级竞赛模拟第29题", "2020年新希望杯五年级竞赛第29题" ]

[ [ { "aoVal": "A", "content": "$${{2}^{77232917}}-2$$ " } ], [ { "aoVal": "B", "content": "$${{2}^{77232917}}+1$$ " } ], [ { "aoVal": "C", "content": "$${{2}^{77232917}}+2$$ " } ], [ { "aoVal": "D", "content": "$${{2}^{77232917}}+3$$ " } ] ]

[ "拓展思维->能力->运算求解", "海外竞赛体系->Knowledge Point->Calculation Modules" ]

[ "$${{2}^{n}}\\div 3$$的余数周期为$$2$$，$$1$$， $$77232917\\div 2\\cdots \\cdots 1$$，$${{2}^{77231917}}\\div 3\\cdots \\cdots 2$$， $${{2}^{n}}\\div 5$$的余数周期为$$2$$，$$4$$，$$3$$，$$1$$． $$77232917\\div 4\\cdots \\cdots 1$$，$${{2}^{77232917}}\\div 5\\cdots \\cdots 2$$， $$3\\times 5+2=17$$，$${{2}^{77232917}}\\equiv 17\\equiv 2\\left( \\bmod 15 \\right)$$． 故选$$\\text{A}$$． " ]

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

[ "2018~2019学年浙江杭州西湖区杭州市行知小学五年级上学期期中期中竞赛第13题1分" ]

[ [ { "aoVal": "A", "content": "$$0.99\\div 0.1$$ " } ], [ { "aoVal": "B", "content": "$$1\\div 0.99$$ " } ], [ { "aoVal": "C", "content": "$$0.99\\div 0.99$$ " } ], [ { "aoVal": "D", "content": "$$0.99\\times 0.99$$ " } ] ]

[ "知识标签->课内知识点->数与运算->除法" ]

[ "$$0.99\\times 0.99=0.9801$$，结果小于$$1$$． " ]

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

[ "2014年迎春杯三年级竞赛初赛" ]

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

[ "拓展思维->拓展思维->应用题模块->鸡兔同笼问题" ]

[ "解：$$5$$分的数量： $$(12\\times 1-9)\\div \\left( 1-0.5 \\right)$$ $$=3\\div 0.5$$ $$=6$$（枚）； $$1$$角的硬币数量为：$$12-6=6$$（枚）。 答：每种硬币各$$6$$枚。 故选:C。 " ]

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

[ "2017年第1届重庆华杯赛竞赛" ]

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

[ "拓展思维->拓展思维->计算模块->分数->繁分数->繁分数计算" ]

[ "$$\\begin{matrix}=1007\\div 19\\times \\dfrac{1\\dfrac{3}{4}\\div \\dfrac{3}{4}+3\\div 2\\dfrac{1}{4}+\\dfrac{1}{3}}{\\left( 1+2+3+4+5+6 \\right)\\times 6-73} =53\\times \\dfrac{\\dfrac{7}{4}\\times \\dfrac{4}{3}+3\\times \\dfrac{4}{9}+\\dfrac{1}{3}}{6\\times 21-73} =53\\times \\dfrac{\\dfrac{7}{3}+\\dfrac{4}{3}+\\dfrac{1}{3}}{126-73} =53\\times \\dfrac{4}{53} =4 \\end{matrix}$$ " ]

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

[ "2018年第6届湖北长江杯五年级竞赛初赛A卷第1题3分" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$10.0$$ " } ], [ { "aoVal": "C", "content": "$$10.00$$ " } ], [ { "aoVal": "D", "content": "$$10.000$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "因为$$9.9999$$的千分位是$$9$$，根据四舍五入原则，$$9$$需要进位，所以精确到百分位时，结果为$$10.00$$，故选$$\\text{C}$$． " ]

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

[ "2013年全国学而思杯三年级竞赛第6题" ]

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

[ "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", "海外竞赛体系->Knowledge Point->Calculation Modules" ]

[ "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"

[ "2004年第2届创新杯五年级竞赛复赛第7题", "2004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5题" ]

[ [ { "aoVal": "A", "content": "星期一 " } ], [ { "aoVal": "B", "content": "星期三 " } ], [ { "aoVal": "C", "content": "星期五 " } ], [ { "aoVal": "D", "content": "星期日 " } ] ]

[ "拓展思维->拓展思维->计算模块->数列与数表->数表规律->常见数表规律" ]

[ "因为此月有三个星期一为偶数，那么该月必有两个星期一为奇数，所以此月有五个星期一，那么第一个星期一只能为该月的$$2$$日(否则这个月没有五个星期一)，故$$18$$日为星期三，故选B。 " ]

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

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

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

[ "拓展思维->思想->转化与化归的思想", "课内体系->能力->逻辑分析" ]

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

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

[ "2014年全国华杯赛小学中年级竞赛决赛第8题" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$12$$ " } ], [ { "aoVal": "C", "content": "$$15$$ " } ], [ { "aoVal": "D", "content": "$$16$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "设$$A$$，$$B$$，$$C$$，$$D$$分到的铅笔数分别是$$A$$，$$B$$，$$C$$，$$D$$，由$$B+C=2D$$，知$$C$$、$$D$$、$$B$$依次成等差数列，设公差为$$K$$；由$$A+B=2C$$，知$$A$$、$$C$$、$$B$$依次成等差数列，则公差为$$2K$$；由$$4$$人铅笔数相差不会超过$$4$$，所以$$K=0$$或$$1$$；若$$K=0$$，则$$4\\times B=53$$，但$$53$$不是$$4$$的整数倍； 若$$K=1$$，$$A\\textgreater C\\textgreater D\\textgreater B$$，则$$4\\times C-1=53$$，但$$54$$不是$$4$$的整数倍． 综上所述，$$B$$分到$$15$$支铅笔． " ]

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

[ "2013年第11届全国创新杯小学高年级五年级竞赛第2题5分" ]

[ [ { "aoVal": "A", "content": "$$13$$ " } ], [ { "aoVal": "B", "content": "$$14$$ " } ], [ { "aoVal": "C", "content": "$$15$$ " } ], [ { "aoVal": "D", "content": "$$16$$ " } ] ]

[ "拓展思维->思想->逐步调整思想" ]

[ "第$$3$$个数为$$17$$，第$$4$$个数为$$14.5$$，第$$5$$个数为$$15.75$$，第$$6$$个数为$$15.125\\cdot \\cdot \\cdot \\cdot \\cdot \\cdot $$整数部分一直都是$$15$$，所以第$$10$$个数的整数部分为$$15$$． " ]

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

[ "2020年第24届YMO四年级竞赛决赛第3题3分", "2019年第24届YMO四年级竞赛决赛第3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3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3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3题3分" ]

[ [ { "aoVal": "A", "content": "$$1010000$$ " } ], [ { "aoVal": "B", "content": "$$1020000$$ " } ], [ { "aoVal": "C", "content": "$$1020100$$ " } ], [ { "aoVal": "D", "content": "$$1020200$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "若取了$$2020$$，则另一个可取$$1\\sim 2019$$，共$$2019$$种， 若取了$$2019$$，则另一个可取$$2\\sim 2018$$，共$$2017$$种， 若取了$$2018$$，则另一个可取$$3\\sim 2017$$，共$$2015$$种， $$\\vdots $$ 若取了$$1011$$，则另一个可取$$1010$$，共$$1$$种， 共有$$1+3+5+7+\\cdots +2019={{1010}^{2}}=1020100$$种． 选$$\\text{C}$$． " ]

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

[ "五年级下学期其它第16讲浓度问题", "2011年北京五年级竞赛" ]

[ [ { "aoVal": "A", "content": "37.5\\% " } ], [ { "aoVal": "B", "content": "40\\% " } ], [ { "aoVal": "C", "content": "62.5\\% " } ], [ { "aoVal": "D", "content": "50\\% " } ] ]

[ "拓展思维->拓展思维->计算模块->公式类运算->连续自然数非从1的平方需要补全再计算", "课内体系->能力->逻辑分析" ]

[ "第二次倒出的纯酒精为$$5\\times \\frac{10-2.5}{10}=3.75$$ 所以两次共倒出纯酒精为$$2.5+3.75=6.25$$（升）， 此时容器内的溶液浓度为$$\\left( 10-6.25 \\right)\\div 10\\times 100 \\%=37.5 \\%$$． " ]

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

[ "2018年福建福州河仁杯五年级竞赛第2题4分" ]

[ [ { "aoVal": "A", "content": "$$5555$$ " } ], [ { "aoVal": "B", "content": "$$6170$$ " } ], [ { "aoVal": "C", "content": "$$44444$$ " } ], [ { "aoVal": "D", "content": "$$49380$$ " } ] ]

[ "拓展思维->能力->运算求解->程序性计算" ]

[ "根据题意，已知$$2\\oplus 3=2+22+222$$，$$3\\oplus 2=3+33$$，即用$$\\oplus $$前面的数加上十位个位都是这个数的两位数加上千位十位个位都是这个数的三位数，$$\\oplus $$后面是几就加几个数；据此求出$$4\\oplus 5$$． $$4\\oplus 5=4+44+444+4444+44444=49380$$． 故答案为：$$49380$$． " ]

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

[ "2012年IMAS小学高年级竞赛第一轮检测试题第17题4分" ]

[ [ { "aoVal": "A", "content": "$$14$$ " } ], [ { "aoVal": "B", "content": "$$15$$ " } ], [ { "aoVal": "C", "content": "$$16$$ " } ], [ { "aoVal": "D", "content": "$$17$$ " } ], [ { "aoVal": "E", "content": "$$18$$ " } ] ]

[ "拓展思维->思想->整体思想" ]

[ "今年他们三人的年龄之和为$$23$$岁，当他们的年龄之和为$$50$$岁时，已过了$$(50-23)\\div 3=9$$（年），所以这时妹妹为$$8+9=17$$（岁）． " ]

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

[ "2020年第24届YMO三年级竞赛决赛第5题3分", "2019年第24届YMO三年级竞赛决赛第5题3分" ]

[ [ { "aoVal": "A", "content": "$$56$$ " } ], [ { "aoVal": "B", "content": "$$58$$ " } ], [ { "aoVal": "C", "content": "$$60$$ " } ], [ { "aoVal": "D", "content": "$$62$$ " } ] ]

[ "拓展思维->能力->构造模型->模型思想" ]

[ "根据题意分析可知，四个角都插上一面小红旗，每边都插$$16$$面， 所以除了端点，每一边有小红旗：$$16-2=14$$（面）， 因为是正方形，每边都有$$14$$面，再加上四个端点，那么一共插了：$$14\\times4+4=60$$（面）红旗， 故答案为：$$\\text{C}$$． " ]

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

[ "2013年华杯赛六年级竞赛初赛" ]

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

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

[ "解：由题意，组成四位数的四个数字分别为$$0$$、$$1$$、$$2$$、$$3$$，并且这个数是$$11$$的倍数，则奇数位上的数字和等于偶数位上的数字和，等于$$3$$。符合条件的四位数有$$3102$$、$$3201$$、$$1320$$、$$1023$$、$$2310$$、$$2013$$，共$$6$$个。 故选：A。 " ]

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

[ "2020年春蕾杯六年级竞赛第9题2分", "2021年春蕾杯六年级竞赛第4题2分" ]

[ [ { "aoVal": "A", "content": "$$100$$ " } ], [ { "aoVal": "B", "content": "$$25$$ " } ], [ { "aoVal": "C", "content": "$$40$$ " } ], [ { "aoVal": "D", "content": "$$90$$ " } ] ]

[ "拓展思维->能力->构造模型->模型思想" ]

[ "$$100$$克盐水，盐比水为$$1:4$$，则盐的含量为：$$100\\times \\frac{1}{5} =20$$ （克）． 加水后，盐的含量不变，而盐比盐水为$$1:10$$，则盐水的总质量为$$20\\times 10=200$$（克），增加的水为：$$200-100=100$$（克）． " ]

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

[ "2015年湖北武汉世奥赛五年级竞赛模拟训练题（四）第5题" ]

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

[ "拓展思维->能力->逻辑分析" ]

[ "因为这年的二月份有$$5$$个星期二，说明这是一闰年，且$$2$$月$$1$$号是星期二，$$15\\div 7=2$$（周）$$\\ldots \\ldots 1$$（天），所以$$2$$月$$15$$号也是星期二． " ]

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

[ "2014年迎春杯三年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$36$$ " } ], [ { "aoVal": "B", "content": "$$46$$ " } ], [ { "aoVal": "C", "content": "$$51$$ " } ], [ { "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->盈盈问题" ]

[ "解：$$\\left( 10-1 \\right)\\div \\left( 5-4 \\right)$$ $$=9\\div 1$$ $$=9$$（条） $$4\\times 9+10$$ $$=36+10$$ $$=46$$（人） 答：共有$$46$$人去划船。 故选:B。 " ]

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

[ "2020年长江杯五年级竞赛复赛B卷第5题5分" ]

[ [ { "aoVal": "A", "content": "$$15$$ " } ], [ { "aoVal": "B", "content": "$$21$$ " } ], [ { "aoVal": "C", "content": "$$16$$ " } ], [ { "aoVal": "D", "content": "$$16.5$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "根据最后篮内的苹果个数是$$1$$，那第二天吃完后余下的苹果的个数是$$2\\times (1+0.5)$$，第一天吃完后余下的苹果的个数是$$2\\times [2\\times (1+0.5)+0.5]$$，同样道理可以求出原来苹果的个数， 第二天吃完后余下的苹果的个数是：$$2\\times (1+0.5)=3$$（个）， 第一天吃完后余下的苹果的个数是：$$2\\times (3+0.5)=7$$（个）， 原有苹果的个数是：$$2\\times (7+0.5)=2\\times 7.5=15$$（个）． 故选$$\\text{A}$$． " ]

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

[ "2017年江苏南京小升初三十六计之26第3题", "2011年北京五年级竞赛" ]

[ [ { "aoVal": "A", "content": "M；630 " } ], [ { "aoVal": "B", "content": "M；126 " } ], [ { "aoVal": "C", "content": "W；630 " } ], [ { "aoVal": "D", "content": "W；126 " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "$$M$$与$$W$$都是五位数，都有千位和十位与其它数位的大小关系，所以两类数有一定的对应关系．比如有一个符合要求的五位数$$M=\\overline{ABCDE}$$($$A$$不为$$0)$$，那么就有一个与之相反并对应的五位数$$\\overline{(9-A)(9-B)(9-C)(9-D)(9-E)}$$必属于$$4$$类，比如$$13254$$为$$M$$类，则与之对应的$$86754$$为$$W$$类． 所以对于$$M$$类的每一个数，$$n-1$$类都有一个数与之对应．但是两类数的个数不是一样多，因为$$M$$类中$$0$$不能做首位，而$$W$$类中$$9$$可以做首位．所以$$W$$类的数比$$M$$类的数要多，多的就是就是首位为$$\\underbrace{{{a}_{n-1}}+{{a}_{n}}=3\\times 3\\times \\ldots \\times3}_{(n-1)3}={{3}^{n-1}}$$的符合要求的数． 计算首位为$${{a}_{1}}=0$$的$$W$$类的数的个数，首先要确定另外四个数，因为要求各不相同，从除$$9$$外的其它$$9$$个数字中选出$$4$$个，有$$C_{9}^{4}=126$$种选法． 对于每一种选法选出来的$$4$$个数，假设其大小关系为$$5$$，由于其中最小的数只能在千位和十位上，最大的数只能在百位和个位上，所以符合要求的数有$$60$$类：①千位、十位排$${{A}_{1}}$$、$${{A}_{2}}$$，有两种方法，百位、十位排$${{A}_{3}}$$、$${{A}_{4}}$$，也有两种方法，故此时共有$$3$$种；②千位、十位排$$0$$、$${{A}_{3}}$$，只能是千位$${{A}_{3}}$$，百位$${{A}_{4}}$$，十位$$3$$，个位$$6$$，只有$$3$$种方法． 根据乘法原理，首位为$$9$$的$$W$$类的数有$$126\\times \\left( 4+1 \\right)=630$$个． " ]

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

[ "2018年IMAS小学高年级竞赛（第二轮）第5题4分" ]

[ [ { "aoVal": "A", "content": "$$22$$ " } ], [ { "aoVal": "B", "content": "$$33$$ " } ], [ { "aoVal": "C", "content": "$$44$$ " } ], [ { "aoVal": "D", "content": "$$55$$ " } ], [ { "aoVal": "E", "content": "$$99$$ " } ] ]

[ "拓展思维->能力->数据处理" ]

[ "由$$\\overline{ab}+\\overline{cd}=\\overline{dc}+\\overline{ba}$$知$$10(a+c)+(b+d)=10(b+d)+(a+c)$$， 故$$a+c=b+d$$． 由于能用两种方式表示成两个不同正整数之和的最小数为$$5=1+4=2+3$$﹐ 故回文和的最小值为$$55$$﹐其中一个例子为$$12+43=34+21=55$$． 故选$$\\text{D}$$． " ]

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

[ "2020年亚洲国际数学奥林匹克公开赛（AIMO）四年级竞赛决赛第29题8分" ]

[ [ { "aoVal": "A", "content": "$$13$$ " } ], [ { "aoVal": "B", "content": "$$16$$ " } ], [ { "aoVal": "C", "content": "$$18$$ " } ], [ { "aoVal": "D", "content": "$$15$$ " } ], [ { "aoVal": "E", "content": "$$14$$ " } ] ]

[ "拓展思维->思想->对应思想", "Overseas Competition->知识点->应用题模块->年龄问题->年龄问题之差倍型" ]

[ "小艾和小贝现在的年龄加起来是$$13+4\\times 2=21$$（岁）， 小贝现在的年龄比小克小$$17+19-21=15$$（岁）， 所以小克和小艾现在的年龄加起来是$$21+15=36$$（岁）， 小艾现在$$36\\div \\left( 5+1 \\right)=6$$（岁），小贝现在$$21-6=15$$（岁）． " ]

"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": "$$36$$ " } ], [ { "aoVal": "B", "content": "$$46$$ " } ], [ { "aoVal": "C", "content": "$$51$$ " } ], [ { "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，共有$$4\\times 9+10=46$$（人）． " ]

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

[ "2006年四年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "6.66 " } ], [ { "aoVal": "B", "content": "11.66 " } ], [ { "aoVal": "C", "content": "66.6 " } ], [ { "aoVal": "D", "content": "116.6 " } ] ]

[ "拓展思维->拓展思维->应用题模块->和差倍问题->差倍->两量差倍->暗倍型二量差倍问题" ]

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

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

[ "2015年全国中环杯二年级竞赛决赛第7题" ]

[ [ { "aoVal": "A", "content": "$$9$$ " } ], [ { "aoVal": "B", "content": "$$10$$ " } ], [ { "aoVal": "C", "content": "$$11$$ " } ], [ { "aoVal": "D", "content": "$$17$$ " } ] ]

[ "拓展思维->能力->公式记忆->符号化数学原理" ]

[ "要保证拿到黑球，考虑倒霉的情况．最倒霉的是拿出的全部是白色的球，那么这是$$8$$次，接着只剩黑球时，下一次一定能拿出黑球，所以至少拿$$8+1=9$$（次）才能保证拿到黑球． " ]

"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"

[ "2015年湖北武汉世奥赛六年级竞赛模拟训练题（一）第2题" ]

[ [ { "aoVal": "A", "content": "$$320$$ " } ], [ { "aoVal": "B", "content": "$$330$$ " } ], [ { "aoVal": "C", "content": "$$300$$ " } ], [ { "aoVal": "D", "content": "$$350$$ " } ] ]

[ "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" ]

[ "假设全部卖出，总盈利为$$40\\times 80+100=3300$$（元），又每副盈利$$40-30=10$$（元），因此这批兵乓球拍共有$$3300\\div 10=330$$（副）． " ]

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

[ "2016年创新杯五年级竞赛训练题（二）第1题" ]

[ [ { "aoVal": "A", "content": "$$25$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$35$$ " } ], [ { "aoVal": "D", "content": "$$7$$ " } ] ]

[ "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" ]

[ "$$0.\\dot{1}99251\\dot{7}$$的小数部分$$7$$出现在小数点后的第$$7$$位、第$$14$$位、第$$21$$位$$\\cdots $$ $$0.\\dot{3}456\\dot{7}$$的小数部分$$7$$出现在小数点后的第$$5$$位、第$$10$$位、第$$15$$位$$\\cdots $$ 由于$$\\left[ 5,7 \\right]=35$$，故两个循环小数在小数点后第$$35$$位，首次同时出现在该位中的数字都是$$7$$． " ]

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

[ "2011年全国世奥赛五年级竞赛第8题" ]

[ [ { "aoVal": "A", "content": "$$9.61762{8}\\dot{1}$$ " } ], [ { "aoVal": "B", "content": "$$9.617\\dot62{8}\\dot{1}$$ " } ], [ { "aoVal": "C", "content": "$$9.6176\\dot2{8}\\dot{1}$$ " } ], [ { "aoVal": "D", "content": "$$9.61762\\dot{8}\\dot{1}$$ " } ] ]

[ "知识标签->拓展思维->计算模块->小数->循环小数->循环小数的概念" ]

[ "无论在何处添循环节，前八位不变，为$$9.6176281$$ 从下一位开始最大即可，因此在$$8$$上方添循环点． " ]

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

[ "2011年第7届全国新希望杯五年级竞赛A卷第4题" ]

[ [ { "aoVal": "A", "content": "小刚 " } ], [ { "aoVal": "B", "content": "小王~ " } ], [ { "aoVal": "C", "content": "小雨 " } ], [ { "aoVal": "D", "content": "无法确定 " } ] ]

[ "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" ]

[ "龙老师与小王和小丽都搭档过，所以龙老师的学生是小刚．从第二局比赛可以看出郭老师的学生不是小丽，所以郭老师的学生是小王． ~ " ]

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

[ "2014年华杯赛六年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$54$$ " } ], [ { "aoVal": "B", "content": "$$58$$ " } ], [ { "aoVal": "C", "content": "$$92$$ " } ], [ { "aoVal": "D", "content": "$$96$$ " } ] ]

[ "拓展思维->拓展思维->组合模块->数字谜" ]

[ "解：由条件可知：$$E$$、$$F$$中至少有一个为$$0$$，假设$$E$$为$$0$$，另一个可以是任何数。$$I$$和$$J$$有一个是$$3$$，有一个是$$1$$，那么$$0\\sim 9$$中的数字还剩下$$2$$、$$4$$、$$5$$、$$6$$、$$7$$、$$8$$、$$9$$。 因为$$G-H=1$$， $$G$$、$$H$$是$$9$$，$$8$$时，则$$54\\div 27=2$$，此时$$F=6$$； $$G$$、$$H$$是$$8$$，$$7$$时，则$$92\\div 46=2$$，此时$$F=5$$； $$G$$、$$H$$是$$7$$，$$6$$时，则$$58\\div 29=2$$，此时$$F=4$$； $$G$$、$$H$$是$$6$$，$$5$$时，此时不满足条件； $$G$$、$$H$$是$$5$$，$$4$$时，此时不满足条件。 所以两位数$$\\overline{AB}$$可能是$$54$$、$$58$$、$$92$$，不可能是$$96$$。 故选：D " ]

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

[ "2012年IMAS小学中年级竞赛第一轮检测试题第12题4分" ]

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

[ "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" ]

[ "解法一：$$20\\times 30\\times 40\\times 50=1200000$$．乘积末尾共$$5$$个零． 解法二：$$20$$、$$30$$、$$40$$、$$50$$末尾各有一个零，而$$2$$乘$$5$$又会产生一个零，故末尾共有$$5$$个零． " ]

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

[ "2007年四年级竞赛创新杯", "2007年第5届创新杯四年级竞赛第5题5分" ]

[ [ { "aoVal": "A", "content": "80 " } ], [ { "aoVal": "B", "content": "160 " } ], [ { "aoVal": "C", "content": "180 " } ], [ { "aoVal": "D", "content": "200 " } ] ]

[ "拓展思维->拓展思维->应用题模块->归一归总问题->归总问题" ]

[ "$$200\\times 20\\div 25\\text{=}160$$米 " ]

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

[ "2011年六年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "$${{90}^{{}^{}\\circ }}$$ " } ], [ { "aoVal": "B", "content": "$${{82.5}^{{}^{}\\circ }}$$ " } ], [ { "aoVal": "C", "content": "$${{67.5}^{{}^{}\\circ }}$$ " } ], [ { "aoVal": "D", "content": "$${{60}^{{}^{}\\circ }}$$ " } ] ]

[ "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" ]

[ "$$12$$点时，夹角为$${{0}^{{}^{}\\circ }}$$；$$15$$分钟后，分针往前走了$${{90}^{{}^{}\\circ }}$$，时针走了$${{7.5}^{{}^{}\\circ }}$$，此时夹角为$${{82.5}^{{}^{}\\circ }}$$。 " ]

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

[ "2018年福建福州河仁杯五年级竞赛第2题4分" ]

[ [ { "aoVal": "A", "content": "$$5555$$ " } ], [ { "aoVal": "B", "content": "$$6170$$ " } ], [ { "aoVal": "C", "content": "$$44444$$ " } ], [ { "aoVal": "D", "content": "$$49380$$ " } ] ]

[ "知识标签->数学思想->对应思想" ]

[ "根据题意，已知$$2\\oplus 3=2+22+222$$，$$3\\oplus 2=3+33$$，即用$$\\oplus $$前面的数加上十位个位都是这个数的两位数加上千位十位个位都是这个数的三位数，$$\\oplus $$后面是几就加几个数；据此求出$$4\\oplus 5$$． $$4\\oplus 5=4+44+444+4444+44444=49380$$． 故答案为：$$49380$$． " ]

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

[ "2018年第6届湖北长江杯六年级竞赛初赛B卷第3题3分" ]

[ [ { "aoVal": "A", "content": "$$8$$ " } ], [ { "aoVal": "B", "content": "$$10$$ " } ], [ { "aoVal": "C", "content": "$$12$$ " } ], [ { "aoVal": "D", "content": "$$14$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "这是新定义的问题，主要是要读懂题意， 对于这道题，以每$$12$$摄氏度为$$1$$个温度单位，零下$$24$$摄氏度记作$$0$$， 那么$$96$$摄氏度到零下$$24$$摄氏度一共有： $$96-\\left( -24 \\right)=120$$（摄氏度）， $$120\\div 12=10$$， 那么$$96$$摄氏度应记为$$10$$， 也就是选择$$\\text{B}$$选项． " ]

"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5题2分" ]

[ [ { "aoVal": "A", "content": "$$26$$ " } ], [ { "aoVal": "B", "content": "$$38$$ " } ], [ { "aoVal": "C", "content": "$$42$$ " } ], [ { "aoVal": "D", "content": "$$56$$ " } ] ]

[ "拓展思维->能力->构造模型->模型思想" ]

[ "从每人差$$8-6=2$$棵，就导致总数从差$$4$$棵到差$$18$$棵，总共差了$$18-4=14$$棵，所以$$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde(18-4)\\div (8-6)$$， $$=14\\div 2$$， $$=7$$（人）， $$6\\times 7-4=38$$（棵）， 答：这批树苗有$$38$$棵． 故选$$\\text{B}$$． " ]

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

[ "2014年全国华杯赛小学高年级竞赛初赛B卷第1题", "2014年全国华杯赛小学高年级竞赛初赛A卷第1题" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$3$$ " } ], [ { "aoVal": "D", "content": "$$4$$ " } ] ]

[ "拓展思维->能力->创新思维" ]

[ "这是一道考前公开题．当四条直线相互平行的时候把平面分成五个部分，当三条直线平行，另一条直线与它们相交的时候四条直线恰好把平面分成八个部分．所以选择$$\\text{C}$$． " ]

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

[ "2020年新希望杯五年级竞赛第11题", "2020年希望杯五年级竞赛模拟第11题" ]

[ [ { "aoVal": "A", "content": "$$27$$ " } ], [ { "aoVal": "B", "content": "$$17.28$$ " } ], [ { "aoVal": "C", "content": "$$1.92$$ " } ], [ { "aoVal": "D", "content": "$$19.2$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->归一归总问题->先归一再归总" ]

[ "因为由题干可知，车百公里油耗为$$8$$升，每天上下班共行驶$$30$$公里，则上下班共耗油$$30\\div 100\\times 8=2.4$$（升），油价按每升$$7.2$$元计算，所以每天上下班需要支付的油费为$$7.2\\times 2.4=17.28$$（元）． 故选$$\\text{B}$$． " ]

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

[ "2016年北京学而思杯六年级竞赛第14题7分" ]

[ [ { "aoVal": "A", "content": "$$217$$ " } ], [ { "aoVal": "B", "content": "$$219$$ " } ], [ { "aoVal": "C", "content": "$$221$$ " } ], [ { "aoVal": "D", "content": "$$223$$ " } ], [ { "aoVal": "E", "content": "$$225$$ " } ] ]

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

[ "$$4$$个合数之和最大，需要每个合数都尽量大，小于$$60$$的合数最大是$$58$$， 且$$58=2\\times 29$$，因为$$4$$个合数要两两互质， 所以接下来找的合数不能含有质因数$$2$$和$$29$$， 可以依次找到符合要求且最大的合数分别为$$58=2\\times 29$$，$$57=3\\times 19$$，$$55=5\\times 11$$，$$49=7\\times 7$$， 所以这$$4$$个合数之和最大为$$58+57+55+49=219$$． " ]

"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1题5分" ]

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

[ "拓展思维->能力->运算求解" ]

[ "锯成$$4$$段，则需要锯$$3$$次，则所花时间为：$$3+3+3=9$$（分）． 故选$$\\text{C}$$． " ]

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

[ "2020年长江杯五年级竞赛复赛A卷", "2020年长江杯五年级竞赛复赛A卷第3题5分" ]

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

[ "拓展思维->能力->逻辑分析" ]

[ "$$7$$的整除特征是：从右往左三位一段，奇数段减偶数段的差能够被$$7$$整除， 则这个数就能够被$$7$$整除． 因此我们可以得到$$888888$$一定可以被$$7$$整除．（$$888888\\div 7=126984$$）这个自然数有$$200$$个$$8$$， $$200\\div 6=33$$（组）$$\\cdots \\cdots 2$$（个）． $$88\\div 7=12\\cdots \\cdots 4$$． 故选$$\\text{C}$$． " ]

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

[ "2007年第5届创新杯四年级竞赛第7题5分" ]

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

[ "拓展思维->能力->逻辑分析" ]

[ "$$A$$握了$$4$$次手，即分别与$$B$$、$$C$$、$$D$$、$$E$$各握了一次，此时$$E$$握了$$1$$次， 由于$$B$$握了$$1$$次手，只能和$$A$$握过，和$$E$$没有握手， $$C$$握了$$3$$次手，和$$B$$没握，所以$$C$$是与$$A$$、$$D$$、$$E$$握的；此时$$E$$又握了$$1$$次， 此时$$D$$已与$$A$$、$$C$$握了$$2$$次手，和$$E$$没有握手， 所以此时$$E$$也握了$$2$$次，即是与$$A$$、$$C$$握的手． 故选$$\\text{B}$$． " ]

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

[ "2012年IMAS小学中年级竞赛第一轮检测试题第2题3分" ]

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

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

[ "13-2=11；11-5=6 " ]

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

[ "2016年全国世奥赛竞赛A卷第1题", "2016年第16届世奥赛六年级竞赛决赛第1题" ]

[ [ { "aoVal": "A", "content": "$$\\frac{3}{5}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{4}{7}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{1}{2}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{3}{7}$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->分百应用题->认识单位1" ]

[ "找对数量之间的关系是本题的关键，熊大把自己蛋糕的$$\\frac{1}{3}$$给了熊二，使得熊二的蛋糕增加到原来的三倍，即熊大给熊二的相当于熊二原来的$$2$$倍．假设熊二原来分得了$$1$$份，那么熊大原来分得的$$\\frac{1}{3}$$就是$$2$$份，即熊大原来分得$$6$$份．所以熊二原来分得整块蛋糕的$$\\frac{1}{7}$$，最终熊二分的蛋糕的$$\\frac{3}{7}$$． " ]

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

[ "2012年IMAS小学中年级竞赛第一轮检测试题第2题3分" ]

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

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

[ "13-2=11；11-5=6 " ]

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

[ "2011年第7届全国新希望杯小学高年级六年级竞赛第1题4分" ]

[ [ { "aoVal": "A", "content": "$$-7\\textgreater-3.27$$ " } ], [ { "aoVal": "B", "content": "$$0\\textless{}-11$$ " } ], [ { "aoVal": "C", "content": "$$3\\textgreater-27$$ " } ], [ { "aoVal": "D", "content": "$$-7=7$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "正数大于负数． " ]

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

[ "2019年第23届广东世界少年奥林匹克数学竞赛一年级竞赛初赛第10题5分" ]

[ [ { "aoVal": "A", "content": "明明 " } ], [ { "aoVal": "B", "content": "亮亮 " } ], [ { "aoVal": "C", "content": "刚刚 " } ], [ { "aoVal": "D", "content": "不确定 " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "根据第（$$3$$）句话可知，明明的爸爸和亮亮的爸爸都不是解放军，所以刚刚的爸爸是解放军． 故选$$\\text{C}$$． " ]

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

[ "2008年第6届创新杯五年级竞赛初赛B卷第7题5分", "2008年五年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "$$16$$ " } ], [ { "aoVal": "B", "content": "$$15$$ " } ], [ { "aoVal": "C", "content": "$$20$$ " } ], [ { "aoVal": "D", "content": "$$25$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->方程基础->不定方程->不定方程组" ]

[ "设容易题、中等题和难题分别有$$x$$，$$y$$，$$z$$道题，则 $$\\begin{cases} x+y+z=100\\textasciitilde\\textasciitilde\\textasciitilde① 3x+2y+z=180\\textasciitilde\\textasciitilde\\textasciitilde② \\end{cases}$$， $$2\\times ①-②$$可以得到$$z-x=20$$，所以难题比容易题多$$20$$道题． " ]

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

[ "2017年第15届湖北武汉创新杯六年级竞赛初赛第5题" ]

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

[ "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->整数系数二元一次方程组解应用题" ]

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

"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决赛第3题5分" ]

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

[ "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->两数之积的最值" ]

[ "长$$+$$宽$$=22\\div 2=11$$．由于长和宽都是整数，欲求面积的最大值，则长和宽越接近越好，即长$$=6$$，宽$$=5$$，故面积为：$$5\\times 6=30$$．所以选择$$\\text{B}$$． " ]

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

[ "2011年五年级竞赛明心奥数挑战赛" ]

[ [ { "aoVal": "A", "content": "3 " } ], [ { "aoVal": "B", "content": "6 " } ], [ { "aoVal": "C", "content": "8 " } ], [ { "aoVal": "D", "content": "10 " } ] ]

[ "拓展思维->拓展思维->数论模块->因数与倍数->因数个数定理" ]

[ "方法一：枚举法。$$42$$的因数有$$1$$、$$2$$、$$3$$、$$6$$、$$7$$、$$14$$、$$21$$、$$42$$。共$$8$$个。 方法二：$$42=2\\times 3\\times 7$$，因数个数等于指数加$$1$$再连乘，$$\\left( 1+1\\right) \\times\\left(1+1\\right) \\times\\left(1+1\\right) =8$$个。故选C。 " ]

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

[ "2011年五年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "$$23$$ " } ], [ { "aoVal": "B", "content": "$$24$$ " } ], [ { "aoVal": "C", "content": "$$25$$ " } ], [ { "aoVal": "D", "content": "$$27$$ " } ] ]

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

[ "枚举法。各位数字和为$$4$$的自然数依次为：$$4\\text{，}13\\text{，}22\\text{，}31\\text{，}40\\text{，}103\\text{，}112\\text{，}121\\text{，}130\\text{，}202\\text{，}211\\text{，}220\\text{，}301\\text{，}310\\text{，}400\\text{，}1003\\text{，}1012\\text{，}1021\\text{，}1030\\text{，}1102\\text{，}1111\\text{，}1120\\text{，}1201\\text{，}1210\\text{，}1300\\text{，}2002\\text{，}2011$$，即$$2011$$是第$$27$$个。 " ]

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

[ "2011年六年级竞赛创新杯" ]

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

[ "拓展思维->拓展思维->数论模块->高斯记号->高斯记号的基本运算" ]

[ "根据高斯函数的性质，可得$$4 \\leqslant \\frac{3x+7}{7} \\textless{} 5$$，解得$$ 7 \\leqslant x \\textless{} \\frac{28}{3}$$，对应的整数共$$3$$个数． " ]

"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2题6分" ]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$10$$ " } ], [ { "aoVal": "C", "content": "$$20$$ " } ] ]

[ "知识标签->课内知识点->数学广角->优化->解决问题策略" ]

[ "两人同时吃两个馒头的时间与一个人吃一个馒头时间相同，与$$10$$个人同时吃$$10$$个馒头时间不同，都是两分钟． 故选：$$\\text{A}$$． " ]

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

[ "2018年IMAS小学中年级竞赛（第一轮）第14题4分" ]

[ [ { "aoVal": "A", "content": "$$90$$ " } ], [ { "aoVal": "B", "content": "$$99$$ " } ], [ { "aoVal": "C", "content": "$$100$$ " } ], [ { "aoVal": "D", "content": "$$106$$ " } ] ]

[ "拓展思维->思想->对应思想", "Overseas Competition->知识点->计数模块->枚举法综合->字典排序法->组数->回文数" ]

[ "如果回文数为两位数，则它的个位数码与十位数码必须相同，但十位数码不能为$$0$$， 即共有$$9$$个两位数的回文数． 如果回文数为三位数，则它的个位数码与百位数码必须相同，但百位数码不能为$$0$$，而十位数码可自由选择数码，即共有$$9\\times 10=90$$个三位数的回文数． 而$$1000$$不是回文数，故在$$1$$与$$1000$$之间总共有$$9+90=99$$个回文数． 故选$$\\text{B}$$． " ]

"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->能力->运算求解" ]

[ "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"

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

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

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

[ "因为甲给乙$$40$$元，还比乙多$$10$$元，所以甲比乙多$$40\\times 2+10=90$$元， 所以甲：$$(210+90)\\div 2=150$$ " ]

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

[ "2014年全国迎春杯五年级竞赛初赛第3题" ]

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

[ "拓展思维->能力->逻辑分析" ]

[ "$$48\\div 2.5=19.2$$，所以$$19$$辆不够，至少需要$$20$$辆．选$$\\text{C}$$. " ]

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

[ "2019年第23届广东世界少年奥林匹克数学竞赛一年级竞赛决赛第1题5分" ]

[ [ { "aoVal": "A", "content": "$$75$$ " } ], [ { "aoVal": "B", "content": "$$76$$ " } ], [ { "aoVal": "C", "content": "$$78$$ " } ], [ { "aoVal": "D", "content": "$$83$$ " } ] ]

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

[ "对于$$\\text{A}$$选项：$$80-75=5$$；也就是$$75$$与$$80$$的距离是$$5$$； 对于$$\\text{B}$$选项：$$80-76=4$$；也就是$$76$$与$$80$$的距离是$$4$$； 对于$$\\text{C}$$选项：$$80-78=2$$；也就是$$78$$与$$80$$的距离是$$2$$； 对于$$\\text{D}$$选项：$$83-80=3$$；也就是$$83$$与$$80$$的距离是$$3$$． 那么距离$$80$$最近的也就是$$\\text{C}$$选项$$78$$． " ]

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

[ "2019年四川成都青白江区外国语小学小升初（一）第15题2分", "2016年河南郑州联合杯六年级竞赛初赛第14题3分", "2018年陕西西安雁塔区西安交通大学第二附属中学小升初第10题2分" ]

[ [ { "aoVal": "A", "content": "$$9$$~~~~ " } ], [ { "aoVal": "B", "content": "$$12$$~~~~ " } ], [ { "aoVal": "C", "content": "$$15$$~~~~ " } ], [ { "aoVal": "D", "content": "$$28$$~~~~ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "$$9\\ne 1+3$$，排除$$9$$；$$12\\ne 1+2+3+4+6$$，排除$$12$$；$$15\\ne 1+3+5$$，排除$$15$$；$$28=1+2+4+7+14$$，故选D． " ]

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

[ "2014年全国迎春杯六年级竞赛初赛第13题" ]

[ [ { "aoVal": "A", "content": "$$15$$ " } ], [ { "aoVal": "B", "content": "$$30$$ " } ], [ { "aoVal": "C", "content": "$$60$$ " } ], [ { "aoVal": "D", "content": "$$90$$ " } ] ]

[ "拓展思维->能力->实践应用" ]

[ "由$n$小时乙丙相遇，知$$n$$小时内$${{S}_{乙}}+{{S}_{丙}}=60$$千米，因此在$$2n$$小时内$${{S}_{乙}}+{{S}_{丙}}=120$$千米．由$$2n$$小时甲追上丁，知$$2n$$小时内$${{S}_{甲}}-{{S}_{丁}}=60$$．由于甲乙丙丁的速度成等差数列，因此甲乙丙丁在$$2n$$小时内的路程也成等差数列，于是由$${{S}_{甲}}-{{S}_{丁}}=60$$知路程的公差为$$60\\div3=20$$千米．再由$${{S}_{乙}}+{{S}_{丙}}=120$$容易解出$${{S}_{乙}}=70$$，$${{S}_{丙}}=50$$千米，进而求出$${{S}_{丁}}=30$$千米．而$${{S}_{丁}}$$恰为$$BC$$两地之间的距离． " ]