Datasets:

---

math-eval

/

TAL-SCQ5K

like 57

[ "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10题5分" ]

[ [ { "aoVal": "A", "content": "2275 " } ], [ { "aoVal": "B", "content": "2175 " } ], [ { "aoVal": "C", "content": "2257 " } ], [ { "aoVal": "D", "content": "2157 " } ] ]

[ "拓展思维->拓展思维->计算模块->比较与估算->大数的估算" ]

[ "由已知等式知，其中$$\\left( 1000a+100b+10c+d \\right)-\\left( 100a+10b+c \\right)-\\left( 10a+b \\right)-a=889a+89b+9c+d=2008$$，其中$$1\\leqslant a\\leqslant 9$$，$$0\\leqslant b$$，$$c$$，$$d\\leqslant 9$$.显然$$a=2$$，于是$$89b+9c+d=230$$，从而$$b=2$$，$$9c+d=52$$，所以$$c=5$$，$$d=7$$，因此，这个四位数$$\\overline{abcd}=2257$$. " ]

[ "2017年IMAS小学中年级竞赛（第一轮）第11题4分" ]

[ [ { "aoVal": "A", "content": "$$56$$ " } ], [ { "aoVal": "B", "content": "$$224$$ " } ], [ { "aoVal": "C", "content": "$$228$$ " } ], [ { "aoVal": "D", "content": "$$876$$ " } ], [ { "aoVal": "E", "content": "$$884$$ " } ] ]

[ "拓展思维->能力->逻辑分析->代数逻辑推理" ]

[ "$$\\Delta \\div 2-2=(222-2)\\div 2$$ $$=220\\div 2=110$$， 因此$$\\Delta =(110+2)\\times 2=112\\times 2=224$$． 故选$$\\text{B}$$． " ]

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

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

[ "拓展思维->拓展思维->组合模块->时间问题->认识钟表" ]

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

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

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

[ "拓展思维->拓展思维->数论模块->整除->整除特征->差系整除特征" ]

[ "因为$$2008006=2006\\times 1000+2006=2006\\times 1001=\\left( 2\\times 17\\times 59 \\right)\\times \\left( 7\\times 11\\times 13 \\right)$$ " ]

[ "2016年第21届全国华杯赛小学高年级竞赛初赛B卷第4题" ]

[ [ { "aoVal": "A", "content": "$$16$$ " } ], [ { "aoVal": "B", "content": "$$20$$ " } ], [ { "aoVal": "C", "content": "$$30$$ " } ], [ { "aoVal": "D", "content": "$$6$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->加减法应用->最值问题" ]

[ "极端思考$$\\frac{4}{15}=\\frac{1}{6}+\\frac{1}{10}=\\frac{1}{5}+\\dfrac{1}{15}=\\frac{1}{4}+\\frac{1}{60}$$，则$$a+b$$最小为$$16$$． " ]

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

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

[ "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天的日期都是偶数，所以该月必有两个星期天的日期为奇数，从而该月有5个星期天，由此可知该月的第一个星期天的日期只能是2号(否则，这个月就没有5个星期天)，从而5个星期天的日期为2，9，16，23，30，故14号为星期五 " ]

[ "2014年全国迎春杯五年级竞赛复赛第9题" ]

[ [ { "aoVal": "A", "content": "$$1512$$ " } ], [ { "aoVal": "B", "content": "$$3024$$ " } ], [ { "aoVal": "C", "content": "$$1510$$ " } ], [ { "aoVal": "D", "content": "$$3020$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "考察是计数问题中的排列组合． $$0～9$$是个数中任意挑选$$5$$个都可以组成``神马数''，$$\\text{C}_{10}^{5}=\\frac{10\\times9\\times 8\\times 7\\times 6}{5\\times 4\\times 3\\times 2\\times 1}=252$$种；在被挑选的$$5$$个数中，最小的放中间，剩下的$$4$$个数进行组合，从中任意挑选$$2$$个可以放在左边或者右边，$$\\text{C}_{4}^{2}=6$$种； 在此一定要注意：$$4$$个数中任选$$2$$个放在左边然后再放到右边数的顺序改变了． 所以共有``神马数''$$252\\times 6＝1512$$个． " ]

[ "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9题3分" ]

[ [ { "aoVal": "A", "content": "$$6003$$ " } ], [ { "aoVal": "B", "content": "$$600.3$$ " } ], [ { "aoVal": "C", "content": "$$6.03$$ " } ], [ { "aoVal": "D", "content": "$$6.003$$ " } ] ]

[ "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" ]

[ "把一个三位小数的小数点忘掉了， 结果所得的整数是$$60.03$$的$$100$$倍即$$60.03\\times 100=6003$$． 原来的小数是$$60003\\div 1000=6.003$$． 故选$$\\text{D}$$． " ]

[ "2018年第22届广东世界少年奥林匹克数学竞赛四年级竞赛初赛第3题6分" ]

[ [ { "aoVal": "A", "content": "$$198$$ " } ], [ { "aoVal": "B", "content": "$$199$$ " } ], [ { "aoVal": "C", "content": "$$200$$ " } ], [ { "aoVal": "D", "content": "$$201$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "根据题意：公差为$$5$$， 所以项数：$$\\left( 1003-13 \\right)\\div 5+1=199$$． 故选：$$\\text{B}$$． " ]

[ "2017年河南郑州豫才杯六年级竞赛" ]

[ [ { "aoVal": "A", "content": "甲 " } ], [ { "aoVal": "B", "content": "乙 " } ], [ { "aoVal": "C", "content": "丙 " } ], [ { "aoVal": "D", "content": "任意一个店皆可 " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "甲店：只需付$$27$$个篮球的钱即可买到$$30$$个篮球，要$$27\\times 25=675$$元； 乙店：$$30\\times 25\\times 88 \\% =660$$元； 丙店：$$25\\times 30=750$$元，减$$7\\times 10=70$$元，花费$$750-70=680$$元． 因为$$660\\textless{}675\\textless{}680$$，所以应到乙店购买． " ]

[ "2016年第16届世奥赛六年级竞赛决赛第11题", "2016年全国世奥赛竞赛A卷第11题" ]

[ [ { "aoVal": "A", "content": "$$190$$ " } ], [ { "aoVal": "B", "content": "$$217$$ " } ], [ { "aoVal": "C", "content": "$$127$$ " } ], [ { "aoVal": "D", "content": "无法构造 " } ] ]

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

[ "我们知道所有的偶数都是合数，除了$$5$$以外，个位为$$5$$的数也都是合数，这$$8$$个数中只有$$1$$、$$3$$、$$5$$、$$7$$、$$9$$放在个位才有可能是质数，所以十位上只能是$$2$$、$$4$$、$$5$$、$$6$$．这四个两位数的和是$$\\left( 2+4+5+6 \\right)\\times 10+1+3+7+9=190$$． " ]

[ "2017年河南郑州联合杯竞赛附加赛第一场第5题2分" ]

[ [ { "aoVal": "A", "content": "$$(a+b)b$$ " } ], [ { "aoVal": "B", "content": "$$a(a+b)$$ " } ], [ { "aoVal": "C", "content": "$$(a+0)b$$ " } ], [ { "aoVal": "D", "content": "$$(a+1)b$$ " } ] ]

[ "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）" ]

[ "$$ab+b=a\\times b+1\\times b=\\left( a+1 \\right)b$$． " ]

[ "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": "$$30$$ " } ], [ { "aoVal": "E", "content": "$$25$$ " } ] ]

[ "拓展思维->拓展思维->组合模块->逻辑推理->体育比赛->2-1-0 积分制" ]

[ "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：$$\\text{D}$$． " ]

[ "2016年全国美国数学大联盟杯小学中年级三年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$100$$ " } ], [ { "aoVal": "B", "content": "$$120$$ " } ], [ { "aoVal": "C", "content": "$$360$$ " } ], [ { "aoVal": "D", "content": "$$960$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "每一份晚餐肉丸比蒜蓉多$$5$$，一共多了$$600$$，因此卖了$$600\\div 5=120$$（份）． " ]

[ "2022年第9届广东深圳鹏程杯四年级竞赛初赛第23题5分" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$4$$ " } ], [ { "aoVal": "D", "content": "$$6$$ " } ], [ { "aoVal": "E", "content": "$$8$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->页码问题" ]

[ "无 " ]

[ "2021年第24届世界少年奥林匹克数学竞赛四年级竞赛复赛（中国区）第14题3分" ]

[ [ { "aoVal": "A", "content": "$$24$$ " } ], [ { "aoVal": "B", "content": "$$18$$ " } ], [ { "aoVal": "C", "content": "$$12$$ " } ], [ { "aoVal": "D", "content": "$$6$$ " } ] ]

[ "拓展思维->拓展思维->计数模块->枚举法综合->字典排序法->组数->数字组数（不可重复）" ]

[ "因为用数字$$0$$，$$1$$，$$2$$，$$3$$可以组成的三位数有：$$120$$、$$102$$、$$210$$、$$201$$、$$310$$、$$130$$、$$301$$、$$103$$、$$230$$、$$203$$、$$320$$、$$302$$、$$123$$、$$132$$、$$213$$、$$231$$、$$321$$、$$312$$，所以一共有$$18$$个不重复的三位数， 故选$$\\text{B}$$． " ]

[ "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17题1分" ]

[ [ { "aoVal": "A", "content": "$$2.3\\times 3\\times 7$$ " } ], [ { "aoVal": "B", "content": "$$13.8\\times 7\\div 2$$ " } ], [ { "aoVal": "C", "content": "$$2\\times 3\\times 7+0.9\\times 7$$ " } ], [ { "aoVal": "D", "content": "$$7\\times 7-7$$ " } ] ]

[ "知识标签->数学思想->转化与化归的思想" ]

[ "$$7\\times 7-7=7\\times \\left( 7-1 \\right)=7\\times 6=42$$，提取公因数后发现$$\\text{D}$$式变为$$7\\times 6$$，而非原式的$$6.9\\times 7$$，所以错误． 故选$$\\text{D}$$． " ]

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

[ [ { "aoVal": "A", "content": "$$37\\frac{19}{20}$$ " } ], [ { "aoVal": "B", "content": "$$28\\frac{19}{20}$$ " } ], [ { "aoVal": "C", "content": "$$38\\frac{19}{20}$$ " } ] ]

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

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

[ "2013年华杯赛六年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$20$$ " } ], [ { "aoVal": "B", "content": "$$51.36$$ " } ], [ { "aoVal": "C", "content": "$$31.36$$ " } ], [ { "aoVal": "D", "content": "$$10.36$$ " } ] ]

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

[ "甲学校花钱： $$56\\times 8.06=451.36$$（元）； 乙学校花钱： $$56\\div \\left( 1+5 \\% \\right)\\times\\left(8.06-0.56 \\right)$$， $$=\\frac{160}{3}\\times 7.5$$， $$=400$$（元）； 乙学校将比甲学校少花： $$451.36-400=51.36$$（元）； 答：乙学校将比甲学校少花$$51.36$$元． " ]

[ "2019年第24届YMO一年级竞赛决赛第3题3分" ]

[ [ { "aoVal": "A", "content": "0,63 " } ], [ { "aoVal": "B", "content": "30,33 " } ], [ { "aoVal": "C", "content": "31,32 " } ], [ { "aoVal": "D", "content": "1,62 " } ] ]

[ "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->构造和一定最值原理" ]

[ "和一定，差小积大 " ]

[ "2006年五年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "6913 " } ], [ { "aoVal": "B", "content": "6914 " } ], [ { "aoVal": "C", "content": "6915 " } ], [ { "aoVal": "D", "content": "6917 " } ] ]

[ "拓展思维->拓展思维->应用题模块->页码问题->数字与页码的对应问题->数字个数" ]

[ "求多位数共有多少个数码，分几个阶段来计算(一位，二位，三位，四位).$$9\\times 1+90\\times 2+900\\times 3+\\left( 2006-999 \\right)\\times 4=6917$$(位) " ]

[ "2020年重庆渝中区重庆市巴蜀中学校小升初（十一）第2题2分", "2018年华杯赛小学中年级竞赛初赛第1题10分", "2018年第23届华杯赛小学中年级竞赛初赛第1题" ]

[ [ { "aoVal": "A", "content": "大于$$1$$ " } ], [ { "aoVal": "B", "content": "小于$$1$$ " } ], [ { "aoVal": "C", "content": "等于$$1$$ " } ], [ { "aoVal": "D", "content": "无法确定和$$1$$的大小 " } ] ]

[ "拓展思维->思想->赋值思想" ]

[ "$$0.2+0.3+0.1=0.16\\textless{}1$$ $$0.99\\times 0.99+0.1=1.081\\textgreater1$$ 又因为小数是连续的，因此可以取到等于$$1$$的$$A$$、$$B$$值，故$$A\\times B+0.1$$的计算结果与$$1$$的大小无法比较． 因为$$A$$，$$B$$都小于$$1$$， 所以无法确定$$A\\times B$$与$$0.9$$的大小关系． ①当$$A\\times B\\textless{}0.9$$，则$$A\\times B+0.1\\textless{}1$$， ②当$$A\\times B=0.9$$，则$$A\\times B+0.1=1$$， ③当$$A\\times B\\textgreater0.9$$，则$$A\\times B+0.1\\textgreater1$$． 故选$$\\text{D}$$． " ]

[ "2020年第1届广东深圳超常思维竞赛五年级竞赛初赛第1题4分" ]

[ [ { "aoVal": "A", "content": "$$56$$ " } ], [ { "aoVal": "B", "content": "$$64$$ " } ], [ { "aoVal": "C", "content": "$$72$$ " } ], [ { "aoVal": "D", "content": "$$81$$ " } ], [ { "aoVal": "E", "content": "$$100$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->数列与数表->数列规律->数列找规律填数" ]

[ "这组数的规律如下：$$4\\times4$$，$$4\\times5$$，$$5\\times5$$，$$5\\times6$$，$$6\\times6$$，$$6\\times7$$，$$7\\times7$$，$$7\\times8$$，$$7\\times8=56$$．故选$$\\text{A}$$． ", "<p>$$16+4=20$$，$$20+5=25$$，$$25+5=30$$，$$30+6=36$$，$$36+6=42$$，$$42+7=49$$，$$49+7=56$$．故选$$\\text{A}$$．</p>" ]

[ "2014年全国迎春杯六年级竞赛复赛第3题" ]

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

[ "拓展思维->拓展思维->数论模块->余数问题->余数问题带余除法->除法中四量关系" ]

[ "除数$$=(472-427)\\div 5=9$$，$$472\\equiv 4(\\bmod9)$$，所以余数是4． " ]

[ "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6题5分" ]

[ [ { "aoVal": "A", "content": "$$60$$ " } ], [ { "aoVal": "B", "content": "$$67$$ " } ], [ { "aoVal": "C", "content": "$$70$$ " } ], [ { "aoVal": "D", "content": "$$76$$ " } ] ]

[ "拓展思维->能力->构造模型->模型思想" ]

[ "设小明今年$$x$$岁，爷爷今年$$y$$岁， $$y-3=4\\left( x-3 \\right)$$， $$y+5=3\\left( x+5 \\right)$$， 解得$$x=19$$，$$y=67$$， 则爷爷今年$$67$$岁． 故选$$\\text{B}$$． " ]

[ "2011年全国学而思杯五年级竞赛第5题", "2011年北京学而思综合能力诊断六年级竞赛第3题", "2011年全国学而思杯四年级竞赛第5题" ]

[ [ { "aoVal": "A", "content": "$$60$$ " } ], [ { "aoVal": "B", "content": "$$90$$ " } ], [ { "aoVal": "C", "content": "$$900$$ " } ], [ { "aoVal": "D", "content": "$$27000$$ " } ] ]

[ "拓展思维->思想->整体思想" ]

[ "$$8$$级和$$6$$级差了$$2$$级，那么30*30=900 " ]

[ "四年级其它", "2012年全国学而思杯三年级竞赛第5题" ]

[ [ { "aoVal": "A", "content": "$$2022$$ " } ], [ { "aoVal": "B", "content": "$$2012$$ " } ], [ { "aoVal": "C", "content": "$$2002$$ " } ], [ { "aoVal": "D", "content": "$$2000$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->公式类运算->平方差公式->平方差公式逆向应用", "Overseas Competition->知识点->计算模块->公式类运算->平方差公式->平方差公式逆向应用" ]

[ "$$\\begin{align}\\& 103\\times 107-91\\times99 \\& =(105-2)\\times (105+2)-(95-4)\\times (95+4) \\&=({{105}^{2}}-{{2}^{2}})-({{95}^{2}}-{{4}^{2}}) \\&={{105}^{2}}-{{2}^{2}}-{{95}^{2}}+{{4}^{2}} \\&=({{105}^{2}}-{{95}^{2}})+({{4}^{2}}-{{2}^{2}}) \\& =(105+95)\\times(105-95)+12 \\& =200\\times 10+12 \\& =2012 \\end{align}$$ " ]

[ "2020年四川绵阳涪城区绵阳东辰国际学校超越杯四年级竞赛第3题2分" ]

[ [ { "aoVal": "A", "content": "$$57$$ " } ], [ { "aoVal": "B", "content": "$$64$$ " } ], [ { "aoVal": "C", "content": "$$71$$ " } ] ]

[ "拓展思维->拓展思维->数论模块->余数问题->余数问题带余除法->除法中四量关系" ]

[ "被除数$$\\div $$除数$$=$$商$$+$$余数；被除数$$=$$除数$$\\times $$商$$+$$余数； 从题干可以知道，一道除法题，商和除数都是$$8$$，余数是$$7$$； 所以被除数$$=8\\times 8+7=71$$． 所以选择$$\\text{C}$$选项． " ]

[ "走美杯四年级竞赛", "走美杯三年级竞赛" ]

[ [ { "aoVal": "A", "content": "$$13$$ " } ], [ { "aoVal": "B", "content": "$$14$$ " } ], [ { "aoVal": "C", "content": "$$12$$ " } ], [ { "aoVal": "D", "content": "$$15$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->年龄问题->年龄问题之差倍型" ]

[ "去年妈妈比小明大$$2$$倍，则去年小明$$26\\div 2=13$$（岁），小明今年$$13+1=14$$（岁）。 " ]

[ "2017年湖北武汉创新杯六年级竞赛邀请赛训练题（三）" ]

[ [ { "aoVal": "A", "content": "$$26$$ " } ], [ { "aoVal": "B", "content": "$$28$$ " } ], [ { "aoVal": "C", "content": "$$30$$ " } ], [ { "aoVal": "D", "content": "$$32$$ " } ] ]

[ "拓展思维->能力->构造模型->模型思想" ]

[ "解设独头鸟$$x$$只，四头蛇$$y$$只，五头龙$$\\text{z}$$只，其方程组 $$\\begin{cases} x+4y+5z=106 3x+2y+5z=98 5x+y+3z=76 \\end{cases}$$解得$$\\begin{cases} x=6 y=10 z=12 \\end{cases}$$ $$6+10+12=28$$（只）. " ]

[ "2018年第22届广东世界少年奥林匹克数学竞赛二年级竞赛初赛第2题6分" ]

[ [ { "aoVal": "A", "content": "$$29$$ " } ], [ { "aoVal": "B", "content": "$$31$$ " } ], [ { "aoVal": "C", "content": "$$37$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "爸爸比小林大：$$39-10=29$$（岁），所以今年爸爸的年龄为：$$29+8=37$$（岁）． 故选择$$\\text{C}$$． " ]

[ "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": "$$92$$ " } ], [ { "aoVal": "B", "content": "$$96$$ " } ], [ { "aoVal": "C", "content": "$$100$$ " } ], [ { "aoVal": "D", "content": "$$104$$ " } ] ]

[ "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(段数大于10)" ]

[ "根据题意分析可知，四个角都插上一面小红旗，每边都插$$26$$面， 所以每一边有小红旗：$$26-1=25$$（面）， 因为是正方形，每边都有$$25$$面，那么一共插了：$$25\\times4=100$$（面）红旗， 故答案为：$$\\text{C}$$． " ]

[ "2016年广东深圳华杯赛小学高年级竞赛冬令营二试第4题" ]

[ [ { "aoVal": "A", "content": "$$9.5$$ " } ], [ { "aoVal": "B", "content": "$$10.2$$ " } ], [ { "aoVal": "C", "content": "$$10.5$$ " } ], [ { "aoVal": "D", "content": "$$11.5$$ " } ] ]

[ "知识标签->拓展思维->应用题模块->列方程解应用题->设而不求" ]

[ "设原来一等奖平均分为$$x$$分，二等奖平均分为$$y$$分． $$10x+20y=\\left( 10-4 \\right)\\times \\left( x+3 \\right)+\\left( 20+4 \\right)\\times \\left( y+1 \\right)$$ $$10x+20y=6x+18+24y+24$$ $$4x-4y=42$$ $$x-y=10.5$$ " ]

[ "2009年第9届全国走美杯四年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$38$$ " } ], [ { "aoVal": "B", "content": "$$20$$ " } ], [ { "aoVal": "C", "content": "$$28$$ " } ], [ { "aoVal": "D", "content": "$$32$$ " } ] ]

[ "知识标签->学习能力->七大能力->逻辑分析" ]

[ "和差倍问题（年龄问题），差不变，设当陈老师与小华一样大时，小华为$$1$$份，则陈老师为$$10$$份，此时年龄差为$$9$$份，所以现在小华为$$10$$份，陈老师为$$19$$份，当小华像陈老师一样大时，小华$$19$$份，陈老师为$$28$$份，此时$$1$$份为$$2$$，所以陈老师今年$$38$$岁． $$（10-1）\\times 3+1=28$$ 学生：$$56\\div 28=2$$（岁） 老师：$$2\\times 10=20$$（岁） $$20+（20-2）=38$$（岁） " ]

[ "2017年河南郑州联合杯竞赛第3题4分" ]

[ [ { "aoVal": "A", "content": "$$2$$~~~~~ " } ], [ { "aoVal": "B", "content": "$$8$$~~~~~ " } ], [ { "aoVal": "C", "content": "$$10$$~~~ " } ], [ { "aoVal": "D", "content": "$$15$$ " } ] ]

[ "拓展思维->拓展思维->计数模块->容斥原理->三量容斥" ]

[ "容斥原理；没有空调也没有高级音响的有$$65-\\left( 45+30-12 \\right)=2$$辆． " ]

[ "六年级其它", "2017年华杯赛六年级竞赛初赛" ]

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

[ "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" ]

[ "由已知设这两个数分别为$$a$$，$$b$$，可得$$7\\times 10 \\textless{} a\\times b \\textless{} 8\\times 11$$，即$$70 \\textless{} a\\times b \\textless{} 88$$，则乘积的整数部分$$M$$满足$$M\\leqslant a\\times b$$，则$$70\\leqslant M \\textless{} 88$$，因此可得整数部分可以取$$70$$到$$87$$的所有整数，共有$$87-70+1=18$$（个），因此选$$\\text{C}$$． " ]

[ "2005年六年级竞赛创新杯" ]

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

[ "拓展思维->拓展思维->应用题模块->比例应用题", "课内体系->知识模块->综合与实践" ]

[ "由达标人数是没有达标人数的$$\\frac{1}{4}$$知： 达标人数占总人数的$$\\frac{1}{1+4}=\\frac{1}{5}$$。又有$$2$$人达标后， 达标人数是没有达标人数的$$\\frac{1}{3}$$， 那么这时达标人数占总人数的$$\\frac{1}{1+3}=\\frac{1}{4}$$。 所以这个班的总人数为$$2\\div \\left( \\frac{1}{4}-\\frac{1}{5} \\right)=40$$（人）， 故而选B。 " ]

[ "2014年IMAS小学中年级竞赛第一轮检测试题第17题4分" ]

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

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

[ "假设王师傅没有打碎玻璃，则他应该拿到$$2\\times 40=80$$元，但他只拿到$$60$$元，所以他打碎了$$(80-60)\\div (2+8)=2$$块玻璃． 故选$$\\text{B}$$． " ]

[ "2014年迎春杯三年级竞赛复赛" ]

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

[ "拓展思维->思想->逐步调整思想" ]

[ "解：依题意可知： 从$$2$$月$$6$$日到$$2$$月$$17$$日为止，一共有$$17-6+1=12$$（天）； 其中有$$2$$个星期六，星期日，工作了$$12-4=8$$（天）. 共完成$$1+3+5+7+9+11+13+15=64$$（道）题. 故选：$$\\text{D}$$. " ]

[ "2014年迎春杯六年级竞赛初赛" ]

[ [ { "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->分解质因数" ]

[ "因为$$2014=2\\times 19\\times 53$$，$$x+1$$全部可能的取值为$$2014$$的$$1$$以外的全部因数，根据因数个数公式，故共有$${{\\left( 1+1 \\right)}^{3}}-1=7$$种。 " ]

[ "2006年四年级竞赛创新杯", "2006年第4届创新杯四年级竞赛初赛A卷第8题" ]

[ [ { "aoVal": "A", "content": "$$16+67$$ " } ], [ { "aoVal": "B", "content": "$$15+68$$ " } ], [ { "aoVal": "C", "content": "$$14+69$$ " } ], [ { "aoVal": "D", "content": "$$17+66$$ " } ] ]

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

[ "观察这组式子可知两加数分别由两组等差数列组成，且和亦为等差数列 第一组:4，5，6，7，\\ldots 公差1; 第二组:2，8，14，20，\\ldots 公差6; 各组和:6，13，20，27，\\ldots 公差7; 因此，当和为83时，可求得项数为$$(83-6)\\div 7+1=12$$所以第12项的算式为:$$\\left[ 4+\\left( 12-1 \\right)\\times 1 \\right]+\\left[ \\left( 12-1 \\right)\\times 6+2 \\right]=15+68$$，选B " ]

[ "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->能力->实践应用" ]

[ "因为分钟的十位最大为$$5$$，故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． 故选$$\\text{C}$$． " ]

[ "2016年创新杯五年级竞赛训练题（一）第2题" ]

[ [ { "aoVal": "A", "content": "$$9610$$ " } ], [ { "aoVal": "B", "content": "$$9300$$ " } ], [ { "aoVal": "C", "content": "$$8600$$ " } ], [ { "aoVal": "D", "content": "$$7000$$ " } ] ]

[ "拓展思维->思想->转化与化归的思想" ]

[ "根据题意有相遇时甲比乙多走了$$310\\times 2=620$$（米），已知每分钟甲比乙多走$$10$$米，那么从出发到相遇时间为$$620\\div 10=62$$（分），形完全程甲需要$$60$$分钟，那么走$$310$$米需要$$2$$分钟，那么甲的速度为$$310\\div 2=155$$（米）每分钟，所以学校到体育馆的距离为：$$60\\times 155=9300$$（米）． " ]

[ "2018年IMAS小学高年级竞赛（第一轮）第11题4分" ]

[ [ { "aoVal": "A", "content": "$$15$$ " } ], [ { "aoVal": "B", "content": "$$18$$ " } ], [ { "aoVal": "C", "content": "$$22$$ " } ], [ { "aoVal": "D", "content": "$$24$$ " } ], [ { "aoVal": "E", "content": "$$28$$ " } ] ]

[ "拓展思维->能力->逻辑分析->代数逻辑推理" ]

[ "由题意知被除数为$$54\\times 18+18=990$$， 因此正确的商为$$990\\div 45=22$$． 故选$$\\text{C}$$． " ]

[ "2017年全国希望杯小学高年级五年级竞赛初赛考前100题" ]

[ [ { "aoVal": "A", "content": "第一种 " } ], [ { "aoVal": "B", "content": "第二种 " } ] ]

[ "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" ]

[ "设收购价为$$a$$元/斤，则第二天``保本''情况下，猪的重量应为 $$0.9\\times 150a\\div \\left[ a\\left( 1+25 \\% \\right) \\right]=0.9\\times 150a\\div 1.25a=108$$（斤） 就是说，及时猪的重量减轻了$$42$$斤，也能``保本''，也比第一种选择好，根据生活经验可知，一夜之后猪的重量不可能减轻$$42$$斤，所以应该果断选择第二种！ " ]

[ "2009年四年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "87只与86只 " } ], [ { "aoVal": "B", "content": "87只与82只 " } ], [ { "aoVal": "C", "content": "80只与87只 " } ], [ { "aoVal": "D", "content": "94只与80只 " } ] ]

[ "拓展思维->拓展思维->计数模块->枚举法综合->枚举法" ]

[ "因为$$\\left( 78+94+86+87+82+80 \\right)\\div \\left( 1+2 \\right)=169$$，又$$87+82=169$$，所以$$87$$只与$$82$$只这两筐是徒弟加工的。 " ]

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

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

[ "拓展思维->拓展思维->数论模块->余数问题->余数问题带余除法" ]

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

[ "2017年第16届春蕾杯二年级竞赛决赛第12题6分" ]

[ [ { "aoVal": "A", "content": "6次 " } ], [ { "aoVal": "B", "content": "7次 " } ], [ { "aoVal": "C", "content": "8次 " } ], [ { "aoVal": "D", "content": "64次 " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "（$$1$$）对折$$3$$次，得到$$8$$个完全相同的长方形； （$$2$$）对折$$4$$次，得到$$16$$个完全相同的长方形； （$$3$$）对折$$5$$次，得到$$32$$个完全相同的长方形； （$$4$$）对折$$6$$次，得到$$64$$个完全相同的长方形； （$$5$$）对折$$7$$次，得到$$128$$个完全相同的长方形； 答：把这张正方形的纸连续对折$$7$$次，可以得到$$128$$个完全相同的长方形． " ]

[ "2018年湖北武汉新希望杯五年级竞赛训练题（五）第6题" ]

[ [ { "aoVal": "A", "content": "$$2521$$ " } ], [ { "aoVal": "B", "content": "$$2529$$ " } ], [ { "aoVal": "C", "content": "$$2539$$ " } ], [ { "aoVal": "D", "content": "$$2547$$ " } ] ]

[ "拓展思维->思想->逐步调整思想" ]

[ "先考虑最大，$$741+852+963=2556$$，再调整相近的两个数字$$4$$和$$3$$，则$$731+852+964=2547$$． " ]

[ "2016年全国小学生数学学习能力测评四年级竞赛复赛第8题3分" ]

[ [ { "aoVal": "A", "content": "$$75$$ " } ], [ { "aoVal": "B", "content": "$$90$$ " } ], [ { "aoVal": "C", "content": "$$105$$ " } ] ]

[ "拓展思维->能力->实践应用" ]

[ "$$9$$点$$30$$分时时针与分针构成的角度为$$30\\times 9-30\\times 5.5=105$$（度）． 故选：$$\\text{C}$$． " ]

[ "2011年广东深圳华杯赛小学中年级竞赛复赛第4题" ]

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

[ "知识标签->学习能力->七大能力->逻辑分析" ]

[ "除以$$3$$的余数为$$1$$，$$1$$，$$2$$，$$0$$，$$2$$，$$2$$，$$1$$，$$0$$，$$1$$，$$1$$，$$2$$，$$0$$，$$2$$，$$2$$，$$1$$，$$0.$$八个一周期 100$\\div8=12\\cdot\\cdot\\cdot\\cdot\\cdot\\cdot4$ 第$$100$$个数是除以$$3$$的余数和第$$4$$个数除以$$3$$的余数一样．所以第$$100$$个数除以$$3$$的余数是$$0$$． 故答案是没有余数． " ]

[ "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": "$$4x-2=68$$ " } ], [ { "aoVal": "B", "content": "$$(4x-2)+x=68$$ " } ], [ { "aoVal": "C", "content": "$$(4x+2)+x=68$$ " } ] ]

[ "拓展思维->能力->实践应用" ]

[ "设小李做了$$x$$个幸运星，则大李做了$$(4x-2)$$个，根据题意列方程：$$(4x-2)+x=68$$，故选$$\\text{B}$$． " ]

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

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

[ "拓展思维->能力->运算求解" ]

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

[ "2003年第9届华杯赛竞赛初赛第8题" ]

[ [ { "aoVal": "A", "content": "$$18$$ " } ], [ { "aoVal": "B", "content": "$$54$$ " } ], [ { "aoVal": "C", "content": "$$16$$ " } ], [ { "aoVal": "D", "content": "$$2$$ " } ] ]

[ "拓展思维->拓展思维->组合模块->抽屉原理->简单抽屉原理" ]

[ "如果不算大、小王，每个花色$$13$$张牌，只需$$14$$张便一定有两张相同点数的牌，加上大、小王，则需要$$16$$张牌． " ]

[ "2008年五年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$8$$ " } ], [ { "aoVal": "C", "content": "$$11$$ " } ], [ { "aoVal": "D", "content": "$$12$$ " } ] ]

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

[ "从$$1\\text{、}2\\text{、}4\\text{、}8$$克中任取两个，其和为$$3\\text{、}5\\text{、}9\\text{、}6\\text{、}10\\text{、}12$$克，从$$1\\text{、}2\\text{、}4\\text{、}8$$克任取两个，大的减去小的，其差为$$1\\text{、}3\\text{、}7\\text{、}2\\text{、}6\\text{、}4$$克，又因为和与差中的$$3$$克$$\\text{、}6$$克重复，所以可称出$$6+6-2=10$$（种）不同的重量。 " ]

[ "2014年迎春杯三年级竞赛初赛" ]

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

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

[ "解：$$2000\\div (4+3+2+1)$$ $$=2000\\div 10$$ $$=200$$（组） 商是$$200$$，没有余数，说明第$$2000$$颗龙珠是$$200$$组的最后一个，是白珠。 答：从龙嘴里吐出的第$$2000$$颗龙珠是白珠。 故选:D。 " ]

[ "2003年五年级竞赛创新杯", "2003年六年级竞赛创新杯", "2003年第1届创新杯五年级竞赛复赛第10题", "2003年四年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "右边的石门 " } ], [ { "aoVal": "B", "content": "中间的石门 " } ], [ { "aoVal": "C", "content": "左边的石门 " } ], [ { "aoVal": "D", "content": "不能确定 " } ] ]

[ "拓展思维->拓展思维->组合模块->逻辑推理->假设型逻辑推理" ]

[ "用假设法，假设右边门是，则左中右都是假的，所以假设不符;假设中间门是，则左中右都是真的，不符;假设左边门是，则左右是真的，中间是假的，符合.所以真正通往宝藏之门是左边的石门，选C. " ]

[ "2004年第2届创新杯六年级竞赛复赛第5题" ]

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

[ "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->构造和一定最值原理" ]

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

[ "2012年IMAS小学高年级竞赛第一轮检测试题第13题4分" ]

[ [ { "aoVal": "A", "content": "星期六 " } ], [ { "aoVal": "B", "content": "星期天 " } ], [ { "aoVal": "C", "content": "星期一 " } ], [ { "aoVal": "D", "content": "星期二 " } ], [ { "aoVal": "E", "content": "星期三 " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "因为每天新开的花朵数刚好等于前一天已开的花朵数，所以最后一天新开的花朵数刚好等于总花朵数的一半．这说明花刚好开完一半的那一天是最后一天的前一天，也就是星期三． " ]

[ "2020年广东广州海珠区广州为明学校卓越杯六年级竞赛初赛第5题2分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{10}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{1}{9}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{1}{11}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{1}{2}$$ " } ], [ { "aoVal": "E", "content": "$$\\frac{1}{8}$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "设水的体积是``$$1$$''， 那么水结成冰体积是$$1\\times \\left( 1+\\frac{1}{10} \\right)=\\frac{11}{10}$$， 冰化成水体积减少$$\\left( \\frac{11}{10}-1 \\right)\\div \\frac{11}{10}=\\frac{1}{11}$$． 故选$$\\text{C}$$． " ]

[ "2018年湖北武汉新希望杯六年级竞赛训练题（二）第16题", "2017年新希望杯六年级竞赛训练题（二）第16题" ]

[ [ { "aoVal": "A", "content": "$$128$$ " } ], [ { "aoVal": "B", "content": "$$146$$ " } ], [ { "aoVal": "C", "content": "$$185$$ " } ], [ { "aoVal": "D", "content": "$$149$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "$$19350=$$奇数$$+$$奇数$$+$$偶数，所以$$3$$个质数中必定有一个是$$2$$； 由于质数中除了$$2$$、$$5$$以外，其他的质数都以$$1$$、$$3$$、$$7$$、$$9$$结尾，因此它们的平方末尾相应为$$1$$、$$9$$、$$9$$、$$1$$，相加后不可能得到$$6$$，而$$5$$的平方个位还是$$5$$，所以其中一个奇数必定为$$5$$．$$19350-4-25=19321$$，$$139$$的平方是$$19321$$，$$2+5+139=146$$． " ]

[ "2006年四年级竞赛创新杯", "2006年第4届创新杯四年级竞赛初赛B卷第1题" ]

[ [ { "aoVal": "A", "content": "30870 " } ], [ { "aoVal": "B", "content": "32900 " } ], [ { "aoVal": "C", "content": "32976 " } ], [ { "aoVal": "D", "content": "10000 " } ] ]

[ "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" ]

[ "$$43210-10234=32976$$，选C " ]

[ "2019年第23届广东世界少年奥林匹克数学竞赛一年级竞赛初赛第9题5分" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$15$$ " } ], [ { "aoVal": "C", "content": "$$16$$ " } ], [ { "aoVal": "D", "content": "$$20$$ " } ] ]

[ "拓展思维->能力->实践应用" ]

[ "除了涛涛前面的人和涛涛本身，剩下的就是排在涛涛后面的人， 所以共有：$$26-10-1=15$$（人）． 故选：$$\\text{B}$$． " ]

[ "2006年第4届创新杯五年级竞赛复赛第3题" ]

[ [ { "aoVal": "A", "content": "$$6913$$ " } ], [ { "aoVal": "B", "content": "$$6914$$ " } ], [ { "aoVal": "C", "content": "$$6915$$ " } ], [ { "aoVal": "D", "content": "$$6917$$ " } ] ]

[ "拓展思维->拓展思维->数论模块->位值原理与进制->数与数字->数、数位、数字的认识" ]

[ "$$1\\sim 9$$，共有$$9$$个数字组成， $$10\\sim 99$$共有$$2\\times 90=180$$个数字组成， $$100\\sim 999$$，共有$$3\\times 900=2700$$个数字组成， $$1000\\sim 2006$$共有$$4\\times 1007=4028$$个数字组成， 所以$$1234567891011121314\\cdots 20052006$$是由：$$9+180+2700+4028=6917$$个数字组成． 则其是$$6917$$位数． 故选$$\\text{D}$$． " ]

[ "2016年全国小学生数学学习能力测评五年级竞赛初赛第8题3分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{8}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{1}{4}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{3}{8}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{5}{8}$$ " } ] ]

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

[ "全部情况有：$$2\\times 2\\times 2=8$$（种）， 最终得到两个正面一个反面可有以下三种情况： 正正反、正反正、反正正， 故所求概率为$$\\frac{3}{8}$$． 故答案为：$$\\text{C}$$． " ]

[ "2018年美国数学大联盟杯五年级竞赛初赛第24题5分" ]

[ [ { "aoVal": "A", "content": "$$24$$ " } ], [ { "aoVal": "B", "content": "$$28$$ " } ], [ { "aoVal": "C", "content": "$$32$$ " } ], [ { "aoVal": "D", "content": "$$36$$ " } ] ]

[ "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" ]

[ "我们知道：$${{12}^{3}}=1728$$； 将$$1728$$分解质因数：$$1728=2\\times 2\\times 2\\times 2\\times 2\\times 2\\times 3\\times 3\\times 3$$； 由于$$6$$个因数各不相同，将质因数进行组合； 所以这六个因数可以是：$$1$$、$$2$$、$$3$$、$$4$$、$$6$$、$$12$$； 那么$$6$$个因数之和是：$$1+2+3+4+6+12=28$$． 故选$$\\text{B}$$． " ]

[ "2012年美国数学大联盟杯六年级竞赛初赛第29题5分(每题5分)" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$15$$ " } ], [ { "aoVal": "C", "content": "$$28$$ " } ], [ { "aoVal": "D", "content": "$$56$$ " } ] ]

[ "拓展思维->能力->数据处理" ]

[ "整数$$a$$除以整数$$b(b\\ne 0)$$除得的商正好是整数而没有余数，我们就说$$a$$能被$$b$$整除，或$$b$$能整除$$a$$． $$a$$称为$$b$$的倍数，$$b$$称为$$a$$的因数． $$12$$的因数有：$$1$$、$$2$$、$$3$$、$$4$$、$$6$$、$$12$$， 它们的和是：$$1+2+3+4+6+12=28$$． 故选$$\\text{C}$$． " ]

[ "2016年华杯赛六年级竞赛初赛" ]

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

[ "拓展思维->拓展思维->行程模块->比例解行程问题->行程中的比例" ]

[ "本题是典型的利用正反比例解行程的问题。首先根据不变量判断是否成正反比例。两次相遇过程中两人的时间相同，路程比等于速度比。两次相遇过程中甲的速度没变，通分比较乙的，即可解决问题。 解：第一次相遇过程中，甲、乙两人的路程之比是$$140:\\left( 300-140 \\right)=7:8$$，时间相同，路程比就是速度比。 第二次相遇过程中，甲、乙两人的路程之比是$$\\left( 300-180 \\right):180=2:3$$，速度比也是$$2:3$$。 在两次相遇问题中，甲的速度是保持不变的。通分得，第一次速度比为$$7:8=14:16$$；第二次速度比为$$2:3=14:21$$。 速度从$$16$$份增加到$$21$$份，速度增加$$1$$米/秒，即$$1\\div \\left( 21-16 \\right)=\\frac{1}{5}$$。 乙原来的速度是$$16\\times \\frac{1}{5}=3.2$$（米/秒）$$=3\\frac{1}{5}$$（米/秒）。 故选：D " ]

[ "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17题1分" ]

[ [ { "aoVal": "A", "content": "$$2.3\\times 3\\times 7$$ " } ], [ { "aoVal": "B", "content": "$$13.8\\times 7\\div 2$$ " } ], [ { "aoVal": "C", "content": "$$2\\times 3\\times 7+0.9\\times 7$$ " } ], [ { "aoVal": "D", "content": "$$7\\times 7-7$$ " } ] ]

[ "知识标签->拓展思维->计算模块->小数->小数乘除->小数乘法运算" ]

[ "$$7\\times 7-7=7\\times \\left( 7-1 \\right)=7\\times 6=42$$，提取公因数后发现$$\\text{D}$$式变为$$7\\times 6$$，而非原式的$$6.9\\times 7$$，所以错误． 故选$$\\text{D}$$． " ]

[ "2006年第4届创新杯四年级竞赛初赛B卷第10题" ]

[ [ { "aoVal": "A", "content": "$$12$$ " } ], [ { "aoVal": "B", "content": "$$15$$ " } ], [ { "aoVal": "C", "content": "$$22$$ " } ], [ { "aoVal": "D", "content": "$$17$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "选项$$\\rm A$$：$$12$$题，按第一种给分方法，最多得$$12\\times 5=60$$分，排除$$\\rm A$$；选项$$\\rm B$$：$$15$$题，按第一种给分方法，最多得$$15\\times 5=75$$分，排除$$\\rm B$$； 选项$$\\rm D$$：$$17$$题，按第一种给分方法，得$$81$$分只有一种方法，$$81=5\\times 15+2\\times 3$$，因此比赛试题最少应该有$$15+3=18$$题，排除$$\\rm D$$．故选$$\\rm C$$ " ]

[ "2021年新希望杯一年级竞赛初赛第14题5分" ]

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

[ "拓展思维->思想->对应思想" ]

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

[ "2016年新希望杯小学高年级六年级竞赛训练题（一）第2题" ]

[ [ { "aoVal": "A", "content": "$$80$$ " } ], [ { "aoVal": "B", "content": "$$90$$ " } ], [ { "aoVal": "C", "content": "$$100$$ " } ], [ { "aoVal": "D", "content": "$$120$$ " } ] ]

[ "拓展思维->思想->方程思想" ]

[ "设甲速度是$$3x$$千米/时，以速度是$$2x$$千米/时，$$600=\\left( 2x+3x \\right)\\times 3$$，$$3x=120$$，即甲的速度是$$120$$千米/时 " ]

[ "2013年河南郑州中原网杯六年级竞赛初赛" ]

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

[ "Overseas Competition->知识点->组合模块->逻辑推理->条件型逻辑推理->神推理", "拓展思维->拓展思维->组合模块->逻辑推理->条件型逻辑推理->连线法" ]

[ "首先，``乙没有坐在$$3$$号椅子上''是错的，就可以判断出乙在$$3$$号椅子上；再看，``乙坐在丙旁边''是错的，可以判断出丙不在$$2$$，$$4$$号椅子上，所以在$$1$$号椅子；最后，``甲坐在乙、丙中间''是错的，可以判断出甲在$$4$$号，故得出丁在$$2$$号椅子上． " ]

[ "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", "2014年全国迎春杯三年级竞赛初赛第13题" ]

[ [ { "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$$(10-1)\\div (5-4)=9$$（条）船，共有$$4\\times 9+10=46$$人． " ]

[ "2010年华杯赛四年级竞赛初赛", "2010年华杯赛三年级竞赛初赛" ]

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

[ "拓展思维->拓展思维->应用题模块->加减法应用->同增同减应用题" ]

[ "考虑到``同增同减差不变''的原则，在剪之后，两条纸带长度的差仍为$$23-15=8$$(厘米). 较长的纸带的长度不少于较短的纸带长度的两倍，故他们长度的差不少于较短的纸带长度的一倍， 较短的纸带不能超过$$8$$厘米，故至少剪下了$$7$$厘米。 " ]

[ "2017年河南郑州联合杯竞赛决赛第10题2分" ]

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

[ "拓展思维->能力->运算求解" ]

[ "$$1+2+\\cdots +7=\\left( 1+7 \\right)\\times 7\\div 2=28$$，$$35-28=7$$，则第$$35$$个数为$$8$$． " ]

[ "2015年第11届全国新希望杯小学高年级六年级竞赛复赛第3题" ]

[ [ { "aoVal": "A", "content": "$$25$$ " } ], [ { "aoVal": "B", "content": "$$35$$ " } ], [ { "aoVal": "C", "content": "$$100$$ " } ], [ { "aoVal": "D", "content": "$$125$$ " } ] ]

[ "拓展思维->能力->实践应用" ]

[ "相互给，总和不变，给完后三人的钱数都是$$1800\\div 3=600$$，望望得到希希的钱后是$$600\\div \\frac{4}{5}=750$$，给希希$$150$$后，星星是$$600$$，那么说明星星给出$$\\frac{1}{3}$$后是$$450$$，那么星星原来是$$450\\div \\frac{2}{3}=675$$（元），那么给了希希$$225$$元，希希得到后有$$600\\div \\frac{3}{4}=800$$（元），那么希希原有$$800-225=575$$（元）．那么望望原有$$1800-675-575=550$$（元），所以希希原来的钱比望望原来的钱多$$25$$元． " ]

[ "2014年第3届广东广州羊排赛六年级竞赛第9题1分" ]

[ [ { "aoVal": "A", "content": "琦琦应该拿金钥匙打开大门 " } ], [ { "aoVal": "B", "content": "琦琦应该拿银钥匙打开大门 " } ], [ { "aoVal": "C", "content": "琦琦应该拿铜钥匙打开大门 " } ], [ { "aoVal": "D", "content": "三把钥匙都不能打开大门，琦琦应该另想办法 " } ] ]

[ "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" ]

[ "本题的思路其实很简单，就是``找矛盾''； 1、三句话只有一句是真的；------题目条件； 2、只要存在一对矛盾则必然一真一假；------矛盾的定义； 3、金钥匙和银钥匙上写的话相互矛盾，根据以上两条，真的那句话一定是其中一句（具体到底是金还是银不重要）； 4、根据$$3$$，则铜钥匙那句一定是假的，但它上面说的本身是否定，否定为假则肯定为真，所以铜钥匙就可以打开大门． 答案选$$\\text{C}$$． " ]

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

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

[ "拓展思维->拓展思维->数论模块->质数与合数" ]

[ "因为$$2008006=2006\\times 1000+2006=2006\\times 1001=\\left( 2\\times 17\\times 59 \\right)\\times \\left( 7\\times 11\\times 13 \\right)$$ " ]

[ "2017年第16届湖北武汉创新杯小学高年级六年级竞赛邀请赛训练题（一）" ]

[ [ { "aoVal": "A", "content": "$$9$$ " } ], [ { "aoVal": "B", "content": "$$10$$ " } ], [ { "aoVal": "C", "content": "$$14$$ " } ], [ { "aoVal": "D", "content": "$$15$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->公式类运算->完全平方公式" ]

[ "根据完全平方公式有$${{\\left( a+1 \\right)}^{2}}={{a}^{2}}+2a+1$$，那么可以得到$${{\\left( a+1 \\right)}^{2}}{{a}^{2}}=2a+1$$，$$2a+1$$是一个奇数，那么所在奇数共有$$9$$个，都可以表示为两个自然数平方之形；若一个偶数能表示为两个平方数之差，都为$${{a}^{2}}{{b}^{2}}=\\left( a+b \\right)\\left( ab \\right)$$，此时$$\\left( a+b \\right)$$与$$\\left( a-b \\right)$$同奇偶性，必同为偶数，那么此时$${{a}^{2}}{{b}^{2}}=\\left( a+b \\right)\\left( ab \\right)$$必为$$4$$的偶数，也就是说$$4$$的偶数才能表示为两个平方数之和，那么$$2000$$，$$2004$$，$$2006$$，$$2012$$，$$2016$$也是可以表示成两个平方数之差的，所以一共有$$9+5=14$$个． " ]

[ "2014年全国迎春杯三年级竞赛复赛第1题" ]

[ [ { "aoVal": "A", "content": "$$18$$、$$6$$ " } ], [ { "aoVal": "B", "content": "$$22$$、$$6$$ " } ], [ { "aoVal": "C", "content": "$$25$$、$$6$$ " } ], [ { "aoVal": "D", "content": "$$25$$、$$26$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "$$2$$、$${ 2 }^{ 2 }$$、$$3$$、$${ 3 }^{ 2 }$$、$$4$$、$${ 4 }^{ 2 }$$、$$5$$、（$${5 }^{ 2 }$$）、（$$6$$ ）、$${ 6 }^{ 2 }$$、$$7$$$$\\cdots$$． " ]

[ "2011年六年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "$$6$$ " } ], [ { "aoVal": "B", "content": "$$0$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$2$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->整数->整数加减->整数加减巧算之位值原理" ]

[ "易知最后的几位为$$0$$，$$0$$的个数由$$5$$和$$2$$的个数决定，在这里显然$$2$$的个数多于$$5$$；我们只需要考虑$$5$$的个数即可。$$1\\times 2\\times 3\\times 4\\times 5\\times \\cdots \\times 21$$中含有$$4$$个$$5$$，而$$343={{7}^{3}}$$，所以商千位上的数字为$$0$$。 " ]

[ "2019年亚洲国际数学奥林匹克公开赛（AIMO）三年级竞赛决赛第2题3分" ]

[ [ { "aoVal": "A", "content": "$$286$$ " } ], [ { "aoVal": "B", "content": "$$265$$ " } ], [ { "aoVal": "C", "content": "$$258$$ " } ] ]

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

[ "原式$$=\\left( 13+40 \\right)+\\left( 16+37 \\right)+\\left( 19+34 \\right)+\\left( 22+31 \\right)+\\left( 25+28 \\right)$$ $$=53\\times 5$$ $$=265$$． $$(13+40)\\times 10\\div 2=265$$． " ]

[ "2015年湖北武汉世奥赛六年级竞赛模拟训练题（四）第6题" ]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{2}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{5}{9}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{4}{7}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{3}{5}$$ " } ] ]

[ "拓展思维->拓展思维->计数模块->统计与概率->概率->加乘原理求概率" ]

[ "会爆开的白粒$$\\frac{2}{3}\\times \\frac{1}{2}=\\frac{1}{3}$$，会爆开的黄粒$$\\frac{1}{3}\\times \\frac{2}{3}=\\frac{2}{9}$$，会爆开的白粒是会爆开的玉米总数的$$\\frac{1}{3}\\div \\left( \\frac{1}{3}+\\frac{2}{9} \\right)=\\frac{3}{5}$$． " ]

[ "2015年华杯赛五年级竞赛初赛", "2015年华杯赛六年级竞赛初赛" ]

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

[ "拓展思维->拓展思维->应用题模块->平均数问题->综合题" ]

[ "六位同学的平均成绩为$$92.5$$分，因此这六人的总分为$$92.5\\times 6=555$$（分）。又最高分为$$99$$，最低分为$$76$$，因此剩下四人的分数和为$$380$$分。要使第三名的同学分数尽可能小，那么得使另外三人的分数尽可能大。而第二名最高为$$98$$分，而第四名第五名分数小于第三名，所以第三名至少为：$$(380-98)\\div3+1=95$$（分），选$$B$$。 " ]

[ "2015年第27届广东广州五羊杯小学高年级竞赛第10题4分" ]

[ [ { "aoVal": "A", "content": "$$5$$、$$6$$ " } ], [ { "aoVal": "B", "content": "$$6$$、$$7$$ " } ], [ { "aoVal": "C", "content": "$$7$$、$$8$$ " } ], [ { "aoVal": "D", "content": "$$8$$、$$9$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "首先可以判断编号为$$1$$，$$2$$，$$3$$，$$4$$，$$5$$，$$6$$，$$7$$的同学说的一定都对， 否则的话，其中说的不对的同学的编号乘以$$2$$后所得的编号也将说得不对， 这样就与``只有编号相邻的两个同学说得不对''矛盾． 因而这个数能被$$2$$，$$3$$，$$4$$，$$5$$，$$6$$，$$7$$整除， 从而也能被$$10$$，$$12$$，$$14$$整除， 即编号为$$10$$，$$12$$，$$14$$的同学说得也对． 于是可以判定编号为$$11$$，$$13$$，$$15$$的同学也对． 现在只剩下$$8$$号和$$9$$号了，所以错的是$$8$$，$$9$$两号， 答案为$$\\text{D}$$． " ]

[ "2017年全国华杯赛竞赛初赛模拟3第6题" ]

[ [ { "aoVal": "A", "content": "$$64.3$$ " } ], [ { "aoVal": "B", "content": "$$63.3$$ " } ], [ { "aoVal": "C", "content": "$$50$$ " } ], [ { "aoVal": "D", "content": "$$70$$ " } ] ]

[ "拓展思维->思想->赋值思想" ]

[ "解法一：这一批种蛋的孵化收益损失$$2\\frac{1}{2}-1\\frac{1}{4}=1\\frac{1}{4}$$（倍），若是全部孵化失败，则要损失$$2\\frac{1}{2}+1=3\\frac{1}{2}$$（倍），所以这批种蛋的孵化率是： $$（1-1\\frac{1}{4}\\div3\\frac{1}{2}）\\times100 \\%=\\frac{9}{14}\\times100 \\%=64.3 \\%$$． 解法二：假设每个种蛋的成本价为$$1$$元，$$100$$个种蛋总成本价$$100$$元．又设$$100$$个种种蛋$$S$$个孵化出来，则孵化出后可卖$$S（2\\frac{1}{2}+1）$$元．于是$$S（2\\frac{1}{2}+1）\\div100=1\\frac{1}{4}+1$$， 因此，这批种蛋的孵化率是$$\\frac{9}{14}\\times100 \\%=64.3 \\%$$． " ]

[ "2018年第22届广东世界少年奥林匹克数学竞赛二年级竞赛初赛第3题6分" ]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$30$$ " } ] ]

[ "拓展思维->拓展思维->组合模块->操作与策略->统筹规划->简单时间统筹问题->时间总和问题" ]

[ "鸡蛋可以一起煮，所以所花的时间是一样的． 故选择$$\\text{B}$$． " ]

[ "2014年第10届全国新希望杯小学高年级六年级竞赛复赛第2题4分" ]

[ [ { "aoVal": "A", "content": "$$57$$ " } ], [ { "aoVal": "B", "content": "$$93$$ " } ], [ { "aoVal": "C", "content": "$$100$$ " } ], [ { "aoVal": "D", "content": "$$114$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "由于$$123$$是$$3$$的倍数，操作过程中得到数一定为$$3$$的倍数，故不可能得到$$100$$选$$C$$． " ]

[ "2019年第24届YMO一年级竞赛决赛第7题3分" ]

[ [ { "aoVal": "A", "content": "$$9$$ " } ], [ { "aoVal": "B", "content": "$$12$$ " } ], [ { "aoVal": "C", "content": "$$13$$ " } ], [ { "aoVal": "D", "content": "$$17$$ " } ] ]

[ "拓展思维->拓展思维->组合模块->抽屉原理->最不利原则" ]

[ "根据题意分析可知，考虑最坏的情况，先摸到的都是其他颜色的球，即白球和红球，把$$3+9=12$$（个）球全摸完然后再接下来摸到的球一定是黑球也就是：$$12+1=13$$（个），故选答案$$\\text C$$． " ]

[ "2004年五年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "56 " } ], [ { "aoVal": "B", "content": "78 " } ], [ { "aoVal": "C", "content": "84 " } ], [ { "aoVal": "D", "content": "96 " } ] ]

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

[ "因为两个两位数的最大公因数为8，可设这两个数分别为$$8a\\text{，}8b$$($$a\\text{，}b$$互质)，那么$$8\\times a\\times b=96$$，即$$a\\times b=12$$，则$$a\\text{，}b$$为$$\\left( 3\\text{，}4 \\right)\\text{，}\\left( 1\\text{，}12 \\right)$$，又因为这两个数皆为两位数，所以这两个数只能为24和32，$$24+32=56$$，选 A. " ]

[ "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": "$$24$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "$$12$$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． " ]

[ "2014年迎春杯三年级竞赛初赛" ]

[ [ { "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->假设型逻辑推理" ]

[ "全对的人不会说自己对的题少于$$3$$，故只有乙、丁可能全对。若乙全对，则排名是乙、丁、甲、丙，与丙所说的``丁对了$$2$$道''是假话相矛盾；若丁全对，则丙的后两句是假话，不可能是第二名，又由丁的``甲考得不如乙''能知道第二名是乙，故丙全错，甲只有``丙考得不如丁''是真话，排名是丁、乙、甲、丙且$$4$$人的话没有矛盾。综上，答案是B。 " ]

[ "2014年IMAS小学高年级竞赛第一轮检测试题第7题3分" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$1$$ " } ], [ { "aoVal": "C", "content": "$$2$$ " } ], [ { "aoVal": "D", "content": "$$3$$ " } ], [ { "aoVal": "E", "content": "$$4$$ " } ] ]

[ "拓展思维->思想->整体思想" ]

[ "将甲、乙两数余数之和$$2+4=6$$继续除以$$5$$，可得其余数为$$1$$．故选$$\\text{B}$$． " ]

[ "2020年新希望杯六年级竞赛初赛（个人战）第4题", "2020年希望杯六年级竞赛（个人赛）第4题" ]

[ [ { "aoVal": "A", "content": "$$11:8$$ " } ], [ { "aoVal": "B", "content": "$$33:55$$ " } ], [ { "aoVal": "C", "content": "$$32:55$$ " } ], [ { "aoVal": "D", "content": "$$41:32$$ " } ], [ { "aoVal": "E", "content": "$$55:32$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "已知$$x:y=4:7$$，$$z:x=3:5$$，那么$$y:x=7:4$$，$$x:z=5:3$$， $$y:x=\\left( 7\\times 5 \\right):\\left( 4\\times 5 \\right)=35:20$$，$$x:z=\\left( 5\\times 4 \\right):\\left( 3\\times 4 \\right)=20:12$$， $$y:x:z=35:20:12$$． 假设$$y=35a$$，$$x=20a$$，$$z=12a$$，其中$$a\\ne 0$$． $$x+y=20a+35a=55a$$，$$z+x=12a+20a=32a$$， 那么$$\\left( x+y \\right):\\left( z+x \\right)=55a:32a=55:32$$． 故选$$\\text{E}$$． " ]

[ "2015年湖北武汉世奥赛小学高年级六年级竞赛模拟训练题（四）第4题" ]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$15$$ " } ], [ { "aoVal": "D", "content": "$$16$$ " } ] ]

[ "拓展思维->拓展思维->计数模块->加乘原理->乘法原理->物品搭配（有特殊要求）" ]

[ "每位同学有两种买法，排除全部是毛笔，则共有$$2\\times 2\\times 2\\times 2-1=15$$（种）． " ]

[ "其它", "2010年北京学而思综合能力诊断三年级竞赛第8题" ]

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

[ "知识标签->学习能力->七大能力->运算求解" ]

[ "$$\\left( 60-44 \\right)\\div 8=2$$，做错$$2$$道题，做对$$8$$道题，对的比错的多$$6$$道． " ]

[ "2017年第15届全国希望杯小学高年级六年级竞赛" ]

[ [ { "aoVal": "A", "content": "第$$2$$个到第$$9$$个正方形的面积之和大 " } ], [ { "aoVal": "B", "content": "第$$1$$个正方形面积大 " } ], [ { "aoVal": "C", "content": "一样大 " } ], [ { "aoVal": "D", "content": "不能判断 " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "第$$2$$个到第$$9$$个正方形的面积之和为$$\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{8}+\\frac{1}{16}+\\frac{1}{32}+\\frac{1}{64}+\\frac{1}{128}+\\frac{1}{256}$$ $$=\\frac{1}{256}\\left( 128+64+32+16+8+4+2+1 \\right)$$ $$=\\frac{255}{256}\\textless{}1$$ " ]

[ "2014年迎春杯五年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "大于$$12.345$$ " } ], [ { "aoVal": "B", "content": "小于$$12.345$$ " } ], [ { "aoVal": "C", "content": "等于$$12.345$$ " } ], [ { "aoVal": "D", "content": "无法确定 " } ] ]

[ "拓展思维->拓展思维->应用题模块->平均数问题->公式类" ]

[ "解：因为$$a$$、$$b$$、$$c$$、$$d$$四个数的平均数是$$12.345$$，$$a\\textgreater b\\textgreater c\\textgreater d$$， 所以$$a$$一定大于$$12.345$$，$$d$$一定小于$$12.345$$， 但是$$b$$的取值无法确定，$$b$$可能大于$$12.345$$，也有可能小于$$12.345$$或等于$$12.345$$。 故选：D。 " ]

[ "竞赛" ]

[ [ { "aoVal": "A", "content": "$$613613$$ " } ], [ { "aoVal": "B", "content": "$$6136.13$$ " } ], [ { "aoVal": "C", "content": "61361.3 " } ], [ { "aoVal": "D", "content": "以上答案都不对 " } ] ]

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

[ "原式 $$=123\\times 958.958-123\\times 1.001\\times 345+877\\times 613.613 $$ $$=123\\times \\left( 958.958-345.345 \\right)+877\\times 613.613 $$ $$=123\\times 613.613+877\\times 613.613 $$ $$=613\\times 1.001\\times \\left( 123+877 \\right) $$ $$=613\\times 1001 $$ $$=613613 $$ " ]