Datasets:

---

math-eval

/

TAL-SCQ5K

like 57

[ "2006年第11届全国华杯赛竞赛初赛第2题" ]

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

[ "拓展思维->能力->逻辑分析" ]

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

[ "2018年湖北武汉创新杯小学高年级六年级竞赛初赛数学思维能力等级测试第3题4分" ]

[ [ { "aoVal": "A", "content": "$$28$$ " } ], [ { "aoVal": "B", "content": "$$35$$ " } ], [ { "aoVal": "C", "content": "$$42$$ " } ], [ { "aoVal": "D", "content": "$$40$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "由于前后总和不变，则变比为原来一$$:$$二$$=8:7=24:21$$，现在一$$:$$二$$=4:5=20:25$$，易知$$4$$份$$=8$$人，$$1$$份$$=2$$人，则二班原有$$2\\times 21=42$$人． " ]

[ "2013年第12届上海小机灵杯小学高年级五年级竞赛初赛第5题1分" ]

[ [ { "aoVal": "A", "content": "出入相补原理 " } ], [ { "aoVal": "B", "content": "等差数列求和 " } ], [ { "aoVal": "C", "content": "十进制计数法 " } ], [ { "aoVal": "D", "content": "四舍五入 " } ] ]

[ "拓展思维->拓展思维->数论模块->位值原理与进制->进制的性质与应用->进制的认识" ]

[ "古时候的原始人捕猎运用现在的数学中的十进制计数法． " ]

[ "2017年第15届湖北武汉创新杯六年级竞赛初赛第8题" ]

[ [ { "aoVal": "A", "content": "$$13$$ " } ], [ { "aoVal": "B", "content": "$$17$$ " } ], [ { "aoVal": "C", "content": "$$11$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ] ]

[ "拓展思维->思想->整体思想" ]

[ "每过一年，两组的年龄差减少$$5-3=2$$（岁），所以需要经过$$(73-47)\\div 2=13$$（年）． " ]

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

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

[ "拓展思维->能力->数据处理" ]

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

[ "2019年第23届广东世界少年奥林匹克数学竞赛六年级竞赛初赛第3题5分" ]

[ [ { "aoVal": "A", "content": "$$288$$ " } ], [ { "aoVal": "B", "content": "$$576$$ " } ], [ { "aoVal": "C", "content": "$$144$$ " } ], [ { "aoVal": "D", "content": "$$200$$ " } ], [ { "aoVal": "E", "content": "$$300$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "先把$$2016$$分解质因数，可得分子是$$2$$的倍数、$$3$$的倍数、$$7$$的倍数时，都不是最简分数，$$1\\sim 42$$中，不是$$2$$，$$3$$，$$7$$倍数的数有$$1$$，$$5$$，$$11$$，$$13$$，$$17$$，$$19$$，$$23$$，$$25$$，$$29$$，$$31$$，$$37$$，$$41$$总共$$12$$个，和为$$252$$，把这$$12$$个数分别加$$42$$，就得到了$$43\\sim 84$$中不是$$2$$，$$3$$，$$7$$倍数的数，和为$$252+12\\times 42$$，以此类推，$$1\\sim 2016$$中不是$$2$$，$$3$$，$$7$$倍数的数的和是$$252+(252+12\\times 42)+(252+12\\times 42\\times 2)+(252+12\\times 42\\times 3)+\\cdots +(252+12\\times 42\\times 42)$$，求出和再除以$$2016$$即可． $$2016=2\\times 2\\times 2\\times 2\\times 2\\times 3\\times 3\\times 7$$， 所以分子是$$2$$的倍数、$$3$$的倍数、$$7$$的倍数时，都不是最简分数， $$1\\sim 42$$中，不是$$2$$，$$3$$，$$7$$倍数的数有$$1$$，$$5$$，$$11$$，$$13$$，$$17$$，$$19$$，$$23$$，$$25$$，$$29$$，$$31$$，$$37$$，$$41$$总共$$12$$个，和为$$252$$． 把这$$12$$个数分别加$$42$$，就得到了$$43\\sim 84$$中不是$$2$$，$$3$$，$$7$$倍数的数， 和为$$252+12\\times 42$$ $$=252+504$$ $$=756$$， 以此类推，$$1\\sim 2016$$中不是$$2$$，$$3$$，$$7$$倍数的数的和是 $$252+(252+12\\times 42)+(252+12\\times 42\\times 2)+(252+12\\times 42\\times 3)+\\cdots +(252+12\\times 42\\times 42)$$ $$=252+(252+504)+(252+1008)+(252+1512)+\\cdots +(252+21168)$$ $$=252+756+1260+1764+\\cdots +21420$$ $$=580608$$， 分母是$$2016$$的所有最简真分数的和是$$\\frac{580608}{2016}=288$$． 故选$$\\text{A}$$． " ]

[ "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3题5分" ]

[ [ { "aoVal": "A", "content": "$$450$$ " } ], [ { "aoVal": "B", "content": "$$485$$ " } ], [ { "aoVal": "C", "content": "$$495$$ " } ], [ { "aoVal": "D", "content": "$$505$$ " } ] ]

[ "拓展思维->思想->转化与化归的思想" ]

[ "$$19+28+37+46+55+64+73+82+91$$ $$=(19+91)+(28+82)+(37+73)+(46+64)+55$$ $$=110+110+110+110+55$$ $$=495$$． 所以选择$$\\text C$$． " ]

[ "2013年IMAS小学高年级竞赛第二轮检测试题第4题4分" ]

[ [ { "aoVal": "A", "content": "$$12$$ " } ], [ { "aoVal": "B", "content": "$$20$$ " } ], [ { "aoVal": "C", "content": "$$48$$ " } ], [ { "aoVal": "D", "content": "$$56$$ " } ], [ { "aoVal": "E", "content": "$$78$$~ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "同时被$$3$$、$$4$$、$$7$$整除意味着能分别整除$$3$$、$$4$$、$$7$$． 整除$$3$$各项数之和为$$3$$的倍数$$2013\\to 2+1+3=6$$，排除$$\\text{B}$$项和$$\\text{D}$$项． 整除$$4$$后两项能整除$$4$$，排除$$\\text{E}$$． 剩余部分代入$$\\text{A}$$项不能整除，故选$$\\text{C}$$． " ]

[ "2017年第8届广东广州羊排赛六年级竞赛第2题1分" ]

[ [ { "aoVal": "A", "content": "$$a+b$$ " } ], [ { "aoVal": "B", "content": "$$2a+b$$ " } ], [ { "aoVal": "C", "content": "$$a+2b$$ " } ], [ { "aoVal": "D", "content": "$$2a+2b$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "肥罗有苹果$$(a+b)$$个，两人共有$$a+(a+b)=(2a+b)$$个． 故选$$\\text{B}$$． " ]

[ "2017年第13届湖北武汉新希望杯六年级竞赛决赛第6题" ]

[ [ { "aoVal": "A", "content": "$$2.4$$~~ " } ], [ { "aoVal": "B", "content": "$$3$$~~~~~ " } ], [ { "aoVal": "C", "content": "$$4.4$$~~ " } ], [ { "aoVal": "D", "content": "$$5$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->工程问题->合作工程问题->接力施工问题" ]

[ "甲、乙$$2$$小时一个周期：$$\\frac{1}{4}+\\frac{1}{6}=\\frac{5}{12}$$，两个周期后还剩下：$$1-2\\times \\frac{5}{12}=\\frac{1}{6}$$，剩下的甲做：$$\\frac{1}{6}\\div \\frac{1}{6}=1$$小时，共需$$2\\times 2+1=5$$小时． " ]

[ "2013年第11届创新杯四年级竞赛初赛第5题6分" ]

[ [ { "aoVal": "A", "content": "乙卖的多 " } ], [ { "aoVal": "B", "content": "甲卖的多 " } ], [ { "aoVal": "C", "content": "甲、乙同样多 " } ], [ { "aoVal": "D", "content": "无法确定谁多 " } ] ]

[ "拓展思维->拓展思维->应用题模块->列方程解应用题->不定方程解应用题" ]

[ "设鱼头重量为$$a$$，鱼身重量为$$b$$． 甲卖鱼所得钱为：$$10a+10b$$； 乙卖鱼所得钱为∶$$9.5a+10.5b$$， 乙卖鱼所得钱$$-$$甲卖鱼所得钱得： $$\\left( 9.5a+10.5b \\right)-\\left( 10a+10b \\right)$$ $$=0.5b-0.5a$$． 因为$$b\\textgreater a$$， 所以$$0.5b-0.5a\\textgreater0$$． 即乙卖鱼所得钱比甲卖鱼所得钱多． 故选$$\\text{A}$$． " ]

[ "2013年小机灵杯三年级竞赛初赛" ]

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

[ "拓展思维->拓展思维->组合模块->逻辑推理->假设型逻辑推理" ]

[ "解：若甲说的是真话，则乙说谎，丙说真话，但丙说甲在说谎，所以矛盾。因此，甲说谎。 甲说谎，则乙说的是真话，丙是说谎的，此时符合。 所以，三人中，甲说谎，乙说真话，丙说谎。 故答案为：B。 " ]

[ "2018年第22届广东世界少年奥林匹克数学竞赛四年级竞赛决赛第2题5分" ]

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

[ "拓展思维->思想->转化与化归的思想" ]

[ "依题意有：$$3\\triangle -2\\square =12$$①，$$3\\square -2\\triangle =2$$②， ①$$\\times 3$$，②$$\\times 2$$得： $$9\\triangle -6\\square =36$$③， $$6\\square -4\\triangle =4$$④， ③$$+$$④得：$$5\\triangle =40$$， 所以$$\\triangle =8$$． 故选择$$\\text{B}$$． " ]

[ "2017年第15届湖北武汉创新杯五年级竞赛初赛第8题" ]

[ [ { "aoVal": "A", "content": "$$99$$ " } ], [ { "aoVal": "B", "content": "$$81$$~ " } ], [ { "aoVal": "C", "content": "$$27$$ " } ], [ { "aoVal": "D", "content": "$$18$$ " } ] ]

[ "拓展思维->思想->转化与化归的思想" ]

[ "记两位数的数字和为$$M$$，并且进了$$k$$次位，则可得$$2M-9k=M$$，得$$M=9k$$，即$$M$$是$$9$$的倍数，即两位数最大为$$99$$，$$M$$最大为$$99$$，$$M$$最小为$$18$$，则差为$$99-18=81$$． " ]

[ "2022年新加坡高级学府数学竞赛（SASMO）三年级竞赛第11题1分" ]

[ [ { "aoVal": "A", "content": "Austin " } ], [ { "aoVal": "B", "content": "Carlos " } ], [ { "aoVal": "C", "content": "Daniela " } ], [ { "aoVal": "D", "content": "Ella " } ] ]

[ "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->找一致（同伙）", "Overseas Competition->知识点->组合模块->逻辑推理->假设型逻辑推理->真假话" ]

[ "因为$$Austin$$和$$Daniela$$说的话是矛盾的，$$Benny$$和$$Carlos$$说的话是矛盾的，而五个孩子中有两个孩子说的真话，$$Ella$$说的一定是假话，真相是巧克力是$$Ella$$吃的. " ]

[ "走美杯三年级竞赛", "走美杯四年级竞赛" ]

[ [ { "aoVal": "A", "content": "因为$$92\\times 8\\approx$$ 720（元），$$92\\times$$ 8\\textgreater720，所以$$92\\times$$ 8\\textgreater720，够 " } ], [ { "aoVal": "B", "content": "因为$$92\\times 8\\approx$$ 800（元），$$92\\times$$ 8\\textless800，所以$$92\\times$$ 8\\textless720，够 " } ], [ { "aoVal": "C", "content": "因为$$92\\times 8\\approx$$ 720（元），$$92\\times$$ 8\\textgreater720，所以$$92\\times$$ 8\\textgreater720，不够 " } ], [ { "aoVal": "D", "content": "因为$$92\\times 8\\approx$$ 800（元），$$92\\times$$ 8\\textgreater800，所以$$92\\times$$ 8\\textgreater720，不够 " } ] ]

[ "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" ]

[ "92\\textgreater90，总价大于$$720$$元，所以不够 " ]

[ "2016年河南郑州联合杯小学高年级五年级竞赛复赛改编" ]

[ [ { "aoVal": "A", "content": "$$50$$． " } ], [ { "aoVal": "B", "content": "$$60$$． " } ], [ { "aoVal": "C", "content": "$$90$$． " } ] ]

[ "知识标签->学习能力->七大能力->逻辑分析" ]

[ "得到猎犬和野兔的速度比是$$6:5$$，即相同时间路程比是$$6:5$$，得到猎犬跑了$$60$$米． " ]

[ "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$$24$$的最大公因数是$$24$$ " } ], [ { "aoVal": "C", "content": "$$24$$和$$30$$的最大公因数是$$6$$ " } ], [ { "aoVal": "D", "content": "互质的两个数都是质数 " } ] ]

[ "拓展思维->拓展思维->数论模块->因数与倍数->公因数与公倍数" ]

[ "$$\\text{A}$$选项：互质的两个数公因数为$$1$$，$$\\text{B}$$选项：$$12$$和$$24$$的最大公因数是$$12$$，$$\\text{D}$$选项：互质的两个数不一定都是质数． " ]

[ "2011年五年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "9 " } ], [ { "aoVal": "B", "content": "15 " } ], [ { "aoVal": "C", "content": "22 " } ], [ { "aoVal": "D", "content": "25 " } ] ]

[ "拓展思维->拓展思维->应用题模块->加减法应用" ]

[ "$$3$$被$$1+2+3=6$$代替，$$5$$被$$1+2+3+4+5=15$$代替，所以$$1+3+5$$显示为$$1+6+15=22$$。 " ]

[ "小学高年级六年级其它2014年数学思维能力等级测试第3题", "2014年第12届全国创新杯六年级竞赛第3题" ]

[ [ { "aoVal": "A", "content": "$$120$$ " } ], [ { "aoVal": "B", "content": "$$118$$ " } ], [ { "aoVal": "C", "content": "$$115$$ " } ], [ { "aoVal": "D", "content": "$$110$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "$${{V}_{}}+{{V}_{}}=1:10$$，追上时狗比狐狸多跑$$90$$米，即狗跑了$$90\\times \\frac{10}{10-1}=100$$米． " ]

[ "2017年第22届湖北武汉华杯赛小学高年级竞赛第6题" ]

[ [ { "aoVal": "A", "content": "$$24$$ " } ], [ { "aoVal": "B", "content": "$$25$$ " } ], [ { "aoVal": "C", "content": "$$30$$ " } ], [ { "aoVal": "D", "content": "$$35$$ " } ], [ { "aoVal": "E", "content": "$$36$$ " } ] ]

[ "拓展思维->拓展思维->数论模块->因数与倍数->因数个数定理->因数个数定理逆应用" ]

[ "因为$$\\frac{20}{10}=\\frac{1+1}{0+1}$$，所以$$n$$的质因数分解式中含有$$0$$个$$2$$． 因为$$\\frac{15}{10}=\\frac{2+1}{1+1}$$，所以$$n$$的质因数分解式中含有$$1$$个$$3$$，另一个质因数有$$4$$个． 所以$$6n$$的因数有$$(1+1)(2+1)(4+1)=30$$个． " ]

[ "2015年第11届全国新希望杯小学高年级六年级竞赛复赛第6题" ]

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

[ "拓展思维->思想->方程思想" ]

[ "设一辆汽车一天运走$$1$$份货物，那么进货的速度是每天$$\\left( 18\\times 8-24\\times 5 \\right)\\div \\left( 8-5 \\right)=8$$份，原有货物$$18\\times 8-8\\times 8=80$$份，设需要$$x$$辆车，有$$10x\\geqslant 80+4\\times 8+6\\times 1.5\\times 8=184$$，那么$$x$$至少是19． " ]

[ "2016年全国华杯赛小学中年级竞赛在线模拟第1题", "2013年全国华杯赛小学中年级竞赛初赛A卷第1题" ]

[ [ { "aoVal": "A", "content": "$$9$$ " } ], [ { "aoVal": "B", "content": "$$11$$ " } ], [ { "aoVal": "C", "content": "$$13$$ " } ], [ { "aoVal": "D", "content": "$$15$$ " } ] ]

[ "拓展思维->思想->整体思想" ]

[ "$$45\\times 40=45\\times 2\\times 20=1800$$，数字和$$1+8+0+0=9$$． " ]

[ "2020年长江杯六年级竞赛复赛B卷第4题5分" ]

[ [ { "aoVal": "A", "content": "$$24$$ " } ], [ { "aoVal": "B", "content": "$$60$$ " } ], [ { "aoVal": "C", "content": "$$90$$ " } ], [ { "aoVal": "D", "content": "$$240$$ " } ] ]

[ "拓展思维->能力->实践应用" ]

[ "后来的售价为原来的： $$(1+30 \\%)\\times 90 \\%=1.3\\times 90 \\%=1.17$$（倍）， 利润为：$$(1.17-1)\\times 6=1.02$$； 增加了： $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde(1.02-0.3)\\div 0.3\\times 100 \\%$$ $$=0.72\\div 0.3\\times 100 \\%$$ $$=2.4\\times 100 \\%$$ $$=240 \\%$$． 故答案为：$$240$$． " ]

[ "2017年第16届湖北武汉创新杯五年级竞赛邀请赛训练题（三）" ]

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

[ "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" ]

[ "每周比赛要分出胜负分差必须在$$2$$分或以上，题中又给出每局比赛分差不超过$$3$$分，故每局比赛的分差只有两种可能：差$$2$$分或$$3$$分．且分差为$$3$$分的那局彭老师得分为$$11$$分，总分差为$$48-39=9$$分，故必有$$3$$场分差为$$2$$分，另一场分差为$$3$$分；即有一场的比分为$$118$$，另两场的总比分为$$3731$$，有以下四种情况：①$$11:9$$，$$11:9$$，$$15:13$$②$$11:9$$，$$12:10$$，$$14:12$$③$$11:9$$，$$13:11$$，$$13:11$$④$$12:10$$，$$12:10$$，$$13:11$$．故一共有$$4$$种情况． " ]

[ "2018年第6届湖北长江杯六年级竞赛初赛A卷第5题3分" ]

[ [ { "aoVal": "A", "content": "一定是质数 " } ], [ { "aoVal": "B", "content": "不一定是互质数 " } ], [ { "aoVal": "C", "content": "没有公因数 " } ], [ { "aoVal": "D", "content": "最小公倍数是他们的积 " } ] ]

[ "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" ]

[ "奇数（$$\\text{odd}$$）指不能被$$2$$整除的整数，数学表达形式为：$$2k+1$$，奇数可以分为正奇数和负奇数． 偶数是能够被$$2$$所整除的整数．正偶数也称双数．若某数是$$2$$的倍数，它就是偶数，可表示为$$2n$$； $$\\text{A}$$．质数是指在大于$$1$$的自然数中，除了$$1$$和它本身以外不再有其他因数的自然数．$$7$$和$$9$$是相邻的奇数，但是$$9$$是合数，不是质数；故$$\\text{A}$$错误； $$\\text{B}$$．互质数为数学中的一种概念，即两个或多个整数的公因数只有$$1$$的非零自然数．公因数只有$$1$$的两个非零自然数，叫做互质数．相邻的两个奇数一定是互质数，故$$\\text{B}$$错误； $$\\text{C}$$．$$7$$和$$9$$是相邻的奇数，它们有公因数$$1$$，故$$\\text{C}$$错误； $$\\text{D}$$．相邻的两个奇数最小公倍数是他们的积，故$$\\text{D}$$正确． 故选$$\\text{D}$$． " ]

[ "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4题2分" ]

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

[ "拓展思维->思想->对应思想" ]

[ "根据题意分析可知，求用$$416$$连续减多少个$$8$$后等于$$0$$， 也就是求$$416$$中有多少个$$8$$， 用除法计算，列式为：$$416\\div8=52$$， 所以减$$52$$次后等于$$0$$． 故选$$\\text{B}$$． " ]

[ "2016年新希望杯小学高年级六年级竞赛训练题（六）第3题" ]

[ [ { "aoVal": "A", "content": "$$45$$ " } ], [ { "aoVal": "B", "content": "$$50$$ " } ], [ { "aoVal": "C", "content": "$$55$$ " } ], [ { "aoVal": "D", "content": "$$60$$ " } ] ]

[ "拓展思维->能力->构造模型->模型思想" ]

[ "设星星一开始摘的桃子数是$$12x$$，兰兰摘的桃子数是$$11x$$．$$\\frac{11x}{12x+40}=\\frac{3}{4}$$，解得$$x=15$$． 星星摘的桃子比兰兰多$$12x+40-11x=55$$个． " ]

[ "2005年五年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "一定是质数 " } ], [ { "aoVal": "B", "content": "一定是合数 " } ], [ { "aoVal": "C", "content": "可为质数，也可为合数 " } ], [ { "aoVal": "D", "content": "既不是质数也不是合数 " } ] ]

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

[ "若$$P$$为奇数，那么$$({{P}^{3}}+5)$$为偶数，又大于$$2$$的偶数都是合数，那么$$P$$是偶数，又$$P$$是质数，所以$$P=2$$，则$${{P}^{5}}+5={{2}^{5}}+5=37$$，选A。 " ]

[ "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$$(10-1)\\div (5-4)=9$$人，共有$$4\\times 9+10=46$$块臭豆腐． " ]

[ "2017年IMAS小学中年级竞赛（第一轮）第9题3分" ]

[ [ { "aoVal": "A", "content": "$$15$$ " } ], [ { "aoVal": "B", "content": "$$17$$ " } ], [ { "aoVal": "C", "content": "$$19$$ " } ], [ { "aoVal": "D", "content": "$$21$$ " } ], [ { "aoVal": "E", "content": "$$23$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "由题意可知这四个连续奇数的平均值为$$72\\div 4=18$$， 因此这四个连续奇数为$$15$$、$$17$$、$$19$$、$$21$$， 即最大奇数为$$21$$． 故选$$\\text{D}$$． " ]

[ "六年级其它", "2004年第2届创新杯六年级竞赛复赛第8题" ]

[ [ { "aoVal": "A", "content": "$$\\frac{5}{6}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{7}{12}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{19}{20}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{9}{10}$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->裂项与通项归纳->分数裂项->分数拆分" ]

[ "$$\\frac{5}{6}=\\frac{1}{2}+\\frac{1}{3}$$，$$\\frac{7}{12}=\\frac{1}{2}+\\frac{1}{12}$$，$$\\frac{19}{20}=\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{5}$$．$$D$$不能被拆分． " ]

[ "2021年新希望杯五年级竞赛初赛第18题5分" ]

[ [ { "aoVal": "A", "content": "$$135$$ " } ], [ { "aoVal": "B", "content": "$$153$$ " } ], [ { "aoVal": "C", "content": "$$513$$ " } ], [ { "aoVal": "D", "content": "$$189$$ " } ] ]

[ "拓展思维->拓展思维->组合模块->操作与策略->操作问题->数字操作" ]

[ "$$645$$是$$6^{3}+4^{3}+5^{3}=216+64+125=280+125=405$$， $$405$$是$$4^{3}+0^{3}+5^{3}=64+0+125=64+125=189$$， $$189$$是$$1^{3}+8^{3}+9^{3}=1+512+729=513+729=1242$$， $$1242$$是$$1^{3}+2^{3}+4^{3}+2^{3}=1+8+64+8=9+72=81$$， $$81$$是$$8^{3}+1^{3}=512+1=513$$， $$513$$是$$5^{3}+1^{3}+3^{3}=125+1+27=126+27=153$$， $$153$$是$$1^{3}+5^{3}+3^{3}=1+125+27=126+27=153$$， 则最后是$$153$$． " ]

[ "2016年新希望杯六年级竞赛训练题（六）第1题" ]

[ [ { "aoVal": "A", "content": "$$5$$ " } ], [ { "aoVal": "B", "content": "$$7.5$$ " } ], [ { "aoVal": "C", "content": "$$10$$ " } ], [ { "aoVal": "D", "content": "$$15$$ " } ] ]

[ "拓展思维->思想->整体思想" ]

[ "李凡一天挖$$\\frac{1}{10}$$，李超一天挖$$\\frac{1}{30}$$，兄弟合挖一天能完成$$\\frac{1}{30}+\\frac{1}{10}=\\frac{2}{15}$$，需要$$1\\div \\frac{2}{15}=7.5$$天． " ]

[ "2009年四年级竞赛创新杯" ]

[ [ { "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->操作问题" ]

[ "全体同学走过后，黑板上的数是$$\\left( 30+5 \\right)\\div 5=7$$，最后一名学生走过之前，黑板上的数是$$\\left( 7+7 \\right)\\div 2=7$$，或$$\\left( 7+14 \\right)\\div 3=7$$，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名学生走过之前，黑板上的数还是7，\\ldots，因此可知，第1名学生走过之前，黑板上的数还是7，即黑板上最初的数是7. " ]

[ "2014年广东广州羊排赛六年级竞赛第8题1分" ]

[ [ { "aoVal": "A", "content": "$$20 \\%$$ " } ], [ { "aoVal": "B", "content": "$$25 \\%$$ " } ], [ { "aoVal": "C", "content": "$$30 \\%$$ " } ], [ { "aoVal": "D", "content": "$$37.5 \\%$$ " } ] ]

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

[ "摸到红球可能性为$$3\\div (9+3)\\times 100 \\%=25 \\%$$． " ]

[ "2013年全国华杯赛小学高年级竞赛初赛C卷第1题", "2017年全国小升初建华入学备考第16题" ]

[ [ { "aoVal": "A", "content": "$$1243$$ " } ], [ { "aoVal": "B", "content": "$$1343$$ " } ], [ { "aoVal": "C", "content": "$$4025$$ " } ], [ { "aoVal": "D", "content": "$$4029$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "方法一：找规律 $$\\frac{{{2}^{2}}}{{{3}^{2}}+1}=\\frac{2}{5}$$、$$\\frac{{{3}^{2}}}{{{4}^{2}}+2}=\\frac{1}{2}=\\frac{3}{6}$$、$$\\frac{{{4}^{2}}}{{{5}^{2}}+3}=\\frac{4}{7}$$、$$\\frac{{{5}^{2}}}{{{6}^{2}}+4}=\\frac{5}{8}$$$$\\cdots$$ 规律找到了，$$\\frac{{{n}^{2}}}{{{(n+1)}^{2}}+(n-1)}=\\frac{n}{n+3}$$． $$\\frac{2013\\times 2013}{2014\\times 2014+2012}=\\frac{n}{m}$$=$$\\frac{2013}{2016}=\\frac{671}{672}$$． 方法二：原式= $$\\frac{2013\\times 2013}{(2013+1)\\times (2013+1)+2013-1}=\\frac{2013\\times2013}{2013\\times 2013+2\\times 2013+1+2013-1}=\\frac{2013}{2016}=\\frac{671}{672}$$． 方法三：原式=$$\\frac{2013\\times 2013}{2014\\times (2013+1)+2012}=\\frac{2013\\times2013}{2013\\times 2014+2014+2012}=\\frac{2013}{2016}=\\frac{671}{672}$$． " ]

[ "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->能力->逻辑分析" ]

[ "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$$\\left( 2+4+5+6 \\right)\\times 10+1+3+7+9=190$$． " ]

[ "2018年第22届广东世界少年奥林匹克数学竞赛二年级竞赛决赛第5题5分" ]

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

[ "拓展思维->能力->逻辑分析" ]

[ "小熊都是早晨提$$4$$桶水，则最后一天会提$$4$$桶水到水缸，所以$$8-4=4$$（桶），而前面的每天只装水：$$4-3=1$$（桶）．所以天数为：$$4\\div1+1=5$$（天）． 故选$$\\text{B}$$． " ]

[ "2019年第23届广东世界少年奥林匹克数学竞赛五年级竞赛初赛第9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" ]

[ "坐在快车上的人和慢车进行相遇运动，则速度和$$=455\\div 13=35$$（米/秒），坐在慢车上的人和快车进行相遇运动时，则相遇时间$$=280\\div 35=8$$（秒）． 故选择$$\\text{B}$$． " ]

[ "2016年河南郑州K6联赛小学高年级六年级竞赛第16题2分" ]

[ [ { "aoVal": "A", "content": "长方体大 " } ], [ { "aoVal": "B", "content": "正方体大 " } ], [ { "aoVal": "C", "content": "同样大 " } ], [ { "aoVal": "D", "content": "无法确定 " } ] ]

[ "拓展思维->拓展思维->组合模块->组合模块最值问题->最值原理在几何中的应用" ]

[ "同样大小的铁皮说明两个铁桶的表面积相等，可以先反过来考虑容积相等的长方体和正方体铁桶，它们的表面积哪一个大．假设两者体积都是$$27$$，正方体表面积是$$3\\times 3\\times 6=54$$；可以拼成一个长$$9$$，宽$$3$$，高$$1$$的长方体，表面积为$$\\left( 9\\times 3+3\\times 1+9\\times 1 \\right)\\times 2=78$$．由此可看出容积相等，正方体的表面积小一些．所以表面积相等，正方体的容积大． " ]

[ "2017年第17届湖北武汉世奥赛五年级竞赛决赛第14题" ]

[ [ { "aoVal": "A", "content": "从一个盒子取一粒豆子 " } ], [ { "aoVal": "B", "content": "从两个盒子各取一粒豆子 " } ], [ { "aoVal": "C", "content": "从一个盒子取两粒豆子 " } ], [ { "aoVal": "D", "content": "从三个盒子各取一粒豆子 " } ] ]

[ "拓展思维->拓展思维->组合模块->抽屉原理->最不利原则" ]

[ "因为都错了．从标有``红、绿豆''的盒子中随便取出一个即可，假如取出来的是一颗红豆，则``红、绿''中都是红，剩下的盒子中，标有``红豆''的盒子肯定是绿豆，标有``绿豆''的肯定是``红绿豆''，所以最少从$$1$$个盒子从取出一粒豆子就可以判断出来． 故选$$\\text{A}$$． " ]

[ "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": "$$150$$ " } ] ]

[ "知识标签->课内知识点->数与运算->数的运算的实际应用（应用题）->百分数的简单实际问题->折扣、成数、税率、利率" ]

[ "$$20\\div \\left( \\frac{5}{4} -1 \\right)=80$$（米），$$80\\div \\left( 1-\\frac{1}{5} \\right)=100$$（米）． " ]

[ "2013年全国学而思杯五年级竞赛A卷第20题" ]

[ [ { "aoVal": "A", "content": "$$0102$$ " } ], [ { "aoVal": "B", "content": "$$0312$$ " } ], [ { "aoVal": "C", "content": "$$2222$$ " } ], [ { "aoVal": "D", "content": "$$0123$$ " } ] ]

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

[ "$$2222$$与$$0000$$的每一位数字均不相同，故$$2222$$一定不是红色，选$$\\text{C}$$． " ]

[ "2005年六年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "小王的年龄的5倍与小谢年龄的3倍相等 " } ], [ { "aoVal": "B", "content": "小王的年龄的4倍与小谢年龄的5倍相等 " } ], [ { "aoVal": "C", "content": "小王的年龄的5倍与小谢年龄的4倍相等 " } ], [ { "aoVal": "D", "content": "小王的年龄的3倍与小谢年龄的5倍相等 " } ] ]

[ "拓展思维->拓展思维->应用题模块->分百应用题" ]

[ "小王的年龄的3倍与小谢的年龄的5倍相等，那么小王的年龄就等于小谢的年龄的$$\\frac{5}{3}$$相等.又他们的年龄相差4岁，所以小谢的年龄为$$4\\div \\left( \\frac{5}{3}-1 \\right)=6$$(岁)，那么小王的年龄为$$6\\times \\frac{5}{3}=10$$(岁).10年后，小谢的年龄为16岁，小王的年龄为20岁，那么小谢的年龄的5倍和小王年龄的4倍相等，选$$B$$ " ]

[ "2022年第九届鹏程杯四年级竞赛初赛第29题5分" ]

[ [ { "aoVal": "A", "content": "$$43$$ " } ], [ { "aoVal": "B", "content": "$$45$$ " } ], [ { "aoVal": "C", "content": "$$46$$ " } ], [ { "aoVal": "D", "content": "$$48$$ " } ], [ { "aoVal": "E", "content": "$$50$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "无 " ]

[ "2011年全国美国数学大联盟杯小学高年级五年级竞赛初赛第29题" ]

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

[ "拓展思维->拓展思维->应用题模块->分百应用题->求分率" ]

[ "$$160\\div400=40 \\%$$． " ]

[ "2017年第15届全国希望杯六年级竞赛" ]

[ [ { "aoVal": "A", "content": "$$150$$；$$150$$ " } ], [ { "aoVal": "B", "content": "$$300$$；$$150$$ " } ], [ { "aoVal": "C", "content": "$$150$$；$$300$$ " } ], [ { "aoVal": "D", "content": "$$300$$；$$300$$ " } ] ]

[ "拓展思维->能力->逻辑分析->代数逻辑推理" ]

[ "$$100$$以内的$$25$$个质数分别为：$$2$$，$$3$$，$$5$$，$$7$$，$$11$$，$$13$$，$$17$$，$$19$$，$$23$$，$$29$$，$$31$$，$$37$$，$$41$$，$$43$$，$$47$$，$$53$$，$$59$$，$$61$$，$$67$$，$$71$$，$$73$$，$$79$$，$$83$$，$$89$$，$$97$$．用$$97$$做分母时，真分数有$$24$$个；用$$89$$做分母时，真分数有$$23$$个；\\ldots；用$$3$$做分母时，真分数有$$1$$个，用$$2$$做分母时，真分数有$$0$$个．所以，满足题意的真分数有$$1+2+3+...+23+24=300$$（个）．因此．满足条件的真分数有$$300$$个．把上面的满足条件真分数的分子、分母颠倒就成了假分数，所以符合条件的假分数也是$$300$$个． " ]

[ "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10题5分" ]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$5.6$$ " } ], [ { "aoVal": "D", "content": "$$6$$ " } ] ]

[ "拓展思维->思想->数形结合思想" ]

[ "此题主要考查了相遇问题，要熟练掌握，注意速度、时间和路程的关系：速度$$\\times $$时间$$=$$路程，路程$$\\div $$时间$$=$$速度，路程$$\\div $$速度$$=$$时间，解答此题的关键是求出两村之间的距离是多少． 首先根据题意，可得甲乙相遇时，甲比乙多行的路程是$$30\\times2=60$$（千米），再根据路程$$\\div $$速度$$=$$时间，用甲乙相遇时行的路程之差除以他们的速度的差，求出甲乙相遇时用的时间是多少，进而求出甲行$$30$$千米用的时间是多少；然后求出两村之间的距离是多少，再根据路程$$\\div $$时间$$=$$速度，用两村之间的距离的$$2$$倍除以甲丙的速度之和，求出丙行了多长时间和甲相遇即可． 甲的速度是每小时行： $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde30\\div (30\\times2\\div12-3.5)$$ $$=30\\div1.5$$ $$=20$$（千米）， 丙和甲相遇用的时间是： $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde20\\times3.5\\times2\\div (20-15+20)$$ $$=70\\times2\\div25$$ $$=140\\div25$$ $$=5.6$$（小时）． 答：丙行了$$5.6$$小时和甲相遇． 故答案为：$$\\text{C}$$． " ]

[ "2016年华杯赛四年级竞赛初赛" ]

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

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

[ "某个数除$$2016$$余$$7$$，于是这个数整除$$2016-7=2009$$，$$2009={{7}^{2}}\\times 41$$，所以$$2009$$共有$$3\\times 2=6$$个约数，其中比$$7$$大的约数有$$4$$个(除了$$1$$和$$7$$)．所以满足要求的除数共有$$4$$个． " ]

[ "2020年新希望杯四年级竞赛决赛（8月）第7题", "2020年新希望杯四年级竞赛初赛（个人战）第7题" ]

[ [ { "aoVal": "A", "content": "$$22.5$$ " } ], [ { "aoVal": "B", "content": "$$20$$ " } ], [ { "aoVal": "C", "content": "$$17.5$$ " } ], [ { "aoVal": "D", "content": "$$25$$ " } ], [ { "aoVal": "E", "content": "$$27.5$$ " } ] ]

[ "拓展思维->能力->构造模型->模型思想" ]

[ "路程$$=$$时间$$\\times$$速度， 设路程为$$30$$千米，那么去时的时间是$$30\\div15=2$$（时）， 回来的时间为$$30\\div30=1$$（时）， 因此往返的平均速度是$$(30+30)\\div(2+1)=60\\div3=20$$（千米$$/$$时）． 故答案为$$20$$． " ]

[ "2017年第16届湖北武汉创新杯五年级竞赛邀请赛训练题（二）" ]

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

[ "知识标签->数学思想->转化与化归的思想" ]

[ "先确定个位，本题中个位只能是$$2$$、$$4$$，$$2$$种选择；再确定十位，$$3$$种选择，$$2\\times 3=6$$（个）. " ]

[ "2017年全国美国数学大联盟杯小学中年级四年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$137$$ " } ], [ { "aoVal": "B", "content": "$$138$$ " } ], [ { "aoVal": "C", "content": "$$139$$ " } ], [ { "aoVal": "D", "content": "$$140$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "除以$$4$$余$$1$$． " ]

[ "2013年全国世奥赛五年级竞赛第17题10分" ]

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

[ "知识标签->数学思想->转化与化归的思想" ]

[ "答案请写在答题卡上 " ]

[ "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5题6分" ]

[ [ { "aoVal": "A", "content": "$$9$$ " } ], [ { "aoVal": "B", "content": "$$12$$ " } ], [ { "aoVal": "C", "content": "$$14$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "根据加乘原理：十位上的数有$$3$$种可选，然后个位只有$$3$$种可选，故$$3\\times 3=9$$（种），故$$\\text{A}$$正确． 故选$$\\text{A}$$． " ]

[ "2013年全国美国数学大联盟杯小学高年级竞赛初赛第15题" ]

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

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

[ "$$2^{3}\\times2^{5}\\times2^{7}\\times2^{11}=2^{3+5+7+11}=2^{26}=4^{13}$$． " ]

[ "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": "$$17$$ " } ] ]

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

[ "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． " ]

[ "2020年第24届YMO五年级竞赛决赛第6题3分", "2019年第24届YMO五年级竞赛决赛第6题3分" ]

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

[ "Overseas Competition->知识点->组合模块->逻辑推理->假设型逻辑推理->真假话", "拓展思维->能力->逻辑分析" ]

[ "对比发现，$$O$$与$$Y$$说的矛盾，相互对立， 则$$O$$与$$Y$$必一对一错， 则$$M$$说假话，则$$M$$会开车，选$$\\text{B}$$． " ]

[ "2020年广东广州羊排赛六年级竞赛第8题3分" ]

[ [ { "aoVal": "A", "content": "$$a\\textgreater b$$ " } ], [ { "aoVal": "B", "content": "$$a=b$$ " } ], [ { "aoVal": "C", "content": "$$a\\textless{}b$$ " } ], [ { "aoVal": "D", "content": "无法确定 " } ] ]

[ "拓展思维->思想->转化与化归的思想" ]

[ "当积一定时，一个因数大，则另一个因数就小， $$\\frac{2}{5}=\\frac{14}{35}$$，$$\\frac{3}{7}=\\frac{15}{35}$$，$$\\frac{2}{5}\\textless{}\\frac{3}{7}$$， 所以$$a\\textgreater b$$， 但是当$$a=b=0$$时，$$\\frac{2}{5}a=0=\\frac{3}{7}b$$， 所以$$a$$和$$b$$的大小无法确定． 故选$$\\text{D}$$． " ]

[ "2015年美国数学大联盟杯六年级竞赛初赛（中国赛区）第34题5分", "2016年全国美国数学大联盟杯小学高年级六年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$46$$ " } ], [ { "aoVal": "B", "content": "$$57$$ " } ], [ { "aoVal": "C", "content": "$$68$$ " } ], [ { "aoVal": "D", "content": "$$79$$ " } ] ]

[ "拓展思维->思想->枚举思想" ]

[ "$$\\text{N}$$是一个两位数．当$$\\text{N}$$被其个位数除的时候，商是$$8$$余数是$$1$$．当$$\\text{N}$$被其十位数除的时候商是$$11$$余数是$$2$$．那么$$\\text{N}$$是几？ 此题最直接的方式，就是四个数试除，答案是$$\\text{B}$$. " ]

[ "2020年长江杯六年级竞赛复赛A卷第3题5分" ]

[ [ { "aoVal": "A", "content": "$$0.5$$ " } ], [ { "aoVal": "B", "content": "$$10$$ " } ], [ { "aoVal": "C", "content": "$$20$$ " } ], [ { "aoVal": "D", "content": "$$25$$ " } ] ]

[ "拓展思维->能力->实践应用" ]

[ "设全程$$a$$千米， 从荆州到长沙的速度是$$a\\div 50=\\frac{a}{50}$$（千米/分钟）， 返回的速度是$$a\\div 40=\\frac{a}{40}$$（千米/分钟）， 返回时速度提高了$$\\left( \\frac{a}{40}-\\frac{a}{50} \\right)\\div \\frac{a}{50}\\times 100 \\%=25 \\%$$． 故选$$\\text{D}$$． " ]

[ "2014年北京海淀区五年级其它", "2014年全国迎春杯四年级竞赛复赛第3题" ]

[ [ { "aoVal": "A", "content": "$$12$$ " } ], [ { "aoVal": "B", "content": "$$24$$ " } ], [ { "aoVal": "C", "content": "$$36$$ " } ], [ { "aoVal": "D", "content": "$$48$$ " } ] ]

[ "拓展思维->能力->实践应用" ]

[ "$$\\left[ 4,6 \\right]=12$$，$$12\\div 6=2$$，$$12\\div 4=3$$，$$12\\div 3-2=2$$，$$12\\times ~2\\times ~2=48$$（米）． " ]

[ "2010年六年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "$$1:3:5$$ " } ], [ { "aoVal": "B", "content": "$$2:3:5$$ " } ], [ { "aoVal": "C", "content": "$$3:5:7$$ " } ], [ { "aoVal": "D", "content": "$$6:10:13$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->浓度问题->浓度基本题型" ]

[ "由于$$2$$种溶液质量不定，混合后的比例应该是有无数种可能。关键点是第二个物质两种溶液比例全是$$\\frac{1}{3}$$，所以新溶液中的比例也是$$\\frac{1}{3}$$。这样排除$$B$$,$$D$$。再分析另两种物质的最大比例$$\\frac{1}{2}$$和最小比例$$\\frac{1}{6}$$。新溶剂的比例不能超过极值。$$A$$中的水比例$$\\frac{1}{9}$$小于最小比例，舍去。所以取$$C$$。 " ]

[ "2019年广东广州羊排赛六年级竞赛第10题3分" ]

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

[ "拓展思维->思想->整体思想" ]

[ "根据余数的性质，和的余数等于余数的和， $$1$$，$$1$$，$$2$$，$$3$$，$$5$$，$$8$$，$$13$$，$$21$$，$$\\cdots $$，$$\\div 4$$余$$1$$，$$1$$，$$2$$，$$3$$，$$1$$，$$0$$，$$1$$，$$1$$，$$2$$，$$3$$，$$1$$，$$0$$， $$1$$，$$1$$，$$\\cdots $$，$$6$$个为一周期，$$2019\\div 6=336$$（组）$$\\cdots \\cdots $$$$3$$（个）， 为周期的第$$3$$个，余数为$$2$$． " ]

[ "2010年中环杯四年级竞赛决赛", "2011年中环杯四年级竞赛初赛", "2011年中环杯四年级竞赛决赛" ]

[ [ { "aoVal": "A", "content": "$$17$$ " } ], [ { "aoVal": "B", "content": "$$3$$ " } ], [ { "aoVal": "C", "content": "$$160$$ " } ], [ { "aoVal": "D", "content": "$$32$$ " } ] ]

[ "拓展思维->拓展思维->计数模块->加乘原理->乘法原理" ]

[ "康康先选择汉堡，共有$$8$$种选择，然后选小吃，有$$4$$种选择，最后选饮料，有$$5$$种选择，根据乘法原理，共有：$$8\\times 4\\times 5=160$$（种）搭配方法。 " ]

[ "2019年第24届YMO一年级竞赛决赛第1题3分" ]

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

[ "拓展思维->能力->逻辑分析" ]

[ "分析可知：$$\\square =\\bigcirc +\\bigcirc =2\\bigcirc $$， 把$$\\square =2\\bigcirc $$代入$$\\bigcirc +\\square +\\square =30$$中得： $$\\bigcirc +2\\bigcirc +2\\bigcirc =5\\bigcirc $$，$$5\\bigcirc =30$$， 所以$$\\bigcirc =30\\div 5=6$$， 那么$$\\square =2\\bigcirc =2\\times 6=12$$， 由此可知，$$\\square $$与$$\\bigcirc $$的差为$$12-6=6$$． 故选$$\\text{C}$$． " ]

[ "2016年第14届全国创新杯五年级竞赛初赛第6题" ]

[ [ { "aoVal": "A", "content": "$$5678$$ " } ], [ { "aoVal": "B", "content": "$$6947$$ " } ], [ { "aoVal": "C", "content": "$$6950$$ " } ], [ { "aoVal": "D", "content": "$$6953$$ " } ] ]

[ "拓展思维->拓展思维->组合模块->图形规律->数量变化" ]

[ "$$1\\times 9+2\\times 90+3\\times 900+4\\times 1016=6953$$． " ]

[ "2019年第7届湖北长江杯六年级竞赛复赛B卷第4题3分" ]

[ [ { "aoVal": "A", "content": "$$4:3$$ " } ], [ { "aoVal": "B", "content": "$$3:4$$ " } ], [ { "aoVal": "C", "content": "$$12:25$$ " } ], [ { "aoVal": "D", "content": "$$25:12$$ " } ] ]

[ "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" ]

[ "本题是一道有关比例问题的求解，熟知路程、速度和时间之间的关系是解题的关键；分析题意，把甲的路程和乙的路程分别看作$$5$$份和$$4$$份，甲和乙的时间分别看作$$3$$份和$$5$$份；然后根据``速度$$=$$路程$$\\div $$时间''分别求出甲和乙的速度，进而再求出最简比即可． 甲的速度为：$$5\\div 3=\\frac{5}{3}$$， 乙的速度为：$$4\\div 5=\\frac{4}{5}$$， 甲乙的速度比为：$$\\frac{5}{3}:\\frac{4}{5}=25:12$$． 故选$$\\text{D}$$． " ]

[ "2019年美国数学大联盟杯竞赛" ]

[ [ { "aoVal": "A", "content": "$$14$$ " } ], [ { "aoVal": "B", "content": "$$16$$ " } ], [ { "aoVal": "C", "content": "$$18$$ " } ], [ { "aoVal": "D", "content": "$$20$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "$$8$$千克的香蕉是$$12$$美元．$$12$$千克的香蕉是多少钱？ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde12\\div 8\\times 12$$ $$=12\\times 12\\div 8$$ $$=144\\div 8$$ $$=18$$（美元）． 故选$$\\text{C}$$． " ]

[ "其它改编题", "2015年湖北武汉世奥赛五年级竞赛模拟训练题（二）第7题" ]

[ [ { "aoVal": "A", "content": "$$0.35$$ " } ], [ { "aoVal": "B", "content": "$$0.45$$ " } ], [ { "aoVal": "C", "content": "$$0.55$$ " } ], [ { "aoVal": "D", "content": "$$0.65$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->小数->小数换元法" ]

[ "设$$a=0.23+0.34$$，$$b=0.23+0.34+0.45$$，原式$$=(1+a)b-(1+b)a=b+ab-a-ab=b-a=0.45$$. " ]

[ "2015年湖北武汉世奥赛六年级竞赛模拟训练题（三）第5题" ]

[ [ { "aoVal": "A", "content": "$$425$$ " } ], [ { "aoVal": "B", "content": "$$426$$ " } ], [ { "aoVal": "C", "content": "$$427$$ " } ], [ { "aoVal": "D", "content": "$$428$$ " } ] ]

[ "拓展思维->思想->转化与化归的思想" ]

[ "$$\\left( 0+2+4+6+8 \\right)\\times 10+\\left( 1+2+\\ldots \\ldots +9 \\right)\\times 5+1=426$$． " ]

[ "2007年第12届全国华杯赛竞赛初赛第6题10分" ]

[ [ { "aoVal": "A", "content": "$$280$$ " } ], [ { "aoVal": "B", "content": "$$270$$ " } ], [ { "aoVal": "C", "content": "$$252$$ " } ], [ { "aoVal": "D", "content": "$$216$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "余下的数之和为：$$55\\times \\frac{7}{11}=35$$，取出的数之和为：$$55-35=20$$， 要使取出的三个数之积尽量大，则取出的三个数应尽量接近， 我们知$$6+7+8=21$$，所以取$$5\\times 7\\times 8=280$$． " ]

[ "2014年华杯赛四年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$11$$点$$35$$分 " } ], [ { "aoVal": "B", "content": "$$12$$点$$5$$分 " } ], [ { "aoVal": "C", "content": "$$11$$点$$40$$分 " } ], [ { "aoVal": "D", "content": "$$12$$点$$20$$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" ]

[ "由于全程是匀速运动，所以从晚$$6$$分追到早$$6$$分，前半程和后半程所需时间是一样的，所以经过中点的时间应该是不变的，就是$$10$$点$$10$$分和$$13$$点$$10$$分的中点$$11$$点$$40$$分。 " ]

[ "2017年河南郑州豫才杯四年级竞赛初赛第15题" ]

[ [ { "aoVal": "A", "content": "每分钟$$300$$字 " } ], [ { "aoVal": "B", "content": "每分钟$$400$$字 " } ], [ { "aoVal": "C", "content": "每分钟$$500$$字 " } ], [ { "aoVal": "D", "content": "每分钟$$600$$字 " } ] ]

[ "拓展思维->能力->构造模型->模型思想" ]

[ "$$1000000\\div 100\\div 20=500$$（字），所以阅读速度至少提升到每分钟$$500$$字的水平． " ]

[ "2020年希望杯二年级竞赛模拟第4题" ]

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

[ "拓展思维->思想->对应思想" ]

[ "后一个数等于前一个数$$\\times$$ 2 " ]

[ "2020年新希望杯六年级竞赛（2月）第13题" ]

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

[ "拓展思维->能力->运算求解" ]

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

[ "2015年湖北武汉世奥赛六年级竞赛模拟训练题（三）第6题" ]

[ [ { "aoVal": "A", "content": "$$105$$分钟 " } ], [ { "aoVal": "B", "content": "$$130$$分钟 " } ], [ { "aoVal": "C", "content": "$$135$$分钟 " } ], [ { "aoVal": "D", "content": "$$150$$分钟 " } ] ]

[ "知识标签->课内知识点->式与方程->数量关系->路程=速度×时间" ]

[ "$$60\\times 4+20=50\\times 5+10$$，去时共走了$$40\\times 5+10=210$$（分钟），则返回时共需走$$210\\div 2=105$$（分钟），$$105\\div 35=3$$，加上休息的时间共需要$$105+15\\times \\left( 3-1 \\right)=135$$（分钟）． " ]

[ "2017年环亚太杯一年级竞赛初赛第4题" ]

[ [ { "aoVal": "A", "content": "28，40 " } ], [ { "aoVal": "B", "content": "28，42 " } ], [ { "aoVal": "C", "content": "30，40 " } ], [ { "aoVal": "D", "content": "30，42 " } ] ]

[ "知识标签->课内知识点->数学广角->找规律->数字排列规律" ]

[ "每个数之间差了一个6 " ]

[ "2015年湖北武汉世奥赛五年级竞赛模拟训练题（一）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->应用题模块->鸡兔同笼问题->假设法解鸡兔同笼->基本型->原型题" ]

[ "假设这$$10$$辆客车都是大客车，则大客车比小客车多坐$$10\\times 100=1000$$（人），比实际多了$$1000-520=480$$（人），这是因为每把一辆小客车看作大客车，就会多算$$100+60=160$$（人），所以小客车一共有$$480\\div 160=3$$（辆），大客车有$$10-3=7$$（辆）． " ]

[ "2003年五年级竞赛创新杯", "2003年第1届创新杯五年级竞赛复赛第8题" ]

[ [ { "aoVal": "A", "content": "7 " } ], [ { "aoVal": "B", "content": "8 " } ], [ { "aoVal": "C", "content": "5 " } ], [ { "aoVal": "D", "content": "1 " } ] ]

[ "拓展思维->拓展思维->应用题模块->周期问题" ]

[ "若a为1到6的整数，则$$\\frac{a}{7}$$化成小数时，其小数部分均为1、4、2、8、5、7 这6个数码组成.由于$$10\\div 7=1\\frac{3}{7}=1+0.\\dot{4}2857\\dot{1}$$，而$$2003\\div 6=333\\cdots \\cdots 5$$， 那么小数部分第2003个数为一个周期数码428571中的第5个为7 " ]

[ "2014年全国学而思杯一年级竞赛第5题" ]

[ [ { "aoVal": "A", "content": "单 " } ], [ { "aoVal": "B", "content": "双 " } ] ]

[ "拓展思维->思想->转化与化归的思想", "Overseas Competition->知识点->应用题模块->分百应用题->认识单位1" ]

[ "四个数中三个单数一个双数，三个单数和是单数，再加一个双数和是单数． " ]

[ "2018年湖北武汉创新杯小学高年级六年级竞赛初赛数学思维能力等级测试第8题4分" ]

[ [ { "aoVal": "A", "content": "$$600$$ " } ], [ { "aoVal": "B", "content": "$$420$$ " } ], [ { "aoVal": "C", "content": "$$350$$ " } ], [ { "aoVal": "D", "content": "$$360$$ " } ] ]

[ "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" ]

[ "由于甲完成任务的$$\\frac{3}{5}$$等于易完成任务的$$\\frac{1}{2}$$，则甲、乙的效率比$$:$$为甲乙$$=\\frac{3}{5}:\\frac{1}{2}=6:5$$，则当乙完成$$50$$个小时，甲完成$$60$$个，则甲的任务量为$$60\\div \\frac{1}{3}=180$$个，则共有$$180\\times 2=360$$个． " ]

[ "2019年广东广州学而思综合能力诊断五年级竞赛第11题12分" ]

[ [ { "aoVal": "A", "content": "$$360$$ " } ], [ { "aoVal": "B", "content": "$$380$$ " } ], [ { "aoVal": "C", "content": "$$390$$ " } ], [ { "aoVal": "D", "content": "$$420$$ " } ], [ { "aoVal": "E", "content": "$$430$$ " } ] ]

[ "拓展思维->拓展思维->数论模块->因数与倍数->因数个数定理->因数个数定理逆应用", "Overseas Competition->知识点->应用题模块->分百应用题->认识单位1" ]

[ "本题是因数个数定理的逆应用，先从$$A$$、$$B$$、$$C$$三个的因数个数开始讨论：$$C$$有$$11$$个因数，$$11$$是质数，所以$$C$$一定是一个质数的$$10$$次方，又因为三个数的和为$$2021$$，小于$$3$$的$$10$$次方，所以$$C$$只能是$$2$$的$$10$$次方，等于$$1024$$，则$$A+B=2021-1024=997$$，$$A$$有$$8$$个因数，则$$A$$的分解质因数形式可以是$$a$$的$$7$$次方、$$a$$的$$1$$次方乘$$b$$的$$3$$次方、$$a\\times b\\times c$$这$$3$$种可能，$$B$$的分解质因数形式可以是$$c$$的$$9$$次方、$$c$$的$$1$$次方乘$$d$$的$$4$$次方，因为$$B$$和$$C$$互质，所以$$B$$不含质因数$$2$$，根据和的大小可以确定$$B$$不等于$$c$$的$$9$$次方，又$$5$$的$$4$$次方等于$$625$$，$$625$$乘$$3$$等于$$1275$$大于$$997$$，所以$$d$$只能等于$$3$$，$$3$$的$$4$$次方等于$$81$$，$$81$$乘$$5$$等于$$405$$，$$997-405=592$$不符合$$A$$的分解质因数条件，$$81$$乘$$7$$等于$$567$$，$$997-567=430$$符合条件，则$$A$$等于$$430$$． " ]

[ "2012年全国创新杯五年级竞赛第1题5分", "五年级其它", "2016年创新杯小学高年级五年级竞赛训练题（四）第4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->计数模块->容斥原理->三量容斥" ]

[ "让三项都会的人最少，即让所有人都只会两项， 则一共会$$48\\times 2=96$$项，$$27+33+40=100$$项， 即至少有$$100-96=4$$人会三项． 构造如下：$$8$$人会游泳$$+$$自行车； $$15$$人会游泳$$+$$乒乓球；$$21$$人会自行车$$+$$乒乓球；$$4$$人全会． " ]

[ "2017年河南郑州豫才杯竞赛第15题", "2016年河南郑州K6联赛六年级竞赛第17题2分", "2017年河南郑州小升初豫才杯第二场第15题" ]

[ [ { "aoVal": "A", "content": "$$A\\textgreater B$$ " } ], [ { "aoVal": "B", "content": "$$A=B$$ " } ], [ { "aoVal": "C", "content": "$$A\\textless{}B$$ " } ], [ { "aoVal": "D", "content": "无法确定 " } ] ]

[ "知识标签->拓展思维->数论模块->位值原理与进制->数与数字->比较大小" ]

[ "$$A-B=9876543\\times 23456789-9876544\\times 23456788$$ $$ ~=9876543\\times \\left( 23456788+1 \\right)-9876544\\times 23456788 $$ $$ ~=9876543\\times 23456788+9876543-9876544\\times 23456788 $$ ~$$=9876543-23456788\\textless{}0 $$ 所以$$A\\textless{}B$$． " ]

[ "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，所以选择$$\\text{C}$$． " ]

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

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

[ "知识标签->课内知识点->数与运算->混合运算->分、小数四则混合运算" ]

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

[ "小学高年级六年级其它2014年数学思维能力等级测试第2题", "2014年第12届全国创新杯小学高年级六年级竞赛第2题" ]

[ [ { "aoVal": "A", "content": "$$20$$ " } ], [ { "aoVal": "B", "content": "$$28$$ " } ], [ { "aoVal": "C", "content": "$$30$$ " } ], [ { "aoVal": "D", "content": "$$36$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "$$\\overline{abc}+a+b+c=429$$，$$101a+11b+2c=429$$， 所以$$a=4$$，$$11b+2c=25$$，$$b=1$$，$$c=7$$， $$4\\times 1\\times 7=28$$． " ]

[ "2018年第23届华杯赛小学中年级竞赛初赛第2题", "2018年华杯赛小学中年级竞赛初赛第2题10分" ]

[ [ { "aoVal": "A", "content": "$$11$$ " } ], [ { "aoVal": "B", "content": "$$12$$ " } ], [ { "aoVal": "C", "content": "$$39$$ " } ], [ { "aoVal": "D", "content": "$$40$$ " } ], [ { "aoVal": "E", "content": "以上都不对 " } ] ]

[ "拓展思维->能力->实践应用" ]

[ "由题目条件可知，反面的数为质数．又知反面数字和相等，故反面三个数依然相差$$12$$，$$9$$． 数学质数有$$2$$，$$3$$，$$5$$，$$7$$，$$11$$，$$13$$，$$17$$，$$23$$． $$2$$，$$11$$，$$23$$依次相差$$9$$和$$12$$，故确定其平均数为$$\\left( 2+11+23 \\right)\\div 3=12$$． " ]

[ "2015年全国AMC小学高年级六年级竞赛8第10题" ]

[ [ { "aoVal": "A", "content": "$$3024$$ " } ], [ { "aoVal": "B", "content": "$$4536$$ " } ], [ { "aoVal": "C", "content": "$$5040$$ " } ], [ { "aoVal": "D", "content": "$$6480$$ " } ], [ { "aoVal": "E", "content": "$$6561$$ " } ] ]

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

[ "翻译：$$1000$$至$$9999$$中有多少个位数字互不相同的四位数？ 组数问题：千位数字有$$1\\sim 9$$共$$9$$种选法，百、十、个位分别有$$9$$、$$8$$、$$7$$种选法，一共有$$9\\times 9\\times 8\\times 7=4536$$（个）数字互不相同的四位数． " ]

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

[ [ { "aoVal": "A", "content": "$$42$$ " } ], [ { "aoVal": "B", "content": "$$43$$ " } ], [ { "aoVal": "C", "content": "$$15\\frac{1}{3}$$ " } ], [ { "aoVal": "D", "content": "$$16\\frac{2}{3}$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->分数->分数运算->分数四则混合运算" ]

[ "原式$$=\\left[ \\left( \\frac{1}{4}+\\frac{1}{5} \\right)-\\left( \\frac{1}{5}+\\frac{1}{6} \\right)+\\left( \\frac{1}{6}+\\frac{1}{7} \\right)-\\left( \\frac{1}{7}+\\frac{1}{8} \\right)+\\left( \\frac{1}{8}+\\frac{1}{9} \\right) \\right]\\times 120-\\frac{1}{3}\\div \\frac{1}{4}$$ $$=\\left( \\frac{1}{4}+\\frac{1}{9} \\right)\\times 120-\\frac{1}{3}\\div \\frac{1}{4}$$ $$=\\frac{13}{36}\\times 120-\\frac{4}{3}$$ $$=\\frac{130}{3}-\\frac{4}{3}$$ $$=42$$ " ]

[ "2016年全国华杯赛小学中年级竞赛在线模拟第3题", "2013年全国华杯赛小学中年级竞赛初赛A卷第3题" ]

[ [ { "aoVal": "A", "content": "丽丽 " } ], [ { "aoVal": "B", "content": "莱克西 " } ], [ { "aoVal": "C", "content": "$$Ellie$$ " } ], [ { "aoVal": "D", "content": "Michelle " } ] ]

[ "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", "Overseas Competition->知识点->组合模块->逻辑推理->假设型逻辑推理->真假话" ]

[ "由于只有一个人说对了，而$$Michelle$$支持丽丽，那么他们俩都错了，所以反对丽丽的$$Ellie$$说对了． " ]

[ "2020年长江杯六年级竞赛复赛A卷第1题5分" ]

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

[ "拓展思维->思想->方程思想" ]

[ "设甲成本为$$x$$元， $$x\\times \\left( 1+\\frac{1}{4} \\right)=120$$， 解得$$x=96$$， 设乙成本为$$x$$元， $$x\\times \\left( 1-\\frac{1}{4} \\right)=120$$， 解得$$x=160$$， 总成本$$96+160=256$$（元）， 总收入为$$120\\times 2=240$$（元）， $$256\\textgreater240$$， 故成本大于收入，亏了． 故选$$\\text{B}$$． " ]

[ "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": "$$20$$ " } ] ]

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

[ "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$$10$$个馒头时间相同，都是两分钟． 故选：$$\\text{A}$$． " ]

[ "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": "$$3.14159\\dot{2}\\dot{6}$$ " } ] ]

[ "知识标签->课内知识点->数的认识->比较大小->小数比较大小->多位小数比较大小" ]

[ "要使循环小数最小，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$$3.\\dot{1}41592\\dot{6}$$． " ]

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

[ "拓展思维->能力->运算求解" ]

[ "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． 故选$$\\text{D}$$． " ]

[ "2018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": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$11$$ " } ], [ { "aoVal": "C", "content": "$$12$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "本来甲乙所处的位置就在$$1$$层， 即甲从$$1$$层到$$4$$层的时间，其实是跑了$$3$$层楼梯的时间， $$\\left( 16-1 \\right)\\div \\left( 4-1 \\right)\\times \\left( 3-1 \\right)=10$$层，$$10+1=11$$层． 故选$$\\text{B}$$． " ]

[ "2014年迎春杯三年级竞赛复赛" ]

[ [ { "aoVal": "A", "content": "$$48$$ " } ], [ { "aoVal": "B", "content": "$$36$$ " } ], [ { "aoVal": "C", "content": "$$24$$ " } ], [ { "aoVal": "D", "content": "$$18$$ " } ] ]

[ "拓展思维->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" ]

[ "解：$$6\\div 3\\times 4$$ $$=2\\times 4$$ $$=8$$（只） $$8\\div 2\\times 6$$ $$=4\\times 6$$ $$=24$$（条） $$24\\times 2=48$$（条） 答：$$2$$只小猪的重量等于$$48$$条鱼的重量。 故选:A。 " ]

[ "2013年全国华杯赛小学高年级竞赛初赛B卷第5题" ]

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

[ "拓展思维->能力->逻辑分析" ]

[ "设一个仓库的稻谷量为``$$1$$''，爸爸、妈妈、阳阳的效率分别是$$\\frac{1}{10}$$、$$\\frac{1}{12}$$、$$\\frac{1}{15}$$， 三人同时运完两仓，需要的时间：$$(1+1)\\div \\left( \\frac{1}{10}+\\frac{1}{12}+\\frac{1}{15} \\right)=8$$（天）；妈妈$$8$$天共搬运了：$$8\\times \\frac{1}{12}=\\frac{2}{3}$$（仓）；妈妈剩下的就是阳阳帮妈妈运的，所以，阳阳帮妈妈运了$$\\left( 1-\\frac{2}{3} \\right)\\div \\frac{1}{15}=5$$（天）． 三人一共搬了： $$(1+1)\\div \\left( \\frac{1}{10}+\\frac{1}{12}+\\frac{1}{15} \\right)$$， $$=2\\div \\frac{1}{4}$$ $$=8$$（天）； 阳阳帮妈妈运的天数： $$\\left( 1-\\frac{1}{12}\\times 8 \\right)\\div \\frac{1}{15}$$ $$=\\frac{1}{3}\\times 15$$ $$=5$$（天）； 答：阳阳帮妈妈运了$$5$$天． 故答案为：$$\\rm C$$． " ]

[ "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": "$$30$$ " } ] ]

[ "知识标签->拓展思维->行程模块->流水行船问题->基本流水行船问题->水速变化" ]

[ "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． " ]