Datasets:

---

math-eval

/

TAL-SCQ5K

like 57

[ "2017年新希望杯小学高年级六年级竞赛训练题（二）第5题", "2018年湖北武汉新希望杯六年级竞赛训练题（二）第5题" ]

[ [ { "aoVal": "A", "content": "$$112$$ " } ], [ { "aoVal": "B", "content": "$$114$$ " } ], [ { "aoVal": "C", "content": "$$120$$ " } ], [ { "aoVal": "D", "content": "$$128$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "$$\\left[ \\frac{200}{2} \\right]+\\left[ \\frac{200}{7} \\right]-\\left[ \\frac{200}{2\\times 7} \\right]=100+28-14=114$$（个）． " ]

[ "2017年第15届湖北武汉创新杯小学高年级六年级竞赛初赛第9题" ]

[ [ { "aoVal": "A", "content": "$$14$$ " } ], [ { "aoVal": "B", "content": "$$23$$ " } ], [ { "aoVal": "C", "content": "$$32$$ " } ], [ { "aoVal": "D", "content": "$$41$$ " } ] ]

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

[ "和有千位，则百位一定向前进一位，和的百位为$$9$$且向前进位，则十位一定向百位进一位，和的个位为$$9$$，则个位一定不向前进位，所以总共进两位，数字之和减少$$9\\times 2=18$$，则$$\\overline{\\square } $$中的数字之和是$$1+9+4+9+18=41$$． " ]

[ "2019年第24届YMO六年级竞赛决赛第7题3分" ]

[ [ { "aoVal": "A", "content": "$$149$$ " } ], [ { "aoVal": "B", "content": "$$244$$ " } ], [ { "aoVal": "C", "content": "$$264$$ " } ], [ { "aoVal": "D", "content": "$$274$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "设第$$n$$级有$${{a}_{n}}$$种登的方式， 则第$$n-3$$，$$n-2$$，$$n-1$$级再分别登$$3$$，$$2$$，$$1$$级可到第$$n$$级， 则$${{a}_{n}}={{a}_{n-3}}+{{a}_{n-2}}+{{a}_{n-1}}$$， 而$${{a}_{1}}=1$$，$${{a}_{2}}=1+1=2$$，$${{a}_{3}}=1+1+2=4$$， 故$${{a}_{4}}=1+2+4=7$$， $${{a}_{5}}=2+4+7=13$$， $${{a}_{6}}=4+7+13=24$$， $${{a}_{7}}=7+13+24=44$$， $${{a}_{8}}=13+24+44=81$$， $${{a}_{9}}=24+44+81=149$$， $${{a}_{10}}=44+81+149=274$$． 故选：$$\\text{D}$$． " ]

[ "2016年第21届全国华杯赛小学高年级竞赛初赛B卷第2题" ]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$15$$ " } ], [ { "aoVal": "C", "content": "$$127$$ " } ], [ { "aoVal": "D", "content": "$$2047$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "选项$$\\rm A$$：$$2^{n}-1=4$$，$$n$$无整数解； 选项$$\\rm B$$：$$2^{n}-1=15$$，$$n$$为$$4$$，但$$n$$不是质数，故舍去； 选项$$\\rm C$$：$$2^{n}-1=127$$，$$n$$为$$7$$，$$127$$不是合数，故舍去； 选项$$\\rm D$$：$$2^{n}-1=2047$$，$$n$$为$$11$$，$$n$$为质数，且$$2047=23\\times 89$$，是合数，满足条件． " ]

[ "2014年全国迎春杯三年级竞赛初赛第8题" ]

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

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

[ "$$2000\\div (4+3+2+1)=200$$． " ]

[ "2017年第22届全国华杯赛小学高年级竞赛初赛第6题10分" ]

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

[ "拓展思维->思想->枚举思想" ]

[ "设四个空依次为$$a，b，c，d$$，即，``这句话里有$$a$$个数大于$$1$$，有$$b$$个数大于$$2$$，有$$c$$个数大于$$3$$，有$$d$$个数大于$$4$$．''根据包含关系，必有$$a\\textgreater b\\textgreater c\\textgreater d\\geqslant 0$$，所以$$a\\textgreater b\\geqslant 2$$，所以至少有$$5$$个数大于$$1$$($$a，b$$以及给出的$$2，3，4$$)，所以$$a\\geqslant 5$$，说明至少有一个数大于$$4$$(这个数是$$a$$)，所以$$d\\geqslant 1$$．从$$d$$开始考虑: 若$$d=1$$，说明只有$$1$$个数大于$$4$$(只能为$$a$$)，且$$a，b，c$$均大于$$1$$，于是有$$a\\textgreater4\\geqslant b\\textgreater c\\geqslant 2$$，则有$$6$$个数大于$$1$$($$a，b，c$$以及给出的$$2，3，4$$)，所以$$a=6$$．由$$a\\textgreater4\\geqslant b\\textgreater c\\geqslant 2$$可知$$b\\geqslant 3$$说明至少有$$4$$个数大于$$2$$($$a，b$$以及给出的$$3，4$$)，所以$$b\\geqslant 4$$，又因为$$b\\geqslant 4$$，所以有$$b=4$$．此时有$$3$$个数大于$$3$$($$a，b$$以及给出的$$4$$），但此时发现$$8$$个数中有$$5$$个大于$$2$$，和$$b=4$$矛盾； 若$$d=2$$，说明有$$2$$个数大于$$4$$(只能为$$a，b$$)，且$$a，b，c$$均大于$$2$$，于是有$$a\\textgreater b\\textgreater4\\geqslant c\\geqslant 3$$，则有$$7$$个数大于$$1$$($$a，b，c$$以及给出的$$2，3，4$$)，$$5$$个数大于$$2$$($$a，b，c$$以及给出的$$3，4$$)，所以$$a=6$$，$$b=5$$．此时无论$$c=3$$或$$c=4$$均满足条件，这时的两组解为$$a=7$$，$$b=5$$，$$c=3$$，$$d=2$$和$$a=7$$，$$b=5$$，$$c=4$$，$$d=2$$． 若$$d=3$$，说明有$$3$$个数大于$$4$$(为$$a，b，c$$)，那么大于$$3$$的数有$$4$$个($$a，b，c$$以及给出的$$4$$)，所以$$c=4$$，但$$c\\textgreater4$$，矛盾，所以$$d=3$$不存在． 若$$d=4$$，说明有$$4$$个数大于$$4$$(为$$a，b，c，d$$)，但$$d=4$$，与$$d\\textgreater4$$矛盾． 显然$$d=5$$不存在，所以共有$$2$$组解． " ]

[ "2020年希望杯二年级竞赛模拟第19题" ]

[ [ { "aoVal": "A", "content": "$$56$$ " } ], [ { "aoVal": "B", "content": "$$74$$ " } ], [ { "aoVal": "C", "content": "$$92$$ " } ], [ { "aoVal": "D", "content": "$$146$$ " } ], [ { "aoVal": "E", "content": "$$188$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "要使得和最小，那么两个因数应该尽可能的小； 如果要求两个因数尽可能的小，那么两个因数十位应该填入较小数， 即这两个因数应该是：$$35+39=74$$， 所以和最小是$$\\text{B}$$． " ]

[ "2015年第27届广东广州五羊杯小学高年级竞赛第8题4分" ]

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

[ "拓展思维->能力->逻辑分析" ]

[ "解析分类讨论，有序列举． 有$$4$$个因数的整数只有两种情况：立方数和两个质因数的乘积． 不难验证，立方数中没有任何一个满足条件， 所以只能是第二种情况． 1．自然数由大到小：$$10=2\\times5$$；$$21=3\\times7$$；$$65=13\\times5$$；$$87=3\\times29$$； 2．自然数由小到大：$$34=2\\times17$$，所以共有$$5$$个． 故选$$\\text{A}$$． " ]

[ "2021年迎春杯小学高年级六年级竞赛决赛C卷第7题10分" ]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$6$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ], [ { "aoVal": "E", "content": "$$9$$ " } ] ]

[ "拓展思维->拓展思维->组合模块->逻辑推理->条件型逻辑推理->神推理", "Overseas Competition->知识点->应用题模块->分百应用题->认识单位1" ]

[ "本题中，通过甲说出的``另外三个数两两互质''，乙丙就知道了自己的数，说明乙、丙、丁三个人的数字，在$$1\\sim 9$$当中是唯一固定的一组三个数明确的互质数即``$$5$$、$$6$$、$$7$$''三个数，而在这三个数中，只有``$$6$$''在$$1$$$$\\sim 9$$当中有且仅有$$5$$、$$7$$两个互质数，所以乙、丙两个人的数为$$5$$、$$7$$或$$7$$、$$5$$，丁为$$6$$． $$1$$$$\\sim9$$各个数字对应的互质数． $$1$$：$$1$$、$$2$$、$$3$$、$$4$$、$$5$$、$$6$$、$$7$$、$$8$$、$$9$$， $$2$$：$$3$$、$$5$$、$$7$$、$$9$$， $$3$$：$$2$$、$$4$$、$$5$$、$$7$$、$$8$$， $$4$$：$$3$$、$$5$$、$$7$$、$$9$$， $$5$$：$$2$$、$$3$$、$$4$$、$$6$$、$$7$$、$$8$$、$$9$$， $$6$$：$$5$$、$$7$$， $$8$$：$$3$$、$$5$$、$$7$$、$$9$$， $$9$$：$$2$$、$$4$$、$$5$$、$$7$$、$$8$$． 由此可见，只有$$6$$在$$1\\sim9$$的范围内有且仅有两个互质数． " ]

[ "2017年四川六年级竞赛排位赛第17题2分", "2020~2021学年福建福州鼓楼区福州教育学院附属第二小学六年级上学期期中第5题1分" ]

[ [ { "aoVal": "A", "content": "$$a ~\\textless{} ~b$$ " } ], [ { "aoVal": "B", "content": "$$b ~\\textless{} ~c$$ " } ], [ { "aoVal": "C", "content": "$$a ~\\textless{} ~c$$ " } ], [ { "aoVal": "D", "content": "$$c ~\\textless{} ~b$$ " } ] ]

[ "拓展思维->能力->抽象概括" ]

[ "因为$$a$$、$$b$$、$$c$$是不同的自然数，$$a\\times \\frac{c}{b} ~\\textless{} ~a$$，则$$ac ~\\textless{} ~ab$$，即$$c ~\\textless{} ~b$$． " ]

[ "2020年第1届广东深圳超常思维竞赛四年级竞赛初赛第2题4分" ]

[ [ { "aoVal": "A", "content": "$${{1011}^{2}}$$ " } ], [ { "aoVal": "B", "content": "$${{1000}^{2}}$$ " } ], [ { "aoVal": "C", "content": "$${{999}^{2}}$$ " } ], [ { "aoVal": "D", "content": "$${{998}^{2}}$$ " } ], [ { "aoVal": "E", "content": "以上都不正确 " } ] ]

[ "拓展思维->思想->转化与化归的思想" ]

[ "中项定理：等差数列的和等于中项的平方． 中项：$$\\left( 1+2021 \\right)\\div 2=1011$$， 所以答案为$$1011\\times 1011$$，即选择$$\\text{A}$$． " ]

[ "2014年IMAS小学高年级竞赛第二轮检测试题第3题4分" ]

[ [ { "aoVal": "A", "content": "$$100$$ " } ], [ { "aoVal": "B", "content": "$$110$$ " } ], [ { "aoVal": "C", "content": "$$117$$ " } ], [ { "aoVal": "D", "content": "$$181$$ " } ], [ { "aoVal": "E", "content": "$$271$$ " } ] ]

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

[ "如果一个两位数是完全平方数，那么这个两位数的所有可能值是： $$16$$，$$25$$，$$36$$，$$49$$，$$64$$，$$81$$， 而$$3+6={{3}^{2}}$$，$$8+1={{3}^{2}}$$， 所以符合条件的两位数为$$36$$，$$81$$， $$36+81=117$$． " ]

[ "2017年全国亚太杯四年级竞赛初赛第10题" ]

[ [ { "aoVal": "A", "content": "$$4444452$$ " } ], [ { "aoVal": "B", "content": "$$4444450$$ " } ], [ { "aoVal": "C", "content": "$$4444450$$ " } ], [ { "aoVal": "D", "content": "$$4444428$$ " } ] ]

[ "拓展思维->思想->转化与化归的思想" ]

[ "根据位值原理，原式$$=4\\times 1111110+2\\times 6=4444452$$ ． " ]

[ "2018年第6届湖北长江杯五年级竞赛初赛A卷第3题3分" ]

[ [ { "aoVal": "A", "content": "$$14$$ " } ], [ { "aoVal": "B", "content": "$$13$$ " } ], [ { "aoVal": "C", "content": "$$12$$ " } ], [ { "aoVal": "D", "content": "$$11$$ " } ] ]

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

[ "因为$$0.\\underbrace{00000025}_{8位}\\times 0.\\underbrace{000004}_{6位}=0.\\underbrace{000000000001}_{12位}$$， 故小数部分有$$11$$个$$0$$， 故答案是：$$\\text{D}$$． " ]

[ "2014年全国迎春杯五年级竞赛复赛第8题" ]

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

[ "拓展思维->能力->数据处理" ]

[ "$$2014$$要变成$$1$$就需要除以一个数，而除数只能是一位数，那么这个除数显然是越大越好．第一次操作$$2014+2=2016$$；第二次操作$$2016\\div9=224$$；第三次操作$$224\\div 8=28$$；第四次操作$$28\\div 7=4$$；第五次操作$$4\\div 4=1$$． " ]

[ "2017年四川六年级竞赛排位赛第16题2分" ]

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

[ "拓展思维->能力->数据处理" ]

[ "$$3+3=6$$，$$6\\div 3=2$$，分子扩大到原来的$$2$$倍，要使分数大小不变，分母也要扩大到原来的$$2$$倍，即$$8\\times 2=16$$，$$16-8=8$$，分母应加上$$8$$． " ]

[ "2020年长江杯五年级竞赛复赛B卷第3题5分" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$12$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ] ]

[ "拓展思维->思想->枚举思想" ]

[ "不能一颗一颗地拿，不能一次全拿出去，而且每次拿的颗数要相等，最后全部拿完，也就是说次数$$\\times $$每次拿的颗数$$=96$$．$$96=1\\times 96$$，$$96=2\\times 48$$，$$96=3\\times 32$$，$$96=4\\times 24$$，$$96=6\\times 16$$，$$96=8\\times 12$$，那么每次拿的颗数可以是$$2$$、$$48$$、$$3$$、$$32$$、$$4$$、$$24$$、$$6$$、$$16$$、$$8$$、$$12$$，一共有$$10$$种拿法． 故选$$\\text{A}$$． " ]

[ "2019年第23届广东世界少年奥林匹克数学竞赛四年级竞赛决赛第4题5分" ]

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

[ "拓展思维->能力->运算求解" ]

[ "$$1$$、这是一道关于方阵的问题，考查的是方阵的知识； $$2$$、由题意知，用围棋子摆成一个三层空心方阵，最内一层每边有围棋子$$8$$个，由于相邻两层每边相差$$2$$个，相邻两层相差$$8$$个，求出最内一层后依次加$$8$$即可。 根据题意分析可知：最内一层棋子个数为：$$\\left( 8-1 \\right)\\times 4=7\\times 4=28$$（个）， 第二层棋子有：$$28+8=36$$（个）， 第三层棋子有：$$36+8=44$$（个）， 所以三层一共有$$28+36+44=108$$（个）． 故选：$$\\text{B}$$． " ]

[ "2015年第11届全国新希望杯小学高年级六年级竞赛复赛第2题" ]

[ [ { "aoVal": "A", "content": "$$13$$:$$00$$ " } ], [ { "aoVal": "B", "content": "$$13$$:$$09$$ " } ], [ { "aoVal": "C", "content": "$$13$$:$$10$$ " } ], [ { "aoVal": "D", "content": "$$13$$:$$15$$ " } ] ]

[ "拓展思维->拓展思维->行程模块->时钟问题->坏钟问题" ]

[ "手表走$$57$$分钟，正常钟走$$60$$分钟，现在手表走$$6$$小时$$20$$分钟，为$$380$$分钟，那么正常钟应该走$$380\\div 57\\times 60=400$$．为$$6$$小时$$40$$分，此时为$$13$$:$$10$$． " ]

[ "2021年新希望杯三年级竞赛初赛第13题5分" ]

[ [ { "aoVal": "A", "content": "只有苹果 " } ], [ { "aoVal": "B", "content": "只有橘子 " } ], [ { "aoVal": "C", "content": "只有草莓 " } ], [ { "aoVal": "D", "content": "香蕉和草莓 " } ], [ { "aoVal": "E", "content": "橘子和香蕉 " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "根据题意分析可知，苹果和草莓都与橘子相邻， 所以可以确定橘子在苹果和草莓的中间， 假设苹果、橘子、草莓这样排列， 则香蕉在草莓的右边，芒果在苹果的左边， 即苹果与芒果一定相邻， 故选：$$\\text{A}$$． " ]

[ "2017年第8届广东广州羊排赛六年级竞赛第4题1分" ]

[ [ { "aoVal": "A", "content": "$$9.237$$ " } ], [ { "aoVal": "B", "content": "$$9.2395$$ " } ], [ { "aoVal": "C", "content": "$$9.2449$$ " } ], [ { "aoVal": "D", "content": "$$9.244$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->小数->小数基础->取近似值" ]

[ "$$9.237\\approx 9.24$$，$$9.2395\\approx 9.240$$，$$9.2449\\approx 9.245$$，$$9.244\\approx 9.24$$． 故选$$\\text{C}$$． " ]

[ "2016年创新杯五年级竞赛训练题（二）第2题" ]

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

[ "拓展思维->能力->运算求解" ]

[ "按照题目的要求将这个数向后写$$10$$位，得到$$1988962486248624\\cdots $$，可以看到从第$$6$$位开始每$$4$$个数字一循环，与$$\\left( 2016-5 \\right)\\div 4=502$$（组）$$\\cdots 3$$（个），所以第$$2016$$位的数字为$$4$$． " ]

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

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

[ "拓展思维->能力->归纳总结->归纳推理" ]

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

[ "2003年第1届创新杯六年级竞赛复赛第9题" ]

[ [ { "aoVal": "A", "content": "$$2274$$ " } ], [ { "aoVal": "B", "content": "$$2003$$ " } ], [ { "aoVal": "C", "content": "$$2300$$ " } ], [ { "aoVal": "D", "content": "$$2320$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "由于$$9+10+11+\\cdots 68=2310$$，由此可知$$2320$$是错误的．由于$$2274$$、$$2003$$、$$2300$$都小于小于$$2310$$，所以减的数较多，由于减一个数，总和里面就要少这个数的$$2$$倍，如减$$2$$，则是$$2310-2\\times 2=2306$$，所以只要是小于$$2310$$．据此分析即 $$9+10+11+\\cdots 68=2310$$，$$2320\\textgreater2310$$，故$$\\text{D}$$错误； $$\\left( 2310-2274 \\right)\\div 2=18$$，$$18\\div 2=9$$，所以在$$9$$前是减号即可，符合题意． $$\\left( 2310-2003 \\right)=307\\textgreater68$$，错误 $$\\left( 2310-2000 \\right)\\div 2=155\\textgreater68$$，错误． 故选$$\\text{A}$$． " ]

[ "2014年迎春杯三年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$150$$ " } ], [ { "aoVal": "B", "content": "$$155$$ " } ], [ { "aoVal": "C", "content": "$$160$$ " } ], [ { "aoVal": "D", "content": "$$165$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->归一归总问题->乘除法应用->乘法应用（顺口溜）" ]

[ "解：由题意，运四次，去四次回三次，吃掉了$$5\\times (4+3)=35$$根，则最多可以运到$$B$$地$$200-35=165$$根。 故选:D。 " ]

[ "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9题5分" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$8$$ " } ], [ { "aoVal": "D", "content": "$$16$$ " } ] ]

[ "拓展思维->能力->逻辑分析->代数逻辑推理" ]

[ "一个数的一半的一半是$$4$$，则这个数的一半为$$4\\times 2=8$$，则这个数为$$8\\times 2=16$$，故选$$\\text{D}$$． " ]

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

[ "拓展思维->思想->整体思想" ]

[ "$$2017$$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$$8$$． " ]

[ "2015年第13届全国创新杯五年级竞赛复赛第1题" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$2014$$ " } ], [ { "aoVal": "C", "content": "$$2015$$ " } ], [ { "aoVal": "D", "content": "$$2016$$ " } ] ]

[ "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" ]

[ "$${{2015}^{2}}-(2015-1)\\times (2015+1)={{2015}^{2}}-{{2015}^{2}}+1=1$$． " ]

[ "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": "$$2$$；$$100$$ " } ], [ { "aoVal": "B", "content": "$$4$$；$$100$$ " } ], [ { "aoVal": "C", "content": "$$8$$；$$100$$ " } ], [ { "aoVal": "D", "content": "$$8$$；$$200$$ " } ] ]

[ "课内体系->知识点->应用题->行程应用题->火车过桥完全过桥", "拓展思维->思想->对应思想" ]

[ "若通过$$860$$米隧道时速度不变则需要$$120\\times 2=240$$（秒），火车过桥速度：$$\\left( 860-320 \\right)\\div \\left( 240-105 \\right)=4$$（米/秒）：火车车身长：$$105\\times 4-320=100$$（米）． " ]

[ "2014年第2届广东广州羊排赛六年级竞赛第4题1分" ]

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

[ "拓展思维->能力->运算求解" ]

[ "共有$$4+6+8+3=21$$（个）球，拿到黄球概率为$$6\\div 21=\\frac{2}{7}$$． " ]

[ "2017年第17届全国中环杯三年级竞赛初赛第5题" ]

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

[ "拓展思维->拓展思维->应用题模块->平均数问题->公式类->加权平均数" ]

[ "$$1\\times 100+102\\div（ 100+1）=2$$． " ]

[ "2017年全国华杯赛小学高年级竞赛初赛模拟" ]

[ [ { "aoVal": "A", "content": "$$96$$ " } ], [ { "aoVal": "B", "content": "$$120$$ " } ], [ { "aoVal": "C", "content": "$$144$$ " } ], [ { "aoVal": "D", "content": "$$240$$ " } ] ]

[ "知识标签->数学思想->对应思想" ]

[ "先捆绑，再插空，$A_{2}^{\\^{2}}\\times C_{3}^{1}\\times A_{4}^{4}=144$种． " ]

[ "2016年第14届全国创新杯五年级竞赛复赛第8题", "2011年全国创新杯五年级竞赛第8题5分" ]

[ [ { "aoVal": "A", "content": "$$350$$ " } ], [ { "aoVal": "B", "content": "$$360$$ " } ], [ { "aoVal": "C", "content": "$$370$$ " } ], [ { "aoVal": "D", "content": "$$380$$ " } ] ]

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

[ "$$4$$小时甲乙共行了$$AB$$，$$3$$小时甲乙共行了$$\\frac{3}{4}AB$$，$$(10+80)\\div \\left( 1-\\frac{3}{4} \\right)=360$$． " ]

[ "2018年第21届世界少年奥林匹克数学竞赛四年级竞赛决赛（中国区）第6题3分" ]

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

[ "拓展思维->能力->公式记忆->符号化数学原理" ]

[ "扑克牌的点数有$$1$ $13$$点，所以要保证$$2$$张牌有相同的点数．最坏的情况就是抽到$$1$ $13$$，这$$13$$张牌，再来一张，就一定和之前的$$13$$张相同，所以是$$13+1=14$$（张）． 故选$$\\text{B}$$． " ]

[ "2019年福建泉州鲤城区泉州师范附属小学三年级竞赛模拟第2题2分", "2019年广西防城港小升初模拟第13题", "2015~2016学年内蒙古乌兰察布丰镇市丰镇实验小学四年级上学期期中第25题5分", "2019年广西防城港小升初模拟第12题" ]

[ [ { "aoVal": "A", "content": "$$111105$$ " } ], [ { "aoVal": "B", "content": "$$1111105$$ " } ], [ { "aoVal": "C", "content": "$$111104$$ " } ], [ { "aoVal": "D", "content": "$$111005$$ " } ] ]

[ "课内体系->能力->运算求解", "拓展思维->拓展思维->计算模块->整数->整数加减->整数加减巧算之凑整法" ]

[ "根据题意，把$$9$$、$$99$$、$$999$$、$$9999$$、$$99999$$看作$$\\left(10-1\\right)$$、$$\\left(100-1\\right)$$、$$\\left(1000-1\\right)$$、$$\\left(10000-1\\right)$$、$$\\left(100000-1\\right)$$，再进行计算． $$9+99+999+9999+99999$$ $$=\\left(10-1\\right)+\\left(100-1\\right)+\\left(1000-1\\right)+\\left(10000-1\\right)+\\left(100000-1\\right)$$ $$=\\left(100000+10000+1000+100+10\\right)-\\left(1+1+1+1+1\\right)$$ $$=111110-5$$ $$=111105$$． " ]

[ "2014年迎春杯三年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$18$$ " } ], [ { "aoVal": "D", "content": "$$27$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->方程基础->等量代换" ]

[ "$$12$$个樱桃的钱可以买$$18$$个苹果，大小是$$1$$个苹果的大小，所以$$1$$个苹果大小的樱桃可以买到$$18$$个苹果，所以$$1$$箱樱桃的钱可以买$$18$$箱苹果。 " ]

[ "2020年长江杯六年级竞赛复赛A卷第4题5分" ]

[ [ { "aoVal": "A", "content": "六折 " } ], [ { "aoVal": "B", "content": "七折 " } ], [ { "aoVal": "C", "content": "七五折 " } ], [ { "aoVal": "D", "content": "八折 " } ] ]

[ "拓展思维->能力->实践应用" ]

[ "假设商品的成本是``$$1$$''，原来希望获得利润$$60 \\%$$，现出售$$70 \\%$$的商品已获得利润$$60 \\%\\times 70 \\%=0.42$$． 剩下的$$30 \\%$$商品将要获得利润：$$0.6\\times 80 \\%-0.42=0.06$$，因此这剩下$$30 \\%$$的商品的售价是：$$1\\times 30 \\%+0.06=0.36$$（元）原来定价是：$$1\\times 30 \\%\\times (1+60 \\%)=0.48$$，$$(0.36\\div 0.48)\\times 100 \\%=75 \\%$$． 故选$$\\text{C}$$． " ]

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

[ "知识标签->课内知识点->数与运算->加法->分数加法->异分母分数加法" ]

[ "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． " ]

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

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

[ "拓展思维->思想->对应思想" ]

[ "开单数次数，改变原来灯的状态，开双数次，不改变原来的状态．总共开的次数为：$$4+3=7$$（次）， 所以改变了原来的状态，原来是开着的， 所以现在不亮． 故选择$$\\text{B}$$． " ]

[ "2012年全国美国数学大联盟杯小学高年级竞赛初赛第30题", "2013年美国数学大联盟杯小学高年级竞赛初赛第30题5分" ]

[ [ { "aoVal": "A", "content": "$$11$$ " } ], [ { "aoVal": "B", "content": "$$22$$ " } ], [ { "aoVal": "C", "content": "$$30$$ " } ], [ { "aoVal": "D", "content": "$$36$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->比例应用题->按比分配" ]

[ "$$66\\div(6+5)=6$$， 花园里有玫瑰$$6\\times6=36$$朵． " ]

[ "2019年第23届广东世界少年奥林匹克数学竞赛五年级竞赛决赛第4题5分" ]

[ [ { "aoVal": "A", "content": "$$9$$ " } ], [ { "aoVal": "B", "content": "$$18$$ " } ], [ { "aoVal": "C", "content": "$$27$$ " } ], [ { "aoVal": "D", "content": "$$36$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "要使这个数能同时被$$9$$，$$25$$，$$8$$整除， 那么这个数一定是$$9$$，$$25$$，$$8$$的公倍数， 首先算出$$9$$，$$25$$，$$8$$的最小公倍数是$$1800$$， 然后可以使用试除法：$$2019999\\div 1800=1122\\cdots \\cdots 399$$， 所以这个七位数为$$2019999-399=2019600$$， 数字和是$$2+0+1+9+6+0+0=18$$． 故选$$\\text{B}$$． " ]

[ "2019年第7届湖北长江杯五年级竞赛复赛A卷第6题3分" ]

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

[ "拓展思维->思想->转化与化归的思想" ]

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

[ "2017年全国美国数学大联盟杯小学高年级六年级竞赛第17题5分" ]

[ [ { "aoVal": "A", "content": "$$66$$ " } ], [ { "aoVal": "B", "content": "$$110$$ " } ], [ { "aoVal": "C", "content": "$$129$$ " } ], [ { "aoVal": "D", "content": "$$176$$ " } ] ]

[ "知识标签->拓展思维->应用题模块->比例应用题->按比分配" ]

[ "停车场只有黑白两种车，白车与黑车的比率是$$3:8$$，黑白车的总数是$$242$$辆．请问黑车比白车多多少辆？ ratio比率； a total of 总数； more．．than 比．．多． 白车数量：$$\\frac{3}{11}\\times 242=66$$辆，黑车数量：$$\\frac{8}{11}\\times 242=176$$辆．黑车比白车多$$176-66=110$$辆． " ]

[ "2019年第7届湖北长江杯五年级竞赛复赛B卷第2题3分" ]

[ [ { "aoVal": "A", "content": "$$48$$ " } ], [ { "aoVal": "B", "content": "$$60$$ " } ], [ { "aoVal": "C", "content": "$$36$$ " } ], [ { "aoVal": "D", "content": "$$64$$ " } ] ]

[ "拓展思维->思想->逆向思想" ]

[ "∵$$12=(2+1)\\times (1+1)\\times (1+1)$$， ∴最小是$$4\\times3\\times5=60$$， $$12$$个因数为：$$1$$、$$2$$、$$3$$、$$4$$、$$5$$、$$6$$、$$10$$、$$15$$、$$12$$、$$20$$、$$30$$、$$60$$． 故选$$\\text{B}$$． " ]

[ "2008年四年级竞赛创新杯" ]

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

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

[ "在数串(列)1，2，3，4，5，6，7，8之间任意添上四个``+''号，可将它分成5段，因而可分成5个整数求和，所以有以下三种情况: ⑴1个四位数和4个一位数的和，此时和的最小值为$$1234+5+6+7+8=1260$$; ⑵1个三位数，1个两位数与3个一位数的和，此时最小值为$$123+45+6+7+8=189$$; ⑶3个两位数与2个一位数的和，此时最小值为$$12+34+56+7+8=117$$. 因此和的最小值是117 " ]

[ "2016年第28届广东广州五羊杯小学高年级竞赛初赛第3题3分" ]

[ [ { "aoVal": "A", "content": "$$11$$点 " } ], [ { "aoVal": "B", "content": "$$11$$点$$30$$分 " } ], [ { "aoVal": "C", "content": "$$11$$点$$36$$分 " } ], [ { "aoVal": "D", "content": "$$11$$点$$45$$分 " } ] ]

[ "拓展思维->拓展思维->计算模块->比和比例->比例->比例尺" ]

[ "$$AB$$距离为$$2\\times 9000000=18000000\\text{cm=180}$$千米， 则需要$$180\\div 50=3.6\\text{h}=3\\text{h}36\\min $$， 则$$8:00+3:36=11:36$$到达． 故选$$\\text{C}$$． " ]

[ "2021年新希望杯五年级竞赛初赛第14题5分" ]

[ [ { "aoVal": "A", "content": "$$40$$ " } ], [ { "aoVal": "B", "content": "$$70$$ " } ], [ { "aoVal": "C", "content": "$$38$$ " } ], [ { "aoVal": "D", "content": "$$56$$ " } ], [ { "aoVal": "E", "content": "$$62$$ " } ] ]

[ "拓展思维->思想->枚举思想" ]

[ "假设阿宝胜，有发以下几种获胜情况（用``$$+$$''表示胜``$$-$$''表示，负） ①连胜四局$$1$$种， $$++++$$， ②四胜一负$$4$$种， $$-++++$$；$$+-+++$$；$$++-++$$；$$+++-+$$ ③四胜两负$$10$$种， $$-\\/-++++$$；$$+-\\/-+++$$；$$++-\\/-++$$；$$+++-\\/-+$$；$$-+-+++$$；$$-++-++$$；$$-+++-+$$；$$+-+-++$$；$$+-++-+$$；$$++-+-+$$． ④四胜三负， （1）甲负三局连负$$4$$种， $$-\\/-\\/-++++$$；$$+-\\/-\\/-+++$$；$$++-\\/-\\/-++$$；$$+++-\\/-\\/-+$$． （2）甲连负两局$$12$$种， $$-\\/-+-+++$$；$$-\\/-++-++$$；$$-\\/-+++-+$$，$$+-\\/-+-++$$；$$+-\\/-++-+$$；$$++-\\/-+-+$$；$$-+-\\/-+++$$；$$-++-\\/-++$$；$$-+++-\\/-+$$；$$+-+-\\/-++$$；$$+-++-\\/-+$$；$$++-+-\\/-+$$． （3）甲负三局全都分开$$4$$种， $$-+-+-++$$，$$-+-++-+$$；$$-++-+-+$$；$$+-+-+-+$$． $$1+4+10+4+12+4=35$$（种）$$35+35=70$$种， 当考虑金猴的情况时与上述一致也为$$35$$种，因此一共$$35+35=70$$种可能． " ]

[ "2017年全国小升初八中入学备考课程", "2008年全国华杯赛竞赛复赛第6题" ]

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

[ "拓展思维->拓展思维->计算模块->分数->分数运算->分数加减" ]

[ "对于这种类型的题目，我们可以采取``反驳''的方法来做，找出每个不成立的案例来，若找不到则正确．①反例：$$\\frac{1}{2}+\\frac{1}{2}=1$$，$$\\frac{4}{5}+\\frac{3}{5}=\\frac{7}{5}$$；④反例：$$\\frac{1}{2}\\times \\frac{3}{2}=\\frac{3}{4}$$，$$\\frac{1}{5}\\times \\frac{8}{5}=\\frac{8}{25}$$． " ]

[ "2018年第8届北京学而思综合能力诊断二年级竞赛年度教学质量第10题" ]

[ [ { "aoVal": "A", "content": "$$20$$ " } ], [ { "aoVal": "B", "content": "$$460$$ " } ], [ { "aoVal": "C", "content": "$$480$$ " } ], [ { "aoVal": "D", "content": "$$720$$ " } ] ]

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

[ "$$2\\times 2\\times 3=12$$（种）． " ]

[ "2017年全国小升初八中入学备考课程", "2014年全国华杯赛小学高年级竞赛初赛B卷第5题" ]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$8$$ " } ], [ { "aoVal": "D", "content": "$$10$$ " } ] ]

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

[ "甲乙的年龄差是$$22-16=6$$岁；当甲$$19$$岁时， $$13$$岁；至少一年前甲$$22$$岁，所以当甲$$19$$岁的时候，此时至少是$$4$$年前的年龄，那么甲今年至少是$$23$$岁；甲$$19$$岁时，丙的年龄是丁的$$3$$倍，假设丁为$$1$$岁，丙为$$3$$岁，此时四人的年龄和至少是$$19+13+1+3=36$$岁；且甲今年的年龄至多为$$19+\\left(72-36 \\right)\\div 4=28$$；所以甲今年的年龄可能是$$23$$，$$24$$，$$25$$，$$26$$，$$27$$，$$28$$；共$$6$$种，所以选$$\\text{B}$$． " ]

[ "2014年IMAS小学中年级竞赛第二轮检测试题第3题4分" ]

[ [ { "aoVal": "A", "content": "$$3219290$$ " } ], [ { "aoVal": "B", "content": "$$321929$$ " } ], [ { "aoVal": "C", "content": "$$342190$$ " } ], [ { "aoVal": "D", "content": "$$34190$$ " } ], [ { "aoVal": "E", "content": "$$32129$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->整数->四则混合运算" ]

[ "$$32\\times 1000+19\\times 100+29\\times 10=34190$$，答案选$$\\text{D}$$． " ]

[ "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（四）" ]

[ [ { "aoVal": "A", "content": "甲获利润较多 " } ], [ { "aoVal": "B", "content": "乙获利润较多 " } ], [ { "aoVal": "C", "content": "两人利润相同 " } ], [ { "aoVal": "D", "content": "无法比较谁多 " } ] ]

[ "拓展思维->拓展思维->应用题模块->经济问题->基本经济概念->已知成本售价求利润" ]

[ "设原成本为$$1$$元/个，甲获利$$15\\times 20 \\% =3$$（元），乙获利$$0.15\\times 24=3.6$$（元）．选$$\\text{B}$$． " ]

[ "1989年华杯赛六年级竞赛初赛" ]

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

[ "拓展思维->拓展思维->组合模块->抽屉原理->最不利原则" ]

[ "如果在最不利的情况下能完成目标，则保证在任何情况下都能完成。最不利的情况是：抽的前$$12$$张是$$4$$种花色各$$3$$张，再抽两张王牌，这时抽第$$15$$张，无论是哪种花色，都能保证凑成$$4$$张牌同一花色。所以至少要抽$$15$$张牌，才能保证有四张牌是同一花色的。 " ]

[ "2011年世界少年奥林匹克数学竞赛六年级竞赛初赛第9题5分" ]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$37$$ " } ], [ { "aoVal": "C", "content": "$$7$$ " } ] ]

[ "知识标签->数学思想->对应思想" ]

[ "本题是鸡兔同笼问题的变式应用． 解决此类问题一般都用假设法，做对一题得$$8$$分，相当于``兔''，做错一题倒扣$$5$$分，相当于``鸡''，通过适当的替换进行解答． 假设$$10$$道题全部做对，则得分为$$10\\times 8=80$$（分），而答错一题比答对一题少得$$8+5=13$$（分），从题目中可以求出小宇一共少得$$80-41=39$$（分），由此可以求出他做错了$$39\\div 13=3$$（道）． $$(10\\times 8-41)\\div (48+5)$$ $$=39\\div 13$$ $$=3$$（道）， $$10-3=7$$（道）． 故答案为：$$7$$． " ]

[ "2009年五年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "$$667$$ " } ], [ { "aoVal": "B", "content": "$$668$$ " } ], [ { "aoVal": "C", "content": "$$669$$ " } ], [ { "aoVal": "D", "content": "$$700$$ " } ] ]

[ "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" ]

[ "项数$$=\\left( 2009-5 \\right)\\div 3+1=669$$ " ]

[ "2018年第12届北京学而思综合能力诊断小学高年级五年级竞赛第6题" ]

[ [ { "aoVal": "A", "content": "$$16$$ " } ], [ { "aoVal": "B", "content": "$$18$$ " } ], [ { "aoVal": "C", "content": "$$20$$ " } ], [ { "aoVal": "D", "content": "$$22$$ " } ], [ { "aoVal": "E", "content": "$$24$$ " } ] ]

[ "拓展思维->能力->构造模型->模型思想", "Overseas Competition->知识点->应用题模块->分百应用题->认识单位1" ]

[ "孙悟空打妖精的工作效率是$$\\frac{1}{30}$$； 猪八戒与孙悟空合作打妖精的工作效率是$$\\frac{1}{24}$$， 则有猪八戒打妖精的工作效率是$$\\frac{1}{24}-\\frac{1}{30}=\\frac{1}{120}$$； 沙和尚、猪八戒与孙悟空合作打妖精的工作效率是$$\\frac{1}{20}$$， 所以沙和尚和孙悟空合作打妖精的工作效率为$$\\frac{1}{20}-\\frac{1}{120}=\\frac{1}{24}$$， 需$$1\\div \\frac{1}{24}=24$$分钟． " ]

[ "2017年河南郑州联合杯竞赛第7题4分" ]

[ [ { "aoVal": "A", "content": "$$206$$ " } ], [ { "aoVal": "B", "content": "$$203$$ " } ], [ { "aoVal": "C", "content": "$$153$$ " } ], [ { "aoVal": "D", "content": "$$211$$ " } ] ]

[ "知识标签->拓展思维->应用题模块->页码问题->数码综合问题" ]

[ "页码问题；$$1\\sim 99$$中数字``$$1$$''出现了$$20$$次，$$100\\sim 199$$中数字``$$1$$''出现了$$120$$个，剩下的$$200\\sim 500$$有$$3\\times 20=60$$（个），$$501\\sim 513$$中数字``$$1$$''出现了$$6$$次，所以一共有$$20+120+60+6=206$$（次）． " ]

[ "2012年美国数学大联盟杯六年级竞赛初赛第32题5分(每题5分)" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$12$$ " } ], [ { "aoVal": "C", "content": "$$14$$ " } ], [ { "aoVal": "D", "content": "$$24$$ " } ] ]

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

[ "根据题意分析可知，假设现在是$$x$$岁，那么两年前的年龄为$$(x-2)$$岁，两年后的年龄为$$(x+2)$$岁，根据等量关系列方程如下： $$(x-2)+(x+2)=24$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde x-2+x+2=24$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde x+x=24+2-2$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde2x=24$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde x=12$$． 即现在$$12$$岁， 故选$$\\text{B}$$． " ]

[ "2016年IMAS小学高年级竞赛第一轮检测试题第2题3分" ]

[ [ { "aoVal": "A", "content": "$$1.2\\times 3.4=12\\times 3.4$$ " } ], [ { "aoVal": "B", "content": "$$0.98\\times 0.99\\textgreater0.99$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{1}{2}-\\frac{1}{3}\\textless{}\\frac{1}{3}-\\frac{1}{4}$$ " } ], [ { "aoVal": "D", "content": "$$10.4\\times 0.1\\textless{}1.04$$ " } ], [ { "aoVal": "E", "content": "$$1.1\\times 1.1\\textgreater1.1$$ " } ] ]

[ "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" ]

[ "$$\\text{A}$$：$$1.2\\times 3.4\\textless{}12\\times 3.4$$ $$\\text{B}$$：$$0.98\\times 0.99\\textless{}0.99$$ $$\\text{C}$$：$$\\frac{1}{2}-\\frac{1}{3}=\\frac{1}{6}$$，$$\\frac{1}{3}-\\frac{1}{4}=\\frac{1}{12}$$，$$\\frac{1}{2}-\\frac{1}{3}\\textgreater\\frac{1}{3}-\\frac{1}{4}$$ $$\\text{D}$$：$$10.4\\times 0.1=1.04$$ " ]

[ "2013年北京迎春杯六年级竞赛", "2013年全国迎春杯六年级竞赛初赛第9题" ]

[ [ { "aoVal": "A", "content": "$$100$$ " } ], [ { "aoVal": "B", "content": "$$110$$ " } ], [ { "aoVal": "C", "content": "$$120$$ " } ], [ { "aoVal": "D", "content": "$$130$$ " } ], [ { "aoVal": "E", "content": "$$140$$ " } ] ]

[ "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", "Overseas Competition->知识点->行程模块" ]

[ "方法一：由第一个条件知甲乙速度比为$$（\\frac { S }{ 2 } +12）：（\\frac { S }{ 2 } -12）$$；再由第二条件知甲乙速度比为$$\\frac { S }{ 2 } ：（\\frac { S }{ 2 } -20）$$，两者相等得$$S=120$$千米． 方法二：设全长为$$2S$$千米，则第一次相遇时，甲走了$$S+12$$千米，乙走了$$S-12$$千米；第二次相遇点走到各自目的地时，甲走了$$S$$千米，乙走了$$S-20$$千米．由于两人速度不变，故以速度比作等量关系列方程可得：$$\\frac{S+12}{S-12}=\\frac{S}{S-20}$$，解得$$S=60$$，故全长为$$2S=120$$千米． " ]

[ "2014年走美杯六年级竞赛" ]

[ [ { "aoVal": "A", "content": "$$10000001111$$ " } ], [ { "aoVal": "B", "content": "$$10000001110$$ " } ], [ { "aoVal": "C", "content": "$$10000000111$$ " } ], [ { "aoVal": "D", "content": "$$100000011001$$ " } ] ]

[ "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->进制间的互化" ]

[ "用除$$2$$倒取余数法，得到$$10000001111$$。 " ]

[ "2018年四川成都小升初公立名校初一开学考（一）第5题5分", "2008年全国华杯赛竞赛初赛第6题", "2008年第13届全国迎春杯竞赛初赛第6题10分" ]

[ [ { "aoVal": "A", "content": "$$a\\textgreater b\\textgreater c$$ " } ], [ { "aoVal": "B", "content": "$$a\\textgreater c\\textgreater b$$ " } ], [ { "aoVal": "C", "content": "$$a\\textless{}c\\textless{}b$$ " } ], [ { "aoVal": "D", "content": "$$a\\textless{}b\\textless{}c$$ " } ] ]

[ "知识标签->学习能力->七大能力->运算求解" ]

[ "由糖水原理可知$$a\\textless{}b\\textless{}c$$． " ]

[ "2017年河南郑州豫才杯六年级竞赛第7题3分" ]

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

[ "拓展思维->拓展思维->应用题模块->经济问题->利息问题" ]

[ "第一年付：$$30000$$元，第二年付：$$5000+90000\\times 0.4 \\% =5360$$（元），第三年付：$$5000+85000\\times 0.4 \\% =5340$$（元），第四年付：$$5000+80000\\times 0.4 \\% =5320$$（元），$$\\cdot \\cdot \\cdot $$，以此类推，第十年付：$$5200$$元． " ]

[ "2014年第12届全国小机灵杯小学高年级五年级竞赛决赛第14题" ]

[ [ { "aoVal": "A", "content": "请在答题卡上作答 " } ] ]

[ "课内体系->思想->逐步调整思想", "拓展思维->七大能力->实践应用" ]

[ "如果橘子的个数扩大到原来的$$3$$倍，那么余下的橘子数量将与余下苹果的相等； 即$$235\\times 3=705$$与$$270$$分别除以大班小朋友的人数，余数相同； 大班小朋友的人数是$$705-270=435$$的因数； 如果橘子的个数扩大到原来的$$2$$倍，那么余下的橘子数量将与余下的梨的相等； 即$$235\\times 2=470$$与$$180$$分别除以大班小朋友的人数，余数相同； 大班小朋友的人数是$$470-180=290$$的因数； 则大班小朋友的人数是$$435$$和$$290$$的公因数； $$(435,290)=5\\times 29=145$$； 大班小朋友的人数是$$145$$的因数：$$1$$、$$5$$、$$29$$、$$145$$ 当大班有$$1$$名小朋友时，$$270\\div 1=270$$，$$180\\div 1=180$$，$$235\\div 1=235$$； 余下$$0$$个苹果、$$0$$个梨、$$0$$个橘子，不符题意； 当大班有$$5$$名小朋友时，$$270\\div 5=54$$，$$180\\div 5=36$$，$$235\\div 5=47$$； 余下$$0$$个苹果、$$0$$个梨、$$0$$个橘子，不符题意； 当大班有$$29$$名小朋友时，$$270\\div 29=9\\cdots \\cdots 9$$，$$180\\div 29=6\\cdots \\cdots 6$$，$$235\\div 29=8\\cdots \\cdots 3$$； 余下$$9$$个苹果、$$6$$个梨、$$3$$个橘子，$$9:6:3=3:2:1$$，符合题意； 当大班有$$145$$名小朋友时，$$270\\div 145=1\\cdots \\cdots 125$$，$$180\\div 145=1\\cdots \\cdots 35$$，$$235\\div 145=1\\cdots \\cdots 90$$； 余下$$125$$个苹果、$$35$$个梨、$$90$$个橘子，$$125:35:90\\ne 3:2:1$$，不符题意； 经检验，大班有$$29$$名小朋友． " ]

[ "2017年北京学而思杯四年级竞赛年度教学质量测评第19题3分" ]

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

[ "拓展思维->思想->逐步调整思想" ]

[ "略 " ]

[ "2014年全国华杯赛小学高年级竞赛复赛C卷第12题" ]

[ [ { "aoVal": "A", "content": "$$400$$、$$208$$、$$160$$ " } ], [ { "aoVal": "B", "content": "$$2848$$、$$400$$、$$160$$ " } ], [ { "aoVal": "C", "content": "$$2848$$、$$400$$、$$208$$、$$160$$ " } ], [ { "aoVal": "D", "content": "$$2848$$、$$400$$、$$208$$、$$190$$ " } ] ]

[ "知识标签->学习能力->七大能力->逻辑分析" ]

[ "设这两个完全平方数分别为$${{a}^{2}}$$和$${{b}^{2}}$$（$$a$$、$$b$$均为整数），则$${{a}^{2}}-{{b}^{2}}=144-39=105$$． 由平方差公式得：$$\\left( a+b \\right)\\left( a-b \\right)=105$$． 考虑到$$105=1\\times 105=3\\times 35=5\\times 21=7\\times 15$$，因此有$$4$$种可能： $$\\begin{cases}a+b=105 a-b=1 \\end{cases}$$，$$\\begin{cases}a+b=35 a-b=3 \\end{cases}$$，$$\\begin{cases}a+b=21 a-b=5 \\end{cases}$$，$$\\begin{cases}a+b=15 a-b=7 \\end{cases}$$ 解得：$$\\begin{cases}a=53 b=52 \\end{cases}$$，$$\\begin{cases}a=19 b=16 \\end{cases}$$，$$\\begin{cases}a=13 b=8 \\end{cases}$$，$$\\begin{cases}a=11 b=4 \\end{cases}$$． 原数为$${{a}^{2}}+39$$，对应$$4$$种可能：$$2848$$、$$400$$、$$208$$、$$160$$． " ]

[ "2019年第23届广东世界少年奥林匹克数学竞赛三年级竞赛初赛第4题5分" ]

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

[ "拓展思维->思想->整体思想" ]

[ "由题意知，用围棋子摆成一个三层空心方阵，最外一层每边有围棋子$$12$$个，由于相邻两层每边相差$$2$$个，则由外向里的两层每边分别是$$(12-2)$$个、$$(12-2\\times 2)$$个，根据``四周的个数$$=$$（每边的个数$$-1$$）$$\\times 4$$''可分别求得这三层棋子的个数，再相加就是所用的总个数，据此解答． 本题关键是求出每层的个数；方阵问题相关的知识点是：四周的个数$$=$$（每边的个数$$-1$$）$$\\times 4$$，每边的个数$$=$$四周的个数$$\\div 4+1$$，中实方阵的总个数$$=$$每边的个数$$\\times $$每边的个数，空心方阵的总个数$$=$$（最外层每边的个数$$-$$空心方阵的层数）$$\\times $$空心方阵的层数$$\\times 4$$，外层边长数$$^{2}-$$中空边长数$$^{2}=$$实面积数． 最外边一层棋子个数：$$(12-1)\\times 4=44$$（个）， 第二层棋子个数：$$(12-2-1)\\times 4=36$$（个）， 第三层棋子个数：$$(12-2\\times 2-1)\\times 4=28$$（个）， 摆这个方阵共用棋子：$$44+36+28=108$$（个）． 故选$$\\text{B}$$． " ]

[ "2013年第25届广东广州五羊杯六年级竞赛第4题5分" ]

[ [ { "aoVal": "A", "content": "星期一 " } ], [ { "aoVal": "B", "content": "星期二 " } ], [ { "aoVal": "C", "content": "星期三 " } ], [ { "aoVal": "D", "content": "星期四 " } ] ]

[ "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->求再过几日是周几问题" ]

[ "$$2013\\div 7=287$$（组）$$\\cdots \\cdots 4$$（个），星期四后的第四天是星期一． " ]

[ "2018年华杯赛六年级竞赛模拟卷" ]

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

[ "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" ]

[ "可以设乙队工作了$$x$$天，列方程得：$$\\frac{30-1-x}{20}+\\frac{x}{35}=1$$，解得：$$x=21$$。 " ]

[ "2016年全国华杯赛小学中年级竞赛初赛第6题" ]

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

[ "拓展思维->能力->运算求解" ]

[ "某个数除$$2016$$余$$7$$，于是这个数整除$$2016-7=2009$$，$$2009={{7}^{2}}\\times {{41}^{1}}$$，所以$$2009$$共有$$6$$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": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$15$$ " } ], [ { "aoVal": "C", "content": "$$20$$ " } ], [ { "aoVal": "D", "content": "$$25$$ " } ], [ { "aoVal": "E", "content": "$$30$$ " } ] ]

[ "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->设数法", "Overseas Competition->知识点->应用题模块" ]

[ "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年河南郑州联合杯竞赛决赛第6题2分", "2021年河南郑州二七区京广实验学校小升初第6题2分" ]

[ [ { "aoVal": "A", "content": "付出$$4.5$$元，找回$$0.1$$元 " } ], [ { "aoVal": "B", "content": "付出$$4.7$$元，找回$$0.3$$元 " } ], [ { "aoVal": "C", "content": "付出$$5.4$$元，找回$$1.0$$元 " } ], [ { "aoVal": "D", "content": "付出$$10$$元，找回$$5.6$$元 " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "B选项不合理． " ]

[ "2019年江苏南京栖霞区南外仙林分校小学部六年级下学期单元测试《正比例和反比例》第29题2分", "2015年全国迎春杯小学高年级竞赛复赛第9题" ]

[ [ { "aoVal": "A", "content": "$$999$$ " } ], [ { "aoVal": "B", "content": "$$1099$$ " } ], [ { "aoVal": "C", "content": "$$2099$$ " } ], [ { "aoVal": "D", "content": "$$3099$$ " } ] ]

[ "课内体系->七大能力->逻辑分析", "拓展思维->拓展思维->数论模块->奇数与偶数" ]

[ "两位数没有符合要求的数，$$99$$、$$100$$亦不符合，故知至少为三位数．两个相邻数数字和都是偶数，说明必有进位，且三位数必然只进$$1$$次位（数字和加$$1$$再减$$9$$），即这两个数是$$\\overline{ab9}$$和$$\\overline{a\\left( b+1 \\right)0}$$，必有$$a+b=9$$和$$a=b+1$$，故这两个数为$$549$$和$$550$$．$$549+550=1099$$． " ]

[ "2006年第4届创新杯四年级竞赛初赛A卷第10题" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$9$$ " } ] ]

[ "拓展思维->能力->数据处理" ]

[ "$$1\\times 1=1$$， $$2\\times 2=4$$， $$3\\times 3=9$$， $$4\\times 4=16$$， $$5\\times 5=25$$， $$6\\times 6=36$$， $$7\\times 7=49$$， $$8\\times 8=64$$， $$9\\times 9=81$$， $$10\\times 10=100$$， 我们可以把每$$10$$个数当作一个整体，它们的和的个位数是$$5$$， 从$$1$$到$$2006$$可以看作$$200$$个上面的整体和剩下$$6$$个数的乘积， 所以$$200$$个整体的和的个位数字应该是$$5$$乘$$200$$的个位，即为$$0$$， 显然剩下的$$6$$个数的个位是$$1$$， 所以最终的个位就应该为$$1$$． 故选$$\\text{A}$$． " ]

[ "2016年新希望杯六年级竞赛训练题（四）第3题" ]

[ [ { "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" ]

[ "甲、乙两队第一阶段完成了$$\\left( \\frac{1}{16}+\\frac{1}{24} \\right)\\times 4=\\frac{5}{12}$$，乙队第二阶段需要$$\\left( 1-\\frac{5}{12} \\right)\\div \\frac{1}{24}=14$$天． " ]

[ "2008年五年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "$$560$$ " } ], [ { "aoVal": "B", "content": "$$605$$ " } ], [ { "aoVal": "C", "content": "$$625$$ " } ], [ { "aoVal": "D", "content": "$$725$$ " } ] ]

[ "拓展思维->拓展思维->计数模块->枚举法综合->字典排序法" ]

[ "两个数中小的数为$$1$$，有$$\\left( 1\\text{，}50 \\right)$$，$$1$$种取法； 同理，小的数为$$2$$，有$$\\left( 2\\text{，}50 \\right)$$，$$\\left( 2\\text{，}49 \\right)$$，$$2$$种取法； 小的数为$$3$$，有$$\\left( 3\\text{，}50 \\right)$$，$$\\left( 3\\text{，}49 \\right)$$，$$\\left( 3\\text{，}48 \\right)$$，$$3$$种取法； $$\\cdots\\cdots$$ 小的数为$$25$$，有$$\\left( 25\\text{，}50 \\right)$$，$$\\left( 26\\text{，}50 \\right)$$，$$\\cdots$$，$$\\left( 25\\text{，}26 \\right)$$，$$25$$种取法； 小的数为$$26$$，有$$\\left( 26\\text{，}50 \\right)$$，$$\\left( 26\\text{，}49 \\right)$$，$$\\cdots$$，$$\\left( 26\\text{，}27 \\right)$$，$$24$$种取法； $$\\cdots\\cdots$$ 小的数为$$49$$，有$$\\left( 49\\text{，}50 \\right)$$，$$1$$种取法。 所以共有$$1+2+3+\\cdots +25+24+\\cdots +3+2+1$$ $$=2\\times \\left( 1+2+3+\\cdots +25 \\right)-25=2\\times \\frac{25\\times 26}{2}-25=25\\times 26-25$$ $$=25\\times 25=625$$（种）不同的取法。 " ]

[ "2015年湖北武汉世奥赛五年级竞赛模拟训练题（三）第5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" ]

[ "将$$15$$拆成四个不同的加数，按最小的加数分类数：$$15=1+2+3+9=1+2+4+8=1+2+5+7=1+3+4+7=1+3+5+6=2+3+4+6$$，共$$6$$种不同的分类． " ]

[ "2014年全国迎春杯四年级竞赛初赛第1题" ]

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

[ "拓展思维->能力->运算求解" ]

[ "解：$$\\rm A$$、$$3\\times 3\\div 3+3$$ $$=3+3$$ $$=6$$ $$\\rm B$$、$$3\\div 3+3\\times 3$$ $$=1+9$$ $$=10$$ $$\\rm C$$、$$3\\times 3-3+3$$ $$=9-3+3$$ $$=9$$ $$\\rm D$$、$$3\\div 3+3\\div 3$$ $$=1+1$$ $$=2$$ 故选：$$\\rm C$$． " ]

[ "2014年全国学而思杯一年级竞赛第10题" ]

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

[ "知识标签->拓展思维->组合模块->逻辑推理->比较型逻辑推理" ]

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

[ "竞赛" ]

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

[ "拓展思维->拓展思维->数论模块->余数问题->余数的性质->余数找规律" ]

[ "$$8$$的若干次方除以$$9$$的余数是以$$8$$、$$1$$为周期的数，$$25$$为奇数，对应余数为$$8$$。 " ]

[ "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．所以选$$\\text{C}$$． " ]

[ "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": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$19$$ " } ], [ { "aoVal": "C", "content": "$$20$$ " } ], [ { "aoVal": "D", "content": "$$40$$ " } ] ]

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

[ "由于第一个机器人回答对，所以说明除了第一个外的机器人，开始时$$19$$个回答对，$$20$$个回答错，第一个机器人根据判定需要改答案，其他机器根据第一个机器人改了答案判断，应该都回答错． " ]

[ "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A卷第9题" ]

[ [ { "aoVal": "A", "content": "40 " } ], [ { "aoVal": "B", "content": "45 " } ], [ { "aoVal": "C", "content": "48 " } ], [ { "aoVal": "D", "content": "50 " } ] ]

[ "拓展思维->拓展思维->应用题模块->页码问题" ]

[ "设撕掉这张的两个页码的和为x，原书有n页，则所有页码和为$$\\left( 1+2+3+\\cdots +n \\right)-x=\\frac{n\\left( n+1 \\right)}{2}-x=1000$$，只有当$$n=45$$时，其所有页码和为$$45\\times 46\\div 2=1035$$，恰好大于1000且最接近1000，故撕掉这张纸的两个页码的和为$$1035-1000=35$$，撕掉的这张的两个页码为17，18.所以这本书原有45页，故选B. " ]

[ "2018年第17届春蕾杯一年级竞赛初赛第14题6分" ]

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

[ "拓展思维->能力->推理推导->言语逻辑推理" ]

[ "推理题目 " ]

[ "2021年鹏程杯六年级竞赛初赛第12题" ]

[ [ { "aoVal": "A", "content": "$$20$$ " } ], [ { "aoVal": "B", "content": "$$18$$ " } ], [ { "aoVal": "C", "content": "$$15$$ " } ], [ { "aoVal": "D", "content": "$$13$$ " } ], [ { "aoVal": "E", "content": "以上都不对 " } ] ]

[ "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" ]

[ "不难列出算式：$$\\left[ 1-\\frac{1}{1+25 \\%} \\right]\\times100 \\%=20 \\%$$， 故选$$\\text{A}$$． " ]

[ "2017年第15届全国希望杯六年级竞赛" ]

[ [ { "aoVal": "A", "content": "存在 " } ], [ { "aoVal": "B", "content": "不存在 " } ], [ { "aoVal": "C", "content": "不能确定 " } ] ]

[ "拓展思维->思想->枚举思想" ]

[ "$$8\\div \\left( 7+1 \\right)=1$$，$$15\\div \\left( 7+8 \\right)=1$$，$$15\\div \\left( \\frac{5}{4}+3.75 \\right)=3$$，$$8\\div \\left( 1.\\dot{2}+0.\\dot{7} \\right)=4$$，$$7\\div \\left( \\frac{2}{5}+1 \\right)=5$$等． " ]

[ "2014年全国创新杯小学高年级五年级竞赛第4题5分", "2016年创新杯小学高年级五年级竞赛训练题（四）第5题" ]

[ [ { "aoVal": "A", "content": "$$26$$ " } ], [ { "aoVal": "B", "content": "$$27$$ " } ], [ { "aoVal": "C", "content": "$$28$$ " } ], [ { "aoVal": "D", "content": "$$29$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "取$$1$$，$$4$$，$$7$$，$$\\cdot \\cdot \\cdot $$，$$79$$，最多有$$(79-1)\\div 3+1=27$$个． " ]

[ "2008年四年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "1957 " } ], [ { "aoVal": "B", "content": "1959 " } ], [ { "aoVal": "C", "content": "2008 " } ], [ { "aoVal": "D", "content": "2009 " } ] ]

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

[ "所得的101个新数之和为: $$2008-1+2-3+4-\\cdots -99+100-101=2008+\\underbrace{1+1+1+\\cdots +1}_{50个1}-101=1957$$ " ]

[ "2014年全国创新杯小学高年级五年级竞赛第5题5分" ]

[ [ { "aoVal": "A", "content": "甲地到乙地的路程为$$8$$千米 " } ], [ { "aoVal": "B", "content": "甲乙两地路程为$$7$$千米 " } ], [ { "aoVal": "C", "content": "甲乙两地路程大于$$7$$千米但不超过$$8$$千米 " } ], [ { "aoVal": "D", "content": "甲乙两地路程最多为$$8$$千米 " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "$$19-7=12$$，$$12\\div 2.4=5$$，故甲、乙两地路程大于$$7$$千米但不超过$$8$$千米． " ]

[ "2018年第21届世界少年奥林匹克数学竞赛四年级竞赛决赛（中国区）第10题3分" ]

[ [ { "aoVal": "A", "content": "$$40$$ " } ], [ { "aoVal": "B", "content": "$$56$$ " } ], [ { "aoVal": "C", "content": "$$64$$ " } ], [ { "aoVal": "D", "content": "$$80$$ " } ] ]

[ "拓展思维->能力->构造模型->模型思想" ]

[ "根据题意分析可知，因为小框装的是中筐的一半且大框装的是小筐的$$5$$倍可知， 大筐装的是中筐的$$5\\div 2=2.5$$倍， 又因为中框比大框少装$$24$$千克， 用差倍问题求出小筐的能装：$$24\\div \\left( 5-2 \\right)=8$$（千克）， 中筐装：$$8\\times 2=16$$（千克）， 大筐装：$$16\\times 2=32$$（千克）， 所以三个筐一共能装：$$8+16+32=56$$（千克）． 故选$$\\text{B}$$． " ]

[ "2018年IMAS小学中年级竞赛（第一轮）第10题3分" ]

[ [ { "aoVal": "A", "content": "$$144$$ " } ], [ { "aoVal": "B", "content": "$$233$$ " } ], [ { "aoVal": "C", "content": "$$234$$ " } ], [ { "aoVal": "D", "content": "$$235$$ " } ], [ { "aoVal": "E", "content": "$$377$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->数列与数表->数列规律->斐波那契数列（兔子数列）" ]

[ "逐月推算，我们可以得到下面的一列数： $$1$$、$$1$$、$$2$$、$$3$$、$$5$$、$$8$$、$$13$$、$$21$$、$$34$$、$$55$$、$$89$$、$$144$$、$$233$$、$$\\cdots $$ 其中第$$13$$个数为$$233$$，所以经过$$12$$个月后共有$$233$$对兔子． 故选$$\\text{B}$$． " ]

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

[ "每个数之间差了一个6 " ]

[ "2010年全国华杯赛竞赛初赛第3题" ]

[ [ { "aoVal": "A", "content": "$$112$$ " } ], [ { "aoVal": "B", "content": "$$168$$ " } ], [ { "aoVal": "C", "content": "$$224$$ " } ], [ { "aoVal": "D", "content": "$$336$$ " } ] ]

[ "课内体系->七大能力->实践应用", "拓展思维->知识点->应用题模块->比例应用题->按比分配" ]

[ "本题原意为两人捞第一个水池内的金鱼，亮亮与红红捞到得金鱼数之比为$$3:4$$，共捞了$$7$$份；这样，第二个水池内涝完后，亮亮和红红所捞到的金鱼数目比是$$5:3$$，共捞了$$8$$份；由于两个水池内的鱼的量是相等的，则找$$\\left[ 7,8 \\right]=56$$． 两个水池内的总份数，均统一为$$56$$份，则在捞第一个水池时，亮亮和红红所捞到的金鱼数目之比为：$$3:4=24:32$$； 捞第二个水池时，亮亮和红红所捞到的金鱼数目之比为：$$5:3=35:21$$． 亮亮第一次捞了$$24$$份，第二次捞了$$35$$份，差了$$11$$份，为$$33$$条，则$$1$$份为$$3$$条． 所以原来每隔水池内的金鱼为：$$3\\times56=168$$． " ]

[ "2008年六年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "24 " } ], [ { "aoVal": "B", "content": "26 " } ], [ { "aoVal": "C", "content": "28 " } ], [ { "aoVal": "D", "content": "30 " } ] ]

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

[ "不妨设$$A\\geqslant B\\geqslant C\\geqslant D$$，则 $$B\\text{+}C\\text{+}D=23\\times 3=69$$ $$A\\text{+}C\\text{+}D=26\\times 3=78$$ $$A\\text{+}B\\text{+}D=30\\times 3=90$$ $$A\\text{+}B\\text{+}C=33\\times 3=99$$ 四个等式相加得$$3\\times \\left( A\\text{+}B\\text{+}C\\text{+}D \\right)=336$$，从而$$A\\text{+}B\\text{+}C\\text{+}D\\text{=}112$$，所以$$A$$、$$B$$、$$C$$、$$D$$的平均数为$$112\\div 4\\text{=}28$$ " ]

[ "2009年第9届全国新希望杯三年级竞赛初赛" ]

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

[ "拓展思维->能力->运算求解" ]

[ "将``$$1$$红$$2$$蓝$$4$$黄''看作一组，共有$$7$$颗，$$32 \\div 7 = 4 \\cdots ~\\cdots 4$$，说明这一组共出现了$$4$$次，余数是$$4$$说明还依次出现了$$1$$颗红珠、$$2$$颗蓝珠、$$1$$颗黄珠，所以第$$32$$颗彩珠是黄色． " ]

[ "2020年希望杯一年级竞赛模拟第2题" ]

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

[ "拓展思维->思想->对应思想" ]

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

[ "2018年四川成都锦江区四川师范大学附属第一实验中学小升初模拟8第3题3分", "2018年全国小学生数学学习能力测评六年级竞赛初赛第8题3分", "2018年湖南长沙小升初数学入学试卷第4题4分" ]

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

[ "拓展思维->思想->对应思想" ]

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

[ "2018年第8届北京学而思综合能力诊断二年级竞赛年度教学质量第1题" ]

[ [ { "aoVal": "A", "content": "$$85$$ " } ], [ { "aoVal": "B", "content": "$$55$$ " } ], [ { "aoVal": "C", "content": "$$53$$ " } ], [ { "aoVal": "D", "content": "$$62$$~ " } ] ]

[ "拓展思维->拓展思维->计算模块->整数->四则混合运算" ]

[ "先算乘除，再算加减，$$8+9\\times 5=8+45=53$$． " ]

[ "2018年第6届湖北长江杯六年级竞赛初赛A卷第1题3分" ]

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

[ "拓展思维->拓展思维->计算模块->分数->分数基础->分数的约分" ]

[ "$${{15}^{2}}=225$$，$${{16}^{2}}=256$$，$${{17}^{2}}=289$$，$${{18}^{2}}=324$$， $$\\text{A}$$. $$\\frac{1}{225}\\times 2=\\frac{2}{225}$$，$$\\frac{1}{256}\\times 2=\\frac{2}{256}=\\frac{1}{128}$$， $$\\frac{1}{289}\\times 2=\\frac{2}{289}$$，$$\\frac{1}{324}\\times 2=\\frac{2}{324}=\\frac{1}{162}$$； $$\\text{B}$$. $$\\frac{1}{225}\\times 3=\\frac{3}{225}=\\frac{1}{75}$$，$$\\frac{1}{256}\\times 2=\\frac{3}{256}$$， $$\\frac{1}{289}\\times 3=\\frac{3}{289}$$，$$\\frac{1}{324}\\times 3=\\frac{3}{324}=\\frac{1}{108}$$； $$\\text{C}$$. $$\\frac{1}{225}\\times 4=\\frac{4}{225}$$，$$\\frac{1}{256}\\times 4=\\frac{4}{256}=\\frac{2}{128}=\\frac{1}{64}$$ $$\\frac{1}{289}\\times 4=\\frac{4}{289}$$，$$\\frac{1}{324}\\times =\\frac{4}{324}=\\frac{2}{162}=\\frac{1}{81}$$； $$\\text{D}$$. $$\\frac{1}{225}\\times 5=\\frac{5}{225}=\\frac{1}{45}$$，$$\\frac{1}{256}\\times 5=\\frac{5}{256}$$， $$\\frac{1}{289}\\times 5=\\frac{5}{289}$$，$$\\frac{1}{324}\\times 5=\\frac{5}{324}$$． 故选：$$\\text{C}$$． " ]

[ "2017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": "$$19$$ " } ], [ { "aoVal": "B", "content": "$$20$$ " } ], [ { "aoVal": "C", "content": "$$21$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "由$$1$$、$$2$$、$$3$$、$$4$$这四个数字组成的四位数有： $$1234$$；$$1243$$；$$1324$$；$$1342$$；$$1423$$；$$1432$$； $$2134$$；$$2143$$；$$2314$$；$$2341$$；$$2413$$；$$2431$$； $$3124$$；$$3142$$；$$3214$$；$$3241$$；$$3412$$；$$3421$$； $$4123$$；$$4132$$；$$4213$$；$$4231$$；$$4312$$；$$4321$$， 所以将它们从小到大的顺序排列，$$4123$$是第$$19$$个． 故选：$$\\text{A}$$． " ]