南外小升初数学英数经典题(2)

1、思维题分享1.Calculate:1 + 12 + 14 + 18 + 161 + 321 + 641 + 1281 + 2561 = _ .1=147a +1340b +1c +12. Ifd, and a,b,c,d are all nonzero natural numbers,so a+b+c+d=_.3. Select 3 consecutive natural numbers from 1 to 2018, so that their product can be divided by 30. How many choice are there?4. Divide 60 chess

2、 pieces into 27 piles, with 1, 2 or 5 pieces in each pile, If the sum of the number of the piles with 2 pieces and 5 pieces are 2 times of the number of the piles with 1 piece, so how many piles with 2 pieces are there?5.Fill two adjacent integers in the blanks._ 109 + 1011 + . + 1718 + 1918 _6.Ther

3、e are 2019 dwarfs, and they are either good people or bad people. They will take part in a party every day. And the number of the dwarfs taking part in the party is the multiple of 3 or 5. Among the people taking part in the party, if the good people are in the majority, all the people in the party

4、will become good people. If the bad people are in the majority, all the people in the party will become bad people. The question is:All the people become good people if the party on the third day ends, whats the number of the good people before the party of the first day at least?7.Tom takes part in

5、 a test six times. The average score of the third and fourth time is 2 score higher than the average score of the first two times and is 2 score lower than the average score of the last two times. If the average score of the last three times is 3 score higher than the average score of the first thre

6、e times, then whats the score of the fourth test?8. Place A is 2.8 kilometers away from Place B. Two dogs keep running between these two places. They run at speeds of 450 and 350 meters per minute respectively. They first run towards each other for 1 minute, and then they turn back and run in opposi

7、te directions for another 2 minutes. Then, they turn back again and run towards each other for 3 minutes, and then turn back and run for 4 minutes They keep running in this way until they meet. So how soon will they meet?解析：1. 1 + 12 + 14 + 18 + 161 + 321 + 641 + 1281 + 2561 = _ .【解析】 原式 = 1 + 12 +

8、14 + 18 + 161 + 321 + 641 + 1281 + 2561 + 2561 - 2561 = 2 - 2561 = 1 256255 .2.若1=147，且 a,b,c,d 均为非 0 自然数，则 a+b+c+d=_.a +13401b +c +1d【解析】1=147=1=1a +13403402 +46b +1147147c +1d340147=2.46,即 a=214746=3.9,即 b=3469=5.1,即 c=591=9，即 d=9所以 a+b+c+d=2+3+5+9=193. 从 12018 中选出连续 3 个自然数，使得它们的乘积能被 30 整除，一共有_种选法

9、【解析】因为是连续选的 3 个数，所以肯定有 3 的倍数和 2 的倍数，所以它们的乘积一定能被 6 整除，所以只要连续三个数字有一个能被 5 整除，那么就可以被 30 整除，而对于每一个被 5 整除的数来说，有 3 种选法：如 3，4、5；4、5、6；5、6、7；12018 有：20185=403.3,所以有 403 个可以被 5 整除的数，每个 5 的倍数对应有 3 种选法，那么就有：4033=1209（种）。4. 把 60 枚棋子分成 27 堆，每堆中的棋子数为 1 枚，2 枚或 5 枚。如果有 2 枚和5 枚的棋子堆数的和是只有 1 枚棋子的堆数的 2 倍，那么恰好含 2 枚棋子的有多少

10、堆?【解析】1 枚：27(2+1)=9(堆），2 枚和 5 枚的和：27-9=18(堆），60-91=51(枚)，2 枚：（518-51）(5-2）=13(堆）。5.在空白处填上相邻两个整数。_ 9+10+ . +17+18 _10111819【解析】此题用扩展法进行解答，为了方便，把中间的式子假设为 A，则A=10-（1+1+ . +1+1） ， 因 为101+1+ . +1+1 1， 所 以 ，1819191011181910119 = 10 -1 A 10 - 1910 10 进而得出结论。 9 109 + 1011 + . + 1718 + 1918 3,所以尽量5人一组，2019=5

11、402+33，最后一次共分402个5人组和3个3人组,每个5人组中有3个好人,每个3人组中有2个好人，第二次聚会后好人至少有4023+32=1212(人)，同理1212=5240+34,第一次聚会后好人至少有2403+42=728(人), 728=5145+31,则最初好人至少有 1453+12=437(人)。7.汤姆参加了六次测验，第三、第四次的平均分比前两次的平均分多 2 分，比后两次的平均分少 2 分。如果后三次平均分比前三次平均分多 3 分，那么第四次比第三次多得几分？【解析】第三、四次的成绩和比前两次的成绩和多 4 分，比后两次的成绩和少 4 分，推知后两次的成绩和比前两次的成绩和多 8 分。因为后三次的成绩和比前三次的成绩和多 9 分，所以第四次比第三次多 98=1（分）。8.A、B 两地相距 2.8 千米，两条狗从 A、B 两地相向奔跑。它们每分钟分别跑450 米和 350 米，它们相