2021年大联盟(Math League)国际夏季五年级数学挑战活动一(含答案).docx

上传人:优****虫 文档编号:96696222 上传时间:2024-03-09 格式:DOCX 页数:16 大小:475.35KB
返回 下载 相关 举报
2021年大联盟(Math League)国际夏季五年级数学挑战活动一(含答案).docx_第1页
第1页 / 共16页
2021年大联盟(Math League)国际夏季五年级数学挑战活动一(含答案).docx_第2页
第2页 / 共16页
点击查看更多>>
资源描述

《2021年大联盟(Math League)国际夏季五年级数学挑战活动一(含答案).docx》由会员分享,可在线阅读,更多相关《2021年大联盟(Math League)国际夏季五年级数学挑战活动一(含答案).docx(16页珍藏版)》请在taowenge.com淘文阁网|工程机械CAD图纸|机械工程制图|CAD装配图下载|SolidWorks_CaTia_CAD_UG_PROE_设计图分享下载上搜索。

1、2021 美国大联盟(Math League)国际夏季数学挑战活动2021 Math League International Summer ChallengeGrade 5, Individual Questions & SolutionsQuestion 1:Sarah is using popsicle sticks to build some grids for her city planning project. She needs 4 popsicle sticks to make a 1 by 1 grid, and 12 sticks to make a 2 by 2 grid

2、 as shown below. How many sticks does she need to make a 10 by 20 grid?Answer: 430 Solution:Analyze the 2 by 2 grid. To create it Sarah used 3 rows of 2 sticks each and 3 columns of 2 sticks each. Similar to it, there would be 11 rows and there should be 20 sticks in each row. So, there are 20 11 =

3、220 sticks (placed horizontally).There would also be 21 columns and there should be 10 sticks in each. Therefore, there are 21 10 = 210 sticks placed vertically.Total = 220 + 210 = 430 popsicle sticksQuestion 2:The sides of the large rectangle are 20 m and 16 m, figure below, not drawn to scale. All

4、 six shaded rectangles are identical. What is the total area of all the shaded regions, in square meters?16Answer: 192 Solution:Find the dimension of the small rectangle. The length: 16 2 = 8 m. The width: 20 8 8 = 4 m. The total area of all the shaded regions: 6 (8 4) = 192 m2.Question 3:If two sid

5、es of a square field were increased by five feet, as seen in the diagram below, not drawn to scale, the area of the field would increase by 245 square feet. Find the area of the original square.Answer: 484 Solution:The area of the shaded square is 5 5 = 25, figure below. The area of each of the two

6、“new” rectangles is (245 25) 2 = 110. Thus the length of the “new” rectangle, or the side length of the original square, is 110 5 = 22. So the area of the original square is 22 22 = 484.Question 4:During practice, Danas six arrows land on the target shown. Each arrow is inside one of the regions of

7、the target. Which of the following total scores is possible: 44, 31, 26, 16? (If an arrow lands inside the smallest circle in the center, the score Dana gets is 7. If an arrow lands inside the second circle, but not inside the smallest circle, the score Dana gets is 5. If an arrow lands inside the l

8、argest circle, but not inside the second circle or the smallest circle, the score Dana gets is 3. Assume no arrow lands on the boundary of any of the three circles.)(a) 44(b) 31(c) 26(d) 16 Answer: (c) Solution:Assume x arrows land inside the smallest circle, and y arrows land inside the second circ

9、le. Then 6 x y arrows land inside the largest circle. And the total score is:7x + 5y + 3(6 x y) = 4x + 2y + 18The total score is an even number, so (b) is not good.The total score is greater than or equal to 18, so (d) is not good.If the total score is 44, then we have 4x + 2y + 18 = 44. 4x + 2y = 2

10、6, 2x + y = 13. But it is not possible that 2x + y = 13, as 0 x 6, 0 y 6, and x + y 6. So (a) is not good.When x = 0 and y = 4 or x = 1 and y = 2, or x = 2 and y = 0, 4x + 2y + 18 = 26.Question 5:There are 1000 people in a room. Each pair either shakes hands or does not.(1) Is it always true that so

11、me two people shook the same number of hands? (Please enter 1 if your answer is Yes, and 0 if your answer is No.)(2) Is it always true that some three people shook the same number of hands? (Please enter 1 if your answer is Yes, and 0 if your answer is No.)(3) Is it always true that some four people

12、 shook the same number of hands? (Please enter 1 if your answer is Yes, and 0 if your answer is No.)(4) Is it always true that some five people shook the same number of hands? (Please enter 1 if your answer is Yes, and 0 if your answer is No.)Note:(1) One doesnt shake his/her own hand.(2) One doesnt

13、 shake the same persons hand more than once.Answer:(1) 1(2) 0(3) 0(4) 0Solution:(1) If someone shook 999 times, then everyone shook at least one time. So the possible numbers are 1 to 999. But there are 1000 people, from Pigeonhole Principle, at least two people shook the same number of hands.(2) No

14、t true.(3) Not true.(4) Not true.Question 6:The numbers 1 through 10 are written in a row. Can the signs “+” and “” be placed between them, so that the value of the resulting expression is 0? (Note: There are 10 numbers, and there are 9 places to place signs.)Answer: No (Yes/No) Solution:1 + 2 + 3 +

15、 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55Changing any “+” to “”, the resulting expression is also an odd number. So the resulting expression is always an odd number.Question 7:Albert and Bernard just became friends with Cheryl, and they want to know when her birthday is. Cheryl gives them a list of 10 possib

16、le dates, figure below. Cheryl then tells Albert and Bernard separately the month and the day of her birthday respectively. (That is Cheryl tells Albert the month of her birthday, and tells Bernard the day of her birthday.)Conversation:Albert: I dont know when Cheryls birthday is, but I know that Be

17、rnard also doesnt know.Bernard: At first I didnt know when Cheryls birthday is, but I know now.Albert: Then I also know when Cheryls birthday is. When is Cheryls birthday?Please enter the month first, followed by the day. When you enter the month, please enter 5 for May, 6 for June, 7 for July, and

18、8 for August.Answer:716Solution:Albert: I dont know when Cheryls birthday is, but I know that Bernard also doesnt know.So Bernard knows that it is not May or June. If it is May, then it is possible that it is May 19. Cheryl will tell Bernard that the day is 19, then Bernard will know her birthday is

19、 May 19. So it is not May. It is not June either.Bernard: At first I didnt know when Cheryls birthday is, but I know now.So her birthday is July 16, August 15, or August 17. The number that Cheryl tells Bernard is 15, 16, or 17.Albert: Then I also know when Cheryls birthday is.If it is August, then

20、Albert will not be able to tell Cheryls birthday. So it has to be July. So the answer is July 16.Question 8:Goal:1. Arrange the six different juices into the following order.2. The order is: 1 red, 2 orange, 3 yellow, 4 green, 5 blue, 6 violet.Rules:1. You may only pour a filled cup into an empty cu

21、p.2. You may not switch the positions of any cups.3. The empty cup must end up on the right.Examples:2 Pours are necessary to solve the following example puzzle, figure below.Pour One, figure below.Pour Two, figure below:For another example below, this example needs four pours.Pour One, figure below

22、.Pour Two, figure below.Pour Three, figure below.Pour Four, figure below.For 6 different juices and one empty cup, there are 5040 (= 7! = 7 6 5 4 3 2 1) possible permutations. Two of them were listed above. One needs 2 pours. The other needs 4 pours.(1) Of all the 5040 possible permutations, how man

23、y permutations are impossible to solve? That is no matter how you pour the juices, you cant arrange the juices in order.Answer: 0 Solution:Every permutation is solvable. In the case of 6 different juices, if we look at every cup fromthe leftmost to the one right before the rightmost, there are at mo

24、st 6 cups that are not fille with the correct juices. Lets call these cups “incorrect cups.” For every incorrect cup, we need at most two pours to make it correct, that is to fill this cup with the correct juice. So every permutation is solvable.(2) What is the largest number of pours you need to so

25、lve any solvable permutation?Answer: 9 Solution:Lets first look at the case when there are only 2 different juices, figure below. The largest number of pours is 3. And it happens when the positions of Juice 1 and Juice 2 are swapped, and the empty cup is on the right.Then lets look at the case when

26、there are 4 different cups, figure below. The largest number of pours is 6. And it happens when the positions of 2 juices are swapped, and the positions of the other 2 juices are swapped too, and the empty cup is on the right.So when the number of juices is n, n is an even number, the largest number

27、 of pours is n 3 .2For 1000001 different juices and one empty cup, there are 1000002! possible permutations. There is a pre-defined order of the 1000001 juices, just as the order for the 6 juices above.(3) Of all the 1000002! possible permutations, how many permutations are impossible to solve? That

28、 is no matter how you pour the juices, you cant arrange the juices in order.Answer: 0 Solution:Every permutation is solvable. In the case of 1000001 different juices, if we look at everycup from the leftmost to the one right before the rightmost, there are at most 1000001 cups that are not fille wit

29、h the correct juices. Lets call these cups “incorrect cups.” For every incorrect cup, we need at most two pours to make it correct, that is to fill this cup with the correct juice. So every permutation is solvable.(4) What is the largest number of pours you need to solve any solvable permutation (fo

30、r 1000001 different juices and one empty cup)?Answer: 1500001 Solution:When the number of juices is n, n is an odd number, the largest number of pours is n - 1 3 + 1.2Question 9:Twelve straightjacketed prisoners are on the death row. Tomorrow they will be arranged in a single-file line, positions de

31、termined randomly, all facing the same direction. The person in the back of the line, who will be referred to as the twelfth person, can see the eleven people standing in front of him; the eleventh person can see the ten people standing in front of him, but not the twelfth person who is standing beh

32、ind him; so on and so forth until you get to the first person in line who cannot see anyone. The warden of this prison likes to play games with his prisoners and has lined everyone up in this fashion so that he can give them an opportunity to gain their freedom.The warden will place, at random, a bl

33、ack or white hat on each prisoners head. The prisoners cannot see the color of the hat they are wearing, but they can see the color of the hats of each prisoner standing in front of them in line. After all of the hats have been placed on the prisoners heads, the warden asks the person in the back of

34、 the line, the twelfth person, “what is the color of the hat on your head?” The prisoner can reply by saying either “black” or “white”, but nothing else. When the prisoner answers the warden, all the prisoners in line can hear the reply. If the prisoner replied with the correct color of his hat, he

35、will be freed from the prison. If he is wrong, he will be executed that night. This game continues until all twelve of the prisoners in line have been asked about the color of their hat. While the game is in progress, no prisoner is allowed to speak, move, or do anything until it is their turn to an

36、swer the question. The warden allows the prisoners to get together in the courtyard the day before this game to come up with a plan to maximize the number of lives they can save. The prisoners come up with a plan that guarantees x prisoner(s) will survive, and 12 x prisoner(s) will each have a 50% c

37、hance to survive. What is the largest possible value of x?Answer: 11 Solution:The prisoners can come up with a plan that will give the twelfth prisoner to act a 50/50 chance at survival and will guarantee that the other eleven prisoners in line all survive. This is how they do it:The prisoners decid

38、e that whoever is standing in the twelfth position and speaks first will simply say whichever color he sees an even amount of. There is no way for this prisoner to raise his chances of survival above 50%, but by saying whichever color he sees an even amount of, every other prisoner can deduce the co

39、lor of their own hat.For example: If the twelfth person says “black”, then the eleventh person in line knows what color hat he is wearing and can correctly state it to the group based on the remaining hats that he can see. If he sees an odd number of black hats in front of him, then he can be sure t

40、hat his hat is black because the person in line behind him saw an even amount. Now, thetenth person in line, knowing that the twelfth person saw an even amount of black hats and knowing that the eleventh person in line is wearing a black hat, can deduce the color of his hat based on the remaining ni

41、ne hats. If the twelfth person in line saw an even amount of black hats, and the eleventh person in line is wearing a black hat, then the tenth person in line knows that if he sees an odd number of black hats, he is wearing a white hat, and if he sees an even number of black hats, he must be wearing

42、 a black hat himself.This system allows for eleven prisoners to guarantee their survival. No so much fun the twelfth prisoner.Question 10:Somewhere in the world there is a small neighborhood full of very eccentric and intelligent people. There are five houses that are each a different color. There i

43、s a person of a different nationality who lives in each house. Each of these five people drinks their own special drink, smokes their own special brand of cigarettes, and has their own special pet. No one has the same pet, smokes the same brand of cigarettes, or drinks the same drink. One of these f

44、ive people owns a pet fish. With just the following information, figure out who owns the cat and who owns the fish:(1) The British man lives in the red house.(2) The Swedish man has a dog for a pet.(3) The Danish man drinks tea.(4) The green house and the white house are next to each other, and the

45、green house is to the left of the white house.(5) The owner of the green house drinks coffee.(6) The person that smokes Pall Mall cigarettes has a bird for a pet.(7) The owner of the yellow house smokes Dunhill cigarettes.(8) The person who lives in the middle house drinks milk.(9) The Norwegian liv

46、es in the first house, counting from left to right.(10) The person who smokes Blend cigarettes lives next to the one that has a cat for a pet.(11) The person who has a horse for a pet lives next to the one who smokes Dunhill cigarettes.(12) The one who smokes Bluemaster cigarettes drinks beer.(13) T

47、he German smokes Prince cigarettes.(14) The Norwegian lives next to a blue house.(15) The person who smokes Blend cigarettes has a neighbor who drinks water.(16) The five houses are positioned in a single-file line.Question (1): Who owns the cat? (Please enter 1 if your answer is Norwegian, 2 if your answer is Swedish, 3 if your answer is German, 4 if your answer is British, and 5 if your answer is Danish.)Answer: 1Question (2): Who owns the fish? (Please enter 1 if your answer is Norwegi

展开阅读全文
相关资源
相关搜索

当前位置:首页 > 教育专区 > 小学奥数

本站为文档C TO C交易模式,本站只提供存储空间、用户上传的文档直接被用户下载,本站只是中间服务平台,本站所有文档下载所得的收益归上传人(含作者)所有。本站仅对用户上传内容的表现方式做保护处理,对上载内容本身不做任何修改或编辑。若文档所含内容侵犯了您的版权或隐私,请立即通知淘文阁网,我们立即给予删除!客服QQ:136780468 微信:18945177775 电话:18904686070

工信部备案号:黑ICP备15003705号© 2020-2023 www.taowenge.com 淘文阁