2022年大联盟(Math League)国际夏季五年级数学挑战活动一（含答案）

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1、2022 Math League International Summer Challenge, Grade 5 (Unofficial version, for reference only)Note: There are nine questions in total. Seven questions are worth 10 points each. Two questions are worth 15 points each. The total points are 100.Question 1 (10 Points)Objective:Help as many ladybugs a

2、s possible land on the leaves.Rules:1. Ladybugs arrive in numerical order: Ladybug 1, Ladybug 2, Ladybug 3, etc.2. You help each ladybug choose whether to land on the left leaf or the right leaf.3. If on one leaf, the number of dots on two ladybugs adds up to the number of dots on a third ladybug, a

3、ll of the ladybugs fly away.In the figure below, you put Ladybug 1 on the right leaf. Then you put Ladybug 2, Ladybug 3, and Ladybug 4 on the left leaf. Then you put Ladybug 5 on the right leaf.Now there is nowhere to put Ladybug 6.If Ladybug 6 lands on the left leaf, all of the ladybugs will fly aw

4、ay, because 2 + 4 = 6. If Ladybug 6 lands on the right leaf, all of the ladybugs will fly away, because 1 + 5 = 6.12You saw how to help 5 ladybugs. What is the largest number of ladybugs that you can help in this case? The answer is 8, figure below.Note: You cant skip any ladybugs. For example, the

5、following is not allowed. You place Ladybug 1 and Ladybug 2 on the left leaf. Then you skip Ladybug 3, and place Ladybug 4 on the left or right leaf. This skipping Ladybug 3 is not allowed. Ladybugs arrive in numerical order, and you must place each of them on either leaf in numerical order. This is

6、 true for all the following questions.(a) If only ladybugs that are numbered with powers of 2 (1, 2, 4, 8, 16, 32, 64, 128, ) are out flying, what is the largest number of ladybugs that you can help? Ladybugs still arrive in increasing order.Note: Please enter 0 if your answer is infinitely many, wh

7、ich means you can place as many ladybugs on the leaves as you want. There is no limit.Answer: 0(b) If only ladybugs that are numbered with square numbers (1, 4, 9, 16, 25, 36, 49, 64, ) are out flying, what is the largest number of ladybugs that you can help? Ladybugs still arrive in increasing orde

8、r.Note: Please enter 0 if your answer is infinitely many, which means you can place as many ladybugs on the leaves as you want. There is no limit.Answer: 0(c) If only ladybugs that are numbered with cube numbers (1, 8, 27, 64, 125, 216, ) are out flying, what is the largest number of ladybugs that y

9、ou can help? Ladybugs still arrive in increasing order.Note: Please enter 0 if your answer is infinitely many, which means you can place as many ladybugs on the leaves as you want. There is no limit.Answer: 0(d) If only ladybugs that are numbered with non-multiples of 5 (1, 2, 3, 4, 6, 7, 8, 9, 11,

10、12, 13, 14, 16, 17, 18, 19, ) are out flying, what is the largest number of ladybugs that you can help? Ladybugs still arrive in increasing order.Note: Please enter 0 if your answer is infinitely many, which means you can place as many ladybugs on the leaves as you want. There is no limit.Answer: 0Q

11、uestion 2 (15 Points)Objective:Find the gold bar in as few weighings as possible.Rules:1. You can use a balance scale to compare the weights of two groups of bars.2. Only one bar is gold, and all other bars are counterfeits (fakes).3. The gold bar is a little heavier than each of the counterfeits.4.

12、 All counterfeits weigh the same.5. The appearances of all bars are identical. The only way to find out the gold bar is using a balance scale.Example 1:There are two bars. One is gold. The other is a counterfeit. The gold bar is a little heavier than the counterfeit. You need one weighing to find ou

13、t the gold bar. In the figure below, bar 2 is the gold bar.Example 2:There are four bars. One is gold. The other three are counterfeits. The gold bar is a little heavier than each of the counterfeits. All counterfeits weigh the same. The minimum number of weighings to guarantee to find out the gold

14、bar is 2, figure below. In the figure below, the gold bar is bar 2.Another way to do this is that you first compare two bars. If one is heavier, you are done. Otherwise, you compare the other two bars. Again, the minimum number of weighings to guarantee to find out the gold bar is 2.(a) There are 8

15、bars. One is gold. The other 7 are counterfeits. The gold bar is a little heavier than each of the counterfeits. All counterfeits weigh the same. What is the minimum number of weighings to guarantee to find out the gold bar?Answer: 2(b) There are 9 bars. One is gold. The other 8 are counterfeits. Th

16、e gold bar is a little heavier than each of the counterfeits. All counterfeits weigh the same. What is the minimum number of weighings to guarantee to find out the gold bar?Answer: 2(c) There are 10 bars. One is gold. The other 9 are counterfeits. The gold bar is a little heavier than each of the co

17、unterfeits. All counterfeits weigh the same. What is the minimum number of weighings to guarantee to find out the gold bar?Answer: 3(d) If you can do at most 1 weighing, the largest number of bars you can start with and still find the gold bar is 3. And here is why.Place bar 1 on the left pan, and b

18、ar 2 on the right pan. If bar 2 is heavier, then the gold bar is bar 2, figure below.If the weighing is balanced, figure below, then the gold bar is bar 3.If you can do at most 2 weighings, what is the largest number of bars you can start with and still find the gold bar?Answer: 9(e) If you can do a

19、t most 3 weighings, what is the largest number of bars you can start with and still find the gold bar?Answer: 27(f) Now we have new rules:a. You can use a balance scale to compare the weights of two groups of bars.b. Only one bar is gold, and all other bars are counterfeits (fakes).c. The gold bar i

20、s either lighter or heavier than each of the counterfeits, but you dont know which.d. All counterfeits weigh the same.e. The appearances of all bars are identical. The only way to find out the gold bar is using a balance scale.There are 4 bars. One is gold. The other 3 are counterfeits. The gold bar

21、 is either lighter or heavier than each of the counterfeits, but you dont know which. All counterfeits weigh the same. What is the minimum number of weighings to guarantee to find out the gold bar?Answer: 2(g) There are 8 bars. One is gold. The other 7 are counterfeits. The gold bar is either lighte

22、r or heavier than each of the counterfeits, but you dont know which. All counterfeits weigh the same. What is the minimum number of weighings to guarantee to find out the gold bar?Answer: 3(h) There are 12 bars. One is gold. The other 11 are counterfeits. The gold bar is either lighter or heavier th

23、an each of the counterfeits, but you dont know which. All counterfeits weigh the same. What is the minimum number of weighings to guarantee to find out the gold bar?Answer: 3(i) There are 13 bars. One is gold. The other 12 are counterfeits. The gold bar is either lighter or heavier than each of the

24、counterfeits, but you dont know which. All counterfeits weigh the same. What is the minimum number of weighings to guarantee to find out the gold bar?Answer: 3Question 3 (10 Points)(1) True or false: Whenever there are 19 girls and 19 boys are seated around a circular table, there is always a person

25、 both of whose neighbors are boys.Note: Please enter 1 if your answer is True, and 0 if your answer is False.Answer: 1(2) True or false: Whenever there are 20 girls and 20 boys are seated around a circular table, there is always a person both of whose neighbors are boys.Note: Please enter 1 if your

26、answer is True, and 0 if your answer is False.Answer: 0Question 4 (10 Points)How many diagonals does a regular 20-gon have?Note: A regular 20-gon is a polygon. It has 20 sides (edges) and 20 angles. All sides have the same length, and all angles are equal in measure.Answer: 170Question 5 (10 Points)

27、Given a regular 180-gon, we construct 180 isosceles triangles on the exterior of the polygon such that each isosceles triangle has an edge of the polygon as its base and has legs formed by the extensions of the two adjacent sides. Compute in degrees the largest angle of one such isosceles triangle.N

28、ote:(1) A regular 180-gon is a polygon. It has 180 sides (edges) and 180 angles. All sides have the same length, and all angles are equal in measure.(2) Figure below shows the base and two legs of an isosceles triangle. The shape of this isosceles triangle in the figure below is not necessarily the

29、shape of the isosceles triangles in this question.(3) Enter the numeric value of your answer only. For example, if your answer is 100 degrees, please enter 100.Answer: 176Question 6 (10 Points)You are a contestant on a game show. There are three closed doors in front of you. The game show host tells

30、 you that behind one of these doors is a million dollars in cash, and that behind the other two doors there are trashes. You do not know which doors contain which prizes, but the game show host does.The game you are going to play is very simple: you pick one of the three doors and win the prize behi

31、nd it. After you have made your selection, the game show host opens one of the two doors that you did not choose and reveals trash. At this point, you are given the option to either stick with your original door or switch your choice to the only remaining closed door.To maximize your chance to win a

32、 million dollars in cash, should you switch? Answer:(a) YesChoices: (a) Yes(b) No(c) It doesnt matter because the probability for you to win a million dollars in cash stays the same no matter if you switch or not.Question 7 (10 Points)One hundred extremely intelligent male prisoners are imprisoned i

33、n solitary cells and on death row. Each cell is soundproofed and completely windowless. There is a separate room with one hundred small boxes numbered and labeled from 1 to 100. Inside each of these boxes is a slip of paper with one of the prisoners names on it. Each prisoners name only appears once

34、 and is in only one of the one hundred boxes. The warden decides he is going to play a game with all of the prisoners. If they win, they will all be let free, but if they lose the game, they will all be immediately executed.The hundred prisoners are allowed to enter this separate room with 100 boxes

35、 in any predetermined order they wish, but each can only enter the room once and the game ends as soon as the hundredth person enters the room. (At any time, only one prisoner is allowed to enter and remain in this room.) Once a prisoner enters the room, he is allowed to open and look inside as many

36、 as X boxes, where X is a positive integer not greater than 100. After a prisoner has opened the boxes (not more than X) and looked inside, he must shut them and leave everything exactly the way it was before he entered. The prisoners are not allowed to communicate with each other in any way. If eve

37、ry prisoner is able to enter the room and open the box that contains his own name, they will all be released from prison immediately! However, if even just one prisoner enters the room, opens X boxes, and does not open the box containing his own name, they will all be executed immediately. Luckily f

38、or the prisoners, the warden has decided to allow the first prisoner in the room to open all one hundred boxes if necessary and switch the two names in any two boxes if he would like to. The first prisoner must shut all the boxes he opened and leaveeverything exactly the way it was before he entered

39、, with the possible exception of the two names he chose to switch.Again, in order to win this game, all one hundred prisoners need to enter the room and open the box with their name in it. The warden allows the prisoners to get together in the courtyard the week before this game begins to discuss an

40、d come up with a plan. The prisoners come up with a plan to guarantee that they would win the game. Of course, if X is big enough, for example X = 100, then they will all be released from prison. What is the least possible value of X?Note:(1) No two or more prisoners share the same name.(2) The firs

41、t prisoner in the room is an exception, meaning that he is able to open all one hundred boxes if necessary and switch the two names in any two boxes if he would like to. Each of the other 99 prisoners is allowed to open and look inside as many as X boxes, where X is a positive integer not greater th

42、an 100. And they cant switch anything or do anything else.(3) No prisoner is allowed to make any kind of mark, sign, or hint. The prisoners are not allowed to communicate with each other in any way.(4) Heres part of their plan: First the prisoners will decide the order in which they will enter the r

43、oom and number each prisoner from 1 to 100, with Prisoner 1 being the first to enter the room and Prisoner 100 being the last. Then, each prisoner must memorize all other prisoners positions and names in line, which they are able to do.Answer: 50Question 8 (10 Points)We say two numbers m and n are c

44、onsecutive perfect squares when m is the square of a positive integer a, and n is the square of (a + 1). For example, 4 and 9 are consecutive perfect squares since 4 is the square of 2, and 9 is the square of 3. The year when a man was born and the year when he celebrated his 87th birthday are conse

45、cutive perfect squares. In what year was he born?Note: When a man was born, he was 0 years old. When a man celebrated his 87th birthday, he was 87 years old.Answer:1849Question 9 (15 Points)How many positive integers less than 1000a) are divisible by 7?b) are divisible by 7 but not by 11?c) are divi

46、sible by both 7 and 11?d) are divisible by either 7 or 11?e) are divisible by exactly one of 7 and 11?f) are divisible by neither 7 nor 11?g) have distinct digits? (Numbers are written without leading 0s. For example, it is 23, not 023.)h) have distinct digits and are even? (Numbers are written without leading 0s. For example, it is 23, not 023.)Note:(1) Digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.(2) “divisible”: for example, 6 is divisible by 3.Answer:(a) 142(b) 130(c) 12(d) 220(e) 208(f) 779(g) 738(h) 373