Some puzzles
I thought the puzzles forum was a bit cold, so...
1A company called a meeting such that the probability of at least two members having a birthday on the same date is higher than 50%. What is the minimal number of members possible present? 2What is the number hidden? (Don't multiply, please.) 847398654*638952=54144706?770608 3Two smart students (X and Y) choose a number and show it to their teacher. The teacher says that the result of the addition of both is either 1994 or 2990. He asks X, do you know Y's number? No. He asks Y, do you know X's number? No. He gives time to think about it, and in the next day... He asks X, do you know Y's number? No. He asks Y, do you know X's number? Yes, I do! :banana: What is X's number? (And Y's?) I didn't solve 4 and 5... 4For what positive integers m and n is it possible to build an melement set of positive integers such that the sum of any n of them is not divisible by n? 5A tweak of the Monty Hall: A show host shows to a guest n closed doors.(n is an integer, greater or equal to 3) Behind one of the doors is a car, the other ones, a goat. The guest picks a door. After that, the show host opens a door (with no car behind it). The guest, then, [b]has to switch[/b] to a door not open yet, and so it goes, until there are only two doors (one chosen and one closed). The guest has to switch and sees if he's won the car or not. a)What's the chance that the guest wins the car? Try to be as objective and simple as possible (with the least number of operations) b)What's the chance when n tends to infinity? You can use only addition, subtraction, multiplication, division, and the number [i]e[/i]. 6 You have a round table, and an infinite number of coins. You play a game with your friend, where you must place coins on the table (no overlaps). Who plays last wins. Would you like to start? Why? (Strategy) Good luck :wink: 
For n°3, IF x and y are integer and > 0
[spoiler]x+y = 1994 or x+y = 2990 x says no => x < 1994 y says no => y < 1994; y>996 x says no => x < 998; x>996 => x = 997 y = 997[/spoiler] 
7 A superstitious girl, when she was numbering her 200 page diary, started from number 1 but jumped every page where the numbers 1 and 3 appeared together, in any order. What was the number she wrote on the last page?

1[spoiler]two :razz:
It's a trick question. If the company calls a meeting of two people who they know have the same birthday, the probability that they have the same birthday is 100%![/spoiler] 
5a: [spoiler]It depends what point in the show you're at.
If you're at the beginning (case a), the chance of winning is: n1 [/spoiler] in [spoiler] n Not choosing a door at the end is equivalent to our guest choosing at the beginning, "I won't try door number m", for some m<=n. Then he tries all the other doors, in any sequence. If the car is behind a door other than m, he wins. So the chance the car was behind m and he loses is 1 in n, so his chance of winning is n1 in n. On the other hand, if the contestant is down to two doors when we start this scenario, (case b), his chances are always 1 in 2.[/spoiler] 5b: [spoiler]In (case a), lim(n>inf)(n1)/n = 1 In (case b), lim(n>inf)1/2 = 1/2 [/spoiler] 7: The command "jot  1 1000 1  grep v "1.*3\3.*1"  head 200" produces all the page numbers. They start at 1, and end at [spoiler]222[/spoiler]. 
[QUOTE=Ken_g6]1[spoiler]two :razz:
It's a trick question. If the company calls a meeting of two people who they know have the same birthday, the probability that they have the same birthday is 100%![/spoiler][/QUOTE] Sorry Ken, it was meant to be serious. [QUOTE=Ken_g6]The command "jot  1 1000 1  grep v "1.*3\3.*1"  head 200" produces all the page numbers. They start at 1, and end at [spoiler]222[/spoiler][/QUOTE] Sorry? I can't get this number. I got [spoiler]213[/spoiler] and my teacher keeps saying that the answer is [spoiler]214[/spoiler]!!! Can you list all the "jumped" numbers? [QUOTE=Ken_g6][spoiler]It depends what point in the show you're at. If you're at the beginning (case a), the chance of winning is: n1 [/spoiler] in [spoiler] n [/QUOTE] Didn't experts said that not switching is best? If so, not switching is ]n1/n . That way, we can conclude that the chance is lower . Maybe if the host opens m... 
Problem 6: [spoiler]Yes, you want to go first. You put your first coin in the middle. Then, whatever your opponent does, you go diametrically opposed to them.[/spoiler]
[spoiler]If they can move, so can you. So you will play last. [/spoiler] [spoiler] Note that this solution doesn't require the table to be round. Only that it have a line of symmetry.[/spoiler] 
[QUOTE=ZetaFlux]Problem 6: Note that this solution doesn't require the table to be round. [spoiler] Only that it have a line of symmetry.[/spoiler][/QUOTE][spoiler] Isn't it actually a Point of Symmetry? An equilteral triangle has a line of symmetry which is he perpendicular bisector of the base. And placement that includes a portion of the line cannot be countered by a symetrically opposite move. Therefore, the last player who is able to place his coin such that it crosses the line would be the winner. But that is not necessarily the first player.[/spoiler]

[QUOTE=fetofs]
Sorry? I can't get this number. I got [spoiler]213[/spoiler] and my teacher keeps saying that the answer is [spoiler]214[/spoiler]!!! Can you list all the "jumped" numbers?[/quote]Jumped numbers: [spoiler]13 31 103 113 123 130 131 132 133 134 135 136 137 138 139 143 153 163 173 183 193 213[/spoiler] Did you mean appeared together [i]contiguously[/i]? Then I get 214. It can't be 213 because that's 2[b]13[/b]! :smile: P.S. This came from your teacher? :ermm: [quote]Didn't experts said that not switching is best? If so, not switching is ]n1/n . That way, we can conclude that the chance is lower . Maybe if the host opens m...[/QUOTE] I assume the guest knows that there is one car, and n1 goats, and that the car is the only goal. Then the guest effectively selects one door he won't open. If the car is behind that door, he loses, but if it's not, eventually he will find it. So his chance of winning is (n1)/n. I think the not switching thing is for cases where there is more than one cartype goal. 
Regarding the Monty Hall problem, in the original problem you have 3 doors and he will always open a door with a goat. By switching you have a 2/3 chance of winning (since your original choice was 1/3). In this version you are forced to switch after that door is opened, giving you a 2/3 chance of winning.
As the number of doors increase, I believe that the probability remains at 2/3 because you are forced to switch after each door is opened, which is unlike the original Monty Hall problem where n2 doors are opened and you then have the option of switching giving you a (n1)/n chance to win if you switch. 
[QUOTE=Ken_g6]J
P.S. This came from your teacher? :ermm: [/QUOTE] Actually, this one I took from my brazilian math competition. Like this one: 8Prove that any square isn't a perfect number. P.S: Oh, these guys are tricky!!!! I never thought about 2[b]13[/b] 
All times are UTC. The time now is 14:30. 
Powered by vBulletin® Version 3.8.11
Copyright ©2000  2021, Jelsoft Enterprises Ltd.