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Old 2005-10-21, 22:38   #1
fetofs
 
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Default Some puzzles

I thought the puzzles forum was a bit cold, so...

1-A 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?

2-What is the number hidden? (Don't multiply, please.)
847398654*638952=54144706?770608

3-Two 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!

What is X's number? (And Y's?)

I didn't solve 4 and 5...

4-For what positive integers m and n is it possible to build an m-element set of positive integers such that the sum of any n of them is not divisible by n?

5-A 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, has to switch 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 e.

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
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Old 2005-10-22, 12:33   #2
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For n°3, IF x and y are integer and > 0
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

Last fiddled with by flava on 2005-10-22 at 12:33
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Old 2005-10-22, 13:32   #3
fetofs
 
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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?
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Old 2005-10-22, 15:27   #4
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1-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%!
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Old 2005-10-22, 21:53   #5
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5a: It depends what point in the show you're at.
If you're at the beginning (case a), the chance of winning is:

n-1
in 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 n-1 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.


5b: In (case a), lim(n->inf)(n-1)/n = 1
In (case b), lim(n->inf)1/2 = 1/2


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 222.
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Old 2005-10-23, 00:20   #6
fetofs
 
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Quote:
Originally Posted by Ken_g6
1-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%!
Sorry Ken, it was meant to be serious.

Quote:
Originally Posted by 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 222

Sorry? I can't get this number. I got 213 and my teacher keeps saying that the answer is 214!!! Can you list all the "jumped" numbers?

Quote:
Originally Posted by Ken_g6
It depends what point in the show you're at.
If you're at the beginning (case a), the chance of winning is:

n-1
in [spoiler] n
Didn't experts said that not switching is best? If so, not switching is ]n-1/n . That way, we can conclude that the chance is lower .
Maybe if the host opens m...

Last fiddled with by fetofs on 2005-10-23 at 00:20 Reason: Couldn't spoilerize :(
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Old 2005-10-23, 00:46   #7
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Problem 6: 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.

If they can move, so can you. So you will play last.

Note that this solution doesn't require the table to be round. Only that it have a line of symmetry.

Last fiddled with by Zeta-Flux on 2005-10-23 at 00:47
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Old 2005-10-23, 02:17   #8
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Quote:
Originally Posted by Zeta-Flux
Problem 6: Note that this solution doesn't require the table to be round. Only that it have a line of symmetry.
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.
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Old 2005-10-23, 04:51   #9
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Quote:
Originally Posted by fetofs
Sorry? I can't get this number. I got 213 and my teacher keeps saying that the answer is 214!!! Can you list all the "jumped" numbers?
Jumped numbers:
13
31
103
113
123
130
131
132
133
134
135
136
137
138
139
143
153
163
173
183
193
213

Did you mean appeared together contiguously? Then I get 214. It can't be 213 because that's 213!

P.S. This came from your teacher?

Quote:
Didn't experts said that not switching is best? If so, not switching is ]n-1/n . That way, we can conclude that the chance is lower .
Maybe if the host opens m...
I assume the guest knows that there is one car, and n-1 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 (n-1)/n. I think the not switching thing is for cases where there is more than one car-type goal.

Last fiddled with by Ken_g6 on 2005-10-23 at 04:53
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Old 2005-10-23, 12:18   #10
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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 n-2 doors are opened and you then have the option of switching giving you a (n-1)/n chance to win if you switch.
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Old 2005-10-23, 17:56   #11
fetofs
 
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Quote:
Originally Posted by Ken_g6
J
P.S. This came from your teacher?
Actually, this one I took from my brazilian math competition. Like this one:

8-Prove that any square isn't a perfect number.

P.S: Oh, these guys are tricky!!!! I never thought about 213
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