Light the first torch by the 2 ends, and the second by one end. When the first torch is finished (after 1/2 hour), light the second end of the second torch.
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Light the first torch by the 2 ends, and the second by one end. When the first torch is finished (after 1/2 hour), light the second end of the second torch.
Fine. So now it's Elrond's turn.
Ummm,,, but they worked. Is there another flaw in this question?? Can't we just forget it, and you tell us the answer!Quote:
Originally posted by SeventhStar
Nope thats not what these operators do.
I'd ask a question now, but I pledged not to :(.
sorry guys yesterday my pc went crazy and i could post so:
so you know what the base numbers are
operators:
i and m mean plus but more correctly
barwoniganalhul = 199 = 1 after the 6th 33
babumbarwonhul = 37 = 4 after the first 33
o and s mean -22+ but more correctly:
babusgutarhul = 48 = 4 after the first 11 from the second 33
t and y mean - 44 +
babutfarokhul = 59 = 4 after the first eleven before the third 33
a and an operator that wasnt given mean before
gautaragutarhul = 64 = 2 before the second 33
so it's not that solarflare isn't correct but he didn't get the idea of the whole thing
Elrond doesn't seem to show up. Anyone else a question?
Well, there's one I have heard of a few years ago and liked it very much. It might be widely known by now, however, I could give it a try.
OK, I'll go ahead:
Four people need to cross a narrow bridge at night.
The bridge will only hold a maximum of two people. A torch must be carried by someone crossing the bridge - else they will fall down. But there is only one torch, it's battery life is limited to 1 hour.
One person takes 25 minutes to cross, another 20 minutes, another 10 minutes and another 5 minutes.
When two people cross the bridge together, they must go at the speed of the slowest.
You have to get all four people across in 1 hour. Throwing the torch back or similar tricks are not allowed.
Sorry for not showing up earlier. :o
The guy of 10 and the guy of 5 cross (10 minutes passed)
The guy of 10 goes back (20 minutes passed)
The guy of 25 and the guy of 20 cross (45 minutes passed)
The guy of 5 goes back (50 minutes passed)
the guy of 10 and the guy of 5 cross (60 minutes passed)
And by the way, I still don't have any question, so fell free to ask.
1. Gabriel 8
2 me and Simon 6 (go Simon only 660 to go :D)
3. Elrond 5
4. Saturno 4
5. solarflare 3
6. John 2
7. dimm_coder, Xplorer, galathaea, doctor luz, gstercken 1
Question:
Two meteorites going in the same direction passed one point simultaneously. The first one was going with a speed of 200km per hour. The second one passed 1km the first hour, 2 the second and N km in the N th hour. Within how many days will the second meteorite reach the first one?
After 399 hours. Time needed so
1+2+3+...+N = N*200
yep 399 is correct
Two persons, Mr S. et Mr P. , repectively know the sun and the product of two integers A and B. A and B are between 2 and 100.
They don't know A and B.
S. et P. meet and start talking :
P. : "I can't find these numbers"
S. : "I knew it"
P. : "Then I have know what they are !"
S. : "Really ?! Then I have know as well !"
What are the values of A and B ?
They are the solution of a second degree ecuation
X*X-(A+B)*X+(A*B)=0
s know the sum and p know the product right?Quote:
Originally posted by Elrond
Two persons, Mr S. et Mr P. , repectively know the sun and the product of two integers A and B. A and B are between 2 and 100.
They don't know A and B.
S. et P. meet and start talking :
P. : "I can't find these numbers"
S. : "I knew it"
P. : "Then I have know what they are !"
S. : "Really ?! Then I have know as well !"
What are the values of A and B ?
I was wondering about this too... But then I noticed that the names 'S' and 'P' were probably not chosen at random... ;)
Yes, and that's all they know at the beginning!Quote:
Originally posted by SeventhStar
s know the sum and p know the product right?
A=(SUM+SQRT(SUM*SUM-4*PROD))/2
B=(SUM-SQRT(SUM*SUM-4*PROD))/2
This means we already have three 6's. I thought I wouldn't have 660 to go. :D :p :p
1. Is this already the complete answer?Quote:
A=(SUM+SQRT(SUM*SUM-4*PROD))/2
B=(SUM-SQRT(SUM*SUM-4*PROD))/2
2. If so, why? :confused:
It is easy to test:
A+B=SUM
A*B=PROD
OK, but the preceding dialogue between P and S suggests that by knowing that S knew that P can't find the numbers, P can figure them out (the actual values!), and with this knowledge, S finds them too. In my opinion, it must have something to do with prime factorials...Quote:
Originally posted by Doctor Luz
It is easy to test:
A+B=SUM
A*B=PROD
You seem to be missing that P and S know the product and the sum *respectively*. IMHO, this means that P knows *only* the product and S *only* the sum, and these values are never communicated between S and P.
You say S knows A and B right? If sum and product are never comunicated I don't know how S can know A and B only knowing A+B. There are infinite posibilities except if one number is 0.
in this case A=sum b=0; PROD=0
@Elrond: Are A and B allowed to be equal?
Quote:
Originally posted by Doctor Luz
You say S knows A and B right? If sum and product are never comunicated I don't know how S can know A and B only knowing A+B. There are infinite posibilities except if one number is 0.
in this case A=sum b=0; PROD=0
No, S knows A+B, not A and B.Quote:
You say S knows A and B right?
No, because A and B are between 2 and 100.Quote:
There are infinite posibilities except if one number is 0.
S not only knows A+B, he equally knows that P can't find the numbers knowing only the product.Quote:
I don't know how S can know A and B only knowing A+B.
NO!!!!Quote:
Originally posted by Doctor Luz
You say S knows A and B right? If sum and product are never comunicated I don't know how S can know A and B only knowing A+B. There are infinite posibilities except if one number is 0.
in this case A=sum b=0; PROD=0
S only knows the result of A+B, he does not know A, B, or A*B.
P only knows the result of A*B, he does not know A, B or A+B.
The only additional thing each of them know if that A and B are between 2 and 100. It means that if A+B = 128, S know that it cannot be something like 112+16, and the if A*B = 256, P knows that it cannot be 128*2.
Then they don't speak any additional word except the ones I have told you.
As far as I know, there's no rule about A and B being equal or different.
You are right, I did not read those details.
I should read more carefully. Sorry
Elrond, this is a hard one! :eek:
Is it OK to think aloud? Just to know if the approach is correct, or if it's somethig completely different.
Yes, it's OK to think aloud. The only risk is that some one takes your ideas to get the correct answer. ;)
I'll face this risk. Let's see:Quote:
Originally posted by Elrond
Yes, it's OK to think aloud. The only risk is that some one takes your ideas to get the correct answer. ;)
1. P states that he can't find the numbers. This means that the A and B cannot be prime, as the solution would otherwise be obvious to P.
2. S states that he knew that. This *could* mean to P (but need not) that S knew that the numbers cannot be prime. He could have deduced that by the fact that the sum is odd (Goldbach's conjecture states that every even integer >= 4 can be written as the sum of two primes. Although it has not generally been proven, it has been proven for numbers < 4*10^14 AFAIK).
3. Assuming the sum is odd, this means that either A or B is odd and the other is even.
At this point, I'm stuck... I feel that there can be made more deductions from the fact that P can't find A and B (what about squares?), but I'm still missing something.
It only means that not both numbers at the same time can be prime.Quote:
Originally posted by gstercken
1. P states that he can't find the numbers. This means that the A and B cannot be prime, as the solution would otherwise be obvious to P.
Of course, sorry. That's what I meant.Quote:
Originally posted by Simon666
It only means that not both numbers at the same time can be prime.
are the numbers 2 and 9
tell me if they are and i'll explain why
2 and 9 are not correct
can one of the numbers be 2Quote:
Originally posted by Elrond
2 and 9 are not correct
@SeventhStar: based on your strategy, I have worked out an algorithm which should work:Quote:
Originally posted by SeventhStar
can one of the numbers be 2
:D :D :DCode:for(int a = 2; a <= 100; a++)
{
for(int b = 2; b <= 100; b++)
{
AskElrondIfCanBe(a,b);
}
}
Nope i thought of a great idea but the promlem is that if one of the nubers can be 2 i have too much answersQuote:
Originally posted by gstercken
@SeventhStar: based on your strategy, I have worked out an algorithm which should work:
:D :D :DCode:for(int a = 2; a <= 100; a++)
{
for(int b = 2; b <= 100; b++)
{
AskElrondIfCanBe(a,b);
}
}
PS. you can remove these '{}' thingies
I have an even more efficient one:
:D:D:DCode:for(int a = 2; a <= 100; a++)
{
for(int b = a+1; b <= 100; b++)
{
AskElrondIfCanBe(a,b);
}
}
Solve first, optimize later! :DQuote:
Originally posted by Simon666
I have an even more efficient one:
:D:D:DCode:for(int a = 2; a <= 100; a++)
{
for(int b = a+1; b <= 100; b++)
{
AskElrondIfCanBe(a,b);
}
}
I like them. They keep the code much cleaner, IMHO. ;)Quote:
Originally posted by SeventhStar
PS. you can remove these '{}' thingies
Commentary: P thinks Hmmm... I know the sum is 24, but that means the numbers could be (2,12), (3,8), or (4,6).Quote:
P. : "I can't find these numbers"
Commentary: S thinks Well of course he can't know it... since the sum is 11, the numbers could be (2,9), (3,8), (4,7), or (5,6). For each possibility, the product could be made from more than one possibility. For example, if it's (2,9), Mr. P would know the product is 18, meaning the numbers could be (2,9) or (3,6).Quote:
S. : "I knew it"
Commentary: P thinks If the numbers were (2,12) then S couldn't have known that, because the sum is 14 so the numbers could have been (3,11), and their product can only be factored one way. If the numbers were (4,6) then S again couldn't have known that, because the sum is 10 so the numbers could have been (3,7), and their product can only be factored one way. If the numbers were (3,8), then their sum is 11, giving S the possibilities (2,9), (3,8), (4,7), or (5,6), whose products can all be factored more than one way. Therefore, the numbers must be (3,8).Quote:
P. : "Then I have know what they are !"
Commentary: S thinks If it were (2,9), whose product is 18, Mr. P could have thought the numbers were (3,6), which would also have fit his statements; since there are two possibilities, how can he know which one? So it can't be (2,9). The same logic holds for (4,7) and (5,6), leaving (3,8) as the only working possibility. Therefore, the numbers must be (3,8).Quote:
S. : "Really ?! Then I have know as well !"
In conclusion, 3 and 8.
but why not 2 and 9
i thought of the same thing and 2 and 9 look the same to me
okay okay I get it... I forgot the last statement smy mistake
again and again I see that I'm as stupid as it gets...
My I'm a stupid **** :mad: :mad: :mad:
Nice idea. But are you sure that no other pair of numbers would fit to the same argumentation?Quote:
Originally posted by solarflare
In conclusion, 3 and 8.
That's true, I haven't proven that no other number pairs work. My reasoning was that if there were two possible answers, then the question would not have implied that there is a single solution. Therefore, I am assuming that there is only one solution. Since (3,8) works, that must be the solution.Quote:
Originally posted by gstercken
Nice idea. But are you sure that no other pair of numbers would fit to the same argumentation?
Not exactly. Of course the question only makes sense if there is a single solution. But if there would be other numbers fitting your argumentation, that would only mean that your argumentation cannot be the correct solution. In this case, there had to be a different kind of reasoning which would unambiguosy lead to a single pair of numbers. The fact that (3,8) "works" is true only in the context of your (possibly wrong) solution.Quote:
Originally posted by solarflare
Since (3,8) works, that must be the solution.
Don't misunderstand me, I am not saying that your solution is wrong. But to prove that it is right, it must be guaranteed that no other pair of numbers "works" in this reasoning.
There must be a flaw in the reasoning or in the question. Anyway, 3 and 8 was not the correct answer.
In the reasoning I have seen, there is one step I still don't understand, and I now think it might be wrong. But it was reducing the different possibilites to just one. It really seems that there are multiple solutions now. Or something is missing in the question, but I can't find what. :(
I guess you'll have to go for another question, and I'll try to find the missing information that implies there is only ONE solution.
If you say which part, I can either clarify it or realize my own mistake.Quote:
Originally posted by Elrond
In the reasoning I have seen, there is one step I still don't understand, and I now think it might be wrong.
That's not correctQuote:
Originally posted by solarflare
Commentary: S thinks If it were (2,9), whose product is 18, Mr. P could have thought the numbers were (3,6), which would also have fit his statements; since there are two possibilities, how can he know which one? So it can't be (2,9). The same logic holds for (4,7) and (5,6), leaving (3,8) as the only working possibility. Therefore, the numbers must be (3,8).