That is a question. The answer is you just asked something, so you are the asker. As I've solved your question, it is my turn. :D :p Nah, just kidding, it is Gabriel's turn.Quote:
Originally posted by saturno7
But... who is the asker right now??
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That is a question. The answer is you just asked something, so you are the asker. As I've solved your question, it is my turn. :D :p Nah, just kidding, it is Gabriel's turn.Quote:
Originally posted by saturno7
But... who is the asker right now??
:D :D :DQuote:
Originally posted by Simon666
That is a question. The answer is you just asked something, so you are the asker. As I've solved your question, it is my turn. :D :p Nah, just kidding, it is Gabriel's turn.
I'll wait for Gabriel's question and I'll try to think of one just in case I answer one today. :)
did you see this:Quote:
That is a question. The answer is you just asked something, so you are the asker. As I've solved your question, it is my turn. Nah, just kidding, it is Gabriel's turn
Quote:
Originally posted by Gabriel Fleseriu
Carry on people. I don't have a good riddle yet :)
hmm... Does Michael Jones have anything to do with our Brad :D ;) :)Quote:
Originally posted by sbrothy
If you like puzzles you should check what these guys are doing on m0rph's C++ forum:
http://www.m0rph.com/bbs/
I Don't know. I suspect not though.
While we wait Gabriel, Do you want me to ask you a new question?
We don't wait for Gabriel so go ahead! ;)
A mathematical entertainment.
I don't know if this is a correct translation but do you know what is a self-describing number?
A self-describing number is the number which its first digit from the left is the number of zeroes which has this number, the second digit is the number of "1" which has the number, the third digit is the number of "2"... and so on.
There are only 7 self-describing numbers (this concept has no sense for numbers with more than 10 digits)
For example
1210
2020
21200
Question:
Could you tell me wich is the only self-describing number with 10 digits.?
P.S. Posting the other 4 self-describing numbers is also welcome. ;)
6210001000
Is that right? :)
Ana.
Nice problem doctor....
Don't know the answer (in fact, I'm not sure I even understand the question) but I'm afraid I'm under pressure here to bow out of this thread.
It's a lovely idea but it's generating 10 - 15 emails an hour which I just can't deal with (at least, not if I want to keep my job!!)
Good luck, riddlers!
John
Yes, Congratulations.
You have been very fast.
Doctor Luz, didn't you forget to say that the sum of the digits must be the number of digits? :confused:
Ask us, Ana...
if this is your question I answer YES and get a point :) don't IQuote:
Originally posted by saturno7
Doctor Luz, didn't you forget to say that the sum of the digits must be the number of digits? :confused:
:D :D :D
That's true but is not necessary. :DQuote:
Originally posted by saturno7
Doctor Luz, didn't you forget to say that the sum of the digits must be the number of digits? :confused:
Score:
1. Saturno, Gabriel and Simon - 3pts
2. Elrond, John, solarflare - 2pts
3. me - 1pt
Asker: Saturno7
Ok, I will ask, but I'm still trying to think of the other 4 self-describing numbers.
Again a silly one, but I like it.
A man goes into a bar and asks for a dry martini with an olive and a plate (or a dish, I don't know how to say it). The barman gives both things to him and he eats the olive and pours the martini on the plate. Then the barman asks him: "you're a fireman, aren't you?".
How did he know? :)
But if it's not necessary all you have to do is add 0's at the end of the numbers you already posted... :confused:Quote:
Originally posted by Doctor Luz
That's true but is not necessary. :D
You are totally right. :o
Forget this, please, I've realized that it's not that easy. Ok, I'll keep on dealing with this while you try to think of my nice fireman... ;) :p :DQuote:
Originally posted by saturno7
But if it's not necessary all you have to do is add 0's at the end of the numbers you already posted... :confused:
mmmm wait a moment, No, you are not right
Then the first digit should change :D
See my post above... ;) :)Quote:
Originally posted by Doctor Luz
mmmm wait a moment, No, you are not right
Then the first digit should change :D
he saved the victim and poored the water ;)Quote:
Originally posted by saturno7
Ok, I will ask, but I'm still trying to think of the other 4 self-describing numbers.
Again a silly one, but I like it.
A man goes into a bar and asks for a dry martini with an olive and a plate (or a dish, I don't know how to say it). The barman gives both things to him and he eats the olive and pours the martini on the plate. Then the barman asks him: "you're a fireman, aren't you?".
How did he know? :)
is this so?
Nooops, it's a nice answer but I'm afraid it's not the right one. Anyway, do you consider that eating the olive is saving the victim? :DQuote:
Originally posted by SeventhStar
he saved the victim and poored the water ;)
is this so?
Are those really correct? I think it depends where you start from:Quote:
Originally posted by Doctor Luz
A mathematical entertainment.
For example
1210
2020
21200
0
10
1110
3110
132110
1113122110 10 digits
311311222110
1
11
21
1211
111221
312211
132221
11133211
31231221
13111213112211
111331121113212221
31232112311312113211
Doctor Luz, I've found them:
321000
42101000
521001000
The next one is the 10 digits self-describing number that I posted a while ago.
Now, does nobody know the answer to my question?. It's easy, I can assure you... :) :p :D
Ana.
ooh did the barman guess by the man's uniform
They ARE correct!Quote:
Originally posted by Simon666
Are those really correct? I think it depends where you start from:
0
10
1110
3110
132110
1113122110 10 digits
311311222110
1
11
21
1211
111221
312211
132221
11133211
31231221
13111213112211
111331121113212221
31232112311312113211
Your example has nothing to do with the initial question. The question was about a number self explanatory. In your example, it is a series, and therefore requires the beginning of the series.
A few things altogether:
- Codeguru is starting to go veeeery slowly again, I don't know if I'll be able to post much more from now on.
- Simon, I think you didn't understand what a self-describing number is, in the number you posted there are lot of 1's and their number don't coincide with the second digit.
- SeventhStar, YOU GOT IT!!!. Yes, it was the uniform the man was wearing. So it's your turn.
Ana. :)
yep CodeGuru is getting slow again verrrrrrrrrrrrrrry slllllllllllllllow;
so Score:
1. Saturno, Gabriel and Simon - 3pts
2. Elrond, John, solarflare, and me - 2pts
Asker: me
question:
you have twelve balls and a pair of scales. All balls are equal in size but one of them is different from the others in weight (you don't know if it's lighter or heavier). With 3(three) measurements with the scales determine which this ball is. I need a 100% secure way. with Maximut three measurements.
answers should be like pseudo code:
i put X balls. if these are heavier do this etc...
15 minutes to post a reply!!!!!Quote:
Originally posted by saturno7
Doctor Luz, I've found them:
321000
42101000
521001000
The next one is the 10 digits self-describing number that I posted a while ago.
Now, does nobody know the answer to my question?. It's easy, I can assure you... :) :p :D
Ana.
321000 NO
42101000 YES
521001000 YES
To Simon: Elrond is right, that's a serie of auto explanatory numbers. Maybe the problem is my bad english.
Oh ****, didn't read the task correct again. I was still thinking of this thread from a while ago.
Put 5 and 5 balls (1 measure)
From those whose weight is more put 2 and 2 (2 measures)
if the weight is the same the different ball is the other (finish)
if not compare the 2 balls wich had more weight before. (3 measures)
Oh, the first one was a typing mistake. I was trying to say 3211000. Is this one right? :)Quote:
Originally posted by Doctor Luz
15 minutes to post a reply!!!!!
321000 NO
42101000 YES
521001000 YES
To Simon: Elrond is right, that's a serie of auto explanatory numbers. Maybe the problem is my bad english.
Now where did I see that one before?Quote:
Originally posted by SeventhStar
Asker: me
question:
you have twelve balls and a pair of scales. All balls are equal in size but one of them is different from the others in weight (you don't know if it's lighter or heavier). With 3(three) measurements with the scales determine which this ball is. I need a 100% secure way. with Maximut three measurements.
answers should be like pseudo code:
i put X balls. if these are heavier do this etc...
Quote:
Originally posted by saturno7
Oh, the first one was a typing mistake. I was trying to say 3211000. Is this one right? :)
YES!
:D ;)
We alredy had such thread early and even had such task, but only about coins. :)
I think, that Simon remember about it ;)
Nevertheless U can have even 27 balls by this way.
9, 9, 9
3, 3, 3
1, 1, 1
There were elso enough more difficult task about coins there.
I don't know if the answer is right, but Doctor Luz has already answered in this thread, a few posts ago. Doesn't this mean that he gets a point and it's his turn? :)
Ana.
I did not read carefully, they are 12 not 10
then 6-6
From the 6 with more weight 2-2, if they have the same weight 1-1 from the other two, else 1-1 frome those which weight is bigger.
read the question CAREFULLY here it is again:
you have twelve balls and a pair of scales. All balls are equal in size but one of them is different from the others in weight (you don't know if it's lighter or heavier). With 3(three) measurements with the scales determine which this ball is. I need a 100% secure way. with Maximut three measurements.
Doctor Luz, your answers are not correct because you DON'T know if the "false" ball is heavier or lighter!
If you start working on the heavy load and if the "false" ball is lighter, you won't find it!
The result being that I still don't have a 100% working solution :(
Wow! Such a long time without an answer!
Will I get a point?...
:D :D :D :D
Ohh, yeh, I was not so carefully!
But nevertheless, We already discass it somewhere in October.
Something like this: We have 3 parts,
a - 4(a1,a2,a3,a4),
b - 4(b1,b2,b3,b4),
c - 4(c1,c2,c3,c4)
Then weight a and b.
1) = , so we have part c (4 balls), and it is not big problem to find what :)
2) <(or >) , for ex. a < b
This means that if the false ball in a - it weight some less, if in b - some havier.
then we have else 3 parts:
1(a1, a2, b1) 2(a3, a4, b2 ) and (b3, b4, c1 ) ( c1 - is a good ball )
weight 1 and 2.
if (1 == 2) then false ball in 3 and it is heavier (b1 or b2)
for ex.1 < 2 (or 1> 2 it is not the reason)
so we have 3 balls only - (a1, a2, b2),
b1, a3, a4 - are good, because othervise we have 2 > 1 ( remember that b - is heavier, a - lighter ).
so we have (a1, a2, b2)
then weight a1 and a2
if == , than b2 - false and it is heavier
a1<a2 we have a1 , because a - lighter
So, that is right solution :)
somehow it misses when ball is b1(heavier) i thinkQuote:
Originally posted by dimm_coder
Ohh, yeh, I was not so carefully!
But nevertheless, We already discass it somewhere in October.
Something like this: We have 3 parts,
a - 4(a1,a2,a3,a4),
b - 4(b1,b2,b3,b4),
c - 4(c1,c2,c3,c4)
Then weight a and b.
1) = , so we have part c (4 balls), and it is not big problem to find what :)
2) <(or >) , for ex. a < b
This means that if the false ball in a - it weight some less, if in b - some havier.
then we have else 3 parts:
1(a1, a2, b1) 2(a3, a4, b2 ) and (b3, b4, c1 ) ( c1 - is a good ball )
weight 1 and 2.
if (1 == 2) then false ball in 3 and it is heavier (b1 or b2)
for ex.1 < 2 (or 1> 2 it is not the reason)
so we have 3 balls only - (a1, a2, b2),
b1, a3, a4 - are good, because othervise we have 2 > 1 ( remember that b - is heavier, a - lighter ).
so we have (a1, a2, b2)
then weight a1 and a2
if == , than b2 - false and it is heavier
a1<a2 we have a1 , because a - lighter
So, that is right solution :)
I'm looking more carefully once more but you look too.
nope my mistake! :rolleyes:
youre right!
you win a point and get to ask a question!
Sorry my I've read it wrong :) ;)
Score
1. Saturno, Gabriel and Simon - 3pts
2. Elrond, John, solarflare, and me - 2pts
3. dimm_coder - 1pt
Asker: dimm_coder
(a1, a2, b1) < (a3, a4, b2 )Quote:
Originally posted by SeventhStar
somehow it misses when ball is b1(heavier) i think
I'm looking more carefully once more but you look too.
I this case b1 cannot be heavier, because we will have (a1, a2, b1) > (a3, a4, b2 ), but we have
(a1, a2, b1) < (a3, a4, b2 )
_______________________________
Ok, but if siriously, as I've mentioned above, I solved this task in October, when we had analog thread, now I've only remembereded it :)
So, don't know can I have point or no ;)
I can give my right to ask question for someone who has interesting question now :)
Because some interesting tasks I knew, I posted already there, and I think that, at least, Simon, Gabrial, Irona, Elrond can know solution for them
:)
If YOU had solved it before, have a point now...
another question.
Two freinds that haven't met since highschool meet accidently on the street. They havent seen or heard eachother from years. Here's a part of their conversation:
'How's life?'
'fine! Thank you! Do you know I have a daoughter?'
'no! What's her name?'
'like her mother's'
'Oh! And how old is the little Jenny?'
How did the friend know the daughter's name?
The friend is not a real friend and is messing around with the other guy's wife? :eek: :D ;) :pQuote:
Originally posted by SeventhStar
Two freinds that haven't met since highschool meet accidently on the street. They havent seen or heard eachother from years. Here's a part of their conversation:
'How's life?'
'fine! Thank you! Do you know I have a daoughter?'
'no! What's her name?'
'like her mother's'
'Oh! And how old is the little Jenny?'
How did the friend know the daughter's name?
sorry but NO :pQuote:
Originally posted by Simon666
The friend is not a real friend and is messing around with the other guy's wife? :eek: :D ;) :p
that way they would've at least heard of each other.
I got it! One is the two friends is a woman called Jenny... They aren't two men, maybe two women or maybe man and woman. :D :D