Well i just checked it for a 100th time and there are no more mistakes. Sorry! Really! :( :(
And I'll post the answer tommorow. If you want me to do it today just tell me...
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Well i just checked it for a 100th time and there are no more mistakes. Sorry! Really! :( :(
And I'll post the answer tommorow. If you want me to do it today just tell me...
I vote for tomorrow at least, as I'd really like to work out some options. However, I haven't been an active member of this thread, so I'd defer to the real participants. I still want to work this out, as I haven't had a good abstract algebra problem to practice on in some time... :)
Starting with G+BC=39, and B+GC=199:Quote:
Originally posted by galathaea
Now I brute force tested the endpoints and came up with two possible solutions: B = 1, G = 33, C = 6 or B = 1, G = 6, C = 33. I have some heuristics for choosing the first one (as solarflare has), but I still don't have a definitive proof for either one.
B+GC=160+G+BC
B+GC-G-BC=160
(G-B)(C-1)=160
(G-B)=160/(C-1) (Eq.1)
also, (again from the original 2 equations)
G+BC+B+GC=238
(G+B)(C+1)=238
(G+B)=238/(C+1)
Now prime factoring 238 gives
238=2*7*17
Since the sum of positive integers is a positive integer, (238/(C+1)) must be a positive integer, meaning C+1 is a factor of 238, meaning C+1=2,7, or 17, meaning C=1,6, or 16.
Letting C=1 makes Eq.1 invalid. Letting C=16 gives (from Eq.1) G-B=160/15, which is not a whole number. The difference of integers is an integer, so this cannot be.
This leaves only C=6. Solving for the other variables gives B=1, G=33. There is no other solution.
As for all the other random stuff in the problem, I have made no progress in the minutes I have spared to work on it. :(
Shame on you... but I'll accept your apology. I think the problem is equally hard with or without the typo ;).Quote:
Originally posted by SeventhStar
Oh my god!!! I'm sososososo sorry!!!!!! I've missed this. there is no 'hutar' it's 'gutar'....
I'm really sorry guys... My I'm a stupid ********. Sorry.
Unfortunately, the correct implication is that C+1 is one of the 8 divisors of 238 (1, 2, 7, 14, 17, 34, 119, or 238) and 33 is not then excluded. This makes me want to keep both options open, which is quite a pain!Quote:
Originally posted by solarflare
Since the sum of positive integers is a positive integer, (238/(C+1)) must be a positive integer, meaning C+1 is a factor of 238, meaning C+1=2,7, or 17, meaning C=1,6, or 16.
Oops!Quote:
Originally posted by galathaea
Unfortunately, the correct implication is that C+1 is one of the 8 divisors of 238 (1, 2, 7, 14, 17, 34, 119, or 238) and 33 is not then excluded. This makes me want to keep both options open, which is quite a pain!
I left you plenty of time to solve the problem and I am going to post the solve. None of you maneged to find a single number for sure... I'm dissapointed :p
So: here it is:
You will find that there are nine numbers shown:
uruk, gutar, ganal, barwon, omar, farok, babu, donze, hul
and these operators:
i m y o s t
Lets take these two numbers:
39 = ganalibarwonhul
199 = barwoniganalhul
(will call the base numbers G B and H for short)
so you noticed that
39+199 = G + BxH + B + GxH
238 = G(1+H) + B(1+h) = (G+B)(1+H) -> 238%(1+H) =0
then (1+H) is one of these numbers :
2 7 17 34 119 so H is one of these 1 6 16 33 118
also
199 - 39 = B+GxH-G-BxH
160 = G(H-1)-B(H-1) -> 160%(H-1) =0
then (H-1) is one of these numbers :
2 4 5 8 10 16 20 32 40 80 so H is one of these 3 5 6 9 11 17 21 33 41 81
six canot be it because then we'll get negative values fo G and B so hul is 33
then
39 = G + B*H -> 33 >B*H > 39 so b can only be 1
barwon is 1
199 = B+G*H = 1+x*33
ganal is 6
You will have to explain this one, because the system does allow 6 from all my calculations and all the calculations you show.Quote:
Originally posted by SeventhStar
six canot be it because then we'll get negative values fo G and B
And I can't wait for the operators explanation... :)
so lets look at
71 = urukigutarhul
71 = U + 33G
33G<71 so G is either 1 or 2 but barwon is 1 so
gutar is 2
that means that uruk is 5
106 = omarifarokhul
106 = O + 33F
33f < 106 so F is either 1 or 2 or 3. 1 and 2 are occupied so
farok is 3
and omar is 7
what we have by now:
barwon = 1, ganal = 6, uruk = 5, gutar = 2, farok = 3, omar = 7
Honestly men I thought that you'll get at least this far... :D
But I have a question for you. Do you want me to proceed with the solve or do you want to try to slove it by yourselves now?
Yes it's true... I've misscalculated :rolleyes:Quote:
Originally posted by galathaea
You will have to explain this one, because the system does allow 6 from all my calculations and all the calculations you show.
And I can't wait for the operators explanation... :)
But if hul was 6 there wouldnt be way to solve this would it...
Quote:
Originally posted by SeventhStar
Yes it's true... I've misscalculated
But if hul was 6 there wouldnt be way to solve this would it...
Do you see the contradiction? :)Quote:
Originally posted by SeventhStar
Honestly men I thought that you'll get at least this far...
Just put me out of my misery please. I want to see the single letters!Quote:
Originally posted by SeventhStar
But I have a question for you. Do you want me to proceed with the solve or do you want to try to slove it by yourselves now?
But galathaea (and with his permission, I'll say I helped ;)) proved that barwon=1!Quote:
Originally posted by SeventhStar
None of you maneged to find a single number for sure.
Hmmm, maybe I'll give this problem another chance, but I must admit, it was a huge let down that we were just supposed to assume that hul naturally was equal to 33.
With the small assumption that hul = 33 that SeventhStar made (it's a pretty serious assumption 7*, you should have had some clarification there in the problem), the solution comes easily with some thought.
Using the tools Simon and galathaea have provided (the organizations in their posts), I have solved the problem! The method is tedious, so I'll just list the translations.
Assume that xx, and zz are integers (doubled to avoid confusion with operators). Now
xxozz = xxszz = xx + zz - 22.
xxyzz = xxtzz = xx + zz - 44.
xxazz = zz - xx.
xxmzz = xxizz = xx + zz.
babu = 4, and donze =11 from observations made with the numbers 37, 233, and 242 involving the 'm' operator.
Using these solutions,
178 = gutaroganalhul
189 = gutaryomarhul
351 = farokiomarodonzehul
372 = farokiganalidonzehul
although there are many different ways to actually write these numbers.
And the final two questions:
barwonadonzesbarwonhulhul = 1077
omarydonzehul = 326.
These translations of the base numbers and operation symbols work for every number listed in the question, but nevertheless I await SeventhStar's confirmation.
Quote:
Originally posted by SolarFlare
I am currently guessing that each operator does the following:
blahoblahh (blah o blahh) = (some constant)*(blah)+(some constant)*(blahh)+(some constant), where the operator used (in this case, 'o') defines the three constants.
But... my solutions fit this pattern exactly! It just so happened that my above quote was a bit complex, because I intended it to cover all solutions in general. Again, I've been mislead!Quote:
Originally posted by SeventhStar
noooo! then it'd be impossible to solve!
I would say that you solved barwon=1! Your post was up while I was working on it, and though I tried real hard not to think about any of the subsequent posts while I was working on the problem, I'm sure it was there leading me.Quote:
Originally posted by solarflare
But galathaea (and with his permission, I'll say I helped ) proved that barwon=1!
And thank you, solarflare, for the solution. I have not had time to work on it since SeventhStar's (incomplete!) posts, but it was driving me a little batty! I finally see the light.
PS: the fact that 'm' = 'i' is alittle annoying, since I had been trying to work it out by assuming (yes, I see the prefix) that the operators were different...
Sory that I couldn't post yesterday but I got a pretty urgent job... Anyway I can explain everything now since the new issue of the magazine that I took the problem from came out today...Quote:
Originally posted by solarflare
But galathaea (and with his permission, I'll say I helped ;)) proved that barwon=1!
Hmmm, maybe I'll give this problem another chance, but I must admit, it was a huge let down that we were just supposed to assume that hul naturally was equal to 33.
There you can find the author's solve.
He sais that knowing that the system's idea is the forming of numbers in numerical intervals should've helped us.
because
39 = ganalibarwonhul actually means "6 after the first 33"
if hul was 6 and ganal was 33 that would've been "33 after the first 6" but that's ridiculous (maybe "3 after the 6th 6")
so that why it's 33 and not 6...
Also the question said that base numbers do not alterate morphologically but that's not said for the operators.
The 'after operator' or 'plus' if you prefer is i only between two consonant and 'm' if one of the surrounding lettrs is a vowel. that's also true for 's' and 'o' and 't' and 'y'
Not all of the base number were given, the autor sais and not all of the operators. But the system was really used by some indians in south america and is not just made up.
That's it. Now does anyone have a good quesytion or shall I proceed from my new magazine. (btw the Bulgarian magazine 'Math Plus Plus' is the best - payed comersial)
1. Gabriel 7
2. me :D 6
3. Simon 5
4. Saturno 4
5. Elrond, solarflare 3
6. John 2
7. dimm_coder, Xplorer, galathaea 1
solarflare, galathaea and simon get an extra point for being the only ones trying to answer this :D Dont be mad guys. I didn't know it would be so hard and controvresial...
Aaaarrrrggghhh. Somebody please stop this guy. He's going to drive everybody nuts and psycho dreaming about numbers and operators.Quote:
Originally posted by SeventhStar
That's it. Now does anyone have a good quesytion or shall I proceed from my new magazine.
be cool :cool: Simon! The problem I had in mind isn't from the 'Mathematical Linguistics' part.Quote:
Originally posted by Simon666
Aaaarrrrggghhh. Somebody please stop this guy. He's going to drive everybody nuts and psycho dreaming about numbers and operators.
But if you like you can post a probem
I am cool. It's chilly here.Quote:
Originally posted by SeventhStar
be cool :cool: Simon!
Oof.Quote:
Originally posted by SeventhStar
The problem I had in mind isn't from the 'Mathematical Linguistics' part.
No thanks.Quote:
Originally posted by SeventhStar
But if you like you can post a probem
I am not sure whether it is a good idea to post questions from a mathematical magazine, actually. The problems should be solvable by mere mortals like us, you know.
okay
who's gonna post then
Does this imply the question you had in mind was of the same difficulty level/time consuming level as the previous one? Otherwise go ahead.Quote:
Originally posted by SeventhStar
who's gonna post then
You have an empty bottle and a straw. How can you lift the botthe by touching it only with the straw and only touching the *inside* of the bottle?
I'll stick my finger in the bottle and lift it. Then with my other hand I'll throw away the straw.Quote:
Originally posted by Gabriel Fleseriu
You have an empty bottle and a straw. How can you lift the botthe by touching it only with the straw and only touching the *inside* of the bottle?
oops didn read carefully I'm wrong ofcourse
A few more info, just to be sure: the bottle is a *normal* bottle made of glass, of about 1 L -- no trick here.
The straw is like any straw you can get at MacDonalds. No glue, no heat, no other stupid tricks. And you're not in space, i.e. there IS gravity.
[EDIT]
For the solution, you can either make a drawing, or, if you have a digital camera, shoot a picture and post it. If no one finds an answer, I will post a photo :)
[/EDIT]
Can be bending the straw carefully? Something like this?
Yep, that's it. You actually have to bent the straw onl yonce. I will post a picture ASAP. :)Quote:
Originally posted by Doctor Luz
Can be bending the straw carefully? Something like this?
Your turn.
Hmmm, I don't think that would work with glass coke bottles and about any glass water bottle in Belgium as they end conically.
My turn? OK. Here is an easy one.
Do you remember the SeventhStar's question about a man with 12 balls ?
Then after solving his problem a friend of him came to his house carring a big DIN A0 piece of paper. A big strange closed figure
was painted with a pencil on the paper. This man must to calculate the total surface of this strange figure. He asked for aid to the SeventhStar's man. There were no rulers, no tape measure, and no drawing tools of any kind. But SeventhStar's man could calculate the surface of the strange figure. HOW?
I would say: weigh the DIN A0 piece, cut out the stupid thing and weigh the two pieces, if you know what the surface of a DIN A0 is you can calculate the surface of the figure.
Yes!
Very easy true?
Your turn.
It was easy but I didn't know I was allowed to cut it out (destructive method). I don't know one, if somebody has a hard one (not barwoniganalhul style), please post it.Quote:
Originally posted by Doctor Luz
Very easy true?
OK, here is a though one:Quote:
Originally posted by Doctor Luz
Your turn.
This morning I left home, I walked one kilometer north and then one kilometer west and finally walked one kilometer south and arrived back at my home. Can you tell me what location(s) my home is located and if you know I live on the northern hemisphere?
ANDQuote:
Originally posted by solarflare
I am currently guessing that each operator does the following:
blahoblahh (blah o blahh) = (some constant)*(blah)+(some constant)*(blahh)+(some constant), where the operator used (in this case, 'o') defines the three constants.
This is still bothering me. Is this now correct, are these indeed the operators? Then you have mislead us all SeventhStar by denying the operators where under the form solarflare mentioned. :rolleyes:Quote:
Originally posted by solarflare
Assume that xx, and zz are integers (doubled to avoid confusion with operators). Now
xxozz = xxszz = xx + zz - 22.
xxyzz = xxtzz = xx + zz - 44.
xxazz = zz - xx.
xxmzz = xxizz = xx + zz.
That's very obvious Simon!
Your house is in Gent, Belgium. Every body knows... ;)
But you must be rich if you have a house one kilometer long :p
:D :D :D
You got me there. ;) Additional info (spoiler?): it is not a location on a specific spot that is the answer, but multiple zones.Quote:
Originally posted by Doctor Luz
Your house is in Gent, Belgium. Every body knows... ;)
But you must be rich if you have a house one kilometer long :p
"Gent" must be the belgian name for the North pole, Simon.... :D
Nope thats not what these operators do. all indicate orderQuote:
Originally posted by Simon666
AND
This is still bothering me. Is this now correct, are these indeed the operators? Then you have mislead us all SeventhStar by denying the operators where under the form solarflare mentioned. :rolleyes:
Can't you at least give an overview of all numbers and operators now that nobody wants to spend time on that any more? :rolleyes::(
Nope, it is not the North Pole, close though. Very close. :DQuote:
Originally posted by Gabriel Fleseriu
"Gent" must be the belgian name for the North pole, Simon.... :D
Sorry i forgot about that... :D I thought I had...Quote:
Originally posted by Simon666
Can't you at least give an overview of all numbers and operators now that nobody wants to spend time on that any more? :rolleyes::(
there they are:
Simon: your house is placed somewhere on a circle with the centre at the North Pole and a radius of 1+1/(2*Pi) km.
Actually multiple circles are possible: 1+1/(2*Pi*n) km, with n=1,2,3,...Quote:
Originally posted by Gabriel Fleseriu
Simon: your house is placed somewhere on a circle with the centre at the North Pole and a radius of 1+1/(2*Pi) km.
But I'll accept it as correct. Your turn. Was it too easy?
Can you continue this seires?
t,o,f,o,f,n,t,s,....
Well, what about
t,o,f,o,f,n,t,s,f,t,f,e,n,s,n,t,t,t,...
Is that enough? :cool:
Guido
Well, we discussed about this kind of things before: you'll have to explain the logic too.Quote:
Originally posted by gstercken
Well, what about
t,o,f,o,f,n,t,s,f,t,f,e,n,s,n,t,t,t,...
Is that enough? :cool:
Guido
Oh, sorry, I forgot:
t,o,f,o,f,n,t,s,f,t... are the initials of
three, one, four, one, five, nine, two, six, five, three...
or
3,141592653...
:p
How the you-know-what-I-mean does one get the idea to look in that direction? Congratulations, I think it is your turn unless Gabriel had something else in mind (doubt it).Quote:
Originally posted by gstercken
Oh, sorry, I forgot:
t,o,f,o,f,n,t,s,f,t... are the initials of
three, one, four, one, five, nine, two, six, five, three...
or
3,141592653...
No objection. His turn.Quote:
Originally posted by Simon666
How the you-know-what-I-mean does one get the idea to look in that direction? Congratulations, I think it is your turn unless Gabriel had something else in mind (doubt it).
OK, here is mine:
Assume a torch which burns exactly one hour when lit at the end. No assumptions about the linearity of the burning process can be made, that is, the remaining length of the torch need not be proportional to the remaining time. How can you measure exctly 3/4h with *two* of those torches?