[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 443
You are missing trading opportunities:
- Free trading apps
- Over 8,000 signals for copying
- Economic news for exploring financial markets
Registration
Log in
You agree to website policy and terms of use
If you do not have an account, please register
1. The statement of the recipient of the work: "I don't know the pair of numbers conceived".
We rely on the implication formula. (a → b). Let us assume that the expression ¬a reads as "not-a" and is a logical operation of negation of the truth of the variable a. Therefore the statement of the first sage for an outside observer should be understood as follows:
If a product is decomposable into multipliers in the only way (a), then Sage A knows the multipliers (b). The sage A rejects the fact that the factors are known (b). Hence, the product obtained by Sage A is not decomposable into factors in a unique way (¬a). [(a → b)&(¬b) => ¬a] It follows directly that by telling sage B the phrase that he does not know a pair of numbers, sage A stated: "The product whispered in my ear by the conceiver is decomposable into multiples in more than one way". So, the information that Sage A gave to Sage B is "I cannot decompose the resulting product into its denominators in one way". Or like this: "The product is decomposable into the factors in more than one way".
2. The statement of the recipient of the sum: "I knew you would answer this way".
In order that Sage B could have foreseen that the product is decomposable into its factors in more than one way without Sage A's answer, he had to understand from the expansion of the sum that the product of any pair of summands cannot be decomposed into factors in more than one way. Now start discarding the variants which contradict this thesis. Take the numbers 2 and 2. The product is decomposed by the single method. So it is not 2 and 2. Take a pair of numbers 2 and 3. Product = 6 can be solved only as 2*3. It means that it is not 2 and 3. Take 2 and 4. Product = 8 decomposes only as 2*4. Then it is not 2 and 4. Continuing in this way, we find the product = 12. This is decomposed into 4*3 and 6*2. So, assumption #1: Sage A got product = 12. If assumption #1 is true, then the phrase "I knew you would answer this way" is true.
Now let's see what the sum is equal to. The numbers are 7 and 8.
Shit, the phone rang, I've got to go. I can't go on with the reasoning, although it's so rigid that you can't escape - it's bound to lead us to the right conclusion. Sorry for running away, but I don't want to lose my train of reasoning either. Therefore, I am writing back here and taking my leave - this problem has struck me in particular!
Let's formalise it.
With the third remark ("Then I know the numbers") A informed B that the information in B's remark "I knew in advance that you could not determine the numbers" was enough to solve the problem.
This was enough for B to solve it, too.
--
Is that clearer? I didn't say anything new, I just spelled out the content of the messages.
Let me try to rephrase it one more time.
1. A: My product consists of more than two factors.
2. B: My sum is only decomposed into such differences that at least one of the resulting two numbers is composite.
. . . . By the way, as you can guess, it is odd and , as you know, less than 100.
3 A: Hmm. This information lets me find the only two factor product which satisfies the restrictions of the problem.
4. B: Yep. I have only one variant of the expansion of the sum which allows you to deduce the solution from the information you have.
--
Does that make more sense?
I think I've found an option.
П=486
С=87
a=81
b=6
I can prove the logic of the dialogue at these numbers, though it's a bit long. Try to disprove it better. It will be easier.
If you can't, I'll explain how I found (my logic) and try to prove the uniqueness of the solution (or disprove it).
// If we disprove uniqueness, it will not make the Wise Men stupider. In their problem uniqueness is present in every case.
// It is absent only on the meta-level (or present too, if we prove it) - in observers, to whom this problem is offered now.
Well, let's get started.
А: ("486 = 2*343 = 3*162 = 6*81 = 9*54 = 18*27. Bummer. Probable amounts - 87, 63, 45." I can't. (Telepathically: "You won't get anything from me, freeloader.")
[Some information A did tell B - but it's too little, especially since his subsequent comment B further refines it, narrowing the search. Probably, in this conversation scenario the information from A is simply useless. He could have kept quiet altogether].
B: ("Sum of 87 = 2+5*17.") (Telepathically: "So what if you're a freeloader? And you are impotent, and you can see it at once. Fuck it, I'll take pity on you a little, you miserable.") I knew you couldn't do it without you.
[B reports for A that the sum of the numbers is 2+ odd_component.]
A: ("Yeah, now I know the likely amounts. Which of my probable sums are those numbers? 87 - yes, 63 - no, 45 - no. That's it, problem solved.") I know the numbers. (Telepathically: "You've got it bad, though. Still a freeloader. Now work hard.")
[And now tells B that of all possible sums only one is " 2+ odd_component".]
B: (Immediately telepathically: "Fuck, you're an asshole. I still have a lot of options. I wish I had a supercomputer...") Boo.
_________________________________
MetaDriver, help me out!
Yes, I can see that in principle B can try to calculate. But it does come out a bit long. He has to go through dozens of variants.
Let's go. B has a total of 87 and the information that A got the only solution. And we really have to work hard.
Let's write out the possible amounts at once: 11,17,23,27,29,35,37,41,47,51,53,57,59,65,67,71,77,79,83,87,89,93,95,97.
87 = 2+85. The product is 170 = 2*85 = 5*34 = 10*17. The probable sums that impotent A would go through for this P are 87, 39, 27. The solution is not singular (the two choices are 87 and 27, not one).
87 = 3+84. П=252 = 2*126 = 3*84 = 4*63 = 6*42 = 9*28 = 14*18. Possible sums are 87, 67, 48, 37, 32. Not singular.
87 = 4+83. П = 332 = 2*166 = 4*83. Possible sum is singular! The numbers are 4 and 83. MD, something doesn't work with the stone flower. Looking further.
87 = 5+82. П = 410 = 2*205 = 5*82 = 10*41. Possible sums are 87, 51. Not a single one.
87 = 6+81. П = 486 = 2*343 = 3*162 = 6*81 = 9*54 = 18*27. The probable sums are 87, 63, 45. The solution is the only one again! But the numbers are yours, i.e. 6 and 81.
Already now B with his last line will not be able to say that he knows the numbers too.
Mathemat:
_________________________________
MetaDriver, help me out!
Yes, I can see that in principle B can try to calculate. But it does come out a little long. He'd have to go through dozens of options.
I've brute-force it. Took me about 12 to 15 minutes.
There are only 43 numbers (pairs) to check. Go for it. !
--
I'm not a sadist. I'm just trying to make you happy. There's still a lot of beauty in there when you test it. But it seems to go all the way through.
See there, on the previous page. I found two solutions. Bummer. Check also (4.83). The only solution comes out there too.
It's not hard to code a check for given P and C, the calculations are simple. The most important thing is to organize a competent search for variants. Is it better to search for them - by given numbers or by P and C?
Well, do we have the right to ask ValS about two solutions he has?
See there, on the previous page. I found two solutions. Bummer. Check also (4.83). There, too, the only solution comes out.
4 and 83 does not work - then A would immediately without question give the correct answer, because he knew that the other two factorizations of 2*166 are greater than 100.
Bleh... ;-Р
See there, on the previous page. I found two solutions. Bummer. Check more (4.83). The only solution comes out there too.
It's not hard to code a check for given P and C, the calculations are simple. The most important thing is to organize a competent search for variants. Is it better to search for them - by given numbers or by P and C?
So, dowe have the right to ask ValS about the two solutions he has? Check it out...
I suggest that at the end (after finding an analytical solution) we should end it, but in a nice way. So that there are two mutually recursive procedures, imitating a dialogue of sages. I already have a draft.
strongly against it zpt offering to finish it zpt we're already on the way tcc
Be.... ;-)
Boo... ;-Р
OK, let's slow down with the answer from ValS. ValS, don't tell me the answer!!!
Next. Keeping the allowable amounts in front of us: 11,17,23,27,29,35,37,41,47,51,53,57,59,65,67,71,77,79,83,87,89,93,95,97.
87 = 7+80. П=560 = 2*280 = 4*140 = 5*112 = 7*80 = 8*70 = 10*56 = 14*40 = 16*35 = 20*28. The probable sums are 87, 78, 66, 54, 51, 48. The solution is not unique.
87 = 8+79. П=632 = 2*316 = 4*158 = 8*79. The probable sum is 87. The solution is singular, but is ruled out by the first comment of A.
87 = 9+78. П=702 (=27*13*2)= 2*351 = 13*54 = 26*27. The probable sums are 67, 53. The solution is not singular.
87 = 10+77. П=770 (=2*5*7*11) = 2*385 = 5*154 = 7*110 = 10*77 = 11*70 = 14*55 = 22*35. The probable sums are 87, 81, 69, 57. The solution is not unique.
87 = 11+76. П=836 (=2*2*11*19) = 2*418 = 4*209 = 11*76 = 19*44 = 22*38. The probable sums are 87, 63, 60. The solution is unique and is not disproved by the first retort A! The numbers are 11 and 76.
I think we're screwed after all. Check the green pair.