[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 433
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It's you, then you :)
No I'm not - I'm trying to show that even a great sage would not be able to cope with 138 combinations. Take at least a product of 42. It could be the numbers 2 and 21, 6 and 7, 3 and 14. A guy who has been told a product equal to a two-digit number is somewhat easy for him. Now let's look at the sums. 2+21=23, 6+7=13, 3+14=17. Having been given one of these sums, the person must decompose it into its summands. 23=2+21, 3+20, 4+19, 5+18, 6+17, and so on. There is no need to go far. I will now give you the sum and Alexei the product of the numbers. The same dialogue will occur between the two of you. If the product is two-digit, you will not be able to unambiguously name the original numbers. Shall we experiment? Well, to make the experiment clean, I'll pack the numbers into a locked text document and post it here on the forum. After your answers, I'll give you the password. The condition is that you do not tell each other the numbers.
I'm afraid that a dialogue like the one in the task won't work.
And let me give my solution (I do not pretend to correctness), and you personally estimate it.)
There's plenty of options like that, Abzascasc. 13, 15, for example. One simple, one composite.
13 you can't... 15 is 3 and 5, okay... but there's not that much in the 2-99 range. We got to narrow it down somehow.
Although... if someone who was told the product was 15, they'd give you the answer without the sum.
By the way, I showed drknn that the proposed option (P=75 and C=28) does not pass.
To Sage A I will now tell ... the product of these numbers.
to Sage B I will tell ... their sum".
A: "I cannot determine the numbers". Consequently, he has more than one way of decomposing the product into its factors.
B: "I knew beforehand that you wouldn't be able to solve the numbers. Hence B guessed that A has more than one pair of numbers.
A: "Then I know the numbers. Hence his opponent's criticism allowed sage A to discard the extra pairs of numbers (if he wasn't lying).
B: "Then I know too".
That's right, 75 and 28 worked as a contraption. They showed that if the presenter had conceived the pair 25 and 3, the problem would have no solutions. And I'm sure that maybe there is a solution. Maybe, but for that to be possible, sage A would have to have the only way to decompose the product into its factors. In that case he would have lied with his first statement. So he would have to get not the product in the ear, but the sum. In that case it would add up - the one who decomposes the sum into its summands would have to say that he really doesn't know the numbers and that would be true. As soon as B says that he foresaw it, A would guess that B has the product in his hand, which can only be decomposed into the sum of the factors. So among his pairs of numbers A would have to choose such a pair whose product has the only way to be decomposed into factors. This is how he recognizes the numbers. But even in this case the last answer of B would be a lie or a joke - like he pretended that until the last moment he did not know the only possible way of dividing the product into factors.
I'm telling you - the problem is not formulated correctly. Abzasc admitted that he didn't create it, he just copied it from another source. That's why there cannot be any claims against him. And most likely, someone once tried to solve this problem and then shared it with people, retelling it in his own words and not really thinking about the construction of a rigid wording of the conditions.
No, the solution is only personal to drknn if he wants it. It's a magnificent task, I haven't given up on it.
Okay. But again, I'll stress that I solved it my way. My buddy solved it by applying sets and got a different answer.
drknn, suggest a variant of sum and product (concrete) that will collapse the problem as incorrect. I can suggest: the sum should be odd (hence the product should be even). I've already proved this rigorously.
Also:
Б: «Я заранeе знaл, что ты не смoжешь опредeлить числа». Следовательно Б догадался что у А больше чем одна пара чисел.
It's not all a consequence. B already knew beforehand that A doesn't recognize the numbers, because he saw their sum and made sure that any decomposition of it into its summands gives at least one composite number. B thereby informs A that the sum can only be equal to one of the numbers 11,17,23,27,29,35,37,41,47,51,53,57,59,65,67,71,77,79,83,87,89,93,95,97. But now A manages to find out everything.
2 ValS: And both are following the script of the conversation of the wise men?
Abzasc admitted that he did not compile it - he simply copied it from another resource.
No, the solution is only personal to drknn if he wants it. It's a magnificent task, I haven't given up on it.
I agree - the solution in person - I didn't give up - I just almost lost my mind from the ambiguity of the problem's condition. So I had to go over and over what can't be. Tasks of this class are problematic tasks. They are also called creative tasks and are defined as a special class of tasks the solutions to which are not visible. In these tasks it is necessary to engage in creative search, narrowing down the range of possible solutions. These are tasks for applying justified hypotheses. I am interested to see the solution, because I have no strength to try to formulate the condition correctly. So that the problem has a real solution. It is good practice, because in life we see the problem and formulate the problem conditions ourselves. So I just got a five-point training today. I am satisfied. But, as they say, everything has to be in moderation. I am waiting for the solution in private.
?
Sorry, I misspoke. ValS suggested the task.