[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 432
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By the way, not a word about the fact that the product is less than a hundred in the problem ))
So now you're getting wise).
It's always on a first-name basis.)
No, I'm not - I'm trying to show that even a great sage couldn't handle 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 don't tell each other the numbers.
You don't say, do you? Well, let's simulate the situation - let's play it open. I tell you the sum = 28. You decompose it into its summands: 26+2 25+3 24+4 You have no other options, because their product is more than a hundred. I give Alexei the product of 75. He decomposes it into its factors: 25*3 5*15. You have three choices, Alexei has two. The dialogue does not allow you to exclude the non-working ones. The task is a failure for both of you. Neither negotiation helped.
Prove me wrong if I am wrong!
I don't understand the question, Abzasc.
2 drknn: OK, let me be A. I know that the product of 75 = 3*5*5. I say the first line. "I don't know the numbers."
Let Valery know the sum, 28. He knows about Goldbach's hypothesis (it is exactly verified for numbers less than 100 :) ) and sees that 28 = 11+17. He cannot say his line that he "knew beforehand" because the numbers 11 and 17 interfere with him, they are both prime.
The conversation has gone the wrong way. P=75 and C=28 do not roll as a solution.
Shall we play some more, drknn? It's useful: now something will become clear to you.
When you can only divide integers without a remainder by one option... 9 you can, 7 you can't...