[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 440
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The task is unequivocally 'golden'.
For the Meta-Sages. ;)
Well I can't do the last step, MetaWizard. (0b 24 numbers, which are comma-separated, I'll tell you separately how.)
If P = 100, then only the numbers 4 and 25 come out. Product=100, Sum=29.
A: (100 = 2*50 = 4*25 = 5*20 = 10*10.) (Possible sums are 52, 29, 25, 20.)
B: (We know, we know, 29 is just a case of A's complete initial insanity.) "And I knew you couldn't do it without you."
A: (Yeah, so since he's so sure before I even opened my mouth, I should pick the sums from the set of 11,17,23,27,29,35,37,41,47,51,53,57,59,65,67,71,77,79,83,87,89,93,95,97. Great - only 29 fits. And only realised as 4*25.) " Know the Numbers.
B: (...don't know...where else info....) "Be-e-e-e-e."
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мудрецы знали что задача однозначно решаема
Heh. I don't know how to use this fact yet... They've been slipped specific numbers, so they're going around them. If they'd given P=72 and C=27, they wouldn't have got anything...
Mathemat:
1) ...... only 29......It's only realised as 4*25.)
2) 'I know the numbers'.
Not true. 20 (= 2*2*5*5) works too.
If it were only 29, then B would really have no more information.
B: (...don't know...where else info....)
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Heh. I don't know how to use this fact yet...
No, 20 is no good. Prove it - or will you do it yourself?
Hint: if the amount is 20, then B, who knows the amount, would not say his phrase "I knew it without you...".
A: (Yeah, so, since he's so sure before I even opened my mouth, I should pick the amounts from the set of 11,17,23,27,29,35,37,41,47,51,53,57,59,65,67,71,77,79,83,87,89,93,95,97. Great - only 29 fits. And only realised as 4*25.) "Know the numbers".
The green numbers are the ones you have to choose A from.
But B needs to be chosen from just four: 100, 99, 98, 97.
;)
Oops. Going to think about it.
P.S. Have you forgotten, by the way, that the product is not limited to 100, but can theoretically reach 2500?
No, 20 is no good. Prove it, or will you drive yourself?
Hint: if the amount is 20, then B, who knows the amount, would not say his phrase "I knew it without you...".
Prove it!
I've disproved it from the contrary, seemingly correctly, but it's very long. Wouldn't it be easier to find a spit in your proof?
// Besides, that would make the problem even more interesting for you.
Here is the proof specifically for 20.
20 = 13+7.
B, who knows the sum of 20, can also assume such numbers - 13 and 7. Both are prime. These are potential multipliers of the unknown product. In this case, on the first retort of A he will no longer say "I already knew that you are up to your neck in searching one-digit expansion", because it is the 13*7 expansion that is one-digit.
That is, with the sum of 20 and the number 91 itself, the expansion is one-digit.
If 1500 is divided by 4 (the maximum of one of the numbers), the resulting quotient is 375, well over 100 (10^2, the maximum product of the remaining two).....
So... Mnagavata! Mnagavata!!!Here is the proof specifically for 20.
20 = 13+7.
B, who knows the sum of 20, can also assume such numbers - 13 and 7. Both are prime. These are potential multipliers of the unknown product. In this case, on the first retort of A he will no longer say "I already knew that you will be up to no good with the one-digit expansion", because it is the 13*7 expansion that is one-digit.
That is, with the sum of 20 and the number 91 itself, the expansion is one-digit.
All right, I've won with 20.
But at 29 he knew for sure. There's no one-size-fits-all decomposition there.