[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 441
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The wise men were told that the sum is 99 and the product is 2450. The solution is only 49*50. And A's first cue will be that he doesn't know. True, the second one won't say his line "And I am without you...".
OK, 97 and 2350 (the numbers are 47 and 50).
Explain. With a total of 29, yes, B still says his phrase. What's the problem?
You tell me this. How did you get to the end at P=100? I'm interested in the last stage.
The wise men were told that the sum is 99 and the product is 2450. The solution is only 49*50. And A's first cue will be that he doesn't know. True, the second one won't say his line "And I am without you...".
OK, 97 and 2350 (the numbers are 47 and 50).
Why? He (A) is given a product of, say, 30*30 = 900. He will not name them. The possible multipliers are (30,30) and (60,15).
But yes, you got me thinking a bit. The problem keeps getting more and more peculiarities. And how did these wise men calculate...?
Why? He (A) was given a product of, say, 30*30 = 900. He will not name them. The possible multipliers are (30,30) and (60,15).
But yes, you got me thinking a bit. The problem keeps getting more and more peculiarities. And how did these wise men calculate...?
Yeah. So did you. Looks like I found the upper limit wrong. I'm off to think.
To understand the essence of recursion, you have to understand the essence of recursion..... ... .. . :)
It's not clear to me at all where B gets the information the second time around (in the last line). Probably about the same as A in the penultimate line. Recursion, recursion again...
P.S. So, in the light of your remarks about large sums something is beginning to emerge.
It's not clear to me at all where B is getting the information the second time around (in the last line). Probably about the same as A in the penultimate line. Recursion, recursion again...
P.S. So, in the light of your remarks about large sums something starts to emerge.
Let's speculate.
1. Sage A knows the product P = X * Y, but does not know X and Y.
Hence:
P is such that it can be represented by more than a single pair of X and Y.
2. Sage B knows the sum C = X + Y, but does not know X and Y.
Hence:
C is such that it may not be represented by the only pair X and Y.
Thus the product of members of any of the pairs, has the property mentioned in item 1 (it follows from the phrase B).
(3) Of all the pairs, only one has the sum corresponding to p.2 (so the wise man A knows what this pair is).
(4) Among all the pairs the sum of their terms gives C, only the product of one has the property mentioned in (1).
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From here I can only see combinatorics.
A: "My number is not decomposable into two prime numbers".
B: "I know because my number is odd."
A: "Then I know the numbers."
B: "Wow, that's interesting. But you could only know this if you have the info that ALL other odd decompositions add up to more than 100. Then I know too..."
Possible answer: Product = 576 (= 3*3*2*2*2) Sum = 73 (64+9)
Numbers : 64 and 9
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It would seem that A could think that the sum could also be 51 (48+3). But then their product of 144 is an even number, and there is no way B could have said in the first remark that he KNEW it was not decomposable into 2 prime numbers... Since B killed this version with his first remark, A was able to solve the problem unambiguously and help B.
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That's how we communicate on the forum .......... so that ideas don't go to lazy fools...
;)
73 does not fit. If this number had been communicated to Sage B as a sum, he, having no information, could not deny the combination of 2 and 71, i.e. the one-digit decomposition of 2*71 = 142 into multipliers. 71 is prime.
Your paraphrase of phrase B is not quite accurate.
Lemma. For B to say his phrase "I knew without you that you wouldn't find a number", n. and e. that the sum communicated to him must be less than 100 and be represented as 2+complete_odd.
Try proving it.
I'm off to bed.