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I'm going to try just one last time, Warp. What is your definition of "conceivable" and/or "inconceivable"? What makes one number conceivable and another number inconceivable? Edit: Also...
thelegendarymudkip wrote:
The first logarithm of googol is 100. I'd say the average person can comprehend 100 quite well. It's also perfectly easy to say how much water it is to weigh googol kilograms. Quite simply, it's a 100x100x100x100x100x100x100x100x100x100x100x100x100x100x100x100x100x100x100x100x100x100x100x100x100x100x100x100x 100x100x100x100x100x100x100x100x100x100x100x100x100x100x100x100x100x100x100x100x100x100 meter 50-hypercube full of water.
Professor Frink (explaining that Homer Simpson has fallen into the third dimension): "Well it should be obvious to even the most dim-witted individual, who holds an advanced degree in hyperbolic topology..."
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Scepheo wrote:
Your capacity to write down the definition of a number in a system of choice is a metric you yourself dismiss: after all, if it were one, the possibility of defining Graham's number would make it conceivable.
I think you are being intentionally obtuse. The size of googol is trivial to explain to the average person. Even the average person has some grasp of a concept like "a 1 followed by a hundred zeros". As I have said several times, it's not a question of the average person understanding the exact size of the number, but a notion of how large it is, in common terms that can be related to common concepts. And of course there are myriads of other ways you could describe the size of a googol (such as describing the diameter of a sphere of water made up of that many molecules, for instance.) Googol is very easy to describe, and it's very easy to compare it to other numbers of similar magnitude. Its magnitude can be grasped. Conceived. Googolplex is admittedly a bit harder to describe, but there nevertheless are relatively simple descriptions of it, like the several examples I quoted from wikipedia. However, Graham's number is so immensely large that it defies any such description. You just can't describe it as "a 1 followed by n zeros" (because you can't express n in a simple manner), or "n times the atoms in the universe" (again, because n can't be easily described) or anything like that. You simply can't compare Graham's number to anything, which means you can't get a grasp of how big it really is, compared to anything. You understand perfectly what I mean, yet you still argue against it. Why?
in which case it's pointless arguing with you.
How exactly did my semi-humorous post become a flamewar?
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Warp wrote:
And of course there are myriads of other ways you could describe the size of a googol (such as describing the diameter of a sphere of water made up of that many molecules, for instance.)
Such a sphere would rival the size of the visible universe, which the layperson has a very poor grasp of. Even the best physicists and cosmologists don't have a very good intuitive understanding for how large the universe is. And you still haven't defined "conceivable" or "inconceivable".
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Warp wrote:
I think you are being intentionally obtuse. [...] You understand perfectly what I mean, yet you still argue against it. Why? [...] How exactly did my semi-humorous post become a flamewar?
All three statements appear to me to indicate that you have completely missed the point of my post, something which is also illustrated by the arguments you bring up against me. Bobo the King keeps asking you for a definition of conceivable or inconceivable for a reason: unless you give reasons for why being able to
Warp wrote:
describe it as "a 1 followed by n zeros" (because you can't express n in a simple manner), or "n times the atoms in the universe" (again, because n can't be easily described)
means a number is conceivable, yet the definition of Graham's number does not count, we can't understand these reasons or argue against them. After all, all are just definitions of numbers in arbitrary systems, so where do you draw the line? What is "a simple manner"?
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Scepheo wrote:
yet the definition of Graham's number does not count, we can't understand these reasons or argue against them. After all, all are just definitions of numbers in arbitrary systems, so where do you draw the line? What is "a simple manner"?
Well, try to explain the magnitude of Graham's number in a simple manner that's easy to understand.
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Bobo the King wrote:
What is your definition of "conceivable" and/or "inconceivable"? What makes one number conceivable and another number inconceivable?
The definition of conceivable is subjective. Suppose it's objective. So, there must be the smallest inconceivable positive integer (mathematical induction). But then, it's quite strange to say that N is inconceivable but N-1 is conceivable, because if we can imagine how big n is, we can surely imagine n+1.
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Warp wrote:
Well, try to explain the magnitude of Graham's number in a simple manner
Scepheo wrote:
What is "a simple manner"?
Unless you actually give me something to work with, like a definition of what you consider to be a "simple" description of a number, I can't respond to your questions and arguments in any satisfactory way. I myself find the description for Graham's number fairly basic: the arrow notation isn't hard to understand and neither is the recursion it's defined with. Sure, I can't grasp the resulting number, but that goes for a googol too.
Amaraticando wrote:
The definition of conceivable is subjective. Suppose it's objective. So, there must be the smallest inconceivable positive integer (mathematical induction). But then, it's quite strange to say that N is inconceivable but N-1 is conceivable, because if we can imagine how big n is, we can surely imagine n+1.
Is it so strange to say that? On average, a human is capable of recognizing only 5 objects as being 5 objects. Even though 6 is only 1 more, by far most people will recognize it as two groups of 3 objects.
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Scepheo wrote:
Is it so strange to say that? On average, a human is capable of recognizing only 5 objects as being 5 objects. Even though 6 is only 1 more, by far most people will recognize it as two groups of 3 objects.
So what happens when they are looking at a prime number of objects? Such as 7, only 1 more than 6. It cannot be split into smaller group, are they incapable of recognising it as 7 objects? I think not.
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thelegendarymudkip wrote:
So what happens when they are looking at a prime number of objects? Such as 7, only 1 more than 6. It cannot be split into smaller group, are they incapable of recognising it as 7 objects? I think not.
Actually, the number is only 3 or 4 objects. And 7 would be recognized as 3 + 4, or 3 + 3 + 1 or what have you. If you still don't believe me, here's the Wikipedia article.
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Scepheo wrote:
Unless you actually give me something to work with, like a definition of what you consider to be a "simple" description of a number, I can't respond to your questions and arguments in any satisfactory way. I myself find the description for Graham's number fairly basic: the arrow notation isn't hard to understand and neither is the recursion it's defined with. Sure, I can't grasp the resulting number, but that goes for a googol too.
Actually I think you are demonstrating my point quite well. You can't come up with a simple, intuitive description of the number, which is what makes it so inconceivable.
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While doing some trigonometry today I came upon a problem I found interestingly worded: Translated to English it says: "Determine the exact value of tan22.5 by using the figure." I realize what the purpose of this problem is, I'm supposed to find that tan22.5 = DE/1 = AE/1 = DE = AE = CB - DB = ((sqrt of 2) - 1), by using various trigonometric identities. But then I thought, since they just want the exact form, why can't I answer like this: tan22.5 = tan22.5 (exact form) Now, this answer of course completely defeats the purpose of the problem, but can you really say that it's an incorrect answer?
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NordicAnomaly wrote:
But then I thought, since they just want the exact form, why can't I answer like this: tan22.5 = tan22.5 (exact form) Now, this answer of course completely defeats the purpose of the problem, but can you really say that it's an incorrect answer?
Perhaps a case of being a bit too literal here? :) Oh, and the reason why AE=CB-DB is because ∠ABC=45°, ∠ACB=45°, and ∠CDE=90°, so ∠CED=45°, and so AE=ED=CD=CB-DB.
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FractalFusion wrote:
NordicAnomaly wrote:
But then I thought, since they just want the exact form, why can't I answer like this: tan22.5 = tan22.5 (exact form) Now, this answer of course completely defeats the purpose of the problem, but can you really say that it's an incorrect answer?
Perhaps a case of being a bit too literal here? :)
Yeah, I suppose. The intended solution here is pretty clear after all. :) It just happened to remind me of a question on a math test which was something like; "Create an equation with the solution x = 5", and someone went ahead and answered "x = 5". While not exactly the kind of answer the teacher was looking for, they couldn't say that it was wrong. And really, why make it more complicated than it has to be? Shouldn't math be about the opposite - to solve problems as efficiently as possible? Edit: Btw, I just remembered a question that my teacher was unable to answer a few years back... maybe someone here wants to give it a shot? Here it is: "How many digits does the number 200^2010 have?" At this point I hadn't been introduced to logarithms, so it should be possible to solve without them.
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There's a story my high school maths teacher told me about our national exams (for graduating high school): there had been a question that was phrased "Can you draw a graph representing this function?", to which a student had answered "yes". In the end, they had to give him full points for the question.
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NordicAnomaly wrote:
Btw, I just remembered a question that my teacher was unable to answer a few years back... maybe someone here wants to give it a shot? Here it is: "How many digits does the number 200^2010 have?" At this point I hadn't been introduced to logarithms, so it should be possible to solve without them.
Without logarithms, huh? All I can come up with is a good approximation :D 200^2010 = (100*2)^2010 = 100^2010 * 2^2010 = 100^2010 * (2^10)^201 since 2^10=1024 is around 1000 we approximate ≈ 100^2010 * 1000^201 = 10^4020 * 10^603 = 10^4623 which has 4624 digits, which is close to the correct result 4626 (2010*log(200) rounded down and +1). (Other approximations could be 2^103≈10^31, 2^196≈10^59, 2^681≈10^205, 2^1166≈10^351, etc. but they are not useful here)
Warning: Might glitch to credits I will finish this ACE soon as possible (or will I?)
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NordicAnomaly wrote:
"How many digits does the number 200^2010 have?" At this point I hadn't been introduced to logarithms, so it should be possible to solve without them.
When we're trying to find the number of decimal digits b^a has, unless b = 10^c, the closest we can get are approximations like the one Masterjun gave. Of course, that's excluding calculating 200^2010 and counting the digits. This is because to find the number of digits, we essentially need to find the log. The way to do this is log(b)*a, since now we have (10^]log(b))^a as a representation of b^a. One of the simpler rules of exponents is that (d^e)^f = d^(e*f). Since we're finding a logarithm in base 10, d = 10. We now need to represent b as (10^e). This is 10^log(b), by definition of logarithms. So now (10^log(b))^c = 10^(log(b)*c). From this, round down and add 1. It's not possible to do unless you can calculate logarithms to a reasonable degree of accuracy. This is (for everyone, I'd assume) only possible if b is already a power of 10. So no, it isn't possible to solve without logarithms unless you're a computer or b is a power of 10.
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Thank you both for the explanations. It's kinda weird, because this question came from a book used in the first year of Swedish upper secondary school, while logarithms aren't introduced until the second year. So either it's an oversight, or you're basically supposed to figure out how to use logarithms by yourself.
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Which 6 countdown numbers would give the greatest possible range of attainable answers?
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This is what Flip is referring to. (Note: The numbers on the cards are 25, 50, 75, 100, and two each of 1-10.)
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I heard this one on Car Talk a little while back. Suppose three guys, Alex, Bill, and Charlie, go out fishing one day. They get up early, pack their stuff up onto the boat, and head out to sea. They spend all day fishing, catching some number of fish, i.e., more than zero. They get back to dock late that night, and decide to just sleep on the boat and wake up early in the morning and go home. In the middle of the night, Alex starts to feel sick and decides he'll go home to have access to some sort of medicine. So he gets up, goes to the fish tank, and counts the number of fish. After he's counted the fish, he finds out that there's no way the fish can be divided into three equal parts, so he throws a fish overboard, and takes a third of the fish home, without waking up the others. Sometime later, Bill gets homesick, and decides that he's going to leave early to be with his wife. So he goes to the fish tank, counts the fish, and notices that the number of fish in the tank can't be separated into three equal groups. So he throws a fish overboard, and takes home a third of the remaining fish, not knowing that Alex has already come through and done the same. Morning comes, and Charlie thinks he's up first. Ready to get home, he goes down to the tank, counts the fish, sees that they can't be separated into thirds, throws a fish overboard, and takes a third of the fish home, without knowing that Alex and Bill have already done the same and aren't on the boat. So, after all this is done, what is the fewest number of fish that could have been in the fish tank initially for this situation to have happened? (Bonus points if you can give me the general solution)
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DiscoRico wrote:
I heard this one on Car Talk a little while back. Suppose three guys, Alex, Bill, and Charlie, go out fishing one day. They get up early, pack their stuff up onto the boat, and head out to sea. They spend all day fishing, catching some number of fish, i.e., more than zero. They get back to dock late that night, and decide to just sleep on the boat and wake up early in the morning and go home. In the middle of the night, Alex starts to feel sick and decides he'll go home to have access to some sort of medicine. So he gets up, goes to the fish tank, and counts the number of fish. After he's counted the fish, he finds out that there's no way the fish can be divided into three equal parts, so he throws a fish overboard, and takes a third of the fish home, without waking up the others. Sometime later, Bill gets homesick, and decides that he's going to leave early to be with his wife. So he goes to the fish tank, counts the fish, and notices that the number of fish in the tank can't be separated into three equal groups. So he throws a fish overboard, and takes home a third of the remaining fish, not knowing that Alex has already come through and done the same. Morning comes, and Charlie thinks he's up first. Ready to get home, he goes down to the tank, counts the fish, sees that they can't be separated into thirds, throws a fish overboard, and takes a third of the fish home, without knowing that Alex and Bill have already done the same and aren't on the boat. So, after all this is done, what is the fewest number of fish that could have been in the fish tank initially for this situation to have happened? (Bonus points if you can give me the general solution)
If I'm not horrifically mistaken, the easiest way to go about this is to think about it backwards. The number of fish charlie sees when he wakes up can be defined as (3n+1) (where n is a whole number, since we're not chopping up fish here), since he chucks 1 fish and is able to take an even third of the remaining fish. For this amount to be left after bill tossed a fish and took a third, we consider that bill left 2/3 of the fish that remained after subtracting a fish. That means that we can write the number of fish he saw as ((3/2)(3n+1))+1 fish when he woke up. Following the same logic, Alex must have seen ((3/2)(((3/2)(3n+1))+1))+1 fish when he woke up. At this point you can do some simple plugging and chugging with n as a whole number until the end result is also a whole number. The smallest value of n that seems to work is 3. This means that Charlie saw 10 fish when he woke up (he tossed 1 and took 3), Bill saw 16 fish when he woke up (he tossed 1 and took 5), and Alex saw 25 fish when he woke up (he tossed 1 and took 8) (this might be wrong, I just did this on a pad of scratch paper at work) (edit: thanks for the interesting distraction from work! Boring week =P)
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Sounds exactly like the newest Numberphile topic: https://www.youtube.com/watch?v=U9qU20VmvaU
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Flip wrote:
Which 6 countdown numbers would give the greatest possible range of attainable answers?
Looking at just the numbers allowed within the game (3-digit numbers), the numbers 1, 6, 7, 9, 10 and 100 allow you to make every number. Disregarding validity within the game, the numbers 5, 8, 9, 50, 75 and 100 win, allowing a whopping 32644 possible numbers to be made. If you (or anyone else) like, I could make files listing all these numbers and a possible way to make them. EDIT: Actually, no need to ask. You can download the first set here, and the second set here.
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Warp wrote:
Sounds exactly like the newest Numberphile topic: https://www.youtube.com/watch?v=U9qU20VmvaU
... Indeed, it does. Although, as I said, I got mine from a puzzler on Car Talk.
z1mb0bw4y wrote:
At this point you can do some simple plugging and chugging with n as a whole number until the end result is also a whole number. The smallest value of n that seems to work is 3. This means that Charlie saw 10 fish when he woke up (he tossed 1 and took 3), Bill saw 16 fish when he woke up (he tossed 1 and took 5), and Alex saw 25 fish when he woke up (he tossed 1 and took 8)
That is correct! The general solution to this problem is fishinitial = 27*n - 2, for all integers n. Extending this to the monkey-coconuts problem, it appears that for problems of this exact structure, for x iterations, the first solution is of the form objectsinitial = xx - x + 1 for all integers x.
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I was recently watching an episode of Yu-Gi-Oh! S01E21 Double Trouble Duel (Part 3). After Yugi and Joey defeat Para & Dox in a duel. They get a truth liar teller riddle where there is two choices. One door leads to freedom; the other leads to being lost forever. Joey asks what most people would ask i.e. Show me your door? That way the liar would have to point to the truth tellers door. However, what I find hard to figure out is how Yugi is able to determine that both of them are lying, and that both doors a lead to the same terrible fate.