How Can We Make 100! Divisible by 12^{49}?

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In summary, the conversation was about a Q&A game where one person asks a math question and others try to answer it. The first correct answer gets to ask the next question. One of the questions was about finding the least number that must be multiplied to 100! to make it divisible by 12^{49}. The correct answer was 12^{49} / 100!. There was also a question about why mathematicians often forget to specify that they require a whole number solution.
  • #246


snipez90 said:
If a group G is the set theoretic union of a family of proper normal subgroups each two of which have only the identity in common, then G is abelian.

Unless I'm missing something, the center can non-trivially intersect at most one of the normal subgroups. If it didn't, then either the center is trivial or two of them have more elements in common. Either way, how can G be abelian? Unless there were only one...
 
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  • #247


snipez90 said:
If a group G is the set theoretic union of a family of proper normal subgroups each two of which have only the identity in common, then G is abelian.

Let the normal subgroups be labeled [itex]G_i[/itex] where [itex]i \in \mathcal{I}[/itex]. Suppose that [itex]a \in G_i, b \in G_j,\ i \neq j[/itex]. Then by the normality of G_i and G_j, [itex]aba^{-1}b^{-1}\in G_i \cap G_j = \{e\}[/itex], so [itex]aba^{-1}b^{-1}=e[/itex] and thus a and b commute.

Now, suppose that [itex]a, b\in G_i[/itex]. Since G_i is a proper subset of G, we have that there is some element [itex]c \in G \setminus G_i[/itex]. Let G_j be the normal subgroup containing c. By the previous paragraph, c commutes with both a and b. Consider the elements ac and bc. If these elements commute, then we have abc² = acbc = bcac = bac², and so ab = ba. So now suppose that ac and bc do not commute. Then by the first paragraph they must both lie in a common subgroup G_k. If k=i, then we would have both a and ac in G_i and hence [itex]c\in G_i[/itex], which contradicts our choice of c. So k≠i. But then [itex]ac(bc)^{-1} = acc^{-1}b^{-1} = ab^{-1} \in G_k \cap G_i = \{e \}[/itex], so [itex]ab^{-1}=e[/itex] and thus a=b, so a and b trivially commute. Thus G is abelian.
 
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  • #249


The statement that "since there are an infinite number of circles in fig. 1 and fig. 3 the total area of all circles of both figures approach equality" is blatantly false. The total area of the circles is not even close to being equal in fig. 1 and fig. 3.
 
  • #250


Care to explain?
 
  • #251


Niivram said:
Care to explain?

I wouldn't expect Citan to explain, as this looks very similar to the https://www.physicsforums.com/showthread.php?t=450364" thread.

Citan is correct, point 2 is false. And I'm not even a mathematician. As a matter of fact, I cannot understand a single thing in this entire thread, except that point 2 is false.
 
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  • #252


Yeah, I'm convinced that Point 2 is wrong. But I'm still confused. You're saying, if you have an infinite number of small circles, and an infinite number of big circles, the area isn't the same right?
 
  • #253


Niivram said:
Yeah, I'm convinced that Point 2 is wrong. But I'm still confused. You're saying, if you have an infinite number of small circles, and an infinite number of big circles, the area isn't the same right?

As I said, I am not a mathematician, and got way over my head linguistically in the troll physics thread. I will not make the same mistake again.

I also believe you've broken the rules of the thread:

Gokul43201 said:
A Q&A game is simple: One person asks a relevant question (it can be research, calculation, a curiosity, something off-the-top-of-the-head, anything ... as long as it's a math question) and other people try to answer. The person who posts the first correct answer (as recognized by s/he who asked the question) gets to ask the next question, and so on.

Let me start this off with a simple number theory problem :

What is the least number than must be multiplied to 100! (that's a factorial) to make it divisible by [itex]12^{49} [/itex] ?

(throw in a brief -couple of lines or so- explanation with the answer)

You need to answer the last question before you can ask your own.

And I'm afraid I'm going to have to unsubscribe from this thread. I read the post(#247) before your original, and can't figure out if it's a question or an answer. :bugeye:

Ciao!
 
  • #254


hehe, Never thought of the rule : D. Sorry.

Ciao!.
 
  • #255


Gib Z said:
O well since I guess I got the last one, I'll just ask again what post 169 asked: How can one find the surface area of a ring?
What do you mean by a "ring"? A torus?
 
  • #256


I may have, but I have a better question now.

If [itex]a_n[/itex] is a monotone decreasing sequence such that [itex] \sum a_n [/itex] converges, show [itex]na_n \to 0[/itex], and then make the generalization that a measure theorist would make.
 
  • #257


Gib Z said:
I may have, but I have a better question now.

If [itex]a_n[/itex] is a monotone decreasing sequence such that [itex] \sum a_n [/itex] converges, show [itex]na_n \to 0[/itex], and then make the generalization that a measure theorist would make.
Well, the proof is quite easy: We must have [itex]a_n\ge 0[/itex] for all [itex]n[/itex], for otherwise we would have a series in which all but finitely many terms are smaller than some fix negative number, and then the series would diverge to [itex]-\infty[/itex]. Assume, to get a contradiction, that [itex]na_n \to 0[/itex] does not hold. This means that there is an [itex]\epsilon>0[/itex] and arbitrary large [itex]m[/itex] such that [itex]ma_m\ge\epsilon[/itex].
Now, there is an [itex]N[/itex] such that [itex] \sum_{n=N+1}^\infty a_n <\epsilon/2[/itex]. Then, we can choose an [itex]m\ge 2N [/itex] such that [itex]ma_m\ge\epsilon[/itex]. Since [itex]a_n[/itex] is monotone decreasing, it folllows that [itex]a_n\ge\epsilon/m[/itex] for all [itex]n\le m[/itex]. Hence, [itex] \sum_{n=N+1}^m a_n\ge (m-N)\epsilon/m \ge\epsilon/2[/itex], which is a contradiction. Thus, [itex]na_n \to 0[/itex].

Generalization? The only one that comes to my mind is: Let [itex]\mu[/itex] be a positive measure on the real interval [itex][a,\infty)[/itex], ant let f be a [itex]\mu[/itex]-integrable on [itex][a,\infty)[/itex] (i.e. the integral over the entire interval is exists and is finite) and monotone decreasing. Then [itex]xf(x)\to 0[/itex] as [itex]x\to\infty[/itex].
 
  • #258


Niivram said:
Care to explain?

The sum of the areas of the circles in figure 1 is bounded by the area of the triangle containing them, which is obviously far less than the area of the central circle in figure 3. You could probably show that the areas of the triangle in fig 1 and the squares in figs 2 and 3 are the least upper bounds for the summed areas of the circles in those respective figures, thus...

fig 1 < fig 2 < fig 3
 
  • #259


I got a challenge for you... This question is answered in two different ways from two different professionals with different backgrounds. I asked a Physics Professor and a Mathematics Professor a question and got two different answers. But, isn't it true there is only one truth in answering a simple question such as this one? Here it is- if I take a distance or an object and cut it perfectly and half, then take one of those halves and cut it perfectly in half, again. And repeat this over and over again. What will happen eventually to the distance or thickness of the object ? The reason I brought this to your attention is because it appears that you pride yourself and/or you have a good understanding of physics, I am assuming. Good luck, I would love to hear your response to this question.
 
  • #260


Inventor, what do you mean with "eventually"? Do you mean that you actually cut an infinite number of times, or just that you cut a finite, but arbitrarily large number of times?
 
  • #261


Inventor 4U2 said:
I got a challenge for you... This question is answered in two different ways from two different professionals with different backgrounds. I asked a Physics Professor and a Mathematics Professor a question and got two different answers. But, isn't it true there is only one truth in answering a simple question such as this one? Here it is- if I take a distance or an object and cut it perfectly and half, then take one of those halves and cut it perfectly in half, again. And repeat this over and over again. What will happen eventually to the distance or thickness of the object ? The reason I brought this to your attention is because it appears that you pride yourself and/or you have a good understanding of physics, I am assuming. Good luck, I would love to hear your response to this question.

Can you elaborate? The question is very unclear, which may explain why you received two different answers. What do you mean by "what will happen to the thickness of the object"? Are you asking what will happen if you halve its length an infinite number of times?
 
  • #262


Inventor 4U2 said:
I got a challenge for you... This question is answered in two different ways from two different professionals with different backgrounds. I asked a Physics Professor and a Mathematics Professor a question and got two different answers. But, isn't it true there is only one truth in answering a simple question such as this one? Here it is- if I take a distance or an object and cut it perfectly and half, then take one of those halves and cut it perfectly in half, again. And repeat this over and over again. What will happen eventually to the distance or thickness of the object ? The reason I brought this to your attention is because it appears that you pride yourself and/or you have a good understanding of physics, I am assuming. Good luck, I would love to hear your response to this question.

nothing, it remains the same..it`s a question similar to the question "which is heavier?" a kilo of cotton or a kilo of nails..lol
 
  • #263


Inventor 4U2 said:
But, isn't it true there is only one truth in answering a simple question such as this one?
Definitely not. Most "simple questions" tend to be the hardest because the terms used are open to interpretation.
if I take a distance or an object and cut it perfectly and half, then take one of those halves and cut it perfectly in half, again. And repeat this over and over again. What will happen eventually to the distance or thickness of the object ?

A more interesting problem would be catalog as many possible answers as we can think of!
  1. An infinite number of divisions of space ("distance") is possible, and the limit is zero.
  2. You are dividing space, but don't have infinite time, so it is impossible.
  3. You are dividing space, and have infinite time, but you can't divide smaller than Plank's distance.
  4. You take shorter time with each subdivision and you can divide Plank's distance, but you can't observe it. So it is moot.
  5. You are dividing a physical object. You can't divide subatomic particles.
  6. Heck, you can't divide an atom perfectly either.
  7. You can't do perfect division, full stop. The question is meaningless.
 
  • #264


Hello "pwsnafu". I see that you pretty much understood what I was asking. I was asking the question using two different scenarios. Cutting a distance in halves until there's nothing left, if possible OR cutting on object in halves until there is nothing left, if possible. Well, you explained it well that by cutting a PHYSICAL object in half over and over again you will eventually get down to subatomic particles which apparently are impossible to cut in half, again physically. I also agree with you when you said "An infinite number of divisions of space is possible, and the limit is zero.". I agree with you 100%. But, why is it so difficult for so many other people including so-called professionals to answer the question properly when I was referring to DISTANCE being cut in half over and over and over and over again, without putting a limit on how many times you can do it over and over. I had to word it that way and if I worded it differently such as saying "over and over, what will happen eventually" it leads people to assume there is an end conclusion which leads them to say "you will run out of space", which I know is absolutely incorrect. That is why I asked the question the way I did but there may have been a better way to ask it. The two persons whom left the other posts above yours agree with that, apparently. It was at Davis, California University where I had asked a Physics and Mathematics Professor the question. And when I asked the two professors the question I used ONLY the scenario using a space or distance and NOT on object. The physics professor responded to my question by saying "Yes, you will run out of space and the two objects will touch each other." Why did he answer the question that way, word for word. Keep in mind I asked if you have two objects and referring to the distance between the two objects that are coming closer to each other. I could not have been more specific to what I was referring to when I asked that Davis University Physics professor. But, I had waited about one half hour for the mathematics professor to return to his class on the campus to ask HIM the exact same question. He responded and said "The two objects will never touch each other no matter how many times you cut the distance between the two objects in half." I would love to see both of those professionals in the same room and at the same time discussing their reasoning for reaching their conclusion to each other! Again, believe me, I did ask both those professors the exact same question using the exact same words. I already knew the answer to my question but, I'm just trying to make some sense out of why I got two completely different answers. I guess sometimes the smarter we think we are, that is assume we are, the more sloppy we get at answering the most basic simple questions that are somewhat easy to answer. With all the above being said and realized, I guess there is another way to share the thought of what happens when you cut a space between two objects over and over continuously and here it is: "There is a way to prove how you can take two objects and make them move toward each other for an eternity but they will never touch each other!" It's called MATHEMATICS. Meaning it can be proved mathematically but NOT physically.
 
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  • #265


The responded differently because they understood the question differently. They had to speculate as to what you meant just like we did, and since they speculated differently, they arrived at different answers.
 
  • #266


Thank you for your comment regarding my choice of words, "eventually". You and others were correct on wondering what I was referring to when I chose to use the word "eventually" when asking the question what happens eventually when you keep cutting a distance between two objects in half over and over again. I assumed my word vocabulary was better than that. The word eventually does not mean to infinity. Rather it is a undetermined amount of time. Oops! Sorry about that.
I've been out of town for four weeks searching for meteorites but ended up finding historic World War II 50 caliber and 20 mm bullets and shells on the desert in Arizona. I would've responded sooner to your post. Thanks and no reply necessary.
 
  • #267


This reminds me of this puzzle: An attractive woman is waiting for a male mathematician who is 10 meters away. As each minute passes, the man moves half the remaining distance closer.

Question: Will the man reach the location woman in any finite time?
Answer: No, but he will get close enough for all practical purposes.
 
  • #268


LCKurtz said:
This reminds me of this puzzle: An attractive woman is waiting for a male mathematician who is 10 meters away. As each minute passes, the man moves half the remaining distance closer.

Question: Will the man reach the location woman in any finite time?
Answer: No, but he will get close enough for all practical purposes.

An infinite number of mathematicians walk into a bar; the first orders a glass of beer, the second orders half a glass of beer, the third orders a fourth of a glass of beer...

The bartender says "You're all idiots!" and pours two glasses of beer.
 
  • #269


Number Nine said:
An infinite number of mathematicians walk into a bar; the first orders a glass of beer, the second orders half a glass of beer, the third orders a fourth of a glass of beer...

The bartender says "You're all idiots!" and pours two glasses of beer.

:rofl:
 
  • #270


How many Lebesgue measurable subsets of the reals are there?
 
  • #271


Jimmy Snyder said:
How many Lebesgue measurable subsets of the reals are there?

i tried counting them, but i gave up after aleph-null...
 
  • #272


Jimmy Snyder said:
How many Lebesgue measurable subsets of the reals are there?

I just came across this:

Let C be the Cantor set and [itex]E = C \times [0,1]^{k-1} \subset R^k[/itex]. Then E is uncountable with cardinality c and with Lebesgue measure zero. So there are 2c subsets of E, each Lebesgue measurable.
 
  • #273


pwsnafu said:
I just came across this:

Let C be the Cantor set and [itex]E = C \times [0,1]^{k-1} \subset R^k[/itex]. Then E is uncountable with cardinality c and with Lebesgue measure zero. So there are 2c subsets of E, each Lebesgue measurable.
This is close, but I am looking for subsets of the reals. So, set k to 1 and you have shown that lower limit is at least 2c. Now just provide an upper limit.
 
  • #274


Jimmy Snyder said:
This is close, but I am looking for subsets of the reals. So, set k to 1 and you have shown that lower limit is at least 2c. Now just provide an upper limit.

But isn't the cardinality of all subsets of the reals ##2^c##, so that is also an upper limit?
 
  • #275


LCKurtz said:
But isn't the cardinality of all subsets of the reals ##2^c##, so that is also an upper limit?
Yes, you have solved it.
 
  • #276


Topic:

Two birds in the tree, the hunter shot one.

Ask:

Only a few were left in the tree? Live or die?

You need to determine the answer
 
  • #277


godsaveme said:
Only a few were left in the tree? Live or die?

What??

Sorry, I cannot parse those sentence fragments.
 
  • #278


DaveC426913 said:
What??

Sorry, I cannot parse those sentence fragments.

I am a chinese,my english is poor.
That may be how many birds in the tree?
 
  • #279


This is a certainty and uncertainty question!
 
  • #280


How to determine?
The number of birds,live or die?
 
<h2>1. How do we make 100! divisible by 12<sup>49</sup>?</h2><p>To make 100! divisible by 12<sup>49</sup>, we need to find the highest power of 12 that is a factor of 100!. This can be found by dividing 100 by 12, which gives us 8. This means that 12<sup>8</sup> is the highest power of 12 that is a factor of 100!. To make it divisible by 12<sup>49</sup>, we need to multiply 12<sup>41</sup> to 100!.</p><h2>2. Can we make 100! divisible by 12<sup>49</sup> without changing its value?</h2><p>Yes, we can make 100! divisible by 12<sup>49</sup> without changing its value by multiplying it with 12<sup>41</sup>. This ensures that the value of 100! remains the same, but it becomes divisible by 12<sup>49</sup>.</p><h2>3. Why is it important to make 100! divisible by 12<sup>49</sup>?</h2><p>Making 100! divisible by 12<sup>49</sup> is important because it allows us to easily perform calculations involving large numbers. By making it divisible by 12<sup>49</sup>, we can break down the calculation into smaller, more manageable parts.</p><h2>4. What is the significance of 12<sup>49</sup> in this context?</h2><p>12<sup>49</sup> is the highest power of 12 that is a factor of 100!. This means that it is the smallest number that can divide 100! without leaving a remainder. It is significant because it allows us to make 100! divisible by a large number, making calculations easier.</p><h2>5. Is there a general rule for making a factorial divisible by a large number?</h2><p>Yes, there is a general rule for making a factorial divisible by a large number. We need to find the highest power of the number that is a factor of the factorial, and then multiply it to the factorial. This will ensure that the factorial is divisible by the large number without changing its value.</p>

1. How do we make 100! divisible by 1249?

To make 100! divisible by 1249, we need to find the highest power of 12 that is a factor of 100!. This can be found by dividing 100 by 12, which gives us 8. This means that 128 is the highest power of 12 that is a factor of 100!. To make it divisible by 1249, we need to multiply 1241 to 100!.

2. Can we make 100! divisible by 1249 without changing its value?

Yes, we can make 100! divisible by 1249 without changing its value by multiplying it with 1241. This ensures that the value of 100! remains the same, but it becomes divisible by 1249.

3. Why is it important to make 100! divisible by 1249?

Making 100! divisible by 1249 is important because it allows us to easily perform calculations involving large numbers. By making it divisible by 1249, we can break down the calculation into smaller, more manageable parts.

4. What is the significance of 1249 in this context?

1249 is the highest power of 12 that is a factor of 100!. This means that it is the smallest number that can divide 100! without leaving a remainder. It is significant because it allows us to make 100! divisible by a large number, making calculations easier.

5. Is there a general rule for making a factorial divisible by a large number?

Yes, there is a general rule for making a factorial divisible by a large number. We need to find the highest power of the number that is a factor of the factorial, and then multiply it to the factorial. This will ensure that the factorial is divisible by the large number without changing its value.

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