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

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SUMMARY

The discussion centers on determining the least number that must be multiplied to 100! to make it divisible by 1249. The correct answer is 6, derived from the factorization of 100! which contains 297 and 348. To achieve divisibility by 1249 (which equals 298 * 349), one must multiply by 6 (21 * 31). The discussion also touches on the properties of nilpotent matrices and the implications of the trace being zero for all powers.

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  • Understanding of factorials and their properties, specifically 100!
  • Knowledge of number theory, particularly divisibility and prime factorization.
  • Familiarity with matrix theory, including concepts of eigenvalues and nilpotent matrices.
  • Basic understanding of complex numbers and their properties in linear algebra.
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  • Study the properties of factorials and their prime factorization techniques.
  • Learn about divisibility rules in number theory, focusing on powers of primes.
  • Explore nilpotent matrices and their characteristics in linear algebra.
  • Investigate the implications of the trace of matrices and its relation to eigenvalues.
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  • #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...
 
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  • #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 E = C \times [0,1]^{k-1} \subset R^k. 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 E = C \times [0,1]^{k-1} \subset R^k. 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?
 
  • #281


Can you help me to translate chinese into english?
 
  • #282


Zero. The other bird flew away.

Edit: Or maybe one, if the bullet didn't knock the first bird off the perch. In any case, the live bird is gone.
 
Last edited:
  • #283


I don't think you can answer questions like that.

There are two birds to begin with, one is shot dead. 1 is left.

1 does not equal a few, as a few is generally equated to mean 5.
 
  • #284


Actually I can answer questions like that. Proof: I just did :biggrin:

I think the "a few" was just a translation failure on the part of our Chinese friend.
 
  • #285


I heard this one before.
He: There were two birds in the yard and I shot one of them. How many were left in the yard?
She: One.
He: No, one. The one that I shot. The other one flew away.
 
  • #286


Okay, guess it's my turn to ask a new one.

What are all the continuous functions f:\mathbb{C} \rightarrow \mathbb{C} such that \forall z,w\in \mathbb{C},\ f(z+w) = f(z)f(w)? Does the answer change if continuous is replaced with measurable?
 
  • #287


jgutierrez218 said:
...a few is generally equated to mean 5.

What?? Where did you get this?

For me, "a few" is three or more.
 
  • #288


2 is a couple. 3 is a crowd. 3 to 7 is a few. 5 to 10 is some. 8 to 15 is several. 15 to 37 is a bunch or if it is something you don't like, then it's many, or even too many if you really don't like it. 30 - 100 is a profusion. 100 - 1000 is a multitude. More than that is a plethora or a surfeit.
 
  • #289


Jimmy Snyder said:
2 is a couple. 3 is a crowd. 3 to 7 is a few. 5 to 10 is some. 8 to 15 is several. 15 to 37 is a bunch or if it is something you don't like, then it's many, or even too many if you really don't like it. 30 - 100 is a profusion. 100 - 1000 is a multitude. More than that is a plethora or a surfeit.

Oh yes, so often do I ask for a crowd of things.
 
  • #290


jgutierrez218 said:
I don't think you can answer questions like that.

There are two birds to begin with, one is shot dead. 1 is left.

1 does not equal a few, as a few is generally equated to mean 5.

Hahaha, have you guys seen the xkcd strip about this sort of thing?

More seriously, I can see "one" being a valid value for "a few," although this is certainly not its most common usage. It would sort of be analogous to the way "some" is used to mean "at least one" in formal logic.
 
  • #291


Ooh, can we bring this back? We did a math trivia type game in math club and I have a few good ones, ranging from basic high school algebra, through analysis and some historical trivia.
 
  • #292


Let me see if I can do ...

Once we have figured out that 100! has 2^97 * 3^48 in it. Factorise 12^19.
It's (2*3*2)^49 = 2^98*3^49. So the number is 2*3 = 6...
 

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