Least multiplier to make 100! divisible by 12^49

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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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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.
 


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.
 


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?
 


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.
 


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
 


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

What??

Sorry, I cannot parse those sentence fragments.
 


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?
 


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.
 
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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.
 


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.
 


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.
 


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

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


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

What?? Where did you get this?

For me, "a few" is three or more.
 


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.
 


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.
 


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.
 


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.
 


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...