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

[snoble]

Unless somebody minds how about we wait until tomorrow morning to see if anyone wants to fill in the details. Otherwise I'll say it's all yours mansi (and I'll post a proof myself for why there are no finite paired sets containing an irrational number)

Steven

 

[snoble]

go ahead and ask a new question mansi. I'll post a solution to my problem some time soon (today or tomorrow)

 

[mansi]

Hi!

Here is a problem I've been struggling with,so it appears real tough to me.just for the record...i haven't (yet) been able to write down a proof for it...

prove that Z + 2^(1/2)Z is dense in R.
(in words the given set is the set of integers + (square root of 2) times the set of integers and R is is the real line)

 

[chingkui]

I am thinking you only need to consider the interval [0,1] and study if sqrt(2)*z (mod 1) is dense in the interval. Then, it will look like a circle (by identifying 1 with 0). Explore the consequence if it is not dense (i.e. prove by contradiction).

 

[mansi]

i know it's a bad idea but looks like this is going to be the end of this sticky...


anyways...i want to work on the problem i posted...chingkui,could you elaborate the circle part...didn't quite get that...

 

[chingkui]

the reason to just look at [0,1] is that if it is dense there, then, it is dense in [n,n+1] (by translation) and you know that the set Z+\sqrt{2}Z is dense in R.

As for n\sqrt{2} mod(1), let's clarify what it means:
\sqrt{2} mod(1)=0.4142...=\sqrt{2}-1
2\sqrt{2} mod(1)=0.8284...=\sqrt{2}-2
3\sqrt{2} mod(1)=0.2426...=\sqrt{2}-4
So, it is just eliminating the integer part so that the number fall between 0 and 1.
This is what I mean by "circle", it is like you have a circle with circumference 1, you start at point 0, go clockwise for\sqrt{2} unit, you will pass the point 0 and reach 0.4142..., and go clockwise for another \sqrt{2}, you will get to 0.8284...

Now, if \sqrt{2}Z (mod 1) is not dense in [0,1], then there are a<b, where a=m\sqrt{2} (mod 1) and b=n\sqrt{2} (mod 1) such that nothing between a and b is a multiple of \sqrt{2} (mod 1). Note that \mu=b-a>0 is irrational.

The set S={c|a<c<b} is a "forbidden region" (meaning that S is free of multiple of \sqrt{2} (mod 1). Then, for any integer k, k\sqrt{2}+S (mod 1) is also forbidden. Pick any integer M>1/\mu, the total length of the M+1 sets S, \sqrt{2}+S,..., M\sqrt{2}+S exceed 1, so, at least two of them, say S_{p}=p\sqrt{2}+S and S_{q}=q\sqrt{2}+S must intersect. If S_{p}!=S_{q}, then a boundary point is inside a forbidden region, wlog, say upper boundary of S_{p} is inside S_{q}, then, since that boundary point is just b+p\sqrt{2} (mod 1), which should not be in any forbidden region, we are left with S_{p}=S_{q}. But this is impossible, since \sqrt{2} is irrational. So, we have a contradiction.

 

[chingkui]

Where is Steven's proof?

 

[snoble]

I have posted the proof to the paired set problem in a separate thread to avoid clutter. Please make any comments about it there.
https://www.physicsforums.com/showthread.php?t=74511

thanks,
Steven

 

[mansi]

here's a proof of the question i had posted...thanks to matt grime(he sent me the proof) and i guess chingkui's done the same thing...so he gets to ask the next question...

Let b be the square root of two, and suppose that the numbers
If nb mod(1) are dense in the interval [0,1), then m+nb is dense in R. Proof, let r be in R since there is an n such that nb and r have as many places after the decimal point in common as you care, we just subtract or add an integer onto nb so that m+nb have the same bit before the decimal point too. Thus m+nb and r agree to as much precision as you care.

So it suffices to consider nb mod 1, the bits just after the decimal point.

now, there is a nice map from [0,1) to the unit circle in the complex plane, which we call S^1


t --> exp(2pisqrt(-1)t)


the map induced on the circle by t -->t+b is a rotation by angle 2(pi)b radians.

it is a well known result in dynamical systems that such rigid rotations have dense orbits if and only if b is irrational, and then all orbits are dense.

the orbit of t is just the images of t got by applying the rotation repeatedly. Thus the orbit of 0 is just the set of all points nb mod(1), whcih is dense.

The proof of density isn't too hard, though you need to know about compactness and sequential compactness, at least in the proof I use.


Let r be a rotation by angle 2pib for some irrational b, then the images of t, namely

t+b,t+2b,t+3b... must all be distinct, otherwise

t+mb=t+nb mod 1 for some m=/=n

that is there is an integer k such that

k+mb=nb, implying b is rational.

thus the set of images of t must all be distinct. S^1 is compact, thus there is a convergent subsequence.

Hence given e>0, there are points t+mb and t+nb such that

|nb-mb|<e mod 1.

Let N=n-m, and let p and q be the points

mb and nb mod 1.


Consider the interval between p and q, and its image under rotating by 2pibN. These intervals are no more than e long, and the cover the circle/interval, hence the forward orbit under rotating by 2pibN is dense, thus so is rotating by 2pib, and hence all forward orbits are dense as required.

 

[chingkui]

Sorry for not posting so long. I have a counting question that is not very difficult (at least after I saw the solution), some of you might know the answer already. If you have already seen it somewhere, please wait 1 or 2 days before posting, let people think about it first. Thanks.

Here is the question:

I was thinking about this problem with a number of friends when we were cutting a birthday cake: How many pieces can be produced at most when we are allowed to cut n times (no need to have equal area/volume).

More precisely, in 2 dimensions, we want to know with n straight lines (extend to infinite at both ends), how many pieces at most can we cut the plane (R^2) into? In this case, the answer is 1+1+2+3+...+n=1+n(n+1)/2.

In 3D, it looks a lot more complicated. Again we have n infinite planes, and we wish to cut R^3 into as many pieces as possible. Does anyone know the answer?

How about in m dimensions? n hyperplanes to cut R^m.

 

[chingkui]

Dear all,

It has been exactly one month since I post the question, is the question too difficult or just not interesting at all?
If anyone is still interested, here is one hint: the formula in R^3 is a recursive formula that actually depends on the formula in R^2. That is as much hint as I can give, and I am almost writing down the solution.
If this still doesn't generate any interest and response in say 2 weeks, I will probably post another question if everyone agree.

 

[mustafa]

I think this recursion might be correct, not sure though:
1+1+2+4+7+11+...+(1+n(n-1)/2) = (n+1)(n^2-n+6)/6

 

[chingkui]

Sorry for replying so late. mustafa is right, and he can ask the next question.
The recursive relation I mentioned is P3(n)=P3(n-1)+P2(n-1)
where P2(n) is the number of pieces with n cut in R^2
and P3(n) is the number of pieces with n cut in R^3
Solving the relations, we can get mustafa's formula.

 

[DeadWolfe]

Well mustafa?

 

[siddharth]

Since mustafa doesn't seem to be online anymore, here is a question to revive this thread

SECOND EDIT:

Let f(\frac{xy}{2}) = \frac{f(x)f(y)}{2} for all real x and y. If f(1)=f&#039;(1) [/itex], Prove that f(x)=x [/itex] or f(x)=0 for all non zero real x.&lt;br /&gt;

 

[matt grime]

Is that the correct question? f(x)=x seems to do the trick a little too obviously.

 

[siddharth]

Is that the correct question? f(x)=x seems to do the trick a little too obviously.

Oops, I should have seen that coming.
I was expecting someone to prove that the function has to be of this form from the given conditions, not guess the answer.

 

[matt grime]

Then add the rejoinder that they must prove that this is the only possible answer (if indeed it is; since i didn't prove it but merely guessed by inspection i can't claim that 'prize'; of course it is explicit that f is differentiable, hence continuous)

EDIT: obviosuly it isn't the only solution: f(x)=0 for all x will do.

 

[D H]

It doesn't work for arbitrary c.

If f(x)=cx then f(\frac{xy}{2}) = c \frac{xy}{2} and \frac{f(x)f(y)}{2} = c^2 \frac{ xy}2. Equating these two yields c=c^2 so the only solutions for c are the two Matt has already found.

 

[siddharth]

Yes, I see that.
Can you prove that the only possible solutions are
f(x)=x
and
f(x)=0
from the given conditions?
Sorry for the poorly worded question.

 

[matt grime]

It doesn't work for arbitrary c.

If f(x)=cx then f(\frac{xy}{2}) = c \frac{xy}{2} and \frac{f(x)f(y)}{2} = c^2 \frac{ xy}2. Equating these two yields c=c^2 so the only solutions for c are the two Matt has already found.

What doesn't work for arbitrary c?

 

[AKG]

From f(2x/2) = f(x), we get that either:

a) f = 0, OR
b) f(2) = 2

From case b), using that f(0) = f(0)f(x)/2 for all x, we get either:

b1) f(0) = 0, OR
b2) f(x) = 2 for all x

b2) is impossible given the condition that f'(1) = f(1), so we have two cases overall:

a) f = 0
b) f(0) = 0 and f(2) = 2

In general, it holds that f(x) = +/- f(-x) since f(xx/2) = f((-x)(-x)/2). In fact, by looking at f((-x)y/2) = f(x(-y)/2), we can make an even stronger claim that either for all x, f(x) = f(-x) or for all x, f(x) = -f(-x).

Note that this gives a solution f(x) = |x| which satisfies the criteria (it is not required that f be differentiable everywhere, only at 1 is necessary) but is neither 0 nor identity.

Suppose f(x) = 0 for some non-zero x. Then for all y, f(y) = f(x(2y/x)/2) = 0, so f(0) = 0, and either f(x) is non-zero for all other x, or f(x) is zero for all other x:

a) f = 0
b) f(0) = 0, f(2) = 2, f is either odd or even, and f(x) is non-zero for x non-zero.

 

[siddharth]

Note that this gives a solution f(x) = |x| which satisfies the criteria (it is not required that f be differentiable everywhere, only at 1 is necessary) but is neither 0 nor identity.

I followed what you said up until this part, but I don't understand what you are saying here. Can you explain it in more detail so that I can understand?

 

[AKG]

siddharth, you asked:

Can you prove that the only possible solutions are f(x)=x and f(x)=0

I'm simply saying that f(x) = |x| satisfies the given criteria [i.e. |xy/2| = |x||y|/2 for all real x, y and f'(1) = 1 = |1| = f(1)] but f is neither of those two solutions you proposed above. The stuff I wrote in brackets in my post #72 was to acknowledge that f(x) = |x| is not differentiable at x=0, but the given criteria only require that f be diff'able at 1, not necessarily everywhere.

 

[NateTG]

Stupid brain. Must read before posting.

 

[chingkui]

So, other than f(x)=x, f(x)=|x| and f(x)=0, is there a forth solution?
In general, we are not even assuming continuity except at x=1, at which point f is also differentiable?

 

[siddharth]

I'm simply saying that f(x) = |x| satisfies the given criteria [i.e. |xy/2| = |x||y|/2 for all real x, y and f'(1) = 1 = |1| = f(1)] but f is neither of those two solutions you proposed above. The stuff I wrote in brackets in my post #72 was to acknowledge that f(x) = |x| is not differentiable at x=0, but the given criteria only require that f be diff'able at 1, not necessarily everywhere.

Ok then, my question was incorrect. So the only solutions are
f(x) = 0 , f(x) = |x| ,ie ( f(x) = x for x>0 and f(x)=-x for x<0 )

I hope this is the final edit to the question. Can you prove that the only solutions are f(x) = 0 and f(x) = |x| from the given conditions?

In general, we are not even assuming continuity except at x=1, at which point f is also differentiable?

I picked this question up from a calculus book and no information is given in the question in the book about it's differentiability at points other than x=1.

But I think it is necessary for the condition that f(x) is differentiable at all points except x=0 to be given.
So you are given that f(x) is differentiable at all points except x=0.

I hope that the question is still clear and I did not confuse everyone too much

 

[matt grime]

I thought the setter was supposed to know the answer so that they could agree upon the first correct response? Without imposing some more conditions I don't see that you can conclude there won't be some more answers. And as you keep changing the question we presume that you don't know the answer, which makes it a little unfair on those attempting it.

If we were to require that the function had a taylor series with radius of convergence infinity about 0 then you could get the answer but that would necessarily exclude AKG's answer.

Didn't the original question ask for f(x) for positive x only? Because then, assuming that the function is smooth in some neighborhood of 1, you can get the answer since every number is nth the power of a number close to 1.

 

[chingkui]

I thought we have a much more interesting question... it would be a lot more fun to assume nothing about continuity and differentiability (except at x=1)... I think we can prove continuity everywhere and differentiability except possibly at x=0 with the assumption of differentiability at x=1.

 

[matt grime]

We have a question with no known answer, that doens't really make it useful for this particular thread.

 

[siddharth]

I thought the setter was supposed to know the answer so that they could agree upon the first correct response? Without imposing some more conditions I don't see that you can conclude there won't be some more answers. And as you keep changing the question we presume that you don't know the answer, which makes it a little unfair on those attempting it.

I understand that I messed up this question, so I'll provide the answer which I was expecting. The reason I had to change the question was that the book I picked it up from did not give the correct answer.
Besides, I wanted to revive interest in this thread.

From the given condition
<br /> f(\frac{x+y}{2}) = [f(x) f(1+\frac{y}{x})]/2 so x \neq 0

Here put y=0.
So,

f(\frac{x}{2})=\frac{f(x)f(1)}{2}

ie,

2f(x)=f(2x)f(1)

Now
f&#039;(x) = \lim_{h\rightarrow 0} \frac{f(x+h) - f(x)}{h}

= \lim_{h\rightarrow 0} \frac{f(\frac{2x+2h}{2}) - f(x)}{h}

= \lim_{h\rightarrow 0} \frac{f(2x)f(1+\frac{h}{x}) - 2f(x)}{2h}

Since 2f(x)=f(2x)f(1)
we have,
=(\frac{f(2x)}{2x}) \lim_{h\rightarrow 0} \frac{f(1+\frac{h}{x}) - f(1)}{\frac{h}{x}}So, we have
f&#039;(x) = \frac{f(2x)}{2x} f&#039;(1)

Since 2f(x)=f(2x)f(1) and f(1)=f&#039;(1)

f&#039;(x)=\frac{f(x)}{x}

f(x)=|cx|

Hence, by substituting back into the original condition, the only solutions are

f(x) = |x| and f(x) = 0

 

[matt grime]

Your 'solution' somhow omits the identity function and appears to claim that |x| is everywhere differentiable, and makes untold assumptions about f

 

[siddharth]

That's why I changed the question, so that f(x) is differentiable everywhere except x=0 was given

So you are given that f(x) is differentiable at all points except x=0.

May I know what untold assumptions are being made?

 

[matt grime]

Firstly I must admit I've given up on rereading the new assumptions you keep making, and secondly you conclude f(x)=|x| is a solution. That is not an everywhere differentiable function, and you *do not conclude* that f(x)=x is a solution. Read your own proof and its conclusion that states the solutions are f(x)=|x| and f(x)=0!

 

[siddharth]

Firstly I must admit I've given up on rereading the new assumptions you keep making, and secondly you conclude f(x)=|x| is a solution. That is not an everywhere differentiable function, and you *do not conclude* that f(x)=x is a solution. Read your own proof and its conclusion that states the solutions are f(x)=|x| and f(x)=0!

I agree, and that is why I messed up the question. Sorry. I changed the question in post #77 so that the solutions are f(x)=0 and f(x)=|x|

 

[matt grime]

But as I observed in my first post f(x)=x is a solution and not one of the ones you derive, or have you altered the conditions so that f(x)=x is not a solution?

 

[siddharth]

But as I observed in my first post f(x)=x is a solution and not one of the ones you derive, or have you altered the conditions so that f(x)=x is not a solution?

No, I did not alter the conditions so that f(x)=x is not a solution. In retrospect, it should have been positive x.
That's a lesson then. Next time, I'll research the question thoroughly before I post it.

 

[chingkui]

You don't even need to assume f(x) is continuous to prove the result. All you need is f(xy/2)=f(x)f(y)/2 and f is differentiable at x=1 with f'(1)=f(1).
With just these, you can prove that f(x)=x or f(x)=|x| or f(x)=0.
Siddharth had proved differentiability of f(x) at every non-zero x in post 81.
You can use similar argument to show f(x) is continuous everywhere including x=0.

 

[matt grime]

How about another question. I heard this on the radio (Chris Maslanka, I think, presents a panel show with puzzles in on BBC radio 4) a few months ago.

There is a tribe of gnomes, each gnome, except the chief gnome, wears either a red or blue hat but doesn't know which (they're stuck on and there are no mirrors and whatever makes sense here). The chief gnome wants to conduct a census with the minimal opf fuss so he can count the number of red and blue hats. The gnomes being sensible people do this without even needing to communicate: one day they walk into the central clearing and form the red group and the blue group and the chief gnome can count them all. How do they do this? Remember, no communictation is needed (no gnome asks the colour of his hat).

It's quite an interesting puzzle even if you don't think it's maths.

 

[D H]

It's quite an interesting puzzle even if you don't think it's maths.

This is a sorting problem, which is an kind of applied mathematics.

Insertion sort: Gnomes arrive one-by-one. The first two gnomes stand next to each other. The third gnome stands between the first two if the third gnome sees two different colored hats, otherwise the third gnome goes to the end of the line (after the second gnome). Each newly arriving gnome inserts himself between the two gnomes with different colored hats or goes to the end of the line if all of the gnomes in line have the same color hat.

Some jostling is needed to make room for the newly arriving gnomes in the insertion sort method. This jostling is a kind of communication, just not talking. I assume that jostling and other physical contact is allowed. Therefore,

Heap sort: The gnomes all meet in the clearing. The chief tells the gnomes they are to take any punches they receive like a gnome. He then tells the gnomes to pummel all the gnomes they see wearing red hats into unconsciousness and pile them in a heap. When the brawl ends, the chief gnomes counts all the standing gnomes (the blue group) and the unconscious gnomes (the red group).

 

[matt grime]

naturally, it being radio 4, the first method was the one used. so ask a question (hoping to breathe life into this thread)

 

[D H]

naturally, it being radio 4, the first method was the one used. so ask a question (hoping to breathe life into this thread)

Sticking with the radio show puzzles, here's one from Car Talk.

Three different numbers are chosen at random, and one is written on each of three slips of paper. The slips are then placed face down on the table. The objective is to choose the slip upon which is written the largest number.

Here are the rules: You can turn over any slip of paper and look at the amount written on it. If for any reason you think this is the largest, you're done; you keep it. Otherwise you discard it and turn over a second slip. Again, if you think this is the one with the biggest number, you keep that one and the game is over. If you don't, you discard that one too.

The chance of getting the highest number is one in three. Or is it? Is there a strategy by which you can improve the odds?

 

[Moo Of Doom]

Turn over the first slip and remember the number. Discard it, then turn over the second slip. If the number on it is higher than the first slip, take it. It has a 2/3 chance of being the largest. If the number on the second slip is smaller than the first slip, you know it's not the largest, so you should go on to the third slip. In this case there's a 1/3 chance. (Unless, of course, you can go back to the first slip, in which case there's a 2/3 chance).
I'm more sure about the strategy than the odds though, seeing as I've never taken a probability class, and haven't bothered reading the book I got about it >.<

 

[D H]

Moo, you got it. Now it's your turn -- ask a question.

 

[Moo Of Doom]

Yarr... I totally don't have a question. I pass this round. Someone take it. >.<

 

[Moo Of Doom]

Well, that didn't work. Alright then. The next question will go to the first person to post a proof of the following:

Let a,b \in \mathbb{N} and S=\{ n \in \mathbb{N}\ |\ n=ax+by for some x,y \in \mathbb{Z} \}.

Prove \inf{(S)}=\gcd{(a,b)}.

(Note: in this case \mathbb{N} does not include 0.)

 

[matt grime]

You mean min, not inf, for what it's worth. There's no need to invoke the unnecessarily opaque inf. The min exists, it is less than the gcd by... and greater because...

 

[Galileo]

It's been a while, so I`ll have a go.
Just let me know if I use something that may not be regarded as known for the purpose of solving this problem.

For given positive integers a,b we can find integers x,y such that ax+by=gcd(a,b). (This is Bézout's Identity, which follows from the Euclidian Algoritm). This is also the smallest positive integer that can be expressed in this way. Thus gcd(a,b) is the minimum of S.

 

[Moo Of Doom]

matt: Yeah, I guess min would be better.

Galileo: I'd say Bézout's Identity is off-limits because it's practically what we want to prove. But I guess finding a proof of that would be no big deal... so, ehm... you got it. Post something! :)

 

[WARGREYMONKKTL]

i don't get it!

There are 50 factors of 100 divisibile by 2, 25 by 4, 12 by 8, 6 by 16, 3 by 32 1 by 64 so the power of two in 100! is 97.

similiarly there are

33 div by 3, 11 by 9, 3 by 27 and 1 by 81 making 48 times 3 divides, so i guess

2*3^50 will do

you said the thing that in the quotation.
i don't really get it
how can you get 100!=2^98*3^49?