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

[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
$$f(\frac{x+y}{2}) = [f(x) f(1+\frac{y}{x})]/2$$ 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
$$f'(x) = \lim_{h\rightarrow 0} \frac{f(x+h) - f(x)}{h}$$

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

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

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

Since $$2f(x)=f(2x)f(1)$$ and $$f(1)=f'(1)$$

$$f'(x)=\frac{f(x)}{x}$$

$$f(x)=|cx|$$

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?

 

[Galileo]

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

Yeah, thought so, that's why I chose not to participate for a while, but I want to revive the thread a bit.
So here's a question that requires no real math knowhow.

Given any five points on a sphere. Show that some four of them must lie on a closed hemisphere.

 

[AKG]

Pick any two points and consider the great circle passing through them. This splits the sphere into two closed hemispheres (that overlap on the great circle), but there are three remaining points. By the pigeonhole principle, one of the hemispheres must contain two of these points. That hemisphere has four, since it has the two just mentioned, plus the original two used to make the great circle.

 

[Galileo]

That is, ofcourse, correct. Your turn AKG.

 

[matt grime]

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

But I didn't say anything like that. Try reading it again, all of it, all the preceding posts too.

 

[WARGREYMONKKTL]

Thanks.
I Have A Question. How Can We Apply Calculus In Physics.
Can You Give Me Somelinks About It Or Else...
Thanks
Hae A Godd Day Everyone!