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

[AKG]

I don't know the answer to this one yet, but I'll post it:

Prove that if an element of a ring has at least two right inverses, it has infinitely many

Recommended steps:

If an element b of a ring has N right inverses (N > 1), then bx = 0 has N+1 solutions

If b has at least N right inverses, it has at least N+1

The following isn't a solution, it's just what I've thought of so far:

It's clear how the first hint implies the second, and how the second proves the desired result. Now this problem was from a combinatorics class, so given N right inverses, there is some way to combine them to get a number of solutions to bx = 0, and we probably need to use some combinatorial argument to count the number of such solutions.

If x, y, z are distinct right inverses, then 0, x-y, y-x, y-z, z-y, x-z, z-x are solutions to the homogeneous equation. We know that at least 0, x-y, and x-z are distinct. We are left with four things, y-x, y-z, z-y, and z-x. And none of these are zero. So if we assume that there are only 3 distinct solutions to the homogenous equation, then we get that the pigeons:

y-x, y-z, z-y, z-x

must fit into the pigeon holes

x-y, x-z

It's also clear that y-x and y-z can't go in the same hole. Likewise, z-y and z-x can't be in the same hole. Now if you put y-x in the first hole, then it contains both x-y and y-x. Hence for every a-b in that hole, b-a is in that hole. This will lead to contradictions. So you must put y-x in the second hole, and y-z in the first, giving:

{0}, {x-y, y-z}, {x-z, y-x}

and z-x, z-y yet to be placed. It's clear, by the contradictions mentioned above which arise by putting a number and it's additive inverse in the same hole, that we must get:

{0}, {x-y, y-z, z-x}, {y-x, z-y, x-z}

Does this give a contradiction? Moreover, how can this be generalized to n > 3?

 

[AKG]

Another thought I had:

If x-y = y-z = z-x, then 3(x-y) = x-y + y-z + z-x = 0. So we can conclude some things:

If c and d are distinct elements of {x, y, z}, then 0 is not equal to c-d, and d-c is not equal to c-d. Moreover, 2(c-d) = d-a, and 3(c-d) = 0.

Again, we're looking for some sort of contradiction here. We're saying that if b has 3 right inverses, we want a contradiction if there are only 3 solutions to bX = 0. If there are only 3 solutions, then it must be that:

the three distinct solutions are 0, x-y, and y-x
x-y = y-z = z-x
3(x-y) = 0

where x, y, and z are the distinct right inverses of b. We want this to give a contradiction. Moreover, we want to generalize this to the case where there are n distinct right inverses, not just n=3.

 

[vishal@physicsforums]

well if you ask for the least number(N belongs to I) i guess it should be some small( very small no.)infact -ve no.

even for comprehendible numbers ( whole nos may be) it should be zero.This is a trivially obtained solution but nevertheless should be correct?

What do you think..!

 

[BoTemp]

This seems to be a bit dead. If I knew anything about rings I'd offer my own answer, but as is I'll just post my own:

Find a general identity for sin^n x for n odd. By identity, I mean rewriting in terms of a sum of single powers of the sine and cosine harmonics (sin(nx), cos(nx).

This problem is more useful than interesting, I realize, but I'm just trying to restart the game.

 

[cliowa]

but I'm just trying to restart the game.

So am I. So I'll give you an easy one:

There is a 1-dim. rod fixed in space in horizontal position. On this rod there are point-like ants, which can move alonge the rod, i.e. in 1 dimension only.
As the rod has a fixed, finite length L, an ant may fall off the rod and end up somewhere else (that doesn't matter here). Once it has fallen off there's no way getting back.

The particular kind of ants we are considering here has the following property: It moves with a speed v (either backwards or forwards) and reverses direction if it hits another ant.

The scenario now is this: There is a certain number N (may be huge or small) of ants on the rod. At some initial point in time t0=0 they are all moving (that implies motion with a speed of v), but it is not known in which direction. They move according to the rule stated above.

These are my questions:
1) What will happen? :-)
2) Is there a point of time t1>t0, when not even one single ant is left on the rod?
3) If yes, what's the maximum time needed for this to happen? What's the minimum?
4) If no, how would one need to modify the setting (i.e. the rules of motion) to make all ants eventually fall off?

I hope you enjoy

 

[AKG]

Yes, the ants will eventually fall off. This is easy to see. The ants closest to the ends will certainly fall off, and this will continue until they all fall off. For any N, I believe the minimum amount of time is 0, and the maximum is L/v.

 

[cliowa]

...and the maximum is L/v.

That needs some explanation, right?
Best regards...Cliowa

 

[AKG]

Actually, it's wrong. With eight ants, you can get them to stay on for 3L/2v at least.

 

[StatusX]

Imagine instead of bouncing off each other, the ants just pass right through each other. If the ants are indistinguishable, there's no way to tell this scenario apart from the original one. So the best they can do is if an ant starts at one end facing the other, which will leave some ant on the rod for time L/v, as AKG originally said.

 

[Gokul43201]

Imagine instead of bouncing off each other, the ants just pass right through each other. If the ants are indistinguishable, there's no way to tell this scenario apart from the original one. So the best they can do is if an ant starts at one end facing the other, which will leave some ant on the rod for time L/v, as AKG originally said.

Neat!

 

[AKG]

Very nice, StatusX!

 

[cliowa]

Very nice, StatusX!

Yeah, that's true. It's why I like the problem. Another nice visualization (basically the same) is the following: Imagine every ant carries with it a flag with a certain color (let them all be distinguishable). Now let the ants exchange flags when they meet. The picture you would see is lots of flags just "walking" straight on, all the way along the rod until they fall off.

 

[StatusX]

Oh, I guess I'm supposed to post another question. Ok, here's one. If $p(x)=a_0 x^{16} + a_1 x^8 + a_2 x^4 +a_3$ and $a_0/16+a_1/8+a_2/4+a_3=0$, show p(x) has a real root.

 

[AKG]

Here's an inelegant solution:

Given the relation between the coefficients, we can write:

$$p(x) = a_0\left (x^{16} - \frac{1}{16}\right ) + a_1\left (x^8 - \frac{1}{8}\right ) + a_2\left (x^4 - \frac{1}{4}\right )$$

Regarding p as a function on R, if it has no real roots then it is strictly positive or strictly negative. So:

$$\mbox{sgn}(p(0)) = \mbox{sgn}(p(1)) = \mbox{sgn}\left (p\left (16^{-1/16}\right )\right ) = \mbox{sgn}\left (p\left (4^{-1/4}\right ) \right )$$

From this, we can deduce:

$$\mbox{sgn}(a_0 + 2a_1 + 4a_2) = \mbox{sgn}(a_0 + 2a_1) = \mbox{sgn}(-a_1 - 2a_2) = \mbox{sgn}(-15a_0 - 14a_1 - 12a_2)$$

If these four expressions really did have the same signs, then for any four positive numbers A, B, C, and D, we could not have:

$$A(a_0 + 2a_1 + 4a_2) + B(a_0 + 2a_1) + C(-a_1 - 2a_2) + D(-15a_0 - 14a_1 - 12a_2) = 0$$

But take (A,B,C,D) = (11,4,16,1)

 

[StatusX]

That looks correct, but I think there's a simpler way.

 

[StatusX]

Ok, forget that one. Here's a little one I came up with screwing around during a boring class today. Find:

$$\sum_{n=0}^{\infty} {\left( \begin{array}{c} n+k \\ k \end{array} \right)}^{-1}$$

EDIT: Ok, Latex is working now.

 

[NateTG]

That's the sum of the reciprocals of the binomial coefficients?

 

[StatusX]

Is this too easy or too hard? Here's a clue: (click to get it. the white text was showing up, so I thought this would be a good way to hide it.)

$$\telescoping \series$$

 

[Moo Of Doom]

the white text was showing up, so I thought this would be a good way to hide it.

White text might still be visible, but color #e9e9e9 text is impossible to read.

 

[StatusX]

Well I'll give the answer and let someone else go if they want:

$$\sum_{n=0}^N \left( \begin{array}{c} n + k \\ k \end{array} \right)^{-1} = \sum_{n=0}^N \frac{n! k!}{(n+k)!}$$

$$= k! \sum_{n=0}^N \frac{1}{(n+k)(n+k-1)...(n+1)}= k! \sum_{n=0}^N \frac{1}{(n+k)(n+1)}\frac{1}{(n+k-1)...(n+2)}$$

$$=k! \sum_{n=0}^N \frac{1}{k-1}\left(\frac{1}{n+1} - \frac{1}{n+k} \right) \frac{1}{(n+k-1)...(n+2)}= \frac{k!}{k-1} \sum_{n=0}^N \left(\frac{1}{(n+k-1)...(n+2)(n+1)} - \frac{1}{(n+k)(n+k-1)...(n+2)} \right)$$

But this is a telescoping series, so we get:

$$\sum_{n=0}^N \left( \begin{array}{c} n + k \\ k \end{array} \right)^{-1} = \frac{k!}{k-1} \left( \frac{1}{(k-1)...(2)(1)} - \frac{1}{(N+k)(N+k-1)...(N+2)} \right) = \frac{k}{k-1} - \frac{k!}{(k-1)(N+k)...(N+2)}$$

Which goes to k/(k-1) as N goes to infinity.

 

[darwid]

Oh, I guess I'm supposed to post another question. Ok, here's one. If $p(x)=a_0 x^{16} + a_1 x^8 + a_2 x^4 +a_3$ and $a_0/16+a_1/8+a_2/4+a_3=0$, show p(x) has a real root.

What was the elegant solution for this one ? I tried for 2 days and couldn't find it ;)

 

[StatusX]

Woa, sorry, missed your post. And, yea, sorry again, I didn't have one in mind, I just assumed there was one. I still think there is, but I haven't found it yet either. I'll get back to you (if you still exist).

 

[whatta]

I take it there is no on-going puzzle here, so I thought I'll post something. Solve this proportion: 11/2 = 3/10, 10/8 = ?/10.

 

[StatusX]

?=6 (mod 13)

 

[whatta]

well it could be, but I had something completely different in mind (hint: / was not supposed to mean division)

 

[shamsoddin]

I think it was better that you specify more equations not just one.(for example two of them).But anyway I think :
?=8

 

[whatta]

yep, that was it! more equations for those who did not figured yet: 1001/2 =11/8, 112/3 = 15/9, etc.

 

[StatusX]

Ok, I see. Yea, more equations would have been a good idea to eliminate other possibilities, since 11/2 = 3/10 is also true in the field Z/13Z (ie, mod 13):

11/2=24/2=12
3/10=120/10=12

Your turn shamsoddin.

 

[shamsoddin]

This is my quastion :
Suppose function F is continuous in [2,4] and differntiatable in (2,4).We have
F(2)=2 and F(4)=4.Prove a point C exists that tangent to the corresponding curve in this point C passes the origin i.e this tangent line includes the origin.

 

[whatta]

cant we say, if it does not exist, all angles have to be either less than or greater than 45 and so integrating it from (2,2) onwards wouldn't ever come to (4,4)?

 

[shamsoddin]

It is not a persist proof and in mathematics persist ones are needed.So I give you some hints :You should define a function and then use Roll's Theorem to solve this problem.
good luck !

 

[StatusX]

Ok, I'll try to get this started again.

Here's one way to prove it, but there's probably a simpler one. For each real number m, let L_m be the line through the origin with slope m, ie, the graph of y=mx. Let A be the subset of R such that for all m in A, L_m meets the graph of f(x). This is just the set f(x)/x, x in [2,4]. Since f(x)/x is continuous on [2,4], it has a min and max.

f(2)/2=f(4)/4=1, so if the min and max both occur at endpoints, f(x)/x is constant, ie, f(x)=x, and we're done. So assume that, say, the max occurs at an interior point c. Then it's easy to see if f'(c) is not equal to this max, we can get a bigger max by moving to a neighbor of c, a contradiction. Thus f'(c) is equal to the slope of the line through the origin meeting f(c), ie, this line is the tangent to f(x) at c.

 

[StatusX]

I think shamsoddin's gone, but I think my proofs ok, so I'll post a problem.

You have a 3x3 cube that you want cut into 27 1x1 cubes. You are allowed to slice it along a plane, stack up the pieces in any way you want, slice it again, and repeat the process. What is the minimum number of cuts required?

 

[TheDestroyer]

Minimum number is 6 cuts, even if the cut cubes was increasing geometrically

First Way to cut is to cut in planes x=1,x=2,y=1,y=2,z=1,z=2 in 3D xyz axes without moving the peaces:

27
18+9 (1)
9+9+9 (2)
6+3+6+3+6+3 (3)
3+3+3+3+3+3+3+3+3 (4)
2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1 (5)
1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1 (6)

This is the simplest way, if some one thought there is a more clever way, let me show you that if we used every cut to cut every remaining peace this will give the same result, see:

27
18+9 (1)
9+9+3+6 (2)
6+3+6+3+2+1+3+3 (3)
3+3+2+1+3+3+2+1+1+1+1+2+1+2+1 (4)
2+1+2+1+1+1+1+2+1+2+1+1+1+1+1+1+1+1+1+1+1+1+1 (5)
1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1 (6)

Solved ;)

Anyone can take my turn :) I don't have questions

 

[Gib Z]

Ahh I don't know if this counts, solve Pell's equation?

solutions to x^2=ny^2 + 1.

I couldn't do it >.<"