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

[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[/color] text might still be visible, but color #e9e9e9[/color] text is impossible to read.

 

[StatusX]

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

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

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

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

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 >.<"

 

[whatta]

x=+-1, n=0, y=any?

 

[StatusX]

TheDestroyer,

How do you know you can't do it in less cuts?

 

[TheDestroyer]

The BEST way to use minimum number of cuts is to interfere every remaining peace from any previous cut in the next cut, you can observe this in the second solution, in every cut every remaining peace from the previous situation is cut also, that means every 18 becomes 9+9, and every 9 becomes 6+3, and every 6 becomes 3+3, and every 3 becomes 1+2, and every 2 becomes 1+1,

I don't think you can cut more that cutting every remaining peace, especially because you said all peaces are size-equaled ;)

Understood? or should I more explain?

 

[StatusX]

That's not a proof. The easy way is just look at the middle cube: it has 6 sides that must be cut, and you can't cut more than one each time, so it takes at least 6 cuts.

 

[TheDestroyer]

LOL, Not a proof?

Your proof is simpler, but that doesn't mean my proof is wrong :),

I use my brain as computer in numerical processing ;), I don't care about the easier or harder way.

I don't have a problem to post, if someone wants to solve my problem Go to the general Math forum and find my post "Cross product in spherical coordinates" and consider it as my problem

 

[StatusX]

Your proof isn't wrong because it's more complicated, it's just not rigorous. What do you mean every 18 becomes a 9+9? Not only have you done nothing to prove this, it's not true, because an 18 can become a 12+6. The point is that kind of reasoning becomes really complicated to make precise in this problem, which is why the easy solution is nice.

 

[TheDestroyer]

I understand, but let me say, 18 can ONLY become 9+9 because the maximum number of peaces in 1 plane is 9, so 12 is not possible, got me?

 

[StatusX]

How about cutting a 3x3x2 block into a 3x2x2 and a 3x1x2?

 

[TheDestroyer]

This choice is useless, because the best way to cut is to select the cut which gives most equal number of peaces,

If we made the cut as you said, that means 3x1x2 will require 3 cuts minimum to become unit peaces, while the 3x2x2 will require 4 cuts minimum,

But if we select the cut which gives 3x3x1 and 3x3x1 from 3x3x2, for every 3x3x1 we will require only 3 cuts for every one to reach the unit cubes... Thats why taking the cut which gives equal number of peaces is always the best choice ;)

ANYTHING ELSE TO DISCUSS ABOUT IT?

 

[whatta]

ANYTHING ELSE TO DISCUSS ABOUT IT?

Well, since God is omnipotent, he's should have no problem to do the job with single cut, so your solution is only suboptimal. You should now spend the rest of your life looking for such a space transform that it only takes 1 cut in that space. The day when you find it, heavens will open and accept your enlightened soul.