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

[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.

 

[whatta]

btw, on Pell equation. I wrote a brute force script (copy paste in html file),

<script>

// find solutions for x^2=ny^2 + 1 in integers for n > 0 by brute force
// this takes lots of time, so be patient and do not abuse task manager

limit = 1234567; // this is when do we stop wasting CPU cycles

for (n = 1; n < 10; n++) {

	x = 1;
	y = 1;

	i = 0;
	while (i++ < limit) {

		l = x * x;
		r = n * y * y + 1;

		if (l==r) {
			alert ("Solution: n="+n+", x="+x+", y="+y);
			break;
		}

		if (l > r) y++;
		if (l < r) x++;
	}

	if (x * x != n * y * y + 1)
		alert ("No solution was found for n="+n+" up to x="+x+", y="+y);
}

alert ("Nothing to wait for :(");

</script>

(edit:ah, I've found the flaw in my script (fixed above), now it actually finds something)

 

[TheDestroyer]

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.

Give me a break ! Did I say anything wrong? or only because I wrote the last sentense in capital mode?

I feeled that StatusX want to proof that there is situation that can destroy my way of solution, That's the reason !

Take it friendly, we are always friends, and must not care about things like this... even if I mistaken, you should pass :)

 

[whatta]

that was supposed to be a joke :/

 

[StatusX]

It looks like this got locked up again. Gibz, exactly what were you looking for? Solving pell's equation doesn't seem like a good brain teaser question, more of an ongoing research project.

And please don't answer a question unless you're willing to post a new one. I don't mean to take over this thread, but I'd like to see it get started again.

Here's another question. Prove that, except for the groups of order 1 and 2, every finite group has a non-trivial automorphism (ie, an isomorphism from the group to itself).

 

[sriperso]

answer to gokul's Question: 2/3
100! contains 98 2's and 49 3's.

 

[pki15]

Here's another question. Prove that, except for the groups of order 1 and 2, every finite group has a non-trivial automorphism (ie, an isomorphism from the group to itself).

Okay, suppose G is non-abelian. Let $$a,b \in G$$ such that $$ab \not = ba$$. Then $$\phi (x) = axa^{-1}$$ is the inner automorphism, and $$\phi(b) = aba^{-1} \not = b$$. So $$\phi$$ is a nontrivial automorphism.

Now if G is abelian, $$G = Z_{p_1^{n_1}} \oplus Z_{p_2^{n_2}} \oplus \dots \oplus Z_{p_k^{n_k}}$$ by the Fundamental Theorem of Finite Abelian Groups. But any automorphism of $$Z_{p_i^{n_i}}$$ gives an automorphism of G, just apply the automorphism to the i-th component of G. But we know that sending a generator of $$Z_k$$ to a different generator (i.e. any element relatively prime to k) gives a non-trivial automorphism. So as long as G has a component not $$Z_2$$, we're done.

So the only remaining case is $$G = Z_2 \oplus Z_2 \oplus \dots \oplus Z_2$$. Now there is an automorphism of $$Z_2 \oplus Z_2$$ that sends (1,0) to (0,1) and (0,1) to (1,0) and (1,1) to (1,1). Extend this to an automorphism of G by applying this automorphism to the first two components and leaving the rest of the components fixed.

This should take care of all the cases I think... Let me know if there's a flaw in the argument somewhere, I didn't check too carefully.

Here's another question:

Prove the 4-color theorem for all planar graphs with at most 12 vertices. (Obviously, don't apply the 4-color theorem or any of its equivalent statements) Also, don't waste your time coloring all such planar graphs, there's a nice argument.

 

[StatusX]

Yea, that's right pki, although for the middle case it's easier to just send elements to their inverses. By the way, does anyone know if this is also true for infinite groups? The only complication would come from non-finitely generated abelian groups where every element has order 2. If these were all isomorphic to an infinite direct product/sum of Z2's, it'd be easy, but I'm not sure they are.

 

[nugae]

A new question

I'm a bit confused about when you post a question here and when you start a new thread, so if anyone wants me to make a new thread out of this, just tell me and I'll do it!

I've been looking at the value N(n) of N that satisfies the equation

$$\sum_{1}^{n}(N-i)^{n}=N^{n}$$

Thus turns out to be

$$N(n)=1.5+\frac{n}{ln2}+O(1/n)$$

where the O(1/n) term is about 1/400n for n>10.

I've verified this by calculation up to about n=1000, using Lenstra's long integer package LIP.

This result is so beautiful and simple that it must be possible to prove it without brute-force calculation. If anyone has any suggestions as to how to begin then I'd be very grateful!

 

[krateesh]

New question !

What is largest possible value for n...when case is----

1000! / 10^n

 

[Gib Z]

Are you sure you put the question correctly? n can be arbitrarily large..

If your expression is for the largest value of that expression, its when n tends to negative infinity.

 

[krateesh]

i mean for what largest value of n we get a positive integral value as the quotient of these two...i think it has sometging to do with progressions...

 

[nugae]

I'm going to answer the question for 1000! / 5^n. You should be able to see that the answer is the same for 1000!/10^n.

Numbers from 1 to 1000 that are divisible by 5 each contribute a factor of 5 to 1000!. That gives us 1000/5 factors of 5 so far.

Numbers from 1 to 1000 that are divisible by 5² each contribute two factors of 5 to 1000!, which gives us 2×1000/5² factors of 5; but everything that is divisible by 5² is also divisible by 5, so all numbers divisible by 5² have already been counted once, which means that we only need to add 1000/5² to our original total of 1000/5.

Same reasoning for 5³: three factors, but we've counted two of them already, so we add 1000/5³ and get a total of 1000/5 + 1000/5² + 1000/5³.

Same reasoning for 5^4 (except that we have to remember to discard fractions, so that 1000/5^4 = 1 and not 1.6)... and then we can stop, because 5^5 > 1000.

So the total you want is

1000/5 + 1000/5² + 1000/5³ + 1000/5^4.

If you want the answer for 123456 rather than 1000, divide 123456 by 5 repeatedly, discarding the fractional part of each answer, until you get to 0, and then add together all the quotients you've calculated.

 

[krateesh]

but in case of 10 we have also got individuals...such as 2*5..4*5...

 

[Moo Of Doom]

Krateesh, for each 10 we have we must have a factor of 5 and a factor of 2. There are 500 even numbers less than 1000, so we have at least 500 factors of 2 (we have way more, actually), but there are only 249 factors of 5 (by nugae's formula) in 1000!. For each of those factors, we have a power of 2, and so we can divide 1000! by 10^249 and still get an integer. Dividing by 10 one more time, though, will result in a non-integer since we have no more factors of 5 in the resulting number.

 

[StatusX]

Browsing through some old threads I found https://www.physicsforums.com/showthread.php?t=119552" I had that I was never able to figure out. It seems pretty interesting. I'll work on it some more, but I thought I'd put it up here to get this going again:

Let R be a ring with the property that x³=x for all x. Prove that R is commutative. Does this generalize for x^k=x?

 

[morphism]

Ah... That's one of two infamous problems in Herstein. (The other being the lcm problem that appears early in the section on group theory. I forgot the actual statement though!)

Here's a relevant webpage. Don't look if you don't want to see the solution.

 

[Diffy]

Hi,

(first post!)

I really have enjoyed this thread. I have got to go back and think about all the problems posted. Some of them I really enjoyed spending time on.

Here is an interesting question (interesting to me, and hopefully to you) that I came up with as soon as I was first introduced to group theory. It started when in the first chapter of my book we were introduced to the group of symmetries of a square. One was to imagine taking a square out of a plane and all the different ways of putting it back in. This of course, defines the dihedral group of order 8.

Imagine all the ways a cube can taken out of a solid 3d space and put back in. Does this form a group the same way a square does? If so what is this group? What is its order?

 

[fantispug]

I proved SatusX's problem (for n=3) using a slightly different method (though it uses similar ideas it was developed independently):

$$a+b=(a+b)^3=a^3+b^3+a^2b+ab^2+b^2a+bab+aba$$
And
$$a-b=(a-b)^3=a^3-b^3-a^2b+ab^2-ba^2+bab-aba$$
Using a^3=a, b^3=b, and adding and subtracting the equations gives
$$a^2b+ba^2+aba=0$$
$$ab^2+b^2a+bab=0$$
This is more useful with a rearrangement:
$$a^2b=-(ba+ab)a$$
$$ab^2=-b(ba+ab)$$
Then
$$ab=a^3b=a(a^2b)=-a(ba+ab)a$$
$$ab=ab^3=(ab^2)b=-b(ba+ab)a$$
Using symmetry from the former equation
$$ba=-b(ab+ba)b=-b(ba+ab)a=ab$$
Generalising for any k is much harder via my method though.

 

[Alan J.]

How can one find the surface area of a ring?

 

[fantispug]

Imagine all the ways a cube can taken out of a solid 3d space and put back in. Does this form a group the same way a square does? If so what is this group? What is its order?

I assume we're just talking about rigid rotations here. Then it is a group:
Identity: Pick it up and put it back without changing it.

Associativity: This is intuitive, but I find it hard to argue. The groupings of the rotations don't matter. One way to see this is that any rotation in 3D space can be performed by multiplying the position vectors by a single 3x3 special orthogonal matrix. Matrices are associative, so all our rotations are associative.

Inverses: Just rotate the cube in the exact opposite way, this is always possible.

If we just turn the cube on the spot (as if it were a square) we get three operations;
r, r^2, r^3,r^4=I. (Take r=rotation clockwise for convenience).
Now we can pick up the whole cube and rotate it towards us. That gives
t,t^2,t^3,t^4=I.
Finally pick up the whole cube and turn it left:
l,l^2,l^3,l^4=I.

The total number of elements in the group, is the same as the total number of different orientations of the cube - for each face it is like a square confined to the plane, giving 4 orientations * 6 faces = 24 elements.

Now the three operations are not independent.
t=l.r.l^-1
r=t.l.t^-1
l=t.r^-1.t^-1

The multiplication table is too long to write out, however I believe the following is an isomorphism with S_4:
$$\phi(r)=(1234)$$
$$\phi(l)=(1423)$$
$$\phi(t)=(1243)$$

I would be interesting to see what happens if you included 4d rotations.

 

[JohannesKepler]

Math, although integral, is useless without the metaphysical ideas of Kepler, Leibniz, Kastner, Gauss, and Riemann. It is impossible to unerstand the causes of motion in the universe, through mathematical manipulations.

 

[kbaumen]

Here's a funny one, though rather simple. Sorry if it has been posted already.

a = 1, b = 1
a = b
a$$^{2}$$ = ab
a$$^{2}$$ - b$$^{2}$$ = ab - b$$^{2}$$
(a + b)(a - b) = b(a - b)
a + b = b
1 + 1 = 1
2 = 1

Where is the flaw?

 

[sutupidmath]

Here's a funny one, though rather simple. Sorry if it has been posted already.

a = 1, b = 1
a = b
a$$^{2}$$ = ab
a$$^{2}$$ - b$$^{2}$$ = ab - b$$^{2}$$
(a + b)(a - b) = b(a - b)
a + b = b
1 + 1 = 1
2 = 1

Where is the flaw?

since b=a=1 it implies that a-b=0, so u cannot acutally divide by a-b, since it is zero, and we all know that division by zero is not allowed, or not defined in the reals.

 

[sutupidmath]

here it goes

integ of dx/x, do it applying integraton by parts. what do u get?

 

[Gib Z]

We get $$I = 1 - I$$, which only seems problematic because in our notation we do not differentiate different indefinite integrals (ie indefinite integrals that differ by an additive constant). The constant on integration for the two different I's in the expression have different constants.