Math Question

[jacob15728]

When I was in fifth grade, I was required to prove or disprove the statement p=np. It took a few minutes and was a bit harder than most of the stuff I'd previously done, but I managed it. How about you guys try and figure it out? I'll post the answer tomorrow to see if you've got it.

 

[spdqbr]

When I was in fifth grade, I was required to prove or disprove the statement p=np. It took a few minutes and was a bit harder than most of the stuff I'd previously done, but I managed it. How about you guys try and figure it out? I'll post the answer tomorrow to see if you've got it.

I have a truly remarkable proof of this fact, which this post is too short to contain. Also, dinner.

 

[qqwref]

When I was in fifth grade, I was required to prove or disprove the statement p=np. It took a few minutes and was a bit harder than most of the stuff I'd previously done, but I managed it. How about you guys try and figure it out? I'll post the answer tomorrow to see if you've got it.

hahaha, nice one

 

[Bryan]

When I was in fifth grade, I was required to prove or disprove the statement p=np. It took a few minutes and was a bit harder than most of the stuff I'd previously done, but I managed it. How about you guys try and figure it out? I'll post the answer tomorrow to see if you've got it.

I just state that it's true for value of n=1 or value of p=0. Then you just point back to the basic axioms.....QED.

 

[Johannes91]

When I was in fifth grade, I was required to prove or disprove the statement p=np. It took a few minutes and was a bit harder than most of the stuff I'd previously done, but I managed it.

Actually, Martin Musatov published recently an informal and highly experimental, unorthadox proof P=NP: http://groups.google.com/group/sci.math/browse_thread/thread/16d3720a3cffc651/37a6a16911e183e9. Several other proofs of it are have also been mentioned in sci.math, so if you had known how to use the search function, you wouldn't have needed to waste those few minutes reinventing the wheel.

 

[ManasijV]

1)
x[1-(1/2)-(1/2)1/3-1/2 1/3 1/4]=7777
x=26664

 

[ManasijV]

2)1/2x +1/2(1/2x + 1/2(1/2x +1/2(1/2x + ... = y
1/2x +1/2(y)=y
x=y
Is it difficult to visualize this?
3)Did you really have to post the number of horses question? When I was in 7th grade I would still solve it mentally i think. (But then I'm from India you are not )
4)quadratic
5)(31+39))/2=35. This is kind of a guess. Still haven't learnt permutations and combinations

 

[qqwref]

When I was in fifth grade, I was required to prove or disprove the statement p=np. It took a few minutes and was a bit harder than most of the stuff I'd previously done, but I managed it.

Actually, Martin Musatov published recently an informal and highly experimental, unorthadox proof P=NP: http://groups.google.com/group/sci.math/browse_thread/thread/16d3720a3cffc651/37a6a16911e183e9. Several other proofs of it are have also been mentioned in sci.math, so if you had known how to use the search function, you wouldn't have needed to waste those few minutes reinventing the wheel.

I'm interested in looking at this but it seems that (a) the entire thread is just a flamewar and (b) every link to his paper has been removed for some reason. Do you have a working copy?

 

[Johannes91]

I'm interested in looking at this but it seems that (a) the entire thread is just a flamewar and (b) every link to his paper has been removed for some reason. Do you have a working copy?

I didn't notice (b) until after I posted. I don't have a copy; I might've checked it when the thread had just started but don't really remember anything about it.

There are a lot of such threads, where someone (possibly a very bored troll) seems to honestly believe that he's found something remarkable, like a proof of P=NP or P≠NP or that reals are countable. They are sometimes quite interesting to follow.

 

[trying-to-speedcube...]

I have no idea if thats right. But it looks good, i wonder if anyone else could do it. Was it hard?

Also, for #5, could you say how you got 35?

I think that the average number that's thrown is (1+2+3+4+5+6)/6 = 21/6=3.5
Because you throw 10 dice the most appearing sum is 3.5*10 = 35

 

[blah]

Here's an interesting math problem to think about. It is not as easy as you think!

Suppose you have N boxes and K objects. The boxes are numbered from 1 to N; for each object, we randomly choose a box number, and write that down. Now we place the objects in the boxes, according to what number each object was given.

Okay, so here's the question: we're going to choose a random object and look in the box corresponding to that object. What is the expected value of the number of objects that will be in that box (i.e. on average, how many objects will be there)?

EDIT: If you can solve that, try this harder version: at the last part, instead of choosing a random object, we're going to choose (at random) one of the boxes that is not empty. What's the expected value of the number of objects in that box now?

I'm not sure if I understand the question correctly, so can I just clarify something, Michael? (Highlight below to read my question.)

We pick a random box, if there's nothing in there, we pick another random box, and we just keep doing this till we hit a box with at least 1 object. Question: What is the expected number of objects in that box? - Is this question equivalent to yours? (This refers to the first "easier" question.)

Also, I'm currently seeing this as NK slots and K objects to be "slotted in" (pardon the expression, side effect of too much Fridriching), am I heading towards the right direction?

Edit: Oops, I didn't notice your edit just now, which (I think) is the exact same thing I'm asking. Coincidence? So I guess your answer to my question is no? Or is it a yes, i.e. the "harder" version in fact the exact same question, and you're just trying to trick us?

 

[qqwref]

What you're talking about, blah, is the harder question. Each non-empty box is equally likely to be chosen. In the easier one, a non-empty box with M objects in it is M times as likely to be chosen as a box with 1 object.

Here's a hint to the first question, to show why it's relatively easy to figure out:

Spoiler

Since we're randomly choosing an object and looking in its box, but the objects are all treated identically, we can just label the chosen object. Then the expected value of the number of objects in that box is one (for the labeled object) plus the expected value of the number of the OTHER K-1 objects that will end up in the given box, which is pretty easy to calculate.

 

[blah]

I find it amusing that I predicted the harder question by misunderstanding the simpler one

And err... Yeah, whatever you said in the spoiler was what first came to my mind. But something told me there was a flaw in that logic but I didn't know what, so I headed in a different direction. Guess I gotta trust my first instincts huh?

 

[Sin-H]

I'm not sure if it's still important, but the taylor series stuff works like this (just one way)
you have to equate the function term with a general polynom a(1)+a(2)x+a(3)x^2+a(4)x^3+...

then insert x=0, and you get a(1).
derive, reducing the polynom's grade by 1. same procedure: x=0, now you get a(2). and so on, until you have your demanded number of terms.

but wait. Since when does a 8th grader know how to find derivatives?

 

[blah]

I'm not sure if it's still important, but the taylor series stuff works like this (just one way)
you have to equate the function term with a general polynom a(1)+a(2)x+a(3)x^2+a(4)x^3+...

then insert x=0, and you get a(1).
derive, reducing the polynom's grade by 1. same procedure: x=0, now you get a(2). and so on, until you have your demanded number of terms.

but wait. Since when does a 8th grader know how to find derivatives?

What's eighth grade equivalent to?

I realize that was a stupid question. How old would the typical eighth grader be then?

 

[Sin-H]

I realize that was a stupid question. How old would the typical eighth grader be then?

hm. As far as I know, an eighth grader in the US is 13-14 years old...

 

[Gparker]

I realize that was a stupid question. How old would the typical eighth grader be then?

hm. As far as I know, an eighth grader in the US is 13-14 years old...

Hes actually 15 because he started school late. And he got this as an extra credit problem as well. I'm not going to show him any of this work or ansewers. I just thought it would be fun if someone else tried.

 

[Lord Voldemort]

Strange teachers...
I mean, do they expect him to be able to work that out himself?
Most people (before calculus) don't even know what a power series is.

 

[jacob15728]

Like I said, it was completely unnecessary. In fact, the sole reason I claimed to solve p=np in fifth grade was to make fun of Gparker, in case nobody noticed. It's generally not cool to lie.

 

[Lord Voldemort]

When did Sin-H lie?
All he did was answer a math question and say that 8th graders are 13-14 years old.