Well, Yea Forums?

Well, Yea Forums?
Also, Kurisu thread

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vixra.org/pdf/1702.0286v1.pdf
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the answer is always 42

What kind of a retarded question is this? What happens after you select a point? does the bug take the shortest path towards B? more information is needed

its because its night time

>>>>>>>>>>>Assume that the bug takes the shortest path from A to the destination point

It would be somewhere between the center of the top and B. Too lazy to figure out what the exact point is.

(0.75,2,0.75), roughly.
The bug takes the shortest path to the point you pick, and you need to maximize the distance.

I could do this when I did math in college. Nodes and shit on matlab. Thankfully I'm a neet now.

but theres only 2 points marked on the cube

Get fucked OP I already got my grade 10 I don't need to keep doing this shit. Take your shitty school homwework assignments and fuck off.

that point is still closer than B

can the bug go under the cuboid? if it can how is the answer not B?

The bug is alive so the question cannot be answered due to it's unreliability.

The destination point is A. Since the bug is already at A, and it takes the shortest path between its origin and target, this means the shortest path A to A is a point. Since the bug never leaves point A, it can never complete the action of reaching point A, since it's already there. Time to complete the action is infinity

no, it's definitely B

brainlet detected
even bigger brainlet detected

you aren't fooling me. A smug anime picture and you calling me a brainlet are not enough to make me doubt my common sense.

Na. Just having fun with my answer. If I could actually solve this I wouldn't be on an anime board 18 hours a day. I'd be working a good job.

It's already on A so the question of longest time is irrelevant or invalidates it.

The answer is 1-(sqrt(2))/2 below point B
You can get to this by assuming that since the shortest distance from B is across the top and down, which is sqrt(2)+2. The only way we can make this distance longer is by going down from B, which would either make the bug move up the side back to B or make it go around the other 2 sides. Since we need to find the maximum distance, we simply need to treat the distance from B as X, and find the value for X that makes the two routes of equal length. So we get 2+sqrt(2)+X=1+1+2-X
Moving stuff around gets 2x=2-sqrt(2), then dividing both sides by 2 and simplifying results in our answer, 1-(sqrt(2))/2 below point B

The shortest distance to B is 2sqrt2

>Time to complete the action is infinity
Null minus null is not infinity.

wouldn't the twos be squared?

Nvm I'm a retard lol.

>you're allowed to move off the lines
The question gives no indication of this.

It never specified that the point is B.

B

>You can get to this by assuming that since the shortest distance from B is across the top and down, which is sqrt(2)+2

Answer: The set of all points inside the cuboid, as well as all other places the bug cannot reach.
Why? Because the distance to an unreachable place is considered ∞ (infinite) and there is no higher distance than that.

Tell me where is says what/where the destination point is.

That doesn't matter. I was just pointing out a blatantly wrong assumption.

A because movement is impossible.

Christina a cute

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And to add to what I said in if it's ment to be some word-salad trick about "At which point [...] should you choose [...]" making the choice cosmetic and therefore meaningless because it doesn't decide the "destination point" but rather just some random unrelated point, then the answer is "the set of all points" because all answers are equal to each other and therefore take the same "longest time" (the "shortest time" as well). Unfortunately it was half-assed and not specified enough.

If you're too retarded to prove this you might as well enrol in women's studies or feminist dance therapy.

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If A is at (0, 0, 0), then X is (2sqrt(2) - 2, 2, -2sqrt(2) + 2).
I think.

Only if they are inside the sqrt.

>completely useless complex shit no one will ever need
>b-but muh space tech which uses basic equations

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what's the answer then

>n=13
>k=2
>no primes that divide both 14 and 15
What did Kurisu mean by this?

arent they though? a^2 + b^2 = c^2?

Christina a best.

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Think again

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>Trying to sound smart while typing out p_i
My eyes! They are bleeding!
If you don't Tex it in, fine, but I'd even take OpenOffice formula editor over this half-assed garbage you presented.

That's where you're wrong

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Yes, but
sqrt(2² + 2²)
=sqrt(8)
=sqrt(4*2)
=sqrt(4)*sqrt(2)
=2*sqrt(2)
=sqrt(2) + sqrt(2)

Sorry guys, but the winner is clearly Kurisu.

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inb4 someone actually tries

someone posted this problem on /sci/ a couple weeks ago and a bunch of idiots including myself were convinced that it really was B for about 50 posts

@channeler and metaloooopa cannot compete with my fluffy wife. This is science.

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Facebook-tier opinion.

So is "at which point should you choose" a typo or is that the "trick of the question"?

She's better than Kurisu and you know it.

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Maybe Yea Forums will get another scientific paper.

oh indeed very clever, i did not even see that reduction

>he can't solve it

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I think the trap is that autists only consider vertices instead of points on the whole surface.

There is nothing to solve. You posted a blatantly incorrect statement.

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>blatantly incorrect
Prove it.

You can't, because you're a retard.

haha I can draw so there's no need for me to know anything else

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n=7
k=2

p_1 = 2; p_2 = 3

Try again.

Well, if I understood correctly then: if n+1is prime, the p_i is equal to n+1,because nothing says that the prime must be lower that the number. If n+1 is not prime, then it ends up being divided by 2 if even, or 3 or 5 if odd. Am I smarter than a feminist therapist dancer?

Isn't this ignoring all the zeros on the critical line and therefore trivial? It should be more like this to be correct

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None of the n+i can be prime, as they are composite numbers by definition. Also, the p_i are distinct.

The idea is that if you have a sequence of non-primal numbers in the form n+1, n+2, n+3, ..., n+k then you can find distinct prime numbers p_i such that p_i divides n+i, for 1

Well I can't think of anything right now but now, but I'll research that when I find the time. Thanks for pointing out something I might not now yet.

Can't you read the problem?

>by this point everyone immediately notices when I try to trick them into proving the Riemann hypothesis
>I know, I'll post something even more difficult but seemingly simpler!
For real, how did you come across this conjecture? I don't think it even has a name, and it's only mentioned in a few papers.

It is B and the distance is √6. Just go straight there.

Wait I just found it on Wikipedia, never mind.
I guess my google-fu is weaker than I thought.

here, almost forgot
>the bug cannot go through
What is quantum tunneling.

Since we're starting from a corner, the problem is symmetrical, so we can bisect the cuboid and work with the three remaining surfaces (diagram 1). We know that the shortest distance from A to another point is always the straight line distance across the surfaces. Also, since for any point not on a colored edge (diagram 2) we can always draw a line through it connecting to an edge, we know that the farthest point has to lie on a colored edge.

From the diagram, we observe that any point on a green/pink edge that's not B will have a shorter distance compared to B. Since B is presumed not to be the answer, the farthest point must lie on the purple edge. For this edge we note that there are up to two possible paths (diagram 3). Their distances are respectively sqrt(x^2 + (2+x)^2) = sqrt(2x^2 + 4x + 4) and sqrt((1+x)^2 + (3 - x)^2) = sqrt(2x^2 - 4x + 10). (Note: the second path only applies when x >= 2.0/3). The overall farthest point is the one with the highest minimum. This occurs at x = 0.75, with a distance of sqrt(8.125). By comparison, point B has a distance of sqrt(8). So the farthest point is (0.75, 0.75, 2) with respect to point A, as pointed out by .

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This proof is totally dependent on the fundamental theorem of arithmetic. Since by this theorem every positive integer has a single unique prime factorization, then by deduction, every positive composite integer has a single unique prime factorization. Therefore, every n+1,n+2, ...,n+k are all composite numbers, therefore each has a single unique prime factorization. Furthermore, since every consecutive set of composite numbers in the form n+1,n+2, ...,n+k has a unique prime number that divides it since they have a single unique prime factorization, according to the fundamental theorem of arithmetic. As a result, your problem is proven to be true since a unique prime divisor makes the following true for all k, where, 1≤i≤k, therefore if n+1,n+2, ...,n+k are all composite numbers, then there are k distinct primes p_i such that p_i divides n+i for 1≤i≤k is true for all n and k. This conclusively proves that your problem is true.

I didn't learn this in middle school

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What does |x| mean?

That does not yet prove that your prime numbers are distinct. Give it a bit more thought, you're almost there.

Sorry, I just copy-pasted this idiot: vixra.org/pdf/1702.0286v1.pdf

apple, orange, cherry = 3, 5, 9
if we equate the cyclic group with Z/15Z, we have that exactly two of 3x, 5x, 9x are congruent modulo 15. That means that either 2x, 6x, or 4x is congruent to 0 modulo 15. We get the only non-zero possibility with 6x, which gives x=5 or x=10. In either case, (3+9=12)*x=0

>What is quantum tunneling.
Physicists don't belong here. Reported.

1

>Also, Kurisu thread
Also, repost thread.

Go fuck yourself Zeno. Go fuck yourself with a rack.

Fuck that gay bug.

Underneath the cuboid