My brain is rotting, Veeky Forums

My brain is rotting, Veeky Forums.

Give me math puzzles...

Other urls found in this thread:

dailyfeynmanlongdivisionpuzzles.blogspot.com/2009/09/feynman-long-division-puzzle-366.html?m=1
twitter.com/AnonBabble

6รท2(1+2)=???

-1/12

Ok bruh
So you got a rectangle divided in smaller rectangles like in pic related for example.

Every smaller rectangle has at least two sides that have a length that is an integer.
Show that the big rectangle has to have at least two sides which lengths are integers too.

>every smaller rectangle
>the big rectangle

you're gonna have to label this diagram, nigga

If a board moves at 10 posts per minute and a given thread gets 1 posts per 5 minutes what's the probability that someone in the thread will get dubs in an hour.

You'll have to /b/ more specific about which board.

The big rectangle = the outer one
the smaller = all the rectangles it is made of

Savage.

There will be 600 total posts in the hour.

Of these posts, at most 60 will be dubs.

Also of these posts, 12 will be posts in the thread.

The probability that one of the posts is dubs is [math]\frac{60}{600}=10\%[/math].

The probability that one of the posts is a thread post is [math]\frac{12}{600}=2\%[/math].

Therefore, the probability that a post is in the thread and dubs is [math](\frac{1}{10})(\frac{1}{50})=\frac{1}{500}=0.2\%[/math].

What do I win?

0%

this board

checked

An explorer walks one mile due south, turns and walks one mile due east, turns again and walks one mile due north. He finds himself back where he started. Where is the explorer?

Hint: There's more than one solution

Nowhere because the earth is flat.

Anywhere because you didn't specify the turning angle.

How about this. So we all know that humans and mammals have a common ancestor. We also have a common ancestor with whales. But here's the interesting thing. We're also supposed to have a common ancestor with spiders, shrimp, centipedes, etc. So the problem is this: how many legs did the common ancestor have to allow the simplest amount of change between species over time?

Example to get you started:
Humans have 2 legs
Mammals have 4 legs
Therefore, the human-mammal ancestor would have 3 legs in order to be the simplest amount of change between the common ancestor and its descendants.
But if you put spiders in there, then that's 3+8/2, which equals 5.5, and you can't have 5.5 legs, so, assuming that there really is a common ancestor, that means something went wrong in the leg calculation.

1
11
21
1211
111221
312211
13112221

What's the next line?

And

I am a six digit number
I am between 400000 and 499999
The sum of my numbers is 30
I have all even numbers
My thousands digit and ones digit are the same
I do not have any zeros;s
My ten thousands digit and tens digit are the same
My 100 thousands digit and my hundreds digit are different
What number am I?

The very tip of the North Pole is one. Not sure about the second, since the South Pole doesn't really work.

I got a whole book of riddles too if you want logic ones or more abstract ones.

Correct so far

that doesn't make any sense, there's no need for the result to be an integer because what we're calculating is just the mid-point between numbers of legs

I assume you the answer to the spiders/humans/mammals one would be 5, since at most you go from 5 to 8 and to 2, but that's shit phrasing

Ever try Feynman Long Division?

A has a constant value throughout. The circles can be any 0-9 digit.

Got this from dailyfeynmanlongdivisionpuzzles.blogspot.com/2009/09/feynman-long-division-puzzle-366.html?m=1

It's the "evolution is a meme" tard bleeding into other threads.

Prove that there is a constant [math]K > 0[/math] such that, for each continuously differentiable [math]f \in \mathcal C^1(\mathbb R_{\ge 0}, \mathbb R)[/math] such that [math]f(0) = 0[/math] and [math]f'\in \mathcal L^2(\mathbb R_{\ge 0})[/math], the following inequality is satisfied: [eqn]\int_0^{+\infty}\left(\frac{f(t)}{t}\right)^2 dt \le K\int_0^{+\infty} f'(t)^2 dt[/eqn]

I think there's more than one of them bud...

Let me live in my fantasy, please.

Similar to:
But maybe a little easier: Cryptarithms. Some are really straightforward, others are a little tricky.

Pic related. Each letter has a corresponding 0-9 value.

On the Little Prince's Planet

Clever, but there is another solution on Earth

almost at the south pole, so the 1 mile east comes back around to the same place

But how do you go south if you're at the south pole? :^)

>almost :^)

Nigga, if you're within a mile of the south pole, then if you try going south one mile, you will end up going past the south pole and going north. Get rekt, kid

>What do I win?

the satisfaction

He's on an oil tanker (big one) that is travelling west

Any point one mile from the south pole. Because there is no concept of east and west directly at the south pole, the dude won't move east at the south pole and simply move back up north.

I think I've got it, I'm just too lazy to actually come up with an exact answer.

You start more than a mile away from the South Pole. So when you go east, you're circling the South Pole. Come around the circle to the same place you started, walk the same line north that you first walked south.

Make sense?

1113213211

cant get the second one.

4_ _ 4_ _

Nice job!

The last digit is 8

makes sense to me. You would need to have a very accurately defined south pole.

>The last digit

wont work then.

4_84_8

30-24=6

6/2=3 not an even number.

he spins for a mile

:^)

There is only one four

My hundred thousands and hundreds digit are different

Dat against the rules, though, bro. He can't change his angle unless specified; he can only move straight along longitude and latitude lines. Such movement doesn't exist at the poles.

I guess you could say that he is moving along an infinitesimally small latitudinal line, but then he would be "spinning" for eternity. He would never cover a mile of distance doing that.

462864

Yeah getting close

4=/=2

Fuck, nevermind I missed a line.

Starting 1 and 1/(2pi) miles above the South Pole.

446646

fucking scrubs

yep

Assume an idealized, perfectly running watch with a sweep second hand. At noon all three hands point to exactly the same spot on the dial. What is the next time at which the three hands will be in line again, all pointing in the same direction? The answer is: Midnight.

The first part of this problem-much the easiest-is to prove that the three hands are together only when they point straight up. The second part, calling for more ingenuity, is to find the exact time or times, between noon and midnight, when the three hands come closest to pointing in the same direction. "Closest" is defined as follows: two hands point to the same spot on the dial, with the third hand a minimum distance away. When does this occur? How far away is the third hand?
It is assumed (as is customary in problems of this type) that all three hands move at a steady rate, so that time can be registered to any desired degree of accuracy.

71.75%

You calculated the probability of randomly selecting a double from that thread out of the whole sample of 600.

Couldn't it also be 464664?

I've got an idea of how to do this graphically- just modeling the position of each hand over time as three periodic functions , then, well... Looking at the graph.

But there's probably a more elegant way to do it, isn't there?

1. Prove that the sum of any amount of positive integers is positive.
2. Prove that the sum of any amount of positive whole integers is a whole integer.
You would be doing math a giant favor OP if you solved these

>this
kek

>There is a cube and an ant is performing a random walk on the edges where it can select any of the 3 adjoining vertices with equal probability. What is the expected number of steps it needs till it reaches the diagonally opposite vertex?

For the last 342 days, I've gone to the bathroom once every day at some time between 2:00 PM and 4:00 PM. If I go to the bathroom at 5:00 PM today but not between 2:00 PM and 4:00 PM, what is the probability that I go to the bathroom after 4:00 PM for the next 27 days?

The numbers are the sum of the values of the symbols in their respective rows and columns.
What is the question mark equal to?

I create a 100% copy of you, particle for particle. Lets call this copy of you Bob. Bob has every thought, memory, belief, etc as you do, as a result of this perfect copy.

I place you and bob in separate, but identical rooms. You or bob do not know which of you is the "original" you. in fact I could kill the original you, and let bob take your place in society, and neither bob nor anyone else would know anything changed.

How do you know if you are Bob or the real you?

25

844644

anywhere on a small spherical surface

A Sugeron needs to fix 3 people but he has only 2
pairs of gloves how dos he protect himself and the people

apple=7
banana=8
strawberry=3
cherry=2
?=25

1 + 1/(2pi) miles from south pole
1 - 1/(2pi) miles from north pole
?

428628
446646
448248
464664
466266
482682
484284

How many automorphisms does the usual 4D cube graph have?

I'm laughing way too hard on this
Is Veeky Forums supposed to be retard?

well it appears that 1/343 is the probability of going after 4 PM. And each day you do it the chances increase. So the first day you'd have a 1/343 and the second day you'd have a 2/344 and so on. So the answer is the product of these probabilities: (28!*343!)/370!

>Prove that the sum of any amount of positive integers is positive

Why mathematicians are so fucking retarded?
Who gave this question? Some high school teacher?

If the fucking shit is positive, where the fuck a negative will come from? And the sign of the integration always follow the sign of the integrated thing

Inside is side A
Outside is side B

Glove pair 1
Glove pair 2

Glove pair 1 is put on
Glove pair 2 is put on over them, with side B facing out.

First surgery renders side B on pair 2 unusable.

He removes pair 2, and turns them inside out as they come off(easily done without contamination), leaving pair 1.
Performs second surgery.
Puts pair 2 back on inside out to reveal a clean, uncontaminated side A, while he is protected from side B by pair 1

I know there's plenty of bs ways to do this, but this way seems fun

Another is to put one glove on left hand, three on right.

After each surgery, the glove on the right is turned inside out as it is removed, and placed over the left

Try learning a language. Helped me. The app Duolingo works your brain every day for 10-30 minutes, depending on what you choose for intensity.

There is a 12 faced dice.
1-4 means no gf
4-8 means no gf
8-12 means no gf
what do

Don't roll it

According to some mathematicians, the sum of all natural numbers is -1/12. Google it then join me on my quest to disprove it.

464664 isn't right on the last one.

Makes no difference because I'm happy without a girlfriend.
Such is married life...

The missing Dollar
>3 mans buy lunch, together they must pay 30$ and everyone pays 10$
>after they left the restaurant the waitress notices that they paid 5$ to much
> so she follows the mans and give them the 5$ they payed to much back.
>she doesn't wants to divide 5/3 so she keeps
2$ and gives every man 1$

so every man payed 9$ 9*3=27
27 + the 2 dollar the waitress kept = 29$
wheres the missing dollar

100%, chek em

The Rothschilds have it, man.

this only ever happened during /pol/ocaust

>mans
>to much
>mans
>to much

kys

>there is no missing dollar
>$30 on table
>wait takes $5
>$25 on table
>$3 in man hand
>$2 in wait hand
>remembering $25+$3+$2=$30

And for those of who still don't believe, the error in your logic is because of the falsehood in the question itself.
>27 + the 2 dollar the waitress kept =29$
The $27 actually includes what the waitress kept.
>math is hard

$27 (the $25 bill and the $2 tip) + $3 ($1 dollar given back to each man) = $30

calculate the surface area of your moms cunt

Ayy, someone make a formula or something desu

inb4 the result is infinite