Tough questions thread

Here is one from a recent (HS) exam that is pleb tier, but still really made me think:

Prove result for [math] \lim_{x\to\infty} (a^{n} +b^{n})^{1/n} [/math]

Well shit, that didn't turn out to be as difficult as I expected.

Fine, here's another one:

A bag contains 10 balls labelled 1 to 10. I draw two balls from the bag and tell Albert the product of the two numbers drawn. I tell Bernard the (absolute value of the) difference between the two numbers, and I tell Cheryl the sum.

Immediately afterwards, the following conversation happens:
Albert: I don't know what the two numbers are.
Bernard: Neither do I.
Cheryl: Neither do I.
Albert: Now I know what they are.
Bernard: So do I.
Cheryl: So do I.

What are they?

>difficult problems in math
HAHAHAHA
BRAINLETS BTFO
AHAHAHAHAHAHA
GO BACK TO /MLP/ FAGGOTS

6 and 9

Wrong.
Suppose you were Albert and you were told that there are two integers, ranging from 1 to 10, whose product is 54.
Would you be able to tell what they are?

I found it hard user.

Reminds me of a simpler version:

Hello Albert and Bernard! I have given you each a different non-negative integer. Who of you has the larger number?

Albert I don’t know.

Bernard I don’t know either.

Albert Even though you say that, I still don’t know.

Bernard And still neither do I.

Albert Alas, I continue not to know.

Bernard And also I do not know.

Albert Yet, I still do not know.

Bernard Aha! Now I know which of us has the larger number.

Albert In that case, I know both our numbers.

Bernard. And now I also know both numbers.

Question: What numbers do Albert and Bernard have?

No solution.

It's not exactly a hard problem if you can brute-force it in a microsecond.

You're retarded. This is an open problem but they expect it to have integer solutions

integers not naturals, I think is the point.

You only tried it with positives, didn't you? Fucking idiot

Minimize [eqn] J(\mathbf{R}) = \frac{1}{2} \sum_{k=1}^{N} a_k|| \mathbf{w}_k - \mathbf{R} \mathbf{v}_k ||^2 [/eqn]

I wouldn't call your problem simpler, but that's probably because I've never seen anything like it before, whereas mine has a straightforward (if tedious) algorithmic solution.

Nonetheless, I've deduced the solution (Albert has 8 and Bernard has 7), though I'm not sure if the logic is completely sound:

Let A and B denote Albert and Bernard's numbers respectively. We will quote Albert and Bernard's dialogue, and intersperse unquoted text for 'common knowledge':

>different non-negative integer
The fact that they are different is important!

>Albert: I don’t know.
If A=1, Albert would know. So A > 1
>Bernard I don’t know either.
If B 2.
>Albert Even though you say that, I still don’t know.
If A 3.
>Bernard And still neither do I.
Mutatis mutandis, B > 4
>Albert Alas, I continue not to know.
A > 5
>Bernard And also I do not know.
B > 6
>Albert Yet, I still do not know.
A > 7
>Bernard Aha! Now I know which of us has the larger number.
B=7 is one possibility, but there is another:
If B=8 then Bernard can also infer A>8 by ruling out A Albert In that case, I know both our numbers.
Albert can eliminate one of B=7 or B=8. The only way he can do this is if either
A=7 or 8. It is common knowledge that A>7, so A=8...
>Bernard. And now I also know both numbers.
...and by elimination B=7.

...and I forgot that 0 is non-negative as well. Which means that the correct solution should be to subtract 1 from everything (i.e., Albert has 7 and Bernard has 6).

Yup your solution is correct (although I guess I did say non-negative, so it should start at 0, so then the answer is 6 and 7)

it's too late! everyone saw your wrong solution

8 and 1 fammm

is this right?

It is! Well done, I think this is the first time I've seen someone solve it.

Hoping someone else will start posting actual grad problems, though.
I'm not a math grad so I only know basic bitch problems that a logically-inclinced 5-year-old could solve.

max(a,b)

a^n+b^n -> max(a,b)^n, i.e. the smaller term ends up being negligible.

I know you said that the pic is a bad example but that's fucking trivial as shit. Give us something that an average undergraduate can't solve easily.

I mean the thing about grad problems is they tend to require knowledge of grad level topics. They may be more difficult, but the really limiting factor is the presumed familiarity with relevant undergrad topics.

I think it's been posted a gorillion times
look closely, the angles are not off.
can be solved using basic geometry (no triggeronomoetry)

It has been a long time since I did some math, so it's difficult "for me".

I have a fixed amount of cash, and I want to determine how much of each I must buy in order to have an equal partition (as much as possible).

Example: 350k cash, item 1 costs 4, item 2 costs 4, and item 3 costs 3

I think it had something to do with two variables, but that obviously fucks everything up (which is logic, you can't solve an equation with two variables). If I try:

350k= 4x + 4x + 3y

Then x= 34.75 - 3/8y

and 350k= 8(34.75 - 3/8y) + 3y goes sideways
I tried to google it but I didn't manage to fidn the right search terms. anyone care to help?

Linear programming?

en.wikibooks.org/wiki/Operations_Research/Linear_Programming

Let [math] n [/math] be an odd integer [math] >2 [/math] and let [math] f(x)\in \mathbb{Q}[x] [/math] be an irreducible polynomial of degree [math] n [/math] such that the Galois group [math] Gal(f/\mathbb{Q}) [/math] is isomorphic to the dihedral group [math] D_n [/math] of order [math] 2n [/math]. Let [math] \alpha [/math] be a real root of [math] f(x) [/math]. Prove [math] \alpha [/math] can be expressed by real radicals if and only if every prime divisor of [math] n [/math] is a Fermat prime.

Thanks mister. Didn't remember it involved a matrix.

Getting a tad bit too complicated, so it'll be faster if I just guess my way through it. Thank you anyway

If 8 and 1 is right then how did Albert know it wasn't 2 and 4?

thats a complex answer you retards

i dunno maybe, I solved by drawing half a 10x10 table with sum, difference and product values. and started discarding accordingly. i did it mechanically so to speak

7,0,-3

>replying to a 15 hour old post

shit i meant -4 instead of -3

[math] 7^3 - 4^3 = 279 [/math]

But user, 0 is both negative and positive, so 0 is indeed negative and your answer is right

I remember once someone solved it or a problem like by pulling out an obscure french theorem.

>both

I'm very bad at maths, but I'm stuck here and I'm going in circles :/
I only used the following rules:
>the sum of a triangle's angles is always 180°
>a flat angle is 180°
>if two straight lines cross each other, two opposite angles have the same value
Maybe I lack some knowledge, or maybe I'm too tired or too stupid.

a^a = e^-(pi/2)

aln(a) = -(pi/2)

start here. The rest should be easy

>The rest should be easy

t. Someone who has never done analysis in his life

t. Someone who probably hasn't done differential equations either

t. Someone who maybe even hasn't even done calculus and has the highschool mindset of "lol just apply log and it is solved xD"

t. Someone who absolutely doesn't know about the Lambert-W function

>Lambert-W function
not a function

>2 or less
Test invalidated for poor grammar.

Couldn't go further than that...

I got down to 1 and 6 or 1 and 8. How does 1 and 6 get eliminated?

you know that your white text is just an interpretation for the green text?

hey retard, if f(a)=a^a, it has a minimum at (1/e, e^(-1/e)). 1/e^(pi/2) is outside the range. so have fun trying to solve it without euler's formula

B, B, C, C, A, C, B, A, C, B

I am not proud of the amount of effort I put into solving this

Finished it
BEF = 30

Here's an impossible question:

d (sum[matmul[X, Y] ^ 2]) / d X ?
X and Y are matrices.

Answer: EnCt21105e5dbfe7170c52499a2437e60f840179c7f951105e5dbfe7170c52499a2439l91YtWieAM
Xm4uxTFh2MG5Neia18rsCLrzDih3aD0RK1AEnoHzktdU=IwEmS
encrypted via encipher.it
will post key after the wrong answer is invariably given.

but its the simplest solution to the problem.
i^i = e^(-pi/2)

kek

do you even euler bro?

ding ding first on thread first to win
heres ur bone

>nobody even gave it a try
Shame on you Veeky Forumstards. The key is "fug:DDD" without quotes.

Is it ab? Brainlet here, but I think it's ab.

It's min(a,b)

t. another brainlet who had to graph it

Yeah, I realized I made a mistake trying to use logarithms on it. Too bad no one has explained how to prove it the symbolic way.

#4 is wrong, check it

[math] max(a,b) \leq (a^n+b^n)^{1/n} \leq (2 \times max(a,b)^n)^{1/n}=2^{1/n}max(a,b) [/math]

Taking limits gives max(a,b) so the limit converges by squeeze theorem. Or at least that's what I said on the exam.

There are no solutions with real numbers because of an application of Bolzano theorem.

That still does not solve the problem because a is on both sides of the equation.

You need to compute the sides and then you will get the angles.
I am not going to do it, I am lazy, but that is the way.

Prove or disprove the following:

A disk can be divided by chords into pieces of equal area, none off which are congruent.

And? there is still a solution.

Can someone verify that [eqn]tan^4x+2tan^2x+1=sec^4x[/eqn]

That's the sup norm

wolframalpha.com

Pi=2??

Be my guest, find it.
Using analysis, it can be easily proven that the solution must be in the imaginary axis.
It is a complex number with no real part.

>Using analysis, it can be easily proven that the solution must be in the imaginary axis.
>It is a complex number with no real part.
the answer has already been stated in the first 3 posts, what are you babbling on about. a=i. use euler's formula mane, it high school shit

How is it wrong? There are exactly 2 occurrences of a question having the same answer as the one after it: Question 1 and Question 3.

I don't even know what I'm talking about

That question is trivial with the Lambert W """"function""""

Here's an easy warm-up question

(sqrt(60) - 6)/4 ?
p-pls no bully

I have one: If the order of a finite group G isn't divisible by 3 and (a * b)^3 = a^3 * b^3 for all a, b in G, prove that G is abelian

P(gold) = 0.3 + 0.2 * P(gold)
P(gold) = 0.3 / (1 - 0.2) = 0.375

-> 37.5%

Wait. Aren't there 2 boxes inside the main one? So shouldn't the 3rd term be P(gold) ^ 2 ?

Anyway, what I did:
P(x) = P(no gold)
P(x) = 0.5 +0.2*P(x)^2
P(gold) = 1 - P(x)

P = 0.3 + 2*0.2P -> 0.6P = 0.3 -> 0.5

isn't it just 30%

Why are you multiplying by 2? What if there are 10 boxes inside the main one?

1/6

>20% chance of containing two treasure chests with the same odds as this one

Yes, but what if the question was
>20% chance of containing 10 treasure chests with the same odds as this one
You clearly shouldn't multiply by 10 in this case

Solve a*ln(|a|) + i*a*t = -pi/2. a = r*e^i*t, so the equation becomes ln(r)*r*e^i*t + t*r*e^i*(t+pi/2) = - pi/2. Assume r = 1, then the equation is t*e^i*(t + pi/2) = - pi/2. We know e^i*(t + pi/2) is real, so t must be -pi/2 + 2pi*k, or pi/2 + 2pi*k. Assuming k = 0, either solution works, giving us a = 1*e^i*pi/2, or 1*e^i*(-pi/2), hence, i^i = e^(- pi/2) = (-i)^(-i).

P = 1-(0.5 + 2*0.5P) = 0.5 - P -> P = 0.25
then.
I don't know why it's not the same shit. Fuck math lmao, lame shit for faggots.

Ah yeah, didn't read carefully enough...

[math]5^5[/math]

ok got it

He didn't know it wasn't 2 and 4; that's why he said
> Albert: I don't know what the two numbers are.

But if it was 2 and 4, Cheryl would have been given the sum 6, for which the only combinations are 1 and 5 and 2 and 4. And if the numbers were 1 and 5, Albert would have been given the product 5, for which 1 and 5 are the only combination.

So when Cheryl, knowing that Albert said he didn't know, says that she doesn't know, Albert knows that Cheryl wasn't given the sum 6 so the answer isn't 2 and 4. Which means that Albert would know that it must be 1 and 8.

Formally, this is a problem in epistemic modal logic (the same field as the "blue-eyed islanders" problem, but that one is much more complex).

This is correct, if you only have one box (confirmed with simulation).

With two boxes, the right answer is something around 0.435

Octave (Matlab) code:
-----------------
main.m:
-----------------
clc; clear; close all;

N = 2E5;
win = zeros(2, 1);

for i = 1:N
win(kek()+1)++;
end

printf("No gold: %d\nGold: %d\nP(gold) = %d\n", win(1), win(2), win(2)/N);
-----------------

-----------------
kek.m
-----------------
function retval = kek()
n = rand();
if n < 0.5
retval = 0;
elseif n >= 0.5 && n < 0.8
retval = 1;
else
retval = kek() || kek();
end
end
-----------------

I got 5/12

If the probability of getting a gold bar from one chest is P, the probability of not getting a gold bar from one chest is (1-P),
the probability of not getting at least one gold bar from two chests is (1-P)^2,
the probability of getting at least one gold bar from two chests is 1-(1-P)^2 = 2P-P^2.

So we have
P = 3/10 + (2/10)(2P-P^2)
= 3/10 + 4P/10 - 2P^2/10
=> 10P = 3 + 4P - 2P^2
=> 2P^2 + 6P - 3 = 0
=> P = (sqrt(15)-3)/2 = 0.43649

Which is the same as . But he forgot to cancel to the lowest form. Or show working.

ISTR the fact that AEB is isosceles (20-140-20) comes into it somewhere.

Google "Hardest Easy Geometry Problem" if you want a solution.