tough questions thread

[Anonymous No.8528499]

really difficult grad problems on math

pic is a poor example of 'difficult', but try to solve it anyway anons

[Anonymous No.8528507]

a=i

[Anonymous No.8528508]

a = i

[Anonymous No.8528641]

a = j

[Anonymous No.8528663]

a = (1/e^(pi/2))^(1/a)

[Anonymous No.8528668]

>>8528499
find integers x,y,z so that x^3+y^3+z^3=33

[Anonymous No.8528913]

I like pic related because it has an elegant algebraic solution that requires neither advanced number theory nor brute force computation.

Let [math]f(x) = \frac{4x + \sqrt{4x^2 - 1}}{\sqrt{2x + 1} + \sqrt{2x - 1} }[/math]. Find the sum f(1) + f(2) + ... + f(60).

[Anonymous No.8528965]

>>8528913
[math] f(x)={\frac 1 2}((2x+1)^{3/2}-(2x-1)^{3/2})[/math]

Let g(x)=(2x-1)^{3/2}/2. Then

[math] f(x)=g(x+1)-g(x)[/math]

so

[math] \sum_{n=1}^Nf(n)=g(N+1)-g(1)[/math]

[Anonymous No.8528979]

>>8528913
I'm assuming it's just partial fractions and then a telescoping sum.

[Anonymous No.8528996]

>>8528913
[math] \frac{1}{2} (121^{\frac{3}{2}}-1) [/math]

[Anonymous No.8529013]

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]

[Anonymous No.8529015]

>>8528965
>>8528979
>>8528996
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?

[Anonymous No.8529017]

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

[Anonymous No.8529029]

>>8529015
6 and 9

[Anonymous No.8529031]

>>8529029
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?

[Anonymous No.8529033]

>>8529015
I found it hard anon.

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?

[Anonymous No.8529039]

>>8528668
No solution.

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

[Anonymous No.8529040]

>>8529039

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

[Anonymous No.8529044]

>>8529039
integers not naturals, I think is the point.

[Anonymous No.8529046]

>>8529039
You only tried it with positives, didn't you? Fucking idiot

[Anonymous No.8529053]

>>8528499
Minimize [eqn] J(\mathbf{R}) = \frac{1}{2} \sum_{k=1}^{N} a_k|| \mathbf{w}_k - \mathbf{R} \mathbf{v}_k ||^2 [/eqn]

[Anonymous No.8529054]

>>8529033
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, Bernard would know. So B > 2.
>Albert Even though you say that, I still don’t know.
If A <= 3, Albert would know. So 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 <= 8.
I.e., either B=7 or B=8.
>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.

[Anonymous No.8529059]

>>8529054
...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).

[Anonymous No.8529060]

>>8529054
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)

[Anonymous No.8529071]

>>8529059
it's too late! everyone saw your wrong solution

[Anonymous No.8529080]

>>8529015
8 and 1 fammm

[Anonymous No.8529090]

>>8529080
is this right?

[Anonymous No.8529095]

>>8529080
>>8529090
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.

[Anonymous No.8529123]

>>8529013
max(a,b)

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

[Anonymous No.8529132]

>>8528499

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.

[Anonymous No.8529137]

>>8529095
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.

[Anonymous No.8529143]

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

[Anonymous No.8529154]

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?

[Anonymous No.8529160]

>>8529154
Linear programming?

https://en.wikibooks.org/wiki/Operations_Research/Linear_Programming

[Anonymous No.8529177]

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.

[Anonymous No.8529181]

>>8529160


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

[Anonymous No.8529640]

>>8529080
>>8529095
If 8 and 1 is right then how did Albert know it wasn't 2 and 4?

[Anonymous No.8529644]

>>8528507
>>8528508
>>8528641


thats a complex answer you retards

[Anonymous No.8529676]

>>8529640 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

[Anonymous No.8529897]

>>8528668
7,0,-3

>replying to a 15 hour old post

[Anonymous No.8529898]

>>8529897
shit i meant -4 instead of -3

[Anonymous No.8529926]

>>8529898
[math] 7^3 - 4^3 = 279 [/math]

[Anonymous No.8529937]

>>8529059
But anon, 0 is both negative and positive, so 0 is indeed negative and your answer is right

[Anonymous No.8530027]

>>8529143
I remember once someone solved it or a problem like by pulling out an obscure french theorem.

[Anonymous No.8530032]

>>8529937
>both

[Anonymous No.8530038]

>>8529143
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.

[Anonymous No.8530048]

>>8528499

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

aln(a) = -(pi/2)

start here. The rest should be easy

[Anonymous No.8530060]

>>8530048
>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

[Anonymous No.8530062]

>>8530060
>Lambert-W function
not a function

[Anonymous No.8530064]

>>8529095
>2 or less
Test invalidated for poor grammar.

[Anonymous No.8530301]

>>8530038
Couldn't go further than that...

[Anonymous No.8530362]

>>8529095
I got down to 1 and 6 or 1 and 8. How does 1 and 6 get eliminated?

[Anonymous No.8530437]

>>8529054
you know that your white text is just an interpretation for the green text?

[Anonymous No.8530444]

>>8530048
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

[Anonymous No.8530496]

>>8529095
B, B, C, C, A, C, B, A, C, B

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

[Anonymous No.8530515]

>>8530301
Finished it
BEF = 30

[Anonymous No.8530528]

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.

[Anonymous No.8530576]

>>8529644
but its the simplest solution to the problem.
i^i = e^(-pi/2)

[Anonymous No.8530601]

>>8530060
kek

[Anonymous No.8530606]

>>8530048
do you even euler bro?

[Anonymous No.8530609]

>>8528507
ding ding first on thread first to win
heres ur bone

[Anonymous No.8530711]

>>8530528
>nobody even gave it a try
Shame on you /sci/tards. The key is "fug:DDD" without quotes.

[Anonymous No.8530736]

>>8529013
Is it ab? Brainlet here, but I think it's ab.

[Anonymous No.8530738]

>>8530736
It's min(a,b)

t. another brainlet who had to graph it

[Anonymous No.8530773]

>>8530738
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.

[Anonymous No.8530779]

>>8530496
#4 is wrong, check it

[Anonymous No.8530791]

>>8530773
[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.

[Anonymous No.8530820]

>>8528499

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

>>8528663
That still does not solve the problem because a is on both sides of the equation.

[Anonymous No.8530838]

>>8530038
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.

[Anonymous No.8530858]

>>8528499
Prove or disprove the following:

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

[Anonymous No.8530938]

>>8530820
And? there is still a solution.

[Anonymous No.8530952]

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

[Anonymous No.8530954]

>>8529013
That's the sup norm

[Anonymous No.8530962]

>>8530952
http://www.wolframalpha.com

[Anonymous No.8530991]

>>8528499
Pi=2??

[Anonymous No.8531069]

>>8530938
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.

[Anonymous No.8531072]

>>8531069
>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

[Anonymous No.8531276]

>>8530779
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.

[Anonymous No.8531439]

>>8531276
I don't even know what I'm talking about

[Anonymous No.8531449]

>>8528499
That question is trivial with the Lambert W """"function""""

[Anonymous No.8531454]

Here's an easy warm-up question

[Anonymous No.8531487]

>>8531454
(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

[Anonymous No.8531491]

>>8531454
P(gold) = 0.3 + 0.2 * P(gold)
P(gold) = 0.3 / (1 - 0.2) = 0.375

-> 37.5%

[Anonymous No.8531496]

>>8531491
Wait. Aren't there 2 boxes inside the main one? So shouldn't the 3rd term be P(gold) ^ 2 ?

[Anonymous No.8531504]

>>8531487
>>8531491
Anyway, what I did:
P(x) = P(no gold)
P(x) = 0.5 +0.2*P(x)^2
P(gold) = 1 - P(x)

[Anonymous No.8531508]

>>8531487
P = 0.3 + 2*0.2P -> 0.6P = 0.3 -> 0.5

[Anonymous No.8531509]

>>8531454
isn't it just 30%

[Anonymous No.8531512]

>>8531508
Why are you multiplying by 2? What if there are 10 boxes inside the main one?

[Anonymous No.8531522]

>>8531454
1/6

[Anonymous No.8531527]

>>8531512
>20% chance of containing two treasure chests with the same odds as this one

[Anonymous No.8531538]

>>8531527
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

[Anonymous No.8531556]

>>8530048
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).

[Anonymous No.8531576]

>>8531538
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.

[Anonymous No.8531600]

>>8531496
Ah yeah, didn't read carefully enough...

[Anonymous No.8531631]

$Testing the LaTeX$

[Anonymous No.8531635]

$$Dble money?$$

[Anonymous No.8531642]

>>8531635
[math]5^5[/math]

ok got it

[Anonymous No.8531668]

>>8529640
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).

[Anonymous No.8531717]

>>8531491
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
-----------------

[Anonymous No.8531718]

>>8531454
I got 5/12

[Anonymous No.8531721]

>>8531454
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 >>8531487. But he forgot to cancel to the lowest form. Or show working.

[Anonymous No.8531752]

>>8529143
ISTR the fact that AEB is isosceles (20-140-20) comes into it somewhere.

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

[Anonymous No.8531761]

>>8529013
There's no information about n, so...

[Anonymous No.8531846]

>>8531717
I'm getting something different.

[Anonymous No.8531871]

>>8531717
Oh jk didn't read your post correctly.

[Anonymous No.8531875]

>>8529143
Recognizing that ABC and BEA are isosceles and that you're allowed to do algebra on lengths makes the problem easy.

[Anonymous No.8531900]

>>8530496
I get the same answer. Brute-forced using Python.
http://pastebin.com/JfYj5xKr

[Anonymous No.8531934]

>>8530362
> How does 1 and 6 get eliminated?
When Bernard says that he knows what they are.

Prior to that, 1,6 and 3,8 are still plausible combinations. Both have the same difference (5), so if Bernard knows the answer, the difference can't be 5.

[Anonymous No.8531978]

>>8531761
its supposed to be limit as n goes to infinity

[Anonymous No.8531987]

>>8531900
great minds think alike :^)

[Anonymous No.8532019]

>>8530444
theres nothing wrong with that method, just assume that the number is complex once you find there are no real solutions, and the look at the case where it is purely imaginary and you quickly find both a set of general and most simple solutions

[Anonymous No.8532209]

>>8530952
That˙s (tan x squared plus one) squared.

So just remember that tan = sin/cos, write the 1 as a fraction with cos x squared in denominator, remember that cos squared + sin squared = 1, and get your result.

[Anonymous No.8532606]

>>8531509
30% to get 1 gold bar, 20% to try again twice.

[Anonymous No.8532611]

>>8531576
try this:
>20% chance of containing 10000000000000000 treasure chests with the same odds as this one.

You should try catching up on some basics on probability theory, before jumping in on harder problems.

[Anonymous No.8533619]

>>8530060
>he uses analysis and algebra
Your only tool is a compass. Fucking pleb.
Who needs those shitty numbers.

[Anonymous No.8533666]

>>8528499
Semi-pleb here.
Why is taking the derivative of both sides to find a not a valid step?

Did this and solved a = e^-1 (incorrect)

[Anonymous No.8533670]

>>8533666
What are you even taking the derivative with respect to? The left side is not equal to the right side for all a, the derivative of the left and right side are not necessarily equal. It's like saying that because x^2 = 1 at x = 1, the the derivative of x^2 equals the derivative of 1.

[Anonymous No.8533676]

>>8533666
first, you should't be taking the derivative to find a. that would simply complicate things.
second, one must be familiar with euler's formula to understand that the answer is i, or (-1)^(1/2)

[Anonymous No.8533677]

>>8533670
Of course, thanks.

[Anonymous No.8533679]

>>8533676
Thanks, I understand that but is there no way other than successive approximation to get a real answer for a?

[Anonymous No.8533680]

>>8533679
Never-mind, can't do that, only answer is i

[Anonymous No.8533697]

>>8528507
i^i has multiple solutions tho

[Anonymous No.8533714]

>>8533697
yes. so read the instructions of the question

[Anonymous No.8533875]

>>8528499
P=NP

[Anonymous No.8534078]

>>8529898
try again

[Anonymous No.8534680]

>>8528499
-1/12

[Anonymous No.8534718]

Let [math]n \ge 2[/math]. Compare [math]n, \sqrt{\phi(n)\sigma(n)}[/math], and [math]\frac{\phi(n)+\sigma(n)}{2}[/math], where [math]\phi[/math] is the Euler totient function and [math]\sigma[/math] is the sum-of-divisors function (ie. it associates to n the sum of its positive divisors)

[Anonymous No.8534861]

>>8530496
7 is wrong, supposed to be A

[Anonymous No.8535274]

>>8534861
Changing the answer to any one question changes the answers to other questions, ending up with a domino effect.

At least two of us have solved it by brute force and discovered that the only solution which doesn't lead to an inconsistency is the one given.

If you think that there's another answer, you need to state the answers to all 10 questions (so that we can point out which one you got wrong).

Clearly, changing Q7 to A alone results in an incorrect solution because the answer to Q3 wouldn't be C.

[Anonymous No.8535278]

>>8534861
> 7 is wrong, supposed to be A
The answer given is B, i.e. that Q2 (B) has the same answer as Q7 (B). Which isn't wrong.

[Anonymous No.8536700]

>>8531487
Maybe I'm being retardeed, but isn't the Quaternions a counterexample to this?

[Anonymous No.8536800]

>>8536700
Yeah nevermind I'm being retarded

[Anonymous No.8537489]

>>8531454
0.999...

[Anonymous No.8537578]

>>8529013

... (a^n + b^n)^(1/n). The thing we are taking the limit of is independent of x...