SuicideFuel Math thread problem (official)

[Med Amine]

That's medicine, not STEM

medicine is a part of science

 

[trying to ascend]

medicine is a part of science

So are geography and sociology, they are social sciences.

But the S in STEM doesn't refer to all sciences

 

[turbosperg]

agepill-teenlovepill=

Over

---

If an oven can bake one cake every hour,

how many cakes can 40 ovens bake in 3 years?

---

 

[Ahnfeltia]

View attachment 844872

 

[Ahnfeltia]

If an oven can bake one cake every hour,

how many cakes can 40 ovens bake in 3 years?

Is that a Holocaust reference or am I way too far gone?

 

[turbosperg]

Is that a Holocaust reference or am I way too far gone?

If you forget the cake in the oven for too long it will surely become holocausted down to ash.

 

[CruxGammata]

why is this still pinned?

 

[trying to ascend]

why is this still pinned?

Because it's the HIGHEST IQ thread on the site

 

[belysning]

when multiplied by 0 the answer is 0

Duh

 

[Puppeter]

Chad and Pepe are playing with a pile of black, red and green stones.
When the game starts, there are many black pieces on the game board
could be any positive integer.

Players take turns. Chad took the first move. There are 2 options for a move:
(1) Place a red or green stone (from an infinite supply) on top of an existing stone pile. At the end of the round, there may not be two tiles of the same color in the stack.
A stack can contain up to three stones.
or
(2) Take out 2 stacks, but the top colors of the two stacks must be the same. Stack height doesn't matter.

If you can't take any action, you lose.
Determine whether either player can use a strategy to guarantee victory, and if so, explain how.
k=6

Tldr

 

[incelpardo99]

Finally, some good fucken high IQ level thread, get them one of the Guidorizzi's books so they can whack their brains in OP

 

[incelpardo99]

Finally, some good fucken high IQ level thread, get them one of the Guidorizzi's books so they can whack their brains in OP

Just remembered this one is brazilian, might be hard to understand the question, maybe a Strang or Stewart might be better

 

[incelpardo99]

Lol like an inverse relationship, if you are good at math chances are you are a professional or are studying in STEM and you are basically a betabuxxx bitch which makes you a volcel for not trying hard enough in your academic/professional area and rotting here instead

When your realize that your qt foid STEM class coleague chooses to be declared volcel but in reality fucks a chad a day just like everyone else but you you might end up in here too

 

[trying to ascend]

When your realize that your qt foid STEM class coleague chooses to be declared volcel but in reality fucks a chad a day just like everyone else but you you might end up in here too

Consider the function f(n), such that f(0) = 0 and f(⌊n/10⌋) + n - 10⌊n/10⌋, how many algorisms does the smallest positive integer m has, such that f(m) = 2015?

 

[incelpardo99]

Consider the function f(n), such that f(0) = 0 and f(⌊n/10⌋) + n - 10⌊n/10⌋, how many algorisms does the smallest positive integer m has, such that f(m) = 2015?

Is the function f(n) defined by f(n) = f(⌊n/10⌋) + n - 10⌊n/10⌋? You didn't make that clear

 

[trying to ascend]

Is the function f(n) defined by f(n) = f(⌊n/10⌋) + n - 10⌊n/10⌋? You didn't make that clear

Yes

 

[incelpardo99]

Yes

def f(n):
if n == 0:
return 0
else:
return f(n // 10) + n - 10 * (n // 10)
m = 1
power = 10
algarism_counter = 1
while True:
if f(m) == 2015:
break
m += 1
if m == power:
power *= 10
algarism_counter += 1
print("m has:", algarism_counter)

Could only count to 9, gave up

 

[trying to ascend]

def f(n):
if n == 0:
return 0
else:
return f(n // 10) + n - 10 * (n // 10)
m = 1
power = 10
algarism_counter = 1
while True:
if f(m) == 2015:
break
m += 1
if m == power:
power *= 10
algarism_counter += 1
print("m has:", algarism_counter)

Could only count to 9, gave up

What is that? Why do you write like a computer?

 

[incelpardo99]

What is that? Why do you write like a computer?

It's code in python for calculating functions with recursive method, but it took so much heap for a long time that I could only see m reaching 9 algarisms, by the way what is the actual answer I'm curious

 

[trying to ascend]

It's code in python for calculating functions with recursive method, but it took so much heap for a long time that I could only see m reaching 9 algarisms, by the way what is the actual answer I'm curious

It's higher than 200 bhai.

Also, why do you use that? YOU CAN'T USE PYTHON AT OLYMPIAD

 

[Grim_Reaper]

def f(n):
if n == 0:
return 0
else:
return f(n // 10) + n - 10 * (n // 10)
m = 1
power = 10
algarism_counter = 1
while True:
if f(m) == 2015:
break
m += 1
if m == power:
power *= 10
algarism_counter += 1
print("m has:", algarism_counter)

Could only count to 9, gave up

I don't think it's anywhere near 9. The highest possible value for n - 10(⌊n/10⌋) is 9, so there's at least 223 algorithms. Idk how you can do this question without brute force, which is already impossible.

 

[incelpardo99]

It's higher than 200 bhai.

Also, why do you use that? YOU CAN'T USE PYTHON AT OLYMPIAD

Didn't know that, sorrie. Also, I reckon floor functions are not differentiable, so I assume you can't use derivatives on this one for m to reach 2015, how do you solve it?

 

[trying to ascend]

Didn't know that, sorrie. Also, I reckon floor functions are not differentiable, so I assume you can't use derivatives on this one for m to reach 2015, how do you solve it?

It's from a high school olympiad, you can solve it without calculus.

I haven't solved it

 

[incelpardo99]

I don't think it's anywhere near 9. The highest possible value for n - 10(⌊n/10⌋) is 9, so there's at least 223 algorithms. Idk how you can do this question without brute force, which is already impossible.

Ye, no heap anywhere near as capable of reaching this result

 

[Monke]

Apes should live comfortable life in the jungle, not this shit!!!!

 

[CountBleck]

Let V be a vector space with dim(V) = n < ∞. Suppose that L: V -> V is a linear map and that 0 is an eigenvalue of L.
Prove that L cannot be surjective.

 

[Grim_Reaper]

Let V be a vector space with dim(V) = n < ∞. Suppose that L: V -> V is a linear map and that 0 is an eigenvalue of L.
Prove that L cannot be surjective.

If 0 is an eigenvalue, then there exists a non-zero vector in the kernel, which means it's nontrivial so dim(L(V)) < n. Therefore, L is not subjective.

 

[CountBleck]

If 0 is an eigenvalue, then there exists a non-zero vector in the kernel, which means it's nontrivial so dim(L(V)) < n. Therefore, L is not subjective.

hey good job. i know it wasnt the most challenging problem, but i just wanted to throw in a cute little algebra problem

 

[Ahnfeltia]

Chad and Pepe are playing with a pile of black, red and green stones.
When the game starts, there are k many black pieces (stone piles) on the game board
k could be any positive integer.

Players take turns. Chad took the first move. There are 2 options for a move:
(1) Place a red or green stone (from an infinite supply) on top of an existing stone pile. At the end of the round, there may not be two tiles of the same color in the stack.
A stack can contain up to three stones.
or
(2) Take out 2 stacks, but the top colors of the two stacks must be the same. Stack height doesn't matter.

If you can't take any action, you lose.
Determine whether either player can use a strategy to guarantee victory, and if so, explain how.
k=6

I think Chad can always force a win. First, Chad takes away 2 stacks. After that, he keeps mirroring Pepe's move. I didn't check super thoroughly, but I think this works.

 

[Ahnfeltia]

Consider the function f(n), such that f(0) = 0 and f(n) = f(⌊n/10⌋) + n - 10⌊n/10⌋, how many algorisms does the smallest positive integer m has, such that f(m) = 2015?

I'll just find the smallest m s.t. f(m) = 2015. Given that the algorithm is already specified, I don't really understand the "how many algorithms" part of the question.

For n < 10, f(n) = n
For 9 < n < 100, f(n) = ⌊n/10⌋ + n - 10⌊n/10⌋ = n - 9⌊n/10⌋
For 99 < n < 1000, f(n) = ⌊n/10⌋ - 9⌊⌊n/10⌋/10⌋ + n - 10⌊n/10⌋ = n - 9⌊n/10⌋ - 9⌊n/100⌋
For 999 < n < 10,000, f(n) = ⌊n/10⌋ - 9⌊⌊n/10⌋/10⌋ - 9⌊⌊n/10⌋/100⌋ + n - 10⌊n/10⌋ = n - 9⌊n/10⌋ - 9⌊n/100⌋ - 9⌊n/1000⌋

The pattern should be clear by now. If n has digits abcd with a > 0, then f(n) = abcd - 9(abc + ab + a) = a + b + c + d. At this stage, it should be easy to convince yourself that f is just the digit sum function.

The smallest number with digit sum 2015 is an 8 followed by 223 9s since 2015 = 223*9 + 8.

@Grim_Reaper @incelpardo99

 

[trying to ascend]

I'll just find the smallest m s.t. f(m) = 2015. Given that the algorithm is already specified, I don't really understand the "how many algorithms" part of the question.

For n < 10, f(n) = n
For 9 < n < 100, f(n) = ⌊n/10⌋ + n - 10⌊n/10⌋ = n - 9⌊n/10⌋
For 99 < n < 1000, f(n) = ⌊n/10⌋ - 9⌊⌊n/10⌋/10⌋ + n - 10⌊n/10⌋ = n - 9⌊n/10⌋ - 9⌊n/100⌋
For 999 < n < 10,000, f(n) = ⌊n/10⌋ - 9⌊⌊n/10⌋/10⌋ - 9⌊⌊n/10⌋/100⌋ + n - 10⌊n/10⌋ = n - 9⌊n/10⌋ - 9⌊n/100⌋ - 9⌊n/1000⌋

The pattern should be clear by now. If n has digits abcd with a > 0, then f(n) = abcd - 9(abc + ab + a) = a + b + c + d. At this stage, it should be easy to convince yourself that f is just the digit sum function.

The smallest number with digit sum 2015 is an 8 followed by 223 9s since 2015 = 223*9 + 8.

@Grim_Reaper @incelpardo99

Correct

 

[CountBleck]

Let A be an n x n matrix with complex entries. If trace(AB)=0 for every n x n matrix B, show that A=0.

 

[Ahnfeltia]

Let A be an n x n matrix with complex entries. If trace(AB)=0 for every n x n matrix B, show that A=0.

Suppose a_ij != 0. Let B = PD where P is a permutation matrix which will swap columns i & j of A and D is a diagonal matrix with d_ii = 1 and d_kk = 0 for k != i. Then AB will be nil save for column i, whose diagonal element will be a_ij. However, a_ij = tr(AB) = 0 is now a contradiction.

 

[CountBleck]

Suppose a_ij != 0. Let B = PD where P is a permutation matrix which will swap columns i & j of A and D is a diagonal matrix with d_ii = 1 and d_kk = 0 for k != i. Then AB will be nil save for column i, whose diagonal element will be a_ij. However, a_ij = tr(AB) = 0 is now a contradiction.

good job ! an elegant solution too. that problem is an old favorite of mine

 

[Grim_Reaper]

I'll just find the smallest m s.t. f(m) = 2015. Given that the algorithm is already specified, I don't really understand the "how many algorithms" part of the question.

For n < 10, f(n) = n
For 9 < n < 100, f(n) = ⌊n/10⌋ + n - 10⌊n/10⌋ = n - 9⌊n/10⌋
For 99 < n < 1000, f(n) = ⌊n/10⌋ - 9⌊⌊n/10⌋/10⌋ + n - 10⌊n/10⌋ = n - 9⌊n/10⌋ - 9⌊n/100⌋
For 999 < n < 10,000, f(n) = ⌊n/10⌋ - 9⌊⌊n/10⌋/10⌋ - 9⌊⌊n/10⌋/100⌋ + n - 10⌊n/10⌋ = n - 9⌊n/10⌋ - 9⌊n/100⌋ - 9⌊n/1000⌋

The pattern should be clear by now. If n has digits abcd with a > 0, then f(n) = abcd - 9(abc + ab + a) = a + b + c + d. At this stage, it should be easy to convince yourself that f is just the digit sum function.

The smallest number with digit sum 2015 is an 8 followed by 223 9s since 2015 = 223*9 + 8.

@Grim_Reaper @incelpardo99

Correct

How many algorithms are there though?

 

[Ahnfeltia]

How many algorithms are there though?

Given that the algorithm is already specified, I don't really understand the "how many algorithms" part of the question.

 

[Ahnfeltia]

good job ! an elegant solution too. that problem is an old favorite of mine

It's a nice problem. I'll be assisting with linear algebra this quarter. Maybe I'll try to sneak in this problem.

 

[CountBleck]

It's a nice problem. I'll be assisting with linear algebra this quarter. Maybe I'll try to sneak in this problem.

damn youre living my dream

 

[Ahnfeltia]

damn youre living my dream

I hope you'll get there then. I also hope it'll bring you more joy than it does me. I'm currently getting a PhD but I don't see myself continuing in academia.

 

[CountBleck]

I hope you'll get there then. I also hope it'll bring you more joy than it does me. I'm currently getting a PhD but I don't see myself continuing in academia.

yea for now as a 3rd year undergrad academia seems like the move. but i imagine that may change once i actually see how brutal it is. i'll still try it out tho. i got nothing to lose

 

[Med Amine]

when you go asian, you never fail an equation

 

[trying to ascend]

How many algorithms are there though?

224

 

[Grim_Reaper]

224

Amazing

 

[Ahnfeltia]

yea for now as a 3rd year undergrad academia seems like the move. but i imagine that may change once i actually see how brutal it is. i'll still try it out tho. i got nothing to lose

My thoughts exactly at the time

 

[IronsideCel]

I hate IQ moggers.
If you answered a correct question in this thread, you must be a sexhaver.

 

[Ahnfeltia]

I hate IQ moggers.
If you answered a correct question in this thread, you must be a sexhaver.

I'm too busy doing math to have sex

 

[Grim_Reaper]

I hate IQ moggers.
If you answered a correct question in this thread, you must be a sexhaver.

@Ahnfeltia probably has foid students asking him to give them better grades, and in-exchange they have sex with him.

 

[IronsideCel]

@Ahnfeltia probably has foid students asking him to give them better grades, and in-exchange they have sex with them.

 

[Ahnfeltia]

@Ahnfeltia probably has foid students asking him to give them better grades, and in-exchange they have sex with him.

only happened thrice thus far

 

[Grim_Reaper]

only happened thrice thus far

@IronsideCel