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File 162191634958.png - (12.94KB , 500x500 , C2_5_000.png )
1001663 No. 1001663 ID: afe7de

There’s going to be guest artists and writers in this thread, so they’ll be credited as they come up, with a final credits at the end of the thread.

CATALYST PART 1: https://questden.org/kusaba/quest/res/986604.html
CATALYST PART 2: https://questden.org/kusaba/quest/res/993796.html

WIKI: https://questden.org/wiki/CATALYST
DISCUSSION: https://questden.org/kusaba/questdis/res/134892.html


Author’s Commentary: This Quest contains 18+ content including violence, sexual content, angry characters, and more. Reader discretion is advised.
304 posts omitted. Last 50 shown. Expand all images
No. 1005157 ID: 094652

You got it right - my idea was to permutate again, but I forgot that stacking similar terms means I counted the same outcome twice. My bad.

>"I feel so grateful that I have the chance to help educate and improve the people of this fine planet."
Mizz Es sus
I'll take the safe bet of five internet points for "non-Angelic Supernatural Faction" (Demon Faction, Reincarnation (standard and/or exalted), or Alien). She's clearly carnal in nature and is advocating the virtues of critical thinking.
No. 1005160 ID: f8fa51

Seconding this answer.
No. 1005162 ID: 871f70

What if we get enough questions right, she comes visit Lyst sometime this coming week to personally congratulate us? Maybe even live in lyst for awhile even?

Support >>1005154

Also, see what happens if you decide to call the number instead of texting it... if you dare
No. 1005218 ID: eedbeb
File 162586641615.png - (12.23KB , 500x500 , p10.png )

Gena: This one’s easy.
Andrea: I’ll take your word for it.
Gena: So you can have either one 1, one 6, or neither. Find the likelihood of each scenario and add ‘em together.
Andrea: You handle this one. I feel like I burned out young.
Gena: The high expectations of you parental figure has made you fall into the DEBILITATING trap of poor executive function.
Andrea: Sure babe, anything you say.

Gena types in 20/27 into her COMM and sends the message. Mizz Es twirls and strikes a pose.

Mizz Es: Gena has done it again! Such a delight, I could do this forever. They say teaching is the most rewarding career and I wholeheartedly agree.

Andrea: You got that recording, right? We need it so people believe our story and don’t think we were sleep deprived or on drugs.
Gena: Weirder things have happened in Lyst, but yeah, I got it.

No. 1005221 ID: eedbeb
File 162586659246.png - (14.50KB , 500x500 , p11.png )

Mizz Es: I’m getting word that our excellent students are from the lovely town of Lyst. This week that area seems to be the hot destination for action of all kinds, and the citizens there could certainly benefit from the virtues of problem solving and critical thinking.

Gena: Okay why hasn’t she dropped the act yet and start talking to us like she can hear us instead of pretending she’s on TV?
Andrea: Respect the production. Is she saying we’re stupid?
Gena: I hope she meant that everyone else here is stupid, and we’re the lone beacons of intelligence and rationality.

You and Gena lapse into SELF-DEPRECATING giggles at the joke.

Andrea: Let’s ask her to just talk to us like normal.
Gena: Hey Mizz Es! One of us has to sleep sometime tonight so are you going to give us math problems forever or do we get a prize or closure or something? You know where we live now, and stuff, so if you want to stop by sometime we’d be happy to show you around.
Andrea: We have a good brothel.
Gena: There is a SUPERB brothel. I can make you a cute dress.
Andrea: I can show you all the hot boys AND girls.

No. 1005222 ID: eedbeb
File 162586661112.png - (8.61KB , 500x500 , p12.png )

You’re not nervous that Mizz Es is going to smite or mind control you anymore and you’re smart enough to SUCK UP to mysterious benevolent figures of extreme power so you eagerly anticipate the woman’s response.

Mizz Es’s smile falters for a moment, then returns with extra force.

Mizz Es: I’ve had a marvelous idea, if Gena and Andrea can correctly answer an exciting gauntlet of FOUR math problems, then they’ll win a special reward!

There’s a canned ‘OOOOooooo’ for emphasis.

Mizz Es: I understand if our guests need to leave, but here’s the first problem if they’d like to participate. Note that not all of these questions will be PROBABILITY.

The first question appears on screen. It reads:

Let L be the set of positive integers y for which 1/y has the repeating decimal representation 0.ststst… repeating, with s and t being different digits. What is the sum of the numbers in L?
No. 1005226 ID: 96c896

ok so any repeating decimal of that form has to include 11 in its denominator, because 1x+10x = 11x and would turn the repeating decimal into a non-repeating decimal. However, not all denominators with 11 as a factor work. In addition, y cannot be greater than 100 since that would return a decimal in the form of 0.00x, which won't work.

We can just test each multiple of 11 up to 99, and it turns out 11, 33, 99 work. The answer is 143.

Hey, do a little more research and find out if anyone got all their questions right. I feel like what she's doing is farming achievement points, as there's no reason to target you two specifically for something nefarious and people she interacts with get out just fine. I feel like she's sad because she either doesn't want the session to end so soon, or she knows she can't visit. Maybe it's a transmission from another dimension!
No. 1005237 ID: 094652

>Special Reward
Ask if the special reward is incriminating evidence of whoever has been torturing the local wildlife with fire. Because seriously.

If that's not possible, you'll settle for everything-expansion.
No. 1005239 ID: 96c896

Oh and I'm fairly sure 11, 33 and 99 work because 1/3 is the only one-digit repeating decimal with a prime denominator between 1 and 10, so you can put that into the denominator to make a new decimal without interfering with 11's pattern.
No. 1005248 ID: f8fa51

Don't ask for a specific reward. I have a feeling Mizz Es already has one in mind and might get annoyed if we ask for something specific.
No. 1005250 ID: e51896


Support, I wanna be surprised by the gift we'll receive instead of asking for a specific reward.
No. 1005287 ID: eedbeb
File 162595028802.png - (12.96KB , 500x500 , p13.png )

You turn to Gena with a solemn expression on your face.

Andrea: I’m counting on you to do these problems cause I like the sound of a special prize. Do you need anything? Coffee, stamina liquor, head?
Gena: MORAL support and some extra scratch paper would be good. Whatever the reward is, I’ll share it with you.
Andrea: Gena Gena, she’s our mouse, if she can’t do it then I understand and am grateful for the work she’s put in so far.

Gena punches you in the shoulder lightly and you bustle around fluffing the couch cushions and making her comfortable. Mizz Es sits down on the desk and absently pulls out a cigarette which makes your panties feel reasonably DAMP.

Gena: Okay so like...it’ll have to have 11 in the denominator, and then we can use numbers that are multiples of 11 ‘cause we’re doing 1/y…

You slide off the couch and slither in front of the TV to watch Mizz Es smoke. If only you could reach through the screen and offer a lighter with a trembling paw. Jerk off material for later. When you called Gena a simp earlier it had been a defense mechanism against the truth. That you are the simp.
No. 1005288 ID: eedbeb
File 162595035692.png - (7.84KB , 500x500 , p14.png )

Mizz Es taps ash on the set floor and not your tits, sadly, which reminds you of the case you’re supposed to be solving at work, the one where someone’s hurting the local wildlife and being a general prick. You don’t want to be rude and ask what the prize is, but you wonder if Mizz Es could HELP fight CRIME using her apparent OMNISCIENCE.

Andrea: Yo teach, you sound like you know what’s going on in Lyst. Can you give me some clues about a crime I’m working on? Some asshole is burning animals in the words to get their rocks off and I can’t figure it out or catch ‘em.

Mizz Es chuckles and wastes more of that delicious looking ash.

Mizz Es: I’m afraid I don’t know much about your sweet little town besides recent hearsay and gossip. You’re a bright young lady, even if you’re not a fan of math I think you can work out this problem.

Andrea: Well, the fires aren’t being caused by gasoline or something so it’s probably an awakened. I don’t really know anyone with fire powers here so that’s a dead end.

Mizz Es: You do know a very interesting young man whose arrival has kicked off several unusual events.

Andrea: And he has a mighty fine dick. He was the one who reported it originally, but maybe he’s learned something new. Yeah, okay I’ll ask him tomorrow.
No. 1005289 ID: eedbeb
File 162595041254.png - (11.14KB , 500x500 , p15.png )

While you’re stewing in thought and unrequited lust, Gena types a few numbers into her COMM calculator.

Gena: Think I got it, only 11, 33, and 99 work, so the sum of those is 143.
Andrea: Go for it.

Mizz Es stubs out her cigarette on her wrist and claps her hands once the answer has sent.

Mizz Es: Starting out strong! We have three questions left before I reveal my prize. Here’s the next question, which is also NUMBER THEORY.

The second question appears on the screen. It reads:
Let C be a subset of {1,2,3,….,30} with the property that no pair of distinct elements in C can have a sum that’s a multiple of 5. What is the largest possible size of C?
No. 1005292 ID: 465a14

That's easy. 5's a prime number so its only factors are 1 and itself. Just avoid having both 1 and 4 or both 2 and 3 in C, which means we can just have it be the set of all integers from 3 to 30.

Of course, possibly something else was meant by "factor" but then that's a problem for the problem-writer, not us.
No. 1005293 ID: f8fa51

Let's do the question as if it were worded as multiples anyway. This is a modulus problem. We can't have any number that is 1 mod 5 with any number that is 4 mod 5, and the same for 2 mod 5 with 3 mod five, and we can only have a single number that is 0 mod 5. So let's say we pick a single 0 mod 5 number, every 1 mod 5 number, and every 2 mod 5 number. That's two-fifths of the set plus 1, which is 13. This is simplified by the fact that the original set is evenly distributed between the modulus groups, and would have involved another step otherwise.
No. 1005294 ID: bc11b8

No. 1005295 ID: 094652

Let's generalize the problem and see if we can work out an induction.
Let A be the set of natural numbers from 1 to a, and let b be a natural number. Let B be any subset such that for all x,y in B_i, x/=y -> MOD(x+y, b) /= 0.
We want to find the particular subset B_i whose sum of all elements is equal to or higher than any B.

There's a trick to this question; she said no pair of distinct elements can be a multiple of 5, which means that 30 can be in the subset because as long as we don't add any other multiple of 5 to the subset,
One possible subset is [26,27,28,29,30]=140, which is the set of numbers greater than the largest number in the set minus 5, which ensures each of the numbers is of a different type of modulo with 5, and thus none of the numbers hybridize. But my intuition suspects that at sufficiently high numbers, skimming the 'top' will not be sufficient to offset the majority of contributions.
Another possible subset is [1,6,11,16,21,26,30], because each number is sufficiently spaced apart evenly that no two numbers reach a multiple of 5. However, 1 isn't expected to be of much use, here.
I suspect the answer is a hybrid of the two questions, dependent on the ratio between a and b.

Maybe the answer is simpler than one might think: if a = 5, what is the largest possible value?
5 is allowed because its modulo is 0 and thus when added to any nonzero modulo will be nonzero.
4 cannot interact with 1
3 cannot interact with 2
So the set with the largest possible sum of elements is [3,4,5] = 12.
Let's test this by induction:
Suppose we select a subset B which consists of only elements x in A up to a such that x mod b <= b/2. We assume that sum of subset B's elements is the largest possible out of all subsets in the limitation that (for all x,y in B_i, x <= a, y <= a, x/=y -> MOD(x+y, b) /= 0) (where a is a multiple of b). Now let's prove that B', which consists of only elements up to a+b, is also the largest of its kind.
B' consists of the elements in B. We have assumed that B is the largest. Now we focus on subset B_5, which consists of elements [a-floor(b)/2...a+b]. We can arbitrarily see that this is the largest possible set of all sets of numbers from a to a+b, because adding any number from the lower levels means sacrificing at least one higher-numbered level. Since B and B_5 make up B', and each is the respective maximum of its range on natural numbers, we conclude that B' is the largest possible.

[3,4,5, 8,9,10, 13,14,15, 23,24,25, 28,29,30] = 240
This takes up a chunk of the largest numbers without actively finding any two which sum to a multiple of 5.
No. 1005296 ID: 96c896

Oh, yep, that's the solution. I'll write out more of it to make it easier for more people to see why:
0 mod 5 is 5,10,15,20,25,30 and any of those can add to any other one in the group to make a multiple of 5. Thus only one can exist.
1 mod 5 would be: 1,6,11,16,21,26. Six numbers, none of which can add to eachother to make a multiple of 5. Same for 2 mod 5: 2,7,12,17,22,27.
3 mod 5 is 3,8,13,18,23,28 which can clearly make a lot of sums with the 2 mod 5 group, and are in fact the only numbers that can make sums with the 2 mod 5 group that add up to a multiple of 5.
Thus, 6+6+1.
No. 1005297 ID: 7f8419

I.e. c is the set
[1,2,5,6,7,11,12,16,17,21,22,26,27] which has 13 elements
No. 1005370 ID: eedbeb
File 162603495563.png - (31.51KB , 500x500 , p16.png )

Gena: I’m getting kind of tired. You doing okay?
Andrea: Big. Mommy. MILKERS.
Gena: Ho-kay, it’s pretty serious then.
Andrea: I need her to step on me with those heels and then spank me with a ruler and tell me I’ve been baaaaaad.
Gena: You can borrow my toys and go masturbate in the other room if you want.
Andrea: A good offer that I may take you up on. In the meantime, can you tell me how to do this problem?
Gena: Oh yeah, sure, so like, you can kind of think it through without knowing any special math stuff, but it helps to use modular arithmetic.
Andrea: Wow, my horniness, completely GONE. The doctors said it couldn’t be done.

Gena rolls her eyes and writes the numbers 10 through 15 on her paper. Despite your teasing you’re still interested and you lean over to look.

Gena: So MODS are like the remainder after you divide the number by the intended target. For example, 10 mod 5 is 0 and 13 mod 5 is 3. This problem is actually pretty nice because our brains are used to dividing by 5s.
Andrea: Right.
Gena: We can have all the numbers in our range that are 2 mod 5 and numbers that are 4 mod 5 and they won’t add to a multiple of 5. We could do the same with the 3s and 1s, but it’s the same answer either way. We can throw one 0 mod 5 in there to get a total of 13 numbers.

No. 1005371 ID: eedbeb
File 162603497447.png - (9.60KB , 500x500 , p17.png )

Mizz Es: That was fast! We’ve done PROBABILITY and NUMBER THEORY, how about some TRIGONOMETRY? Don’t be afraid to use a calculator or refresh yourself on SIN, COS, and TAN.

Andrea: Why couldn’t it have been normal GEOMETRY. I love CURVES.

A helpful diagram of a triangle appears alongside the next problem. It reads:

Maple and Kazu live 2 miles apart. One day Maple looks due north from her house and sees a hero flying across the sky. At the same time Kazu looks due west from his house and sees the same hero. The angle of elevation of the airplane is 30 degrees from Maple’s position and 60 degrees from Kazu’s position. What is the hero’s altitude, in miles?

You can choose to go suffer in bathroom while Gena is doing this problem.
No. 1005374 ID: 67181a

The answer is (the square root of 3)/2. O dont have a convenient way to type a square root symbol.

30/60/90 triangles have 1/square root of 3/2 ratios of side length, so the triangle can be broken down into two rifht triangles with a perpendicular line from the hypotenuse that intersects the right angle point representing the hero. This makes two more 30/60/90 triangles, with hypotenuses of either 1 or the square root of 3; and breaking it down from either gives the same answer for the hero's height.
No. 1005375 ID: 67181a

I misread the question! Didnt notice that they werent looking at a hero between them, so in the interest of bullheadedly solving this trig problem without remembering any trig i made a chart to help me keep everything organized.

To be as precise as possible, the hero is flying at an altitude of "the square root of 1.2" miles
No. 1005376 ID: f8fa51

Let's call the angle of elevation between Kazu and the hero K (60°), the angle of elevation between Maple and the hero M (30°), the distance between Maple and Kasu MK (2 mi.), and the hero's elevation e.

Using tanθ = O/A, we can say MH=e/tan(M) and KH=e/tan(K). Using a²+b²=c² we can say that e²(1/tan(M)²+1/tan(K)²)=MK² and therefore that e²=MK²/(1/tan(M)²+1/tan(K)²). Substituting values, we get e² = (2 mi.)² / (1/√⅓² + 1/√3²) = 4 mi.² / (3 + ⅓) = 4 mi.² / (10/3) = 12/10 mi.², therefore e = √6/5 mi. (approximately 1.0954 mi.)
No. 1005377 ID: 96c896

Ok, so, this is a 3d trig problem. We know the angles of two triangles oriented perpendicular to the ground, but no distances of their sides. We also know a third triangle's hypoteneuse-- the one that's parallel to the ground-- and that triangle shares sides with the other two.

I got h=sqrt(6/5) miles. We don't even need trig functions because the 30/60/90 triangle is so nice and the Pythagorean theorem doesn't care about angles for the one triangle we don't know the angles for.

Detailed steps:
A 30/60/90 triangle has a ratio of x,xsqrt(3),2x for its sides. The x is opposite the 30, the xsqrt(3) is opposite the 60, the 2x is opposite the 90.
We want the height of the hero, so we label that h. We use Maple's triangle first, which has h as the side opposite the 30. Thus the side opposite the 60, which is the distance along the ground to the hero, is hsqrt(3). We don't care about the hypoteneuse.
In Kazu's triangle the side opposite the 60 is h, so we need to do a bit of basic algebra to find out the side opposite the 30. h=xsqrt(3) so x=h/sqrt(3). Thus, the side we're looking for is h/sqrt(3).
Now, Pythagorean theorem. a^2+b^2=c^2. We now know a and b, and c is 2 miles. [hsqrt(3)]^2 + [h/sqrt(3)]^2 = 2^2.
3h^2 + (h^2)/3 = 4
(10h^2)/3 = 4
h^2 = 12/10 = 6/5
h= +/-sqrt(6/5) but it's distance so it's positive.
No. 1005378 ID: f8fa51


Ah, in this post I forgot to mention that I had also labelled the horizontal distance between Kazu and the hero KH and the horizontal distance between Maple and the hero MH.
No. 1005379 ID: bc11b8

one mile
No. 1005380 ID: bc11b8

lol woops, 1.0954451150103322269139395656016
No. 1005382 ID: 094652

Argh, I messed up the previous problem by not reading the question right and it wasn't even the right answer for max sum![3,4,8,9,13,14,18,19,23,24,28,29,30] is the correct solution to "what is the largest set" AND "what is the set whose sum is the largest".

I'll verify:

There are a lot of unknowns in this problem; Primary of which is: where the heck are Mark and Kazu, anyway?
Say that the height of the airplane (yep) is y, which we need to evaluate for this problem. We know that the angle at which Mark sees the - wait, we assume that this takes place on a worldplate or something, right? Because the curvature of a ball-shaped planet would require higher-level mathematics to solve, and more importantly, the actual size and shape of the planet otherwise this problem is impossible anyway airplane is 30 degrees. This means, if the airplane shoots a marker perfectly downwards, the distance between Mark and the marker (heh), which we define as m_n, is m_n = y/tan(30 degrees), where tangent represents the ratio of a right-triangle's angle based on the distance of the opposite side's length divided by the adjacent non-hypotenuse side's length - in this case opposite is y and adjacent is m_n. Similarly, the distance between kazu and the marker is k_w = y/tan(60 degrees). Now we see that there is a right triangle whose sides are m_n, k_w, and 2 (the Euclidian distance between Mark and Kazu's houses). simply apply Pythagoreas' theorem and solve for y:
m_n^2+k_w^2 = 2^2 -> y^2/(tan(30deg))^2+y^2/(tan(60deg))^2 = 4. Tangent of 30 degrees is 1/sqrt(3), Tangent of 60 degrees is sqrt(3), so y^2/(1/3)+y^2/(3)=4 -> y^2*(10/3) = 4 -> y^2 = 12/10 = 6/5 -> y = sqrt(6/5). So yeah, you got it.
No. 1005411 ID: eedbeb
File 162611207880.png - (11.54KB , 500x500 , p18.png )

Gena is totally immersed in drawing triangles and you stumble to the bathroom to spray yourself with cold water and collect your nerves. The shower has one of those convenient extendable nozzles and you scream your frustration to the Goddess as you deal with the fact that some people are probably magic ILLUSIONS and will never have sex with you.

Some of that screaming may involve the stress that’s come with Cat’s arrival. Not that there’s anything wrong with the guy, but there’s definitely been a shift in Lyst since he got here. You don’t have Iraphena’s foresight, but you’ve lived here long enough that you can feel the atmosphere shifting. Willamina’s awakening, Isabella getting more unstable, Cannie LOSING a fight, and now a mysterious lady is on Gena’s TV.

You gargle some shower water and slick back your fur. Every night, you’re slowly left alone as your friends go to sleep.
No. 1005412 ID: eedbeb
File 162611210804.png - (10.83KB , 500x500 , p19.png )

By the time you collect yourself and return to the living room, Mizz Es is already explaining the next problem.

Mizz Es: I’m positively quivering with excitement. We’ve reached the last problem! I don’t want to make this too easy for you so we’re going back to PROBABILITY with a good helping of ALGEBRA thrown in. This problem includes the FLOOR function, which takes a real number x and gives the greatest integer less than or equal to x. For example the floor of 2.3 would be 2.

Gena: Cool, that makes sense. Andrea, you alive?
Andrea: I get the feeling that we’re not the protagonists, but we’re important enough that one of us might die for the hero’s plot development.
Gena: Bitch, I have been doing math problems for an HOUR, do not pull this PHILOSOPHY BULLSHIT on me I will have a breakdown, LMAO.
Andrea: Forget I said anything, get that prize gurl.

The LAST PROBLEM appears on screen:

Real numbers a and g are chosen at random, independently and uniformly from the interval (0,1). What is the probability that the floor of log2(a) is equal to the floor of log2(g)?
No. 1005415 ID: ed0fb9

Andrea: going down your line of thinking, start silently wondering if Gena and your character getting killed in GLM VI by the hands of the Zeroes League is a form of FORESHADOWING
No. 1005417 ID: fd4d13

I don't know how to properly notate this, but I THINK my reasoning is sound. Log2(1/2) is -1, so the log of any number in the set between 1/2 and 1 will also floor to -1. Similarly, Log2(1/4) is -2, so all numbers between 1/4 and 1/2 would return a floor of -2. This continues with the denominator for A or G being doubled each time, so log2(1/8) is -3, log2(1/16) is -4, and each time all numbers between this key fraction and the one before it is the same as the remainder.

So what we really need is to check the odds of both a and g falling within a given range. So the odds of both floors being -1 is (1/2)^2, the odds of both floors being -2 is (1/4)^2, the odds of both floors being -3 is (1/8)^2.

This means that this is the sum of an infinite series from n=1 to n=infinity for (1/(2^n)^2), which I believe is the same as 1/(4^n) if you simplify the exponents. And the infinite series of 1/(4^n) from n=1 to n=infinity (no idea how to notate that typing) sums to 1/3.

I'm pretty confident in 1/3 as the answer to this, but if one of the math folks wants to correct me I'd change my tune
No. 1005418 ID: 094652

>Andrea: I get the feeling that we’re not the protagonists, but we’re important enough that one of us might die for the hero’s plot development.
Eh, you're fine. The rest of the world is so very screwed though.
Unless you're talking about almost everyone being doomed to mortality no matter what, in which case you may begin bunny screeching.

Yeah log was never my strong suit, but your logic seems sound.

... Hey wait a minute.
Lyst... Maple... Kazu... ag? Rezan clan Makag is going to invade Lyst?
No. 1005422 ID: e51896

It'd be funny if after they are done watching this and they check their recording if the television channel was just noise static


If Andrea is thinking about foreshadowing, maybe she can consider this:

Gena's and Andrea's character in GLM VI went inside a post office and met a male postal worker named Greg inside who was somewhat uncomfortable with their characters' lewd state of undress, but Greg came around to it and they even got him to join in their nudity after persuading him by tearing his clothing off.
It is similar to how Cat was uncomfortable to Andrea's advances for awhile when he was on the clock, but came around to it after Andrea convinced him to let her give him a blowjob with Gena.

She could consider There was a moment where Greg took Gena and Andrea into their delivery vehicle which lead them to them being stopped by the Zeroes league and ended in Andrea's character dieing. It's also important to note that Greg even had a mana pistol just like Cat.

Maybe Andrea's time in the RATIOLATRY group meetings taught her to pick up on weird patterns to think up ridiculous theories like that.
No. 1005424 ID: 96c896

There's a mistake here. 1/((2^n)^2) simplifies to 1/(2^2n). Fortunately that ALSO converges to 1/3.

Otherwise it looks correct to me!
No. 1005436 ID: f8fa51

Starting from the result that our probability is the sum of the infinite series 1/2²ⁿ. 1/2²ⁿ further simplifies to 1/4ⁿ. To give the full explanation for how to sum this, this is a geometric series (a + ar + ar² + ..) with a=1/4 and r=1/4.

From wikipedia:
>If |r| < 1, the terms of the series approach zero in the limit (becoming smaller and smaller in magnitude), and the series converges to the sum a / (1 - r).

a / (1 - r) = 1/4 / (1 - 1/4) = 1/4 / 3/4 = 4/12 = 1/3
No. 1005443 ID: 96c896

Oh right I forgot that x^yz = x^zy = (x^y)^z = (x^z)^y due to the commutative property of multiplication, so 2^2n=4^n.
No. 1005517 ID: eedbeb
File 162620345225.png - (11.68KB , 500x500 , p19a.png )

Andrea: I have no idea what that says.
Gena: Ah crap, I don’t think I know how to do it either. I’m bad with logs.
Andrea: She said before that we could look stuff up to help. As long as we don’t look up the actual problem it should be fine.

Gena sighs and rubs her eyes before squinting at her COMM. You keep as quiet and still as possible so you don’t disturb her.

Gena: This range means that the exponent of 2 is going to be negative and the floor will make it more—er—negative.
Andrea: Totally.
Gena: And it breaks into ½, ¼, 1/8 and so on, and the probability the floors are the same as the range squared.
Andrea: I believe every word you are saying.
Gena: Then we get an infinite geometric sequence, and there’s an equation for the sum of those, let me find it…
Gena: The answer’s 1/3!

No. 1005518 ID: eedbeb
File 162620346703.png - (8.81KB , 500x500 , p19b.png )

Mizz Es wipes a genuine tear from her eye and sniffs loudly.

Mizz Es: That’s correct! I’m glad you two got to work together on this one, it really warmed my heart to see everyone’s strengths.

Her face suddenly darkens, as if half the lights in the studio went out. Her abnormally red pupil stands out and sure, rationally it could still be a mod, but you suspect something more MAGICAL.

Mizz Es: You’re both lovely girls who have worked very hard. I’m not supposed to…interact with my students outside of class but in the interest of your continued education I’ll be making an exception.

You hold your breath. It’s happening. Holy shit it’s happening. You’re gonna smash.

Mizz Es: This is your prize. If either of you are in DANGER, give me a call at the number I provided earlier. It will not work for anyone else and there will also be a PRICE for my help. I assure you, however, that the exchange will be worth it. Good luck.

Darn. Foiled again.
No. 1005519 ID: eedbeb
File 162620348120.png - (12.21KB , 500x500 , p20.png )

The TV turns itself off. You and Gena sit staring at the dark screen in the dimly lit room for a while, stewing with your COMMS heavy in your pockets.

Gena: Well it’s late. I had a good time hanging out with you and I’m glad you enjoyed the games. Same time next week?
Andrea: Yep, that’s the plan. If we’re still around by then at the rate things are going.

You don’t move from the couch and instead tap nervous rhythms on the floor with your toes. Anxiety is a piece of shit and your thoughts start spiraling into pathetic self-defeatist paranoia.

Andrea: She wasn’t…mad at us for winning, right?
Gena: No, no I don’t think so. She was trying to spook us so we don’t call her for silly stuff.
Andrea: Would casual sex count as—you know what, I’m self-aware enough to know the answer to that question.
Gena: She’s probably super busy beaming into people’s homes and giving them math problems. Can’t just show up whenever we want, even if we are her favorites.
Andrea: Yeah.

No. 1005520 ID: 465a14

Pretty expected, really, and still cool overall. Math demons or whatever. Neat. At this point I think the only thing left is to find some toys or boytoys or go to the brothel for those blue balls.
No. 1005522 ID: 094652

You two are now hornteased so deeply that your angel and demon halves are busy screwing each other. So give yourselves a quickie.
No. 1005525 ID: 4240ff

When Andrea leaves, she should go to the woods and find Maple if Gena is too tired
No. 1006079 ID: afe7de
File 162685919119.png - (8.70KB , 500x500 , C2_5_122.png )

The night is over, it’s 3AM and you can tell Gena wants to get some shuteye. It’s around this time you begrudgingly pack up. You get a nice long hug from Gena before she smacks your ass and you leave the apartment. You stand by the balcony and listen to your surroundings, looking around. The Cicadas are chirping as they tend to do at this time of year and it adds to the feeling of melancholy you normally get at this time of night. It’s a little muggy out and slightly warm, you already knew it was going to be hot tomorrow even if the news didn’t tell you. It REALLY FUCKING SUCKS to have to deal with this all the time. The lack of sleep, the lack of awake friends. Yeah you could just make some Internet friends but that only goes so far, you never really liked texting as a form of communication, preferring to do stuff irl.

You’re startled from your reverie by the sound of coughing. You look around seeing no one and remember that it’s your text tone, it’s subtle, distracting, and makes others think someone was in the room. It’s really just a super subtle prank Gena got you into but you love the reaction it gets outta kin sometimes. You pull out your phone and see that it’s a message from RANDOM. Probably seeing if you’re going tomorrow.

Random: Yo, coming to the meeting tomorrow?
You: Yeah, same time as always?
Random: Yup, though you can come early if you want, I’m gonna be there early to set up.
You: Kay, I’ve got nothing going on till then anyway.
Random: Cool, see ya there bunny cop

You hang up the phone, not deigning a further response. He’s like you in that way, curt and businesslike over the phone whilst being animated and full of energy in reality. You had a thing for him once, but now you’re more just friends without benefits. You like listening to his rants and theories, and although you aren’t that smart, sometimes you have insights he misses. You let out another sigh

You: Why are kin so fucking WEIRD about stuff sometimes.
You: You’d have thought with all my free time I could at least fake being smart or whatever
You: But noooooooo, my stupid idiot pea brain just fucking sucks and thinks it’s all BORING SHIT

No. 1006081 ID: afe7de
File 162685921978.png - (85.29KB , 500x500 , C2_5_123.png )

You push back the memories, you don’t want to cry or think about sad shit right now. You need a distraction, so you head back to your room. Time to jerk off for a few hours. The door slams as the camera zooms out above the town, and eventually settles on the moons.

No. 1006082 ID: afe7de
File 162685923044.png - (21.48KB , 500x500 , C2_5_124.png )

AUTHORS NOTE: Hey y’all, just checking back in! Sorry for the slight delay, went to visit my parents and was away from my computer for a week. I hope you enjoyed this weird and wild experiment. I wasn’t sure how y’all would take to it, but I and the other authors involved had fun with it. Here’s some parting words from those authors right now!

Poltergeist Ethanoic Acid - PABE segment
You suggestors are surprising. I really didn't expect you guys to pick a third option to have Kelsey work independently over joining the Heroes League or staying with PABE... and I would not have it any other way, thanks for that! :D

Thank you all for being invested in the PABE storyline, and thanks EDMANGO for letting us be a part of this project. Who knows what the future will hold for the Plantimals tv series, Kelsey, and Mizz Es? Wherever they end up, remember that your choices were the CATALYST that shaped their paths =P

Donut - Plantimals Segment
You’re all awesome!

Tippler - Mizz Es’s Math corner
i had no idea there were a bunch of nerds here
that impressed me a lot and made it fun to find new problems
i thought the sexy lady wouldn't be enough to get them to do math
but turns out they just wanted to do the math

I really just wanted to make a neat little collab where I experimented with several ideas that I and they had floating around in our heads. I was wanting to flesh out the world a little bit and plant seeds of things to come, Donut wanted to make an action heavy sequence, Poltergeist Ethanoic Acid wanted to test run an expedited schedule with a new quest idea, and Tippler wanted to trap you with math problems. I’m surprised so many of you took to the math problems with such fervor and creativity. I shouldn’t be cause y’all are a clever bunch, but I am anyway. I’ll probably do more collabs like this in the future with other quest authors if I’m invited or I plan another one because I like how it turned out.

Next up is thread 3, we’ve got a suite of new characters to meet and potential dungeonmates to make so I hope that’ll catch your interest. Logistically the dungeon run is a little intimidating to me and I’m not exactly sure how I want to handle it with so many characters, but I don’t want that to influence you all, if you wanna go in with the full squad of 8, or with as little as 2, you totally can and I won’t stop you, my interests are in telling an interesting or at least mildly compelling story, I’ll work around any weird issues or corner cases like that. But don’t expect everyone to be invincible or anything. I won’t design any 1 hit murder scenarios that aren’t painfully obvious, but 1 hit KO situations could definitely happen.

That’s about all from me, I’ll be doing a recap for the first post, getting everyone up to speed on all that’s happened. There’s a lot and I want new readers to only have to sift through a small amount of crunch to be able to catch up if they’re not willing to read or skipped the previous chapters (especially the intermission since I know many people just don’t read intermissions). Oh, and I’m switching to an on and off schedule, one day I’ll update this and the other I’ll update SHARDS or whatever it is that I’m working on. Maybe I’ll take one day off a week too, I dunno yet. I worked myself ragged the past month or so and im ready to take it at least a little easier haha.

Thanks for reading! -EDMANGO

No. 1006083 ID: 094652

How could we not pick walking away after that psyche-destroying lambaste from a cursed rubber duck? Great story, hope to see more adventures of Beamjill Skunkwoman.


I'm just sorry I didn't get more than one of the answers! Thanks for letting me use my rotting math skills.
No. 1006086 ID: 18f653

These little stories were fun to read. You all did very well in adding to the catalyst story and understanding its themes.

For the PABE storyline, the fight was fun even if it was short and I liked Flicker as a character. I hope to see her again to find out if her psyche will get better in the future and if she can mend her relationship with her father and with others she had hurt in the past. I think we took the step in the right direction in getting her treated after that duck damaged her mentally by having her go her own way, and I want to see her get better in the future.

I'm normally not a fan of puns, but the excessive amount of puns in plantimals just made it so ridiculous to the point where I couldn't help but find it hilarious. Plus, the way the characters dealt with their enemies were really creative. Dog Dog was my favorite in this. MVP!

Mizz Es was really secksay. I'm kinda sad I could not participate since I'm bad at math, but I'm glad the others pulled through and got all the answers right and got us someone who could save Andrea's and Gena's lives. Mizz Es is a really interesting character that I would love to see again to find out more about with how mysterious she is. (Though sadly, seeing her again might involve Andrea and Gena getting into intense danger which I want to avoid, perhaps there are other ways to see her again though.)

I'm hoping to see that poodle moth again in the future, and I'm already thinking about how we can deal with that Clinic situation (maybe have Cat and Will look into the address that Trey left at the post office for some clues on where Tina is to capture? Or use the dagger artifact to phase through the package and see what he is delivering without damaging the box?)

Though overall, I think my favorite segment was the Brain Worms 5 segment. It was really silly and I like how creative we got with our suggestions as we added to the story.

Looking forward to the third thread!
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