Made it about a million times harder when it had to be when simple multiplication would have gotten her to the same answer... But her determination to take it seriously and do the damn algebra.. that's why she's our oshi.
she doing it by the book to remember the "know how", in case it's not x + y = 5 but x+ y = 343 and 2x+ 4y = 1928 then you aren't gonna spend all day to replace x and y one by one.@@alanepithet2931
Ina and Hololive as a whole are making me realize something: introducing a math side puzzle is a player trap. When the player (or at least a streamer) sees it, terror and brain-frying will ensue even if the puzzle is not mandatory for progress. I love this way too much. 😂
She could have just kept going at 2x + 20 - 4x = 16, at 4:30 mark, which will become -2x = -4, which is same as 2x = 4, or x = 2. It would've been done there, but the negative confused her for a moment and she went a roundabout way of solving for y instead... lol.
Nr. of Chocobos 5 14 18 22 26 30 4 *_12_* 16 20 24 28 3 10 *_14_* 18 22 26 2 8 12 *_16_* 20 24 1 6 10 14 *_18_* 22 1 2 3 4 5 Nr. of Antelopes here's also a table showcasing the correlation between the Nr. of certain stock and their respective number of legs, incase anyone wanted to have a visual repesentation (the bold & cursive numbers are all those, who's total count of stock adds up to 5)
You can either use the x or the Y to substitute into the original equation and utll work out anyway. The thing about math is one method can form muktiple ways to find an answer.
I could just hear the *Truly immense sizzling of Tako brains* as Ina relentlessly tried to figure out the math problem without giving up. Proud of my Ina for getting the answer. 😂❤
In her defense, Tako's brain sizzling is just an everyday occurrence. They are, after all, Ina's emergency ration. Especially when she runs out of cookies...
My God, when she suddenly began taking it so seriously because it's a book of sums I was reminded for the billionth time why she's my oshi, she's so damn cute...
I always got into trouble in early math classes with word problems like this because I'd just get the answer in my head without using the formulas the teachers wanted me to.
Issue is, that works for simple problems like this, but when things get harder, especially when the solutions aren’t whole numbers, you’re gonna wish you had those formulas.
I can smell the smoke coming out from her brain, lol. props to you, Vaan, for going the extra mile on the edits by following each of her steps, great work! thanks for the clip!
Once you discount the antelope with 1 remaining leg you have 12 legs in 4 animals. And the only way to split that between 2 and 4 legs is 2 each. Thus, 2 Chocobos.
Maybe unconventional, but setting up the system of equations as an augmented matrix and row-reducing also works and is pretty fast to do on paper (does the same thing as elimination, which is just more efficient than substitution in this case). So, the system 2x + 4y = 16, x + y = 5, gives the augmented matrix, [2 4 | 16] [1 1 | 5] and doing Gaussian elimination, R_1 = R_1 / 2 [1 2 | 8] [1 1 | 5] R_2 = R_1 - R_2 [1 2 | 8] [0 1 | 3] R_1 = R_1 - R_2 [1 0 | 2] [0 1 | 3] Which finally just gives x=2 and y=3 from the row-reduced matrix. I also like this method since we can know for sure that our system only has one unique solution; we can see that the non-augmented matrix is just the identity matrix, which means it is invertible and our the vectors should be linearly independent.
Matrices has always been a fast way to solve equations of multiple variables It's not unconventional either, since you will be frequently using it when dealing with large datasets or modern AI stuff
I didn't watch until the end because I wanted to try out the problem myself. So Basically you just need to hit 13, right? If we use the book's logic you can assume that aside from the one antelope that got mangled, the farmer has more than two of them as Chocobos only have 2 legs so they'll never exceed a minimum total of 6 legs. So that would mean you need a minimum of two antelopes to reach 8 legs and 2 Chocobos to make it to 12 then add in the last antelope with the single leg to reach 13 legs. The answer is 2 Chocobos. EDIT: Ina got the same answer! Though, my 5th Grade Math Teacher would probably shoot me if she knew my solution lmao
No need to hit 13, you only care about 12. And that's between 4 animals. So, if all 4 are chocobos, then they could only reach 8 legs max. So we go from biggest fractions, i.e. antelopes. 1 is not enough, but 2 fit perfectly. Counting the one-legged antelope we ignored, that makes 3 of them. And 2 chocobos. I only spend ~1min solving this problem while watching the vod, including the time it took to verify my solution. Celes in chat figured it out pretty quickly, too.
@@AntonYadrov777 They are basically the same thing. Just expressed differently. Let chocobos = x, antelopes = y. Given total legs (ignoring the 1-legged atenlope) = 12 and x+y = 4. If x = 4 and y = 0, total legs = 2x4 + 4x0 = 8 If x = 3 and y = 1, total legs = 2x3 + 4x1 = 10 If x = 2 and y = 2, total legs = 2x2 + 4x2 = 12 The way you did it (and described) is just brute forcing and substituting x and y with number, instead of solving the equations. It's only faster because the numbers are small and the question is easy.
10 years ago,I don't understand algebra,substitution or whatever it's called. Now,with a bachelors degree,I still don't understand. I lost ina when she finished writing 2x+4y =16
She's having to construct a 2x2 system of equations. Given that she can't resolve the number of each animal simply based on legs, she makes a connected equation based solely on animals: x+y=5. She must then find the solution to each variable that satisfies BOTH equations at once. There are several ways of doing this including substitution method, and elimination method.
@@TwinklesTheChinchilla Pretty much. Point being, you need at least as many equations as there are variables. Any less, and the most you can do is consider one variable a constant.
Props to Ina... I think the majority of the people I know wouldn't even recognise that technique let alone be able to perform it. If there's any criticism it is that she should have solved for what she wanted (x) rather than solving for y and then use that to work out x but that is a nitpick and she got the right result in the end so it's all good.
Either 2x+4y=16 and x+y=5 and we are searching for x or 2x+4y=12 and x+y=4 and we are again looking for x You can also change around the x and y so that it is 4x+2y=16 with x+y=5 and 4x+2y=12 with x+y=4 and we are looking for y.
I immediately thought if the right solution, but I feel she simply wanted to practice algebra classically after a while. It’s like when those Veritasium videos pop up in my recommended
🧑: Andy, your math grade's getting better this week! Mind sharing with the class how did you manage it? 🧒: Thanks, Mr. Smith. A certain VTuber gets me interested.
My first thought was just "turn one of the into the other and you're done with the calculation"...then Ina hits me that math ain't suppose to be easy most of the time lol
Two minutes in i’ll do it myself for fun Y = 4 X= 2 Y= 2X So 13 = x+ 2x +1 So 12= 3x 3=x Maximum there are three antelopes with one missing 3 legs and minimum there are two with one missing 3 legs 8+1 +x2= 13 X2= 4 Two chocobos and three antelopes 13=4+1+2x 8=2x So 4 chocobos and two antelopes Re read the question, there are five animals in total in the barn so there is two chocobos and three antelopes in the barn
The way I figured it out was “ok 1 animal is missing 3 legs which means its an antelope. And there are 12 remaining. 4+4+4 is 12 but that only accounts for 4 animals total. So there are 2 chocobos. Because 4+4+2+2 is 12, making 5 animals total.”
can solve it via linear algebra by inverse the matrix [[2,4], [1,1]] to get [[ -0.5, 2], [0.5, 1]] then multiply by transpose of [16, 5] to get [x, y]. Since the original matrix is 2 by 2 square matrix, thus invertible. [[ -0.5, 2], [0.5, 1]] * [16, 5] get you x = 2, y = 3. plug back in original formula to check, you get 2(2) + 4(3) = 16 and (2) + (3) = 5. It's more scalable way of solving the problem as number of variable increases, it become way harder to separate the variable than just inverting the matrix. Although occasionally if the matrix is not fully rank, you might get negative number as a result. Which doesnt always make sense, since you can't really have negative number of limb in the context of this problem.
I still didn't knew how I passed my Algebra and Calculus class with B+ score despite having no understanding of it before and after and only during the classes that my brain somehow managed to work math
A different (arguably easier) solution is to do something like a probability table. Combos of animals that equal 5. So 5 and 0 of each, 1 and 4 of each, 2 and 3 of each. Then just do arithmetic to find the one solution there must be that add up to 16 legs and has at least 1 antelope. Always fun to see how people approach solving these kinds of simple problems.
>devoured three of the limbs on one of his antelopes So we can assume there's at least 2 antelopes, who have a combined 5 legs after the leg eater had a nibble. That leaves 8 legs that can be divided over either 4 chocobos, or 1 additional antelope and 2 chocobos.
The numbers are small enough that the problem can be solved using logic, but I wanted to try expressing the solution mathematically so here's my attempt at doing that. I also want to account for the injured antelope all throughout the equation because I'm pretending that the variables can't be simplified using logic. First, let's represent some variables like so: x = total number of chocobos y = number of antelopes (uninjured) 1 = number of antelopes (injured) x + y + 1 = 5 = total number of animals To find the total number of chocobos, we need to solve for the two missing variables x and y. To do that, we need to come up with two equations that include them both. The first can simply come from the total number of animals: x + y + 1 = 5 x + y = 4
I just started counting up from the number of antelopes. 1 antelope is 1 leg, 12 chocobo legs left. 2 antelope is 5 legs, 8 chocobo legs left. 3 antelopes is 9 legs, 4 chocobo legs left... which is 2 chocobos. Even cavemen like me can count.
A lot of math for a really simple solution. There are five animals. They can either be antelopes or chocobos. Has to have 16 legs in total. 4x4 is 16, that's not five so that's not going to work. 4x3 is 12; 2x2 is 4; 4+12 is 16=16 what we're looking for. There has to be 3 antelopes and 2 chocobos in order for this to work. It mentions that the legs were cut but that doesn't matter in this math problem, that's a diversion created to stump people as the real question was "How many chocobos are in the barn" which is 2. Another idea is to actually use the bottom piece and just get rid of the antelope all together so that you're left with 4. Then it becomes 12 legs. 3x4 is 12, we need 4 so that's not it. 2x4=8; 2x2 is 4; 4+8=12. This, again, leaves you with 2 chocobos.
TIL you can put an apostrophe before some number-y text in Excel to avoid the program from treating it as an expression. Recreational spreadsheets! Matt Parker would be so so proud.
There is an efficient way to go about this kind of problem if it only has two kinds of animal: 1. Subtract the one-legged one. So the remaining are 4 animals with 12 legs. 2. Assume that all are 2 legged. This gives you 4x2=8 3. We are missing 4 legs. Replacing a Choco with a Gazelle adds 2 legs, so we have to replace 2 of them. 4. Check: 2 Chocos and 2 Gazelles = 4+8 legs = 12 legs. Plus the gnawed on Gazelle, this makes 5 animals with 13 legs. Done!
5 Ill animals. 13 remaining limbs. x + y = 5 2x + (4y-3) = 13 y < 5, and unless there are zero chocobos, y < 4 That leaves 3 possible solutions that can be quickly plugged in and tested. Further algebra will also give the solution, but once you have that few testable solutions with such a simple equation, it can make just as much sense to just plug them in and test if your algebra is rusty.
I found a shortcut different from anything I've read in the comments so far (though not necessarily better) The leg counts being 2 and 4 makes things really easy, since you can always just replace one antelope with two chocobo and have the same number of legs. Start by assuming they're all antelope. 16 legs total requires four of them (with one missing three of its legs to hit 13). Four is the wrong number of total animals, so replace one of them with two chocobos. Now you have 3 antelopes and 2 chocobos and still 13 legs. Five is the right number, so you're done. But if you'd needed more, you could just keep replacing one by one until you got there. All told the whole process took me about a minute, and I needed no paper or anything.
![VTubers React to Elementary School Math [13 POVs]](http://i.ytimg.com/vi/2AhDSWmOWQ0/mqdefault.jpg)
I can think in a fastest way to solve this problem...bring me Ollie.
Ollie would've set up a system of equations lickety split and reached the end result even faster than that.
@@DinnerForkTongue Ollie probably will solve this from gauss equation. Would you want it?
I think the implication was more that he can solve the problem visually by moving around individually detachable zombie limbs 😅
@@graywolf6441using gaussian elimination for this is like using a precision laser cut to cut your bread 😂😂😂
@@ワースト002 Solving this with Gauss was about as quick and hassle-free as using the insertion method
Student: "Why do I need to learn this?"
Teacher: "Imagine you are an idol..."
I lost it 😂😂😂😂😂
Although she got confused and did the less efficient substitution, it was otherwise a textbook solution :)
Nice rhyme
Made it about a million times harder when it had to be when simple multiplication would have gotten her to the same answer...
But her determination to take it seriously and do the damn algebra.. that's why she's our oshi.
she doing it by the book to remember the "know how", in case it's not x + y = 5 but x+ y = 343 and 2x+ 4y = 1928 then you aren't gonna spend all day to replace x and y one by one.@@alanepithet2931
Right? I first got it to x=8-2y and just plugged the rest in from there
Ina and Hololive as a whole are making me realize something: introducing a math side puzzle is a player trap. When the player (or at least a streamer) sees it, terror and brain-frying will ensue even if the puzzle is not mandatory for progress.
I love this way too much. 😂
I watch Kidnap streams for this only.
even Ollie lol
@@irkanorphyn Ollie skips the panic phase and goes straight into brain frying mode.
Somehow the older gen’s VTubers shriek in agony when I saw more simple math questions.
She could have just kept going at 2x + 20 - 4x = 16, at 4:30 mark, which will become -2x = -4, which is same as 2x = 4, or x = 2. It would've been done there, but the negative confused her for a moment and she went a roundabout way of solving for y instead... lol.
To avoid the negatives you could go:
2x + 20 -4x = 16
20 - 2x = 16
16 + 2x = 20
2x = 4
X = 2
what I thought she got x=10 at that point, and was also wondering what went wrong...
I feel Like the question is wrong. Need simpler words.
Nr. of Chocobos
5 14 18 22 26 30
4 *_12_* 16 20 24 28
3 10 *_14_* 18 22 26
2 8 12 *_16_* 20 24
1 6 10 14 *_18_* 22
1 2 3 4 5 Nr. of Antelopes
here's also a table showcasing the correlation between the Nr. of certain stock and their respective number of legs, incase anyone wanted to have a visual repesentation
(the bold & cursive numbers are all those, who's total count of stock adds up to 5)
You can either use the x or the Y to substitute into the original equation and utll work out anyway. The thing about math is one method can form muktiple ways to find an answer.
I could just hear the *Truly immense sizzling of Tako brains* as Ina relentlessly tried to figure out the math problem without giving up.
Proud of my Ina for getting the answer. 😂❤
In her defense, Tako's brain sizzling is just an everyday occurrence.
They are, after all, Ina's emergency ration. Especially when she runs out of cookies...
My God, when she suddenly began taking it so seriously because it's a book of sums I was reminded for the billionth time why she's my oshi, she's so damn cute...
> "You can solve this without algebra"
> Proceed to do algebra, but with words and not variable
lmao
your ability to use spreadsheets would put some of my students to shame, amazing work
Ina derailing a stream to do a math problem is one of my favourite parts about her streams. Reminds me of the 30 min long Zelda physics question
I always got into trouble in early math classes with word problems like this because I'd just get the answer in my head without using the formulas the teachers wanted me to.
Issue is, that works for simple problems like this, but when things get harder, especially when the solutions aren’t whole numbers, you’re gonna wish you had those formulas.
I can smell the smoke coming out from her brain, lol. props to you, Vaan, for going the extra mile on the edits by following each of her steps, great work! thanks for the clip!
Once you discount the antelope with 1 remaining leg you have 12 legs in 4 animals. And the only way to split that between 2 and 4 legs is 2 each.
Thus, 2 Chocobos.
That was how I did it.
yup that's absolutly correct no fancy pants equasions required here
Brains diff 😂
Like in games we say caryy diff and stuff. But its ina its ok
You don't even need to discount the ampulope, the only way it can work even with 16 legs is 2 chocobos and 3 antelope.
@@Elzzaw -Yeah, but 2&2 is easier to find at a glance.-
4:30 she just needed to keep going. The negatives would have cancelled out. 😿
She tried her best
The length of this clip already fills me with dread...
Maybe unconventional, but setting up the system of equations as an augmented matrix and row-reducing also works and is pretty fast to do on paper (does the same thing as elimination, which is just more efficient than substitution in this case).
So, the system 2x + 4y = 16, x + y = 5, gives the augmented matrix,
[2 4 | 16]
[1 1 | 5]
and doing Gaussian elimination,
R_1 = R_1 / 2
[1 2 | 8]
[1 1 | 5]
R_2 = R_1 - R_2
[1 2 | 8]
[0 1 | 3]
R_1 = R_1 - R_2
[1 0 | 2]
[0 1 | 3]
Which finally just gives x=2 and y=3 from the row-reduced matrix.
I also like this method since we can know for sure that our system only has one unique solution; we can see that the non-augmented matrix is just the identity matrix, which means it is invertible and our the vectors should be linearly independent.
Love me some linear algebra ❤
How to use matrix? I don't understand.(I don't learn it yet at school)
Matrices has always been a fast way to solve equations of multiple variables
It's not unconventional either, since you will be frequently using it when dealing with large datasets or modern AI stuff
@@diniza Me neither. I am very curious to know how they work
@@Tyler-bp4md
2x + 4y = 13 legs (+ 3 eaten legs) = 16 legs originally
x + y = 5 animals
x = 5 - y
2(5 - y) + 4y = 16
10 - 2y + 4y = 16
10 + 2y = 16
2y = 6
y = 3 antelopes
X = 5 - y = 5 - 3 = 2 chocobos
That's basically what the matrix is saying
Ina stop my braincell hurts
Who knew Cloove was such a math wiz.
Tako's come to Ina specifically to not use their brain.
What is this betrayal.
Since Jim from accounting died, everyone's felt the squeeze.
Wait. Jim is still dead? I thought AO-chan brought him back last week.
@@ellskilltaker1267 She did? I swear the paperwork for that did not reach the Resurrection Department's inbox.
@@DinnerForkTongue Did you eat the papers for that?
@@jordanholloman5907 No. Just the Department of Departments' cookie invoice.
Don't look at me like that, it was a hungry day!
4 marks in the bag 👏👏👏👏 We're on our way to A+ takos
I didn't watch until the end because I wanted to try out the problem myself.
So Basically you just need to hit 13, right? If we use the book's logic you can assume that aside from the one antelope that got mangled, the farmer has more than two of them as Chocobos only have 2 legs so they'll never exceed a minimum total of 6 legs.
So that would mean you need a minimum of two antelopes to reach 8 legs and 2 Chocobos to make it to 12 then add in the last antelope with the single leg to reach 13 legs.
The answer is 2 Chocobos.
EDIT: Ina got the same answer! Though, my 5th Grade Math Teacher would probably shoot me if she knew my solution lmao
Yeah you can just use logic lmao
No need to hit 13, you only care about 12. And that's between 4 animals. So, if all 4 are chocobos, then they could only reach 8 legs max. So we go from biggest fractions, i.e. antelopes. 1 is not enough, but 2 fit perfectly. Counting the one-legged antelope we ignored, that makes 3 of them. And 2 chocobos. I only spend ~1min solving this problem while watching the vod, including the time it took to verify my solution. Celes in chat figured it out pretty quickly, too.
I mean, the logical way of finding the solution is basically the same as working it out algebraically like Ina did
@@falsefingolfin yet it's way faster, as you can see. And it works via fractions, not finding two unknown variables algebraically.
@@AntonYadrov777 They are basically the same thing. Just expressed differently.
Let chocobos = x, antelopes = y.
Given total legs (ignoring the 1-legged atenlope) = 12 and x+y = 4.
If x = 4 and y = 0, total legs = 2x4 + 4x0 = 8
If x = 3 and y = 1, total legs = 2x3 + 4x1 = 10
If x = 2 and y = 2, total legs = 2x2 + 4x2 = 12
The way you did it (and described) is just brute forcing and substituting x and y with number, instead of solving the equations. It's only faster because the numbers are small and the question is easy.
10 years ago,I don't understand algebra,substitution or whatever it's called.
Now,with a bachelors degree,I still don't understand.
I lost ina when she finished writing 2x+4y =16
Maybe if she'd chosen *a* and *c* for variables, it'd make more sense. It's what I did to not lose perspective.
x 2 legged chocobos plus y 4 legged antlope equals 16 legs before leg eater sustraction of 3.
Bruh, how can u get bachelors degree without basic algebra? Did your study didn't have statistics?
She's having to construct a 2x2 system of equations. Given that she can't resolve the number of each animal simply based on legs, she makes a connected equation based solely on animals: x+y=5. She must then find the solution to each variable that satisfies BOTH equations at once. There are several ways of doing this including substitution method, and elimination method.
@@TwinklesTheChinchilla Pretty much. Point being, you need at least as many equations as there are variables. Any less, and the most you can do is consider one variable a constant.
Props to Ina... I think the majority of the people I know wouldn't even recognise that technique let alone be able to perform it.
If there's any criticism it is that she should have solved for what she wanted (x) rather than solving for y and then use that to work out x but that is a nitpick and she got the right result in the end so it's all good.
the instant number/math show up my brain already start dead-ing
props to editor🎉
I just woke up my mind is not ready for this
*_I love the kind of woman that will actually just outsmart me_*
I only started watching to escape the confines of reality, not be reminded of it
I love how comment sections turn into discussions about the formulas in Math on each member
As an asian, good job Ina, you've done the first question, only 39 more. Oh, rememeber, it's a 60-minute test
At 6:45 Ina is correct that the equation is -2y+4y=6 however I am just the one being weird😔
Man's translated a clip into a spreadsheet, the heck.
Two Chocobos in the barn.
Did it by trial and error.
I remembered when I gotten A+ in Math in the finals before.
I don't think I'm confident in doing that again now :v
my brain already starting to hurt
She is one of my people, she has to attempt any math problem she sees, like a vampire counting salt grains.
Either 2x+4y=16 and x+y=5 and we are searching for x
or 2x+4y=12 and x+y=4 and we are again looking for x
You can also change around the x and y so that it is 4x+2y=16 with x+y=5 and 4x+2y=12 with x+y=4 and we are looking for y.
Ahh, this is a simultaneous equation. Classic high school level algebra.
I always thoughts in my head that is what ancient people problem, using number and letter
I immediately thought if the right solution, but I feel she simply wanted to practice algebra classically after a while. It’s like when those Veritasium videos pop up in my recommended
Im glad im not the only one that is so slow at doing word problems... I guesss this is a sign to do my college algebra homework...
FF got a nice BGM, and conbined with Ina's relaxing voice is such a chill time
The brain neurons alone could have powered her channel. So many takos brains were on overdrive helping her figure out the problem
2 chocobo and 3 antelopes one of which lost 3 of its legs.
🧑: Andy, your math grade's getting better this week! Mind sharing with the class how did you manage it?
🧒: Thanks, Mr. Smith. A certain VTuber gets me interested.
Now do 2x2.
@@teilzeitberndruclips.net/video/pjYTC2Djijw/видео.htmlsi=-XL3AuFt6YI8x2C2
My first thought was just "turn one of the into the other and you're done with the calculation"...then Ina hits me that math ain't suppose to be easy most of the time lol
top notch editing
the gears in ina´s brain are turning, not fast but they turn
she passed the math
please stop! we only have one total braincell!
So happy to see someone else use spreadsheets as math notebook, she is so proficient in that she uses the ' automatically
Our lovely priestess being a massive nerd is something I didn’t expect.
Better call Ollie!
I love their add this kind side read 😂😂
Gahdamn it I opened RUclips to escape from my maths homework and _you bring me this?!_
I'm so proud of her. I'm not sure how many others would even try to solve it.
I was hovering over a hxh clip on another tap and thought the audio was on when I heard the music
I like how you pulled out an excel for this. Also to quote a chatter "what is that burnt smell"
I wish I knew what she is doing, but my two-cell brain can't handle this.
That's sum book you got there Priestess, if that's the power of her single brain cell, I can just imagine how her full power would be.
Pain tako. I guess it's been too long since the priestess has done any algebra 😂
Two minutes in i’ll do it myself for fun
Y = 4
X= 2
Y= 2X
So
13 = x+ 2x +1
So
12= 3x
3=x
Maximum there are three antelopes with one missing 3 legs and minimum there are two with one missing 3 legs
8+1 +x2= 13
X2= 4
Two chocobos and three antelopes
13=4+1+2x
8=2x
So 4 chocobos and two antelopes
Re read the question, there are five animals in total in the barn so there is two chocobos and three antelopes in the barn
I'm so proud of her man, not all en members can do that
when asian make a game 👍
She made the incredible world of mathematics sound boring for a while there.
The way I figured it out was “ok 1 animal is missing 3 legs which means its an antelope. And there are 12 remaining. 4+4+4 is 12 but that only accounts for 4 animals total. So there are 2 chocobos. Because 4+4+2+2 is 12, making 5 animals total.”
my dude you gotta give us some lessons on those sexy excel skills.
loll sory ina i did that one within 2 minutes without noting down
Man, I so like how you use sheets to illustrate
Thanks for the clip
can solve it via linear algebra by inverse the matrix [[2,4], [1,1]] to get [[ -0.5, 2], [0.5, 1]] then multiply by transpose of [16, 5] to get [x, y]. Since the original matrix is 2 by 2 square matrix, thus invertible. [[ -0.5, 2], [0.5, 1]] * [16, 5] get you x = 2, y = 3. plug back in original formula to check, you get 2(2) + 4(3) = 16 and (2) + (3) = 5. It's more scalable way of solving the problem as number of variable increases, it become way harder to separate the variable than just inverting the matrix. Although occasionally if the matrix is not fully rank, you might get negative number as a result. Which doesnt always make sense, since you can't really have negative number of limb in the context of this problem.
this reminded me to study for my finals
Not the first math stream funnily enough.
Holy shit the journey.
I still didn't knew how I passed my Algebra and Calculus class with B+ score despite having no understanding of it before and after and only during the classes that my brain somehow managed to work math
She found the answer almost immediately and then kept going for some bizarre reason
線形代数の知識を使いました
she so cute, and smart :3
as a math tuition teacher, i understand the pain the write down all the steps with pc
A different (arguably easier) solution is to do something like a probability table. Combos of animals that equal 5. So 5 and 0 of each, 1 and 4 of each, 2 and 3 of each. Then just do arithmetic to find the one solution there must be that add up to 16 legs and has at least 1 antelope. Always fun to see how people approach solving these kinds of simple problems.
>devoured three of the limbs on one of his antelopes
So we can assume there's at least 2 antelopes, who have a combined 5 legs after the leg eater had a nibble. That leaves 8 legs that can be divided over either 4 chocobos, or 1 additional antelope and 2 chocobos.
Thing I hate about math is not finding the answer, it’s explaining how the hell I got there.
She must be really bad at math. I'm not the best at math, but I had this figured out just after reading the sentence.
Spreadsheets, my beloved
Wait, there's a HoloEN member who's actually good at math? Impossible.
The numbers are small enough that the problem can be solved using logic, but I wanted to try expressing the solution mathematically so here's my attempt at doing that. I also want to account for the injured antelope all throughout the equation because I'm pretending that the variables can't be simplified using logic.
First, let's represent some variables like so:
x = total number of chocobos
y = number of antelopes (uninjured)
1 = number of antelopes (injured)
x + y + 1 = 5 = total number of animals
To find the total number of chocobos, we need to solve for the two missing variables x and y. To do that, we need to come up with two equations that include them both. The first can simply come from the total number of animals:
x + y + 1 = 5
x + y = 4
I just started counting up from the number of antelopes.
1 antelope is 1 leg, 12 chocobo legs left. 2 antelope is 5 legs, 8 chocobo legs left. 3 antelopes is 9 legs, 4 chocobo legs left... which is 2 chocobos.
Even cavemen like me can count.
Let's go brainTako; i love math and i just like her in games, i need to resolved by myself🤣🤣
A lot of math for a really simple solution.
There are five animals. They can either be antelopes or chocobos. Has to have 16 legs in total. 4x4 is 16, that's not five so that's not going to work. 4x3 is 12; 2x2 is 4; 4+12 is 16=16 what we're looking for. There has to be 3 antelopes and 2 chocobos in order for this to work. It mentions that the legs were cut but that doesn't matter in this math problem, that's a diversion created to stump people as the real question was "How many chocobos are in the barn" which is 2.
Another idea is to actually use the bottom piece and just get rid of the antelope all together so that you're left with 4. Then it becomes 12 legs. 3x4 is 12, we need 4 so that's not it. 2x4=8; 2x2 is 4; 4+8=12. This, again, leaves you with 2 chocobos.
That kinda hurt at 4:35 - she was right there and second-guessed herself.
I’m so proud of her
antelopes used to be everywhere qq
I basically got it once she did 2x+4y=16.
2 chocobos with 2 legs each plus 3 antelopes with 4 legs each.
well , what can we say , she always found a way to take our sanity
I get what it's asking me to do. I hate how much inferring I had to do to get to that point though. It fails as a paragraph, I think.
TIL you can put an apostrophe before some number-y text in Excel to avoid the program from treating it as an expression.
Recreational spreadsheets! Matt Parker would be so so proud.
There is an efficient way to go about this kind of problem if it only has two kinds of animal:
1. Subtract the one-legged one. So the remaining are 4 animals with 12 legs.
2. Assume that all are 2 legged. This gives you 4x2=8
3. We are missing 4 legs. Replacing a Choco with a Gazelle adds 2 legs, so we have to replace 2 of them.
4. Check: 2 Chocos and 2 Gazelles = 4+8 legs = 12 legs. Plus the gnawed on Gazelle, this makes 5 animals with 13 legs. Done!
Boom, great work fellow tako
Precisely what I meant in other comments. Basic fraction-based solution, finger arithmetic.
If stocks meant to be all the creatures the answer is 2
And they DARE to say the video games are not educated. kek
5 Ill animals.
13 remaining limbs.
x + y = 5
2x + (4y-3) = 13
y < 5, and unless there are zero chocobos, y < 4
That leaves 3 possible solutions that can be quickly plugged in and tested. Further algebra will also give the solution, but once you have that few testable solutions with such a simple equation, it can make just as much sense to just plug them in and test if your algebra is rusty.
I found a shortcut different from anything I've read in the comments so far (though not necessarily better)
The leg counts being 2 and 4 makes things really easy, since you can always just replace one antelope with two chocobo and have the same number of legs.
Start by assuming they're all antelope. 16 legs total requires four of them (with one missing three of its legs to hit 13).
Four is the wrong number of total animals, so replace one of them with two chocobos.
Now you have 3 antelopes and 2 chocobos and still 13 legs.
Five is the right number, so you're done. But if you'd needed more, you could just keep replacing one by one until you got there.
All told the whole process took me about a minute, and I needed no paper or anything.
Exact same thing I did except I started with 12 legs and substituting upwards until I hit 5 animals.