You're Ready for a Maths Olympiad if you Can Solve this one...
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- เผยแพร่เมื่อ 29 มิ.ย. 2023
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The video: • I'm Ruining my Entire ...
Today we solve a diophantine equation x+xy+y=55 using two very different, but awesome attempts :D Enjoy :3
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The video: th-cam.com/video/uWWcGHUN3QE/w-d-xo.html
CHALLENGE ACCEPTED (I'm a 6th grader lmaoo9
IM ALMOST THERE (getting 53.4)
WOOOOOOOO IM MATHING AN ANSWER IS 13 AND 3 🎉🎉🎉🎉🎉🎉🎉🎉 IM JUST GRADE 6
I broke it down using tree method and slowly offsetting my 2 numbers
Wht's that on Ur t-shirt ?
y=0, x=55. Where's my million dollars??
genius
💀
x = 0 , y = 55. Where's mine
$x=n, y=\frac{55-n}{n+1}$ solves it, $\forall x\in\mathbb{Z}\setminus\{-1\}$. You didn't solve it hard enough.
Of these: x=0 (y=55),1 (y=27),3 (y=13), 6 (y=7),-2 (y=-57),-3 (y=-29),-5 (y=-15),-8 (y=-9) are the only distinct solutions where |x|
@@cxpKSip🤓
The method of "completing the rectangle" is taught in the Singapore's mathematics curriculum not to solve diophantine equations but to factorise quadratice equations.
Haven't done Diophantine equations in a while so this was really fun and refreshing to watch :)
Okkkk I admit the visualization was actually really really good.
Also the sponsor-segway was just butter smooth.
Segway is a trademark brand name for an electric vehicle with two wheels. The rider stands on a base between the wheels and holds onto a bar at waist height.
I tried myself before watching the video, and the way i solved this was similar:
x + xy + y = 55
x(y + 1) + y = 55
Let a = y + 1, so y = a - 1:
x * a + a - 1 = 55
ax + a = 56
a(x + 1) = 56
Let b = x + 1:
ab = (2^3) * 7
the values of a and b that satisfy the equation are all the divisors of (2^3) * 7, thus:
a or b = 1, 2, 4, 7, 8, 14, 28, 56
To find x and y we just need to substract 1 to the values of b and a, so:
x or y = 0, 1, 3, 6, 7, 13, 27, 55.
Mitochondria are the powerhouse of the cell. Also, the phrases 'this blows' and 'this sucks' are both opposite and identical at the same time.
Non sequitur.
Flammable = inflammable, confirmed
Alternative title: Why succ is my favorite function
The only math teacher who greets everyone with Satanism. This is peak math content
Without doing a lot of math:
Since it is the sum of two numbers plus the product of two numbers and the answer is an integer, the two numbers also have to be integers.
Also either both numbers are positive or both are negative, otherwise the result would also be negative.
Also neither number can be -1, as it would cause the equation to collapse to -1 = 55, which is obviously not correct.
Some valid results (not all of the valid ones):
0 and 55 → obvious, as the 2 of the 3 terms become 0 and the last one has to be 55
1 and 27 → not as obvious, the equation becomes 1 + 1*27 + 27 = 1 + 2*27 = 1+ 54 = 55
6 and 7 → splitting 55 into the product and the sum of roughly equal numbers, the numbers have to be somewhere close to 7, as 7² + 2*7 = 49 + 14 = 63
Seeing, that I found 6 solutions, with all positive numbers, there has to be at least double, probably triple, that.
Dude, this is so cool!
I didn't realize you could factor it at the start, so I just did x+y(x+1)=55 because why not, at which point it becomes more obvious that you can add 1 to both sides and factor. Really cool visualization!
I came up with y = (55 - x) / (x + 1), then I noticed it is a function and started testing values
f(0) = 55 ==> x = 0; y = 55;
f(1) = 27 ==> x = 1; y = 27;
...
It is a bit dumb way to solve it, but hopefully I've got smarter with Flamys sAtAnIç explanation. Thanks!
This is called the programmer's method. You don't have to think and still you have the fastest answer assuming you have your computer already booted up in front of you with some simple language like BASIC, Python, Matlab, etc.
@@richardbloemenkamp8532wrong. The fastest answer is saying y=0 x=55.
@@himanbam True, but that is an incomplete answer if the question is to find all the integer solutions to the equation. If the question is to find an integer solution then I agree of course.
I took the same approach, but I took it one step further. When the rquation looks like y = (55 - x) / (x + 1), you can take the one out of the right side, turning a nasty looking rational into an easy-to-understand transformation of the function f(x) = 1 / x. When you take the one out, you end up with y = ( 56 / (x + 1) ) - 1. From there, it's easier to see how finding integer solutions is related to the prime factorization of 56, and then find the solutions
Complete the rectangle. Very cool !
Wait... you can use Brilliant for Chemistry too?!
Well... I think I'ma check it out.
The way i did it: x(1+y) +y=55 => x= 55-y / (1+y) =>( 56 - (1+y) )/(1+y) = 56/(1+y) -1 and then you just gotta impose 56/(1+y) to be integer
Awesome visual method, thank you. I will never forget this. #FeedTheAlgorithm
I was working on this before and one of my best results was
x+xy+y=n is equivalent to
(2x+y+1)^2=(y-1)^2 + (2x)^2 + (2√n)^2
in your case
(2x+y+1)^2= (y-1)^2 + (2x)^2 + 220
from the Pythagorean diagonal in 3D
And i stopped right there i couldn't devlope the results to something useful 😅
Maybe it helps👍
Unironic question: Wouldn't x= 15 y= 2.5 achieve the same effect? If you were to sub those numbers into the formula you'd get 55..
@@mob6804the question asks the integer solutions, there are infinitely many rational ones
The equation on his shirt is actually based.
Tau > Pi. And that's a fact.
⊤(τ>π)∴⊤(τ>π)∵⊤(τ>π) Is logically sound, because it's logically sound... Tautology for the sake of Tautology is still Tautology.
nice one ty
that satenic intro was gold
5 mins after he uploaded
the moment I saw x + xy + y = 55 I instantly thought of this: xy + x + y + 1 = (x+1)(y+1)
yeah as you can guess from the original equation, we have (x+1)(y+1) = 56 (here we add 1 to both sides)
if you are looking for x and y in forms of integers, well the rest of the problem is pretty self-explanatory
56 = 1.56 = 2.28 = 4.14 = 8.7 (in this case we're looking for the values of x + 1 and y + 1 , which applies for both pairs of (x ; y) and (y ; x) as they play the same role in the equation regardless of the order of x and y)
I mean, it didn't even take me minutes. It took me seconds
@@jevinliu4658 I commented this after he uploaded 5 mins, didn't even watch the video
maybe I didn't state it clearly enough that I was very early but whatever
I was only able to get one solution from this (with the exception of x=0 and y=55), x=27 and y=1, because I got lazy.
I got it by doing X + xy + y = 55, then rewriting that as x + xy = 55 -y, then I divided both sides by y to get this x/y +x =55/y - y/y, then I noticed that -y/y would give me -1 for a constant, so next up was (x/y)+x = 55/y - 1, then I multiplied both sides by y, giving x + xy = 55-1, which became, x+ xy = 54.
I concluded since I removed y and 55 became 54, y had to equal 1.
So, I plugged y into the equation I simplified the starting equation to, x+x(1)=54, then I got 2x= 54, and resultingly attained x=27.
Then I tested the results and got 27 + 27(1) + 1 = 27 + 27 + 1 = 27 + 28 = 40 + 15 = 55.
Can you spot where you made a mistake? Hint: your first step you divided by y. Your second step you multiplied by y, so you should have just got back to the original equation without having developed anything new.
@@zunaidparker Ah! I just caught onto it now that you mentioned it, thanks!
It's called the Simon's Favorite Factoring Trick. Man it takes me back to those olympiad days😌
Blast from the AoPS past
X is 6 and y is 7...very simple
X is 7 and y is 6...?
I found 16 solutions (if x and y are integers), let me know if I’m wrong! :D
x = 0, y = 55
x = 1, y = 27
x = 3, y = 13
x = 6, y = 7
x = 7, y = 6
x = 13, y = 3
x = 27, y = 1
x = 55, y = 0
x = -2, y = -57
x = -3, y = -29
x = -5, y = -15
x = -8, y =-9
x = -9, y = -8
x = -15, y = -5
x = -29, y = -3
x = -57, y = -2
Thanks from colombia papa flammable maths
$x=n, y=\frac{55-n}{n+1}$ solves it, $\forall x\in\mathbb{Z}\setminus{-1}$. This comes from taking various integer inputs for x and then working out what y would have to be for that to equal 55. I guess what I did was this, if I am to put this on an exam:
Suppose |x|
Thanks for video I was able to do this
Sounds like a lot of commenters here just solved the equation and commented before watching the rest of the video, and those guys missed out, big time, because the visual solution given is really elegant.
And also it feels like my brain has expanded a little bit after watching it.
Saw the thumbnail and brute forced 6 and 7 in my head, but your way is cool too, maybe I'll watch it later
0:05 lol
Man you need to make some fresh videos on diophantine equations mod and such... I am so out of touch from these things for over an year now... I t would be fun to go to basic math for a while. Thanks papa flammy
Simon says: Add 1 to both sides.
Finally you wear a t-shirt that has a correct equatiin
I suppose, that x,y are natural numbers , no zero.
If x= 2n & y=2k then Is there the contradiction.
For x=2n, y=2k+1 se get also contradiction of type z+2zp+p=3
Only x=2n+1 And y=2k+1 Is possible
I take x=1;3;5;7.....
Your solution Is nice.
Just make a shifted hyperbola to (-1,-1) and you will get your points
I've always been keen to take the algebraic approach to everything, but after teaching others I've started to recognize the importance of thinking graphically and geometrically and how it can really simplify problems and improve understanding, like you've shown with completing the rectangle. Want to solve x^2 > x? Old me: Subtract x, factor out x, solve x(x-1) = 0 and check intervals to see where x(x-1) > 0. New me: WTF are you doing fool, just look at the graph and see where the parabola is above the line lol.
Squaring usually gives bigger numbers, except fractions, they give smaller numbers when squared. So my intuition based guess for when x² > x is whenever x > 1 (if it's 1 squaring would be equal)
Edit: I forgot about negative numbers. Squaring a negative always results in a positive, so x < 0 is also another part of the solution. So the full solution is (∞, 0) ∪ (1, ∞)
Challenge Accepted.
Disclaimer: if you can solve this you are in the 100% percentile within the olympiad community, this is SFFT
Real, just factor x or y then add 1 both sides
Ok, it took me a minute to realise, that we just needed to use some distributivity. That is why practice is important.
Ok, integer solutions, now I have to look at the problem again.
ok natural number solutions can be bruteforced, you only have to check 55 numbers, if they are integers.
Why sometimes chalk scratches on board
Lol, so I saw the thumbnail and grabbed pen and paper. Ended up with y = (55 - x) / (x + 1) . "leave the solving for all rational values as an exercise for the reader" helped explain why we didn't just hop into that and, instead, did the FOIL route lol. Makes sense, and you got the integer components; I guess? I dunno, there wasn't any thing else that I saw which led me to want to find out particular things about the thing besides figure out what X and Y were... And I found out what Y is when we pick an X so... I am just watching to learn Satanism, I guess.
Once you have y = (55 - x) / (x+1), you then want x to appear only in the denominator, so you remove the x from the top by adding and subtracting 1, like so: y = -1 + (55 - x + x + 1)/(x+1) = -1 + 56/(x+1), then you have exactly what you get in the video.
No one:
Greeks when they have to solve an equation:
x+xy+y=55
x(y+1)+y=55
x(y+1)+y+1=55+1
(x+1)(y+1)=56
56=1*56=2*28=4*14=7*8
WLOG take x
My secret friend is better than your secret friend.
i factored to (x+1)(y+1)
6,7 or 3,13 or 1,27
Schauen deine Schüler deine Videos und falls ja, sprechen die dich dann zb auf neue Videos an oder so?^^
13:11 Forgot 'from brilliant'?
Wtf is the start of this omg u r ceazy
But more seriously i forgot how to do this kind of problem, agh
Bro I got x=335 and y= -5/6 and when I plugged it in I got 55.
X & y both odd numbers or one odd the other even.
Consider x=y then both are odd .. the solution between 7 and 5 .. both did not solve the equation .. so it is 6 and 5 or 6 and 7 which it is.
easy, x=i and y=27-28i
Welcome satin the founder of I
I got that x= 6 and y=7 due to x+y=55-x. Only 12 but am i ready
Thtat is exactly me
6,7
Tau's becoming a thing,
The equation on your t shirt is wrong!
x = -5, y = -15
I tried to solve this myself before watching the video and came up with the same visual method to factor it since my instinct was to try visualize the equation as an area.
From there on I rearranged the equation to express **x** in terms of **y** :
(x + 1) * (y + 1) = 56
x + 1 = 56 / (y + 1)
x = 56 / (y + 1) - 1
Then I realized that the only way **x** could be integer is if **(y + 1)** divides 56.
The positive integers that divide 56 are its prime factors (1, 2, 7) and all the products
Gotta be the best solution I've seen so far, well done
x=6 y=7
Where's my invitation?
i solved it
brute force best method
This is a basic number theory problem, what is the deal?
Already solved it 😈 (I am lying)
B I G B E T
What is this intro bro💀
I have found (1, 27), (3, 13), (6, 7) pretty quickly by guessing. :p
Edit: of course I missed (0, 55), who would count 0 as a natural number anyway?
Sorry!
my hyperbolic cosine and sine solution doesn't work, since u're only considering integers; u should swap x and y by a and b in the thumbnail, u deceived me u brat!
Your shirt bothers me
me too
Succ(x)Succ(y)=Succ(55)
A very succ says full simplification indeed. I hope it won't lead to any off-by-one errors.