*Over on my Channel you can find the Outtake ( **ua-cam.com/video/mfaSDC4eIog/v-deo.html** ) as well as my first attempt, the Alternative Solution! **ua-cam.com/video/uaFZ69NLRBU/v-deo.html* Make sure to support the channel by watching all the relevant videos and to consider subscribing to Flammy 2 if you haven't done so already! =)
Worst "how to peel a boiled egg" video I've ever seen. You don't even look good peeling the egg. Come to think of it, I don't think you followed the celebrated 13 step process of egg peeling. What has happened to UA-cam!?
I always messed up the “and” and “or” symbols, but here is a little trick I have learned to remember them: the “and” symbol is like an A for “and”. It sounds stupid, but for me it works very well
I think that when water is poured on top of "and", the water will flow on both sides, and for "or", the symbol will fall on one side and the water flows on that side only or maybe the symbol doesnt fall and the water, again, flows on both sides!
My earlier solution was a specific specific solution. There can be innumerable solution sets to the problem if we consider real values for a and b. The general solution values for a = (1/2)*[ 3 * k + sqrt( 9 * k^2 - 8072 * k)] and b = (1/2)*[ 3 * k - sqrt( 9 * k^2 - 8072 * k)] for any value of k greater than 897.
Sir, alternatively, u could have done, taking 1/b to RHS once then cross multiply, We get 2018b = (3b - 2018)a Similarly with 1/a, 2018a = (3a - 2018)b On multiplying those 2 eqn, we get 2018² = (3a - 2018)(3b - 2018) N rest as u solved!
Why didn't you just check the modulus by 3? Each factor on RHS (3a-2018,3b-2018) is 1 mod 3, while the only "prime factors mod 3" (factors which are 1 mod 3 themselves but any subfactors are not) of 2018^2 on the LHS is 4*1009^2, so since we have to distribute 3 possible "prime factors" between a and b, 2^3 or 8 solutions exist
You know, this would be much easier, if you simply write a=c*x, b=d*x where x=gcd(a,b) (positive). Then c,d are coprime. 1/a+1/b=3/2018 is (by rearranging) equivalent to (2018-3b)a=-2018b, so a| 2018b and by symmetry b|2018a. Hence, cx|2018dx and dx|2018cx. Dividing by x and remembering that c and d are coprime yields c|2018 and d|2018. So, c and d can both only be in the set {-2018,-1009,-2,-1,1,2,1009,2018}. On the other hand, 1/a+1/b=(c+d)/(cd x)=3/2018, so 3|c+d and as soon as we have c,d, we can simply solve for x by rearranging to x=2018(c+d)/(3cd) [note that since c,d|2018 and gcd(c,d)=1, we get cd|2018, so this is an integer iff 3|c+d] W.l.o.g, we can say |c|0 we have c>0. Taking into account that c and d are coprime, we have the following pairs (c,d) left to check: (1,-2018): Impossible, since -2017 is not divisible by 3 (1,-1009): possible, since -1008 is divisible by 3 and indeed, with x=2*1008/3=336*2, hence a=336*2, b=-336*2*1009, we get 1/a+1/b=3/2018 (1,-2) impossible, since -1 is not divisible by 3 (1,-1) obviously here 1/a+1/b=0 regardless of x (1,1) not divisible by 3 (1,2) possible and x=1009 yields a=1009, b=2018 and 1/a+1/b=3/2018 (1,1009) not divisible by 3 (1,2018) possible and x=673 yields a=673, b=673*2018 (2,-1009) not divisible by 3 (2,1009) possible with x=337, a=674, b=1009*337 (for bigger c there is no possible d) So, all solutions are the following: (336*2,-336*2*1009), (1009,2018),(673,673*2018),(674,1009*337), (-336*2018,336*2),(2018,1009),(673*2018,673),(1009*337,674) (Note that this method works for all problems of the form 1/a+1/b=y/z with y/z in lowest terms if the prime factorization of y and z is known: we always get y|c+d and c|z, d|z)
@@avisingh0009 I think the solution of Michael is very less time comsuming and good for learning new tricks too. While this video is rather aimed at a more general public, Michael prefers to use some good tools which are very helpful.
I did the simplification multiplying both sides by 2018 ab and then by 3. Then added 2018^2 to both sides to get (3a - 2018)(3b - 2018) = 2018^2 = 2^2 1009^2 and solved all cases where for instance 3 a-2018=1,3 b-2018=2018^2 which gives (673, 1358114) 3 a-2018=2,3 b-2018=2 1009^2 and so on, getting 8 solutions (-678048,672), (672,-678048), (673,1358114), (674,340033), (1009,2018), (2018,1009), (340033,674), (1358114,673)
I don't get why you took such a roundabout approach to get to the factorization at 9:45, It can be directly written by observation from the original equation.
Physicist to mathematician: Why do you have to call it commutative? You can say I have 2 equations and I am multiplying them together. I understood the symmetry part. If you ask me I would write a program for all integers. It seems like you have found a better way. Checking for multiples of 3. That move got me. I did not think about that. Thanks. Hmmmm.
Believe me or not... The similar problem was given to 5th and 6th graders in India (age (10-12)). The name of examination was NMTC (Gauss Contest) Level 2. The main point was... The calculation was lot easier... It was 1/m + 1/n = 3/17. This was the problem. Can you imagine? Btw, I found a direct formula for these questions. For 1/x + 1/y = 1/a (x-a)(y-a)= a² Later you're smart...
This might be this first one I get to solve lol (except for the fact that I only read the thumbnail and looked for positive a and b, thus missing one solution)
You made an error in the way you copied the question, the question is not taking place in Z x Z, rather N x N. Copied directly from the exam it says "Find all ordered pairs (a,b) of *positive integers* for which...", i.e., you should not be considering values of a or b that are less than 0.
It must be that you were watching a video with those views and then went here and the video was still loadig so you have the views of the previous ideo you were watching
u were criticizing India's education system for giving Arithematic mean & geometric mean problems hidden in Algebra questions. but I feel this to b more stupid to get through long calculations rather than linking different areas in maths together.
Why do you post so much garbage comments on every video ? I really like reading comment section for additional insights and u spoil it for many like us. Its ok to have some fun but have some respect for the forum
*Over on my Channel you can find the Outtake ( **ua-cam.com/video/mfaSDC4eIog/v-deo.html** ) as well as my first attempt, the Alternative Solution! **ua-cam.com/video/uaFZ69NLRBU/v-deo.html*
Make sure to support the channel by watching all the relevant videos and to consider subscribing to Flammy 2 if you haven't done so already! =)
ok
papa plz say yeet in your next vidya, I will give you internet points plz.
ua-cam.com/video/TKm_sJ14KRk/v-deo.html
k
@@PapaFlammy69 thanku
Worst "how to peel a boiled egg" video I've ever seen. You don't even look good peeling the egg. Come to think of it, I don't think you followed the celebrated 13 step process of egg peeling. What has happened to UA-cam!?
;_;
war of the maffs
I always messed up the “and” and “or” symbols, but here is a little trick I have learned to remember them: the “and” symbol is like an A for “and”. It sounds stupid, but for me it works very well
problem is the german "und" ^^'
I did the relationship with Union and Intersection when you work with inequalities.
x > 3 ∧ x < 5 → x > 3 ⋂ x < 5
x > 1 ∨ x > 3 → x > 1 ⋃ x > 3
@Flamable Maths I think of it as like a pointy n in “and” (and “und”). Like fish ‘n’ chips.
I think that when water is poured on top of "and", the water will flow on both sides, and for "or", the symbol will fall on one side and the water flows on that side only or maybe the symbol doesnt fall and the water, again, flows on both sides!
The “or” symbol comes originally from the Latin “vel”(same way that the & symbol comes from a stylised “et”). Makes it easy for me to remember.
Ur patience level is ultimate in this question ✌
Nope! I won't say that. I'm ready to do any calculation if I get an Olympiad problem correct.
There's the 145 iq papa flammy
17:47 also all the solutions where a and b are switched :)
My earlier solution was a specific specific solution. There can be innumerable solution sets to the problem if we consider real values for a and b. The general solution values for a = (1/2)*[ 3 * k + sqrt( 9 * k^2 - 8072 * k)] and b = (1/2)*[ 3 * k - sqrt( 9 * k^2 - 8072 * k)] for any value of k greater than 897.
a = infinity, b = (3/2018)^-1, nailed it
13:25
Everything makes sense if you think about it *hard enough* .
Sir, alternatively, u could have done, taking 1/b to RHS once then cross multiply,
We get 2018b = (3b - 2018)a
Similarly with 1/a,
2018a = (3a - 2018)b
On multiplying those 2 eqn, we get 2018² = (3a - 2018)(3b - 2018)
N rest as u solved!
Why didn't you just check the modulus by 3?
Each factor on RHS (3a-2018,3b-2018) is 1 mod 3, while the only "prime factors mod 3" (factors which are 1 mod 3 themselves but any subfactors are not) of 2018^2 on the LHS is 4*1009^2, so since we have to distribute 3 possible "prime factors" between a and b, 2^3 or 8 solutions exist
If I may say so, you are a great math teacher!
Pretty cool solution, thank ou for sharing this fellow mathematician :)!
8:28 I have no idea what's going on but I'm just pleased that he's happy
Thumbs up for explaining why some of the solutions are eliminated!
You know, this would be much easier, if you simply write a=c*x, b=d*x where x=gcd(a,b) (positive). Then c,d are coprime.
1/a+1/b=3/2018 is (by rearranging) equivalent to (2018-3b)a=-2018b, so a| 2018b and by symmetry b|2018a.
Hence, cx|2018dx and dx|2018cx. Dividing by x and remembering that c and d are coprime yields c|2018 and d|2018.
So, c and d can both only be in the set {-2018,-1009,-2,-1,1,2,1009,2018}.
On the other hand, 1/a+1/b=(c+d)/(cd x)=3/2018, so 3|c+d and as soon as we have c,d, we can simply solve for x by rearranging to
x=2018(c+d)/(3cd) [note that since c,d|2018 and gcd(c,d)=1, we get cd|2018, so this is an integer iff 3|c+d]
W.l.o.g, we can say |c|0 we have c>0. Taking into account that c and d are coprime, we have the following pairs (c,d) left to check:
(1,-2018): Impossible, since -2017 is not divisible by 3
(1,-1009): possible, since -1008 is divisible by 3 and indeed, with x=2*1008/3=336*2, hence a=336*2, b=-336*2*1009, we get 1/a+1/b=3/2018
(1,-2) impossible, since -1 is not divisible by 3
(1,-1) obviously here 1/a+1/b=0 regardless of x
(1,1) not divisible by 3
(1,2) possible and x=1009 yields a=1009, b=2018 and 1/a+1/b=3/2018
(1,1009) not divisible by 3
(1,2018) possible and x=673 yields a=673, b=673*2018
(2,-1009) not divisible by 3
(2,1009) possible with x=337, a=674, b=1009*337
(for bigger c there is no possible d)
So, all solutions are the following:
(336*2,-336*2*1009), (1009,2018),(673,673*2018),(674,1009*337), (-336*2018,336*2),(2018,1009),(673*2018,673),(1009*337,674)
(Note that this method works for all problems of the form 1/a+1/b=y/z with y/z in lowest terms if the prime factorization of y and z is known:
we always get y|c+d and c|z, d|z)
In fact, the method works for all problems of the form k/a+l/b=y/z with given integers k,l,y,z (y,z in lowest terms, k and l not equal to 0)
MIchael Penn also has a great solution of this
@@avisingh0009 I think the solution of Michael is very less time comsuming and good for learning new tricks too. While this video is rather aimed at a more general public, Michael prefers to use some good tools which are very helpful.
Link?
@@vikranttyagiRN ua-cam.com/video/2bB_Ec0jmCI/v-deo.html here bro.
@@vikaskalsariya9425 thanks
Yoooo I remember this problem
ah finally something my smol brain can easily understand
This is indeed a genious method...this video is one of the best of urs🙌🙌
i guess you could say this is 🆎rilliant question
ababou???
I did the simplification multiplying both sides by 2018 ab and then by 3.
Then added 2018^2 to both sides to get
(3a - 2018)(3b - 2018) = 2018^2 = 2^2 1009^2
and solved all cases where for instance
3 a-2018=1,3 b-2018=2018^2
which gives (673, 1358114)
3 a-2018=2,3 b-2018=2 1009^2
and so on, getting 8 solutions
(-678048,672),
(672,-678048),
(673,1358114),
(674,340033),
(1009,2018),
(2018,1009),
(340033,674),
(1358114,673)
I took the exam that year! Yes!
what you proved here is a nice relation between the geometric mean GM(a,b) = sqrt(ab) and the harmonic mean HM(a,b) = 2/(1/a + 1/b)
nice solution.teaches a new method
I don't get why you took such a roundabout approach to get to the factorization at 9:45, It can be directly written by observation from the original equation.
Where did you get that clock from?
It's an exclusive for my channel. Only available till tomorrow ^^ teespring.com/flammable-math-clock
@@PapaFlammy69 Phew so good thing I looked today
Definitely! :D
If you know Simon's Favorite Factoring Trick, you can skip the video to 9:14
13:00 it was around 12 minutes long
9^2 views in 9 minutes 😳
Cuadratic logic :D
I remember this question, I got it wrong lol. Thanks Flamma pappy you thicc boi :)
I remember solving this one a couple years ago. I can’t remember exactly what method I used but it definitely wasn’t this nice.
what is the intro music called?
Putnam 2018 A1: Flammable Maths vs Micheal Penn
See his video:ua-cam.com/video/2bB_Ec0jmCI/v-deo.html
What is ZxZ
So smart.😮
Michael Penn did this problem already. Are you doing a different method
Bois don't copy methods
@@chhabisarkar9057 haha nice one
What is the definition of problem...
MM very good papa thank you
just use python? i mean the math approach booil down to computation so why not just use python
Love from 🇭🇰
Physicist to mathematician: Why do you have to call it commutative? You can say I have 2 equations and I am multiplying them together. I understood the symmetry part. If you ask me I would write a program for all integers. It seems like you have found a better way. Checking for multiples of 3. That move got me. I did not think about that. Thanks. Hmmmm.
Good evening fellow mathematicians coz it's 630 pm in India
Hello smart PPL
Believe me or not...
The similar problem was given to 5th and 6th graders in India (age (10-12)). The name of examination was NMTC (Gauss Contest) Level 2. The main point was... The calculation was lot easier... It was 1/m + 1/n = 3/17.
This was the problem. Can you imagine?
Btw, I found a direct formula for these questions. For 1/x + 1/y = 1/a
(x-a)(y-a)= a²
Later you're smart...
This might be this first one I get to solve lol (except for the fact that I only read the thumbnail and looked for positive a and b, thus missing one solution)
wtf, this is actually amazing no baiteroni
0:13 Competitive *_METH_* problems
I could not figure this out...): looks like pappa has to hold my hand and walk me through the problem once again):
:'(
notation 3a = 2019 AND 2017 freaked me out
When you don´t know what to comment and you just comment I don´t know what to comment
Shut up carl
Something is wrong with this video but i can’t get it
What do you mean?
this comment was made before watching:
is the solution b=2018 a=1009
_oh you need to find all ordered pairs_
no one:
me: how the fuck was this video released one hour ago and the top comment from flammable maths 2 is two days old?
Damn I would not find this solution out in time lmao
You made an error in the way you copied the question, the question is not taking place in Z x Z, rather N x N. Copied directly from the exam it says "Find all ordered pairs (a,b) of *positive integers* for which...", i.e., you should not be considering values of a or b that are less than 0.
Easy solution „und“ is „unten“ open and „oder“ is „oben“ open. Isn’t German a beautiful language.
The u in und is the killer for me here, since u looks like ∨ lel
can someone tell me how cos(x)=1 in the clock?
By the fundamental theorem of engineering lmao
Michael penna video was also real guuci
I am having a deja vu...
Why?
@@PapaFlammy69 How can a student do that in exam!! Btw, how much time is given per question?
@@PapaFlammy69 I have seen the question on a thumbnail of another video. Forgot which,but not yours. Didn't watch said video though.
a before b, except after... no, wait, that's something else.
Whoa. so cool!
Question: What happened to my 3000 apples?
Nothing. If you have 3000 apples and you dont substract any apples, then you have 3000 apples.
Bruh, UA-cam is so broken that it's showing me that this video has 30,624,950 views like wha?
It must be that you were watching a video with those views and then went here and the video was still loadig so you have the views of the previous ideo you were watching
use the gOUgU tHeOReM
Arithmethicc xD lol
u were criticizing India's education system for giving Arithematic mean & geometric mean problems hidden in Algebra questions. but I feel this to b more stupid to get through long calculations rather than linking different areas in maths together.
??? a=1009, b=2018, what is the problem? What is the video about? I do not watch it, the solution is trivial.
This is not the only solution...
nice intro.
I’m so early that papa hasn’t hearted all the comments yet
You had this video unlisted before Niw you un(un listed it)
Why do you post so much garbage comments on every video ? I really like reading comment section for additional insights and u spoil it for many like us. Its ok to have some fun but have some respect for the forum
@@caesar_cipher >:(
arethmethicc phew...
Smort
9^2 views in 9 minutes
9 views per minute in the first 9 minutes
NOICE
Only 81 views i don´t get it
81 views it´s Frozen
3ab-2018a-2018b=0 -> 9ab-3*2018a-3*2018b=0 --> (3a-2018)(3b-2018)=(2018)^2 .....
Eat Joshi
666 views
No, actually the views are 6969 views which is OBVIOUSLY my fav no.
@@Kdd160 Why?
@@carlosdecabodelavega3660 :))
@@Kdd160 I guess you saw my video on the proof that a negative times a negative is a positive right?
@@carlosdecabodelavega3660 sorry I did not see it yet I am watching it now
This video is mathematically CuRSeD
The first dislike
bro just made sfft so much harder than it had to be