To be pedantic, there are thirty answers to the question as originally stated, but the other 27 are permutations of those given in the video. E.g. for the first set {1,1,1,2,5}, you could also have {1,1,5,1,2} and this would be a valid answer to the *original question.* The restriction that a ≤ b ≤ c ≤ d ≤ e was a restriction used as a tool to work help work it out, but we need to later remove this restriction again to actually answer the question itself and find *all* possible solutions. We have to do this as the question as stated does not say to discount permutations. a=1, b=2 is not the same as a=2, b=1. We can do this quite simply by saying "any permutation of these is allowed" and if we wanted the number of possible solutions, we just calculate 5 choose 2 for the first and third solutions written at 11:28 and 5 choose 3 for the middle solution. All of these are 10, showing there are thirty total answers to the original question. I won't bother to write them all out lol. Edit: There are actually 40, as I forgot to take into account the fact that you can swap the 2 and 5 for each of the 10 "choices" I calculated for it above.
Great exercise in algebraic reasoning. It shows the power of step by step logic. My mind is always in a hurry and wants to try a hypothesis rather than thinking through the problem. That is a habit with which I need to struggle.
Your voice is so warm and calming! I'm a young guy, and number theory questions always sort of irritate me, but this delighted me very much! I'm going to be more open to these problems from now on! Thanks
Giving this a go: Without loss of generality, assume a≤b≤c≤d≤e. Divide through by abcde, so: s=1/abcd+1/abce+1/abde+1/acde+1/bcde=1 Also note 1/abcd≥1/abce≥...≥1/bcde If abcd>5 then 1/abcde (2)(2): d=e=3, solution: (1,1,1,3,3) (1)(4): d=2, e=5, solution: (1,1,1,2,5) (-1)(-4), (-2)(-2), (-4)(-1): d ≤ 0. So there are three solutions up to reordering, and 40 in total: (1,1,1,2,5) (20 solutions) (1,1,1,3,3) (10 solutions) (1,1,2,2,2) (10 solutions)
Very nice. I love how you explained that kind of problem. Your calm voice is perfect for teaching. I'm leaving a sub and will check your previous videos. Thank you for that content!
Some advice for formality, write the natural numbers except 0 explicitly as N/{0} since whether or not 0 is part of N is a matter of definition. Also, make sure that it is clear that the *e is not part of the equality, either by seperating it with e.g. a semikolon (abcd = 1*1*1*2; * e= 5) or (more commen but you need to leave space earlier) with an underbrace. If you don't do that your solution path includes a+b+c+d=1*1*1*2*e=5, which is wrong.
You are clearly the best math instructor on the Internet. Again I would ask you to give us a brief bio of yourself, education, where do you teach if you do etc. Every time you start a video and convinced that can possibly understand when you get done. Brought to my attention interesting aspects of math. Keep it up tell something about yourself. Thanks and have a great 2025
Bad dictation. It should say You are clearly the best math instructor on the Internet. Again I would ask you to give us a brief bio of yourself, education, where you teach if you do, etc. every time you start a video I am convinced I can’t possibly understand it yet because of your teaching elegance I find that I do. Let us know more about yourself, for gosh sake don’t stop making these videos and have a great 2025.
@@raymondseligman7003 he's nice, but he's not always rigorous or consistent in his train of thought. But he is sincere and you can feel genuine care and passion in his words. Just remember to never stop learning. Or else...
@@suyunbek1399 I don't know maybe I'm being too personal, but I would love to have some idea of his biography, where he learned all this, is he teaching, etc. The one thing I saw said that he had a degree in the culinary arts, which is not exactly the same as complicated integration problems. I'm just wondering, but he is entitled to his own privacy. Have a really good new year.
@@raymondseligman7003 idk I just got here. I found his channel from the video about finding derivative of a sin function from the first principles. Actually the day before that I learned almost all by myself how to derive the chain rule formula from the first principles. It was really cool. He keeps inspiring me with number theory problems. Hopefully he gets to the point where he will explain how the residue theorem from complex analysis is derived. And how it is connected to the idea of differentiation. It probably won't even be enough for me to understand it all because the topic truly is complex.
0 is a part of N, so {0,0,0,0,0} is correct, if 0 must not be taken we are in N* Update : N is not well defined, today the both are possible, and initialy they didn't take it but the definition can contain it.
@fx.paperboy no, N always start with '0' {0,1,2,...}, they are no rule in this context who say 'no trivial 0 answer' or any other formulation, so it miss some parts of desmonstration for (2 or 3 lines for explain), or use N* in thz sentence who start with '1' {1,2,3,...}, the '*' after the sets N, Z, R means without '0'.
That is correct. It's easy to see that we could redo the problem assuming wlog that a ≥ b ≥ c ≥ d ≥ e. That would give the solution set { (2,2,2,1,1), {5,2,1,1,1), (3,3,1,1,1) }. Any of the permutations of the inequalities will yield a permutation of those solutions.
With complex, real or even rational variables we can always choose four arbitrary values and calculate the fifth one. (With the exception that the product of the four chosen values must not be one)
@@MrGeorge1896 unless of course you somehow end up with ABCD = 1 and A + B + C + D = 0. Like in the case of ±1/sqrt(2) ± i/sqrt(2). Free infinitely many solutions
strictly speaking of the question. it's in natural numbers so complex numbers aren't included. But if we talk about real or complex solution there might be a relation solvable with calculus or smth. it will be an infinity of answers how like 1x 5 = 2 x 2.5 = 3 x (1,667)
If you want to restrict the set to natural numbers times i there don't seem to be any solutions. This is because the sum must be greater than 0 (specifically 5+5i or at least one component must be greater), but for five terms where a,b,c,d,e,f,g,h,j,k are natural and i^2=-1, I can't find any solutions where: Real component((a+bi)(c+di)(e+fi)(g+hi)(j+ki)) > 0 Imaginary component((a+bi)(c+di)(e+fi)(g+hi)(j+ki)) > 0 are both true. Now I haven't proven this, so maybe wait for someone else to verify, but this seems correct to me.
@@alansmithee419 Natural complex numbers is a very strange set. I don't know whether to include the natural numbers themselves because they have imaginary component 0. For the sake of my argument I'll just assume Natural Complex numbers means all complex numbers with natural (or zero) real part and natural (or zero) imaginary part. Already by inspection the same three solutions for the real naturals can be adapted into the imaginary naturals: (i, i, i, 2i, 5i) (i, i, i, 3i, 3i) (i, i, 2i, 2i, 2i) And since we've added 0 to make things consistent, there's a new solution there too (0, 0, 0, 0, 0) Now for the rest of them. Consider the arguments of A, B, C, D, and E The argument of ABCDE is the sum of the arguments of A, B, C, D, and E The argument of (A + B + C + D + E) is somewhere between all five of them. CASE 1: 0° < arg(A) + arg(B) + arg(C) + arg(D) + arg(E) < 90° Then arg(ABCDE) = [sum of args] > arg(whatever the largest one is) >= [arg of sums] = arg(A+B+C+D+E) Contradiction! CASE 2: 360° < [sum of args] < 450° Then: arg(ABCDE) = [sum of args] - 360° = arg(smallest) + (arg(next) - 90°) + ... you get the idea < arg(whatever the smallest one is)
You're amazing. I minored in math and understood most of this. The only part I fail to understand is why the third option will not work (1*1*1*5). You said the answer needs to be bigger than 5.
He showed that a=1, b=1, c=1, d=5 would mean e had to be 2 (check: 1*1*1*5*2=10 and 1+1+1+5+2=10). But that contradicts the assumption that d ≤ e, so it simply duplicates the solution a=1, b=1, c=1, d=2, e=5 that he already found.
Nice solution. There is only one tiny detail that bothers me: Using set notation would imply numbers can only occur once in the solutions, making the results ambiguous, like {1,1,1,2,5}={1,2,5}={1,2,5,5,5}. I would prefer (ordered) tuples as results, i.e. (1,1,1,2,5) etc. But this is only nitpicking.
It's a multiset, which often uses the same notation. An ordered tuple implies that (5,2,1,1,1) is a different solution from (1,1,1,2,5). Granted, the ordering constraint fixes that, so that's more a nitpick than anything.
I like the way you present the solution! Awesome! I would prefer at the end if you would add that all permutations of these solutions are correct since the w.l.o.g. assumption you made was yours and not the problem's.
Nice way to solve it. I did it a little bit more complicated in dividing both sides of the equation by abcde. Then you get 1/bcde+1/acde+1/abde+1/abce+1/abcd=1. From here you can see that it is impossible to get 1 if all the five summands are less than 1/5. Therefor at least one of them needs to be equal or greater than 1/5. Without loss of generality we can say 1/abcd is the one, we then get all the solutions by creating all the permutations at the end. That gives us abcd b+c+d+e=0. Since b, c, d and e can not be negative the only possibility in that case is b=c=d=e=0. That gives us (0;0;0;0;0;0) as the only additional solution.
@@enneitesamoht1603 I did some research. We both are wrong. There are different definitions used and some of them consider 0 as a natural number and some don't. I wasn't even aware that definitions exist where 0 is not a natural number, because when I was a kid we learned it at school that way.
@@SG49478 Yeah, when I'm doing proofs in proof assistants, be usually define (inductively) N = Z | S N, and assign semantics of Z as 0 and S n as (n + 1).
Without watching: Wlog, a≤b≤c≤d≤e Therefore, abcde = a+b+c+d+e ≤ e+e+e+e+e = 5e Since e>0, abcd≤5 Split into cases: If abcd = 1, then a=b=c=d=1. So, 4+e = 1e => 4=0. Contradiction. Next, abcd=2, then (a,b,c,d) = (1,1,1,2). So 5+e = 2e and e=5. If abcd = 3, then (a,b,c,d) = (1,1,1,3) So 6+e = 3e and e=3. If abcd = 4, there are 2 options: 1. (a,b,c,d) = (1,1,1,4) And then 7+e = 4e, 7=3e =>
There's a mistake in the notation at the end: A solution set with more than 1 variable cannot be a set of numbers. It has to be a set of tuples. {(1,1,1,2,5), (1,1,1,3,3), (1,1,2,2,2), and their permutations}
Well you explicitely found 3 solutions and eliminated all others hence these 3 are "all". Basically the reason is that a.b.c.d must be between 1 and 5 and you found ALL a,b,c,d that satisfy this condition.
How can you have 1x1x1x2 in 5:45 cause in that scenario you have d that is bigger than e. and if you want to do that, dont you also have to change the 1 and 5 so if you use 2 than you would have to have 2< 1x1x1x2< 10
Seems like u mistaken it although his explanation was clear, first remember that we assume that a≤b≤c≤d≤e(because this is always true, since a, b, c, d, e are constants so we can just switch those spots for the correct order), nowwe have 11 in N are 2 3 4 and 5, 4 can be rewrite as 2.2, so abcd have 5 possible solutions, one more thing that he forgot to set the a=b=c=d=e=0, it's also a solution
Just looking at it without going through the video, if any variable is 0 and the rest are positive and negative numbers which balance to zero, the equation is satisfied. There is no maximum, and there are infinite solutions. Ah. Natural numbers. That was not stipulated in the thumbnail.
i tried this one in my head and i got it within a few seconds: a = -b c = -d e = 0 a,b,c,d can be all n ex: 2.5 + (-2.5) + 1.5 + (-1.5) + 0 = 2.5(-2.5)(1.5)(-1.5)(0) 0 = 0
It is usually said to be referring to investment in generosity, because the Bible often has a principle of sowing and reaping. Throw out your seed and broadcast good wherever you can. You don't know where your seed will grow, but where it does grow, you will get a big return.
At 4:10, there can be a = b = c = d = e = 0 but you said e is not 0 so I guess you don't count this one bcz it's not the maximum anyways, it's the minimum 🙂🙂 Edit: I'm sorry it's all solutions so you gotta count a = b = c = d = e = 0 but still thanks!
You're the Bob Ross of Mathematics, man.
True
Tx for sharing your knowledge
Thank you. Your kindness is appreciated.
To be pedantic, there are thirty answers to the question as originally stated, but the other 27 are permutations of those given in the video.
E.g. for the first set {1,1,1,2,5}, you could also have {1,1,5,1,2} and this would be a valid answer to the *original question.*
The restriction that a ≤ b ≤ c ≤ d ≤ e was a restriction used as a tool to work help work it out, but we need to later remove this restriction again to actually answer the question itself and find *all* possible solutions. We have to do this as the question as stated does not say to discount permutations. a=1, b=2 is not the same as a=2, b=1.
We can do this quite simply by saying "any permutation of these is allowed" and if we wanted the number of possible solutions, we just calculate 5 choose 2 for the first and third solutions written at 11:28 and 5 choose 3 for the middle solution. All of these are 10, showing there are thirty total answers to the original question. I won't bother to write them all out lol.
Edit: There are actually 40, as I forgot to take into account the fact that you can swap the 2 and 5 for each of the 10 "choices" I calculated for it above.
Exactly; WLOG is used solely for solving, so it's important to realize these permutations.
Great exercise in algebraic reasoning. It shows the power of step by step logic. My mind is always in a hurry and wants to try a hypothesis rather than thinking through the problem. That is a habit with which I need to struggle.
Your voice is so warm and calming! I'm a young guy, and number theory questions always sort of irritate me, but this delighted me very much! I'm going to be more open to these problems from now on! Thanks
Never stop learning cuz those who stop learning stop living. Love it ❤❤❤
Those who stop learning... vote. 😢
my favorite threat
Who?
I practiced this one a few times until I got it all digested! Thanks!
One of the best math channels out there. Glad you’re doing this!
I rarely bother commenting on videos, but this one was really beautiful. Subbed
Excellent! Well prepared exercise. Simple and beautiful. Straight to the solution without detours or distractions.
Giving this a go:
Without loss of generality, assume a≤b≤c≤d≤e. Divide through by abcde, so:
s=1/abcd+1/abce+1/abde+1/acde+1/bcde=1
Also note 1/abcd≥1/abce≥...≥1/bcde
If abcd>5 then 1/abcde
(2)(2): d=e=3, solution: (1,1,1,3,3)
(1)(4): d=2, e=5, solution: (1,1,1,2,5)
(-1)(-4), (-2)(-2), (-4)(-1): d ≤ 0.
So there are three solutions up to reordering, and 40 in total:
(1,1,1,2,5) (20 solutions)
(1,1,1,3,3) (10 solutions)
(1,1,2,2,2) (10 solutions)
Being a Nigerian who loves maths and always wanted to go to the IMO, i must say you are doing great Prime Newron ❤
Wonderful question, nice solution using inequalities.
Thanks for excellent video.
Btw, you write in nice clear, unambiguous, and crazily -accurately-horizontal! No ruler required haha
Thank you 😊
Very nice. I love how you explained that kind of problem. Your calm voice is perfect for teaching. I'm leaving a sub and will check your previous videos. Thank you for that content!
Some advice for formality, write the natural numbers except 0 explicitly as N/{0} since whether or not 0 is part of N is a matter of definition. Also, make sure that it is clear that the *e is not part of the equality, either by seperating it with e.g. a semikolon (abcd = 1*1*1*2; * e= 5) or (more commen but you need to leave space earlier) with an underbrace. If you don't do that your solution path includes a+b+c+d=1*1*1*2*e=5, which is wrong.
I enjoyed this video a lot. Youre voice is so calming 😭
Your handwriting on the blackboard is soooo satisfying
"Who stops learning stops living" love it 🙂
Never stop learning those who stop learning have stopped living.
You got good handwriting, man! I love the blackboard in the soft chalk. And the math is good too! Thanks for the videos
You are clearly the best math instructor on the Internet. Again I would ask you to give us a brief bio of yourself, education, where do you teach if you do etc. Every time you start a video and convinced that can possibly understand when you get done. Brought to my attention interesting aspects of math. Keep it up tell something about yourself. Thanks and have a great 2025
Bad dictation. It should say
You are clearly the best math instructor on the Internet. Again I would ask you to give us a brief bio of yourself, education, where you teach if you do, etc. every time you start a video I am convinced I can’t possibly understand it yet because of your teaching elegance I find that I do. Let us know more about yourself, for gosh sake don’t stop making these videos and have a great 2025.
@@raymondseligman7003 he's nice, but he's not always rigorous or consistent in his train of thought.
But he is sincere and you can feel genuine care and passion in his words. Just remember to never stop learning. Or else...
@@suyunbek1399 I don't know maybe I'm being too personal, but I would love to have some idea of his biography, where he learned all this, is he teaching, etc. The one thing I saw said that he had a degree in the culinary arts, which is not exactly the same as complicated integration problems. I'm just wondering, but he is entitled to his own privacy. Have a really good new year.
@@raymondseligman7003 idk I just got here. I found his channel from the video about finding derivative of a sin function from the first principles. Actually the day before that I learned almost all by myself how to derive the chain rule formula from the first principles. It was really cool. He keeps inspiring me with number theory problems. Hopefully he gets to the point where he will explain how the residue theorem from complex analysis is derived. And how it is connected to the idea of differentiation. It probably won't even be enough for me to understand it all because the topic truly is complex.
1 1 1 2 5
1 1 1 3 3
1 1 2 2 2
Get the formula out that a
This solution seems cleaner than the first go around but that is just based on my fuzzy memory. Very logical development.
There's something just so relaxing about this guy that just spreads happiness
I love your channel, please continue!
That was amazing! Definitely subscribing!
11:35 the way bro said "bye bye" and then immediately disappeared took me out 😆
why is (0,0,0,0,0) not a solution as well? 0 is not part of N?
Ye 0 is not part of N
0 is a part of N, so {0,0,0,0,0} is correct, if 0 must not be taken we are in N*
Update : N is not well defined, today the both are possible, and initialy they didn't take it but the definition can contain it.
@Eclairiuss so here in this context, what's N? I thought N was natural numbers starting from 1,2,3.
@fx.paperboy no, N always start with '0' {0,1,2,...}, they are no rule in this context who say 'no trivial 0 answer' or any other formulation, so it miss some parts of desmonstration for (2 or 3 lines for explain), or use N* in thz sentence who start with '1' {1,2,3,...}, the '*' after the sets N, Z, R means without '0'.
@Eclairiuss I see. That's new info to me. Thanks for the explanation.
All permutations of the solutions you present are solutions to the original problem i think.
That may be a valid point since it was wlog. I am not sure.
@@PrimeNewtons 0+0+0+0+0=0*0*0*0*0
That is correct. It's easy to see that we could redo the problem assuming wlog that a ≥ b ≥ c ≥ d ≥ e. That would give the solution set { (2,2,2,1,1), {5,2,1,1,1), (3,3,1,1,1) }. Any of the permutations of the inequalities will yield a permutation of those solutions.
Wonderful explanation. Keep our learning journey on.❤❤
love the video! if all of the variables were complex, how would this affect the solution set? would it still be solvable?
With complex, real or even rational variables we can always choose four arbitrary values and calculate the fifth one. (With the exception that the product of the four chosen values must not be one)
@@MrGeorge1896 unless of course you somehow end up with ABCD = 1 and A + B + C + D = 0. Like in the case of ±1/sqrt(2) ± i/sqrt(2).
Free infinitely many solutions
strictly speaking of the question. it's in natural numbers so complex numbers aren't included. But if we talk about real or complex solution there might be a relation solvable with calculus or smth. it will be an infinity of answers how like 1x 5 = 2 x 2.5 = 3 x (1,667)
If you want to restrict the set to natural numbers times i there don't seem to be any solutions.
This is because the sum must be greater than 0 (specifically 5+5i or at least one component must be greater), but for five terms where a,b,c,d,e,f,g,h,j,k are natural and i^2=-1, I can't find any solutions where:
Real component((a+bi)(c+di)(e+fi)(g+hi)(j+ki)) > 0
Imaginary component((a+bi)(c+di)(e+fi)(g+hi)(j+ki)) > 0
are both true.
Now I haven't proven this, so maybe wait for someone else to verify, but this seems correct to me.
@@alansmithee419 Natural complex numbers is a very strange set. I don't know whether to include the natural numbers themselves because they have imaginary component 0.
For the sake of my argument I'll just assume Natural Complex numbers means all complex numbers with natural (or zero) real part and natural (or zero) imaginary part.
Already by inspection the same three solutions for the real naturals can be adapted into the imaginary naturals:
(i, i, i, 2i, 5i)
(i, i, i, 3i, 3i)
(i, i, 2i, 2i, 2i)
And since we've added 0 to make things consistent, there's a new solution there too (0, 0, 0, 0, 0)
Now for the rest of them. Consider the arguments of A, B, C, D, and E
The argument of ABCDE is the sum of the arguments of A, B, C, D, and E
The argument of (A + B + C + D + E) is somewhere between all five of them.
CASE 1: 0° < arg(A) + arg(B) + arg(C) + arg(D) + arg(E) < 90°
Then
arg(ABCDE) = [sum of args] > arg(whatever the largest one is) >= [arg of sums] = arg(A+B+C+D+E)
Contradiction!
CASE 2: 360° < [sum of args] < 450°
Then:
arg(ABCDE) = [sum of args] - 360°
= arg(smallest) + (arg(next) - 90°) + ... you get the idea
< arg(whatever the smallest one is)
I was looking for solution of these types of problems. Thanks for it😊
You're amazing. I minored in math and understood most of this. The only part I fail to understand is why the third option will not work (1*1*1*5). You said the answer needs to be bigger than 5.
He showed that a=1, b=1, c=1, d=5 would mean e had to be 2 (check: 1*1*1*5*2=10 and 1+1+1+5+2=10). But that contradicts the assumption that d ≤ e, so it simply duplicates the solution a=1, b=1, c=1, d=2, e=5 that he already found.
Great stuff, keep going!
Wonderful stuff!! Keeps my brain active and alive! Thanks for posting!!
I really enjoyed this problem. Do you recommend any books to help learn these kind of problems?
Very good stuff!
Very interesting! Thanks for sharing!
Love from India bruv.
never stop learning, because life never stops teaching -- me
Never stop learning.
Those who stop learning stop living.
I am a new fan of this quotation.😅😊
Mathematics 💜✨
Very nice♥️
Very enjoyable.
Nice solution. There is only one tiny detail that bothers me: Using set notation would imply numbers can only occur once in the solutions, making the results ambiguous, like {1,1,1,2,5}={1,2,5}={1,2,5,5,5}. I would prefer (ordered) tuples as results, i.e. (1,1,1,2,5) etc. But this is only nitpicking.
It's a multiset, which often uses the same notation. An ordered tuple implies that (5,2,1,1,1) is a different solution from (1,1,1,2,5). Granted, the ordering constraint fixes that, so that's more a nitpick than anything.
Python user😒
Beautiful, these ideas that give intelligence
Damn it's so simple yet so brilliant, love it!
You're the best at math among all the Black people I've met.
this made my day. Thank you so much for posting.
For 1 < abcd ≤ 5, isnt 1 * 1 * 2 * 3 also a valid solution?
That gives 6 bruh
We can put a
Thank you from the France great video
Nice one
Somehow a much smaller solution set than I expected.
great question, can i make a video to Portuguese audience using your solution? giving you a credit of course.
Brilliant content! Thank you so much!
Man ur things are so nice
Nice solution.
I like the way you present the solution! Awesome! I would prefer at the end if you would add that all permutations of these solutions are correct since the w.l.o.g. assumption you made was yours and not the problem's.
love you man... from india
Amazing! A wonderful appetizer for 2025! God bless you!
I like his diction and handwriting, apart from logical thinking and pedagogy
Nice way to solve it. I did it a little bit more complicated in dividing both sides of the equation by abcde. Then you get 1/bcde+1/acde+1/abde+1/abce+1/abcd=1. From here you can see that it is impossible to get 1 if all the five summands are less than 1/5. Therefor at least one of them needs to be equal or greater than 1/5. Without loss of generality we can say 1/abcd is the one, we then get all the solutions by creating all the permutations at the end. That gives us abcd b+c+d+e=0. Since b, c, d and e can not be negative the only possibility in that case is b=c=d=e=0. That gives us (0;0;0;0;0;0) as the only additional solution.
0 is most definitely not a natural Number. Normally there would be a 0 under N.
@@enneitesamoht1603 I did some research. We both are wrong. There are different definitions used and some of them consider 0 as a natural number and some don't. I wasn't even aware that definitions exist where 0 is not a natural number, because when I was a kid we learned it at school that way.
@@SG49478 Yeah, when I'm doing proofs in proof assistants, be usually define (inductively) N = Z | S N, and assign semantics of Z as 0 and S n as (n + 1).
I got smarter after this video, thx man :)
There is actually one step missing in your solution. The assumption you are making is a
BEAUTIFUL SOL.N SIR
Without watching:
Wlog, a≤b≤c≤d≤e
Therefore, abcde = a+b+c+d+e ≤ e+e+e+e+e = 5e
Since e>0, abcd≤5
Split into cases:
If abcd = 1, then a=b=c=d=1. So,
4+e = 1e => 4=0. Contradiction.
Next, abcd=2, then (a,b,c,d) = (1,1,1,2). So 5+e = 2e and e=5.
If abcd = 3, then (a,b,c,d) = (1,1,1,3)
So 6+e = 3e and e=3.
If abcd = 4, there are 2 options:
1. (a,b,c,d) = (1,1,1,4) And then 7+e = 4e, 7=3e =>
There's a mistake in the notation at the end: A solution set with more than 1 variable cannot be a set of numbers. It has to be a set of tuples.
{(1,1,1,2,5), (1,1,1,3,3), (1,1,2,2,2), and their permutations}
This was a beautyful Q.E.D.
nice one
I didn't understand anything, but it's very interesting.
can you explain how this also proves that those are "all" the solutions?
Well you explicitely found 3 solutions and eliminated all others hence these 3 are "all".
Basically the reason is that a.b.c.d must be between 1 and 5 and you found ALL a,b,c,d that satisfy this condition.
Awesome!!!!
This was beautiful
How can you have 1x1x1x2 in 5:45 cause in that scenario you have d that is bigger than e. and if you want to do that, dont you also have to change the 1 and 5 so if you use 2 than you would have to have 2< 1x1x1x2< 10
Seems like u mistaken it although his explanation was clear, first remember that we assume that a≤b≤c≤d≤e(because this is always true, since a, b, c, d, e are constants so we can just switch those spots for the correct order), nowwe have 11 in N are 2 3 4 and 5, 4 can be rewrite as 2.2, so abcd have 5 possible solutions, one more thing that he forgot to set the a=b=c=d=e=0, it's also a solution
mind blowing thank you very much
I wonder if there if there is a similar way to find general solutions for all natural n,a1,a2,a3,...,an a1+a2+a3+...+an = a1*a2*a3*...*an
WELL DONE
Mathematics ❤❤
sir you are great
there's a solution (0,0,0,0,0) having e < a+b+c+d+e was wrong cause a+b+c+d can be equal to 0 (that's why it was missed)
Just looking at it without going through the video, if any variable is 0 and the rest are positive and negative numbers which balance to zero, the equation is satisfied. There is no maximum, and there are infinite solutions.
Ah. Natural numbers. That was not stipulated in the thumbnail.
what about the other possibilities because of the restriction
They turn out to be permutations of the three solutions found. That's what "without loss of generality (wlog)" implies.
i tried this one in my head and i got it within a few seconds:
a = -b
c = -d
e = 0
a,b,c,d can be all n
ex: 2.5 + (-2.5) + 1.5 + (-1.5) + 0 = 2.5(-2.5)(1.5)(-1.5)(0)
0 = 0
INTEGRAL.
Happy new year 🎉
Solved it using perfect numbers
Nice
It was super!
I don't understand sir 🙏. "cast your bread upon many water: for after many days, you will find it".
It is usually said to be referring to investment in generosity, because the Bible often has a principle of sowing and reaping. Throw out your seed and broadcast good wherever you can. You don't know where your seed will grow, but where it does grow, you will get a big return.
1 1 1 2 5
Took me 5 seconds of thinking about it.
Wow that’s cool
6:21 My dude only erasing one of the equals bars and writing the other one underneath😎
That's really great!!!!! Congrats 👏👏
Mulțumesc !
is zero a natural number?
Historically, no; nowadays, yes.
No
@@putriwachid1848 Historically, no; nowadays, yes.
@@boguslawszostak1784 Historically: sometimes. Nowadays: no -- at least not in the English-speaking world.
I grew up with zero as a whole number, but the natural (I.e., counting) numbers start from one.
I don't know if your naturals include zero (I do include it). In that case, 0,0,0,0,0 is also a solution.
0:57
At 4:10, there can be a = b = c = d = e = 0 but you said e is not 0 so I guess you don't count this one bcz it's not the maximum anyways, it's the minimum 🙂🙂
Edit: I'm sorry it's all solutions so you gotta count a = b = c = d = e = 0 but still thanks!
0 is not natural number
like the other comment says, 0 is not a natural number and the question says a,b,c,d,e all belong to the set of natural numbers.
@stokmlnes-flame2025 A matter of (viewpoint) convention...
@@jige1225 what do you mean
Never stop learning 📖
you can rewrite the equation to a=(b+c+d+e)/(bcde-1)