@@MichaelPennMath what if you dont think of the mod stuff because you were not taught that and always forget it..do you know of another solution? Thanks..
yeah, rly not a whole lotta explaining going on tbh but thats just the content, doesnt say anywhere that the problem would be explained in the video, only solved :P
IN MANY WEBSITES THERE ARE 6 ORDERED PAIRS, BUT THERE WOULD BE INFINITE ♾ SOLUTIONS IF WE WOULD TAKE THE VALUE OF (2018k AND 2018z) IN DECIMALS TO MAKE THE SUM 3. GOOD DAY🙂
Never heard that Sting was this good at maths ! Seriously, as an "international viewer", I appreciate the 100% clear and articulate voice, the not too cluttered blackboard, the straight to the point explanation. Thank you very much, master !
My fav part is how he ends abruptly, "so that's the end of the video." Spectrums of light, burn bright into that cold dark night! Michael Penn, you are the man with the math plan!
Another solution: From the original equation we get: (a+b)/ab = 3/2018 Let k be a natural number such that: i/ a + b = 3k ii/ ab = 2018k The second equation (ii) can also be written: ab = 2*1009*k, we will then decompose further: Let z and t be two natural numbers satisfying the following: k = z*t, and t //a, and z // b, (// = divides) Since a and b are symmetric, we will make our reasoning around a (and it would apply to b in equal measure). We have two possibilities: XX- 2 and 1009 both divide a XXXX- only 1009 divides a (the other two possibilities simply amount to swapping a and b) XX- 2 and 1009 both divide a a = 2018*t b = z Let us replace this in equation (i), we get: 2018*t + z = 3*t*z This means that: t // z => z = p*t (p a natural number) z // 2018*t Thus p // 2018 => p is in the set {1, 2, 1009, 2018} => z is in the set {t, 2*t, 1009*t, 2018*t} We get: a = 2018*t b is in the set {t, 2*t, 1009*t, 2018*t} k is in the set {t², 2*t², 1009*t², 2018*t²} The first equation (i) is always satisfied, so we check the second one (ii): ------ 2018*t + t = 3*t² => 2019*t = 3*t² => t = 673 We replace and get the first solution: SOL1: a = 2018*673 b = 673 ------ 2018*t + 2*t = 6*t² => 2020*t = 6*t² => t = 1010/3, which is not a natural number, so no solution here ------ 2018*t + 1009*t = 3*1009*t² => 3*1009*t = 3*1009*t² => t = 1 We replace and get the second solution: SOL2: a = 2018 b = 1009 ------ 2018*t + 2018*t = 3*2018*t² => 2*2018*t = 3*2018*t² => t = 2/3, which is not a natural number, so no solution here XXXX- only 1009 divides a a = 1009*t b = 2*z Let us replace this in equation (i), we get: 1009*t + 2*z = 3*t*z This means that: ff/ t // 2*z => 2*z = p*t ffff/ z // 1009*t => 1009*t = q*z => 2018*t = q*p*t => q*p = 2018 The possible values of p (or q) are thus in the set {1, 2, 1009, 2018}. Replacing in (ff) we get: 2*z = b = t or 2*z = b = 2*t or 2*z = b = 1009*t or 2*z = b = 2018*t Thus: b is in the set {t, 2*t, 1009*t, 2018*t} While k is in the set {t²/2, t², 1009*t²/2, 1009*t²} Following the same procedure as for XX, we get: ------ 1009*t + t = 3*t²/2 => 2020*t = 3*t² => t = 2020/3, which is not a natural number, so no solution here ------ 1009*t + 2*t = 3*t² => 1011*t = 3*t² => t = 373 We replace and get the second solution: SOL3: a = 1009*373 b = 373 ------ 1009*t + 1009*t = 3*1009*t²/2 => 4*1009*t = 3*1009*t² => t = 4/3, which is not a natural number, so no solution here ------ 1009*t + 2018*t = 3*1009*t² => 3*1009*t = 3*1009*t² => t = 1 We replace and get the second solution: SOL4: a = 1009 b = 2018 This is simply SOL2 swapped So we have 6 solutions, SOL1, SOL2 and SOL3 and their swapped counterparts.
Alternative solution: Rewrite the equation as 1/C*(A+B)/AB=3/2018 with A and B coprime (the original variables correspond to a=AC, b=BC, C=gcd(a,b) ). Then A+B and AB are coprime, so AB divides 2018 and 3 divides A+B. Conversely for every such pair we can find a unique C making the equation hold.
The answer looks more complicated than it is because of the way it is presented. A more natural way of approaching it would be to rewrite it as: 1/a = 3/2018 - 1/b. That is, 1/a = (3b-2018)/2018b. Similarly, 1/b = (3a-2018)/2018a. Multiply the two equations and you get: 1/(ab) = (3a-2018)(3b-2018)/2018^2(ab), That is: (3a-2018)(3b-2018) = 2018^2
Forgive me if I'm being ignorant but I see no improvement here. He actually got to that last equation you've given in about 3 lines which isn't too bad. So yea ... Struggling to see how your different ( but valid ofc ) method helps in anyway?
Yh this method is bit better if you can't see how to split it into a binomial form like in the video. This person just re-ordered things to get 2 equations and then make it into one equation to factor it. I can understand why he thinks it's more natural. I think it is as well.
Different approach but the same level of intricacy. Appreciate it though. I will say his skipping of step did make things complex but what's a problem without a challenge. If he showed all the reasoning and steps it would have been no different than spoon feeding and then it would be called learning rather than understanding.
You can reduce the primes mod 3 getting -1, -1, 1 and 1. Then note you need both the -1s together to get a 1; aka, 4. Then the 1009s can be 0, 1 or 2 of them with the 4. Just a little less "list all possibilities and eliminate".
Amazing growth as a presenter in a year. Here is a bit to fast, more looking to notes etc. A year later its much improved, and current are just polishing.
A quick solution for integer values of a and b. The right hand side is 3/2018. This can be written as (2+1)/2018 which when expanded becomes (2/2018) + (1/2018), that is (1/1009) + (1/2018), which follows a = 1009 and b = 2018 or a = 2018 and b = 1009.
This is how I solved it, though with a few more steps. It helps to start with smaller numbers, like 1/4 + 1/2 = 3/4. This can mean that if 1/4 is a, and 1/2 is b, then a = 2018 and b = 1/2a, or 1009
Looks very similar to the parallel resistances rule lol. 1/((1/a)+(1/b)) = 2018/3 and for parallel resistances, a general rule is that if you have an n ohm resistor and a 2n ohm resistor, their combined resistance is 2n/3. In this case, you could set 2n = 2018 which gives you the 1009 and 2018 solution easily. Not very rigorous but a neat pattern!
Yup because after moving `a`s and `b`s to one side and constants to the other, you get (a+b)/(a·b) = 3/2018 which is pretty much a harmonic mean of `a` and `b`. This way you can find one possible solution by geometric methods. Replacement resistance of two resistors in parallel is also their harmonic mean, or you can think of it as conductances instead, to find out that the replacement conductance is the sum of the conductances of these two resistors (which makes perfect sense: the amount of current through both resistors is the sum of the currents through the first and the second resistor in parallel). Harmonic means also work for calculating focal lengths of lenses, so harmonic means are quite common pattern in nature ;) You find it pretty much everywhere where you work on inverses of some quantities that sum up together.
@@bonbonpony If 1/a +1/b = a/(ab)+b/(ab) = (a+b)/(ab)=3/2018. This will give you complex solutions with a=3/2 ± i√(8063)/2 and b = 3/2 ∓ i√(8063)/2. Of course the original question asks for all natural numbers, so complex solutions which are also negative and fractions don't count at all😁; still it was fun.
@@davidgriffin79 More like 1/a +1/b = b/(ab)+a/(ab). Also I'm not sure where did you get the complex solutions from. But considering that there's only one equation and two unknowns, infinitely many pairs of numbers would work, including your complex ones.
@@bonbonpony 1/a +1/b = b/(ab) + a/(ab), that is multiply 1/a by b/b and 1/b by a/a to get (a+b)/(ab). There are two equations if we assume numerators and denominators are identical; in such instance you have two equations which will give a and b as complex conjugates.
@@davidgriffin79 My first correction was about your mistake in getting those two fractions to common terms. This time you did it correctly. As for your assumption about the numerators and denominators being pairwise equal, there's a catch: they can have a common factor, let's say `C`. Then you have `C·(a+b) = 3` and `C·(a·b) = 2018. So even if you solve for `a` and `b` the way you described, there's still some freedom coming from the factor `C` (as there _should_ be, since we have 2 unknowns, `a` and `b`, but only 1 equation). In other words, for fractions, the equality holds not just for `2/3 = 2/3`, but also for `4/6 = 2/3`, or `6/9 = 2/3`, etc. So the assumption that numerators are equal and denominators are equal is flawed.
The third solution on the blackboard should be a=2018 and b=1009, not 109. And that’s actually what one could easily see as solution without all the fuss. 1/2018 + 1/1009 =. 1/2018 + 2/2018 = 3/2018.
Great video. I would advise that you slow down a bit - not in what you cover, but just the speed of your voice. It comes off a bit breathless and nervous. This is good content. I'm glad to see people making such detailed and informative videos about math.
He’s not saying it’s fast, he’s saying that the speech is not natural. Give yourself time to breath. I do the same thing and have to remind myself to slow down.
I would assume that it’s because the equation was multiplying (which modulo can kinda relate to it ? Maybe) and there a,b were multiplied by 3 sooo... I also have no idea
Well, you had "(3a - 2018)(3b-2018) = 2018*2018" and we know a and b need to be integers. This means that you can assign (3a-2018) and (3b-2018) to any combination of numbers that form 2018*2018. If you add 2018 to any combination, you need to be able to divide by 3, hence the "1 mod 3". Dunno if that makes any sense but it does in my head
Using congruence classes or modulo is actually very common to solve diophantine equations (integer solutions) that requires matching factors. 3 is chosen since it eliminates the unknown a and b. Otherwise you would not be able to determine the congruence of each factor.
I tried all factor pairs, but some of them don't give integer solutions when you divide by 3. So that's a way to find the "valid" factor pairs without trial and error. Btw, he forgot the factor pair 2018*2018, but this one also doesn't give an integer solution.
4:20 When looking at factor pairs, it's not mentioned in the video but 1009 = 1 mod 3, so obviously adding it or its square to either factor doesn't change whether or not the factor equals 1 mod 3. So since 2 = 2 mod 3 and 2^2 = 3 mod 3, you just have to consider the combinations involving either 2^2, 1009 and/or 1009^2 (which is ultimately what's derived in the video.)
IN MANY WEBSITES THERE ARE 6 ORDERED PAIRS, BUT THERE WOULD BE INFINITE ♾ SOLUTIONS IF WE WOULD TAKE THE VALUE OF (2018k AND 2018z) IN DECIMALS TO MAKE THE SUM 3. GOOD DAY🙂
We can work also a bit differently. Let's assume that b>a (symmetry) so there is a positive integer r such that b=a+r. We plug this and we get a quadratic equation with respect to a with integer coefficients. So discriminant must be a perfect square for solution to exist from where we can do a nice factoring and get some easy cases. We also check for a=b and we get permutations too
Well, I paused the video and tried to work it out myself. I found two of the three pairs. Also, shouldn't the first pair be 1358114 and 673, rather than 1354114 and 673?
Am I missing an obvious reason why (2*1009)(2*1009) isn't a factor pair considered in that list? It doesn't affect the answer because 2*1009 = 2 mod 3, so that pair gets rejected anyway, but should it have been checked?
Why does wolfram alpha show 1/673 + 1/1354114 - 3/2018 = 1000/459760295249 instead of zero (my calculator shows the same thing)? Also, why does 2018*673 + 2018*1354114 not equal 3*673*1354114 (2018a+2018b=3ab (from about 0:56 in the video)?
Heya, loved the video, but I spotted a little something that the other comments don't seem to have pointed out. I am sure it's a typo or something, but the b for solution pair is 1358114.
5:15 the original pair 2018·2018 is missing, but since 1009 = 1 (mod 3), all valid pairs must pack the 2's together anyway (must have one factor odd and the other divisible by 4)
I’m struggling to understand the logic. I thought it was to make sure the factors would have an integer solution when plugged back into (3a-2018) but that’s not the case. If we have a prime number P which satisfies mod(p,3)=1 that does not mean mod(p+2018,3)=1. Try 31 for example. Can you explain the logic to me?
To quickly filter out the pairs that won't work. (3a-2018)(3b-2018) must have the same remainder as the right part. It's a neat trick to save time solving several more equations, but it's not crucial.
To solve each equation you have to add the number in each set of parentheses to 2018. Since 2018 is congruent to 2, mod 3, whatever we add to 2018 must be congruent to 1, mod 3, in order to total a number that is evenly divisible by three. That is why both sets of parentheses in his left-hand column both had to be congruent to 1, mod 3.
08:40 i was waiting for that so bad i had the idea that you may not put them into consideration , Great job i did this with my students and i was surprised that 1 of 43 students did answer it like super shocked
Why is it that the second item in your factors pairs list was not congruent to one, mod 3? They all multiply to the same number... Oh I see, each set of parentheses has to be congruent to one, mod 3. Why is that important though? Oh I see now. To solve each equation you have to add this number in parentheses to 2018. Since 2018 is congruent to 2, mod 3, whatever we add to 2018 must be congruent to 1, mod 3, in order to total a number that is evenly divisible by three. I think what confused me was the fact that he said the entire left side was congruent to 1, mod 3, instead of each binomial separately. Seems to me it would have been more clear if he had emphasized the fact that each binomial was that way. And the reason that is the case is, if you have a number evenly divisible by three, and you subtract a number congruent to 2, mod 3, then you will get a number congruent to 1, mod 3.
Yes, for those of us for whom modulo arithmetic was not on our math syllabus that was confusing but your explanation makes sense. It is impressive that these problems require so many different techniques.
You don't really need that step at all. Just try all the factor pairs and you will find quickly that (2) (2 * 1009^2) doesn't result in integer solutions (a = 2020 / 3). I think it was just a technique to pre-filter the cases to try.
825 and 3643 are the 2 closest non solutions I have seen. They work out to 2018.000224 ish. That is darn close. "Found" via computer programming with small "epsilon" (error allowance).
Great video, this question took me about 5 hours using a far more convoluted method and this really helps me learn how to do it more efficiently. Just one question: am I right in saying you missed out the factor pair (2•1009)×(2•1009)? That wouldn't have worked anyway but just to be sure I understand the concept, would that have been a valid factor pair?
IN MANY WEBSITES THERE ARE 6 ORDERED PAIRS, BUT THERE WOULD BE INFINITE ♾ SOLUTIONS IF WE WOULD TAKE THE VALUE OF (2018k AND 2018z) IN DECIMALS TO MAKE THE SUM 3. GOOD DAY🙂
if you dont have to prove it you can just see that its two ones over a number and adds to 3/ a number so that a and b one can be equaled to the number and the other denominator divided by two to create 2 and 1+2=3
I used simple algebra and got only two solutions (1009,2018) and the swap of this pair and when I checked the end of the video all my conciet vanished. I will still count it as a win for me. Sorry couldn't understand the problem because I am yet to study modulus. But still a big thanks as these videos have helped to understand numbers and patterns better.
You could try to divide it by all the primes up to sqrt(1009) and there aren't that many. But I suspect the presenter just has the first n primes memorized, for some value of n :)
What's the justification for the mod trick? Surely it's backed by a theorem, with these mod conditions as premises, concluding that the answers it gives are exhaustive. I have no idea what theorem proves that would be called, and I'd love to know.
I _think_ the reasoning goes like this: The equation at that point had two factors that multiply together to form 2018^2. He was just looking for those two factors to solve the problem. He used "they're both =1(mod 3)" to filter the search down a bit, since he knew the results would have to satisfy that.
He did that to show that both 3a-2018 and 3b-2018 had to be 1 mod three. This allowed him to eliminate some of the factor pairs as he did in the next step. 1 mod 3 just means that you get a remainder of 1 when you divide that number by three, ie 3a-2018=-2018=3(-673)+1=1 mod 3. Hope this helps!
Guys, I saw the thumbnail and I immediately thought it would just be a=2018 and b=1009=(2018/2) so that when you add the numerators become (2+1)/2018? I think that the set of all natural numbers is the restriction. but still.
If n is congruent to a mod b, it means that the remainder of n/b is the same as the remainder of a/b. Usually a is picked between 0 and b-1 because it's more practical. In that case, the remainder of n/b is exactly a.
Wished my dad had sent me to this kindergarten...
Zu Yao Teoh Lmao
why do you want this guy to touch you during nap time?
check out on my channel for more maths problem
@@dieharddougie who wouldn't?
I prefer a happy ending with my juice box.
Dame he lost me at "ok"
My Calc 2 teacher in a nutshell
@@kevboom2088 why is he or she in a nutshell ...
@@coolmodad he says “ok” during lectures every 30 seconds lol
damn*
@@coolmodad stuck
Putnam problems are always fascinating. Thanks for sharing.
Thanks, I am going to try and add a bunch of Putnam solution in the next few days to motivate our students for the upcoming exam!
@@MichaelPennMath what if you dont think of the mod stuff because you were not taught that and always forget it..do you know of another solution? Thanks..
@@MichaelPennMath ruclips.net/video/TKm_sJ14KRk/видео.html
@leahcim38 hahah very underrated comment
@@leif1075 Flammable Maths has an alternative solution I think on his YT. I didn’t see it but I believe he does not use modular arithmethic.
This guy is a world champion at looking at solutions and presenting them like anyone can
yeah, rly not a whole lotta explaining going on tbh but thats just the content, doesnt say anywhere that the problem would be explained in the video, only solved :P
lol
IN MANY WEBSITES THERE ARE 6 ORDERED PAIRS, BUT THERE WOULD BE INFINITE ♾ SOLUTIONS IF WE WOULD TAKE THE VALUE OF (2018k AND 2018z)
IN DECIMALS TO MAKE THE SUM 3.
GOOD DAY🙂
@@sanjusingh92443 You .. can't they are natural ..
You say that, but this is an easy problem.
Note: for a = 673, we find b = 1358114. Thank you so much for the great work you do.
Never heard that Sting was this good at maths !
Seriously, as an "international viewer", I appreciate the 100% clear and articulate voice, the not too cluttered blackboard, the straight to the point explanation. Thank you very much, master !
My fav part is how he ends abruptly, "so that's the end of the video." Spectrums of light, burn bright into that cold dark night! Michael Penn, you are the man with the math plan!
8:15 b is supposed to be 1009. Just a typo since you said it right tho.
Yea, I though I had found an additional solution. the easiest one!.
@@chucktodd7329 yeah bro this aint special tbh
Another solution:
From the original equation we get: (a+b)/ab = 3/2018
Let k be a natural number such that:
i/ a + b = 3k
ii/ ab = 2018k
The second equation (ii) can also be written: ab = 2*1009*k, we will then decompose further:
Let z and t be two natural numbers satisfying the following: k = z*t, and t //a, and z // b, (// = divides)
Since a and b are symmetric, we will make our reasoning around a (and it would apply to b in equal measure). We have two possibilities:
XX- 2 and 1009 both divide a
XXXX- only 1009 divides a
(the other two possibilities simply amount to swapping a and b)
XX- 2 and 1009 both divide a
a = 2018*t
b = z
Let us replace this in equation (i), we get:
2018*t + z = 3*t*z
This means that:
t // z => z = p*t (p a natural number)
z // 2018*t
Thus p // 2018 => p is in the set {1, 2, 1009, 2018} => z is in the set {t, 2*t, 1009*t, 2018*t}
We get:
a = 2018*t
b is in the set {t, 2*t, 1009*t, 2018*t}
k is in the set {t², 2*t², 1009*t², 2018*t²}
The first equation (i) is always satisfied, so we check the second one (ii):
------ 2018*t + t = 3*t² => 2019*t = 3*t² => t = 673
We replace and get the first solution:
SOL1:
a = 2018*673
b = 673
------ 2018*t + 2*t = 6*t² => 2020*t = 6*t² => t = 1010/3, which is not a natural number, so no solution here
------ 2018*t + 1009*t = 3*1009*t² => 3*1009*t = 3*1009*t² => t = 1
We replace and get the second solution:
SOL2:
a = 2018
b = 1009
------ 2018*t + 2018*t = 3*2018*t² => 2*2018*t = 3*2018*t² => t = 2/3, which is not a natural number, so no solution here
XXXX- only 1009 divides a
a = 1009*t
b = 2*z
Let us replace this in equation (i), we get:
1009*t + 2*z = 3*t*z
This means that:
ff/ t // 2*z => 2*z = p*t
ffff/ z // 1009*t => 1009*t = q*z => 2018*t = q*p*t => q*p = 2018
The possible values of p (or q) are thus in the set {1, 2, 1009, 2018}. Replacing in (ff) we get:
2*z = b = t or 2*z = b = 2*t or 2*z = b = 1009*t or 2*z = b = 2018*t
Thus: b is in the set {t, 2*t, 1009*t, 2018*t}
While k is in the set {t²/2, t², 1009*t²/2, 1009*t²}
Following the same procedure as for XX, we get:
------ 1009*t + t = 3*t²/2 => 2020*t = 3*t² => t = 2020/3, which is not a natural number, so no solution here
------ 1009*t + 2*t = 3*t² => 1011*t = 3*t² => t = 373
We replace and get the second solution:
SOL3:
a = 1009*373
b = 373
------ 1009*t + 1009*t = 3*1009*t²/2 => 4*1009*t = 3*1009*t² => t = 4/3, which is not a natural number, so no solution here
------ 1009*t + 2018*t = 3*1009*t² => 3*1009*t = 3*1009*t² => t = 1
We replace and get the second solution:
SOL4:
a = 1009
b = 2018
This is simply SOL2 swapped
So we have 6 solutions, SOL1, SOL2 and SOL3 and their swapped counterparts.
There are infinite solutions
i ain’t reading allat
@@sharmola bro graduated from walmart
Respect for the effort to type all that😎
@@trevoraustinjr It was no chore, but a pleasure! 😎
Just fascinating, I have no words to describe the complexity and elegance of this test and its proofs!
Hey this actually helped me solve a related problem on project Euler (#157) that I was struggling with! I love your videos man :)
Alternative solution: Rewrite the equation as 1/C*(A+B)/AB=3/2018 with A and B coprime (the original variables correspond to a=AC, b=BC, C=gcd(a,b) ). Then A+B and AB are coprime, so AB divides 2018 and 3 divides A+B. Conversely for every such pair we can find a unique C making the equation hold.
This is the kind of channel I've been looking for and stumbled upoun by chance.
Keep it up by various IMO,Putnam etc.. Questions.
The answer looks more complicated than it is because of the way it is presented. A more natural way of approaching it would be to rewrite it as: 1/a = 3/2018 - 1/b. That is, 1/a = (3b-2018)/2018b. Similarly, 1/b = (3a-2018)/2018a. Multiply the two equations and you get: 1/(ab) = (3a-2018)(3b-2018)/2018^2(ab), That is: (3a-2018)(3b-2018) = 2018^2
Forgive me if I'm being ignorant but I see no improvement here. He actually got to that last equation you've given in about 3 lines which isn't too bad. So yea ... Struggling to see how your different ( but valid ofc ) method helps in anyway?
Yh this method is bit better if you can't see how to split it into a binomial form like in the video. This person just re-ordered things to get 2 equations and then make it into one equation to factor it. I can understand why he thinks it's more natural. I think it is as well.
what
Different approach but the same level of intricacy. Appreciate it though.
I will say his skipping of step did make things complex but what's a problem without a challenge. If he showed all the reasoning and steps it would have been no different than spoon feeding and then it would be called learning rather than understanding.
The American Mathematical Monthly publishes my solution for this problem, so seeing this gives me warm feelings 😂
congratulations! you must be really clever
@@alfred1920 it's a combination of talent and hard work.
Took me a while to catch up... thanks for taking me along on the trip!
My school
In class :
x + y = 2 ; x - y = 4
In Test :
x/4 + y/5 = 6 ; x/8 + y/9 = 3
In Exam :
What's the answer
@@devanshsavansha9061 y=0, x=24
double the second one and you get x/4+(2y)/9=6,
subtract x/4+y/5=6,
y(2/9-1/5)=0 therefore y=0
x/4=6, x=24
@@ClikcerProductions 👍
Just missed out the factor pair (2*1009)*(2*1009) but they aren't equivalent to 1 mod 3 anyway.
Yes I checked that too 😁
Yes I also noticed that
You can reduce the primes mod 3 getting -1, -1, 1 and 1. Then note you need both the -1s together to get a 1; aka, 4. Then the 1009s can be 0, 1 or 2 of them with the 4. Just a little less "list all possibilities and eliminate".
Amazing growth as a presenter in a year. Here is a bit to fast, more looking to notes etc. A year later its much improved, and current are just polishing.
Lovely, I didn't know such a channel existed, +1 subscriber
1/1009 + 1/2018 isn't an answer? I don't know any other solutions to that yet
Yes it is, he forgot a "0" in the third solution of b
how did u find it
@@limbslap1917 1/2018 + 2/2018 = 3/2018. :-)
Limbslap One can also state the problem as:
№ 1.1 - 1/𝒂 ⊕ 1/𝒃 = ³⁄₂₀₁₈ (original)
№ 2.1 - 𝒃 = 𝒌𝒂 … and sub in
№ 1.2 - 1/𝒂 ⊕ 1/𝒌𝒂 = ³⁄₂₀₁₈ and move the '𝒌' about
№ 1.3 - 𝒌/𝒌𝒂 ⊕ 1/𝒌𝒂 = ³⁄₂₀₁₈ … reduce
№ 1.4 - (𝒌 ⊕ 1) / 𝒌𝒂 = ³⁄₂₀₁₈ … rearrange
№ 1.5 - 2018( 𝒌 ⊕ 1 ) = 3 𝒌𝒂 … expand
№ 1.6 - 2018𝒌 + 2018 = 3 𝒌𝒂 … get the 𝒌's on same side
№ 1.7 - 2018 = 3 𝒌𝒂 - 2018 𝒌 … isolate 𝒌's
№ 1.8 - 2018 = 𝒌( 3𝒂 - 2018 ) … then rearrange again
№ 3.1 - 2018 / (3𝒂 - 2018) = 𝒌
For 3.1 to be positive, 3𝒂 must be greater than 2018
№ 4.1 - 3𝒂 > 2018
№ 4.2 - 𝒂 > 2018 ÷ 3
Factoring 2018 = ( 2 × 1009 ) - not terribly helpful, but the divised result is
№ 4.3 - 𝒂 > 672.66666…
Well, then how to find the trivial solution? … LOOK AT IT …
№ 5.1 - x/y = 𝒂/𝒃 … therefore
№ 5.2 - x = 𝒂 and y = 𝒃
So, abstracting № 1.4:
№ 5.3 - (𝒌 ⊕ 1) / 𝒌𝒂 = 3 ÷ 2018 then
№ 5.4 - (𝒌 ⊕ 1) = 3
№ 5.5 - 𝒌𝒂 = 2018
The only '𝒌' that satisfies this is '2'
№ 5.4 - (2 ⊕ 1) = 3 and
№ 5.5 - 2 × 1009 = 2018
Making (𝒌 = 2) and (𝒂 = 1009) the trivial integer factors, with
№ 5.6 - 𝒃 = 𝒌𝒂 = 2 × 1009 = 2018 so,
№ 5.7 - 1/𝒂 ⊕ 1/𝒃 = ³⁄₂₀₁₈ becomes
№ 5.8 - ¹⁄₁₀₀₉ ⊕ ¹⁄₂₀₁₈ = ³⁄₂₀₁₈
While I'm not competent to do the modulo arithmetic, nor the number-field theory completeness logic, a simple PERL program yielded:
𝒌 = -1009.00000, 𝒌𝒂 = -678048.00000 … result - 0.00149
𝒌 = 2018.00000, 𝒌𝒂 = 1358114.00000 … result - 0.00149
𝒌 = 504.50000, 𝒌𝒂 = 340033.00000 … result - 0.00149
𝒌 = 2.00000, 𝒌𝒂 = 2018.00000 … result - 0.00149
𝒌 = 0.50000, 𝒌𝒂 = 1009.00000 … result - 0.00149
𝒌 = 0.00198, 𝒌𝒂 = 674.00000 … result - 0.00149
All the known solutions. At least integer ones. There are WAY more non-integer solutions, for sure.
⋅-⋅-⋅ Just saying, ⋅-⋅-⋅
⋅-=≡ GoatGuy ✓ ≡=-⋅
Robert Lynch
This is just a simple hyper-parabolic equation which can be plotted on graph so the number of (a,b) pair is infinite.
A quick solution for integer values of a and b. The right hand side is 3/2018. This can be written as (2+1)/2018 which when expanded becomes (2/2018) + (1/2018), that is (1/1009) + (1/2018), which follows a = 1009 and b = 2018 or a = 2018 and b = 1009.
My thoughts exactly when I first looked at it
This is how I solved it, though with a few more steps. It helps to start with smaller numbers, like 1/4 + 1/2 = 3/4. This can mean that if 1/4 is a, and 1/2 is b, then a = 2018 and b = 1/2a, or 1009
Looks very similar to the parallel resistances rule lol. 1/((1/a)+(1/b)) = 2018/3 and for parallel resistances, a general rule is that if you have an n ohm resistor and a 2n ohm resistor, their combined resistance is 2n/3. In this case, you could set 2n = 2018 which gives you the 1009 and 2018 solution easily. Not very rigorous but a neat pattern!
Yup because after moving `a`s and `b`s to one side and constants to the other, you get (a+b)/(a·b) = 3/2018 which is pretty much a harmonic mean of `a` and `b`. This way you can find one possible solution by geometric methods. Replacement resistance of two resistors in parallel is also their harmonic mean, or you can think of it as conductances instead, to find out that the replacement conductance is the sum of the conductances of these two resistors (which makes perfect sense: the amount of current through both resistors is the sum of the currents through the first and the second resistor in parallel). Harmonic means also work for calculating focal lengths of lenses, so harmonic means are quite common pattern in nature ;) You find it pretty much everywhere where you work on inverses of some quantities that sum up together.
@@bonbonpony If 1/a +1/b = a/(ab)+b/(ab) = (a+b)/(ab)=3/2018. This will give you complex solutions with a=3/2 ± i√(8063)/2 and b = 3/2 ∓ i√(8063)/2. Of course the original question asks for all natural numbers, so complex solutions which are also negative and fractions don't count at all😁; still it was fun.
@@davidgriffin79 More like 1/a +1/b = b/(ab)+a/(ab). Also I'm not sure where did you get the complex solutions from. But considering that there's only one equation and two unknowns, infinitely many pairs of numbers would work, including your complex ones.
@@bonbonpony 1/a +1/b = b/(ab) + a/(ab), that is multiply 1/a by b/b and 1/b by a/a to get (a+b)/(ab). There are two equations if we assume numerators and denominators are identical; in such instance you have two equations which will give a and b as complex conjugates.
@@davidgriffin79 My first correction was about your mistake in getting those two fractions to common terms. This time you did it correctly.
As for your assumption about the numerators and denominators being pairwise equal, there's a catch: they can have a common factor, let's say `C`. Then you have `C·(a+b) = 3` and `C·(a·b) = 2018. So even if you solve for `a` and `b` the way you described, there's still some freedom coming from the factor `C` (as there _should_ be, since we have 2 unknowns, `a` and `b`, but only 1 equation).
In other words, for fractions, the equality holds not just for `2/3 = 2/3`, but also for `4/6 = 2/3`, or `6/9 = 2/3`, etc. So the assumption that numerators are equal and denominators are equal is flawed.
Wow, that also was a final problem of a High School contest at my home country.
Find a condition on two primes p
You solved it better than I had solved, so thumb up. Thank you.
Simon’s Favorite Factoring trick comes in handy so many times in these problems
The third solution on the blackboard should be a=2018 and b=1009, not 109. And that’s actually what one could easily see as solution without all the fuss.
1/2018 + 1/1009 =. 1/2018 + 2/2018 = 3/2018.
Great video. I would advise that you slow down a bit - not in what you cover, but just the speed of your voice. It comes off a bit breathless and nervous.
This is good content. I'm glad to see people making such detailed and informative videos about math.
He’s not saying it’s fast, he’s saying that the speech is not natural. Give yourself time to breath. I do the same thing and have to remind myself to slow down.
This channel needs more subscribers.
What told you to do the mod3? That seemed like hand waving to me.
I would assume that it’s because the equation was multiplying (which modulo can kinda relate to it ? Maybe) and there a,b were multiplied by 3 sooo...
I also have no idea
Well, you had "(3a - 2018)(3b-2018) = 2018*2018" and we know a and b need to be integers.
This means that you can assign (3a-2018) and (3b-2018) to any combination of numbers that form 2018*2018.
If you add 2018 to any combination, you need to be able to divide by 3, hence the "1 mod 3".
Dunno if that makes any sense but it does in my head
@@danielcantoreanu yeah you are 100% right
Using congruence classes or modulo is actually very common to solve diophantine equations (integer solutions) that requires matching factors. 3 is chosen since it eliminates the unknown a and b. Otherwise you would not be able to determine the congruence of each factor.
I tried all factor pairs, but some of them don't give integer solutions when you divide by 3. So that's a way to find the "valid" factor pairs without trial and error.
Btw, he forgot the factor pair 2018*2018, but this one also doesn't give an integer solution.
awesome, but did he explain why he decided to use mod arithmetic?
The ending was absolutely brutal 😂😂😂 love it
4:20 When looking at factor pairs, it's not mentioned in the video but 1009 = 1 mod 3, so obviously adding it or its square to either factor doesn't change whether or not the factor equals 1 mod 3. So since 2 = 2 mod 3 and 2^2 = 3 mod 3, you just have to consider the combinations involving either 2^2, 1009 and/or 1009^2 (which is ultimately what's derived in the video.)
2^2 =1 mod 3, not 3
IN MANY WEBSITES THERE ARE 6 ORDERED PAIRS, BUT THERE WOULD BE INFINITE ♾ SOLUTIONS IF WE WOULD TAKE THE VALUE OF (2018k AND 2018z)
IN DECIMALS TO MAKE THE SUM 3.
GOOD DAY🙂
wrong. 2 = 2 mod 3 and 2^2 is 1 mod 3. He was also wrong. (2mod)(2mod3)=4mod3=1mod3
@@zinzhao8231 Yeah, I apparently made a typo above, I meant to say 2^2 = 1 mod 3.
We can work also a bit differently. Let's assume that b>a (symmetry) so there is a positive integer r such that b=a+r. We plug this and we get a quadratic equation with respect to a with integer coefficients. So discriminant must be a perfect square for solution to exist from where we can do a nice factoring and get some easy cases. We also check for a=b and we get permutations too
why do you hide the background with kids cartoons? you don't have to :)
How would you have time to do some of this arithmetic on the exam?
These are big numbers
Why was 3 chosen as modulo? So that every multiple of and b would be valid?
why the mod 3 thing?
Whats the word for describing someone who speaks thourgh his nose?
Good job man! I think you should upload some Usamo solutions too. Keep it up!
Hi sir can u tell me which book best for theory how can approach to solve putnam problam and i have not experince in high school olympiad competition
Well, I paused the video and tried to work it out myself. I found two of the three pairs. Also, shouldn't the first pair be 1358114 and 673, rather than 1354114 and 673?
that's what i got too
Under what name can i learn more about that modulo thingie ?
I need to review mod3. Is that Capt Kirk's shirt?
How could 3a -2018 be 1 mod 3 ?
How
Am I missing an obvious reason why (2*1009)(2*1009) isn't a factor pair considered in that list? It doesn't affect the answer because 2*1009 = 2 mod 3, so that pair gets rejected anyway, but should it have been checked?
My question is, how do you find a and b at the end without a calculator
Why does wolfram alpha show 1/673 + 1/1354114 - 3/2018 = 1000/459760295249 instead of zero (my calculator shows the same thing)? Also, why does 2018*673 + 2018*1354114 not equal 3*673*1354114 (2018a+2018b=3ab (from about 0:56 in the video)?
So good that this is the easy problem from that year.
wouldn't the congruence be -1 modulo 3 or 2 modulo 3?
I couldn't figure it out, although it seemed to me factorization might be involved at some point. No wonder I didn't get accepted at MIT. :-)
Can you explain.... how 3a-2018 is 1 mod 3
Man, you are genius! Explained hard Putnam question easily
Heya, loved the video, but I spotted a little something that the other comments don't seem to have pointed out. I am sure it's a typo or something, but the b for solution pair is 1358114.
5:15 the original pair 2018·2018 is missing, but since 1009 = 1 (mod 3), all valid pairs must pack the 2's together anyway (must have one factor odd and the other divisible by 4)
Where did the 2018 squared come from?
Can you tell me about Putnam exam?? I am from India and I didn't know about it..please tell me more..and what type of numerical is this??
Hey michael, do you think you could explain the risch algorithm?
Wow, nice crisp presentation, thanks!
I didn't get why he was looking at congruence (mod 3) until he was halfway through listing off the factor pairs.
I’m struggling to understand the logic. I thought it was to make sure the factors would have an integer solution when plugged back into (3a-2018) but that’s not the case. If we have a prime number P which satisfies mod(p,3)=1 that does not mean mod(p+2018,3)=1. Try 31 for example. Can you explain the logic to me?
To quickly filter out the pairs that won't work. (3a-2018)(3b-2018) must have the same remainder as the right part.
It's a neat trick to save time solving several more equations, but it's not crucial.
Me either.
To solve each equation you have to add the number in each set of parentheses to 2018. Since 2018 is congruent to 2, mod 3, whatever we add to 2018 must be congruent to 1, mod 3, in order to total a number that is evenly divisible by three. That is why both sets of parentheses in his left-hand column both had to be congruent to 1, mod 3.
David Spain why did he choose 3 specifically
Your videos are great!
I miss the "And that's a good place to stop"
isn't factor pair (2*1009,2*1009) missing here even though it is not a solution?
08:40 i was waiting for that so bad i had the idea that you may not put them into consideration , Great job i did this with my students and i was surprised that 1 of 43 students did answer it like super shocked
Why must the factor pairs are 1 (mod 3)
Why is it that the second item in your factors pairs list was not congruent to one, mod 3? They all multiply to the same number...
Oh I see, each set of parentheses has to be congruent to one, mod 3.
Why is that important though?
Oh I see now. To solve each equation you have to add this number in parentheses to 2018. Since 2018 is congruent to 2, mod 3, whatever we add to 2018 must be congruent to 1, mod 3, in order to total a number that is evenly divisible by three.
I think what confused me was the fact that he said the entire left side was congruent to 1, mod 3, instead of each binomial separately. Seems to me it would have been more clear if he had emphasized the fact that each binomial was that way. And the reason that is the case is, if you have a number evenly divisible by three, and you subtract a number congruent to 2, mod 3, then you will get a number congruent to 1, mod 3.
Yes, for those of us for whom modulo arithmetic was not on our math syllabus that was confusing but your explanation makes sense. It is impressive that these problems require so many different techniques.
Why do we need to factor after noticing that each side is 1 mod 3?
i was following along, one of the pairs that fail to be both mod 1 but wasn't listed is (2*1009)(2*1009)
Thanks!
Can u make videos on iit jee advanced maths
What is the intuition for looking at factor pairs modulo 3? Why did you decide to take that step?
You don't really need that step at all. Just try all the factor pairs and you will find quickly that (2) (2 * 1009^2) doesn't result in integer solutions (a = 2020 / 3).
I think it was just a technique to pre-filter the cases to try.
825 and 3643 are the 2 closest non solutions I have seen. They work out to 2018.000224 ish. That is darn close. "Found" via computer programming with small "epsilon" (error allowance).
Great video, this question took me about 5 hours using a far more convoluted method and this really helps me learn how to do it more efficiently. Just one question: am I right in saying you missed out the factor pair (2•1009)×(2•1009)?
That wouldn't have worked anyway but just to be sure I understand the concept, would that have been a valid factor pair?
You are correct. That was valid and he forgot it, but luckily the factors are congruent to 2 mod 3, so they don't produce any solution
IN MANY WEBSITES THERE ARE 6 ORDERED PAIRS, BUT THERE WOULD BE INFINITE ♾ SOLUTIONS IF WE WOULD TAKE THE VALUE OF (2018k AND 2018z)
IN DECIMALS TO MAKE THE SUM 3.
GOOD DAY🙂
if you dont have to prove it you can just see that its two ones over a number and adds to 3/ a number so that a and b one can be equaled to the number and the other denominator divided by two to create 2 and 1+2=3
I used simple algebra and got only two solutions (1009,2018) and the swap of this pair and when I checked the end of the video all my conciet vanished. I will still count it as a win for me.
Sorry couldn't understand the problem because I am yet to study modulus. But still a big thanks as these videos have helped to understand numbers and patterns better.
what does 1 Mod 3 mean?
Can you solve by 3/2020?
Why not just do it with (a+b) = 0 mod 3, and a, b are factors of 2018?
I didn't get the factoring pairs step so far, as well as the 1mod3 u said :_((((
Lovely way of solution👍.Want more such videos
I do at least one Putnam solution each week! Here is a playlist:
ruclips.net/p/PL22w63XsKjqxM_TMRRhdASDi7ur9Thwcw
I’m Japanese, but I can enjoy your video.Thank you!
What does 1mod3 mean?
Can we write 2018^2 as (-1009^2)*(-4)?
finally, my egyptian fraction program i wrote for school has a use
This is in fact a very standard manipulation, known in the AoPS community as "Simon's favorite factoring trick".
What does AoPS mean?
@@_cytosine, it means "Art of Problem Solving"
I like Complete the product. It is more descriptive, but like AoPS a ton. Tremendous resource!
What does "one mod 3" mean?
For Solution Number 3 he said "1009" and wrote "109". Missing something?
When working in paper, how would you determine that 1009 is a prime?
You could try to divide it by all the primes up to sqrt(1009) and there aren't that many. But I suspect the presenter just has the first n primes memorized, for some value of n :)
Can somebody point me to a textbook that explains that trick with the mod?
What's the justification for the mod trick? Surely it's backed by a theorem, with these mod conditions as premises, concluding that the answers it gives are exhaustive. I have no idea what theorem proves that would be called, and I'd love to know.
I _think_ the reasoning goes like this: The equation at that point had two factors that multiply together to form 2018^2. He was just looking for those two factors to solve the problem. He used "they're both =1(mod 3)" to filter the search down a bit, since he knew the results would have to satisfy that.
I am new to the channel , And I couldn't Understand the step where he reduced the expression in terms of mod3.
Can anyone explain ? :)
He did that to show that both 3a-2018 and 3b-2018 had to be 1 mod three. This allowed him to eliminate some of the factor pairs as he did in the next step. 1 mod 3 just means that you get a remainder of 1 when you divide that number by three, ie 3a-2018=-2018=3(-673)+1=1 mod 3.
Hope this helps!
that was fun to watch. thanks
2018 is 2 mod 3 , isn't it?
can you do more trigonometry
Guys, I saw the thumbnail and I immediately thought it would just be a=2018 and b=1009=(2018/2) so that when you add the numerators become (2+1)/2018?
I think that the set of all natural numbers is the restriction. but still.
"So that's the end of the video"
Whaaaaaat?!?!
If I don't hear "and that's a good place to stop", my brain doesn't know it's time to shut down.
i thought the solutions were 1009 and 2018, clearly there were more!
Shiva Ranganathan bruh they legit said to list ALL solutions
@@raspberryb1664 I knew there were more, I kinda worded that badly. If I could rephrase that I'd say I only found 2 solutions
@@raspberryb1664 there are infinite solutions to this equation
@@sharmola no
@@sharmola Not over the natural numbers.
What does the mod bit mean? Never got taught it at school and I'm now at uni lol
If n is congruent to a mod b, it means that the remainder of n/b is the same as the remainder of a/b. Usually a is picked between 0 and b-1 because it's more practical. In that case, the remainder of n/b is exactly a.
what’s mod3?
what is 1 mod 3?