Putnam Exam | 2018: A1

[@cornucopiahouse4204]

Wished my dad had sent me to this kindergarten...

[@alexdixon265]

Zu Yao Teoh Lmao

[@dieharddougie]

why do you want this guy to touch you during nap time?

[@TDSONLINEMATHS]

check out on my channel for more maths problem

[@tashikrtv6878]

@@dieharddougie who wouldn't?

[@shadole1867]

I prefer a happy ending with my juice box.

[@coleabrahams9331]

This guy is a world champion at looking at solutions and presenting them like anyone can

[@skivvytv6229]

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

[@murtithinker7660]

lol

[@sanjusingh92443]

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🙂

[@beyondhumanrange6196]

@@sanjusingh92443 You .. can't they are natural ..

[@mark9463]

@henk7747 ur mom is easy

[@cycklist]

Putnam problems are always fascinating. Thanks for sharing.

[@MichaelPennMath]

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!

[@leif1075]

@@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..

[@rakeshbothra639]

@@MichaelPennMath kzbin.info/www/bejne/inzQkKaAZpl-iM0

[@greggould2457]

@leahcim38 hahah very underrated comment

[@tomatrix7525]

@@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.

[@coolmodad]

Dame he lost me at "ok"

[@Kozen001]

My Calc 2 teacher in a nutshell

[@coolmodad]

@@Kozen001 why is he or she in a nutshell ...

[@Kozen001]

@@coolmodad he says “ok” during lectures every 30 seconds lol

[@Videogamewrestling22]

damn*

[@lukyboi_4450]

@@coolmodad stuck

[@MedHamaidia]

Note: for a = 673, we find b = 1358114. Thank you so much for the great work you do.

[@speedsterh]

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 !

[@Ideophagous]

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.

[@sharmola]

There are infinite solutions

[@slimbean4272]

i ain’t reading allat

[@YoungPhysicistsClub1729]

@@sharmola bro graduated from walmart

[@trevoraustinjr]

Respect for the effort to type all that😎

[@Ideophagous]

@@trevoraustinjr It was no chore, but a pleasure! 😎

[@paulkohl9267]

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!

[@rcnayak_58]

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.

[@ethanf8059]

My thoughts exactly when I first looked at it

[@GoldenBoi507]

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

[@ChronusZed]

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.

[@Bodyknock]

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.)

[@reeeeeplease1178]

2^2 =1 mod 3, not 3

[@sanjusingh92443]

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🙂

[@zinzhao8231]

wrong. 2 = 2 mod 3 and 2^2 is 1 mod 3. He was also wrong. (2mod)(2mod3)=4mod3=1mod3

[@Bodyknock]

@@zinzhao8231 Yeah, I apparently made a typo above, I meant to say 2^2 = 1 mod 3.

[@jcnot9712]

8:15 b is supposed to be 1009. Just a typo since you said it right tho.

[@chucktodd7329]

Yea, I though I had found an additional solution. the easiest one!.

[@wigidyv-cee5102]

@@chucktodd7329 yeah bro this aint special tbh

[@SuperShank76]

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

[@kelzangtobgyel3887]

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?

[@siamsama2983]

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.

[@ryantk343]

what

[@fahad-xi-a8260]

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.

[@elenekarangozishvili1194]

The American Mathematical Monthly publishes my solution for this problem, so seeing this gives me warm feelings 😂

[@alfred1920]

congratulations! you must be really clever

[@dontfeelcold]

@@alfred1920 it's a combination of talent and hard work.

[@sergiogiudici6976]

Find a condition on two primes p

[@duncanfisher4346]

Hey this actually helped me solve a related problem on project Euler (#157) that I was struggling with! I love your videos man :)

[@Daniel-nl3ug]

Just missed out the factor pair (2*1009)*(2*1009) but they aren't equivalent to 1 mod 3 anyway.

[@mcwulf25]

Yes I checked that too 😁

[@vgtcross]

Yes I also noticed that

[@adamnevraumont4027]

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".

[@labelus2569]

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!

[@bonbonpony]

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.

[@davidgriffin79]

@@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.

[@bonbonpony]

@@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.

[@davidgriffin79]

​@@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.

[@bonbonpony]

@@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.

[@RRV01]

My school In class : x + y = 2 ; x - y = 4 In Test : x/4 + y/5 = 6 ; x/8 + y/9 = 3 In Exam :

[@devanshsavansha9061]

What's the answer

[@ClikcerProductions]

@@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

[@devanshsavansha9061]

@@ClikcerProductions 👍

[@Danieldsamaral]

Just fascinating, I have no words to describe the complexity and elegance of this test and its proofs!

[@prajwaldeepkamble6617]

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.

[@reinerwilhelms-tricarico344]

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.

[@bobsonny]

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.

[@stephenbeck7222]

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.

[@jerzybaranowski]

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.

[@joshuachristiangud3327]

1/1009 + 1/2018 isn't an answer? I don't know any other solutions to that yet

[@allenhuang3454]

Yes it is, he forgot a "0" in the third solution of b

[@limbslap1917]

how did u find it

[@carl13579]

@@limbslap1917 1/2018 + 2/2018 = 3/2018. :-)

[@robertlynch7520]

​ 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 ✓ ≡=-⋅

[@herrinecus]

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.

[@MichaelPennMath]

I am so excited this video is getting lots of views! I have a playlist devoted to solutions to Putnam problems and try and add about one new Putnam video per week: kzbin.info/aero/PL22w63XsKjqxM_TMRRhdASDi7ur9Thwcw Also, this was my first Putnam video ever and the more recent Putnam videos are much better! I also have videos to support full courses in Elementary Number Theory, Abstract Algebra, Multivariable Calculus, and Differential Equations. Take a look and subscribe if you see something you like!

[@Deeds_of_Love]

I did it in a different way, albeit a bit more convoluted: 1/a + 1/b = (a+b)/ab, then a+b=3k, and ab=2018k, k€N. Eliminate b and solve for a and you get a solution with a root - but the term under the root must be a perfect square for k to be natural, let's call it n^2, n€N. Then solve for k and you get a solution with a root of the term (9n^2+4036^2). This once again must be a perfect square. Equate it to (3n+d)^2 (d€N) and solve for n: n = 2*2*2*1009*1009/3d - d/6 You can see that for n to become natural, d must divide the numerator*2. There's 15 possible factors of 2*2*2*2*1009*1009, many of which can be eliminated right away because n would be negative. Try out the others and you get the same 3 solutions.

[@richardrobertson1886]

What told you to do the mod3? That seemed like hand waving to me.

[@zForce4]

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

[@danielcantoreanu]

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

[@ayoubsaleh2031]

@@danielcantoreanu yeah you are 100% right

[@DrTaunu]

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.

[@juttagut3695]

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.

[@fahad-xi-a8260]

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.

[@kenhaley4]

Error at 6:32: b should be 1358114. And at 7:55: b should be 1009 (which was said, but 109 was written on the board).

[@GiannisMasterio]

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

[@conanichigawa]

This channel needs more subscribers.

[@Arfonfree]

Took me a while to catch up... thanks for taking me along on the trip!

[@HeraldoS2]

Wow, that also was a final problem of a High School contest at my home country.

[@pranav7471]

Lovely, I didn't know such a channel existed, +1 subscriber

[@DancingRain]

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?

[@justinvillafuerte8738]

that's what i got too

[@JonathonV]

I didn't get why he was looking at congruence (mod 3) until he was halfway through listing off the factor pairs.

[@drdale104]

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?

[@AlexBesogonov]

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.

[@mathissupereasy]

Me either.

[@davspa6]

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.

[@mzzzchael]

David Spain why did he choose 3 specifically

[@SteveMathematician-th3co]

Man, you are genius! Explained hard Putnam question easily

[@ChiranjibPatra]

I have an intutive solution to this problem like 3 on RHS can be split into 2 +1, (1+2) no other combination exist as whole number so we can write 1/2018 +2/2018 = 3/2018 or 2/2018 + 1/2018 i.e the values of a , b can be swapped , so by comparing with the given equation 1/a + 1/b =3/2018 so a=2018 and b=1009 .. we can extended this solution considering decimal sets as well.. Cheers!!!

[@josephshort1908]

Simon’s Favorite Factoring trick comes in handy so many times in these problems

[@cmilkau]

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)

[@samueldeandrade8535]

6:26 corrections: 1354114 should be 1358114, so the fourth or middle digit is 8, not 4 the last one, a=2018 and b should be b=1009. This is the "trivial solution".

[@davspa6]

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.

[@f12mnb]

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.

[@ShivaBrass]

i thought the solutions were 1009 and 2018, clearly there were more!

[@raspberryb1664]

Shiva Ranganathan bruh they legit said to list ALL solutions

[@ShivaBrass]

@@raspberryb1664 I knew there were more, I kinda worded that badly. If I could rephrase that I'd say I only found 2 solutions

[@sharmola]

@@raspberryb1664 there are infinite solutions to this equation

[@matko8038]

@@sharmola no

[@AndyZach]

@@sharmola Not over the natural numbers.

[@Walczyk]

i was following along, one of the pairs that fail to be both mod 1 but wasn't listed is (2*1009)(2*1009)

[@Icenri]

Thanks!

[@MyOneFiftiethOfADollar]

No need for the modular arithmetic. Complete the product(They call it SFFT in Maryland) to get 3a-2018 = 2018^2 and b - 2018/3 = 1/3. This gives a=1,358,114 and b=673. Similar for other two solutions. Think you may have typoed the 1,358,114 in video.

[@andreypetukhov4250]

I see no need for any modules here, it’s enough to check five variants when you’re at the step where you have the two brackets, because (3a-2018)(3b-2018)=2018^2, but the task is brilliant, thank you

[@superneenjaa718]

2018 having prime factors 2 and 1009. Thus, the only ways to get 2018 as LCM of two natural numbers (1, 2018), (2, 2018), (2, 1009), (1009, 2018). To get x/2018 as a sum of two numbers of the given form, LCM of a,b have to be 2018. Only the last pair mentioned before gives us 3/2018. So (a,b)= (1009, 2018). This way you can solve it completely mentally, as I did.

[@Ensivion]

I didn't come up with a solution but doing the factoring then using modular arithmetic was really cool, I had come up with two equations, two unknowns and a free parameter, k because you could multiply the right hand equation by k/k and that would be a solution. I was going to subtract the equations getting a+b = 3k, ab = 2018k => ab - a - b = 2015k => (a-1)(b-1) - 1= 2015k I knew that you could use modular arithmetic to deal with this free parameter but couldn't get it to work.

[@km-pw2fi]

I did the exact same thing you did. I even named the third variable k too... Got stuck as well :)

[@km-pw2fi]

I did the exact same thing you did. I even named the third variable k too... Got stuck as well :)

[@jjcadman]

"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.

[@lasalleman]

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. :-)

[@FajorMuckup]

why the mod 3 thing?

[@gregy1570]

awesome, but did he explain why he decided to use mod arithmetic?

[@AdityaKumar-ij5ok]

why do you hide the background with kids cartoons? you don't have to :)

[@Mateusz-Maciejewski]

You solved it better than I had solved, so thumb up. Thank you.

[@davidjames1684]

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).

[@elrichardo1337]

This is in fact a very standard manipulation, known in the AoPS community as "Simon's favorite factoring trick".

[@_cytosine]

What does AoPS mean?

[@luana_coelho]

@@_cytosine, it means "Art of Problem Solving"

[@MyOneFiftiethOfADollar]

I like Complete the product. It is more descriptive, but like AoPS a ton. Tremendous resource!

[@Deristrome]

My question is, how do you find a and b at the end without a calculator

[@heldercomp]

Set gcd(A,B) = 1 and a = gcd(a,b).A b = gcd(a,b).B Then (A+B)/{gcd(a,b).AB} = 3/2018 It’s easy to prove that: 2018 = 2x1009 and 1009 is prime (A+B)/AB is irreducible Then clearly AB = 1, 2, 1009 or 2018 Since 3 divides A+B, only possibilities are: AB = 2 AB = 2018 Just try the possibilities and the solutions will be: (a, b) (1009, 2018) (673, 1.358.114) (674, 340.033) and all opposite pairs (b, a), of course.

[@Amk-ks9mh]

rewrite the equation as 2 · 1009(a + b) = 3ab, and note that 1009 is prime, so at least one of a and b must be divisible by 1009. If both a and b are divisible by 1009, say with a = 1009q, b = 1009r, then we have 2(q + r) = 3qr. But qr≥q+rforintegersq,r≥2,soatleastoneofq,ris1. Thisleadstothesolutions q = 1,r = 2 and r = 1,q = 2, corresponding to the ordered pairs (a,b) = (1009,2018) and (a, b) = (2018, 1009). In the remaining case, just one of a and b is divisible by 1009, say a = 1009q. This yields 2 · 1009(1009q + b) = 3 · 1009qb, which can be rewritten as 2 · 1009q = (3q − 2)b. Because the prime 1009 does not divide b, it must divide 3q − 2; say3q−2=1009k. Then1009k+2=3·336k+k+2isdivisibleby3,so k ≡ 1 (mod 3). For k = 1, we get q = 337, a = 1009 · 337, b = 2q = 674. For k = 4, we get q = 1346, a = 1009 · 1346, b = q/2 = 673. We now show there is no solution with k > 4. Assuming there is one, the corresponding value of q is greater than 1346, and so the corresponding b= 2q ·1009 3q − 2 s less than 673. Because b is an integer, it follows that b ≤ 672, which implies 1≥ 1 > 3 , contradicting 1+1= 3 . b 672 2018 a b 2018 Finally, along with the two ordered pairs (a, b) for which a is divisible by 1009 and b is not, we get two more ordered pairs by interchanging a and b.

[@omintionpg3d652]

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

[@MarceloGondaStangler]

U can say that a+b=3=-B, ab=2018=C, and x²-3x+2018 =0, and solve from Bhaskara

[@limbslap1917]

lmao i did the same thing but i didn't get a final answer just turning around what did u do after it?

[@marylucchampel]

That woud not work. It would only give you one of the three solution... All you can say is that a+b=3k and ab=2018k but you are going nowhere if you follow this route. You need to use congruences...

[@sharmola]

@@marylucchampel there are infinite solutions...

[@sharmola]

@@limbslap1917 since delta is negative you need to multiply the square root of delta with i and that's all

[@coleabrahams9331]

1:44 Apart from looking at the solution, why did you multiply by 3 and not 2 for example?

[@Flaitastic]

If I understood correctly, it's because 3*3 = 9, which is a perfect square AND a multiple of 3, therefore making the equation symmetric by having both sides be almost equal (being the letter the only difference). Having the equation be symmetric usually helps a lot to solve this kind of problems

[@呂永志]

7：55 obviously, b should be 1009, not 109.

[@srmpenedo]

呂永志 just a miswriting, as he said 1009.

[@icespirit]

7:55

[@OMGclueless]

When working in paper, how would you determine that 1009 is a prime?

[@amtep]

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 :)

[@shubhankarbansod8437]

Can you explain.... how 3a-2018 is 1 mod 3

[@vintageempirical5950]

Putnam is like 🔥🔥🔥. Difficult to even move, such is our Jermaine and Adv. For us but they r not difficult like Putnam but difficult for us,at the time of giving papers we find that sum very difficult to do,and after giving out the paper or after the Ques were told it looks easy,same is with you it's easy when it's out......at the time of its released day it was like......🔥🔥🔥🔥🔥a hell to move.......

[@rushyendranathreddynalamal2030]

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.

[@dtkstupid12345]

How could 3a -2018 be 1 mod 3 ?

[@samoiddintemirov7866]

How

[@iain5615]

Given that 1 + 1 = 3 on the nominator line then either a or b is half the value of the other. As such if a = 2018, then b = 1009 or vice versa. There are in reality many possible solutions.

[@iooooooo1]

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?

[@youssefelkhou9310]

talking about Modulo 3 is a big jump for students to be honnest

[@1337vIKz]

I was able to follow quite well until he started talking about modulo and I completely lost him. We never really talk or use about modulo on our courses though :/

[@youssefelkhou9310]

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

[@billcosby5524]

wouldn't the congruence be -1 modulo 3 or 2 modulo 3?

[@crazyvideos235]

Can u make videos on iit jee advanced maths

[@randomcubing7106]

I miss the "And that's a good place to stop"

[@bilbag911]

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)?

[@michaelashcraft8569]

Thank God I have Math Block, never used anything past add, or, subtract, ever, and I'm 70ys old!!!

[@opsrpt]

The ending was absolutely brutal 😂😂😂 love it

[@JDRudie]

How could he tell so fast that everything was congruent to 1 mod 3 in that equation?

[@matteopriotto5131]

3a and 3b are of course congruent to 0. 2018 is congruent to 2, therefore -2018 is congruent to 1. Finally, all perfect squares that aren't congruent to 0 are congruent to 1 mod 3 because if n is congruent to 1, n^2 is congruent to 1*1=1 and if n is congruent to 2, n^2 is congruent to 2*2=4, which is also congruent to 1. You can get quite fast calculating this stuff once you get used to it.

[@amtep]

A trick specifically for mod 3 is that you can add up all the digits and take that result mod 3. So in 2018, 1 + 8 is 9 so you can ignore it, leaving the 2.

[@kamaljain5228]

it is more elementary than this. assume a, b has a common factor of r, a = rA, and b = rB, the equation simplifies to (A+B)/AB = 3r/2018. It gives 2 easy conclusions. A+B is a multiple of 3, and AB is a divisor of 2018. Without loss of generality assume, A a = 673, b = 1358114 if A = 2, B = 1009, r = 337 => a = 674, b = 340033

[@はにわ-r7p]

I’m Japanese, but I can enjoy your video.Thank you!

[@yusefiguess]

There is something I don't understand about this problem. We have 1/a + 1/b = 3/2018 That means that: (a + b) / ab = 3 / 2018 Can't we just solve the system a + b = 3 ; ab = 2018 ? This system has no real solution, therefore no integer solution.

[@MichaelPennMath]

Fractions can be equal while having unequal parts. Ex: 1/2=2/4. You are right that we can't solve a+b=3 and ab=2018, but a+b/ab can reduce to 3/2018.

[@wintersfan]

You Can't Do Basic Number Theory Lol Why Are U Even Here

[@MichaelPennMath]

@@wintersfan Let's try and be nice. Every question adds something important to the discussion.

[@yusefiguess]

@@wintersfan I never said that I didn't think of another solution that actually worked. I just had this little doubt in the back of my mind

[@aitihyamandal9039]

@@yusefiguess you cant do that as there would be a common constant

[@currator4861]

Why must the factor pairs are 1 (mod 3)

[@littlemithsunshine]

you could just change 3/2018 to 1/2018 + 2/2018. then 2/2018 simplified to 1/1009, and you get a and b...

[@andrewedson7010]

Small mistake (3a - 2018) is equivalent to 2 (mod 3) and (3b - 2018) is equivalent to 2 (mod 3) but when you multiply the two the product will be equivalent to 1 (mod 3)

[@matrixate]

I just looked at it and knew that either a or b had to be half the other. This is kind of funny that more than 3 seconds was needed to solve this. 1/2018 + 2/2018 = 3/2018 but simplify the second fraction to 1/1009. Not sure who wouldn't go about it this way.

[@AzureAsura]

it's easy to find the first solution, as you said it takes three seconds. What is difficult is finding all the solutions, and moreover, proving that those are the only ones and that there are no more.

[@ayties5113]

For Solution Number 3 he said "1009" and wrote "109". Missing something?

[@cyberpunk_edgerunners]

2018 is 2 mod 3 , isn't it?

[@danielrhouck]

You missed one of the factor pairs for 2²×1009²: (2×1009)(2×1009). Fortunately the factors arenʼt both congruent to 1 mod 3.

[@ahmedaghadi8281]

What do mean by 1 mod 3?

[@yeojunnam]

it means that if you divide the number with 3, the remainder is 1

[@ahmedaghadi8281]

@@yeojunnam Thanks! That comment is 3 years old and now I already know this. 😂 Your reply brought me to my 3 years old comment.

[@yeojunnam]

@@ahmedaghadi8281 Haha! I didn't expect to get a reply this fast. I learned this a just few weeks ago in school, so I just wanted to share!

[@ahmedaghadi8281]

@@yeojunnam That's great! All the best dude!

[@yeojunnam]

@@ahmedaghadi8281 You too!

[@vaibhavtulsyan7276]

What is the intuition for looking at factor pairs modulo 3? Why did you decide to take that step?

[@fizzley]

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.

[@db0182]

Could somebody explain what is “mod 3” please 😀

[@uzytkownik9616]

If you have a≡b(mod n) it means, that (a - b) is divisible by n, where n is natural and a,b are integers.

[@himeshviews7622]

@@uzytkownik9616 thanks

[@jjcadman]

Mod is short for modulo. It's a type of arithmetic that looks at the remainder, after dividing one integer by another. The answer provided above is correct, but I find it easier to think of it as "remainder math": 6 mod 2 = 0 because when you divide 6 by 2, the remainder is 0. 7 mod 2 is 1 (7 ÷ 2 = 3 with remainder of 1). Quite often in problems, we only care about remainders. Modulo arithmetic is often considered to be a representation of how many revolutions of a particular size you can make, and it helps determine if you come back to where you started, or -- if not -- how far through the next revolution you are.

[@detectivejonesw]

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?

[@matteopriotto5131]

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

[@sanjusingh92443]

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🙂

[@30CurryShivank]

I refuse to believe that a Putnam question uses Simon’s favorite factoring trick. Mathcounts state this year and AIME had harder questions than this☠️

[@orelyosef5060]

Whats the word for describing someone who speaks thourgh his nose?

[@coppertones7093]

finally, my egyptian fraction program i wrote for school has a use

[@bryanbazilauskas8673]

Why can’t you set a+b=3 and ab=2018 and then graph and look for an intersection? It doesn’t work but I do not understand why.

[@cheva1]

how can you tell if something is congruent to 1mod3

[@discrete3511]

say that a number (let’s call it a) is congruent to 1 mod 3 , it means u can add a multiple of 3 to get from the number a to one(and vice-versa). For example, 5 is congruent to 1 mod 3, because -5+ (2*3) = 1. Notice we were able to go from -5 to 1 by adding 3 a number of times ( in our case, we had to add 3 two times)

[@pushpendrajaiswal3117]

I only got the last value...that was easy btw...and I faded the way he did them. Hats off. But I have a doubt and that is This question involves 2 variables but in one equation so probably there's supposed to be the infinite solutions of a and b...? Does this mean that out of all the infinity numbers that we have . there's only these 3 pairs of natural numbers that will go with this equation and rest aren't the natural ones..craaaazzyy