Can you solve this fraction question from China?

This question is from the Hua Luogeng math competition in China. I was told it was for elementary school students (ages 10-12). There are many interesting ways to solve this question! Special thanks this month to: Daniel Lewis, Kyle, Lee Redden, Mike Robertson. Thanks to all supporters on Patreon! / mindyourdecisions
0:00 problem
1:05 method 1
5:36 method 2
7:22 method 3
9:48 method 4
Joshua Brown’s blog post (method 4)
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[@akiraincorrect4968]

I simply subtracted 1, inverted the fractions and it was clear that a is less than 2, so it's 1 and so on. Very similar to method 2.

[@DrJohn123]

That's exactly my method, too! I thing it's even simpler or at least more straightforward than method 2.

[@lorenzosotroppofigo1641]

​@@DrJohn123 It's exactly the second method. You still need to do the math or you won't know what B and C are if you don't have the exact result. I did it the same way btw

[@lack7773]

I am also joining this club 🤗

[@VulcanDoodie]

yeah same club, I leared that in a video of numberphile where they showed wich real number is the least possible to approximate with a continuous fraction. spoiler alert it is Phi

[@thefireyphoenix]

hey same

[@MichaelPiz]

Man oh man! I got something in base 13 with a π, two _i's,_ and an _e._ There was also a tetrahedron and a short, 43-year-old Peruvian named Paco.

[@frax5051]

Lmao.

[@tommyb6611]

ah yes, the multi spacial transwarp solution. You've discovered how to time travel by applying this result of yours in wolframalpha

[@adarshbagri7531]

Fking legend 😅

[@joeschmo622]

I just did it the easy way. All positives, so no negatives. 17/10 = 1 + blah, so blah = 7/10. 1 / blah = 7/10, so blah = 10/7. a + blah = 10/7, so a=1 and blah = 3/7. 1/blah = 3/7, so blah = 7/3. b + blah = 7/3, so b=2 and blah = 1/3. 1/blah = 1/c = 1/3, so blah = 3 and c=3. QED.

[@ryotoiii7945]

Love the blahs, 10/10

[@savitatawade2403]

but you forgot to take the limit of blah as blah approaches 17/10

[@defined.9629]

I don't understand the second line. Why do we do 1 / blah = 7/10?

[@joeschmo622]

@@defined.9629 You want to find a.

[@savitatawade2403]

@@defined.9629 did you see the question?

[@LarzB]

In the first solution you go too far. At the point of 7/10 = (bc+1)/(abc+a+c), you just start with bc+1=7. so bc=6. Substitute this in the denominator equation makes 6a+a+c=10. 7a+c=10. a and c are positive and you clearly see that a=1 and c=3 to make 10. with c=3 you know that b=2. So no cross multiplication.

[@Canadian_Teemo]

Literally the first result when you search for the competition online: " The competition, which began in 1986, is considered one of the toughest math competitions for primary and junior high school students in mainland China"

[@abaddon6078]

lol this channel exaggerates china's avg math level so much

[@justingolden21]

​@@abaddon6078yeah. I figure it's probably the Internet that exaggerates it and has these problems on viral posts, and then he finds the problems and thinks "this'll be a good video" so I don't think it's really on him

[@hassanalihusseini1717]

@@abaddon6078 Yes, I never met a Chinese who knew maths really. Much exaggerated. Same with India.

[@abaddon6078]

​@@hassanalihusseini1717 no one asked for your sarcasm but ok

[@joinfortherapy6470]

I know only 1 Chinese that is bad at math and it's the same guy who failed his 12 yr finals

[@miche027]

after i saw the problem i instantly solved it by method #2 and of course thought about method #1, but other two are really interesting

[@winklethrall2636]

I used method 2 and pretty much solved it in my head before you started going over method 1. I'm going to assume that how the students were expected to solve it, avoiding any algebra.

[@shadowofheaven3279]

Your comment got copied

[@gmar89541]

yeah i did maths competitions and there's generally a tight time limit, so finding the efficient way to do it is actually really important. We couldn't have more than one person per year on a team (so you couldn't weight it with all older kids), but it actually helped. Often the younger kids would think of the faster methods bc they hadn't learned all the higher level maths that would make the longer methods possible

[@angeluomo]

I used method #2. The first method was crazy complicated. I didn't even think of the 3rd and 4th methods.

[@RajveerSingh-vf7pr]

2:57 In the long method, you don't need to cross multiply, that way, way less steps, and not guessing the value of A to be 1,and not checking the other possibilities of a, b and c... At 2:57 the numerator (bc+1) can be safely assumed to be 7 Hence we can substitute the same in the denominator which is 10... After that, it's a piece of cake

[@gregoryknapen9133]

The reason is that bc+1 and a(bc+1)+c have no common factors and since 7/10 is in reduced form, we can conclude that bc+1=7 and a(bc+1)+c=10.

[@JdeBP]

Now what we need, I think, is a video from Prime Newtons explaining _why_ the Euclidian division and division into squares methods work.

[@yurenchu]

The Euclidean division method (method 3) is essentially the same as method 2 (subtract the largest whole number from the fraction, which means subtracting the denominator from the numerator a whole number of times; then "flip" the remaining fraction (upside-down), and repeat). The division into squares method (method 4) is essentially the same as the Euclidean division method (from the larger side of the rectangle, subtract the shorter side a whole number of times; then rotate the remaining smaller rectangle by 90 degrees, and repeat). Euclidean method: 17 - 10 * *1* = 7 10 - 7 * *1* = 3 7 - 3 * *2* = 1 3 - 1 * *3* = 0 ==> the coefficients of the continued fraction representation of 17/10 are *1* , *1* , *2* , *3* .

[@RGP_Maths]

Yes, but the point is that neither method 3 nor 4 solved the problem. He just did Euclid's algorithm numerically, and then diagrammatically, and then said "Hey look, these numbers are the same as we got when we solved the problem previously." That is not a solution! There was no attempt to link these calculations back to the original question, nor to demonstrate that the values of a, b and c had not appeared merely by coincidence, nor even that they are a unique solution set. Of course, this could be done, but that would make method 3 or 4 far longer than shown here. Method 2 is the only sensible way to solve this question.

[@frendlyleaf6187]

My favourite method still has to be the finite continued fraction one, the problem is a nice introduction to them, and I feel that most students would probably use that method as it is pretty straightforward. If anyone is curious about continued fractions then there are some good Mathologer videos on the topic, discussing the infinite continued fraction form of famous constants and functions. Ramanujan also did a lot of famous work on infinite continued fractions which is pretty cool.

[@elpme16]

Algebra will arrive at the equation 10/7 = a +(1/(b+1/c)). a can't be 0, and if a > 1 then we have 1/(b+1/c) is negative, which is impossible because b and c are positive. Thus a = 1. We can simplify further to b+1/c = 7/3. b can't be 0, if b > 2 then we have 1/c is negative, which is impossible because c is positive. b can't be 1 because that implies that c is not an integer. Thus b = 2 which implies c = 3. The sum a + b + c = 6.

[@acarbonbasedlifeform70]

I got to that point 3:50 and got stuck there for the same reason as always - I must've been absent from math class at school when they taught us it was okay to "let's just say a = 1". I always thought that bruteforcing an equation was a big "no-no". Maybe that's why I struggled so much with maths...

[@Junieper]

you can formally prove that a must equal 1 by saying if a =>2 it’s too big

[@FritzTheCat_1030]

You can make an assumption like that as long as you can recognize and eliminate all other possible solutions to the problem.

[@Vertraic]

The full process would be to show that if a, b, and c are positive integers, then a MUST be 1, as the fractional part being added to it is less than one, and the total value is between 1 and 2. So you have an integer, plus some positive proper fraction is between 1 and 2, so the integer must be 1. Same process gets you b and c afterwards.

[@marcelomagallon]

I used to tell my students "guessing the solution is a valid method to solve problems, but you get 0 points if you guess wrong". That is a less formal way of saying given that you know that a, b and c are integers "what if a is equal to 1, can I get a consistent solution walking down that path?" It's the difference between "a blind guess" and "an educated guess".

[@issamzreik]

That's because it's a Diophantine equation (an equation you want to find integers solutions) so you might discover somthing if you tried special values

[@JqlGirl]

While methods 3 and 4 are great, I don't think it's obvious that they correlate to the original problem.

[@PC_Simo]

0:52 I just roll the continued fraction up from the bottom, by turning the mixed number: b + 1/c, into a full fraction: (bc+1)/c, invert it into: c/(bc+1), and repeat. In the end, I get: 17/10 = (ac+2)/(bc+1); thus ac + 2 = 17, and bc + 1 = 10. Assuming a, b, and c are positive integers, we get: ac = 17 - 2 = 15, and: bc = 10 - 1 = 9. Since ac = 15, and bc = 9 are both divisible by 3 (their only common factor) and c, c = 3. Then: a = 15/3 = 5, and: b = 9/3 = 3. So: (ac = 17 - 2 = 15) ∧ (bc = 10 - 1 = 9) -> (a = 5) ∧ (b = 3) ∧ (c = 3). *EDIT:* I probably messed up, with the actual calculations, trying to do it all in my head, without even any pen and paper; but the idea is right.

[@maxhagenauer24]

If you subtract the 1 over, you get 7/10 = 1 / (a + 1 / (b + 1 / c )) so you can multiply top and bottom by 7 so the denominators are equal so 10 = 7(a + 1 / (b + 1 / c ). And dividing 7 and noticing that 10/7 is 1 + 3/7 so a must be 1 and the rest is 3/7, and doing some of the same stuff to the rest, you can easily find b and c.

[@Patrik6920]

..its even easier ... Edit: added some steps for clarity .. [a, b and c are positive integers] consider 17/10 = 1 + ( 1 / ( a + 1 / ( b + 1 / c ) ) -> 1 + 7/10 = 1 + ( 1 / ( a + 1 / ( b + 1 / c ) ) -> 10/7 = a + 1 / ( b + 1 / c ) -> 7/7 + 3/7 = a + 1 / ( b + 1 / c ) give a = 7/7 = 1-> 6/3 + 1/3 = b + 1 / c gives b = 6/3 = 2 -> 1/3 = 1/c gives c = 3 Note: if m/n = 1/k -> n/m = k -> n=mk

[@maxhagenauer24]

@@Patrik6920 So, exactly what I did.

[@Patrik6920]

@@maxhagenauer24 Yes..

[@GurjeetSingh-Gur_Jeet_Ke]

​@@Patrik6920I too applied the same method and could answer in 37 seconds without using pen and paper.

[@Patrik6920]

@@GurjeetSingh-Gur_Jeet_Ke ..thats greate, but i like the teaching aspect .. hopefully someone learned something new .. and i got a bit better explaining..

[@erikaz1590]

when I did it before watching, I did method 2. Glad to see i was able to do it

[@JTGaffley]

1 + 2 + 3 is 6... I did method 2 mentally. It expects a large number understanding from young children, woah!

[@zekaizengin3904]

Extremely wonderful! thank you!

[@user-hu9rd7jk8v]

Really fantastic task. I am so impressed.

[@slothbearanonymous]

I did it slightly differently, following the first method until 2:57 only. I then equated the top and bottom of the fractions: 7=bc+1 10=a(bc+1)+c Subbing the first equation into the second: 10=7a+c If 7a

[@JoydeepDas-xe5zs]

I calculated the answer mentally.

[@aliasmask]

That's really interesting. Good way to solve how to fill a space with the minimun number of squares.

[@Darisiabgal7573]

17 / 10 can be written as 10/10 + 7/10 We can remove 1 simplifying the equation. If 1/u = 7/10 then u = 10/7 Therefor 10/7 = a + 1/(b + 1/c) 10/7 = 3/7 + 7/7 and since a must be positive integer a must be 1 since we cannot have the next integer 2 a = 1 3/7 = 1/(b+ 1/c) Therefore 7/3 = b+ 1/c which is 6/3 + 1/3 Thus b must be equal to 2 leaving 1/3 = 1/c Thus a + b + c = Sum(1,2,3) = 6

[@wyattstevens8574]

I was familiar with continued fractions, so it just jumped out at me- and I noticed that a=1, b=2, and c=3. 1, 2, 3- coincidence that this was for tweens? I think NOT!

[@zekt98arius]

Paused at the 0:52 mark to mental sums this: 17/10 = 1 + 1/(a+(1/(b+1/c))) Subtract 1 from both sides, 1/(a+(1/(b+1/c))) = 7/10 Inverse both fractions, a+(1/(b+1/c)) = 10/7 Since a+(1/(b+1/c)) b and c are positive integers, a < 2 --> a = 1 1/(b+1/c) = 3/7 Inverse this fraction equation, b + 1/c = 7/3 Either b = 1, 1/c = 4/3 and c = 3/4 (reject), Or b = 2, 1/c = 1/3 and c = 3 (solution)

[@theonlymegumegu]

loved the illumination in method 2!

[@joyneelrocks]

The easiest solution to develop would be something similar to the 2nd method. If you just look at: 10/7 = 1 + 3/7 = a + c/(bc + 1) Since a, b, c are positive integers, a = 1, and c/(bc + 1) = 3/7. It’s pretty obvious that c = 3 & b = 2 is solution. Therefore a + b + c = 1 + 2 + 3 = 6.

[@okaro6595]

2:45 Here one can see that b=2 and c=3 or vice versa. Then one can substitute the 7 in the denominator to get 7a + c = 10. Clearly a=1 and C=3.

[@mata2723]

Nice solutions !

[@dafyddthomas6897]

Positive Integers so I brute forced it: the Fraction must = 7/10 A = 1, B =1, C = 1, then F = 2/3 A = 1, B = 1, C =2, then F = 3/5 A = 1, B = 1, then F = (C+1)/(2C+1); 2C +1 can NEVER = 10 A = 1, B =2, then F = (2C +1)/(3C+1) = 7/10 for C=3

[@chasg5648]

I'd really like to see you expand on method 4, specifically why it is a valid approach and how someone would know to deploy it. And, how did the first person discover that method?

[@oliverfiedler8502]

It's somehow reassuring to know that even you haven't seen everything yet ;-) the last method works even "better" graphically, if you start in a corner and then draw lines at a 45 degree angle to the edges, the squares then appear automatically...

[@ashutosh452]

I did something similar to method 2. First, I subtracted 1 from both sides. Then, I inverted the LHS and the RHS. Now, I separated the LHS, 10/7, into 1 + 3/7. So, I got a as 1. Now, one more inversion. b+1/c = 7/3 --> (bc+1)/c = 7/3. c = 3, b = 2. Thus, a+b+c = 6.

[@AlienRelics]

I don't know why you said that method two involved a lot of calculations. I did it before watching. I solved it in something that was a variation of your second method. I did it all in my head without pencil or paper.

[@engineboy_1449]

Presh, I did my research and according to that this competition is for high schoolers of grade 10-12 i.e. if we talk about age group then it might be 15-18...I also cross checked with Chat-GPT...it says the same thing apparently

[@davidgillies620]

Just clear denominators to get (a b c + a + c)/(b c + 1) = 10/7 => b c = 6 => 7 a + c = 10 => a = 1, c = 3, b = 2.

[@ambika90s]

a=1, b=1 c=3/4 didn't take into account that all need to integers. In that case it will be 1, 2, 3.

[@experimentingalgorithm1546]

Nah method 2 was immediate click for me

[@kenbob1071]

Yup. I did it in my head using method 2 before clicking on the video. Easy.

[@gafjr]

Same here.

[@gropius6070]

Solution 28b (wherein some kids in Chinese elementary math classes might begin to squirm uncomfortably): Take the original equation, subtract 1, take the reciprocal, subtract a, multiply by 7. We now have • 10-7a = 7c/(bc+1) Since b and c are positive, the RHS is positive. Therefore the LHS is also positive, and 10-7a > 0 where a is a positive integer implies a = 1. Substitute a=1 back into our bullet point equation above, subtract 1, take the reciprocal, subtract b, and multiply by 3. We then have • 7-3b = 3/c Using a similar argument, the RHS is positive since c is positive. The LHS must be positive as well, and 7 - 3b > 0 where b is a positive integer implies b ∈ {1, 2}. If b = 1, 4 = 3/c is impossible if c is an integer. if b = 2, then c = 3. Therefore (a,b,c) = (1,2,3) is our (only) solution.

[@TaylorRen]

Hua Luogeng (1910/11/12-1985/06/12) was a famous Chinese mathematician.

[@4br4x4s9]

On the first solution, when you get 7/10 you can inverse it and you get a loose a, which must be 1 since something positive plus a positive integer is less than 2. Now you have 2 simple equations and 2 variables.

[@Austin101123]

I started by subtracting 1 from each side and then using reciprocals to find possible a, b, c in similar manner to version 2.

[@zzeroxxero]

I got the answer in less than 15 seconds. But then again, I'm not in elementary school. These kids are amazing.

[@pramodsingh7569]

Thanks

[@spookyaction3236]

I did method 5: Logic, because I was lazy. Minus 1 both sides, then times 10 both sides, then times by the right hand denominator both sides: 7a + (7 / (b + 1/c)) = 10 Since all numbers are positive, a = 1, or else the term 7a would more than 10. Similar process for the others. 7 / (b + 1/c) = 3 7 = 3b + 3/c Since 7 is a whole number, and 3b has to be a whole number, the term 3/c can't be a fraction, so c has to = 3. Then B was easy from there.

[@vallisparmentier9764]

Nice. While it may be implicit in your argument, in response to the > the term 3/c can't be a fraction, so c has to = 3 assertion, an adversary might object that 3/c is a whole number when c = 1 as well. This case is easily refuted [3 must divide (7 - 3/c)], but may be worth mentioning.

[@spookyaction3236]

@@vallisparmentier9764 good point, forgot I could divide by 1 lol. Might have been thinking they were different intergers or something (that's just me mathsplaining though)

[@MangoNutella]

This one was pretty easy. I solved it using method 5 😅. I just subtracted 1, extended the right fraction by 7 so I could regard the denominators. It was clear that a would have to be 1 as a greater a multiplied by 7 would give something greater than 10. Then I subtracted 7, so 3 had to be equal to 7/(b+1/c) and the only positive integers to satisfy this equation are b = 2 and c = 3 as 2 + 1/3 is 7/3, 7s cancel out and we end with 3 = 3.

[@Wildcard71]

1 = 10/10 so for the fraction remains 7/10. The nominator needs to have a value of 10/7 Given the condition of positive integers, a can only be 1. (Equasion: a=1) 1/(b+1/c) = 3/7 b+1/c=7/3 Only b=2 can leave c as an integer as well (c=3). a+b+c=6

[@joliving]

I just started by subtracting one from each side and then cross-multiplying and repeating until you have all three.

[@puzzlesometime341]

2:57 no need to cross muiltiply, 7/10 is an irreducible fraction so bc + 1= 7n and a(bc + 1) + c = 10n. Hence, a7n + c = 10n => c = 10n - a7n => c = n(10 - a7). we can easily see that a cannot ≥ 2 and thus a = 1 ,c = 3n. Because bc + 1 = 7n => b3n + 1 = 7n => 1 = 7n - b3n => 1 = n(7 - b3) => n = 1 ; b = 2 ; c = 3 a + b + c = 1 + 2 + 3 = 6 (q.e.d)

[@roberttelarket4934]

Usually these competitions lead to easy solutions. First thing to try is a = 1, b = 2, c = 3 which turns out to be a solution.

[@aprendiendoC]

Sure but without having a method that you work out, you wouldnt get all the marks

[@morganlaleure8037]

Je ne connaissais pas du tout la 3è méthode, merci ! Et la 4è est super parlante pour un enfant, je la ressortirai !

[@jdnoflegend9719]

i tried the 1st method initially, gave up halfway coz i cant see how i can get a+b+c directly from that. i kept wondering if i can find a+b+c without finding each a, b, c individually. then i tried method 2 and had to make the conclusion a=1 & got b=2 & c=3. seems like there is no beautiful shortcut to get a+b+c directly without finding a, b, c

[@smylesg]

11:10 There's something golden about this method, I just can't put my finger on it. And it's not just because the answer contains the first four in the Fibonacci sequence (whose ratios of consecutive members converges to The Golden Ratio).

[@Ray-of-Hope720]

Wow 😮

[@doomhammer3022]

method 2 is just *chef's kiss*

[@sorsocksfake]

17/10=1 + 1/(a + 1/(b + 1/c))) || both -1 7/10=1/(a + 1/(b + 1/c))) || invert 10/7=a+1/(b + 1/c) || If a>1, then 1/(b+1/c) would have to be negative. This is not possible with positive integers, nor can a 17/10=1 + 1/(1 + 1/(7/3)))=> 17/10=1 + 1/(1 + 3/7)=>17/10=1 + 1/(10/7)=> 17/10=1 + 7/10=> 17/10=17/10 checked

[@user-iq6cc3df3l]

This looks very similar to one of Ramanujan’s infinite fractions. I just read the book “The Man Who Knew Infinity” - I think that’s the title - and it’s very interesting. You probably need to least a basic math understanding to make it through because there are a lot of included equations and ideas. For the life of me I have no idea how he came up with a lot of his formulae - I think that’s the correct plural form - though. It’s like he just pulled them out of a hat. But I guess mathematicians are still trying to prove a lot of his equations today, 100 years later. And even PhDs struggle with his ideas.

[@user-um4xe9lr7v]

الطريقة الاولى كانت لتنتهي في خطوتين: لدينا: 7a+c=10 و منه a=1 لان a>0 و لا يمكن ان يكون اكبر من 1،و منه c=3. لدينا ايضا bc+1=7 و منه b=2. و أخيرا: a+b+c=6.

[@ArcheoLumiere]

Wouldn't a method where you take the whole from the known fraction, subtract the whole from both sides, take the reciprocal of both sides, and repeat be essentially method 2?

[@TearDropFan2763]

im a chinese elementary student, i solved this mentally, but usually these problems are only used in advanced math classes/olympiads and stuff

[@olorinistar9903]

I used method 2, but 3 and 4 are super cool

[@suchaluch5615]

2 Questions: @03:37 We divide by 7ab + 7 -10b Why can you do this? Wouldn't you have to proove that this term isn't zero??? Generally: How would you prove, that this is the only integer solution by Method 2,3,4?

[@natashok4346]

The best is 2.method and if we suppose not big numbers, then good method is 4.method too.

[@ignatiuswong4271]

Is solving the equation after 2:52 necessary? I mean: it is obvious that (bc+1) =7 so bc =6 a(7)+c =10 given a,b,c∈Z+ a can only be 1 else 7a>10 and c will be

[@billy.7113]

I don't think the first method will get full credits in a competition. That is simply NOT smart enough. 🙂😀

[@yurenchu]

The first method is the only one that also inherently proves that (a,b,c) = (1,2,3) is the _only_ possible solution.

[@alexdc8548]

Can this polynomial be solved using the quadratic formula ? : (2b^3+b^2)*x^2 + (b+1)^2*x + 1 = 0

[@vinnusouriyal4623]

Amazing

[@casey653]

I wonder if china has "Can you solve this x from america?"

[@AlienRelics]

What is a woman? 😂😂😂

[@yusufdenli9363]

Third method was very nice 👍

[@KupoPallo]

as a 33 year old math teacher i felt a bit embarrassed that i didn't think of using method 2, instead I proved that A has to be 1 and went from there.

[@algorithminc.8850]

A good problem to reinforce thinking about a problem before doing a problem ...

[@esportsready8361]

Me who used decimals to get the answer:IM SPECIAL

[@SuperTtKiller]

Impressive that Chinese students Had to solve this in English language

[@liamfryers8039]

Let's talk about the last one - by far the most interesting. I think you skipped the ' why we can see it this way' part of the problem. So 17/10 is 'un multiplying' hence the original problem is 17*10. But why would you do that? Then, there is the whole matter of finding squares. Why squares? Could it be that 1 is something divided by itself(more un-multiplying)? So the first '1' is the 10x10 square. Now we have a remainder. Applying the same reasoning to the remainder yields another 1 plus its remainder. We continue until there is no remainder - 3. But why do the squares representing the remainder appear as 1 over something? Could it have something to do with the definition of a continuing fraction? This is another example of a really smart person who 'gets math' but doesn't seem to understand that we mere mortals would like to see the insight, not just the manipulation. Please think of us next time.

[@tagore_14]

I did exactly the same as Method 2

[@devondevon4366]

a + b + c = 1 + 2 + 3 = 6 17/10 = 1 + 7/ 10 1/ 7/10 = 10/7 10/7 = 1 + 3/7 . Hence a = 1 and 1/(b + 1/c ) = 3/7 Hence b + 1/c = 7/3 (the recriprocal) 7/3 = 2 and 1/3 Hence b = 2 and 1/c = 1/3 Hence, c= 3 a = 1 b=2 and c=3 1 + 2 + 3 = 6 Answer

[@HarimauYap]

Singapore questions don't even give you the form - they just ask you to write it in terms of continued fractions...

[@WhiteGandalfs]

4 roads to Rome. Impressing! I, of course, took just another one, and even more complicated than the first shown in the video. But even that one led to Rome.

[@DergaZuul]

Only first method strictly proving there is no other solutions, even though it is presented as hardest one and not recommended. All other methods are just fancy guess and when you add checks for no other solutions become pretty much same as 1st one. Comments already have slightly easier ways to do logic from 1st method and still show uniqueness of solution.

[@Starshine2007]

nice job. 🙂

[@Arvak60]

I tried the 1st method and it seemed... tricky !... Then, the 2nd method just popped out of my mind like an evidence ! ^^"

[@GaurangAgrawal2]

This was really fun to solve. Ans is 6

[@razvilka]

a=2, b=-2, c=4 also is a solution in Z

[@kelzabish827]

1st impression (before watching vid) is that C can't =0, and 1/c cannot = -B, (which cannot = -A).

[@tanelkagan]

I saw 1, 1, 2 and 3 and for a moment was *soooo* hoping that if the fraction were to be continued indefinitely using the Fibonacci numbers, you'd end up with some cool number, maybe even Phi. Alas, it wasn't to be.

[@brree3984]

The 4th method is not magic, if you see the book by 遠山 啓, this method is mentioned in Chapter 1, where Euclidean algorithm is illustrated. 😆

[@tagore_14]

Method 2 Gang 👇

[@hillaryclinton1314]

Subtrack one from each side.. 7/10=stuff and go from there

[@RohitKulan]

Here's an American math Olympiad question for the same age group for comparison. A 7×7 square is marked off into 49 1×1 squares. The small squares along the edge of the large square are painted blue. How many squares are painted blue?

[@AlienRelics]

6x4 = 24 Or 7x2 + 5x2 = 24 Don't count corner pieces twice.

[@munthirzikre2401]

magic I don't write comments on youtube but this video is fire

[@StereoSpace]

They teach this in preschool in Bolivia.

[@bernhardkrickl5197]

Why does method 4 work? That would be neat to know.

[@play_with_Vpp]

a = 1, b = 2, c = 3, solved under 1 minute.

[@soufianehemza4720]

using method 2 I get a=3 b=6 c=1

[@Archduke_Astatos]

Im not gonna give an exact answer, but using a calculator I could guess and check pretty quickly since you could just take away 1 from both sides and make it easy enough to guess over and over. Im not good at math, but im good at shenanigans.

[@rogerphelps9939]

Very contrived. The long way will work in all circumstances.

[@paulgreen9059]

Method 4 reminds me of how the ancient Egyptians did division.

[@adw1z]

Ah continued fraction expansions and convergents, a key part of number theory and in quantum computing such as Shor’s factoring algorithm edit: this was such a hassle. The only good way to do it is method 2: 17/10 = 1 + 7/10 = 1 + 1/(10/7) = 1 + 1/(1+3/10) = …

[@jeremy6384]

Its not in chinese!

[@KipIngram]

This is childishly easy to solve. Note that at the top level we have 1.7 = 1 + 1/K1. So, that's 1/K1 = 0.7, or K1 = 1/0.7 = 1.4285... and we see a = 1. Now, that is equal to 1 + 1/K2. So solve again for K2: K2 = 0.4285 = 1/2.33333... and we get b = 2. Finally, 1/c = 0.3333 and we see c = 3. Easy peasy.