@@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
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
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.
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.
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
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"
@@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
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
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* .
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.
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.
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...
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.
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".
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
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.
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.
..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
@@GurjeetSingh-Gur_Jeet_Ke ..thats greate, but i like the teaching aspect .. hopefully someone learned something new .. and i got a bit better explaining..
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.
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?
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
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)
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!
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.
Read my mind so well! One can convert any fraction into a sequence of positive integers by repeatedly splitting it into its integer and its fractional part and taking the reprocipical of the later one. With that, repeat the process.
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.
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.
@@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)
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
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
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
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.
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.
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
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?
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.
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...
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).
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
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)
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.
Love these videos! Keep them coming! There is a much faster way to do it in your head though, takes only seconds. I'll walk you through it: 17/10 is the same thing as 1 + 7/10. Therefore, the term with the a, b, and c must equal 7/10. Since there is a 1 in the numerator, it means the denominator, which has all the a, b, and c terms, must be equal to the reciprocal or 10/7. We use the first trick again and note that 10/7 = 1 + 3/7. So we know a = 1 and the b,c term which is 1/(b + 1/c) must be = 3/7. Hence, the denominator is 7/3, i.e., b + 1/c = 7/3. But 7/3 = 2 1/3. And since b and c have to be integers too, b = 2 and c = 3. You are basically just breaking up a fraction into mixed numbers several times and using the reciprocal to very rapidly finding the answer 1+2+3 = 6 without even having to use a writing instrument. Pretty cool questions! Can't wait to give this to my 12 year old. Let's see what he comes up with... (One final comment: I was too lazy to pick up a pen and that's sometimes the beauty of math, that laziness helps you find the most simple and elegant solutions) 🙂
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
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?
This was cool. Why did you say method 4 was like magic? It makes sense. It’s roughly equivalent to methods 2 & 3. Creating the squares inside the rectangle is just a visual representation of finding the greatest common factor. It was intuitive and easy to follow, but there was nothing mysterious or magical about it.
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.
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.
The 5th method. You subtract 1 for both sides and invert both sides. Since the numbers must be positive integers then a must be 1. Then continue doing same thing and you find b and c. I don't think elementary school students can solve this problem though.
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
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.
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.
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.
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?
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.
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.
Asking for the sum rather than the individual values of a, b, and c makes me wonder if there's a method that can find the sum without having to find the individual values....
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.
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.
17/10 = 1+ [I won't write this out...] 17/10 = (abc+a+c+bc+1)/(abc+a+c) | cross multiplication 17abc+17a+17c = 10abc+10a+10c+10bc+10 | -10abc ; -10a ; -10c 7abc+7a+7c = 10bc+10 | factor out 7 and 10 7*(abc+a+c) = 10*(bc+1) | ÷7 ; ÷10 (abc+a+c)/(bc+1) = 10/7 Here I got 2 equations: 1: abc+a+c = 10 2: bc+1 = 7 | -1 bc = 6 b, c ∈ N\{0} => 1: b=1, c=6 2: b=2, c=3 3: b=3, c=2 4: b=6, c=1 Tried all 4 cases for equation 1 => only case 2 works => a=1, b=2, c=3 I saw some smarter ways to solve this problem in the comment section, befor I posted this comment. Yeah I still tried without help and it worked :)
