@@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.
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
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
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
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?
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.
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.
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.
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..
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)
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.
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 have been totally unable to understand continued fractions for over 40 years, I shunned them because they were literally total gibberish to me, and thanks to your careful visual iteration I have finally had that WOW moment I have been yearning for all this time 😍 I love you and I want to have your babies 🥰
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
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).
number = integer part + fractional part= integer + 1 / (1 / (fractional part)). The simplest method is to recursively repeat: Number = integer part + 1 / (1 / fractional part). 1 / fractional part is greater than 1, so it has an integer part, and you can repeat this step
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!
Solving math problems (at least school math problems) is 20% having the right tools in your toolbox (having learned them), 10% STILL having the right tools in your toolbox (not having forgotten how to do them properly), and 70% picking the correct tool for the problem at hand, which also requires remembering that it is in your toolbox.
I did it in my head. It's much easier than it seems. a=1, b=2, c=3. 17/10 = 1 + 7/10, and the fraction is expressed as inversed. So 7/10 = 1 / (10/7). The same repetitive idea all along. 10/7 = 1 + 3/7. Now 3/7 is also 1 / (7/3). And 7/3 can be written as 2 + 1/3. So a=1, b=2, c=3.
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.
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)
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.
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
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
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) 🙂
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?
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.
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...
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.
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.
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 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)
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
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
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
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
I did this problem differently. So 17 over 10 as a decimal is 1.7. So 1.7 is equal to 1 plus the rest of the equation. So everything after the plus sign has to be equal to .7. 1 divided by .7 is 1.42857142857 and 1 divided by 1.42857142857 is .7. Since a has to be a positive integer, a is equal to 1. 1 divided by .42857142857 is 2.3 repeating. So b is equal to 2. So 1 divided by c has to equal .3 repeating. 1 divided by 3 is .3 repeating so c is equal to 3.
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.
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.
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.
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.
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.
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.
As the other replier said, it's indeed the same as the Euclidean method (method 3): from the longer side, we subtract the shorter side a whole number of times as most as we can; then we rotate the resulting smaller rectangle by 90 degrees and do it again; and we keep repeating this process until there is no rectangle left.. - - - So we start with a 10-by-17 rectangle. We can remove a 10-by-10 rectangle only *once* ; the new horizontal length becomes 17 - 1*10 = 7 . What remains is a 10-by-7 rectangle. Rotate it by 90 degrees. Now we have a 7-by-10 rectangle. We can remove a 7-by-7 rectangle only *once* ; the new horizontal length becomes 10 - 1*7 = 3 . What remains is a 7-by-3 rectangle. Rotate it by 90 degrees. Now we have a 3-by-7 rectangle. We can remove a 3-by-3 rectangle *two* times; the new horizontal length becomes 7 - 2*3 = 1 . What remains is a 3-by-1 rectangle. Rotate it by 90 degrees. Now we have a 1-by-3 rectangle. We can remove a 1-by-1 rectangle *three* times; the new horizontal length becomes 3 - 3*1 = 0 . What remains is no rectangle. So the procedure stops. The whole numbers of subtraction are respectively 1, 1, 2, 3 ; so those are the coefficients of the continued fraction representation of 17/10 . in comparison: The Euclidean method (method 3): 17 - 10 * *1* = 7 10 - 7 * *1* = 3 7 - 3 * *2* = 1 3 - 1 * *3* = 0 ==> the coefficients of the continued fraction representation are *1* , *1* , *2* , *3* .
Some simple logic with fractions and this is maybe 10 seconds to solve. After doing it in my head, I dragged to 5:04 and about freaked out at the work in progress. Then to the end to verify I was correct, then skipped viewing as I know it would just be frustrating. EDIT: saw the blah blah solution - yeah, like this.
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.
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?
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.
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.
That's exactly my method, too! I thing it's even simpler or at least more straightforward than method 2.
@@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
I am also joining this club 🤗
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
hey same
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.
Lmao.
ah yes, the multi spacial transwarp solution.
You've discovered how to time travel by applying this result of yours in wolframalpha
Fking legend 😅
I checked you solution and it was actually a fat guy named pico from cili. But he was standing about 7 time further than the tetahedron
You forgot that when you exclaim Paco's name, you needed to take the exclamation mark and apply a factorial to one of the i's
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.
Love the blahs, 10/10
but you forgot to take the limit of blah as blah approaches 17/10
I don't understand the second line. Why do we do 1 / blah = 7/10?
@@defined.9629 You want to find a.
@@defined.9629 did you see the question?
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 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.
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
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"
lol this channel exaggerates china's avg math level so much
@@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
@@abaddon6078 Yes, I never met a Chinese who knew maths really. Much exaggerated. Same with India.
@@hassanalihusseini1717 no one asked for your sarcasm but ok
I know only 1 Chinese that is bad at math and it's the same guy who failed his 12 yr finals
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...
you can formally prove that a must equal 1 by saying if a =>2 it’s too big
You can make an assumption like that as long as you can recognize and eliminate all other possible solutions to the problem.
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
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?
I used method #2. The first method was crazy complicated. I didn't even think of the 3rd and 4th methods.
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.
Now what we need, I think, is a video from Prime Newtons explaining _why_ the Euclidian division and division into squares methods work.
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.
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.
Your comment got copied
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
@@Patrik6920 So, exactly what I did.
@@maxhagenauer24 Yes..
@@Patrik6920I too applied the same method and could answer in 37 seconds without using pen and paper.
@@GurjeetSingh-Gur_Jeet_Ke ..thats greate, but i like the teaching aspect .. hopefully someone learned something new .. and i got a bit better explaining..
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)
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.
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.
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 have been totally unable to understand continued fractions for over 40 years, I shunned them because they were literally total gibberish to me, and thanks to your careful visual iteration I have finally had that WOW moment I have been yearning for all this time 😍 I love you and I want to have your babies 🥰
1 + 2 + 3 is 6... I did method 2 mentally. It expects a large number understanding from young children, woah!
Pretty simple, you work from top to bottom and get 1, 2, & 3 which sum to 6.
when I did it before watching, I did method 2. Glad to see i was able to do it
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
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).
Wow 😮
number = integer part + fractional part= integer + 1 / (1 / (fractional part)).
The simplest method is to recursively repeat: Number = integer part + 1 / (1 / fractional part).
1 / fractional part is greater than 1, so it has an integer part, and you can repeat this step
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!
Solving math problems (at least school math problems) is 20% having the right tools in your toolbox (having learned them), 10% STILL having the right tools in your toolbox (not having forgotten how to do them properly), and 70% picking the correct tool for the problem at hand, which also requires remembering that it is in your toolbox.
I did it in my head. It's much easier than it seems. a=1, b=2, c=3.
17/10 = 1 + 7/10, and the fraction is expressed as inversed. So 7/10 = 1 / (10/7). The same repetitive idea all along. 10/7 = 1 + 3/7. Now 3/7 is also 1 / (7/3). And 7/3 can be written as 2 + 1/3. So a=1, b=2, c=3.
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.
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.
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)
That's really interesting. Good way to solve how to fill a space with the minimun number of squares.
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.
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
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
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) 🙂
loved the illumination in method 2!
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?
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.
Really fantastic task. I am so impressed.
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...
Extremely wonderful! thank you!
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.
The last method is so magical!
I got the answer in less than 15 seconds. But then again, I'm not in elementary school. These kids are amazing.
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.
While methods 3 and 4 are great, I don't think it's obvious that they correlate to the original problem.
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?
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.
I calculated the answer mentally.
Method 2 is best. The most lucid.
I started by subtracting 1 from each side and then using reciprocals to find possible a, b, c in similar manner to version 2.
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)
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.
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
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.
Sure but without having a method that you work out, you wouldnt get all the marks
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
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
Your solution #4 seems to be a special case. What if the 1's in the equation were different digits?
Nice solutions !
im a chinese elementary student, i solved this mentally, but usually these problems are only used in advanced math classes/olympiads and stuff
Why does method 4 work? That would be neat to know.
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
Can this polynomial be solved using the quadratic formula ? : (2b^3+b^2)*x^2 + (b+1)^2*x + 1 = 0
which app do you use to creat math expression in video can you help me thanks
I wonder which solution is the one the kids are supposed to have used.
I did this problem differently. So 17 over 10 as a decimal is 1.7. So 1.7 is equal to 1 plus the rest of the equation. So everything after the plus sign has to be equal to .7. 1 divided by .7 is 1.42857142857 and 1 divided by 1.42857142857 is .7. Since a has to be a positive integer, a is equal to 1. 1 divided by .42857142857 is 2.3 repeating. So b is equal to 2. So 1 divided by c has to equal .3 repeating. 1 divided by 3 is .3 repeating so c is equal to 3.
Solved it without pen and paper from the thumbnail. Answer is yes.
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.
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.
Great video! Method seems to be the same as 3, but presented differently
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.
I just started by subtracting one from each side and then cross-multiplying and repeating until you have all three.
Hua Luogeng (1910/11/12-1985/06/12) was a famous Chinese mathematician.
Singapore questions don't even give you the form - they just ask you to write it in terms of continued fractions...
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.
Fabulous!
The best is 2.method and if we suppose not big numbers, then good method is 4.method too.
I used method 2, but 3 and 4 are super cool
Je ne connaissais pas du tout la 3è méthode, merci ! Et la 4è est super parlante pour un enfant, je la ressortirai !
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.
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
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.
Can anyone explain the logic behind the last method the 'visual magical squares'
Visual *division algorithm
It is exactly the same as the euclidean method.
@@mittarimato8994 there's something mysterious🔮 about that square method 🧿
@@Ray-of-Hope720 instead of "how many times 7 fits into 10?", it asks "how many times 7x7 fits into 10x10?"
As the other replier said, it's indeed the same as the Euclidean method (method 3): from the longer side, we subtract the shorter side a whole number of times as most as we can; then we rotate the resulting smaller rectangle by 90 degrees and do it again; and we keep repeating this process until there is no rectangle left..
- - -
So we start with a 10-by-17 rectangle. We can remove a 10-by-10 rectangle only *once* ; the new horizontal length becomes 17 - 1*10 = 7 . What remains is a 10-by-7 rectangle. Rotate it by 90 degrees.
Now we have a 7-by-10 rectangle. We can remove a 7-by-7 rectangle only *once* ; the new horizontal length becomes 10 - 1*7 = 3 . What remains is a 7-by-3 rectangle. Rotate it by 90 degrees.
Now we have a 3-by-7 rectangle. We can remove a 3-by-3 rectangle *two* times; the new horizontal length becomes 7 - 2*3 = 1 . What remains is a 3-by-1 rectangle. Rotate it by 90 degrees.
Now we have a 1-by-3 rectangle. We can remove a 1-by-1 rectangle *three* times; the new horizontal length becomes 3 - 3*1 = 0 .
What remains is no rectangle. So the procedure stops.
The whole numbers of subtraction are respectively 1, 1, 2, 3 ; so those are the coefficients of the continued fraction representation of 17/10 .
in comparison:
The Euclidean method (method 3):
17 - 10 * *1* = 7
10 - 7 * *1* = 3
7 - 3 * *2* = 1
3 - 1 * *3* = 0
==> the coefficients of the continued fraction representation are *1* , *1* , *2* , *3* .
Some simple logic with fractions and this is maybe 10 seconds to solve. After doing it in my head, I dragged to 5:04 and about freaked out at the work in progress. Then to the end to verify I was correct, then skipped viewing as I know it would just be frustrating. EDIT: saw the blah blah solution - yeah, like this.
As I see here, there are always new things to learn... 😊
method 2 is just *chef's kiss*
Me who used decimals to get the answer:IM SPECIAL
Impressive that Chinese students Had to solve this in English language
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.
That sounds like what I did, too. And very possible, just to do in your head. There is no need to write down any calculations.
Any reason they ask for the sum?
Third method was very nice 👍
RIP to all those who could not solve a problem for 10 year olds.
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?
6x4 = 24
Or 7x2 + 5x2 = 24
Don't count corner pieces twice.
I solved like this.
7/10 = 1/{a+1/(b+1/c)}=7/[7*{a+1/(b+1/c)}]
10 = 7a + 7/(b+1/c)
3 = 7/(b+1/c)
3(b+1/c) = 7
therefore b=2, c=1
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. 😆
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.
I wonder if china has "Can you solve this x from america?"
What is a woman? 😂😂😂
Nah method 2 was immediate click for me
Yup. I did it in my head using method 2 before clicking on the video. Easy.
Same here.