Thank you for this great episode. This problem is really awesome! And I tell you why. Being also interested in music theory and tuning systems, I found a strong relation to this domain. Translated into the language of music, this equation asks: How can I divide an octave (which is the frequency ratio of 2 to 1) into three simple intervals where the nominator is one more than the denominator. In just intonation, such intervals are: octave = 2/1, fifth = 3/2, fourth = 4/3, major third = 5/4, minor third = 6/5, major second = 9/8 or 10/9, minor second = 16/15 or 17/16. For example, the solution a=4, b=5 and c=3 leads to the factors 5/4, 6/5 and 4/3, which are a major third, a minor third and a fourth. If I start at C3, the following notes are E3, G3 and C4. This is a major chord. Great. Another example, the solution a=2, b=4 and c=15 maps to the intervals of a fifth, a major third and a minor second. Starting at C3 the following notes are G3, B3 and C4. For me, there is a heavenly connection between mathematics and music.
@@gustavoheras557 Because that would cause the first term to be 2 and so the other terms would have to be 1 total, and the only way to do that is if b and c were infinity, which is not a real number.
А и Б - точные или натуральные единицы, а С - это кислота от кислорода и снова хим реакция и 9.Надеюсь , понял как мои отстающие одноклассники, а меня ругали, что издеваюсь над преподом. Мне нужно обдумать, что сказал этот из Бауманки.
@@jamesnewcomer5963 “… no reason to write out all the permutations” - If you're returning a solution to the Olympiad problem, you should: you won't get the full score otherwise because it would look like you had forgotten your WLOG assumption. You may try writing in words next to the solution that any permutation is also a solution, but IRL I wouldn't take the risk.
I like very much the way you teach. You experiment with possibilities, instead of being smarty and immediately presenting the solution. In this way, the student feels like participating in the solution instead of passively watching it. You're a great tracher.
De acuerdo con el comentario. Es odioso leer a tantos otros que al parecer preferirían videos sobre problemas con este grado de dificultad siendo resueltos en dos minutos y usando solo la mitad de la pizarra, como si se tratara de un concurso. Dejen a los matemáticos hacer verdadera matemática, por favor.
4/3 x 5/4 x 6/5. The frequency ratios between pairs of notes in the sequence G C E G (C major with an extra G in the bass.) The major chord has the frequency ratio 4:5:6 (or 20:25:30) The minor chord has the frequency ratio 1/6:1/5:1/4 (or 20:24:30) (this is based on pure mathematical, "just" chords. To be able to play in all keys without retuning, modern fixed-pitch instruments are tuned in equal temperament, which means all semitones are exactly 2^(1/12) so 4:6 is very well approximated by 2^(7/12) and 4:5 is rather crudely approximated by 2^(4/12)=2^(1/3)
@@cykkm I am 67 and a mathematician by education since 1981 (but never worked as a mathematician) . I found it eyes opening. Thought provoking task and a great teacher. Thank you very much, Teacher!
You are a great teacher and i love your enthusiasm! Here's how i did it: If you set a=b=c=x and solve, you get smth around x=3.8. To balance it, if you make one number larger, you have to make another smaller. One can use this principle to get upper bounds for the smallest of the numbers. Here it means that all solutions must contain at least one 1, 2 or 3. It can't contain a 1 since that's already 2 for one factor. And it can't be (2, 2) or (2, 3) (or (3, 2), to make that clear) for two factors since that's already larger than 2. So now we have lower bounds restrictions for b depending on a. If you set a=2 and b=c=x, you get smth around x=6.5. Repeating the balancing principle and using the restrictions we figured out, you can now set b∈{4,5,6} and solve for c for each case. You get the solutions: (2, 4, 15), (2, 5, 9), (2, 6, 7). Now to set a=3 and b=c=x, you get smth around x=4.4. Using the balancing principle and restrictions again, we can set b∈{3,4} and get the solutions (3, 3, 8), (3, 4, 5) So we have five distinct sets of natural numbers as solutions and we know all other solutions are permutations of those. Figuring out the permutations is kinda trivial.
I was going through it a few times confused about how could you possibly just not write out the balancing description. And then I realized you just counted down to A. Nice proof :D
I love your "nice" comment every time you get a solution! I found your videos a week or so ago and I cannot stop watching. I know I have always loved math, but these videos have stimulated my love for it. Thank you for providing so many problems for us to learn along with you.
Equivalent to 1 +a+b+c+ a b+a c +b c - a b c = 0 Let a=p,b=p+1,c = p+2 Substitute and by some algebra 6+ 9p +3 p^2 -p(p^2 +3 p+2) = 0 3(p^2 + 3p +2)-p(p^2 +3p +2) = 0 (3-p)(p +1)(p +2) = 0 For natural number solutions p must be positive. p = 3 Then a = 3,b= 4,c = 5 There are five more permutations of a,b,c
Multiplying by abc => (a+1)(b+1)(c+1) = 2abc. (Note that a, b, & c can't all be even, and if a = b = c /in R, then a ~= 4 as 5^3 = 125 ~= 128 = 2*4^3. So we're looking for small whole numbers whose product, abc, is somewhat near 64, e.g. a=b=c=~4 and a4, assuming a
Other solution: (a+1)(b+1)(c+1)=2abc. For decomposition, it needs to be for example {(a+1)=b,(b+1)=c,(c+1)=2a} this set of equations leads to a=3 b=4 c=5. There's some other way to factorize, which gives negative or non integer results. The other factorization which gives result is a=b and {(a+1)^2=c,c+1=2a^2}, which leads to a=b=3,c=8.
Hey. Very entertaining problem. Thx! I suggest a quick path: consider [(a+1)/a] ⋅ [(b+1)/b] ⋅ [(c+1)/c], and then use an _ansatz_ where the denominator of the middle term cancels the numerators of the first and third terms (something like that is bound to happen). So try b = (a+1)(c+1), then the problem is to find a, c such that (a+1)(c+1)+1 = 2ac, (i.e.) ac + a + c + 2 = 2ac, or ac - (a + c) = 2. Can we find two positive integers where the product equals the sum plus 2. Sure. Small cases gives you a = 2, c = 4 immediately, and b = 15. We get: (1+1/2)(1+1/15)(1+1/4) = (3/2)(16/15)(5/4) = 16/8 = 2.
@3:37: cubr(2) is irrational and 1+1/a is rational. Further, eliminating the case "a = b = c" is actually not necessary. @4:57: As the denominator of a fraction with a "positive" numerator gets bigger and stays the same sign, the fraction gets smaller. @8:43: A correct argument is: cubr(2) > 1.25, since (1.25)^3 = 1.953125. Therefore, a = b - 3 >= a - 3 = -1. Hence, if c - 3 and b - 3 are negative, they both must be -1, which doesn't work. @19:02: 6 could have two negative integer factors. But c - 2 >= b - 2 >= a - 2 = 1.
I agree that this is a very interesting problem. While I enjoyed your presentation, it nagged at me that there must be a more number theoretic solution. This is what I found. (1+1/a)(1+1/b)(1+1/c) = (a+1)(b+1)(c+1)/abc = 2 We know that for c>1, it does not divide c+1. So we can safely assume that a, b and c divide the alternate terms. So that: (a+1)(b+1) = ck; (a+1)(c+1) = bm; and (b+1)(c+1) = an. It can be a little difficult to see at first. In the first equation, we know that c doesn't divide c+1 but the factors of c must divide a+1 and/or b+1. The 'k' represents the amount left over. Multiplying: [(a+1)(b+1)(c+1)]^2 = abc * kmn. Since we know [(a+1)(b+1)(c+1)]/abc = 2, if we divide both sides by (abc)^2 kmn/abc= [2]^2=4 so kmn = 4abc Interesting, but how does that help us? Looking back at the equations in , we can find the ratios (a+1)(b+1)/(a+1)(c+1) = ck/bm = (b+1)/(c+1) so kc(c+1) = mb(b+1) and similarly, kc(c+1) = na(a+1) and mb(b+1) = na(a+1) This implies t = kc(c+1) = na(a+1) = mb(b+1) All the values are related. Can we find t? What if we multiply? t^3= kmn*abc*(a+1)(b+1)(c+1) = [kmn/abc] * (abc)^3 * [(a+1)(b+1)(c+1)/abc] and substitute '2' using and then t^3 = 4 * (abc)^3 * 2 = 8 * (abc)^3 = (2abc)^3 so that t = 2abc Now we are very close since can look at the relations in na(a+1) = t = 2abc so n(a+1) = 2bc = na + n and from n + an = n + (b+1)(c+1) = n + bc + b + c + 1 = 2bc so n = bc - b - c - 1 But how do we use n? Going back to an(a+1)/abc = 2 changes to a = 2bc/n - 1 = 2bc/(bc - b - c - 1) -1 since a>0. Further this rotates for all the terms so b = 2ac/(ac - a - c - 1) -1 and c = 2ab/(ab - a - b - 1) -1 This now defines all the solution sets. We can test it using a known solution (3,4,5) = (a, b, c) a =3 so b = 2*3*5/(3*5-3-5-1) - 1 = 30/(15-9) - 1 = 5-1 =4 and c= 2*3*4/(3*4-3-4-1)-1 = 24/(12-8) - 1 = 6 - 1 =5 For a = 1, the values are negative but still works. (1, b, -b-1) => 2*(b+1)/b*[-b/-(b+1)] = 2 For a = 2, b = 4c/(2c-2-c-1)-1 = 4c/(c-3) -1 here c=5 or 7 gives b = 9 and 6 3/2*6/5*10/9 = 2 and 3/2*8/7*7/6 = 2 This doesn't guarantee that every number 'a' has a solution. a=13 b = 26c/(13c-13-c-1) - 1 = 26c/(12c-12)-1 there is no value for c that would make this work. Every solution set should satisfy .
My first try to solve this was to multiply both side by abc: (a+1)(b+1)(c+1) = 2abc, but this didn't lead nicely to anywhere. Then I did exactly the same thing that was done in this video to solve for the possible values of a = 2 or 3 (given a ≤ b ≤ c). This gives (1+1/b)(1+1/c) = 4/3 when a = 2 (1+1/b)(1+1/c) = 3/2 when a = 3 From there my solution was different from this video. I solved the equations above as mini versions of the original problem: Since b ≤ c, (1+1/b) ≥ √k, where k = 4/3 when a = 2 and k = 3/2 when a = 3. This gives the possible values for b and the corresponding values of c are calculated from b: c = 3*(b+1)/(b-3) when a = 2, b = 4, 5, 6 c = 2*(b+1)/(b-2) when a = 3, b = 3, 4 This gives the solutions when a ≤ b ≤ c. All the solutions are all the permutations.
What I did is similar to what you did, but instead of solving for c afterwards, I already had the ranges of b ready and just tried them all out. So I still had (1 + 1/b)(1 + 1/c) = 4/3 if a = 2 and 3/2 if a = 3 I checked and if a = 2 then b = 2, 3, 4, 5, or 6. If a = 3 then b = 3 or 4 Then I just tried every setting of a and b available and solved for c instead of coming up with a general form of c So the solutions are: (2, 4, 15) (2, 5, 9) (2, 6, 7) (3, 3, 8) (3, 4, 5)
I did the same thing and it actually led to the solutions quite nicely. Because you realize that (a+1)(b+1)(c+1) = 2*a*b*c means that adding one to the numbers just introduces another 2 in prime factorization. So at least one of the numbers must be uneven (or else the left side of the equation is uneven), and obv. none of the numbers can be 1. Checking for a = 3 gives 2*(b+1)*(c+1) = 3*b*c. It is again easy to see that neither b nor c can be equal to two. Checking b = 3 gives c = 8 and b = 4 gives c = 5. It remains to prove that there are no further solutions however
There's a great tool related to this kind of problems. Whenever a_n is a sequence of real numbers, we consider Sn to be the sum of the terms of the sequence to the n'th power, for example S2=a_0^2+a_1^2... We consider Pn to be a product that generalizes this: P2= a_0*a_1+a_0*a_2.... +a_1*a_2.... Pn is the sum of the different products of n different terms of the sequence. Finally we have this relation: n*Pn=Σ(-1)^(k+1)*P(n-k)*Sk for k between 1 and n, where P0=1
Also, Pn is proportional to the n'th derivative of the polynomial c*(1-x/a_0)*(1-x/a_1)... Which sometimes lets you find infinte sums of powers, if you know the derivatives of the function in question
Pd. You can prove that Σ 1/k^3 = pi^3/32 for k= 1,-3,5,-7,9.... ( if i remember correctly) using this tool, although to be rigorous you would use something like the Weierstrass factoring theorem for the function cos(pi*(1-x)/4)
The only problem, though, is that you may not know an approximate value of cube root of 2 since you mustn't use a calculator in olympiads (btw approximations aren't exact values or constraints so may sometimes lead to the loss of solutions). Instead, I'd recommend to assume that a>=4. Then 4
Excellent Problem, I got the 3,4,5 but missed the 3,3,8 solution but was suspicious that there might be other solutions but was spending too much time on it. I wonder was this problem given under a test scenerio where time was limited? It is very challenging but a great exercise in logic. Unlike some of these other Olimpiad videos which are much less interesting.
If you dont know cube root of two you can calculate it with pencil and paper method 2^(1/3)= 1.25... 1 1000| 304 (Triple square of current approximation and append square of last digit of next approximation) 3 * 2 (Triple current approximation shifted one position to the left and multiply by last digit of next approximation) 364*2 (add them up and multiply by last digit of next approximation) 1000 728 (then subtract from remainder) 272000|43225 36 * 5 43225 180 45025 272000|45025*5 225125 46875000
That was an amazing way to explain how to solve it. Would it be possible to implement the same techniques for the following question or how would you solve this: x + 1/x + y + 1/y - z - 1/z = 2 where x > y > z
The only comment I want to make is, people might not know the cube root of 2. Another way to plugging in is to use binomial, we get that cbrt(2) = (1+1)^(1/3)
I really enjoyed the organized and clear explanation of this problem! Broadening the solution space to all integers, notice that two or three of a, b, and c cannot be negative. Otherwise, if all three of them were negative, we'd have a product of three zero or positive fractions each less than 1, which could not multiply to 2, and if two of them were negative, 1 + 1/a would have to equal or be greater than 8, also impossible. Nevertheless, without loss of generality, we could modify the assumption of a being smallest to a being less than or equal to the other positive integer b, retaining 1 + 1/a as the largest factor in the product, leading to the same bound on a. Then, with the same technique, not including permutations, we can find three new solutions: (a, b, c) = (1, 2, -3), (2, 2, -9), and (1, 3, -4).
We are waiting for your full course on calculus. The Lenard online courses come to mind. Viewing his calculus lessons could explain what we need. Understanding what you are doing is more important than just being able to perform an operation. People only remember so much. Knowing how to get to solutions solves the memory problem. We are talking about starting with the fundamental theorem of calculus and moving through multivariable calculus. Perhaps I should speak for myself. I could greatly benefit from a review of the basics and learning all the rest of calculus. Thank you for your assistance.
I tried the simpler problem with just two variables. Solns are 2, 3 and 3, 2. Before watching the video, I expect a factorization like (a - 1)(b - 1)(c - 1) = something
Interesting way to go about the question. I managed to find 2 unordered triples of solutions using the following method. (1+1/a)(1+1/b)(1+1/c)=2 1 + 1/b + 1/a + 1/ab + 1/c + 1/bc + 1/ac + 1/abc=2 ac + a + bc + b + c + 1 = abc - ab a(c+1) + b(c+1) + 1(c+1) = ab(c-1) (a+b+1)(c+1) = ab(c-1) Assume that for some value(s) of a, b and c, c+1=ab Then, c=ab-1 Therefore, a+b+1=ab-2 a+b=ab-3 By inspection, it is easily observed that (2,5,9) and (3,3,8) are solutions to this equation, and hitherto to the original equation (which is easily verified by substituting into the original equation.)
Imagine the denominator of the previous term cancels the numerator of the next term. Then the equation can be written as (K+1)/K * (K+2)/(K+1) * (K+3)/(K+2) = 2. which gives 3,4,5 Similarly We can have (K+1)/K * (K+1)/K * (K*K)/(K*K - 1) = 2 This gives 3,3,8
Aparantly simple looking problem becomes quite complex.your logic assumptions leads to great solution.explained so nicely that I understood every thing.Great teaching art.
Interesting how there is 3^3 solutions. I wonder if a similar problem can be constructed to have 4^4 solutions. What do you set it equal to? (1+1/a)product through d =?
So. You can expand this to 4 variables in the same kind of product and symmetrical structure. To save everyone time you bound a to be 5,4,3,2,1 and b the same but a=\= b so b can be 4,3,2,1. I’m counting out how many combos this is
The question should be: “find all solutions of …” as otherwise just giving one solution would solve the equation. Is it really necessary to write out all the permutations? This seems somewhat OTT. Surely, giving just one permutation for each solution and then sufficing the phrase, “and all permutations” should be sufficient, or even something like {a, b, c} = {2, 4, 15}, etc.
It is possible to see. If you want one number to be as large as possible, you need to make the others as small as possible. If you figure out the lower bounds for the first two numbers, you get the upper bound for the third. (1, n) doesn't work for any n (2, 2) doesn't work (2, 3) doesn't work (2, 4) could work (3, 3) could work Any other pair would be larger on average. Trying the two possible solutions, you get 15 and 8 as possible upper bounds, meaning that the true upper bound is the larger of those values, 15. I still wouldn't brute force from here. There would be 14!/(3!*(14-3)!) =(14*13*12)/(3*2) =7*13*4 =364 possible ways to choose 3 numbers out of 14 possibilities ignoring the order.
How about this perspective. By turning them into improper fraction we get a form like (a+1)/a … , then observe that when the denominator and numerator can cancel each out, that is when a+1/c=2, a=b+1, c=b+1,then we have our answer, by solving1 this linear equation system, we can get our desired answer (3,4,5)
Really enjoyed it. You're a heck of a good teacher. But is it really necessary to enumerate all the permutations? W.L.O.G., (3,4,5) is the same as any of the six arrangements of the same three numbers. What I find fascinating is, when I look at (2,4,15), (2,5,9), (2,6,7), (3,4,5), and (3,3,8) all being solutions to the same equation, I cannot intuitively feel why or how they could all be solutions. Especially when I consider (2,5,9) and (3,3,8), there are no "handles" I can grasp in order to leap from one solution domain to the other solution domain. Yet, both these solutions simply exist in nature, and are perhaps comprehendible by others far more intelligent than me. That is a humbling feeling!
If a=1 then the other two factors should multiply to 1. Then by reducing the equation b+c=-1 which gives as much combinations of b and c as you want...
Good teaching which is fulsome , very clear and easily listened to. We did not say " not equal to" , so (1+1/3)(1+1/3)(1+1/8) = (4/3)*(4/3)*(9/8)= (16/9)*(9/8) =2 also gives us a possible solution if a=b is allowed. a=3, b=3, c=8. This came at 20.00 However a=b=c is covered at 3.00. I missed 3 sets of solutions out of the 5 sets . i become more humble! 9/27 is 33+1/3% aaw!
Very good presentation! With some simple arithmetic I was able to show a and b could not both be 2, and not all three of a, b, c could be greater than 4. Then by writing out the possible answers I found the 3,4,5 solution. I would have found the other solutions eventually, but your way is faster.
There are only 5 solutions.... the permutation stuff is not necessary since a, b, c are arbitrary naming we can assign the three numbers in each of the solutions to a, b, c so that a is
Before watching video: x/y * z/u *r/q = 2 neccesarily y= x-1 u=z-1 q=r-1 x * z * r = 2 (x-1) (z-1)(r-1) suppose all the same number x^3 = 2 (x-1)^3 y = x^3-2(x-1)^3 Plug in values of x till you find where the output flips from positive to negative, vise versa, or valid x answer. 4->5 goes from positive to negative. (implied, 5, 5, 5 is negative) decrease a random number of 555 (since it is closer than 4,4,4 to 0), to try and make it go back positive. 5, 5, 4 is slightely positive. Take opposite approach on not the 4, to try and get closer to 0 again. try 5, 6, 4 1 answer found. To put it back into its orginal form, you subtract 1 from each value. 4,5,3 Now start from the other side, and repeat the process to find more examples, and also dont give up simply for finding an answer. Okay i'm done watching the video. This is genuis. Wow I feel like my solution is barbaric in comparison. incredible!
if we are dealing with natural numbers then special techniques are available. The first step is to get rid of all the fractions by multiplying both sides by abc (a+1)(b+1)(c+1] = 2abc Now it is possible that large numbers are involved but most likely it will be small numbers. so we could guess that there is no factorisation involved. a+1=b, b+1=c, c+1=2a so the equation now becomes (a+1)(a+2)(a+3)= 2a(a+1)(a+2) hence a+3=2a, a=3,b=4,c=5 lhs =4*5*6 =120 rhs=2*3*4*5=120 so that is a solution, is it unique? well if any of a,b,c are different, then at least one of them must be lower and at least one higher. try a=2 3(b+1)(c+1)=4bc 3bc +3(b+c+1)=4bc 3(b+c+1)=bc so at least one of b,c is a multiple of 3. without loss of generality let b=3k 3(3k+1+c) =3kc 3k+1+c=kc so 1+c must be a multiple of k, say mk , c= mk-1 (3+m)k =k(mk-1) 3+m=mk-1 4=m(k-1), m=1 or 2, or 4 try m=1 k=5 ,b=15,c=4 3(b+c+1)=60=bc ok try m=2 k=3 b=9 c=mk-1=5 3(b+c+1)=45=bc, ok try m=4, k=2 b=3k=6, c=mk-1=7 lhs=3*7*8 rhs=2*2*6*7 ok so the solution sets for a,b,c that i have found are 3,4,5 2,5,9 2,4,15 2,6,7 try 3,3,c lhs=4*4*(c+1) rhs=18c=16c+16 c=8 so 3,3,8 is a solution. do we really need to look at all the permutations, can’t we just say c>=b>=a and the solutions are 2,4,15 2,5,9 2,6,7 3,3,8 3,4,5
"Now it is possible that large numbers are involved but most likely it will be small numbers. so we could guess that there is no factorisation involved." This is where you leave the realm of mathematics and enter the realm of fantasy.
Uau! Que questão linda! A solução dada foi mais linda ainda. Parabéns! Brazil - Julho 2024. Wow! What a beautiful question! The solution given was even more beautiful. Congratulations! Brazil - July 2024.
Dear master, fantastic logical way of solving problems,made my eyes opening after viewing your video, thx u so much However, u hv verified a=b=c is not true as a,b & c are natural numbers. Hence the solution (3,3,8) and its permutations is invalid , total solution is 24 instead of 27, pls advise and correct me if I'm wrong
This question tests the application of equations. It is known that (1+(1/a))(1+(1/b))(1+(1/c))=2. Then (a+1)/a×(b+1)/b×(c+1)/c=2. That is (a+1)(b+1)(c+1)=2abc. Expand it: abc+ab+ac+bc+a+b+c+1=2abc. Therefore, ab+ac+bc+a+b+c+1=abc. Since a, b, and c are all positive integers, Therefore, at least one of a, b, and c is greater than or equal to 2. Assume a≥2, b≥1, c≥1, Then, ab≥2, ac≥2, bc≥1, a≥2, b≥1, c≥1. Therefore, ab+ac+bc+a+b+c+1≥2+2+1+2+1+1=9. And abc≥2×1×1=2. Therefore, ab+ac+bc+a+b+c+1>abc. This is contradictory to ab+ac+bc+a+b+c+1=abc. Therefore, it is impossible for a, b, and c to all be greater than or equal to 2. Then, only one of a, b, and c can be 2, and the other two are both 1. Let's set a=2, b=1, c=1. At this time, (a+1)(b+1)(c+1)=3×2×2=12, 2abc=2×2×1×1=4. Satisfies (a+1)(b+1)(c+1)=2abc. In summary, a=2, b=1, c=1 is the solution of the equation (1+(1/a))(1+(1/b))(1+(1/c))=2.
@@vinceturner3863 let's re-examine this step-by-step: The original equation is: a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c) The given values are: a = 2, b = 1, c = -1 Plugging these values into the left-hand side (LHS): LHS = a ÷ (b + c) = 2 ÷ (1 + (-1)) = 2 ÷ 0 = undefined Plugging the values into the right-hand side (RHS): RHS = (a ÷ b) + (a ÷ c) = (2 ÷ 1) + (2 ÷ (-1)) = 2 + (-2) = 0 Comparing the LHS and RHS: LHS = undefined RHS = 0 Therefore, the original equation is not satisfied, as the LHS is undefined and the RHS is 0. So you are correct - the given values of a = 2, b = 1, c = -1 do not satisfy the original equation, as the LHS evaluates to an undefined value, not 6 as stated. The solution values do not make the LHS equal 6 instead of 2. Thank you for catching my mistake!
Maybe I got lucky, but I solved this in five minutes. First, a little easy algebra: (1 + 1/a)(1 + 1/b)(1 + 1/c) = 2 ((a+1)/a) ((b+1)/b) ((c+1)/c) = 2 (a+1)(b+1)(c+1) = 2abc And then... how about, just match terms. This works if a+1=b, b+1=c, and c+1=2a. So you just start trying consecutive numbers until you find a sequence a, a+1, a+2 where the last number is once less than twice the first number. Obviously this is 3,4,5. And it just works. (4)(5)(6) = 2(3)(4)(5), naturally. Maybe I got lucky that that worked, but it was a heck of a lot easier than this 20-minute answer.
Have you found all of the solutions? Finding a partially correct answer tends to be a heck of a lot easier than finding the full solution set. Oh and by the way, I used the same approach as you!
I started writing the equation on a easy way. (a+1)(b+1)(c+1)=2abc then I thought... It would be "cool" if we could simplify some factors, so... I suposed b=c-1 (a+1)(c-1+1)(c+1)=2a(c-1)c (a+1)(c+1)=2a(c-1) ac+a+c+1=2ac-2a 3a+c+1=ac a and c "must" be small. so... try and try... a=3, c=5 works... (3,4,5) but I cannot probe that those are the unic solutions and I do not know why I started with b=c-1 and not a=c-1
so I solved this problem thinking there would only be 1 solution What I did was first change it to ((a+1)/a)((b+1)/b)((c+1)/c)=2 and ((a+1)(b+1)(c+1))/(abc)=2. So I tried to think of it as 2 of the factors in the denominator and 2 in the numerator will cancel out and the last factor in the numerator would be double the last factor in the denominator. so WLOG, I decided b+1=a and c+1=b so that simplifying it gets (a+1)/c=2 you can get a=c+2 and a+1=2c from these equations. then with substitution, you get 2c-1=a and so c+2=2c-1, so c=3. then using b+1=a and c+1=b, you get b=4 and a=5 to get the 3,4,5 solution ik in general this isn't a good method, but it worked here
2 books for my fellow "I like not to involve calculus" problem solvers: Ivan Niven - Maxima and Minima Without Calculus; and Michael Steele: The Cauchy-Schwartz Master Class.
0 doesn't belong to Natural numbers and besides, 1 + 1/0 is not a number. a (or b or c) couldn't be 1 as then the term in brackets would be 2. You'd then have: (2)(number > 1)(number > 1) ...which is greater than 2.
(1+1/a)*(1+1/b)*(1+1/c) = 2 (a+1)*(b+1)*(c+1)/(a*b*c) = 2 (a+1)*(b+1)*(c+1) = 2*a*b*c From here you can just think your way to the answer. You need three numbers such that adding one to each one doubles their product. It's not to hard to suss out that 3, 4, and 5 work. 3*4*5 is 60, and 4*5*6 is 120. Plug those into the original formula - i works. A possibly more direct way to see this is to re-write the original problem like this: ((a+1)/a)*((b+1)/b)*((c+1)/c) = 2 and note that (6/5)*(5/4)*(4/3) has the 4's and 5's cancel, so you're just left with 6/3, which is 2.
the way I did it before seeing yours is like first making the equation (a+1)(b+2)(c+1)/abc=2 and then this will be true if the numerators can be divided by denominator that means a+1=b, b+1=c and c+1=2a and then solve for equation from this three equations for three variables we have and we get 3, 4, 5 as solution for this.
Funny... I just wanted to see if I could spot some kind of pattern that arises with 'random' choices. So, I started with: a=3, b=4, c=5. Ding, ding, ding, ding.... So, I got an answer in the first 10 seconds. Positively Olympic!
