I dont think its mentioned anywhere that a and b belongs to integers. Only thing it states is that a,b>0 which means they are positive and therefore can exist as positive real nos. too. Sure, they are the only valid "whole" no. Solutions. But there are infinite solutions here. We cant solve this until the question states that a,b belongs to natural numbers
You can't assume the values of a or b this way. Fractions are ratios. (1/13)=(2/26)=(4/52)... Why did you also assume that a=b when it clearly doesn't state this?
@@roulam3001Obviously a+b=52 is a potential solution here and can be checked to be correct. However, as we saw in the video, it isn’t the only solution.
In fact, we have unlimited solution by make one variable as depend to another variable. Lets A is independent variable, then B=f(A). In this case, we can choose B=n×A or another equation. For exampke n=1 and B=1×A=A, then 1/A+1/B=1/A+1/A=1/13 2/A=1/13 or A=26, B=26 n=2, B=2×A=2A 1/A+1/B=1/A+1/2A=3/2A 3/2A=1/13 A=39/2, B=39 .... And so on..
It has infinite solutions. 1/a + 1/b = 1/13 1/a = 1/13-1/b a>0, =>1/a >0, => 1/13 - 1/b >0 => b>13 ||rly a > 13 a>13 and b >13 Hence take any value of a or b greater than 13,.put in the first equation and get the value of other variable
@@БелАлексСогласен. Даже по ролику видно , что в условии было упоминание , что решение должно быть в целых числах. В противном случае 169 можно разложить на множители чуть ли не бесконечным количеством пар.
From the thumbnail and the opening frame of the video, there's an easy, obvious answer: a = b = 26; a + b = 52. The problem becomes interesting only when a couple conditions are imposed, neither of which is stated in either of those 2 places: a ≠ b; a,b ∈ ℤ i.e., that a & b are unequal integers. Or, that the problem is to find all (positive) integer solutions. Or maybe we're just supposed to find all positive solutions. 1/a + 1/b = 1/13 Solve for b in terms of a: 1/b = 1/13 - 1/a b = 1/(1/13 - 1/a) = 13/(1 - 13/a) = 13a/(a - 13) which has infinitely many real positive solutions; for all of them, a & b are both > 13. Fred
Yeah, there’s sort of an assumed set of conditions that aren’t stated in the video: 1. a&b are both integers 2. The question is to find all solutions At the started, it’s stated that a & b are positive (>0) so i don’t count that one as a mistake. One of the written solutions is such that both values are 26, so a can clearly equal b.
I am really glad you posted this reply because I really thought I was stupid for not working the problem exactly like the video. I also chose to put b in terms of a at first as that is how I was taught to deal with multivariable equations. Anyway, this reminds me that it's OK to think outside the box and experiment with other methods so long as I maintain the relationships.
Anyone who repeats the calculation for case-02 instead of just invoking symmetry, or who repeats the lengthy calculations that turn a-13=13 into a=26 to find that b-13=13 leads to b=26, deserves some severe punishment…
If a and b are integers and a = b, the problem is trivial: 1/26 + 1/26 = 1/13, in the same way that 1/4 + 1/4 = 1/2, etc. Continuing with integers, use this formula for 1/a + 1/b = 1/c, which is 1/(a + 1) + 1/((a +1)(c)) = 1/c 1/(13 + 1) + 1/((13 + 1)(13)) = 1/13 1/14 + 1/182 = 1/13 (Obviously, the commutative property applies.) Try the formula on 1/a + 1/b = 1/5. I stumbled into this formula through empirical observation. A formal proof of the formula, assuming it exists, is beyond my math skills. If someone has a proof, please share it.
Nice! Yes in the trivial case, 1/c =1/c(1/2 + 1/2) = 1/2c + 1/2c so let a=b=2c. In the other two cases (both the same except a and b swap values like you mentioned), 1 = (c^2+c) / (c^2+c) ( since we know c > 0) = (c(c+1) / (c^2+c)). Then multiplying both sides by 1/c, 1/c = c+1 / (c^2+c) = (c / (c^2+c)) + (1 / (c^2+c)) = 1/(c+1) + 1/(c(c+1)). Since c is an integer, c+1 and c(c+1) are integers so let a = c+1 and b = c(c+1).
If you solve the system for a you get b= 13 a/(a-13). Now you just plug in values for a in f(a) = a+ 13a /(a-13) to find natural numbers for a+b . This system has of course infinitely many solutions .
You are correct that a+b=a+ 13a /(a-13). But if we are only working with positive integers for a and b then we need 13a to be divisible by a-13 to avoid fractions. I only see 2 solutions, a=14 and a=26.
This can be solved very fast in this way: 13(a+b)=a*b -->a=13b/(b-13) --> case 1) a is integer if b-13=1 => b=14 , a=182=>a+b=196 case 2) b-13=13k=>a=13(k+1)/k=>that is integer only with k=1 => a=26,b=26=>a+b=52
13 prime and if a and b integers then ab/13 implies either or both divisible by 13 so set A =n13 then 13nb/13=13n+b solve for b; b=13*(n/n-1) …n/(n-1) integer if n=2 A=2*13 =26 and from original equation b also 26 A+B=26 Also if n=14 then 13*(14/13)=14 so b=14 and a= 182 A+B=196
There is a simple general expression for this, was commonly asked in the exam CAT which i am preparing. A/x+B/y=1/C => (x-AC)(y-BC)=ABC². One can easily change the reciprocal form into factors.
13*(a+b) = a*b In generally, we can let a=13*n, where n is Nature Num 13*(13*n+b) = 13*n*b 13*n+b = n*b b = n*13/(n-1) is integer n = 2 or 14 a+b = 52 or 196
Which two numbers multiply to 13 and add to 1??? They do not exit as real numbers. They can not both be positive . They can not both be negative. So their product cannot be greater than zero. So no real pair of numbers can be a and b if a+b=1 AND ab=13
a + b = a b/13, ① a + 26 = 2a ⇒ 26 + 26 = 52 = a + b 👈 ② a + 39 = 3a ⇒ 19½ + 39 = 58½ ③ a + 52 = 4a ⇒ 17⅓ + 52 = 69⅓ ④ a + 65 = 5a ⇒ 16¼ + 65 = 81¼ ⑤ a + 78 = 6a ⇒ 15+3/5 + 78 = 93+3/5 a + 182 = 15a ⇒ 14 + 182 = 196 = a + b 👈 1/26 + 1/26 = 1/13 ⇒ a + b = 52 1/14 + 1/182 = 1/13 ⇒ a + b = 196 Should have started with 1/14 + 1/b = 1/13 👈
If a or b are not necesserly integers, the equation gives us : a=13b/(b-13) So a+b=b²/(b-13), so it's not constant and might take any value smaller than 0 or higher than 676...
If a,b is an integer , than you don't need any special procedure, you can take a= 14,15,16... anything then find b or vice-versa. if a.b is not an integer given in question then its a good approach.
1/a+1/b=1/13 we get 1/a=1/13-1/b = (b-13)/13b or a = 13b/(b-13) = 13+169/(b-13) for divisibility it's obvious that b-13=1,13,169 and get a =13+169, 13+13, 13+1 so a+b=13+1+13+169=196, 13+13+13+13=52, 13+169+13+1=196.
If 1/a +1/b = 1/3, what is a+b? Well, 1+1=2, so this is a simplified fraction, so that means we have to multiply 13 by 2 which we get 26. Which means that: 1/26 + 1/26= 2/26=1/13 Which means that a and b are both equal to 26, meaning that a+b=52
I first changed the equation in a + b = ab / 13 looked at it for some time asking what if a = b? Then you get 2a = a.a / 13 so 2 = a / 13 and a = b = 2 . 13 = 26 Check: 1/26 + 1/26 = 2/26 = 1/13 ! So the answer is: a + b = 52
Нельзя так приравнивать дроби. Вы теряете другие значения. Например: a/b = 2/5. По вашей логике a=2, b=5, но это не так a=2, 4,6,8.... b=5,10,15,20... Вот и в разобранном уравнении вы потеряли значения a=26 b=26.
A lot of work here. Explain why this isn’t so: 1/a+1/b=1/13, combining fractions b+a/ab=1/13. Numerator: a+b=1 Denominator: ab=13. Why isn’t a+b =1 as shown in numerator?
Thats not exact as a way if thinking because two fractions can be equal without necessitating any of their terms to be equal, ie 1/2 and 3/6. What is necessary for a/b=c/d is ad=bc and that is how you solve this
The given solution cheats, since it never said that a and b are integers. If a is given, then b=13a/(a-13) and both are positive for a>13. If you are searching for integral solutions, then from the equation b(a-13)=13a you get the solutions by using divisibility criteria.
Nice Math Olympiad Algebra Equation: 1/a + 1/b = 1/13, a, b > 0; a + b =? 1/a + 1/b = (a + b)/(ab) = 1/13, 13(a + b) = ab, 13 is a prime number 1/13 > 1/a ≥ 1/b > 0, b ≥ a or 1/13 > 1/b ≥ 1/a > 0, a ≥ b b = a; 13(a + b) = 26a = a², a² - 26a = a(a - 26) = 0, a > 0, a - 26 = 0; a = 26 = b a + b = 26 + 26 = 52 b > a; a, b must be multiple integers of 13; Let: b = 13a, a ϵ ℤ 13(a + 13a) = 13(14a) = ab = 13a², a(a - 14) = 0, a - 14 = 0; a = 14, b = 13(14) = 182 a + b = 14 + 182 = 196 a > b; Let: a = 13b, b ϵ ℤ 13(13b + b) = 13(14b) = ab = 13b², b(b - 14) = 0, b - 14 = 0; b = 14, a = 13(14) = 182 a + b = 182 + 14 = 196 Answer check: a = b = 26: 1/a + 1/b = 1/26 + 1/26 = 2/26 = 1/13; Confirmed a = 14, b = 182; a = 182, b = 14: a + b = 1/14 + 1/182 = 1/182 + 1/14 = (1/14)(14/13) = 1/13; Confirmed Final answer: a + b = 52 or a + b = 196
nothing say neither a or b must be a whole number .. this is one soultion .. 1/a+1/b = 1/13 -> (a+b)/ab = 13/13^2 -> a+b=13, ab=13^2 ... √a * √b = 13 .. infinite many solutions
Video takes too long watching you writing everything out, would be so much easier to either use your voice to describe the steps or have them appear as type
Hello ,everybody ! Greetings! Thanks , respects and considerations to all of you ...! Let me say that , here everything is just wonderful but uncomplete ( sorry to underline) !!! We have an infinity of solutions for this equation ( or equality ) . Let : a not null , b not null, k not null ,1/a+1/b=1/13 and a=k*b (or b= k*a) . Solving the system : 1/a+1/b=1/13 and a=k*b we get : 13(a+b)=a*b 13(k*b+b)=k*b*b 13(k+1)b- k*b*b=0 b(13*k+13-k*b)=0 and b not null 13*k+13-k*b=0 From here , let find k K*b-13*k=13 K=(13)/(b-13) Fron there: a =k*b=(13b)/( b-13) b not egal to 13 , b >0 and a>0 , that means that b belongs to the interval ] +13; + infinity [ Now for any b belonging to the given interval we have : a+b=b(k+1)=( b*b)/(b-13) Thanks ! Fron a former African and Senegalese Student in Moscow, Russia!
Tried to guess and plug things in, lol, didn't come up with anything...at any rate, solving, one gets 13(a+b) = ab, etc, which means one of a or b is a multiple of 13, etc (of course, lol, assuming they are integers, lol)...so just go through multiples of 13...try one after the other...a=13 doesn't itself work, lol, if I didn't make a mistake, then a=26, a=39, etc, hopefully isn't a millionth multiple of 13, lol, should ger there eventually (it could be negative too, lol, so one could go in the positive direction, then in the negative (well, one way to do that would be to try 13, then -13, lol, etc, for each positive a, try the negative counterpart))...didn't actually try this, so it might not be practical, lol...
From which statement did you conclude, that the solution should consist of natural numbers only? There is no such statement. But as there is 1/13 on the right side of the equation, it can be fairly be assumed, that the equation is an equation between rational numbers (or any other finite or infinite ring or field of numbers which include something like 1/13. And then you may have a different number of solutions or infinitely many solution. Why is this so important? Simply because this claims to be taken from the “Math Olympiad” and not the “Olympiad of Calculations”.
If this was a 10 point question, you would probably get 1 point for finding the 52, 2 points for getting the 196. All ten points if you determine some paramatized form of all solutions.
@@bobh6728 You can try to solve the equation for one set of numbers at a time only. 1.) Let’s try to solve 1/a+ 1/b = 1/13 in Z2 = {0,1} (Integers module 2): Lets start, trying to solve 1/a = x for a. The only candidates for a solution are 0 and 1 (That are the only elements in Z2). Assume a =0. Than (1/0)•0 = 1 = x•0 = 0. Contradiction.! Therefore a can’t be 0. And b can’t be 0 too by the same argument. Therefore, there is only a=b=1 left. But 1/1 + 1/1 = 2 = O = 1/13 = 1/1 = 1. Contradiction! Result: There is no solution of the equation 1/a + 1/b in Z2. 2.) Let’s try to solve 1/a + 1/b = 1/13 in Z3 = {O,1,2} (Integers module 3): In Z3 we have 1/13=1/(4•3+1)=1/1=1. So we have to solve 1/a+1/b=1. In Z3 we have 0+0=0, 0+1=1, 0+2=2, 1+1=2, 1+2=0 and 2+2=1. Case 1: 1/a+1/b = O+1 (or 1+0). Assume a=0, then there must be some x in Z3, so that (1/0)•0 = 1 = x•0 = 0. Contradiction! Hence a (and b) can’t be 0 Case 2: 1/a + 1/b = 2 +2 = 1. Then 1/a = 2 and 1/b =2 must hold, which is equivalent to (1/a)•a = 1 = 2•a. This equation is wrong for a=0 and a=1, but it holds for a=2 (and b=2 respectively. Result: There is exactly one solution in Z3, namely a=b=2. And, as we have seen in the video, there are 3 solutions in Z+ (positive Integers) (2 solutions being symmetrical) and, as we have seen in the comments, there are infinitely many solutions in Q (rational numbers). And so on. From the mathematical point of view, the problem was formulated incomplete, which should not be the case in an Olympiad. If the problem is formulated incomplete, it is up to the problem solver, to complete it in some suitable way. And if I complete it and solve it for example in Z2, the the right answer is “This equation has no solution”. And the assessment should be “10 points”. But I am afraid that “Zero points” will be imore likely. From the engineers point of view, an infinite number of solutions is most likely, as most engineers assume all numbers to be real numbers. But I can’t see any motivation to look for solutions in the positive integers only, as it is done in the video. It’s just arbitrary.
I dont think its mentioned anywhere that a and b belongs to integers. Only thing it states is that a,b>0 which means they are positive and therefore can exist as positive real nos. too. Sure, they are the only valid "whole" no. Solutions. But there are infinite solutions here. We cant solve this until the question states that a,b belongs to natural numbers
That is trick of youtuber, they intentionally forget it, so viewer will make comments, their video will have more interaction
@@sachuvan makes sense. It's their source of income after all
1/13 = 2/26 so 1/26 + 1/26 = 2/26 so a and b are both 26 so a+b = 52
You can't assume the values of a or b this way. Fractions are ratios. (1/13)=(2/26)=(4/52)...
Why did you also assume that a=b when it clearly doesn't state this?
@@roulam3001 it doesnt say anywhere that a and b cant be equal
@@roulam3001Obviously a+b=52 is a potential solution here and can be checked to be correct. However, as we saw in the video, it isn’t the only solution.
The idea of this question is there is not only one answer. With your way of solving this question, it is hard to find other answers.
You can’t also assume that both numbers are integer. There is no such claim in the beginning so the whole video solution is wrong 😎
1) a=b. 1/a+1/a=1/13. a=26. b=26. a+b=52. 2)1/a+1/b=1/13. 13(a+b)=ab. 13a=ab-13b. 13a=b(a-13). b=13a/(a-13). a-13=1. a=1+13=14. b=13a/1=13X14/1=182. a+b=196.
" b=13a/(a-13). a-13=1." Why a- 13= 1 ?
In fact, we have unlimited solution by make one variable as depend to another variable.
Lets A is independent variable, then B=f(A). In this case, we can choose B=n×A or another equation.
For exampke n=1 and B=1×A=A, then
1/A+1/B=1/A+1/A=1/13
2/A=1/13 or A=26, B=26
n=2, B=2×A=2A
1/A+1/B=1/A+1/2A=3/2A
3/2A=1/13 A=39/2, B=39
....
And so on..
@@sie_khoentjoeng4886 Nice Approach 👍
It has infinite solutions.
1/a + 1/b = 1/13
1/a = 1/13-1/b
a>0, =>1/a >0, => 1/13 - 1/b >0
=> b>13
||rly a > 13
a>13 and b >13
Hence take any value of a or b greater than 13,.put in the first equation and get the value of other variable
Я думаю в условии оригинальной задачи было упоминание, что корни должны быть целые, но автор ролика решил умолчать об этом факте.
@@БелАлексСогласен. Даже по ролику видно , что в условии было упоминание , что решение должно быть в целых числах. В противном случае 169 можно разложить на множители чуть ли не бесконечным количеством пар.
@@БелАлекса=1/(1/13-1/b), где b любое не равное 0 число, в том числе комплексное.
Количество решений бесконечно.
You can multiply both sides by 13ab and get 13b+13a=ab right away. Thank you for the video. Nice.
From the thumbnail and the opening frame of the video, there's an easy, obvious answer: a = b = 26; a + b = 52.
The problem becomes interesting only when a couple conditions are imposed, neither of which is stated in either of those 2 places:
a ≠ b; a,b ∈ ℤ
i.e., that a & b are unequal integers. Or, that the problem is to find all (positive) integer solutions.
Or maybe we're just supposed to find all positive solutions.
1/a + 1/b = 1/13
Solve for b in terms of a:
1/b = 1/13 - 1/a
b = 1/(1/13 - 1/a) = 13/(1 - 13/a) = 13a/(a - 13)
which has infinitely many real positive solutions; for all of them, a & b are both > 13.
Fred
Yeah, there’s sort of an assumed set of conditions that aren’t stated in the video:
1. a&b are both integers
2. The question is to find all solutions
At the started, it’s stated that a & b are positive (>0) so i don’t count that one as a mistake.
One of the written solutions is such that both values are 26, so a can clearly equal b.
I am really glad you posted this reply because I really thought I was stupid for not working the problem exactly like the video. I also chose to put b in terms of a at first as that is how I was taught to deal with multivariable equations. Anyway, this reminds me that it's OK to think outside the box and experiment with other methods so long as I maintain the relationships.
Anyone who repeats the calculation for case-02 instead of just invoking symmetry, or who repeats the lengthy calculations that turn a-13=13 into a=26 to find that b-13=13 leads to b=26, deserves some severe punishment…
Its because of folks like you that kids struggle with math.
Even without it, and played on double speed, it just drags on. Not fun drag in this case. 😁
If a and b are integers and a = b, the problem is trivial: 1/26 + 1/26 = 1/13, in the same way that 1/4 + 1/4 = 1/2, etc.
Continuing with integers, use this formula for 1/a + 1/b = 1/c, which is 1/(a + 1) + 1/((a +1)(c)) = 1/c
1/(13 + 1) + 1/((13 + 1)(13)) = 1/13
1/14 + 1/182 = 1/13
(Obviously, the commutative property applies.)
Try the formula on 1/a + 1/b = 1/5.
I stumbled into this formula through empirical observation. A formal proof of the formula, assuming it exists, is beyond my math skills.
If someone has a proof, please share it.
Nice! Yes in the trivial case, 1/c =1/c(1/2 + 1/2) = 1/2c + 1/2c so let a=b=2c. In the other two cases (both the same except a and b swap values like you mentioned), 1 = (c^2+c) / (c^2+c) ( since we know c > 0)
= (c(c+1) / (c^2+c)). Then multiplying both sides by 1/c, 1/c = c+1 / (c^2+c) = (c / (c^2+c)) + (1 / (c^2+c)) = 1/(c+1) + 1/(c(c+1)). Since c is an integer, c+1 and c(c+1) are integers so let a = c+1 and b = c(c+1).
@@dinnybam2057 Well done.
@@Zj-Aka thanks:)
a,b must be positive integers
There are only 2 solutions since 13 is prime number.
For real numbers there are many solutions.
Nice
Thanks 😊
Where does it say they must be whole numbers?
on the screen
nowhere
This is so painful to watch
If you solve the system for a you get b= 13 a/(a-13). Now you just plug in values for a in f(a) = a+ 13a /(a-13) to find
natural numbers for a+b . This system has of course infinitely many solutions .
Don't forget that a > 0 and b > 0.
You are correct that a+b=a+ 13a /(a-13). But if we are only working with positive integers for a and b then we need 13a to be divisible by a-13 to avoid fractions. I only see 2 solutions, a=14 and a=26.
This can be solved very fast in this way: 13(a+b)=a*b -->a=13b/(b-13) --> case 1) a is integer if b-13=1 => b=14 , a=182=>a+b=196
case 2) b-13=13k=>a=13(k+1)/k=>that is integer only with k=1 => a=26,b=26=>a+b=52
Very good. For case 1) you are using that b-13 is a factor of 13 => b=14.
13 prime and if a and b integers then ab/13 implies either or both divisible by 13 so set A =n13 then 13nb/13=13n+b solve for b; b=13*(n/n-1) …n/(n-1) integer if n=2 A=2*13 =26 and from original
equation b also 26 A+B=26
Also if n=14 then 13*(14/13)=14 so b=14 and a= 182 A+B=196
There is a simple general expression for this, was commonly asked in the exam CAT which i am preparing. A/x+B/y=1/C => (x-AC)(y-BC)=ABC². One can easily change the reciprocal form into factors.
Infinite solutions of a,b as a,b>0 and belongs to Real numbers
If a+b=S then,
For all 0
The equation Y^2-XY+13X=0, with every X equal or higher than 52, admits solutions Ymin(X)>0 and Ymax(X)>0 (with Ymin
let 1/a=x(just to check)
1/b=x
x+x=1/13
or,2x=1/13
or,x =1/26
1/a=1/b=1/26
a+b=26+26=52😅
La condición inicial no especifica que A y B sean enteros. Solo se estableció que sean positivos. Al no hacerlo, existen infinitas soluciones.
13*(a+b) = a*b
In generally, we can let a=13*n, where n is Nature Num
13*(13*n+b) = 13*n*b
13*n+b = n*b
b = n*13/(n-1) is integer
n = 2 or 14
a+b = 52 or 196
yeah thast what I I did as we can see eiher 13 divides b or 13 divdes a
Thank you professor for good solution.
You are welcome 🤗
Beautiful
Thanks a lot 😊
b = (n + square root from n to the second degree - 52n) : 2, n is 52 or any another higher number
S=b²/b-13 proof
1/a + 1/b = 1/13
Therefore
a=13b/b-13
And also
(a+b)=s= ab/13
= b*(13b/(b-13))/13
= b²/b-13
Define a valid domain observing a,b
Therefore b belongs to (0,13)....
I just understand
1/a+1/b=1/13 | -1/a
1/b=1/13-1/a=(a-13)/(13a) | 1/y
b=(13a)/(a-13)
a+b=
a+(13a)/(a-13)=
(a^2-13a)/(a-13)+(13a)/(a-13)=
a^2/(a-13)
Another and shorter solution 1/13=(1+13)/13(1+13)=(1/13×14)+ (13/13×14)=(1/182)+ (1/14). Therefore, a+b=196.
Great 👍
Система. а+b=1
ab=13
Which two numbers multiply to 13 and add to 1??? They do not exit as real numbers.
They can not both be positive . They can not both be negative. So their product cannot be greater than zero. So no real pair of numbers can be a and b if a+b=1 AND ab=13
Nice.....
Thanks☺️
a + b = a b/13,
① a + 26 = 2a ⇒ 26 + 26 = 52 = a + b 👈
② a + 39 = 3a ⇒ 19½ + 39 = 58½
③ a + 52 = 4a ⇒ 17⅓ + 52 = 69⅓
④ a + 65 = 5a ⇒ 16¼ + 65 = 81¼
⑤ a + 78 = 6a ⇒ 15+3/5 + 78 = 93+3/5
a + 182 = 15a ⇒ 14 + 182 = 196 = a + b 👈
1/26 + 1/26 = 1/13 ⇒ a + b = 52
1/14 + 1/182 = 1/13 ⇒ a + b = 196
Should have started with 1/14 + 1/b = 1/13 👈
a=12 and b=-153 is another solution finally a+b= -141
Brilliant approach SIR
Are you a teacher by profession 😊
If a or b are not necesserly integers, the equation gives us :
a=13b/(b-13)
So a+b=b²/(b-13), so it's not constant and might take any value smaller than 0 or higher than 676...
If a,b is an integer , than you don't need any special procedure, you can take a= 14,15,16... anything then find b or vice-versa. if a.b is not an integer given in question then its a good approach.
1/a + 1/b = 1/13
ab× (1/a + 1/b = 1/13)
b + a = ab/13
= (1/13)×(a)(b)
Good work ❤
As great , thank you , we enjoyed it .
Thanks a lot 😊
1/a+1/b=1/13 we get 1/a=1/13-1/b = (b-13)/13b or a = 13b/(b-13) = 13+169/(b-13) for divisibility it's obvious that b-13=1,13,169 and get a =13+169, 13+13, 13+1 so a+b=13+1+13+169=196, 13+13+13+13=52, 13+169+13+1=196.
13(a+b)=ab, (a+b)>= 2✔️ab
(a+b)²/4>=ab =>> (a+b)²/4>=13(a+b) => (a+b)/4 -13=0 => a+b=13x4=52
sẽc a=-b
x/13+y/13=1/a+1/b x+y=1 x/13=......
1/n=1/n+1+1/n(n+1) 1/13=1/13+1+1/13(13+1) 1/13=1/14+1/182 14+182=196
Nice Approach 👍
Multiplying by 13ab, you can rewrite as (a - 13)(b - 13) = 169 - two solutions: a=b=26, a=14, b=182
Опять раскладываем только на целые множители. Где это оговорено?
Нигде. Восточная хитрость.
Two variables in one equation can not be solved separately.🎉🎉
Okay Boss 🙂
Kya ye Olympiad ka question hai bhai. Such an interesting question
Pta nhi bhai🫣 Olympiad ha yaa ni.
But yes its interesting 🙂
If 1/a +1/b = 1/3, what is a+b?
Well, 1+1=2, so this is a simplified fraction, so that means we have to multiply 13 by 2 which we get 26.
Which means that: 1/26 + 1/26= 2/26=1/13
Which means that a and b are both equal to 26, meaning that a+b=52
a+b=1
Why u r doing complicated things, it can be solved by just simple observation... We can clearly see That a=b=26, 1/26+1/26=1/13. There fore, a+b=52 👍
👎🏽 Clearly that is not the only answer.
@bobh6728 then give more answers👍
@@bobh6728 Obviously, the video doesn't have all the answers either, just a few random ones.
Actually, the video is missing infinitely many answers.
@@adarshkar4529 196, from a, b being 32 and 164.
You can do it in your head. W.L.O.G you can set a = b and sould get the same answer. Then you can get a = b = 26.
Alright Boss 😊
Bro really knows how to make a clickbait out of easy problems 🤣
0.5/13+0.5/13=1/a+1/b
A rather guessworking without a real generic value
I first changed the equation in a + b = ab / 13 looked at it for some time asking what if a = b? Then you get
2a = a.a / 13 so 2 = a / 13 and a = b = 2 . 13 = 26 Check: 1/26 + 1/26 = 2/26 = 1/13 !
So the answer is: a + b = 52
in about 5 seconds 1/26 + 1/26 = 2/26 = 1/13. Fast forward 10 mins in the video oh that was correct.
How did that take 10 mins ?
easiest solution: a = 13, b = infinity
@@clympsarchery that is only the case for the limit as x approaches infinity of 1/x. 1/infinity is undefined.
Wrong
@@aqlimursadin5948 nope. 1/infinity = 0. 0 + 1/13 = 1/13 duh
@@dinnybam2057 its undefined so i define it as 0 🤣🤣
I got 52
1/a + 1/b = 1/13
1/a = 1/13 - 1/b
1/a = (b-13) / 13b
then
1 = b-13 and. a = 13b
b = 14 , a = 182
Нельзя так приравнивать дроби. Вы теряете другие значения. Например: a/b = 2/5. По вашей логике a=2, b=5, но это не так a=2, 4,6,8.... b=5,10,15,20...
Вот и в разобранном уравнении вы потеряли значения a=26 b=26.
You could say it was diophantine equation
n=ab，a+b≡1，1是一个示性数，也即：a+b可以等于任何数N，与n等于多少无关。
In Korea, indefinite equation learned in the first year of high school. Easy~😊
Alright Boss 😊
My first thought was a+b = (a*b)/13 and that's it
A lot of work here. Explain why this isn’t so: 1/a+1/b=1/13, combining fractions b+a/ab=1/13. Numerator: a+b=1 Denominator: ab=13. Why isn’t a+b =1 as shown in numerator?
Thats not exact as a way if thinking because two fractions can be equal without necessitating any of their terms to be equal, ie 1/2 and 3/6. What is necessary for a/b=c/d is ad=bc and that is how you solve this
@ Thanks. Makes sense.
1/a + 1/b = 1/13
a + b = ab/13
a = b = 26
How to know to add 169
To make perfect factors 😉
The given solution cheats, since it never said that a and b are integers. If a is given, then b=13a/(a-13) and both are positive for a>13. If you are searching for integral solutions, then from the equation b(a-13)=13a you get the solutions by using divisibility criteria.
Alright Boss 😊
Nice Math Olympiad Algebra Equation: 1/a + 1/b = 1/13, a, b > 0; a + b =?
1/a + 1/b = (a + b)/(ab) = 1/13, 13(a + b) = ab, 13 is a prime number
1/13 > 1/a ≥ 1/b > 0, b ≥ a or 1/13 > 1/b ≥ 1/a > 0, a ≥ b
b = a; 13(a + b) = 26a = a², a² - 26a = a(a - 26) = 0, a > 0, a - 26 = 0; a = 26 = b
a + b = 26 + 26 = 52
b > a; a, b must be multiple integers of 13; Let: b = 13a, a ϵ ℤ
13(a + 13a) = 13(14a) = ab = 13a², a(a - 14) = 0, a - 14 = 0; a = 14, b = 13(14) = 182
a + b = 14 + 182 = 196
a > b; Let: a = 13b, b ϵ ℤ
13(13b + b) = 13(14b) = ab = 13b², b(b - 14) = 0, b - 14 = 0; b = 14, a = 13(14) = 182
a + b = 182 + 14 = 196
Answer check:
a = b = 26: 1/a + 1/b = 1/26 + 1/26 = 2/26 = 1/13; Confirmed
a = 14, b = 182; a = 182, b = 14:
a + b = 1/14 + 1/182 = 1/182 + 1/14 = (1/14)(14/13) = 1/13; Confirmed
Final answer:
a + b = 52 or a + b = 196
nothing say neither a or b must be a whole number .. this is one soultion ..
1/a+1/b = 1/13 ->
(a+b)/ab = 13/13^2 ->
a+b=13, ab=13^2 ... √a * √b = 13 .. infinite many solutions
It has infinite solutions
Fool
Take a = 39 and b = 39/2
a+b = 58.5
Пока лучший вариант решения задачи.
Why do you not consider a factorization into (-1)x(-169) or (-13)x(-13) ? Yes, this will lead to solutions with ab
a=b 1/a+1/a=2/a=1/13=2/26 a=26 a+b=52
😮
분모통일시키면 ab=13 a+b=1바로나오자나
a=1-b
b(1-b)=13
-b^2+b-13=0
이차방정식풀어서 b값 뽑으면 끝아님?
The problem statement should state that a and b are natural numbers, not a>0, b>0. Currently, a=15, b=97.5, a+b=112.5 are also possible answers.
This is just one example, you can get an infinite number of answers.
I don't know English, I translated it with Google, if there are any mistakes in the sentences, please forgive me.
13
If a+b =13 and ab =169 , there exists no pair (a,b) for which 1/a +1/b = 1/13
Video takes too long watching you writing everything out, would be so much easier to either use your voice to describe the steps or have them appear as type
Noted 😊
a=1 ; b=(- 12)/13
1+(-12)/13= 1/13
Məntiqlə (loqiq) götürəndə məxrəc 0< 1/a=1/b=1/13.
Must set condition for a and b real or whole number
Sol :
1/a+1/b=1/13
(1-1/2+1/2) 1/13
--> (1/2/+1/2) 1/13
--> 1/26+1/26
a=26, b=26
a+b=52
Ans：a+b=52
for reference only
1/14. 1/(13*14)
Hello ,everybody ! Greetings!
Thanks , respects and considerations to all of you ...!
Let me say that , here everything is just wonderful but uncomplete ( sorry to underline) !!!
We have an infinity of solutions for this equation ( or equality ) .
Let : a not null , b not null, k not null ,1/a+1/b=1/13 and a=k*b (or b= k*a) .
Solving the system : 1/a+1/b=1/13 and a=k*b we get :
13(a+b)=a*b
13(k*b+b)=k*b*b
13(k+1)b- k*b*b=0
b(13*k+13-k*b)=0 and b not null
13*k+13-k*b=0
From here , let find k
K*b-13*k=13
K=(13)/(b-13)
Fron there: a =k*b=(13b)/(
b-13)
b not egal to 13 , b >0 and a>0 , that means that b belongs to the interval ] +13; + infinity [
Now for any b belonging to the given interval we have :
a+b=b(k+1)=( b*b)/(b-13)
Thanks !
Fron a former African and Senegalese Student in Moscow, Russia!
Great effort and brilliant approach Boss 🙂
Thanks for sharing your precious feedback 😌
1/14+1/（14×13）=1/13
則14+182=196
Why will this not work? Express with an LCD: (a+b)/ab=1/13 => by inspection, a+b=1.
Alright Boss 😊
13+13=13*13/13 => 26 = 13 => 26 = (50% of result) => 52 = (100% of result)
a + b = 26 Cuz a=b=13
I think u forgot to state that a and b are natural numbers otherwise you can add more cases in 5:06 for example: 169=338×(1/2)
Multiple value of and b
He tried to introduce an "indeterminate Equation" problem. But he forgot to provide the conditions - a, b are integers.
Why do you assume that a and b are integers?
You should state that as part of the problem.
(1/a)+(1/b)=1/13
(a+b)/ab=1/13
[k(a+b)]]/kab=1/13; k≠0
k(a+b)=1
kab=13
when k=1,
a+b=1 =>b=1-a
ab=13
a(1-a)=13
a-a²=13
a²-a+13=0
♤=-51
♤½=i(51½)
a = (1/2)+[±i(51½)/2]
b = (1/2) -[±i(51½)/2]
ka= kr + ks*z
when k=1
r=(1/2)
s=i(51½)/2
the same for b:
kb= ku - kv*z
when, k=1
u=(1/2)
v=-i(51½)/2
u=r =1/2
s=-v = i(51½)/2
=>a= ku+kv*z
when, k=1
a+b=1
(u+vz)+(u-vz)=1
2u=1
2(1/2)=1
1=1. ok
ab=13
(u+vz)(u-vz)=13
u²-(vz)²=13;
u²=(1/2)²=(1/4)
(vz)²= v²z²= v²(-1)= -v²= -[(51½)/2]²
=> (vz)²= -51/4
ab=13
(u+vz)(u-vz)=13
u²-(vz)²=13
(1/4) -(-51/4)=13
(1/4)+(51/4)=13
(1+51)/4=13
52/4=13
26/2=13
13=13. ok
a=ku + kvz
b=ku - kvz
then,
possible solutions are:
(with k € C, k ≠ 0 )
a=k/2 + ki√(51/4)
b=k/2 - ki√(51/4)
The most obvious answer is 52 just by looking.
Много лишних записей.и тупой перебор вариантов в конце.
Tried to guess and plug things in, lol, didn't come up with anything...at any rate, solving, one gets 13(a+b) = ab, etc, which means one of a or b is a multiple of 13, etc (of course, lol, assuming they are integers, lol)...so just go through multiples of 13...try one after the other...a=13 doesn't itself work, lol, if I didn't make a mistake, then a=26, a=39, etc, hopefully isn't a millionth multiple of 13, lol, should ger there eventually (it could be negative too, lol, so one could go in the positive direction, then in the negative (well, one way to do that would be to try 13, then -13, lol, etc, for each positive a, try the negative counterpart))...didn't actually try this, so it might not be practical, lol...
th-cam.com/video/JF_cp-izoTQ/w-d-xo.htmlsi=kNhtLS9oObQdBEY1
it is equivalent to ask: x = y/13, x>0, y>0, then x=?
If you didn’t specify a must be larger than 0… or b must be larger than 0… there are infinitely many answers…
th-cam.com/video/JF_cp-izoTQ/w-d-xo.htmlsi=kNhtLS9oObQdBEY1
👍
I see such type of problems of fitting LHS to a fraction of RHS. Repeat it and match. No exact theory.
Alright Boss 😊
1/2+1/2=1
1×2+1×2/2×2=4/4=1
1/13=
1*X+1*Y=(1*X)+(1*Y)/X*Y=
(1*X)+(1*Y)/13,
X+Y/X*Y=
1/13
2 1= 3/2 1.5
2 2= 4/4 1
2 3= 5/6
2 4= 6/8 3/4
2 5= 7/10
2 6 = 8/12 2/3
2 7= 9/14
10/16 5/6
11/18
12/20 6/10 3/5
13/22
14/24 7/12
15/26
16/28 8/14 4/7
.a+b/a×b=1/13.
I know answer but what you mean my answer for you.
Try solve it yourself
But de l'équation : gagner de l'argent avec la publicité.
Okay Boss... The thinking of yours is the best
From which statement did you conclude, that the solution should consist of natural numbers only? There is no such statement. But as there is 1/13 on the right side of the equation, it can be fairly be assumed, that the equation is an equation between rational numbers (or any other finite or infinite ring or field of numbers which include something like 1/13. And then you may have a different number of solutions or infinitely many solution.
Why is this so important?
Simply because this claims to be taken from the “Math Olympiad” and not the “Olympiad of Calculations”.
If this was a 10 point question, you would probably get 1 point for finding the 52, 2 points for getting the 196. All ten points if you determine some paramatized form of all solutions.
@@bobh6728 You can try to solve the equation for one set of numbers at a time only.
1.) Let’s try to solve 1/a+ 1/b = 1/13 in Z2 = {0,1} (Integers module 2):
Lets start, trying to solve 1/a = x for a. The only candidates for a solution are 0 and 1 (That are the only elements in Z2).
Assume a =0. Than (1/0)•0 = 1 = x•0 = 0. Contradiction.! Therefore a can’t be 0. And b can’t be 0 too by the same argument.
Therefore, there is only a=b=1 left. But 1/1 + 1/1 = 2 = O = 1/13 = 1/1 = 1. Contradiction!
Result: There is no solution of the equation 1/a + 1/b in Z2.
2.) Let’s try to solve 1/a + 1/b = 1/13 in Z3 = {O,1,2} (Integers module 3):
In Z3 we have 1/13=1/(4•3+1)=1/1=1. So we have to solve 1/a+1/b=1.
In Z3 we have 0+0=0, 0+1=1, 0+2=2, 1+1=2, 1+2=0 and 2+2=1.
Case 1: 1/a+1/b = O+1 (or 1+0). Assume a=0, then there must be some x in Z3, so that (1/0)•0 = 1 = x•0 = 0. Contradiction! Hence a (and b) can’t be 0
Case 2: 1/a + 1/b = 2 +2 = 1. Then 1/a = 2 and 1/b =2 must hold, which is equivalent to (1/a)•a = 1 = 2•a. This equation is wrong for a=0 and a=1, but it holds for a=2 (and b=2 respectively.
Result: There is exactly one solution in Z3, namely a=b=2.
And, as we have seen in the video, there are 3 solutions in Z+ (positive Integers) (2 solutions being symmetrical) and, as we have seen in the comments, there are infinitely many solutions in Q (rational numbers).
And so on.
From the mathematical point of view, the problem was formulated incomplete, which should not be the case in an Olympiad.
If the problem is formulated incomplete, it is up to the problem solver, to complete it in some suitable way. And if I complete it and solve it for example in Z2, the the right answer is “This equation has no solution”. And the assessment should be “10 points”. But I am afraid that “Zero points” will be imore likely.
From the engineers point of view, an infinite number of solutions is most likely, as most engineers assume all numbers to be real numbers.
But I can’t see any motivation to look for solutions in the positive integers only, as it is done in the video. It’s just arbitrary.
@@bobh6728 13b + 13a = ab
b = 13a / (a - 13)
a+b = a + 13a/(a-13) = a^2 / (a - 13)
Not really worth 10 marks.
a = b = 26 works
1/a + 1/b = 1/13
1/a = 1/13 - 1/b
1/a = (b-13) / 13b
then
1 = b-13 and. a = 13b
b = 14 , a = 182
No integer condition answer infinite
th-cam.com/video/JF_cp-izoTQ/w-d-xo.htmlsi=kNhtLS9oObQdBEY1
Clickbait