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
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.
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
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.
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.
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).
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
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
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
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
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
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…
What is the value of an And b? How can it be an equation if doesn’t even describe anything meaningfully. And it can’t solve for anything meaningful if you don’t know the meaning of the variables. How do you even know the sum then.
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.
Я эту задачу в уме решил за 10 секунд. Нам не важно знать сколько будет по отдельности A и B. Нам нужно знать их сумму. Для простоты будем считать что A=B. Тогда A=B=(13×2)=26. (Также как и 1/2=1/4+1/4). Поэтому А+В=26×2=52.
@КириллХ-з4в Ты не внимательно читал моё решение. Мне без разницы какие будут а и б. Хоть дробные. Для простоты решения я взял их равными. Ибо их сумма константа.
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 😎
You can multiply both sides by 13ab and get 13b+13a=ab right away. Thank you for the video. Nice.
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.
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.
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
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 ?
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.
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
Система. а+b=1
ab=13
La condición inicial no especifica que A y B sean enteros. Solo se estableció que sean positivos. Al no hacerlo, existen infinitas soluciones.
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)
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.
a,b must be positive integers
There are only 2 solutions since 13 is prime number.
For real numbers there are many solutions.
I just understand
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:)
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.
Good work ❤
1/a + 1/b = 1/13
ab× (1/a + 1/b = 1/13)
b + a = ab/13
= (1/13)×(a)(b)
Music's good
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.
Where does it say they must be whole numbers?
on the screen
nowhere
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
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
Опять раскладываем только на целые множители. Где это оговорено?
Multiplying by 13ab, you can rewrite as (a - 13)(b - 13) = 169 - two solutions: a=b=26, a=14, b=182
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
13
Why do you not consider a factorization into (-1)x(-169) or (-13)x(-13) ? Yes, this will lead to solutions with ab
The answer could be a=39, b=39/2, but there are infinitely many solutions
13+13=13*13/13 => 26 = 13 => 26 = (50% of result) => 52 = (100% of result)
Много лишних записей.и тупой перебор вариантов в конце.
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 ?
My first thought was a+b = (a*b)/13 and that's it
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.
Multiple value of and b
a=b 1/a+1/a=2/a=1/13=2/26 a=26 a+b=52
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.
easiest solution: a = 13, b = infinity
@@mathy-mathy-maths 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 🤣🤣
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…
😮
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
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
The most obvious answer is 52 just by looking.
Bro really knows how to make a clickbait out of easy problems 🤣
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.
интересен только конец с множителями
If you didn’t specify a must be larger than 0… or b must be larger than 0… there are infinitely many answers…
ua-cam.com/video/JF_cp-izoTQ/v-deo.htmlsi=kNhtLS9oObQdBEY1
(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)
👍
What is the value of an And b? How can it be an equation if doesn’t even describe anything meaningfully. And it can’t solve for anything meaningful if you don’t know the meaning of the variables. How do you even know the sum then.
a=26,b=26 puff~
You could say it was diophantine equation
No integer condition answer infinite
ua-cam.com/video/JF_cp-izoTQ/v-deo.htmlsi=kNhtLS9oObQdBEY1
😲
Is this completing the square method? I think the steps are incorrect tho
Why do you assume that a and b are integers?
You should state that as part of the problem.
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...
ua-cam.com/video/JF_cp-izoTQ/v-deo.htmlsi=kNhtLS9oObQdBEY1
4:53 ??, please
Doesn't this solution imply that we're looking for natural solutions?
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.
And why not (a-13)(b-13)=13*13? There was no a≠b, so 1/26+1/26=1/13, so a+b=52
Я эту задачу в уме решил за 10 секунд. Нам не важно знать сколько будет по отдельности A и B. Нам нужно знать их сумму. Для простоты будем считать что A=B. Тогда A=B=(13×2)=26. (Также как и 1/2=1/4+1/4). Поэтому А+В=26×2=52.
вывод - решил неверно, ибо в условии абсолютно нигде не фигурирует что A и B - натуральные
@КириллХ-з4в Ты не внимательно читал моё решение. Мне без разницы какие будут а и б. Хоть дробные. Для простоты решения я взял их равными. Ибо их сумма константа.
@@АлександрМихин-д4ы дадада, решай и дальше матешу подбирая одно решение и полностью игнорируя что их еще может какое то количество
but you did had to clickbait poeple Einstein did had any business with this
6 case is true:
(1, 169); (13, 13); (-1, -169); (-13, -13); (169, 1); (-169, -1)
solve it again! you have made mistake!
Far out. 10 minutes to solve something that takes 2 seconds?
14+182=196
182+14=196
26+26=52
太牛逼了！
a=b=26 so a+b = 52. By inspection.
А ПРИ ЧЁМ ТУТ эЙНШТЕЙН?
This is so painful to watch
Really painful to watch
Clickbait
Where does 169 come from ?
😮