Russian l can you solve this exponential problem?? l Olympiad Mathematics
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- Опубликовано: 5 фев 2025
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27+8=35 3^3+2^3=35 n=3
Intuitively, ask, what two perfect cubes, when added up, equal 35. The answer is 27 and 8. 27 is 3^3 and 8 is 2^3. X=3. Don't overthink things.
I agree. Since the problem restricts answers to positive integers, just starting with 2 and checking each possibility is way quicker then the long method shown in this video.
Plus, in order to know to multiple n by 3/3 at 0:30 , you have to already know the n = 3.
you have to still prove thats the only solution otherwise its a big fat 0 in a descriptive test
@@aisolutionsindia7138 :You are right. 3 is here an "evident solution" but Is it unique? It is not a big deal to demonstrate that. Knowing that the square function is increasing for natural numbers we just need to verify that the result is >35 for n=4 and
Yep lol... Figured it out in my head lol
Tenkyou very much
Amazing sir
I don´t agree completely with the assumption that (X+Y)*(X^2-YX+Y^2) = 1*35. Why not (X+Y)*(X^2-YX+Y^2) = 5*7 or (X+Y)*(X^2-YX+Y^2) = 7*5, for example. We can not force a result for an equation!!!
I have to agree with people saying that the solution is much simpler by inspection. Since n is a positive integer (as stated in the question (Z+)), it’s quite simple to verify that 3 solves the problem, and it is unique for positive integers.
I have been to a Russian Orthodox wedding, complete with high mass, in traditional Russian Liturgy, accompanied by a full choir and orchestra playing culturally appropriate antique instruments, and, of course, all in the wedding party were wearing ethnically perfect, hand-embroidered, 18th century regalia complete with jewelry of the period, and it was less complicated than his solution to this equation! 🙀😹
Why it somehow related to maths videos, has no word about ruZzian at all
3^3+2^3=27+8=35 so n=3 just by looking at it without needing any complicated calculation
Zo‘r
n ∈ Z+ => n35, test n=1,2,3 =>n=3
It’s crap like this that turns young students against learning proper mathematics! 😾
Grrst video
Thanks to you dear brother, and, always, add a beautiful song so that things become easy !
Achei fantástico!
Three, by inspection.
Still, this puzzle and its RUclips have a valuable lesson for it: some fools are easily distracted, perhaps carrying with them the thought that they are doing something rarified or worthwhile as they go.
The spelling gadget wants me to spell it with an e, "rarefied," which strikes me as inane. Checking with Google, I'm a bit surprised to find that Oxford want the idiotic "rarefied," which would obviously be the nonexistent word pronounced rair-fide, while the often shoddy Merriam-Webster crew agree with me on "rarified."
Much quicker to use logs. Done in 3 lines.
yes read my reply
Very good!
n2.
If n is a positive integer, n=3.
I thought that you could simply take logs both sides so it becomes n log 3 + n log 2 = log 35 which simply becomes n(log 3+log 2)= log 35: thus n =log 35/((log 3+ log2) but what is (nez+)??? that I don't understand what that is all about!
But he log of 3^n + 2^n is *NOT* n log 3 + n log 2
Решается в уме за две минуты! Подбором чисел , 2 - не подошло, а три - получилось ! Ура 👍
А где решение, что данный ответ единственное решение?
Это не математика, получается, а угадывание.
That’s exactly how I did it! In two seconds! 😺
What does the video title mean? I know each word, but they make no sense together
How long did that take? I did it in my head instantly and fast-forwarded to the end of the video to check if my solution was correct. It was, and I'm no mathematician.
3
@@DavisRimu to be accurate, you need to show there's no other solutions, but n=3
@@lukaskamin755 OK, but in a practical situation, if there is one for that equation, if 3 meets the need...I'm no mathematician!
@@DavisRimu of course, my opinion this whole solution is meaningless, it's not quite accurate and it's all based on assumption that we already guessed the solution, but it's actually much easier to show by inspection,as n can't be greater than 3.
Бравооо
Hello, i try to resolve 3^n+2^n=13, here n=2 but cant find algebraic solution, can you get the solution.
If you also know that n needs to be an integer, use that d(3^n+2^n)/dn > 0 for all values of n, and thus, m > n 3^m+2^m > 3^n+2^n
And then just go scanning by using that as an additional guidance, as well as the proof that if you find a solution that that is the only possible one.
How I'd scan:
3^n > 2^n for n > 1, so don't start at an n > log(13)/log(3)
3*(3^2=9)>13 i.e 3^2=9 is the biggest integer power of 3 smaller than 13 => let's try n=2; 3^2+2^2=9+4=13
Trivial. By inspection, n=3
Why unnecessary extra sound ?
Why not:
3^n + 2^n = 27 + 8
3^n - 27 + 2^n- 8 = 0
3^n - 3^3 + 2^n - 2^3 = 0
n - 3 + n - 3 = 0
2n = 6
n = 3
Because it's wrong, you receive the correct answer occasionally , you cannot simply put down the exponents with no regards to the base value, which is different here 2 and 3. You have two zeroes and they sure add up to zero, but why are you sure there're no other ways to receive zero in that exponential equation, that you came to?
@@lukaskamin755 But aren't the bases the same? One difference is 3 with 3. And the other difference is 2 with 2.
@stevesilva2780 a+b=0, doesn't imply a=0 and b=0. It would be true only if both a and b =>0, which is not clear. Anyhow we cannot take exponential equation with various bases and simply bring those numbers down, that's not a valid conversion
Am I only one, bothered with x, y defined thorough fractional powers, while assuming x to be integer? It's actually true only for dividends of 3. And as many mentioned when equating a product only limited amount of cases are regarded, there's more than 2 of them.
What of the right side is a number way bigger than 35? Still making every assumption?
Dlm memecahkan masalah tentu dicari yg termudah.
Untuk kasus soal ini metode menebak nilai dg mengujinya tentu lebih singkat.
Matematika adl alat bantu untuk memudahkan solusi masalah bukannya mempersulit..!!!
Нетрудно заметить ,что значение 3 удовлетворяет уравнению.Рассмотреть случаи,когда n больше 3 и когда n меньше 3 и на основе сложения неравенств с одинаковыми знаками заключаем,что лнвая часть иои больше правой или меньше правой.И все подобные неравенства легко решаются.
3*n+3*n>35. N>=3, 3*n
this music is extremely annoying
Assume n is an integer. A reasonable guess because the question is _obviously contrived_ using an integer n.
Then:
Try n = 1. 3 + 2 < 35.
Try n = 2. 9 + 4 < 35.
Try n = 3. 27 + 8 = 35. Bingo!
Why all the nonsensical messing about with algebra for _15 minutes_ ??
Suppose n>3 so 3^n + 2^n > 35 witch is false so n=< 3 from this is easy to find that n=3
It makes absolutely no sense assuming that n will be divisible by 3. This is a wild guess. Unless you already know! Right?
Then you make further guesses like the 1 * 35...
Lengthy process..
n = 3.
27 + 8 = 35.
😊 Try and error with 1, 2 and 3. It can be solved in 20 seconds.
Yeah, but that's missing the point. Just about anyone can guess a trivial case.
Fifteen minutes to explain a solution that took me under 5 seconds of mental math!?
agreed
Good folk music
Unfathomable to most people except maths professors.
You do not need to be a maths professor: I disagree.
onel look at the equation instantly gives the answer as 3 there is no need for an elaborate the steps to arrive at the solution
There is a need.... think about it.
Olympiad Math? Sure?
Who said that x and y must be integers ?
Устно можно решать 3
Why is 1 = 3/3? Why not n/n? Because you wanted it to be so? Or did you suppose that n maybe is equal 3? How? Then you could just suppose that 3^3 + 2^3 = 35. Why did you suppose that 1 is equal to 3/3? Huh?
2^n≤3^n; y
n=3
3^3+2^3=27+8=35
Could simply be solved by taking log of both sides.
And how are you going to simplify log(3^n+2^n)=log(35) into something like n=... ?
@@Apollorion: n log 3 + n log 2 = log 35 which simply becomes n(log 3+log 2)= log 35: thus n =log 35/((log 3+ log2)
@ But log a + log b isn't log (a+b), but log (ab) instead.
why 3d power, why not 2d ?
1 плюс 1 не равно 35
3 плюс 2 не равно 35
9 плюс 4 не равно 35
27 плюс 8 равно 35
Ответ n=3
Completely unnecessary solution. This kind of problem is known to be solved by summing powers; 3^3 +2^3=35, n=3.
"solved by summing powers" Is that higher math? Sometimes college-level algebra is needed.
Solved by inspection. Sometimes « lower math » is faster
You have to get it practically
@@MustefaKaso yes, an analytic solution is desirable
N=3
А если 3^n+2^n=35,1?
Change the music! Just too much.
зачем раскладывать условия и ответ на а и б если после стрелки слева ответ уже гатов нада только падставить 2 и три с условия
нам нада наити н ен ето степень выже написали что степень ето 3 или вас ваше собственое решение неустраивает вот как вы себя ненавидете
вместо формулы вы бы лучше написали с чего вы взяли что нада 1/3 в на три а не одна 4 ть на четыре еслиб вы ето указали вы поняли что пример прастои в катором ненада некакие формулы я вабше считаю что формулы вабше ненужны каждый все может решить сам а ети формулы вранье что ктото за вас знает как решать
вам некажетса страным в математике говорят мы знаем как вы правильно палучите ответы на задачи и алегархи говорят мы знаем как палучить ответ за вас вы сами верите что ответ может быть сложным а после етих формул так и становитса но сама фраза решать значит что с заданаи задачи выдолжны палучить прастои ответ чего непалучитса па формуле я был на екзамене па математике пака я неначел работать не кто са всеи паралели несмог неодну формулу вспомнить вот и решаите нада ли учитса ими пользаватса если их невспомнить самастаятельно не как
а когда вспомнили формулы начелся бардак примерно как в украине тагда все стали что то делать ну как в расии стало но ето всеже бардак из за которого ведутса воины тоесть формула неодна непамагла решить не одну задачу некаму заработать денег вазможно но непалучить ответ на решение
я думаю нада просто троику как большее число вазвадить встепнь когда ответ станет немного меньше тритсати пяти можно начать расписвать пример рас учителя так любят расписавать но я перевым делам паспешил бы папробвать вазвести и двоику в туже стпеньно ето патаму что я спешу решать интересно ето работа можно и паписать па бумаге
7
Ответ за одну секунду n=3
3^3+2^3=27 +8=35
This unnecessarily complicated resolution risks disgusting students. Why don't you show a simpler way to solve this equation? It could be done in few rows:
3 is here an "evident solution" but Is it unique? It is not a big deal to demonstrate that. Knowing that the square function is increasing for natural numbers we just need to verify that the result is >35 for n=4 and
There is no square function in 3^n+2^n=35 if n isn't even.
On the other hand though, 3^n and 2^n are both exponential functions with a base grater than 1, and thus indeed always increasing for all real numbers, which includes the natural numbers. If you wish to prove that 3^n+2^n is always increasing just differentiate it and check that for all it's domain the derivative is positive.
( y = 3^x + 2^x => dy/dx = (3^x)ln(3) + (2^x)ln(2) > 0 for all real values of x .)
35= 27+ 8= 3³+ 2³ n=3 ?
Correct, and because 3^x, 2^x as well as their sum 3^x+2^x are continuously increasing functions, is that solution the only one.
Good
Too many unessential steps. You can solve it in 5 steps or less
3
What's happening here
>> 3
Ок
N =2 is wrong
Phương Pháp không thuyết phục
n is 3 hhhhh
無聊
n=3
n=7
3
N=3
n=7
3
3
n=3
3
n=7
3
n=3
3
N=3
n=3
n= 3