Russian l can you solve this exponential problem?? l Olympiad Mathematics

[@CBordages]

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.

[@tedr.5978]

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.

[@tedr.5978]

Plus, in order to know to multiple n by 3/3 at 0:30 , you have to already know the n = 3.

[@aisolutionsindia7138]

you have to still prove thats the only solution otherwise its a big fat 0 in a descriptive test

[@gillesdelbreil5414]

@@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

[@charlziedouglas-mo7uc]

Yep lol... Figured it out in my head lol

[@ΒασιληςΑρετακης-μ2ψ]

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.

[@lidiaak]

Решается в уме за две минуты! Подбором чисел , 2 - не подошло, а три - получилось ! Ура 👍

[@BrunoRodriguezMushanya]

Thanks to you dear brother, and, always, add a beautiful song so that things become easy !

[@mister_123]

Good folk music

[@custodiolemoslemos2145]

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!!!

[@eng.md.helaluddin5849]

Why unnecessary extra sound ?

[@pietergeerkens6324]

Fifteen minutes to explain a solution that took me under 5 seconds of mental math!?

[@ryanchiang9587]

n = 3. 27 + 8 = 35.

[@xusky9802]

3*n+3*n>35. N>=3, 3*n

[@marcociot]

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...

[@TheDavidlloydjones]

Three, by inspection. Still, this puzzle and its KZbin 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."

[@arsaluisnubar1074]

3^3+2^3=27 +8=35

[@narsinhapotdar7215]

Good

[@DavisRimu]

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.

[@garygornak]

3

[@rclrd1]

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_ ??

[@Walter_Carnielli]

Completely unnecessary solution. This kind of problem is known to be solved by summing powers; 3^3 +2^3=35, n=3.

[@余儒炫]

N=3

[@ВячеславКулешов-ж9ж]

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

[@ТигрМудрый]

3

[@gillesdelbreil5414]

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

[@Giovanni-o2o]

N =2 is wrong

[@아무개-u2i]

n is 3 hhhhh

[@簡欽慧]

無聊

[@ЭдуардНазарян-в6щ]

3

[@ВалентинГорелов-э6в]

n=3

[@NakazziLillian-ti2xr]

n=7

[@chinnu8038]

3

[@mauriceferraresi1776]

N=3

[@nihalatar3437]

n=7

[@amorzarai8094]

n=3