Ukraine l Very Nice Olympiad Math Problem l Tricky Solution

Hello my Wonderful family 😍
Trust you're doing fine 💕
If you like this video on how to solve this nice Math Problem, like and Subscribe to my channel for me exciting videos 💥
Harvard University Entrance Question
Math Olympiad, Algebra, Simplification, Germany Math Olympiad,Math problem-solving, Grade 8 Maths, Grade 9 Maths, Nice Square Root Simplification, Japanese Can you solve, Imo Class 7, Imo Olympiad Class 7, Advanced math concepts,Math competition,Waec, Find the Value of X,Math Olympiad strategies,Challenging Math problems, How to Differentiate, How to Integrate, Thailand junior math Olympiad Problem,
Calculus,Algebraic expressions,Mathematics education,Math Olympiad questions, Math short trick, best youtube chennal for maths olympiad, competitive math, imo olympiad, only genius can solve this math problem, Impossible math problem, math magic tricks, new math tricks 2024,how to solve math problem quickly, russian mathematics,competitive math, german Olympiad math problem,math magic tricks mind reading,math and science Olympiad,Real number Olympiad, imo algebra problem, Math Olympiad Traning, Hardest Olympiad Problem,International math Olympiad prepration,best youtube channel for maths olympiad,Math Olympiad hardest problem,Math Olympiad channel,state level maths olympiad, math Olympiad elementary school problem,Indian national mathematical olympiad,genius cerebrum math Olympiad,ai math Olympiad,science and math Olympiad,olympiad study,national math Olympiad questions,international mathematics olympiad, maths olympiad for college students, Exponents and powers olympiad questions,math olympiad lectures,Mathmastertv
#matholympiad​#education#algebra​#math​#simplification​#Exponents​#vedicmath​#viralmathproblem​ #howto​#mathematics​#viral​ #mathematicslesson​#calculus​#harvard#mathfocus#statistics#exponentialfunction#mathchallange#advancemath#vedicmath #Mathmastertv

[@nevkon]

Вообще крайне странное решение. Сказано что искомые значения являются положительными числами. Тогда можно решить например относительно искомого a: принимаем b равным любым положительным числом строго больше 6/7, тогда a=5*b/(7*b-6). Поскольку второго условия (уравнения) нет, то это и будет ответом ведь какое бы число b вы не взяли всегда будет a>0 удовлетворяющий условию. Частный случай - b=3, a=5*3/(7*3-6)=15/15=1 или b=1, a=5*1/(7*1-6)=5/1=5. Можно даже дробные значения брать и будет найдено решение. ps. Для тех кто не понял - если b=6/7 то a=5*(6/7)/(7*6/7-6)=5*(6/7)/0 ! Соответственно для a ограничением вниз будет являться значение 5/7.

[@yuriyhorobey2609]

Why olways so complicated solution for primitive "problems"? 5 is prime number so _a_ can be either 1 or 5. 6 has three dividers 1, 2 and 3. Next is primitive strait forward and obvious.

[@thierrygermain5182]

Bravo. A little laborious to get to (7a - 6) (7b - 5) = 30. Since we operate in Z, we must also try ( -1)(-30), (-2)(-15) ... but this does not give integers for a and b.

[@yuriyhorobey2609]

We operate on Z+

[@ajitandyokothakur7191]

Although your No. 4 results are noninteger and therefore rejected, the actual results you obtained are wrong. Check carefully. Dr. Ajit Thakur (USA).

[@thierrygermain5182]

Tell me more, I am interested, dear colleague Doctor as I myself am the author of many textbooks in Switzerland. Thank you in advance.

[@Haideralilughmani]

😮