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Math Olympiad | A Nice Algebra Problem
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4 сен 2024

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Комментарии : 13

@VictorPensioner Месяц назад

Obviously (5/a) 3 Therefore (5/a) + (6/b) < 7 when b > 3 and any a It means b can be 1, 2 or 3 1) b = 1 => (5/a) + 6 = 7 => a = 5 and (b=1, a=5) is solution 2) b = 2 => (5/a) + 3 = 7 - no solution 3) b = 3 => (5/a) + 2 = 7 => a = 1 and (b=3, a=1) is solution All solutions are b = 1, a = 5 b = 3, a = 1

@ajitandyokothakur7191 Месяц назад

Rigorous and nice. As one of the commentators pointed out, the solutions could also be found graphically. However, your method is ideal for grade school algebra students who may not follow modulo arithmetic and possibly the concept of graphical solution. Your method is ideal for them Dr. Ajit Thakur (USA).

@superacademy247 Месяц назад

Thank you Dr. Ajit Thakur for 👍💯 your compliment

@user-hf8dm9ug9j Месяц назад

Подбором за минуту оба решения легко. Думал с комплексными числами задача-там как бы ещё есть решения.

@YAWTon Месяц назад

Use modulo arithmetic in ℤ/7: from 5/a+6/b=7 it follows that a≡5b mod 7. b must be less than 4 (because the LHS decreases when a,b increase, while the RHS increases, and 5/1+6/4 There are 2 solutions in ℤ⁺: (a,b)=(5,1) and (a,b)=(1,3).

@superacademy247 Месяц назад

Thanks 👍💯 for your support and input

@benyseus6325 Месяц назад

Some school jurisdictions do not teach modulo arithmetic so this rigorous solution might be better for them

@Dr_piFrog Месяц назад

Quick solutions just by observation (a,b) ---> (5,1) and (1,3)

@ronbannon Месяц назад

Rewrite as b = 6a/(7a-5), graph this rational function, and you will quickly see that only three cases are possible: b = 1, 2, 3. Test them, and you see the only two work: (1, 3) or (5, 1).

@superacademy247 Месяц назад

Awesome 💯👍😎 approach

@boguslawszostak1784 Месяц назад

I do not like tricks It is easier to multiply both side by 7ab 7*7ab - 7*5b - 6*7a=0 (7a-A)(7b-B)=7*7ab - 7aB-7bA+ab B=5 A=6

@nitrodizon Месяц назад

lol you dont allow comments that show that your whole video is just a waste of time?who on earth would do such a calculation to find a and b mate?

@superacademy247 Месяц назад

It's a diophantine equation in which positive integer solutions are sought. I do allow critiques. Go through my videos, you'll see a brunch of viewers expressing discontent to some of my methods