Math Olympiad | A Nice Exponential Problem | 90% Failed to solve
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Пікірлер: 55
1) It's not difficult to solve mentally, but I used a little trick that made it even easier: 3^X + 3^Y + 3^Z = 3^5 * 91 2) I divide everything by 3^5, resulting in: 3^X-5 + 3^Y-5 + 3^Z-5 = 91 3) Now, even a student in the initial grades can solve it, as it has to be 1 + 9 + 81 That is, X = 5; Y = 7 ; Z = 9 Bingo from Brazil!!!
@vijaymaths5483
3 ай бұрын
Excellent sir!! Thanks for watching :)
@user-it6fh7hy6t
3 ай бұрын
у@@vijaymaths5483А сам почему не додумался до такого решения? Для настоящего математика оно единственное, достойное внимания.Я решил точно таким же способом за 3 минуты вместе с записью.Начинай учиться у знающих людей,блогер.
@user-xu6lt7wg9r
3 ай бұрын
I solved this question by the same way with you
@gheorgheraileanu3332
2 ай бұрын
Qa
@observadorlife
2 ай бұрын
Boa observação
The most difficult part is factorising 22113 = 3⁵ * 91. So we get x=5, otherwise the ratio dividing LHS by 3^x, equal to 1+3^(y-x)+3^(z-x) would be divisible by 3, which it isn't, given that both y-x and z-x are positive. All the rest can be done mentally! 1 + 3^(y-x) + 3^(z-x)=91 => 3^(y-x) + 3^(z-x)=90 => 3^(y-x) * (1 + 3^(z-y)) = 3² * 10, so (for similar reason) y-x=2 => y=7. Then 3^(z-y)+ 1 = 10 => z-y = 2 => z=7. It would be more challenging to replace < with ≤, i.e. x≤y≤z. Since all three can't be equal to each other, the remainder of 1 + 3^(y-x) + 3^(z-x) by 3 can be either 1 or 2, but not 0, which means that still x=5 is the only value for x. Than again the remainder of 1 + 3^(z-y) could be either 1 or 2, so still (5, 7, 9) is the only integer solution... unless I am wrong.
1st step even - odd factorisation is okay! When 3^x =3^7 ,it was lengthen by converting the product again into a compound number. For quicker result, it could have been arrived as below: (3^y + 3^z) = 3^5×90 =3^5(9 + 81) 3^5(3^2 + 3^4) = 3^7+ 3^9 . Thus: x =5 ,y =7 & z =9.
@ganeshdas3174
3 ай бұрын
Read 3^x =3^7 as 3^5
@tarahomsafari9932
3 ай бұрын
@@ganeshdas3174چ
3^( y-x) + 3^(z-x) = 90 = 3^2 +3^4 y-x=2, y=7 z-x=4, z=9 Решение получается короче
After x=5 and x^5=243 we shloud divide 22113/243 and substitute y' = y-x and z'=z-x wich gives us an easier equation to solve : 1+3^y'+3^z'=91 => 3^y'(1+3^(z'-y'))=90 => y'=2 => y=2+5=7 and finally 3^(z'-2) = (10-1) => 3^(z'-2)=9 => z'-2=2 => z'=4 => z=x+4=9.
Here is a simple solution Analyzing last digits: of 3^n follow a cycle of 1, 3, 9, 7. Therefore, the sum of the last digits of 3^x, 3^y, and 3^z must be 3 to reach the last digit of 22113 (which is 3). Possible digit combinations: To achieve a sum of 3, the last digits can be formed in two ways: 3 + 7 + 3 or 1 + 9 + 3. Ruling out options: 3^1 + 3^3 + 3^9 wouldn't work because their last digits sum to 9 + 7 + 7 = 23. Valid solution: This leaves us with the combination 1 + 9 + 3, which corresponds to 3^5 + 3^7 + 3^9. Therefore, the correct solution to the equation 3^x + 3^y + 3^z = 22113, considering x x = 5, y = 7, z = 9
@user-sw2gl7jd6n
3 ай бұрын
Есть более простые решения без всяких комбинаций.
X= 9, y= 7, z= 5
3^x +3y +3^z = 22113 (3^4) * (3^(x -4) + 3^(y -4) + 3^(z-4)) = 81* 273 3^(x -4) + 3^(y -4) + 3^(z-4) = 273 3^(x -4) + 3^(y -4) + 3^(z-4) = 243 + 27 + 3 3^(x -4) + 3^(y -4) + 3^(z-4) = (3^5) + (3^3) + (3^1) X-4=5 so x =9 Y-4 = 3 so y=7 Z-4 = 1 so z=5
@othmaneksir4001
Ай бұрын
Z>y>x So z=9 y=7 and x=5
9, 7, 5 This was easy because these cutesy KZread questions always seemed to be designed to have nice integer answers. And this one was particularly simple to attack - I just found the largest power of 3 that's less than 22113 - it's 9. Then I subtracted that off and repeated - that yielded 7. And what was left after subtracting that off was just 3^5. Couldn't have been easier.
Sir can you tell me how can I strong my geometry for IOQM as my Geometry is weak ?
@vijaymaths5483
3 ай бұрын
For Geometry- Just go through the theorems, write them somewhere , understand the proof and pen them down through your understanding and not by copying. If you find the excircle part a headache leave that for the time-being. Do them at the end. For Algebra- Understand the method and directly go into problem solving. Practice INMO past year problems given at the end of book. Though solutions are not available in the book they are available all over the internet. Here are my uploaded Olympiad geometry problems video links given below : kzread.info/dash/bejne/qaGOuaySmauomc4.html ****************************************************************************** kzread.info/dash/bejne/gqx2srl-k7rae6Q.html ****************************************************************************** kzread.info/dash/bejne/oHqH3JSJlMfUpcY.html Also you can check my playlist of uploaded videos. Good luck:)
x = 5; y = 7; z = 9
so, here at time stamp 6.52, we have 1+ 3^y-x + 3^z-x = 91,, so we need here sum of 91. which can only be obtained by 1+9+81. it implies that y-x = 2, and x=5, so y = 7,, similarly z-x= 4, so z = 9
I find it astonishing a base conversion problem, base 10 to 3, is beyond the ability of so many, 91%. Base 10 , 22113, converts to base 3, 1010100000. There are various ways of doing it : as a programming exercise, a base conversion program would use a base 10 to 3 table. For this example the relevant entries are base 10 3 is base 3 10, base 10 10 is base 31, base 10 100 is base 3 1021 base 10 1000 is base 3 101001 base 10 10000 is base 3 11121121. Convert, using a base 10 to base 3 equivalents, the decimal number, multiplying by the value of each decimal digit, and adding in base 3. Today the most common base conversions are base 2 to 10, vice versa, used for all computer calculations.
2024-03-29: Must be some rule about permutations !? Otherwise, why can’t x, y, & z = any arrangement (permutation) of 7, 9, & 5?
Excellent sir
Fantastic method used for solving this beautiful problem 👍
@vijaymaths5483
3 ай бұрын
Thank you !!
Thank you author for shedding some light.
@vijaymaths5483
3 ай бұрын
Welcome😊
(9; 7; 5) et on fait une permutation pour avoir 6 solutions.
X:5 Y:7 Z:9
@vijaymaths5483
3 ай бұрын
Correct 💯
Great Explanation👌
@vijaymaths5483
3 ай бұрын
Thanks 🙂
Thank you for your solition.It Will be better if you let 3^5.90 like this.good dayı..
@vijaymaths5483
3 ай бұрын
Ok, thank you
Thank you your solition.It would be letter you let 3^5.90 good dayı...
@vijaymaths5483
3 ай бұрын
Thank you so much 🖍
22113₁₀=1010100000₃ → z=9, y=7, x=5
@Dana-ur2pv
2 ай бұрын
Спасибо Вам за прекрасное разъяснение решения,мне около 80 лет,я всегда любила математику.Обычно я переписываю и самостоятельно решаю,а сегодня увидела , что а правой части уравнения стоит огромное число и не решилась решать,а оказалось очень просто,такое доступное объяснение,что кто даже не разбирается и тот человек поймет.Еще раз благодарю Вас,успехов в Вашей трудовой деятельности.
@lifewatery7472
15 күн бұрын
This solution, like the computer Binary! Good!
श्री मान जी 91=7 x 13 factor हो सकता है क्यों नही किए
13 times 7= 91
Nice👏
@vijaymaths5483
3 ай бұрын
Thank you 🙏
👍👍👍
@vijaymaths5483
3 ай бұрын
🙏
Better
これは22113の素因数分解
91 è ancora divisibile per 7 e per 13!
(xyz+2xyz-3)
@vijaymaths5483
3 ай бұрын
🤔
just 1 min 😂
@vijaymaths5483
Ай бұрын
ohhh you are faster than the calculator isn't it Mr.Genius ??
@cosmolbfu67
Ай бұрын
@@vijaymaths5483 yep just practic a lot