Japanese | Math Olympiad | A Nice Algebra Problem
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- čas přidán 27. 05. 2024
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It takes about eight seconds to see that that's 2^18.
Definitely not an Olympiad level problem.
Yeah, pretty obvious that the one and everything past the one goes away immediately
Now solve that in your head. I got 262144 from 512^2
I know the powers of two instantly from direct memory up through 2^16, but I usually then have to calculate a bit. I've just never bothered to learn the higher powers. I spent my career working in the digital hardware area, but by the time powers higher than 65536 started to become "more relevant" I had moved on to other facets of the work, so those beyond 16 just never got "burned in" like the lower ones did.
I can give them all to you in hexadecimal, though. 🙂 0x40000. There's a reason we made so much use of hex.
@@KipIngram I'd bet you know 2^20 (one MB)
for computer person - after you get to 2^18 = 2^10 ( 1K ) * 256 - multiply that - easier than all that stuff after 2^18 in this video. Or 2^18 =- 2^20 (1M ) /4. both give the correct answer and easier.
It's easier to multiply 2¹⁶ = 65.536 with 4.
I get this to be 64 ie 2^6
I think this problem is more interesting if instead of solving for the square root of 2^6^2^1^5^9 you solve for the square root of 2^6^2^1^4^4 which has an answer with same digits as the problem
First learn the basis of math, THEN try and post this video again...
correct, please!
Before watching:
Alright, if you have like a^b^c^d^e^f^g, you'll typically calculate f^g first, then e^(f^g), then d^(e^(f^g)), and so on,.
With that in mind, we can ignore the outermost 3 exponents we see. Why? Because those last 3 are 1^(5^9)), and 5^9 is a real number, and 1^X = 1 for real X.
So what this actually meansi s that we're looking at √ 2^(6^2). But square root is simply an exponent of 1/2.
Thus, √(2^(6^2)) = (2^((1/2)6^2)) = 2^(36/2) = 2^18.
Solution is 2^18. Because we're not solving for X, but simply evaluating a square root, we do not include the negative root in this case.
Most math classes will allow you to leave the term 2^18 as-is. However, if your class is currently doing a unit on *evaluating* exponents, you may have to multiply it out. Quickest way without knowing the higher powers of 2 (suppose you only know as far as...2^12 = 4096) would be to do (2^9)(2^9) = 512*512
= 1024
+ 5120
+256000
= 262144
Honestly 4096*64 is probably going to go faster, but I went with what popped into my head.
The exponentiation is not associative: usually when we write in the canonical up- indexed way a^b^c we conventionally intend (a^b)^c= a^(bc) and not a^(b^c) ; the same convention extends for multiple exponents if wrote without paranthesis:a^b^c^d= ((a^b)^c)^d= (a^(bc)) ^d=(a^(bcd)) and so on. Thus in this case we should obtain the square root of 2^(540)=2^270; In the other hand, this solution uses the opposite convention, i.e. square root of 2^(6^(2^(1^(5^9))))=square root of 2^(6^(2^(1)))= square root of 2^(6^2)=square root of 2^36=2^18
Zebras and 🦒 giraffe are having some positive corelation 😃😄😁😄😃😃😃
oml bruh you can straight up cancel out the 1^5^9 and get sqrt(2^6^2) = 2^18 its that simple dont overcomplicate things take it easy
They're just going step by step for novices who might not see it as quickly. There's other channels for more advanced math stuff.
YES!
2^(0,5*6*2*1*5*9) =2^(270) - i guess - i was wrong and rely bad at math and al those u-tube commies r foking geniuses///
U COMPLICATED IT MORE...
Forget about the top three numbers. One to the power of anything equals one.
Forget powers of 1: result is always 1. 2 power 1 is 2. 6 power 2 = 36. But you are in a square root, and a square root of any number power 2 is this number. Then, square number of 2 power 36 is 2 power 18. Then, the result is 2 power 18 = 512 x 512
Piece cake!
Good
Just put 2 in place of 1 and see the magic of infinity.
2^18
Math olympiad level ? I cannot believe it.
U can solve without assuming 5^9=x one sight easy question
Why is everyone thinking of 2^18 , I think its 2^36 , since there is whole underroot therefore it will be 9^(1/2).
x^(n/k) = ᵏ√xⁿ ; x ≥ 0
9^(1/2) = ²√9¹ = 3
√2³⁶ ≠ 3
It would have been nicer if it came out to 262159
In 5 sec, i get the answet 2^18
The key is number 1
❤❤❤❤❤❤❤
√2³⁶ = 2¹⁸
= 2¹⁶ · 4
= 65.536 · 4
= 262.144
2^18
The key is you need to recognize that 1
😮
Is there any need after final square root is done?
А как вам такой вариант:
1^5^9=х=1^(5^9)=1
2^х=2^1=2
2^6=y
тогда получается
(у^2)^½=у=2^6=64
Что скажете? ❤
2^9×2^9 = 4^9 = 4^5×4^4 = 1024×256 = *262144* 💪🤙🏆🤣🇪🇺🇨🇭🇮🇹🇪🇸
Если 1^а=1, то 5^9, вообще не играет роли и не зачем уделять внимание ей
Yeah that’s what I figured. Lol.
I solved and found 2^18 in 5 seconds
🇰🇭🇰🇭🇰🇭🇰🇭🇰🇭🇰🇭🇰🇭🙏🙏🙏👍ល្អណាស់ good👍
CM Maan is right.Needs reelection for
Parliament.
512*512 using quadratic equation is funny. Maybe you know ANOTHER way to multiply numbers? 😂😂😂
this music is from electrolab channel
Wrong solution..the powers should be multiplied an not rooted...andswer is 2^270
Hic est 2^18 responsum.
MY calculs does 513...
And not so longtime and complicated...
√(2^36) = (√2)^36 = 2^18 = 4^9 = (4^3)^3 = 64 * 64 * 64
Equals 1.
2¹⁸
64
👍😊🇺🇸✌️
What about 29=x
Parabéns! Complicado NÃO, extenso! Longo! Tem como fazer menos Longo?
Square root from 262144 is 512. The final answer 512.
2^(18)
It is really not that difficult to just multiply 512 x 512.
Yes - takes a few seconds to do in your head.
10 seconds to 2^18=262144. Yes I know my powers of 2 by heart
@@johnnyragadoo2414 1 ^ 2 ^ 9 = 1. 2^9 is even, 9^2 is odd. You just fooled your maths completely
@@johnnyragadoo2414 ln(1+2+3)=ln(1)+ln(2)+ln(3) is a singularity, not a démonstration ln is associative. 2^4=4^2 is also a singularity, not a proof power is commutative
@@johnnyragadoo2414 10^4^2=10^16=10000000000000000 Did you write ANYTHING correct?
@@johnnyragadoo2414 back to the question, you have a 1 in the ladder of powers, therefore you must ignore what comes after. 2^6^2^1= 2^6^2=2^36. Apply square root, 2^(36/2)=2^18=262144
@@johnnyragadoo2414 You must read power from right to left 1 ^ 5 ^ 9 = 1 ^ (5 ^ 9) = 1. Result 262144. Period. And if you want to change the rules of math, try with prime numbers and figure out (2^3)^5 and 2^(3^5) are not the same. Result 262144
8
No !!!!!!!
Quan fas (1⁵)⁹, dius que és igual a 1 i no és cert !!!!!
Tractes l'1 com si fos Base i no ho és -----> és un exponent!!!
No és correcta
El correcta és
V(((((6)²)¹)⁵)⁹)=
V6⁹⁰= 6⁴⁵
Непросто не пойти за всеми! ❤
2^18😅
19
2の270乗じゃないの？分からん
2 power x
E o 5^9 ?
29
262 144
xdd it takes like 1 second to see the answer sqrt(2^36) = 2^18
Ну там же есть 1 среди степеней, начиная с неё можно просто все стереть. Остается √2³⁶ = .....
How 2 power 6 power 2 can be 2 power 36. Power get multiplied so it should be 2 power 12 rather than 2 power 36. Sq root of it will be 2 power 6.
Wrong-you perform successive exponentiations from the top down. 2^6^2 should be interpreted as 2^(6^2), or 2^36. The square root of that is then 2^(36/2) = 2^18.
Please note. I was in error. Mea culpa. In hindsight I wish I'd stood on my desk and shouted LIAR! to any math teacher who told me categories in PEMDAS are always, always, always evaluated left to right in absence of parentheses.
Exponentiation is evaluated right to left.
I'll leave my comment below to document what I'm talking about. Please understand my statement below is wrong.
@@ThreePointOneFou Are you saying that 2^6^2 is not the same as 2^12? Even in this video a^b^c is represented as the same as a^(b*c).
Evaluate exponentiation from the bottom up. Otherwise 2^6^2 is 2^36 and not equal to 2^(6*2).
@@johnnyragadoo2414 Successive exponentiations are supposed to be evaluated from _right_ to _left_ when written horizontally, to reflect the top-down rule that applies to exponentiation. (Ideally, exponentiation wouldn't be written like that, but those are the limits we're stuck with in commentbforums without mathML or similar support.) The problem shown should be evaluated as though it were written 2^(6^(2^(1^(5^9)))).
@@ThreePointOneFou Then 2^6^2^1^5^9 wouldn't equal 2^(6*2*1*5*9).
As it shows in the video, a^b^c = a^(b*c).
So, which is it? Should a^b^c = a^(b*c)? I think it should.
Correct, 2^6^2^1^5^9 is not equal to 2^(6*2*1*5*9). And no, a^b^c should be evaluated as a^(b^c).
The video does everything correctly, apart from a bit of slightly sloppy writing when showing the rule that (a^b)^c = a^(b*c).
When evaluating exponents, the exponent must be fully determined before raising the base to that power-in other words, 6^2^1^5^9 must be calculated _before_ raising 2 to that power. Thus, 2^1^5^9 must be calculated before raising 6 to _that_ power, meaning 1^5^9 must be evaluated before raising 2 to _that_ power, etc. In other words, exponents must be calculated from the top down, or from right to left in horizontal notation.
it is 2^18
No, it's not. It's 2^270. Exponents are not associative. They can be rearranged, but not regrouped.
Meh ... took me less than 5 seconds.
2 power 270
Not the right answer.
Did you do it right? You work from the top down. You'd calculate x = 5^9, and then 1^x. But that will just be 1, because 1^ is 1. Then we have 2^1, which is 2. Then 6^2, which is 36. Then 2^36. Then we take the square root of the whole thing, which is 2^18.
You do not start at the bottom 2 and work your way up.
The answer is 2^18.
I got a result 2^18
Before i watch the result
@@bmkdz Same here. Took about eight seconds looking at the opening screen. The 1^5^9 just goes away.
Song name plz ? ❤
La tete vide by BERTYSOLO
1
Nice trick
))))))
Was this olympiad for 4th graders? It is ridiculous to spend 6.5 minutes on this stupid problem
All wrong. The result is 2^540
太吓人了，简单的很：32.
Пояснення не правильне, відповідь 2^270
buy a stylus please!!!
Please speak rather than adding useless music, try to verify the source and having something challenging will help people.
Trur❤
No, on the contrari, Music is better than talk
Full error, answer incorrect! This lie
Annoying music
Can we PLEASE stop with the common core bullshit? These competitions don’t give time to do all of that extra bullshit
Its boring. Could have been shortened.
Es correcto
The actual video is to find the value of 2^18
2^ 18
2^18
2^18