more simple in a few seconds : 222^333 vs 333^222 = (222^3)^111 vs (333^2)^111 so we want to compare 222^3 vs 333^2 ... obviously the first one is vastly greater, even if we lower it to 200^3 = 8000000 and increase the other one to 400^2 = 160000 ... no brainer :) no calculator needed in a few seconds
@randaalzifahri6987
Жыл бұрын
Exactly what I thought
@numb0126
Жыл бұрын
Nice one bro👀
@bipolarminddroppings
Жыл бұрын
Even thats way more than you need to know the answer. 333>222. Done.
@DenCato
Жыл бұрын
@@bipolarminddroppings lol except that's wrong
@snatcherofpeachs
Жыл бұрын
@@DenCatoI'm assuming they are referring to the exponents.
@edzehoo
Жыл бұрын
ChatGPT 3.5 be like : "Yes, I can see my mistake now. You are right. 1 is indeed greater than 3."
@namansharma2721
28 күн бұрын
This is so real 😭😭🤣🤣
@ericlindholm9482
Жыл бұрын
Seems really complicated. 222^333 or (222*1.5)^222 222^333 or (222^222)*((1.5)^222)) divide both sides by 222^222 222^111 or (1.5)^222 222^111 or ((1.5)^2)^111 take 111th root of both sides 222 or 2.25 Should take less than 30 seconds, even if you go slowly.
@sandeep.696.antegra
Жыл бұрын
This looks way too easy🤝
@jonfrankle
Жыл бұрын
Nice.
@slomojohnjoshi5990
Жыл бұрын
Power 🔥🔥🔥
@prakashm1468
Жыл бұрын
Much much faster and easier infact!!!
@shadeblackwolf1508
10 ай бұрын
Very nice, very short, and no invalid steps.
@thePronto
Жыл бұрын
When one exponent is so much larger than the other, the solution is intuitive.
@josephmanno4514
Жыл бұрын
Yep. I was just thinking, "Uh ... common sense?"
@livelaurent
Жыл бұрын
@@josephmanno4514 Yah that's exactly what I thought too. The first exponent is so much bigger that there was no doubt in my mind.
@stvrob6320
Жыл бұрын
In a real test I doubt there would be enough time to prove it, and you would be expected to take an intuitive guess.
@johnhaller5851
Жыл бұрын
@@stvrob6320I’m sure the problem was to prove the answer, not to use intuition.
@samar5992
Жыл бұрын
Intuition is WRONG for numbers which are smaller than 2.718. Don't need to solve such questions even. Just know y = x^(1/x) increase from 0 to e=2.718, and then decrease forever as x increase. Calculate maxima of x^(1/x) to prove it. So for 222^333 vs 333^222 is same comparison as 222^(1/222) vs 333^(1/333), now they are in the form of x^(1/x). As 222 and 333 are both greater than e=2.718, hence for smaller then value of x greater will be x^(1/x). Hence 222^(1/222) > 333^(1/333) Hence 222^333 > 333^222 Question is trickier when numbers are similar like 2023^2022 vs 2022^2023. Easy if you know this function y = x^(1/x)
@guillermotell2327
Жыл бұрын
Take the power 1/111 on both sides (a monotonous operation), then divide by 111^2 and you are left with 888 > 9.
@alecearnshaw9651
Жыл бұрын
Lovely problem, thank you!!!! The exact demonstration is great, and shows just how much stronger a larger exponen is!
@loicdearaujo8557
10 ай бұрын
as soon as the base is greater than 1, in absolute value.
@carmanragatz
Жыл бұрын
Simple rule of thumb, if Y>X, then X^Y is always bigger than Y^X with the only exceptions being: if X=1; or if X=2 and Y=3; or if 0 or negative numbers are involved. This is also under the assumption that they are whole integers and not decimals.
@niveditabhaskar8193
Жыл бұрын
No you fool, this happens because the graph of x^1/x has maxima at x=2.71...
@jonfrankle
Жыл бұрын
Almost -- there's one more exception to rule out. For all pairs of positive integers (x, b) except (3, 1), (3, 2), 𝒂𝒏𝒅 (4, 2), if x > b then b^x > x^b.
@carmanragatz
Жыл бұрын
@@jonfrankle I did forget about (4,2). That one they are equal. Anything (X,1) will be the reverse just because 1^X is always 1. A similar thing occurs with (X,0) because X^0 is always 1 and 0^X is always 0. I think the (3,2) set is the really interesting exception just because they seem like somewhat “normal” numbers yet they are the complete opposite to the rule while (4,2) is the turn around point.
@carmanragatz
Жыл бұрын
@@abigmonkeyforme That almost works. The only issue there though is that you exclude other scenarios wherein X^Y > Y^X for when X=2 as opposed to being greater than 2. For example, if X=2 and Y=5, X^Y (32) will still be greater than Y^X (25). And again, you can’t say the rule applies for all integers 2 or more because of the Y=4 and X=2 scenario.
@jonfrankle
Жыл бұрын
@@abigmonkeyforme @carmanragatz2995 Right - I was looking for the easiest way to say it, but *did* want to cover all the scenarios that follow the rule, which is why I landed on the 3 exceptions instead of the simpler rule that doesn't provide guidance for b=2 scenarios.
@adityamittal4357
Жыл бұрын
Exponentiating more times will obviously be larger than multiplying by a larger number fewer times. 2^10 = 1024 vs. 10^2 is only 100. No brainer for anyone who understands compounding.
@pengbertuuu
Жыл бұрын
Let f(x) = ln(x)/x, f’(x)=(1 - ln(x))/x^2 < 0 if x > e. Therefore f(x) is decreasing if x > e. If b > a> e, we have ln(b)/b < ln(a)/a => a*ln(b) < b*ln(a). Therefore, b^a < a^b
@jitendrapatil3276
Жыл бұрын
Physics and Mathematics are my favourite subjects. Now I am in IT field so hardly use it. Thanks for uploading the video.
@tt128556
Жыл бұрын
It's not the destination but the journey that separates math olympians from casuals.
@Nobody-hs9cl
Жыл бұрын
Quite complicated. Set x = 111 and split all the 222 and 333 to 2x and 3x - this makes ist much easier to see. (2x)^(3x) or (3x)^(2x) cancel out x in the exponents (2x)³ or (3x)² 2³ * x³ or 3² * x² 8 * x * x² or 9 * x² cancel out x² and set x = 111 888 or 9
@MikeAnn193
Жыл бұрын
Interesting. I chose the correct answer intuitively, right at the beginning, because I'm very aware of the power of exponents (no pun intended). Even 2 to the 10th power is 1024, if I did the math correctly in my head. I've been doing that mental exercise for years, as it fascinates me, and it also explains a lot of computer technology. So the much bigger exponent of 333 was the clear answer. But I never would have been able to do all the math you did, to prove it. Maybe 40 years ago, but I've forgotten a lot of what I've learned about working with exponents. 🙂
@babyrazor6887
Жыл бұрын
Bingo! It seemed obvious without the extra penmanship.
@luisgutierrez8047
Жыл бұрын
@@babyrazor6887it's "obvious" but the hard part in math is that " you have to show your work"/write a proof 🤮
@taragnor
Жыл бұрын
Yeah, if the numbers were closer it'd be a lot harder, but given those kind of numbers, it's pretty much intuitive to favor the exponent, which would be massively larger than the other side.
@committedtohealth6082
Жыл бұрын
Seems overcomplicated, only one line is needed: 333^222 < 1000^222 = 100^333 < 222^333 Equality is clear as both terms in the middle equal 10^666
@eragonshurtugal4239
Жыл бұрын
Yeah as soon as you deside to just round down the smaller and/or round up the larger base this question really becomes rather trvivial to proof, for example 222^333 > 10^333 =1000^330 > 333^222
@cryptojihadi265
Жыл бұрын
Or just realize that 50% more exponent is wayyyy more than 50% more base.
@BlackStarASMR
Жыл бұрын
As a mathematician I needed less than a second to know the answer. The first number is 222 multiplied 333 times with itself, resulting in a number greater than 200^333. The second number is 333 multiplied only 222 times with itself, resulting in a number less than 400^222. Therefore first number > 200^333 >> 200^222 * 2^222 = 400^222 > second number. Easy. It's obvious that the first number has several hundred more digits than the second number. Or in other words: Exponential growth is so much faster than Linear growth.
@lakshay-musicalscientist2144
11 ай бұрын
"As a mathematician " lmao
@AloneStroller
Жыл бұрын
If we compare a^b vs b^a, a and b both greater than 0, a^b will always be greater than b^a starting with a ≥ e and b>a. E is Euler number.
@andrewsimoes1425
Жыл бұрын
The easiest is just take a log function on both sides. Log( 333^222) = 222*Log333 Log (222^333) = 333*Log222 It now becomes incredibly obvious that 333*Log222 is much greater than 222*Log333. Therefore 222^333 is much greater than 333^222
@MsLeober
11 ай бұрын
Inequality 222^333>333^222 is equivalent to 333 ln 222>222 ln 333 or ln 222/222>ln333/333. Consider function f(x)=ln x/x, x>1. f(1)=0, f(x)>0, max f(x)=1/e for x=e. Hence f(x) for x>e decreases. Then ln 222/222>ln 333/333.
@armanavagyan1876
Жыл бұрын
Please more videos like this)
@DanielMartinez-ss5co
Жыл бұрын
If you know the logarithmic curve, you could answer easily that 333*log222 >222*log333 without doing calculus
@БранимирМилошевић
Жыл бұрын
I did it using logarithm where you take the log and move the exponent out in front to multiply... if you take log 10 of each, then you have 10^x = 222, where x is something more than 2.. and you have 10^y = 333, where y is something definitely less than 3. That means that 333*x is more than 666 and is going to be more than 222*y which is less than 666.
@sassora
Жыл бұрын
I did similar, let k * LHS = RHS then k = 2/3 * (< 3)/(>2) so k < 1 meaning LHS > RHS
@ДаниилКадетов-и9з
11 ай бұрын
Simpler solution: 222^333=(222^(3/2))^222. So you only need to compare 222^(3/2) with 333. Lets calculate 100^(3/2): (10^2)^(3/2)=10^3=1000. 1000 is greater than 333, so 222^(3/2) would be greater than 333 as well.
@MrPullela
Жыл бұрын
a lot simpler method : raise both sides to 1/666. the comparison then is sq root of 222 vs cube root of 333. the first is around 15 the second is around 7
@ani_meme
11 ай бұрын
Explanation based on graph of x^1/x would be so accurate Also after e (2.71) [graph] every order pair A>B ---> B^A > A^B
@jasont2986
Жыл бұрын
Why is this a math Olympiad problem. Anyone with a basic understanding of exponents comes to the correct answer immediately.
@als2cents679
2 ай бұрын
Most of these problems of the form a^b or b^a which is bigger can be solved as following. a^b OR b^a which is greater => a^(1/a) OR b^(1/b) Now consider that the function x^(1/x) is continuous in the range x > 0 and has a maximum at x = e. This means that: a^(1/a) > b ^(1/b) when e 859^857, because e
@DaveMiller2
Жыл бұрын
If you understand exponents, you can see right away that 222^333 is much much larger than 333^222. You don't need to go through all this to find the answer.
@joeybulford5266
Жыл бұрын
I just did a quick 333^2 and 222^3. Got the obvious answer real quick.
@michaelterry1000
Жыл бұрын
@@joeybulford5266 That is what I did and also got the correct answer, but since getting the correct answer is 50/50 I am not sure that my/our logic is valid.
@stvrob6320
Жыл бұрын
@@michaelterry1000 its not really a 50/50 guess. It would only be a close guess if the exponents were much closer together.
@nathan87
Жыл бұрын
@@michaelterry1000it is valid. Just rewrite as (222^3)^111 or (333^2)^111. To make things even simpler, then consider 200^3=8e6 => 222^3>8e6 or 400^2=1.6e5 => 333^2 < 1.6e5.
@brianmathews2926
11 ай бұрын
Yes but my math teacher never accepted “obviously it’s this”
@pourpourr6042
Жыл бұрын
My solution is this Let's assume that 222^333 > 444^222 > 333^222 (222^111)*(222^222)>(2^222)*(222^222) we devide with (222^222) , positive so the direction of ">" does not change 222^111>2^222 (2^111)*(111^111)>(2^111)*(2^111) we devide with (2^111) , positive so the direction of ">" does not change 111^111>2^111 which is true
@chesscomsupport8689
Жыл бұрын
I like how it specifies "No calculator allowed" ... As if any calculator could calculate something with that many digits.
@Trip_Fontaine
Жыл бұрын
A calculator can calculate the logarithms though, which is why this problem is much easier with a calculator. Just take the logarithm (any base) of both sides and simplify. Then instead of comparing 222^333 and 333^222, you just compare 333 * log (222) with 222 * log (333).
@hasangarmarudi2178
Жыл бұрын
Rule of thumb, when the difference of two numbers are this significant (222 and 333) the larger power always wins. With smaller numbers, it should be calculated though
@eric_welch
Жыл бұрын
what if we used the trick of multiplying by 1?? e^ln(222^333) and e^ln(333^222). this shows the first as e^(333ln(222)) and e^(222ln(333)) and clearly the first is larger (both natural log values are only off by about 0.5, which means we can easily see 333 times that value >>> 222 times that value (they need to add Latex/Math-mode in KZitem comments already)
@sanoop692
Жыл бұрын
Skip to 4:10
@sanjeevmalhotra1332
Жыл бұрын
The function x^(1/x) is decreasing function. So, 222^ (1/222) > 333^(1/333). So, 222^ 333> 333^222. QED.
@eragonshurtugal4239
Жыл бұрын
Not as you havent shown that 222^(1/222) > 333^(1/333) implies that 222^333 > 333^222 (or that x^(1/x) is decreasing for that matter, even if this is rather trivial)
@Crux161
Жыл бұрын
Thinking in terms of the exponent values, the higher exponent should be the higher value. But just to be sure, here’s wolframalpha’s take for proof: Input 333^222 < 222^333 Result: True Difference: -21654117651367595674009303657504323451466064914053983428477671597822265157425685169561874841163256524075141514993489271555967858593066483917829919872733053700297314467289961844505803918784072744460746150445290514796744806235666656463757561016326155941765556826391277242311562899554210498326317066987727891721491660170728677705120359453946342657060495870208537814080428324934794582292339287108921868192024055479059272163841575902928071702174917080532587652484872724858075079774643725094299749679596447224332568671839204836367873548042295543610459585864626586060837280959255187118765654057873263905931012827464919403842627899820588753832982738888429884234296151293320178512094081917814791963490128930557118501667719587215403175978003101914955394494861341712053758332430625675793173863
@perryv
Жыл бұрын
I'd use a different approach to prove that 222^333 > 333^222. Let a, b, and n be positive numbers. If b < a, then b^n < a^n . Now 222 < 333, therefore 222^222 < 333^222. Also, 222^333 = 222^222 * 222^111. Stated differently, 222^333 = 222^222 * w, where w = 222^111. Now we can state the original problem as follows: Which is bigger, 222^222 * w? Or 333^222? Let us assume: 222^222 * w > 333^222 Now we ask, which values of w will satisfy the above inequality? Rearranging it, we get: w > 333^222/222^222, or w > (3/2)^222 Is 222^111 one of the valid values of w? Let's substitute it to w. 222^111 >? (3/2)^222 222^111 >? (3/2)^(2*111) 222^111 >? (9/4)^111 Since 222 > 9/4, then 222^111 > (9/4)^111 Therefore, 222^111 is a valid value of w which satisfies the inequality 222^222 * w > 333^222. And therefore: 222^222 * 222^111 > 333^222, or 222^333 > 333^222
@adokoka
Жыл бұрын
To determine which of 222^333 or 333^222 is greater, it is easier to use logarithms to make the problem manageable. Let's denote: A = 222^333 B = 333^222 To make this comparison manageable, we can use natural logarithms. Taking the natural logarithm of both expressions, we get: ln(A) = 333 * ln(222) ln(B) = 222 * ln(333) We can now compare the two logarithmic results: If ln(A) > ln(B), then A > B If ln(A) < ln(B), then A < B On calculating the values: ln(A) approximately equals 333 * 5.4027 = 1799.9991 ln(B) approximately equals 222 * 5.8081 = 1289.8182 From the above calculations, it's clear that ln(A) > ln(B). This implies that A > B. In conclusion, 222^333 is greater than 333^222.
@silvermack3
11 ай бұрын
Just take the log base 10 of each number. log(222^333) = 333*log(222) > 333*log(100) = 333*2 = 222*3 = 222*log(1000) > 222*log(333) = log(333^222). log is an increasing function so we are finished.
@tylerbakeman
11 ай бұрын
Easier intuition: If y = x + 1 - both are natural numbers, Then [ x^3 ] > [ y^2 ]: [ 3^3 = 27 ] > [ 4^2 = 16 ] [ 4^3 = 64 ] > [ 5^2 = 25 ] … the difference changes more and more, so [ 200^3 ] > [ 200^2 ] *much greater than* 222 and 333 are about equal to 200 in this context, so this is fine. [ 2^3 ] is not > [ 3^2 ] still follows the same trend
@Ynook
Жыл бұрын
I think we have enough information about the bigger number around 2:03. 8^111 is smaller than 9^111, but 111^333 is vastly bigger than 111^222. But I understand the need to go through all the math.
@meadbert
Жыл бұрын
I started by factoring out 222^222 leaving 222^111 and (3/2)^222 I then raised each side to the 1/111 power leaving 222 and (3/2)^2 and 222 is more than 9/4.
@gibbogle
10 ай бұрын
It's much easier to show if you start by setting x = 111. You end up with 222^333 > 333^222 if x > 9/8.
@mikeg2538
Жыл бұрын
It is the exponent that really makes the difference over base multiplier. 3 to 5th power is much higher than 5 to 3rd power.
@mc4ndr3
11 ай бұрын
Was about to leap to the conclusion that a bigger base wins, but then remembered that 25 < 32.
@dailymoonpie
Жыл бұрын
Higher exponent generates higher number. This can also be proven by plotting a graf
@aureissimus
Жыл бұрын
graph
@pendyalanarsimharao4083
11 ай бұрын
There's an easier method using geometry and slope of a line concept take the graph of lnx the slope of the line connecting origin and (222,ln222) is more than slope of line connecting origin and (333,ln333) these can be understood from the graph Now converting that logarthmic inequality into exponential inequality we get this inequality in question
@juhajaako5456
Жыл бұрын
Before jumping into anything we find that 222=2*3*37 and 333=3*3*37. It's rather straightforward after that.
@michaelbell5984
Жыл бұрын
I like this approach as by using prime factors, it simplifies things and makes a generalised approach to solving such problems.
@eragonshurtugal4239
Жыл бұрын
@@michaelbell5984 not realy as it just changes the question (to 2^333 or 3^222, i mean yeah the first one passes the eye test for beeing larger but that doesnt proofs anythink)
@marlow769
3 ай бұрын
Thanks for making that 3 times as long as it has to be. When I was in college, I was convinced that I had met some of the worst math teachers ever that made everything way harder than it had to be…until now.
@reapers8861
6 күн бұрын
My solution There are two exponential numbers 222^333 and 333^222 Well it can be solved easily Lets manipulate 222^333 into 222^222*222^111 and 333^222 into 222^222*3/2^222 Now lets cancel 222^222 from both numbers Therefore we get 1.222^111 2.3/2^222 Respectively. Now' 222^111>3/2*222 Because 3/2*222=(3/2^2)^111=9/4^111 So, If we cancel the exponent from both side (possible because both side has the same exponent) we get 222 and 9/4 Therefore it is proved that my solution is valid. Edit:Thank you.
@C_Rodica
4 ай бұрын
You have 2 curves with their respective branches going up. These will intersect only once and then go their separate ways. You only need to know then which grows faster. And that is how you find the answer without engaging into much calculation.
@MKSMKS-dg2rz
Жыл бұрын
222^333 = (222^3) ^111 333^222 = (333^2) ^111 Essentially we are comparing (8x111^3) and (9x111^2) Answer is obvious.
@noahtr6238
Жыл бұрын
Without even looking into details I pick exponential 333, as the simple fact is that similar range numbers, like single or double or triple or quadruple, etc. exponential 333 is 111 times more than 222 is should be a higher number.
@jacobwiren8142
10 ай бұрын
Remember guys, 2^99 is HALF AS BIG as 2^100, so there was no chance 333^222 was bigger than 222^333. One hundred and eleven more multiplications was practically guaranteed to leave the other half in the dust...
@boaragile82
11 ай бұрын
Simple if basics are applied: 222^3 is more than 100^3, while 333^2 is less.
@keithterry2169
Жыл бұрын
Obviously it's the former. Exponents are powerful.
@markwraith
11 ай бұрын
It’s like me asking you what is bigger, an ant or an elephant. And you whip out your tape measure.
@johnmartin2813
Жыл бұрын
Since 2^3 < 3^2 I am tempted to say that the former is smaller than the latter. But further investigations incline me to the opposite point of view. E.g. 2^6 >> 6^2.
@bipolarminddroppings
Жыл бұрын
With larger numbers, the exponent makes such a huge difference.
@paulortega5317
Жыл бұрын
Same question as 222^(1/222) vs 333^(1/333). X^(1/X) decreases and approaches 1 for X > e. Therefore 222^(333) > 333^(222)
@faustomadebr
Жыл бұрын
You have a number 50% smaller, but multiplied 100 times more...
@killjoyxvi
Жыл бұрын
Clearly, many commenters who claim, 'this is easy,' 'intuition,' 'I'm a mathematician, so I know right away that...,' or 'no need to solve,' may not have experience in math olympiads or even in math competitions at school. It is important to note that those kinds of answers typically won't earn you any points. Judges expect every step to be justified. If you happen to insist on 'proving' something based solely on your intuition or a brief explanation, it's worth mentioning that during the judging process, contestants who rely on guesswork are usually at a disadvantage. Judges often award more points to those whose proofs or explanations are elegant, clear, and well-structured.
@RyanLewis-Johnson-wq6xs
Ай бұрын
222^333>333^222 I don’t need to think about it.
@karolkurek9201
Жыл бұрын
x=111, we have (2x)^(3x) or (3x)^(2x) --> 8^x * x^(3x) or 9^x * x^(2x) --> Divide both sides by 9^x * x^(3x) > 0 and we have (8/9)^111 or (1/111)^111 and now is obvious.
@HR-yd5ib
11 ай бұрын
222^333 = 222^222 * 222^111 222^222/333^222 = (2/3) ^ 222 = ((2/3)^2) ^111 = 0.444^111 since 222 * 0.4444 > 1 it follows that the extra 222^111 far outweighs the difference in the bases.
@ilhantezel6125
Жыл бұрын
If b is greater than a then a.^b is greater than b.^a for a, b are not equal 0 or 1
@suryatejakatha4498
Жыл бұрын
Apply a log base10 , problem gets solved, LHS = 333 log222; RHS = 222log333 log222 ~ 2.3; log 333~ 2.5 LHS = 333*2.3; RHS = 222*2.5 Divide by 111; LHS = 6.9 RHS = 5 LHS>RHS Hope it helps
@Malayali93
Жыл бұрын
And how do you propose to find value of Log 222 or log 333 without calculator in a fast way? Solution given by @aumotion (top comment) is the simplest
@suryatejakatha4498
Жыл бұрын
@@Malayali93 If we are good enough to use a log, please be assured that an approx value of the log can be thought of in mental calculations. Keeping that aside, agree that the comment answer mentioned is the simplest to understand, while the log is the easiest way to do it in my view.
@leosjuranek3827
Жыл бұрын
@@Malayali93 Přibližně log 222 = 2 až 3, log 333= 2 až 3 protože mají tři řády. velikost levé a pravé strany určují exponenty 333>222😀
@adyanto4043
11 ай бұрын
More simple (222)^333 make it smaller = (10^2)^333 (333)^222 make it larger = (10^3)^222 Both comes out to be same value i.e. 10^666 And we know (222)^333 > (10)^666 So by simple deduction (222)^333 is bigger than (333)^222.
@dassadec
11 ай бұрын
Intuitively picked the larger exponent one, the base number wasn't different enough for it to matter
@dunalar9454
Жыл бұрын
Why to waste so much energy if you can only reduce to 2e3 vs 3e2 => 8
@cryptojihadi265
Жыл бұрын
It's fairly obvious as increasing the base number by 50% is insignificant to increasing an exponent by 50%
@epicman6827
11 ай бұрын
A difference in exponents tends to just be worth more than the regular number especially at these scales
@odinakauduma-stupendousmat749
Жыл бұрын
My intuition was wrong! I chose the second. I love this . Please could you tell me the name of the board you use and the pen?
@mdahnafalfee196
3 ай бұрын
LKLogic : Calculators not allowed!! Calculators : Math Error
@MrGbruges
Жыл бұрын
You can apply log with base 222 and base 333 and log properties
@tagnetorare5401
11 ай бұрын
You see questions like this, almost always choose the one with higher exponent.
@gzaf187
8 ай бұрын
Just looking at it without calculating, it’s pretty easy to see that 222^333 is much much greater than 333^222
@antaresmaelstrom5365
11 ай бұрын
I went: 222^(222+111) = 222^222 x 222^111 vs (222 x 1.5)^222 = 222^222 x 1,5^222 so divide both by 222^222 Result: 222^111 vs 1.5^222 means 222^111 vs 2.25^111 therefore left side bigger
@jiqinghuang685
Жыл бұрын
take log with base 222, LHS = 333, RHS = 222*log(333) with base 222, log(222*1.5) = log(222) + log(1.5) with base 222, it is less than 1.5,then done
@marksantiago9841
11 ай бұрын
A difference of 100 in an exponent is much much higher than just the difference of 100 in its base form. The 222^333 is greater. I think the only integers that has a greater value if this comparison is made are 1^2 / 2^1 and 2^3 / 3 ^2
@SeenE-kg9nj
11 ай бұрын
Yeah, it's not even close. The first number is almost 100 times bigger
@skateordiee
10 ай бұрын
You can’t use a calculator, but can’t you just ask Siri? I don’t recall them saying no phones either lol! 😂
@midnick2159
Жыл бұрын
solving a math problem w a pen is max confidence
@AMAR-pc6ht
Жыл бұрын
I just look log and ended up that pretty bigger value in left side so yes, left side is pretty much larger than right. :P
@pascaltomasovic3586
10 ай бұрын
If I'm at the math Olympiad I answer this in less than one second. The answer should be obvious to everyone able to calculate this without having to calculate it.
@carpballet
Жыл бұрын
333 is only 30% bigger than 222. (Not even an order of magnitude). But in exponents, that 100+ extra powers is huge.
@krishnanunnimadathil8142
Жыл бұрын
Wrong. 333 is 50% greater than 222.
@stvrob6320
Жыл бұрын
@@krishnanunnimadathil8142 Better to avoid the use of "percent" altogether and just say 333/222=1.5 I think 'Percent' is the English language's attempt to undermine mathmatics.
@cryptojihadi265
Жыл бұрын
Oops
@bakasaksham3630
10 ай бұрын
Just draw the graph of x^1/x and memorise , then these problems based on x^1/x is a piece of cake
@lateralusDan
Жыл бұрын
I'm even surprised that 222^230 isn't bigger than 333^222
@mitchd949
Жыл бұрын
In 2 seconds you should be able to see the left is greater than the right. Forgetting the exponents at all, 333 is only 1.5 times 222, then seeing that the left is raised to 333 and the right is raised to 222, it's not even close!?
@recursiveidentity
4 ай бұрын
exponents, even if just a few number higher expand, well exponentially, so the higher exponent is the obvious choice
@lebirdking
Жыл бұрын
X^(1/x) is a decreasing function for x>e. So raise both to the power of 1/(222*333). You will be comparing 222^(1/222) vs 333^(1/333). First one is bigger than the second.
@lebirdking
Жыл бұрын
This came in IIT JEE in 1988. Which is bigger? e^pi or pi^e. That was harder because e was a boundary condition.
@AotO_DJ
10 ай бұрын
What do you mean? Every y raised to x = less than 1 is a decreasing function. I don’t get the e.
@fahadansari7576
Жыл бұрын
More simply, if you count the number of digits in 222to the power 333 versus number of digits in 333 to the power 222 , then obviously the one on the left is a bigger number (from Pakistan)
@felixiduh5286
9 ай бұрын
This will get to (2)^3 × (111)^3 and (3)^2 × (111)^2 Obviously 8 × 111^3 is greater than 9× 111^2
@Lightmane
Жыл бұрын
I knew the answer before I watched the video, since the three hundred and thirty third power is WAY MORE than the two hundred and twenty second power. Simple.
@reh3884
Жыл бұрын
That a trivial problem. Exponents grow MUCH faster than bases (i.e., 2^n >> n^2). The first one is greater.
@CommentaryTeam1
Жыл бұрын
^333 likely to be bigger based on proximity of numbers.
@sookgarden23
Жыл бұрын
Thanks so much
@GeriatricMillenial
11 ай бұрын
This one is significantly easier if you understand orders of magnitude. You can just look at it at a glance and answer. 333 is many orders of magnitude higher than 222. I might give it a thought if they were close, but they are easy different here.
@vannoo67
Жыл бұрын
Clearly with an exponent that is half again the size of the other one and the bases are of the same order of magnitude, it is going to be massively larger. Thanks for going through the workings, but it was hardly necessary.
@Xoretre
Жыл бұрын
111^333 is actually greater than 333^222. It can be proven in 3-5 steps that (111^333)/(333^222) = (111/9)^111. Since the proof is trivial, it will be left as an exercice for the reader
@jacob.munkhammar
Жыл бұрын
I did it this way, in 4 steps. Expand the powers to expose 111: 222^3^111 or 333^2^111 Take the 111th root: 222^3 or 333^2 Expand the factors to expose 111: 111^3 x 2^3 or 111^2 x 3^2 Divide by 111^2 to get the answer: 111 x 2^3 or 3^2 111 x 8 or 9 888 or 9
@bipolarminddroppings
Жыл бұрын
I did it in one step. 222^333 or 333^222 333 > 222 Done. Its exponents. These arent even in the same order of magnitude.
@ParasGupta-jn5hn
29 күн бұрын
First one 222³³³ is greater than 333²²².... Reason :- We can write as:- (222)³)111... Which is 888)¹¹¹ simple....
@MrAchum
Жыл бұрын
I just used my power of guessing to solve it. Great time saver, but terrible thing to rely on.
@oov55
Жыл бұрын
solved in 4 seconds. Because there is a very simple alternative.
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