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
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.
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.
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.
@@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.
@@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.
@@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.
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)
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
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.
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
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.
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. 🙂
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.
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
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.
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.
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
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
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
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.
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
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
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.
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
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 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.
@@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.
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
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
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.
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.
It was beautiful that we couldn’t take more than one step at a time! 111^333 * 2^333 vs 111^222 * 3^222 37^111 * 3^111 * 2^333 vs 3^222 37^111 * 2^333 vs 3^111 (37/3)^111 * 2^333 vs 1 So 222^333 is greater! I had a hunch that the larger exponent was going to be bigger, but it’s cool to see it in math :)
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.
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 TH-cam comments already)
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.
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
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)
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.
Can also divide both sides again by 3^111 because 111^111=(3*37)^111 and as a result: (8*37)^111=296^111 > 3^111 Not necessary but makes it even more obvious
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).
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.
I knew right away that the first number was VASTLY larger....not even close. It would make a more interesting question, if the numbers were closer in value.
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.
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.
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.
@@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)
WAY too complicated... The problem is equivalent to "is 222^3 larger than 333^2". Elevating both to the power of 111 doesn't change their ordering. 222^3 = 2^3 x 111^3 333^2 = 3^2 x 111^2 Again, multiplying by 111^2 doesn't change the ordering, so we can drop it. Now you have to compare 2^3x111=888 and 3^2=9. Done.
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.
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.
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.
Why so complicated?! You have 3 = 2 × 1.5 So 333^222 = (1.5)^222 × 222^222. You can then divide by 222^222 on both sides and have to compare 222^111 to (1.5)^222 Given (1.5)^222 is (2.25)^111, and 222 being greater than 2.25, you have your solution.
No calculations needed, base numbers are not much different (200 and 300). The first one will multiply itself 333x and the second 222x...that just ridiculous simple. The first number is much bigger.
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.
@@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.
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
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!?
If (222/333)^2 * 222 > 1 then 222^333 is greater than 333^222. And (222/333)^2 * 222 is greater than 1, so 222^333 is greater than 333^222. Rest of the terms are just symmetry. Everything else is just to confuse you...
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)
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
So much ink wasted. It's (222^(1.5))^222 compared to (222*1.5)^222. That is (222 * sqrt(222))^222 compared to (222*1.5)^222. What now remains to compare is sqrt(222) to 1.5.
Clearly this is just evaluating (2x)^(3x) with (3x)^(2x), for x = 111. Instantly, (8^x)(x^3x) being compared with (9^x) (x^2x),or Directly comparing x^x with (9/8)^x By heart, anyone should easily conclude that : x^x > (1.125)^x, for x = 111. Thus 222^333 > 333^222
@@eragonshurtugal4239 mAyBe, afterall, there's a reason fof your failure in gettin decent amount of subscribers. Ever since, (8^x)*(x^3x) and (9^x)*(x^2x), any thinking humanbeing would know that IF x = 1, then the term x^3x and x^2x will be the SAME! Whilst here, x = 111. do you NOT know that 111 ≠ 1?? get oFF your meds! you're tripping!
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
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.
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
222 raised to 333 is approximately 2.155 x 10^620. 333 raised to 222 is approximately 1.345 x 10^406. Therefore, 222 raised to 333 is significantly larger than 333 raised to 222.
Although my first method was what is being explauned, this can be done with common sense by finding the difference between the numbers and powers and reducing large numbers to small ones for logic.
It was supposed to do without calculation! You did calculation! I did it without at about 30 seconds no paper or pen. You just do logarithm and you see that log for 333 is less of an factor than 333 is to 222.
@@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.
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
Exactly what I thought
Nice one bro👀
Even thats way more than you need to know the answer.
333>222. Done.
@@bipolarminddroppings lol except that's wrong
@@DenCatoI'm assuming they are referring to the exponents.
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.
This looks way too easy🤝
Nice.
Power 🔥🔥🔥
Much much faster and easier infact!!!
Very nice, very short, and no invalid steps.
Take the power 1/111 on both sides (a monotonous operation), then divide by 111^2 and you are left with 888 > 9.
ChatGPT 3.5 be like : "Yes, I can see my mistake now. You are right. 1 is indeed greater than 3."
This is so real 😭😭🤣🤣
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.
No you fool, this happens because the graph of x^1/x has maxima at x=2.71...
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.
@@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.
@@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.
@@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.
Lovely problem, thank you!!!! The exact demonstration is great, and shows just how much stronger a larger exponen is!
as soon as the base is greater than 1, in absolute value.
When one exponent is so much larger than the other, the solution is intuitive.
Yep. I was just thinking, "Uh ... common sense?"
@@josephmanno4514 Yah that's exactly what I thought too. The first exponent is so much bigger that there was no doubt in my mind.
In a real test I doubt there would be enough time to prove it, and you would be expected to take an intuitive guess.
@@stvrob6320I’m sure the problem was to prove the answer, not to use intuition.
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)
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
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.
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.
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
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
Or just realize that 50% more exponent is wayyyy more than 50% more base.
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.
"As a mathematician " lmao
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. 🙂
Bingo! It seemed obvious without the extra penmanship.
@@babyrazor6887it's "obvious" but the hard part in math is that " you have to show your work"/write a proof 🤮
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.
Physics and Mathematics are my favourite subjects. Now I am in IT field so hardly use it. Thanks for uploading the video.
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
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.
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.
I did similar, let k * LHS = RHS then k = 2/3 * (< 3)/(>2) so k < 1 meaning LHS > RHS
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
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
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
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
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.
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
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
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.
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
This seems more genuine, though maths makes me sleepy.
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.
I just did a quick 333^2 and 222^3. Got the obvious answer real quick.
@@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.
@@michaelterry1000 its not really a 50/50 guess. It would only be a close guess if the exponents were much closer together.
@@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.
Yes but my math teacher never accepted “obviously it’s this”
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
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
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.
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.
If you know the logarithmic curve, you could answer easily that 333*log222 >222*log333 without doing calculus
It was beautiful that we couldn’t take more than one step at a time!
111^333 * 2^333 vs 111^222 * 3^222
37^111 * 3^111 * 2^333 vs 3^222
37^111 * 2^333 vs 3^111
(37/3)^111 * 2^333 vs 1
So 222^333 is greater!
I had a hunch that the larger exponent was going to be bigger, but it’s cool to see it in math :)
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.
It's not the destination but the journey that separates math olympians from casuals.
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 TH-cam comments already)
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.
With larger numbers, the exponent makes such a huge difference.
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
The function x^(1/x) is decreasing function. So, 222^ (1/222) > 333^(1/333).
So, 222^ 333> 333^222.
QED.
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)
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)
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.
Why to waste so much energy if you can only reduce to 2e3 vs 3e2 => 8
Higher exponent generates higher number. This can also be proven by plotting a graf
graph
Can also divide both sides again by 3^111 because 111^111=(3*37)^111 and as a result:
(8*37)^111=296^111 > 3^111
Not necessary but makes it even more obvious
I like how it specifies "No calculator allowed" ... As if any calculator could calculate something with that many digits.
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).
It is the exponent that really makes the difference over base multiplier. 3 to 5th power is much higher than 5 to 3rd power.
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.
This came in IIT JEE in 1988. Which is bigger? e^pi or pi^e. That was harder because e was a boundary condition.
What do you mean? Every y raised to x = less than 1 is a decreasing function. I don’t get the e.
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.
I knew right away that the first number was VASTLY larger....not even close. It would make a more interesting question, if the numbers were closer in value.
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.
222^333 = (222^3) ^111
333^222 = (333^2) ^111
Essentially we are comparing (8x111^3) and (9x111^2)
Answer is obvious.
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.
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.
Before jumping into anything we find that 222=2*3*37 and 333=3*3*37. It's rather straightforward after that.
I like this approach as by using prime factors, it simplifies things and makes a generalised approach to solving such problems.
@@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)
WAY too complicated... The problem is equivalent to "is 222^3 larger than 333^2". Elevating both to the power of 111 doesn't change their ordering.
222^3 = 2^3 x 111^3
333^2 = 3^2 x 111^2
Again, multiplying by 111^2 doesn't change the ordering, so we can drop it.
Now you have to compare 2^3x111=888 and 3^2=9.
Done.
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.
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.
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.
If b is greater than a then a.^b is greater than b.^a for a, b are not equal 0 or 1
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?
Please more videos like this)
Why so complicated?!
You have 3 = 2 × 1.5
So 333^222 = (1.5)^222 × 222^222.
You can then divide by 222^222 on both sides and have to compare 222^111 to (1.5)^222
Given (1.5)^222 is (2.25)^111, and 222 being greater than 2.25, you have your solution.
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
No calculations needed, base numbers are not much different (200 and 300). The first one will multiply itself 333x and the second 222x...that just ridiculous simple.
The first number is much bigger.
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.
333 is only 30% bigger than 222. (Not even an order of magnitude). But in exponents, that 100+ extra powers is huge.
Wrong. 333 is 50% greater than 222.
@@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.
Oops
exponents, even if just a few number higher expand, well exponentially, so the higher exponent is the obvious choice
222 and 333 are very close to 200, so this is basically 200^333 vs 200^222.
You can apply log with base 222 and base 333 and log properties
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
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
Obviously it's the former. Exponents are powerful.
333 raise to power 222 is greater.
2×2×2=8 n 3×3= 9 so on multiple multipication we will get 333 to power 222 greater.
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!?
Simple if basics are applied: 222^3 is more than 100^3, while 333^2 is less.
If (222/333)^2 * 222 > 1 then 222^333 is greater than 333^222.
And (222/333)^2 * 222 is greater than 1, so 222^333 is greater than 333^222.
Rest of the terms are just symmetry.
Everything else is just to confuse you...
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)
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
Yeah, it's not even close. The first number is almost 100 times bigger
So much ink wasted. It's (222^(1.5))^222 compared to (222*1.5)^222. That is (222 * sqrt(222))^222 compared to (222*1.5)^222. What now remains to compare is sqrt(222) to 1.5.
Why not compare 2 raised to 3, which is 8, versus 3 raised to 2, which is 9. Or 1 raised to 2, equalling 1, versus 2 raised to 1, equalling 2.
Clearly this is just evaluating
(2x)^(3x) with (3x)^(2x), for x = 111.
Instantly, (8^x)(x^3x) being compared with (9^x) (x^2x),or
Directly comparing x^x with (9/8)^x
By heart, anyone should easily conclude that :
x^x > (1.125)^x, for x = 111.
Thus 222^333 > 333^222
Than plot in x =1 in your initial formular... and also none of your steps after creating your formular make any obvuis sense
@@eragonshurtugal4239 mAyBe, afterall, there's a reason fof your failure in gettin decent amount of subscribers.
Ever since,
(8^x)*(x^3x) and (9^x)*(x^2x),
any thinking humanbeing would know that IF x = 1, then the term
x^3x and x^2x will be the SAME!
Whilst here, x = 111.
do you NOT know that 111 ≠ 1??
get oFF your meds!
you're tripping!
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
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.
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
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.
Why is this a math Olympiad problem. Anyone with a basic understanding of exponents comes to the correct answer immediately.
I don't think a calculator will help you anyway unless you can distinguish between different variants of buffer overflow.
A difference in exponents tends to just be worth more than the regular number especially at these scales
This will get to (2)^3 × (111)^3 and (3)^2 × (111)^2
Obviously 8 × 111^3 is greater than 9× 111^2
Or do a rough, quick count in your head and realise that 200^3 is much bigger than 300^2.
222 raised to 333 is approximately 2.155 x 10^620.
333 raised to 222 is approximately 1.345 x 10^406.
Therefore, 222 raised to 333 is significantly larger than 333 raised to 222.
Although my first method was what is being explauned, this can be done with common sense by finding the difference between the numbers and powers and reducing large numbers to small ones for logic.
It's fairly obvious as increasing the base number by 50% is insignificant to increasing an exponent by 50%
^333 likely to be bigger based on proximity of numbers.
It was supposed to do without calculation! You did calculation!
I did it without at about 30 seconds no paper or pen. You just do logarithm and you see that log for 333 is less of an factor than 333 is to 222.
222^333>333^222 I don’t need to think about it.
solving a math problem w a pen is max confidence
Intuitively picked the larger exponent one, the base number wasn't different enough for it to matter
I'm even surprised that 222^230 isn't bigger than 333^222
Just looking at it without calculating, it’s pretty easy to see that 222^333 is much much greater than 333^222
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
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
@@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.
@@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😀