Japanese Math Olympiad Problem | A Nice Math Problem : Comparison

Math Olympiad Question

[aumotion]

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 = and increase the other one to 400^2 = 160000 ... no brainer :) no calculator needed in a few seconds

[edzehoo]

ChatGPT 3.5 be like : "Yes, I can see my mistake now. You are right. 1 is indeed greater than 3."

[thePronto]

When one exponent is so much larger than the other, the solution is intuitive.

[alecearnshaw]

Lovely problem, thank you!!!! The exact demonstration is great, and shows just how much stronger a larger exponen is!

[guillermotell]

Take the power 1/111 on both sides (a monotonous operation), then divide by 111^2 and you are left with 888 > 9.

[ericlindholm]

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.

[Nobody-hscl]

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

[davemiller]

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.

[carmanragatz]

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.

[jitendrapatil]

Physics and Mathematics are my favourite subjects. Now I am in IT field so hardly use it. Thanks for uploading the video.

[alscents]

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 <= a < b or b < a <= e (both are on the same side of e)
Since,
e <= 222 < 333
222^(1/222) > 333^(1/333)
222^333 > 333^222


857 and 859 are twin primes and very close. If I ask you to solve, which one is greater
857^859 OR 859^857, what are you going to do?
Using the method mentioned above you know the answer is going to be 857^859 > 859^857, because e <= 857 < 859

[pengbertuuu]

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

[AloneStroller]

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.

[MikeAnn]

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. 🙂

[MsLeober]

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.

[committedtohealth]

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

[tt]

It's not the destination but the journey that separates math olympians from casuals.

[MrPullela]

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

[eric_welch]

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 Youtube comments already)

[czennqt]

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.