Comparing Two Gigantic Numbers

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[@gauravraj3945]

Thanks

[@user-gq4vk8iy3c]

This is really cool.

[@scottleung9587]

Very cool!

[@vincehomoki1612]

2^24 = 16,777,216.

[@narfharder]

True color nerds be like

[@matheusjahnke8643]

I think I've seen quite a few with those problems;
Stirling approximation: log(n!)~ n log n
(using log as the natural logarithm)
(more like log(n!)=n log n - n + O(log(n))).... but we aren't going to be so careful)
instead of comparing 2^(24!) and (2^24)!... compare their logs
A=log (2^(24!))=24! log 2
B=log ((2^24)!)=2^24 log (2^24)=24(2^24)log 2
turns out 24! > 24(2^24)... so A = 24! log 2 > 24(2^24)log 2 = B
(to do that you can do the same reasoning used on the video... 23! > 2^23)

[@scottbilger9294]

Is there a solution for 2^x! = (2^x)! ?

[@stephenlesliebrown5959]

Yes indeed, the graphing site desmos shows us where 🙂

[@scottbilger9294]

@@stephenlesliebrown5959 Thank you

[@HoSza1]

x=1. no graphing needed for that

[@yoav613]

Nice and very easy😊

[@robertveith6383]

Thumbs-down on your "very easy" comment. If it were "very easy," it would not be taking the host seven minutes, 40 seconds to explain his method in the video.

[@yoav613]

@@robertveith6383 well,wehen i say very easy i mean it is very easy to me.i can explain it in 1 min: (2^24)!(2^24)!.

[@robertveith6383]

​​​​@@yoav613-- Your third and fourth lines are not clear in the writing. I think they are clear to you in your mind.

[@129140163]

5:37 The exact value of 2^24 is 16,777,216.
Therefore (2^24)! is 16,777,216!
So the comparison can be written as 2^(24!) vs 16,777,216!.

[@user-gq4vk8iy3c]

1st comment