Comparing 999! and 500^999 in Two Ways

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[trynalive]

If we sum the numbers we multiply in each one:
1+2+...+999 = 999*500
500+...+500 = 999*500

If we know a + b = c
Then max of a*b is when they are equal(c/2)
Which means 500^999 is clearly greater

[bargiona]

My method was a bit cheap, but fast. I multiplied 499*501. This is the biggest resulting pair possible in the 999! among the 500 combinations proposed in method 1, and it is still slightly smaller than 500*500. Thus, it must be smaller than 500^999.

[pcklop]

I think I found the simplest solution: the number of base 10 digits a number has is just given by the ceiling of the log of the number. 500^999 has (rounding up) 999log(500)=2697 digits. The number of digits of 999! can also be found using properties of logs. 999! has a number of digits equal to the sum from 1 to 999 of log(n), which is (rounding up) 2565. 500^999 has over 100 more digits base 10, so it is larger.

[jacobfrancobandiero]

Another interesting solution to this problem can be found by taking the natural log of both summations. Using basic logarithmic rules ln(500^999) simplifies out to 999*ln(500) which is equal to 6208.4. For ln(999!) you can use a simplified version of Sterling's approximation which shows that ln(N!) is approximately [ N*ln(N) ] - N. Which approximates ln(999!) to 999*ln(999) - 999 which is equal to 5900.8, which shows 500^999

[iqskirby]

When I was younger, one of the things I noticed was that 5^2 was greater than 4•6, and that this pattern, where 5 is n, 4 in n-1, and 6 is n+1, would always hold true. When looking at this problem, after enough thought, that inequality came back to my head, to which I determined 500^999 was greater than 999! because of the similar elements of a higher number being multiplied by its opposite pair relative to the center. This matches with the first method, which was explained more elegantly in the video.

[songwaikit]

I actually used a different approach, I did 999!/500^999 and I found that 999/500 x 1/500 &, 501/500 x 499/500 are decimals. Therefore every other multiplied pairs are decimal indicating that the denominator is larger than the numerator.

[Imadthebest]

For c>0 and x=/=0; (x-c)(x+c)=x^2-c^2<x^2.
Re-write each number in the multiplciation from 1 to 999 in 999! as 500 +/- c.
Product < than product with two factors with identical c's being replaced by 500 each.
Repeat 498 other times for each c's. You get 500^999 > 999!.

[nnirr]

1:25 duotrigintillion -> duo = 2, triginti = 30, the 32th million. (Just like billion is the 2nd million and trillion is the 3rd million).
With each new -illion there are 3 zeros that needs to be added. So 32 * 3 = 96 zeros
Add another 3 zeros for the initial million (cause it has 6 zeros insread of 3) and you get 99. So in order to get 100 zeros you need 10 duo-trginti-llion

[gregwochlik]

Before the start of the video, I fired up Python (Programming language) and calculated these numbers. The factorial came out to have 2564 digits, while the exponential 2696. I used the stock-standard math library in Python.

[jacrispy]

You can also do this using growth rates.
If you compare x! to 500^x, you find that the latter has a higher growth rate than the other. So, at x=999,
x! < 500^x

[hobbyist]

I went with 999! being equivalent to 500(500+1)(500-1)...
Which, when you FOIL each pair out, you get 500 times itself 999 times, but minus a bunch of numbers. 500^999 on the other hand has no subtraction, thus it's greater.

[danielhudon]

I think method 1 is more intuitive but the last step of method 2 makes it very elegant - nice!

[elwind]

500^999 because intuitively the more centralized your factors of equal sum, the larger the product. Works well in strategy board games when you have multipliers on points and you want to centralize your factors in the product to get the most points.

[MOMO-PLAYGAMES]

The board before i sleep during math: 2:00
The board when i wake up: 6:50

[theabbie]

sterlings formula gives n! = (n/e)^n for large n, though 999 might not be considered large, it's a very good approximate,

taking logs on both sides,

log (999!) = log ((999/e)^999) = 999log(999/e) = 2561 approx.

log (500^999) = 999 log (500) = 2696 approx.

thus, 500^999 is bigger

[aryanmaheshwari]

Thanks for making this. I left maths as a subject to study 4 years ago, and I needed to be reminded how beautiful it is.

[RoderickEtheria]

With 999!, you get 499 non-squares with the same perimeter as the 499 squares you can form with 500^999, after dividing one of the 500s from both calculations. The 499 squares are bigger than the 499 non-squares if you can add their length and width and have equal totals on both sides. 999!<500^999, easily seen immediately.

[slickytail]

Another nice method: take the log of both and then apply Jensen's inequality. You get 999 log(500) vs log(1)+log(2)+...+log(999). If you let U follow a uniform distribution on the integers from 1 to 999, then you have that E(U) = 500. So we're comparing log(E[U]) and E[log(U)]. Since the logarithm is concave, log(E[U]) is bigger by Jensen's inequality.

[ITR]

My half-solution:
First observe the functions (x\*2-1)! and x^(x\*2-1)
From x_0 to x_0+1, the first one has the relation y_1 = y_0\*(x_0+1)\*(x_0+2)
Magically know that because their exponential nature, one of the functions will always be steeper than the other.

Observe that both functions equal each other when x=1, and that the second is larger when x=2, and since neither have decreased in value, that means the second always will be larger. Thus 500^999 is larger.

[jmart]

Great demonstration!
By the way, the number of digits can be estimated using Stirling's approximation for big values of n.

ln(n!)=n*ln(n)-n

ln(999!)=999*ln(999)-999

ln(999!)= 5901

log(999!)= 5901*log e = 2563