Solving A Nice Factorial Equation

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PLAYLISTS 🎵 :

[Barteqw]

Take care, warm hugs, hope you will get better soon!

[허공]

n! = 8*(2^(k-3) - 1)
n! which is 8 * odd number is only when n = 4 or n = 5.

from morocco thank you...i pray ALLAH to heal you and give you health and wealth

[mikecaetano]

The music works better with the math when slowed to 0.75 playback speed. Especially this music. It's like silent movie music. Ragtime. I half expect a jump cut to a damsel in distress tied up on some train tracks. Buster Keaton gets schooled in factorials and abstract algebra!

[christophebosquilloncrypto]

Hey old friend, get well soon!!! (of course, enjoyed the exercise, as always, best mental training ever ;-)

[zrthus]

I did this slightly differently. Starting from the step where you have n! = 2^3 * (2^k-1), remember that no integer multiplied by an even number can ever result in an odd product. So, because (2^k-1) is odd, 2^3 tells us exactly how many nonzero powers of 2 we traverse in this factorial-- 2^1 ^ 2^2 = 2^3. So n must be less than 6, because otherwise the even term would be greater than 2^3. n must greater than or equal to 4, because of the even term. n cannot be 4 because then k would be 0, and then n! would be a negative number, which cannot be. So n must be 5.

[yoav]

Nice.but infact, youכ do not even need to check n=7 or n=6 since for n>=6 n! Divides 16 so n!+8=8(mod16)

[georgesbv]

wrong direction ... you'll get fast one of 4! or 5! to be the only that divide 8, but not 16

[roberttelarket]

You've lost your voice but haven’t lost your mind!

[rey-dqnx]

still can’t speak. This is like Charlie Chaplin silent movie. 🤣🤣🤣But I still enjoy your video. Get well soon!

[prollysine]

5!=120, 120+8=128, 128=2^k, n, k = 5, 7, / & , 4, 5, ... took over... /,

[Don-Ensley]

problem

N! + 8 = 2ᴷ

Note that 0! Is 1. Also 8 = 2³.
Another form is

N! = 2ᴷ-2³
N! = 8 [2⁽ᴷ⁻³⁾ - 1]

Now K >= 4 because K = 3 says n! = 0 which is not true for any N value. K = 4 is the first for possible solutions which value gives us

N! = 8

But N = 3 says

6 = 8,

Which is untrue. The next higher N = 4 says

24 = 8.

From this we conclude that there are no solutions for K = 4.

As K increases from 4, we can devise a table of values to examine 8 [2⁽ᴷ⁻³⁾ - 1] for similarity to factorials.

K 8 [2⁽ᴷ⁻³⁾ - 1] N
--
4 8 none
5 24 4
6 56 none
7 120 5
8 248 none
9 504 none
10 1016 none
11 2040 none
12 4088 none
13 8184 none
14 16376 none
15 32760 none

It turns out that these are all the solutions I could find.

answer

(K, N) ∈ { (5, 4), (7, 5) }