Let n be a positive integer greater than 5. Show that the sum

Why? That's dumb lul

Because you're not a brainlet

Trivial (prime factorization).

>>9092575
sum of the divisors of n! is 1 + 2 + 3 + ... + n which is always lower than n! for every n > 5

>>9092596
>sum of the divisors of n! is 1 + 2 + 3 + ... + n
imagine being this wrong

>>9092596
>his sum of divisors of n! doesn't even include n!
Why did you even post?

>>9092618
there is nothing mentioned about factoring

>>9092618
so wrong you look and see numbers

>>9092625
>there is nothing mentioned about factoring
What?

>>9092627
>so wrong you look and see numbers
What?

>>9092575
Do your own homework OP.

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>>9092575
What a headache...
I'll keep trying.

>>9092575
https://en.wikipedia.org/wiki/Perfect_number

https://en.wikipedia.org/wiki/Euclid%E2%80%93Euler_theorem

>>9092672
Close but no cigar

>>9092697
Find me an n! that is a power of 2 times a prime.
It's done enough for me.
>>9092672
>>9092659

>>9092575
>>9092697

P.S. n can be an integer greater than 3
I just expanded your theorem lol.

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Probably proved through induction? Too tired rn to try. Wrote a quick python script - the pattern seems to hold.
Left side is n, right side is divisor-sum(n!)%(n!)

>>9092809
see
>>9092672

>>9092630
IMAGINE BEING AT COMPUTERS
SO FAT YOU LOOK AND YOU SEE FOOD

>>9092672
>>9092672
>https://en.wikipedia.org/wiki/Perfect_number
>https://en.wikipedia.org/wiki/Euclid%E2%80%93Euler_theorem
How exactly does this solve OPs problem? He asked about divisibility, not equality.