Thread: Why large factorials loose small base number 2 powers while they grow?

1! = 1 = 2^0
2! = 2 = 2^1
3! = 6 = 2^1 + 2^2
4! = 24 = 2^3 + 2^4
5! = 120 = 2^3 + 2^4 + 2^5 + 2^6
6! = 720 = 2^4 + 2^6 + 2^7 + 2^9
7! = 5 040 = 2^4 + 2^5 + 2^7 + 2^8 + 2^9 + 2^12
8! = 40 320 = 2^7 + 2^8 + 2^10 + 2^11 + 2^12 + 2^15
9! = 362 880 = 2^7 + 2^8 + 2^11 + 2^15 + 2^16 + 2^18
10! = 3 628 800 = 2^8 + 2^9 + 2^10 + 2^11 + 2^12 + 2^14 + 2^16 + 2^17 + 2^18 + 2^20 + 2^21
11! = 39 916 800 = 2^8 + 2^10 + 2^12 + 2^16 + 2^21 + 2^22 + 2^25
12! = 479 001 600 = 2^10 + 2^11 + 2^12 + 2^13 + 2^14 + 2^15 + 2^18 + 2^19 + 2^23 + 2^26 + 2^27 + 2^28
13! = 6 227 020 800 = 2^10 + 2^11 + 2^14 + 2^15 + 2^19 + 2^21 + 2^24 + 2^25 + 2^28 + 2^29 + 2^30 + 2^32

So, smallest power of base number 2 is:
2^0 for 1!
2^1 for 2! and 3!
2^3 for 4! and 5!
2^4 for 6! and 7!
2^7 for 8! and 9!
2^8 for 10! and 11!
2^10 for 12! and 13!

Is this a universal feature? Can we be sure that, say, 100! contains no 2^1 or 2^2 components?
What is the explanation for this phenomenon?
Smallest components also seem to appear as pairs, so that an even numbered factorial and its bigger neighbor, odd numbered factorial, both contain the same smallest component.
Is this also a universal phenomenon and what is the explanation for it?

This is just a consequence of binary arithmetic isn't it ?

For starters, think base 10. If we have 100 and multiply by any other number the result will have at least two zeros at the end. If we multiply 1000 by any other number, the product wll have at least three zeros at the end, and so on.

The same sort of thing happens when numbers are written in binary, base two.

For example 4! (base 10) = 24 = 11000 (base 2).

Multiplying this by any number (base 2), the product will have at least three zeros at the end.

So, 5! = 5*4! = 120 (base 10) = 101 * 11000 = 1111000 (base 2).

That number at the end is

Yes, that's a pretty obvious result. There are either n/2 or (n-1)/2 (depending on whether n is even or odd) even numbers less than n each of which has at least one factor of 2. Similarly, there are either n/4, (n-1)/4, (n- 2)/4, or (n-3)/4 numbers less than n that has at least one factor of 4 and so another factor of 2. That is, the larger a number is, the more factors of 2 numbers less than it will have and so the more factors of 2 n! will have. And, once you write n! in as sums of powers of 2 the higher that lowest power will be.

Thank you both of you!