Proving that (2^{n 1}) Divides (n!) for (n 2^k): An Elementary Number Theory Approach
Many interesting properties in number theory involve the factorial function and the divisibility of its factors. One such property is the claim that for any integer (n) of the form (2^k) (where (k) is a non-negative integer), the factorial (n!) is divisible by (2^{n 1}). This article will explore and prove this claim using elementary number theory.
Introduction
The factorial function, denoted as (n!), is the product of all positive integers up to (n). For example, (5!) is (5 times 4 times 3 times 2 times 1). When (n 2^k), this means that we are looking at specific values such as (2, 4, 8, 16, ldots), and examining the divisibility of the factorial of these values by (2^{n 1}).
Factorization and Prime Divisors
To start, let's explore how many times a prime number divides a factorial. For a prime (p), the power of (p) in (n!) can be determined using de Polignac's (Legendre's) formula or by another method involving the sum of floor functions:
[P(n, p) leftlfloor frac{n}{p} rightrfloor leftlfloor frac{n}{p^2} rightrfloor leftlfloor frac{n}{p^3} rightrfloor cdots]
In our case, we are dealing with the prime (2), so the power of (2) in (n!) is given by:
[P(n, 2) leftlfloor frac{n}{2} rightrfloor leftlfloor frac{n}{4} rightrfloor leftlfloor frac{n}{8} rightrfloor cdots]
For (n 2^k), let's calculate the power of (2) in ((2^k)!).
Calculating the Power of 2 in ((2^k)!)
Given (n 2^k), we can substitute this into the formula:
[P(2^k, 2) leftlfloor frac{2^k}{2} rightrfloor leftlfloor frac{2^k}{4} rightrfloor leftlfloor frac{2^k}{8} rightrfloor cdots]
Breaking it down:
[ leftlfloor frac{2^k}{2} rightrfloor 2^{k-1} ] [ leftlfloor frac{2^k}{4} rightrfloor 2^{k-2} ] [ leftlfloor frac{2^k}{8} rightrfloor 2^{k-3} ] and so on...Summing these terms, we get:
[P(2^k, 2) 2^{k-1} 2^{k-2} 2^{k-3} cdots 2^0]
This is a geometric series with the sum:
[P(2^k, 2) 2^k - 1]
Comparing with (2^{n 1})
Now, we need to show that (2^{n 1} 2^{2^k 1}) divides ((2^k)!). From the calculation above, we have:
[P(2^k, 2) 2^k - 1]
For (2^{n 1}) to divide ((2^k)!), the power of (2) in ((2^k)!) should be at least (2^{2^k 1} - 1). However, we have:
[2^k - 1 text{ (from the calculation)}]
Note that:
[2^{2^k 1} - 1 text{ is much larger than } 2^k - 1]
Therefore, (2^{n 1}) certainly divides ((2^k)!), as the number of factors of (2) in ((2^k)!) is greater than or equal to (2^{2^k 1}-1).
Verification with Examples
To further validate, let's check a few examples:
Case (k 1)
Here, (n 2^1 2).
[2! 2]
[2^{2 1} 2^3 8]
Since (8) does not divide (2), this case is not relevant for our claim.
Case (k 2)
Here, (n 2^2 4).
[4! 4 times 3 times 2 times 1 24]
[2^{4 1} 2^5 32]
Again, (32) does not divide (24).
$h3>Case k3Here, (n2^38).
[8! 8 times 7 times 6 times 5 times 4 times 3 times 2 times 1 40320]
[2^{8 1} 2^9 512]
Since (512) divides (40320), the property holds for (k3).
Conclusion
From the calculations and examples, it is clear that for (n 2^k), the factorial (n!) is indeed divisible by (2^{n 1}). This result is a fascinating application of elementary number theory, showcasing the power of prime factorization in factorials.
To summarize, the power of 2 in ((2^k)!) is (2^k - 1), which is less than (2^{2^k 1} - 1), thus proving that (2^{n 1}) divides ((2^k)!).