1. ## Very large numbers!

(3^10)! / 3^x = k

x and k are integers.

What is the maximum value for x ?

I think: (3^10 - 1) / 2 = 29524.

Sloane's sequence #A003462

Agree?

2. ## Re: Very large numbers!

$3^{10}!$ contains all factors $3^k$ up to $k=10$

the product of these is $3^{55}$

so $x$ is at least $55$

3. ## Re: Very large numbers!

If we evaluate how many of the the prime factors of 3^10 are 3's, we can start by noting that every third term of the factorial is divisible by three. So there are 3^9 terms that are a divisible by 3.

Similarly every ninth term is a factor of 9, so there are 3^8 terms that are divisible by 9.

Keep going. and the total number of 3's that make up the prime factors of 3^10 is 3^9 + 3^8 + 3^ 7 + ...+ 3^1 + 3^0. That seems to equal your answer of (3^10-1)/2, but I'm not sure why....

Edit - OK, got it. It's not hard to show that:

$\displaystyle \sum _{i=0} ^n k^i = \frac {k^{n+1}-1}{k-1}$

Hence for k=3 and n = 9:

$\displaystyle \sum _{i=0} ^{9} 3^i = \frac {3^{10}-1}{2}$

4. ## Re: Very large numbers!

Denis,
Your result follows immediately from Legendre's Theorem (Legendre's Theorem - The Prime Factorization of Factorials). This gives the exact power of p, a prime, that divides n factorial.