1. ## Factor Large Factorial

How could I factor 30! into prime factors?

Can you show the steps please...

2. Originally Posted by jzellt
How could I factor 30! into prime factors?

Can you show the steps please...

Prime factorize $\displaystyle 1,\cdots,30$

3. Hello, jzellt!

I'll use the "greatest integer function": .$\displaystyle [x]$

Factor 30! into prime factors.
30! contains $\displaystyle 2^{26}$
. . How do we know this?

Every 2nd term is even, contains a factor of two.
. . There are: .$\displaystyle \left[\frac{30}{2}\right] \:=\:15$ factors of two.

But every 4th term has a factor of 4 $\displaystyle (2^2)$, which contributes an additional two.
. . There are: .$\displaystyle \left[\frac{30}{4}\right] \,=\,7$ more factors of two.

And every 8th term has a factor of 8 $\displaystyle (2^3)$, which contributes yet another two.
. . There are: .$\displaystyle \left[\frac{30}{8}\right] \,=\,3$ more factors of two.

Also every 16th term has a factor of 16 $\displaystyle (2^4)$, which ... etc.
. . There are: .$\displaystyle \left[\frac{30}{16}\right]\,=\,1$ more factor of two.

Hence, $\displaystyle 30!$ contains: .$\displaystyle 15 + 7 + 3 + 1 \:=\:26$ factors of two.

In a similar fashion, we find the number of threes: .$\displaystyle \left[\frac{30}{3}\right] + \left[\frac{30}{9}\right] + \left[\frac{30}{27}\right] \:=\:10 + 3 + 1 \:=\:14$

And the number of fives: .$\displaystyle \left[\frac{30}{5}\right] + \left[\frac{30}{25}\right] \:=\:7$

And the number of sevens: .$\displaystyle \left[\frac{30}{7}\right] \,=\,4$

. . and so on . . .

Therefore: .$\displaystyle 30! \;=\;2^{26}\cdot3^{14}\cdot5^7\cdot7^4\cdot11^2\cd ot13^2\cdot17\cdot19\cdot23\cdot29$

4. Thank you for a superb explanation on how to factor into prime numbers. I've been reading on internet for three days and didn't get it until now.

I just want to add the following comparison with 10!
(since I personally didn't understand your explanation fully and I was a bit suspicious as to where number 6 went - (in the calculation of 10!))

(1*)2*3*4*5*6*7*8*9*10
2*2*2*2*2*2*2*2*3*3*3*3*5*5*7

Both sequences can be eliminated with each other like in a crossword, and it is super-cool.

I guess the prime factors of a factorial really explains why where is no square prime in the factorial of n!
The last prime is always single.

But is there an ordinary square in any n! (?) And if so, would that be possible to puzzle the prime numbers to a square like in the graphical rectangle representation of the fibonacci sequence?(but in a square?)