Find all prime numbers that divide 50

[Math100]

Homework Statement

Find all prime numbers that divide 50!.

Relevant Equations

None.

Proof:

Note that all primes less than 50 will divide 50!,
because each prime is a term of 50!.
Applying the Fundamental Theorem of Arithmetic produces:
Each term k of 50! that is non-prime has a unique prime factorization.
Since 48, 49 and 50 are not primes,
it follows that all primes $\leq47$ divide 50!.
Therefore, all primes $\leq47$ divide 50!.

 

[fresh_42]

$50!$ is so large, why can't there be prime factors above $50$?

 

[Math100]

I haven't thought about that.

 

[Delta2]

I haven't thought about that.

In my opinion there aren't any prime numbers larger than 50 that divide 50! but you need to provide a justification for that.

The fundamental theorem of arithmetic and the definition of factorial will aid you towards that.

 

[Mark44]

because each prime is a term of 50!.

Technically, a term is an expression that is part of a sum. I think what you meant was each prime less than 50 is a factor of 50! .

 

[Math100]

In my opinion there aren't any prime numbers larger than 50 that divide 50! but you need to provide a justification for that.

The fundamental theorem of arithmetic and the definition of factorial will aid you towards that.

So is my proof correct or not? Do I need to revise anything?

 

[fresh_42]

So is my proof correct or not? Do I need to revise anything?

It is correct as all primes less than 48 divide 50!. But you need to prove, too, that no other prime divides 50!.

Think about the definition: A prime is a number that divides already at least one factor if it divides a product.

 

[Math100]

By definition, factorial of a number is the function $n!=(n-1)!$ that
multiplies the number by every natural number below it.
Note that all primes less than $50$ will divide $50!$,
because each prime less than 50 is a factor of $50!$.
Applying the Fundamental Theorem of Arithmetic produces:
Each term k of $50!$ that is non-prime has a unique prime factorization.
Since 48, 49 and 50 are not primes,
it follows that all primes $\leq47$ divide $50!$.
Therefore, all primes $\leq47$ divide $50!$.

 

[Mark44]

By definition, factorial of a number is the function $n!=(n-1)!$

No on two counts: 1) n! = n * (n - 1)!, and 2) what you've written is not a function.

 

[Delta2]

Can you prove, using the fundamental theorem of arithmetic, that 50! can be written as a product of primes that are less than 50 (ok less than 48 to be more precise). Then according to the fundamental theorem again this prime factorization is the only factorization , hence 50! does not contain primes larger than 50.

You have prove it actually using your proof, you just don't state it explicitly.

 

[Math100]

$n!=n\times(n-1)\times\dotsb\times1$

 

[Delta2]

Each term k of 50! that is non-prime has a unique prime factorization.

This statement of yours has actually an implication that proves what we want.

 

[Math100]

This statement of yours has actually an implication that proves what we want.

What should be the implication after this statement then?

 

[Delta2]

What should be the implication after this statement then?

That the number x=50! has a unique prime factorization where all primes present are smaller than 50.

 

[phinds]

That the number x=50! has a unique prime factorization where all primes present are smaller than 50.

A prime factorization means that ALL of the factors are primes. Are all of the factors in 50! primes?

 

[Delta2]

Are all of the factors in 50! primes?

No but I didn't say that. I said that in the unique prime factorization of 50! all primes are smaller than 50.

 

[phinds]

No but I didn't say that. I said that in the unique prime factorization of 50! all primes are smaller than 50.

But what I am saying is that 50! does not HAVE a "prime factorization" because it cannot be expressed as a multiple of only primes.

 

[Delta2]

But what I am saying is that 50! does not HAVE a "prime factorization" because it cannot be expressed as a multiple of only primes.

It can't be, then the fundamental theorem of number theory is wrong, we know that every natural number can be expressed as a multiple of only primes.

 

[phinds]

It can't be, then the fundamental theorem of number theory is wrong, we know that every natural number can be expressed as a multiple of only primes.

Oh, I get it. The non-prime factors all have prime factorizations. My bad.

 

[Mark44]

Oh, I get it. The non-prime factors all have prime factorizations. My bad.

For a simpler example than 50!,
$6! = 2 * 3 * 4 * 5 * 6 = 2 * 3 * 2^2 * 5 * 2 * 3 = 2^4 * 3^2 * 5$
We can express any natural number as a product of primes or primes raised to powers. The representation is unique up to the order in which the factors appear. For the reason that @Delta2 gave, 6! couldn't possibly have any prime factors larger than 6.

 

[Delta2]

Oh, I get it. The non-prime factors all have prime factorizations. My bad.

Yes that's right. So every one of the fifty factors of 50! can be expressed as a product of primes that each is smaller than 50 (because the factor itself is smaller than 50) and the whole thing is a prime factorization of 50! , hence it is the unique prime factorization according to the fundamental theorem.

 

[PeroK]

But what I am saying is that 50! does not HAVE a "prime factorization" because it cannot be expressed as a multiple of only primes.

The trick is that for each prime $p$ less than 50 we have:

One factor for every multiple of $p$; another factor for every multiple of $p^2$; another for every multiple of $p^3$ etc. E.g. for $p = 2$ the total number of factors in $50!$ is:

$25 + 12 + 6 + 3 + 1 = 47$

And:$$50! = 2^{47} \times 3^{22} \times 5^{12} \times 7^8 \times 11^4 \times 13^3 \times 17^2 \times 19^2 \times 23^2 \times 29 \times 31 \times 37 \times 41 \times 43 \times 47$$And, in particular, the last $12$ digits of $50!$ are all zeroes! You get this by looking at the powers of $5$ (as there are always more powers of $2$).

 

[fresh_42]

One can also simply work with the definition of a prime, no fundamental theorem of arithmetics is necessary.
\begin{align*}
p\,|\,50!=50\cdot 49! &\Longrightarrow p\,|\,50 \text{ or } p\,|\,49!\\
p\,|\,49!=49\cdot 48! &\Longrightarrow p\,|\,49 \text{ or } p\,|\,48!\\
&\ldots\\
p\,|\,2!=2\cdot 1! &\Longrightarrow p\,|\,2 \text{ or } p\,|\,1!
\end{align*}
Thus $p\,|\,50! \Longrightarrow p\,|\,50\text{ or }p\,|\,49\text{ or }\ldots p\,|\,2.$