# Prove $10^{\dfrac{5^n -1}{4}} \Huge\vert (5^n)!$ for $Z^+$

• MHB
• karush
In summary, the conversation revolves around proving the formula $10^{\dfrac{5^n -1}{4}} \Huge\vert (5^n)!$ for all positive integers $n$. The vertical bar is interpreted as "divides into evenly" and the factorial symbol is confirmed to mean the factorial operation. The conversation also mentions Legendre's formula and using mathematical induction to prove the formula.
karush
Gold Member
MHB
prove
$10^{\dfrac{5^n -1}{4}} \Huge\vert (5^n)!$
for all $Z^+$

ok this was sent to me on email but thot I could solve it but ?

first I assume the vertical bar means such that

not sure if ! means factorial or not negative

anyway curious

$a \mid b$ means $a$ divides $b$.

The following formula may help.

Legendre's formula: For any prime number $p$ and any positive integer $m$, let
${\displaystyle \nu _{p}(m)}$ be the exponent of the largest power of $p$ that divides $m$. Then

$${\displaystyle \nu _{p}(m!)=\sum _{i=1}^{\infty }\left\lfloor {\frac {m}{p^{i}}}\right\rfloor}$$

ok i have never tried that
are you sure the vertical bar means a/b?

karush said:
ok i have never tried that
are you sure the vertical bar means a/b?
Not a/b it's a divides into b evenly. For example, 3|15 and 5|15.

-Dan

oh
can we prove this just by setting n=1

karush said:
oh
can we prove this just by setting n=1
I don't know what you mean.

It's true for $n=1$ because $10$ divides $120$.

karush said:
oh
can we prove this just by setting n=1
From there use Mathematical Induction. It should be rather easy but I haven't really spent much time thinking about it. (ie. If it works for n = 1 then you always have that factor.)

-Dan

## What is the statement being proven?

The statement being proven is that for all positive integers $n$, the expression $10^{\dfrac{5^n -1}{4}}$ divides the factorial of $5^n$.

## What does the notation $Z^+$ mean?

The notation $Z^+$ refers to the set of positive integers, also known as the natural numbers.

## How can this statement be proven?

This statement can be proven using mathematical induction. We will start by showing that the statement is true for $n=1$, and then assume it is true for some arbitrary positive integer $k$. Using this assumption, we will then show that the statement is also true for $n=k+1$. This will prove that the statement is true for all positive integers.

## Why is mathematical induction used to prove this statement?

Mathematical induction is a commonly used method for proving statements about integers. It involves showing that a statement is true for a base case, and then showing that if the statement is true for some integer $k$, it must also be true for the next integer $k+1$. This allows us to extend the truth of the statement to all positive integers.

## Is there a specific formula or theorem being used in this proof?

Yes, this proof uses the theorem that if a statement is true for some integer $k$, and if it can be shown that the statement being true for $k+1$ follows logically from the statement being true for $k$, then the statement is true for all positive integers.

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