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

[karush]

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

 

[MountEvariste]

$a \mid b$ 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

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

 

[karush]

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

 

[topsquark]

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

 

[karush]

oh
can we prove this just by setting n=1

 

[MountEvariste]

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 $120$.

 

[topsquark]

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