last non-zero digit of factorial

We will show how to compute the last non-zerodigit of the factorial of a number from itsdigits without having to compute the factorial itself.

Let $L(n)$ denote the last non-zero digit of $n$ in base 10.We note some basic properties of $L$ which can easilybe checked:

• For all $n$, we have $L(10n)=L(n)$.

• If $L(m)\ne 5$ and $L(n)\ne 5$, then $L(mn)=L(L(m)L(n))$.

We also tabulate the values of $L(n!)$ for small values of $n$:

$L(0!)$$=$$1$

$L(1!)$$=$$1$

$L(2!)$$=$$2$

$L(3!)$$=$$6$

$L(4!)$$=$$4$

$L(5!)$$=$$2$

$L(6!)$$=$$2$

$L(7!)$$=$$4$

$L(8!)$$=$$2$

$L(9!)$$=$$8$

$L(10!)$$=$$8$

Next, we make two less trivial observations:

For all positive integers $n$, we have $L(n!)\ne 5$.