## Task 4 - 070614

The factorial of a natural number $n \geq 2$ (written as $n$ !) is the product of all natural numbers from 1 to $n$.

For example, $3!=1 \cdot 2 \cdot 3=6$ and $4!=1 \cdot 2 \cdot 3 \cdot 4=24$.

Determine the digit on which the sum $s=3!+4!+5!+6!+7!+8!+9!+10!$ ends!
The first factorials of the sum end with $3!=6$ and $4!=24$ on 4.

5! and all higher factorials always end in 0, as they contain the factors 2 and 5. The sum s thus has, since $4+6=10$, the zero as the last digit.