smc

Let $69=a$. Therefore, the equation becomes $a^5+5a^4+10a^3+10a^2+5a+1$. From Pascal's Triangle, we know this equation is equal to $(a+1{)}^5$. Simplifying, we have the desired sum is equal to $70^5$ which can be prime factorized as $2^5\cdot 5^5\cdot 7^5$. Finally, we can count the number of factors of this number. $6\cdot 6\cdot 6=\boxed{216}$.