Factorial meets Minimum

Start by finding the amount of $2^{x}$ in $80!$ , it will follow the $\sum_{n=1}^{6} \lfloor \frac{80}{2^{n}}\rfloor$ , which equals to $40+20+10+5+2+1 = 78$ , therefore $80! = \alpha \times 2^{78}$ where $\alpha$ cannot be divided by 2

To make $80!$ not divisible by 8, we will try to add the multiple of $2^{n}$ into $\alpha$ , which gives $80! = 4\alpha \times 2^{76}$ And we cannot add anymore of $2^{n}$ as it will make the number divisible by 8, therefore the minimum value so as to make $80!$ not divisible by 8 is $2^{76}$ , which, when compute, gives $a+b = 2+76 =78$