How many zeros at the end of multiplication of natural numbers from 10 to 75?
What i did: $$\left[\frac{75}{5^1}\right]+\left[\frac{75}{5^2}\right]=18=\text{Zeros at the end of 75!}$$ $$\left[\frac{10}{5^1}\right]=2=\text{Zeros at the end of 10!}$$ $$18-2=16$$
But correct answer is $17$.
The multiplication of the numbers 10 to 75 is the same as $10*11*12...73*74*75$ which can be represented as: $$x=\frac{75!}{9!}$$ Sterlings approximations states: $$n!\approx\sqrt{2\pi n}\left(\frac{n}{e}\right)^n$$ so we know that: $$x=\frac{\sqrt{2\pi *75}\left(\frac{75}{e}\right)^{75}}{\sqrt{2\pi*9}\left(\frac{9}{e}\right)^9}=\sqrt{\frac{75}{9}}*75^{75}*9^9*e^{66}\approx2.89*10^{140}*10^{9}*10^{29}=2.89*10^{178}$$ This is my best approximation of the answer, I cant think of a way that nobody else has shown of working out the zeros on the end.