# How many consecutive zeros to the left of the decimal does $125^{14} \times48^{8}$ have?

If $$125^{14}\times48^{8}$$ were expressed as an integer, how many consecutive zeros would that integer have immediately to the left of the decimal point?

I tried:

$$125^{14} \times 48^{8} = (1.25\times10^2)^{14}\times(4.8\times10)^8$$ $$= 1.25^{14}\times10^{28}\times4.8^8\times10^8$$ $$= 1.25^{14}\times4.8^8\times10^{36}$$

then I could actually calculate $$1.25^{14}$$ on the calculator given and simplified to $$\approx2.27\times10$$. For $$4.8^8$$ it came to $$\approx2.8\times10^5$$. At this point I mistakenly thought the answer would have 42 zeros but that's of course wrong.

What is the fastest way to solve this?

Well the number of zeroes in base $$10$$ is simply the number of factors $$10$$ in the number itself. In your case:
$$125^{14}\times 48^8=5^{42}\times 2^{32} \times 3^{8}=10^{32}\times 5^{10}\times 3^{8}$$
So the answer is $$32$$.
You should instead go like $$125^{14} \times 48^8 = 5^{42} \times 2^{32} \times 3^{8} = 10^{32} \times 5^{10} \times 3^{8},$$ so there should be 32 zeroes.