How many terminal zeroes does $40^8 \cdot 75^{18}$ have? That is, when the digits of $40^8 \cdot 75^{18}$ are written out, how many $0$s app

Thanks Chris,

I am going to do it too    

$$\\40^8\times75^{18}\\\\
=10^8\times 4^8\times (\frac{3}{4}\times 100)^{18}\\\\
=10^8\times 4^8\times \frac{3^{18}}{4^{18}}\times 100^{18}\\\\
=10^8\times \frac{3^{18}}{4^{10}}\times 100^{10}\times 100^{8}\\\\
=10^8\times 100^{8}\times 3^{18}\times 25^{10}\\\\
=10^8\times 10^{16}\times 3^{18}\times 25^{8}\\\\
=10^{24}\times 3^{18}\times 25^{8}\\\\
$25 to the 8 will end in a 5$\\
$3 to the power of anything does not end in a 0 $\\
$The last digits of powers of 3 are 3,9,7,1,3,9,$\\
$There are 4 numbers in the pattern. 18=2mod4$\\
$Hence 3 to the 18 ends in a 9$\\
so\\
3^{18}\times 25^{8} $ will end in a 5$\\\\$$

So there will be 24 trailing zeros - just like CPhill said   

Notice that 40^8  =  (4 * 10)^8   =  (4^8) * (10^8).......so this will have 8 trailing zeroes

And notice that 4^8  =  2^16 

Now, notice that 75 = (3 * 5 * 5)   .....so......  (75)^18 =  (3 * 5 * 5)^18  = [ 3^18 *  5^18 *  5^18] =

[3*18 * 5^18 * 5^2 * 5*16]  =  [3^18 * 5^20 * 5^16]

So    2^16 * 75^18   =    2^16 * [ 3^18 * 5^20 * 5^16]   =  [ 3^18 * 5^20] * 2^16 * 5^16  =

[ 3^18 * 5^20]  *  (2 * 5)^16  = [ 3^18 * 5^20] * (10)^16

Notice that the first product in the bracket doesn't add any zeros at all to our calculations.......thus.....

[ 3^18 * 5^20] * (10)^16   ....has 16 trailing zeros.....

And when this is combined with the  8 trailing zeros in the first part, we get 24 trailing zeros in all......