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 appear at the end?

$$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 appear at the end?$$

Mellie
Jul 1, 2015

#2**+10 **#### So there will be 24 trailing zeros - just like CPhill said

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$\\\\$$

Melody
Jul 2, 2015

#1**+10 **

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......

CPhill
Jul 1, 2015

#2**+10 **

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

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$\\\\$$

Melody
Jul 2, 2015