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# 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

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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?

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

Mellie  Jul 1, 2015

#2
+91450
+10

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

Melody  Jul 2, 2015
Sort:

#1
+80983
+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
+91450
+10

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

Melody  Jul 2, 2015

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