#1**0 **

*How many terminal zeroes does 40^8 * 75^18 have?*

Well, 40 to the 8th will have 8 terminal zeros. However many terminal zeros in the base, there will be that many times the exponent in the answer. examples: 10^{2}=100 10^{3}=1,000 10^{4}=10,000 100^{2}=10,000 1,000^{2}=1,000,000

As regards the second factor, any number ending with 5 no matter to what exponent you raise it, the answer will always end with a 5. examples: 5^{2}=25 15^{2}=225 25^{2}=625 105^{2}=11,025 etc. So the second factor adds no terminal zeros to the product.

Multiplying the two factors together, there are 8 terminal zeros in the product.

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Guest Mar 24, 2019

edited by
Guest
Mar 24, 2019

#2**0 **

Note that we can write 40^8 * 75^18 as

( 4 * 10)^8 * ( 3 * 5^2)^18 =

( 2^3 * 5)^8 * ( 3 * 5^2)^18 =

(2 ^3)^8 * (5)^8 * 5^36 * 3 ^18 =

2^24 * 5^44 * 3^18 =

2^24 * 5^24 * 5^20 * 3^18

(2 * 5)^24 * 5^20 * 3^18 =

5^20 * 3^18 * ( 2 * 5) ^24

[5^20 * 3^18 ] * (10)^24 which has 24 terminal zeroes

CPhill Mar 24, 2019

#3**+1 **

I'm the Guest who posted the first answer. I see how I made a mistake. 40^{8} is 65,536 x 10^{8}.

The 6 at the end of 65,536 times the 5 at the end of 75^{18} will produce at least one additional terminal zero.

I need go no further than that to realize the approach to my answer was simplistic, and, worse, it was wrong.

I'm mortified to make such an elementary arithmetic oversight, and I apologize if my blunder led anyone astray.

__ edited to add:__ Thank you, CPhill, for posting the correct answer.

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Guest Mar 24, 2019

edited by
Guest
Mar 24, 2019