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

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:  10²=100  10³=1,000  10⁴=10,000  100²=10,000  1,000²=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: 5²=25  15²=225  25²=625  105²=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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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

I'm the Guest who posted the first answer.  I see how I made a mistake.  40⁸ is 65,536 x 10⁸. 

The 6 at the end of 65,536 times the 5 at the end of 75¹⁸ 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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