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Find the number of trailing zeros in 25^5 * 150^4 * 2008^3.

 May 7, 2020
 #1
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It's 4.00272916e25 so you can do it know :D

 May 7, 2020
 #2
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the number of trailing zeros is determined by how many powers of \(10\) are in the final product. \(25\) has two powers of \(5\)\(150\) has two powers of \(5\) and one power of \(2\), and \(2008\) has three powers of \(2 \). once we count the exponents, we have a total of nine powers of \(2\) and eighteen powers of \(5\), which gives us a total of nine powers of \(10\), giving us \(\boxed{9}\) trailing zeros.

 May 7, 2020
 #3
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 25^5 * 150^4 * 2008^3 = 40,027,291,593,750,000,000,000,000

 

As you can clearly see, there are 13 trailing zeros.

 May 7, 2020

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