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If I expand 25*24*23* ... *12*11*10, how many zeros are there at the end of the number I get?

 Feb 14, 2021
 #1
avatar+268 
-1

This website explains it well --> https://www.purplemath.com/modules/factzero.htm

25*24*23* ... *12*11*10 = 25! / 9!

 

25! has 6 trailing zeros. 

 

9! has 1 trailing zero. 

 

When you divide two numbers with trailing zeros, you can cancel some out. Since 25! has 6 trailing zeros and 9! has 1, there will be 5 trailing zeros left. 

 

However, you will have more trailing zeros since the numbers in 9! makes up a lot of the numbers in 25!. 

 

You need to find the numbers that give you 5s. 25, 20, 15, and 10 gives you fives. 
Now we need numbers that give 2s. 24, 22, 20, 18... but we can assume there are more 2s than 5s. This means:

 

You get 5 + 4 + 3 + 2 = 14 trailing zeros, plus the 5 from when we cancelled out zeros which gives you a total of 19 trailing zeros, which is your answer :)

 

 Feb 14, 2021
 #3
avatar+268 
0

I just realized my answer is really wrong.. Nice answer guest! 

Logarhythm  Feb 14, 2021
 #2
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+1

productfor(n, 10, 25,n)==42,744,736,671,436,800,000 - 5 trailing zeros!.

 Feb 14, 2021

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