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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?

 May 13, 2022
 #1
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productfor(n, 10, 25, n)==427,447,366,714,368,00,000

 May 13, 2022
 #2
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We can use the trailing zeroes trick. 

 

We can divide 25! by each power of 5. 

 

25/5=5. But since we don't include the first multiple of 5, there would be four fives. So this would get 4 zeroes.

 

Now we can divide by 25, or 5^2. This gets an additional one more zero.

 

This then gets 5 trailing zeroes.

 May 14, 2022

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