SOLUTION: How many 9s are there at the end of the number (60!-50!-1)? (n!=n*(n-1)*(n-2)*...3*2*1)

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Question 1188402: How many 9s are there at the end of the number (60!-50!-1)?
(n!=n*(n-1)*(n-2)*...3*2*1)
Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!

# of 0's at the end of 60!
  60/5 = 12
  12/5 =  2
 -----------
         14

# of 0's at the end of 50!
  50/5 = 10
  10/5 =  2
 -----------
         12

There are 14 trailing 0's in 60! and only 12 in 50!, so 60!-50! has 12 trailing 0's.

That means 60!-50!-1 has 12 trailing 9's.

Using the pari calculator, available online: 60!-50!-1 = 8320987112741390113862247981509986337141564440296477108071636127449087 999 999 999 999

which indeed has 12 trailing 9's
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