SOLUTION: Let f(n) = the number of zeros at the end of n! when n! is multiplied out. Find: f(12) f(34)

Question 169479: Let f(n) = the number of zeros at the end of n! when n! is multiplied out.
Find:
f(12)
f(34)
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Let f(n) = the number of zeros at the end of n! when n! is multiplied out.
Find:
f(12) = 12! = 12*11*10*9*8*7*6*5*4*3*2*1
Find the number of times 10 is a factor.
--------------------------
Change factors to prime number form:
f(12) = 2^2*3 * 11 *2*5*3^2*2^3 *7*2*3 *5* 2^2 *3*2*1
f(12) = 2^10 * 3^4 * 5^2 * 7
The prime-factor form has two tens which is 2^2*5^2, so only two zeroes at the end
of 12!
---------------

f(34)= (2*17)*(2^4)*(3*5)*(2*7)*13*(12!)
12! contributes 2 zeroes
The other factors contribute 2*5 or one zero
So, f(34) has a total of three zeroes at the tail end.
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Cheers,
Stan H.

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