document.write( "Question 169479: Let f(n) = the number of zeros at the end of n! when n! is multiplied out.\r
\n" ); document.write( "\n" ); document.write( "Find:\r
\n" ); document.write( "\n" ); document.write( "f(12)\r
\n" ); document.write( "\n" ); document.write( "f(34) \n" ); document.write( "
Algebra.Com's Answer #125278 by stanbon(75887)\"\" \"About 
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.
\n" ); document.write( "Find:
\n" ); document.write( "f(12) = 12! = 12*11*10*9*8*7*6*5*4*3*2*1
\n" ); document.write( "Find the number of times 10 is a factor.
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\n" ); document.write( "Change factors to prime number form:
\n" ); document.write( "f(12) = 2^2*3 * 11 *2*5*3^2*2^3 *7*2*3 *5* 2^2 *3*2*1
\n" ); document.write( "f(12) = 2^10 * 3^4 * 5^2 * 7
\n" ); document.write( "The prime-factor form has two tens which is 2^2*5^2, so only two zeroes at the end
\n" ); document.write( "of 12!
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\n" ); document.write( "f(34)= (2*17)*(2^4)*(3*5)*(2*7)*13*(12!)
\n" ); document.write( "12! contributes 2 zeroes
\n" ); document.write( "The other factors contribute 2*5 or one zero
\n" ); document.write( "So, f(34) has a total of three zeroes at the tail end.
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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