Find the consecutive zeros at the end of the following numbers?
1)72!
2)57*60*30*15625*4096*625*875*975

• consecutive zeros at the end of 72! = 72/5 +72/25 = 14+2=16
  
  consecutive zeros at the end of 57*60*30*15625*4096*625*875*975=
  consecutive zeros at the end of 10*2*3*3*10*5^6*2^12*5^4*7*5^3*39*5^2 =
  consecutive zeros at the end of 10*2*10*5^6*2^12*5^4*5^3*5^2=
  consecutive zeros at the end of 10^2 *(2^13)*(5^13)=
  consecutive zeros at the end of (10^2)*(10^13) = 15 zeros
• 12 years agoHelpfull: Yes(5) No(0)