find the number of consecutive zero at the end of 1!*2!*3!*4!*5!*-------*50!

• 262 zeroes
  no of zeroes in 50!=12
  no of zeroes in 49!=10
  no of zeroes in 48!=10
  .
  .
  .no of zeroes in in 45!=10
  similarly
  no of zeroes in 49! to 45!=10*5
  no of zeroes in 44! to 40!=9*5
  no of zeroes in in 35! to 39!=8*5
  .
  .
  no of zeroes in20! to 24!=4*5 as no 25 has 2 multiple of 5
  .
  .
  no of zeroes in 5! to 9!=1*5
  total no= 12 + ((10*11)/2)*5-25=262
• 9 years agoHelpfull: Yes(3) No(0)
• 12 zeros
  short cut
  50/5=10
  10/5=2
  total 10+2= 12 zeros
• 9 years agoHelpfull: Yes(0) No(3)
• appala can u pls explain ur trick
  
• 9 years agoHelpfull: Yes(0) No(0)
• We can clearly see that in the above expression, the no. of 2's is greater than the no. of 5's.. So we need to count the no. of 5's here.
  then we get, 5! * 10! * 15! * 20! * 25! * 30! * 35! * 40! * 45! * 50!
  = (5/5) * (10/5) * (15/5) * (20/5) * ((25/5) + (25/25)) * ((30/5) + (30/25)) * ((35/5) + (35/25)) * ((40/5) + (40/25)) * ((45/5) + (45/25)) * ((50/5) + (50/25))
  = 1 * 2 * 3 * 4 * 6 * 7 * 8 * 9 * 10 * 12 = 8709120 No.s of zeros will be there... but I think there will be '+' instead of '*' in the question
• 9 years agoHelpfull: Yes(0) No(5)