Q. Find the number of zeros in 27!

• no of zeros in 27! = 27/5 +27/25 = 5+1=6
• 11 years agoHelpfull: Yes(3) No(0)
• to figure out how many zeros there will be at the end of n!
  1.Divide n by 10 and determine a quotient which we'll call q and a remainder which we'll call r.
  The number of zeros at the end will be:
  @ 2q if r < 5
  @ 2q + 1 if r>-5
  
  For 27!
  27/10 = 2 Remainder 7
  so the number of zeros at the end of 27! will be 2*2 + 1 or 5
• 11 years agoHelpfull: Yes(0) No(1)
• 23
  from 5 factorial we get 0 s.
• 11 years agoHelpfull: Yes(0) No(1)
• How about 100!
  
  If we divide 100 by 5, the answer is 20. Now, what is the reason why 100 was divided by 25 in the solution above?
  
  In 100!, we have factors that are multiples of 25. If we divide them by 5, (25, 50, 75, 100)/5 we get (5, 10, 15, 20), still multiples of 5. This means that we still have four 5's that needs an even pair, so we need to divide again by 5. Now, since we have already counted the 5’s in the first division, we need to count the second set of 5’s. Instead of dividing again by 5, we divide them by 25. That is the reason why we also divide 100 by 25.
  
  100 factorial
  
  So, the solution will be
  
  displaystyle frac{100}{5} + frac{100}{25 }= 20 + 4 = 24 zeroes!
  
  Large Numbers
  
  In larger numbers, we divide the factors by 5 three times (5^3), four times (5^4), and so on, or simply powers of 5. For example, in 130!, 125 is a factor, so we divide it by 5, 5^2 and 5^3. That is, we get the sum of the integer quotients of
  
  displaystyle frac{130}{5} + frac {130}{25} +frac {130}{125} = 32.
• 7 years agoHelpfull: Yes(0) No(0)