SOLUTION: Determine how many zeros are at the end of the numerals for the following numbers in base 10. 10! I understand that there is 1-10, 1-5, and 1-2 so 10*5*2=100 so there's 2 zeros bu

Algebra ->  Decimal-numbers -> SOLUTION: Determine how many zeros are at the end of the numerals for the following numbers in base 10. 10! I understand that there is 1-10, 1-5, and 1-2 so 10*5*2=100 so there's 2 zeros bu      Log On


   

Question 248696: Determine how many zeros are at the end of the numerals for the following numbers in base 10.
10! I understand that there is 1-10, 1-5, and 1-2 so 10*5*2=100 so there's 2 zeros but I don't get the 100! or the 1000!
Please explain to me how we can quickly find the number of zeros for any factorial.
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
In every set of 10 consecutive integers, there will be a number that ends in zero, another that ends in a 2 and a third that will end in a 5. The number that ends in 10 will create a zero at the end of the factorial. And the numbers that end in 2 and 5 will create another zero at the end of the factorial. (Remember that factorials are multiplications and multiplication is commutative and associative. This allows us to "pair up" the numbers that end in 2 and 5 together.) So every set of ten consecutive integers will create 2 zeros at the end of the factorial.

So to figure out how many zeros there will be at the end of n! (n factorial):
  1. Divide n by 10 and determine a quotient which we'll call q and a remainder which we'll call r.
  2. The number of zeros at the end will be:
    • 2q if r < 5
    • 2q + 1 if r+%3E-+5

For example:
For 27!
27/10 = 2 Remainder 7
so the number of zeros at the end of 27! will be 2*2 + 1 or 5
For 32!
32/10 = 3 Remainder 2
so the number of zeros at the end of 32! will be 2*3 or 6