Question

There are k zeros at the end of 34!=34.33.32...4.3.2.1. What is the value of k?

Answer 1

Think about where the zeros can come from
1! = 1 , 2! = 2 , 3! = 6 , 4! = 24, 5! = 120 <-- our first zero
..... 10! = 3628800 <-- our 2nd zero

that is, a zero comes from a number ending in 2 multiplied by a number ending in 5, or by a number ending in 0
so the zeros would be caused by
the following multipliers:
(2x5)x10x(12x15)x20x(22x25)x30
--0----0-----0----0-----0-----0

there will be 6 zeros, k = 6

however!!!!
I just looked at
https://www.wattpad.com/4855-a-list-of-factorial-math-constants

and 34! has 7 zeros
I then looked back at 24! and 25!
24! has 4 zeros and follows my argument above
but it ends with 36 before its 4 zeros
we then multiply by 25 to get 25!
but 36 x 25 = 900 , which produces 2 zeros instead of just the expected 1
So I do have a flaw in my argument.

So there are actually 7 zeros for 34!

Answer 2

Each zero involves a factor of 10
There are lots more 2's than 5's
So, count the factors of 5 in 1-34:

5,10,15,20,25*2,30 = 7 fives, so 7 zeros