factorial problem

any product of 2 and 5 will yield a nuber with 0 as units digit.
So here you should count the number of 5's and 2's
there are a lot of 2's, so it is easer to find the number of 5's. there are less 5's then 2's.

in
1!x2!x3!....x49!x50!
after 5! you have one five in each number, this gives us 46 zeros
after 25! you have two five in each number, this gives us additional 25 zeros
So I think the answer should be 46+25=71 zeros

50/5+50/5^2=12

cartera wrote:50/5+50/5^2=12

Your reasoning is correct for the problem
How many zeroes are there in 50!


But we have
How many zeroes are there in 1!x2!x3!....x49!x50!
It is the product of 50 consecutive factorials
1!x2!x3!x4! = 1*1*2*1*2*3*1*2*3*4

50! has 12 5’s
45! – 49! each have 10 5’s (50)
40! – 44! each have 9 5’s (45)
35! – 39! each have 8 5’s (40)
30! – 34! Each have 7 5’s (35)
25! – 29! Each have 6 5’s (30)
20! – 24! Each have 4 5’s (20)
15! – 19! Each have 3 5’s (15)
10! – 14! Each have 2 5’s (10)
5! – 9! Each have 1 5 (5)
1! – 4! Have 0 5’s (0)


I get 262 0's.

Also check out this posting:

https://www.manhattangmat.com/forums/how ... t5008.html

Thanks a lot guys. I got an idea of what to do!