Question

It is given that 2007! = 2007x2006x2005x...x3x2x1. How many trailing zeros are there in 2007! ?

Answer 1

remainder of [2007/5] + remainder of [2007/25] + remainder of [2007/125] + ...

Answer 2

remainder or the floor value????

Answer 3

in general summation of
yep .. sorry.

Answer 4

goes up with 5^n ... and 5^n should be less than 2007

Answer 5

Explanation would be appreciated.

Answer 6

i was going to ask this question, but then i found out!
in wikipedia.

Answer 7

Wow. I have the solution but I don't understand.

Answer 8

5 ! = 1 zeros
10! = 2 zeros
15! = 3 zeros
20! = 4 zeros
25! = 6 zeros ??? <-- ??
30! = 7 zeros

Answer 9

6 is correct for 25!

Answer 10

first two types of zero
5*2 = 10 <-- this gives one zero
the other zero comes from
10 <--- this gives one zero

the other type of zero
25*4 = 100 <--- extra zero at the multiple of 25

Answer 11

because there are two 5's

Answer 12

multiple of 5 give zeros .. just try to extract them.

Answer 13

[2007/5] => no. of 1 zero
[2007/25] => no. of double zeros from multiplication
[2007/125] => no. of triple zeros from multiplication
[2007/625] => no. of four zeros from multiplication

^ right?

Answer 14

yep!!

Answer 15

[2007/25] => no. of zeros from double 5 multiplication
[2007/125] => no. of triple zeros from triple 5 multiplication
[2007/625] => no. of four zeros from quadruple 5 multiplication

Answer 16

That looks better... Thanks!

Answer 17

one 5 get's stripped by [x/5] ... same goes for higher multiple of 5