Show that number N! can terminate in exactly 1,2,3,4,or 6 zeros but never 5 zeros?

zeros at the end of n! come from factors that end in either 0 or 5

so the first zero shows up in 5!
e.g.
1!=1
2!=2
3!=6
4!=24
5!=120

that stays there until we reach 10! which is 3628800
punching in 15! in the Google search line gives
1 307 674 368 000
(after that it goes to scientific notation)

so critical values that cause more zeros are
20!, 25!, 30! etc.
let's look at 20!
20! = 1x2x3x4x5x6x7x8x9x10x11x12x13x14x15x16x17x18x19x20
the factors that cause zeros are
2x5x10x15x20 which is 30 000, so 20! ends in 0000

that will stay like that until 24! which still ends in 0000
but did you notice that all end with an even number before the last string of zeros,

Somebody has actually gone through the effort of printing out all results of the factorials from
1! to 1000! (believe it or not)
(Broken Link Removed)

look at the end of 24!
it looks like this 62.....360000
to get 25! we multiply that by 25
just consider the end result 36x25 or 900
so we are adding 2 zeros this time, and that "shows" it.

A jar contains some red and some yellow jelly beans. If a child ate 1 red jelly bean, 1/7 of the remaining candies would be red. If instead the child ate 5 yellow jelly beans, 1/6 of the remaining candies would be red.

what is the answer to 1x2x3x4x5x6x7x8x9x10x11x12x13x14x15x16x17x18x19x20 please

1x23x4x5x66x7x8x9x10x11x12x13x14x15x16x17x18x19x20

To determine the number of zeros at the end of N!, where N is a positive integer, we need to count the number of factors of 5 in N!.

Let's start by analyzing the factors of 5 in the numbers between 1 and N. Every multiple of 5 contributes one factor of 5, while multiples of 25 contribute an additional factor of 5, multiples of 125 contribute yet another factor of 5, and so on.

Now, consider the prime factorization of N!:

N! = 2^a * 3^b * 5^c * ...

The number of factors of 5 in N! is represented by the exponent c in the prime factorization.

To find the number of zeros at the end of N!, we need to determine the power of 5 in the prime factorization.

Let's consider each case:

Case 1: Exactly 1 zero
In this case, we need to find the smallest power of 5 that results in a factor of 5 in N!. Since 5^1 = 5, we can conclude that N! must be divisible by 5 and has exactly 1 zero at the end.

Case 2: Exactly 2 zeros
To have 2 zeros at the end, we need a factor of 100 in N!. In order to get a factor of 100, we require 2 factors of 5. Therefore, N! must be divisible by 25, resulting in 2 zeros.

Case 3: Exactly 3 zeros
To have 3 zeros at the end, we need a factor of 1000 in N!. To achieve this, we require 3 factors of 5. Hence, N! must be divisible by 125, which gives us exactly 3 zeros.

Case 4: Exactly 4 zeros
To have 4 zeros at the end, we need a factor of 10000 in N!. For this, we require 4 factors of 5. Therefore, N! must be divisible by 625, resulting in exactly 4 zeros.

Case 5: Exactly 6 zeros
To have 6 zeros at the end, we need a factor of 1000000 in N!. This requires 6 factors of 5. Hence, N! must be divisible by 3125, resulting in exactly 6 zeros.

Case 6: Exactly 5 zeros
To have 5 zeros at the end, we would need a factor of 100000 in N!. This requires 5 factors of 5. However, we cannot satisfy this condition because the previous case (exactly 4 zeros) already requires 4 factors of 5. Therefore, it is not possible for N! to have exactly 5 zeros at the end.

In summary, N! can terminate in exactly 1, 2, 3, 4, or 6 zeros at the end, but never 5 zeros.