how many zeros does 1000000 factorial have?

Question

Please show work!!! lol

Answer 1

I assume you are asking about trailing zeros

Every multiple of 5 contributes a zero.
Every multiple of 25 contributes a second zero
Every multiple of 125 contributes a third zero
Every multiple of 625 contributes a fourth zero
etc.

floor(1000000/5) + floor(1000000/25) + ... floor(1000000/5^n)
There are:
200,000 multiples of 5
40,000 multiples of 25
8,000 multiples of 125 (5^3)
1,600 multiples of 625 (5^4)
320 multiples of 3,125 (5^5)
64 multiples of 15,625 (5^6)
12 multiples of 78,125 (5^7)
2 multiples of 390,625 (5^8)
= 249,998 trailing zeros

To help visualize this, let's try a smaller number like 10!

This is 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10

10 will give you a trailing zero.
5 x 2 will give you a trailing zero.
All the rest won't give you any more zeroes.
So for 10! the answer is 2 trailing zeroes.

Now try 25!
1 x 2 x 3 x 4 x 5 x 6 x ... x 23 x 24 x 25
Again
2n x 5 --> a multiple of 10 --> one zero added
n x 10 --> a multiple of 10 --> one zero added
2n x 15 --> a multiple of 10 --> one zero added
n x 20 --> a multiple of 10 --> one zero added
4n x 25 --> a multiple of *100* --> two zeroes added

The number of multiples of 5 --> 5
The number of multiples of 25 --> 1
Total trailing zeroes in 25! is 6.

So following the logic forward to 1 million factorial the answer is 249,998 trailing zeroes.

Answer 2

It's hard to say how many total zeroes are in the final number, but you can at least find the number of zeroes it will end with. This is the same as asking how many times can we factor a "10" out.

How many times can we factor a 5 out, and a 2 to go with each 5? Note that in the series 1, 2, 3, ... 1000000, every other number is even, and every 5th number is a multiple of 5. So we should have more than enough 2s to match up with with the 5s.

Factoring out a 5 from each number, this means you can get 200,000 fives:

(1*2*3*4* 5 *6*7*8*9* 10 *11*12...*1000000) =
(5^2000,000) * (1*2*3*4* 1 *6*7*8*9* 2 *11*12*13*14* 3 *16*17...)

But this still leaves you more fives, because for example 25/5 = 5, and 125/5 = 25, etc. So factor out a 5 from every 25th number inside. This gives 40,000 more fives. Keep this going. Factor out a five from every 5^3 = 125th number. That's 8000 more. Factor out a five from every 5^4 = 625th number. That's 1600. Factor out a five from every 3,125th number. That's 320. Then 64 for 5^6. In the case of multiples of 5^7, there are 12 occurances (even though 1000000 isn't divisible by 5^7), and 2 for 5^8. The value of 5^9 is greater than one million, so we've now factored all the 5s out.

This gives:
200,000 + 40,000 + 8000 + 1600 + 320 + 64 + 12 + 2 = 249998 fives.

We have more than enough even numbers to pair with these, so the answer is the number ends in 249,998 zeroes.

Answer 3

I assume you mean terminal zeros (at the end of the number) cause that's easier than total zeros. The number of terminal zeros of a number is determined by the greatest powers of 2 and 5 the number is divisible by. More precisely, if a number's prime factorization includes a 2^x and a 5^y term, then its number of terminal zeroes will be min(x,y) where min(x,y) represents the lesser of the integers x and y.

Convince yourself that 1000000! will be divisible by a lower power of 5 than by a power of 2 (i.e. y < x). Then all we have to do is to find y. That will represent the number of terminal zeros. Define {x} to represent the greatest integer less than or equal to x.

y = {1000000/5} + {1000000/25} + {1000000/125} + ... + {1000000/(5^n)} + ...

Essentially we count all multiples of 5, then the extra powers of 5 provided by multiples of 5^2, then the extra powers of 5 provided by multiples of 5^3, etc.

You will find that the largest value of n for which {1000000/(5^n)} is nonzero is n = 8. After some computation we find that y = 249998 which I believe is our answer.

Answer 4

249,998
Because the formula for trailing zeros, given x! is:

Sigma of [Floor[5^(-k)*x]] From k=1 to Floor[ln[x]/ln[5]]
Where ln[] is the natural log, Floor[] is the floor function and Sigma represents summation

Answer 5

Alot haha