1000! supposedly ends with 249 zeros. i have to prove why :-/ can anyone help me?

Answer 1

Note that the number of zeros in a number is equal to the number of times 10 is a factor of that number. This is in turn equal to the lesser of the number of 5s in the number's prime factorization and the number of 2s in the same. Since there will obviously be many more 2s than 5s in the prime factorization of n! for any n, we need only find the number of 5s in the prime factorization of 1000!.

Now, by definition, 1000! = 1000 * 999 * 998 * 997... * 3 * 2 * 1. We have:

Each multiple of 5 appearing in the product will contribute at least 1 5 to the prime factorization. There are 200 multiples of 5 less than or equal to 1000.

Each multiple of 25 appearing in the product will contribute a second 5 to the prime factorization. There are 40 multiples of 25 less than or equal to 1000.

Each multiple of 125 appearing in the product will contribute a third 5 to the prime factorization. There are 8 multiples of 125 less than or equal to 1000.

Finally, each multiple of 625 appearing in the product will contribute a fourth 5 to the prime factorization. There is exactly one multiple of 625 less than or equal to 1000, namely 625 itself.

So the number of 5s in the prime factorization is (# of factors contributing at least 1 5) + (# of factors contributing at least 2 5s) + (# of factors contributing at least 3 5s) + (# of factors contributing at least 4 5s) = 200 + 40 + 8 + 1 = 249. There are thus 249 5s in the prime factorization of 1000!, and thus 249 zeros in 1000!.

Answer 2

Think about it. There are 500 even numbers between 0 and 1000, so we have plenty of 2s to play with. All we need now is 249 5s.

There are 200 multiples of 5 (including 5 itself).
40 multiples of 25 (5^2)
8 multiples of 125 (5^3)
And finally 625.

Voila!

Answer 3

If one try to fine the largest power of 5 which can exactly divide 1000!, it would be 249.

But how?

Integer part of(1000/5+1000/5^2+1000/5^3+1000/5^4)
=Integer part of(1000/5+1000/25+1000/125+1000/625)
=200+40+8+1=249

Number of 2s would be much larger than number of 5s.

Answer 4

I'd start by counting the number of factors in 1000! that have multiple zeros:

1000 has 3
900 has -2
800-2
etc
100-2
then count the factors with 1 zero:
990-1
980-1
etc
890
880
etc
all the way down to 10

Then add them all up.

Answer 5

1000! is 1 * 2* 3*4*...*1000. I would just multiply it out. No other way to prove it. Use a calculator, I guess.

Answer 6

1000.0(248 more 0s)??