The expression of n! is read "n factorial" and means n(n-1)(n-2)(n-3)(n-4)(n-5)...(3)(2)(1). Thus, 6! means 6 x 5 x 4 x 3 x 2 x 1=720 and 10! means 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1=3,628,800. Notice that 6! ends with one digit of zero and 10! ends with two digits of zero. How many digits of zero does 5000! end with?
Please explain each step!
Thank you!
2007-04-29 06:17:43 · 5 answers · asked by Danielle N 1 in Science & Mathematics ➔ Mathematics
Every 5th. number is divisible by 5, every 25th number is divisible by 25...
So we get between 0 and 5000:
1000 Numbers are divisible by 5
200 by 25
40 by 125
8 by 625
1 by 3125
5 provides one five in the prim factorization, 25 provides an additional 5 for the prim factorization (because it provides 2, but one is already in it, because 25 is also divisble by 5),...
So in total we get, a: = 1000 + 200 + 40 + 8 + 1 = 1249 with a = the power of 5 in the prim factorization. We also have surely more '2's than '5's, so 1249 will also be the highest exponent of 10 with 10^1249 divides 5000!.
So we get a total of 1249 zeros.
2007-04-29 06:29:32 · answer #1 · answered by Anonymous · 0⤊ 0⤋
OK, this is a little confusing to explain, but here goes. If a number has a zero on the end, it's divisible by 10. For every time you can divide it by 10, there's another zero (so if it ends in three zeros, it's divisible by 10^3, if it ends in 10 zeros, it's divisible by 10^10 and so on). Ten factors into 2*5, so the idea is to find out how many factors of two and five there are in 5000! Whichever you have less of, there are as many factors of ten as of that number.
So first of all, there are 2500 numbers between 1 and 5000 that are divisible by 2. Each of them brings AT LEAST one factor of two to 5000! If a number is divisible by a higher power of 2, it will bring more than one factor of two. 1250 of them are divisible by 4, which brings another factor of 2. 625 are divisible by 8: another factor of two. 312 of these are divisible by 16, 156 by 32, 78 by 64, 39 by 128, 19 by 256, 9 by 512, 4 by 1024, 2 by 2048 and 1 by 4096.
So the total number of factors of two in 5000! is 2500 + 1250 + 625 + 312 + 156 + 78 + 39 + 19 + 9 + 4 + 2 + 1 = 4995.
Now, we do the same thing for factors of 5. There are 1000 numbers between 1 and 5000 that are divisible by 5. 200 are divisible by 25, 40 by 125, 8 by 625, and 1 by 3125. So the number of factors of 5 in 5000! is 1000 + 200 + 40 + 8 + 1 = 1249.
Therefore the number of factors of 2*5 in 5000! is 1249. So 5000! must end in 1249 zeros.
2007-04-29 06:36:27 · answer #2 · answered by Amy F 5 · 0⤊ 0⤋
1 zero from 2*5=10,
and 1 zero from 10
that is 2 zeros comes from 10!
22*25
30
32*35
40
....
Think in this way.
2007-04-29 06:27:51 · answer #3 · answered by iyiogrenci 6 · 0⤊ 1⤋
You can use Stirling's formula to calculate N! for very large numbers.
2007-04-29 06:36:52 · answer #4 · answered by Robert L 7 · 0⤊ 0⤋
q? Sorry. I'm not smart.
2007-04-29 06:20:39 · answer #5 · answered by Anonymous · 0⤊ 1⤋