MaTH!!! :) How many zeroooos??? HELP!!! :)?

Question

The expression n! is read "n factorial" and means n(n-1)(n-2)(n-3)(n-4)(n-5)...(3)(2)(1). Thus, 6! means 6x5x4x3x2x1=720 and 10! means 10x9x8x7x6x5x4x3x2x1=3,628,800. NOTICE: that 6! ends with one digit of zero and that 10! ends with two digits of zero. How many digits of zero does 5000! end with? Please explain your answer/reasoning and show your work! Thank you! Your help is greatly appreciated!!! :)

Answer 1

I don't think there's a formula, but you can analyze this a bit.

You can see why 6! is divisible by 10. It has 2 and 5 as factors. Thus, 5! and every number n! where n > 5 will be divisible by 10. There will always be at least one zero at the end.

Now, for a number to be divisible by 100 (so it has 2 zeroes), it has to have 2 squared and 5 squared as factors. You have to see 2 at least twice and 5 at least twice. You won't see that until 10! Among other factors, you'll have 2 * 2 * 5 * 5 = 100. So, you know that 10! (and larger) will end in at least 2 zeroes.

You can expand on this. 15! has 5 three times. So, you have 125 times everything else. At this point, you have several 2s, so don't worry too much about those. 125 * 8 = 1000. Do you have at least 8 2s in there? Sure thing. You can pull enough 2s out of 2, 4, 6, and 8. So, 15! and higher all have at least three zeroes.

20! has 5 four times. That gives you a factor of 625. Multiply that out by 16 to get 10,000. So, 20! and higher all have at least 4 zeroes.

You would be tempted to say that going up by 5 would give you an extra zero. At 25!, things get a little hairy. Now you have 5 six times (5*1, 5*2, 5*3, 5*4, 5*5). This gives you 15625. Multiply this by 2^6 to get 1,000,000. You know that 25! and higher will have at least six zeroes (instead of five).

So, that's the trend I see. Count the number of 5s you can get out of your number. You're going to have significantly more 2s than 5s, so you can be assured to get enough 2s to get a value that is of the form 10^n.

Counting your 5s can be tricky. You have to examine each number up to 5000. Heck, even the number 5000 has four fives in it. But 4995 only has one 5 in it. You have to consider all factors that have 5s in them. This includes 100 (two 5s), 450 (three 5s), and 4375 (four 5s).

That may be more tedious than it's worth. An Excel spreadsheet could help out.

Edit: Incidentally, I did whip up an Excel worksheet and counted 1248 5s within 5000!. We can see with the first six instances that the number of 5s in a factorial will be the number of zeroes at the end of the final product. So, one may be tempted to say that 5000! ends with at least 1248 zeroes. You are welcome to see this simplistic Excel sheet. Shoot me a message with your e-mail address, and I can attach it. It's not fancy, but it can save you from duplicating the work.

Answer 2

10! has 2 zeros, 1 for the factor of 10, 1 for the 5 times something even, so you get 2 zeros for every 10. Check that 20! has 4 zeros on the end (need at least a TI-89 for that).

5000/10 = 500
500•2 = 1000 zeros.

Uff. 30!, which by my theory has 6 zeros, actually has 7. What's up with that? Good luck. Your teacher might be happy with a reasoned guess.

Answer 3

I agree with Supermatt's answer. Another zero will be introduced every time you get a 5 multiplied by the previous even number and every time you get a 10 itself. None of the other numbers can combine to give a zero. Therefore you will get another zero every time n increases by 5.

Edit. Philo's right. Every time a power of 5 like 25 is introduced you get an extra zero. I think that the answer for 5000! will be 1000 plus 1 extra for 25, 2 for 125, 3 for 625, and 4 extra for 3125 so a total of 1010.

Further edit. No, there's many more than that. All the multiples of these numbers will introduce extra zeros as well. I give up.

Further edit. No, why should I give up. As anonymous above says, you only really need to count the number of 5's as there are plenty of even numbers to combine with them.
Each multiple of 5 introduces one zero and we know that there are 1000 of them.
Each multiple of 25 will introduce an extra zero not already counted in the mutiples of 5 and there are 5000/25 = 200.
Each mutiple of 125 will introduce yet one more zero and there are 5000/125 = 40 of them.
Each multiple of 625 will introduce one more zero not already counted and there are 5000/625 = 8 of them.
finally 3125 will introduce an extra 5 not already counted.
So I think that the final answer is
1000 + 200 + 40 + 8 + 1 = 1249.
I hope that no-one proves me wrong.

Answer 4

I calculated 15! and got three zeroes, so my conjecture is that every 5, you add a zero. So 5000! would have 1000 zeroes (5000/5=1000). I don't know if you could verify this mathematically, but it makes sense.