How many zeros are there in 486 factorial ?

Or it can be written as ''how many zeros are there in 486x485x484x483........3x2x1 ?''

Guest Jul 29, 2014

#3**+13 **

There is a method for determining this.......I'm still working on the reason for why it works....maybe I'll "see the light" later on. But here it goes......

1. Take the number (without the factorial part) and divide it by 5....discard any "remainder' ...so we have..

486/5 = 97.2 = 97

2. Divide the number by succesive powers of 5 that are greater than 1 until we get a "decimal" answer..add all the whole number parts of the results...once we get a "decimal' answer only, we can stop.

3. Therefore, the next power of 5 is 5^{2} = 25. So dividing 486 by 25, we get....19.44 = 19

4. And the next power of 5 = 5^{3} = 125 . And dividing 486 by 125, we have ..... 3.888 = 3

5. Dividing by the next power of 5, (5^{4} = 625), we have..... .7776, and we can stop !!!

So adding all the "whole number" results, we have ...

97 + 19 + 3 = ..... 119 zeroes.....

And there you go......I hate to do things "cookbook style" without having a total understanding of the * reasons*, but, hey...my old brain just doesn't function as well as it used to!!!!

CPhill
Jul 29, 2014

#1**+3 **

Are you wondering how many *place values* are in 486!

or are you wondering how many *number 0's* are in 486!?

I'm not sure if there is a formula to figure this out. I'm sure someone else will post an answer if there is.

I just used calculator soup to find the answer to this. It came out as about 2.402100606 E+1096, so there is **1097 place values **in this number!

The actual number was:

2402100606040589897197638086812797478301774839698665408613794022064502171

2518840982308751981712465229369425164366582252171023331162043658051984062

4955161430625609777000662996493304475232844897531215829462618230008773245

9562592399661153941811053144742421950056894832360367109370125276844433536

9987691048876820773666915202799969234335595330244478731901731737649718146

7399043171966919599315049217413943054852886295473386999403534138407124356

68282206351234254228296347122352788666816045729459170141461529693975081529

76877727071387333558312656416000708181152244328938858913125585365193365947

23189794317070888603528608142248745385164063445311629158230374380269703037

12887468317198282228519107731747348865547753641363534630126790776107563531

23585248202964912004294789482833376070012019153790713089829434762858788417

63101185921443105189669526251030144425248142842777321009919321887669526986

29066830153676333357763330510392498169225574493952371758993101732461354082

15825359031542575267840000000000000000000000000000000000000000000000000000

0000000000000000000000000000000000000000000000000000000000000000000

If you really want you can count out all these zeros, but that seems rather time consuming!

NinjaDevo
Jul 29, 2014

#3**+13 **

Best Answer

There is a method for determining this.......I'm still working on the reason for why it works....maybe I'll "see the light" later on. But here it goes......

1. Take the number (without the factorial part) and divide it by 5....discard any "remainder' ...so we have..

486/5 = 97.2 = 97

2. Divide the number by succesive powers of 5 that are greater than 1 until we get a "decimal" answer..add all the whole number parts of the results...once we get a "decimal' answer only, we can stop.

3. Therefore, the next power of 5 is 5^{2} = 25. So dividing 486 by 25, we get....19.44 = 19

4. And the next power of 5 = 5^{3} = 125 . And dividing 486 by 125, we have ..... 3.888 = 3

5. Dividing by the next power of 5, (5^{4} = 625), we have..... .7776, and we can stop !!!

So adding all the "whole number" results, we have ...

97 + 19 + 3 = ..... 119 zeroes.....

And there you go......I hate to do things "cookbook style" without having a total understanding of the * reasons*, but, hey...my old brain just doesn't function as well as it used to!!!!

CPhill
Jul 29, 2014