Anyone here has (or is) a speed calculator that can display 25 digits?

iamfriendly · 2010-02-14 11:01:25

So i am encounter with the question of:

How many 0s are there at the end of the # that is the product of the first 150 POSITIVE intergers (i.e. 1,2,3,.....150)?

anyone know?

or have a method that probably work?

'Cause I don't want to calculate for a week or so

Last edited by iamfriendly (2010-02-14 13:40:15)

iamfriendly · 2010-02-14 11:03:11

i am guessing about 17?

iamfriendly · 2010-02-14 11:04:25

20?

bobbym · 2010-02-14 11:05:33

Hi;

That product is called a factorial. The product of the first 10 consecutive positive integers is 1*2*3*4*5*6*7*8*9*10 = 10!

For 150!.

Use this well known method;

Where the bracket denotes the floor function, so there are 37 zeros at the end of 150!

150! = 57133839564458545904789328652610540031895535786011264182548375833179829124845398393126574488675311145377107878746854204162666250198684504466355949195922066574942592095735778929325357290444962472405416790722118445437122269675520 000 000 000 000 000 000 000 000 000 000 000 000

iamfriendly · 2010-02-14 11:38:46

Thanks for the help, but why 125, 25, and 5 as denominators?

bobbym · 2010-02-14 12:26:36

Hi;

Since the 10, 100, 1000 ... are what puts the zeros at the end of the factorial and 10 is 5 * 2 and gives 1 zero, 100 is 5*5 *2*2 and gives 2 zeros,
1000 is 5*5*5 *2*2*2 and gives 3 zeros, we only need to find the number of pairs of 5 and 2. Since there will always be as many or
more 2's as 5's, we really only need to find the number of ,5's, 5*5's, 5*5*5's etc.

Also the floor function truncates a number down to an integer,

iamfriendly · 2010-02-14 13:23:37

Hey!

I searched floor funtion, so basically it just "round it down to the whole number" as in my words

Is that somewhat correct?

iamfriendly · 2010-02-14 13:36:39

I think i get it, thanks again for your attention and time, and i presume this will work on fractorial? what is this formula called?

bobbym · 2010-02-14 18:08:26

Hi;

The process of taking the product of the first n consecutive integers is called n factorial.
It is written as n!

1*2*3*4 is called 4 factorial or 4! = 24

1*2*3*4**6 is called 6 factorial or 6! = 720

That technique will always work on any factorial.

Here is one for you, how many zeros are on the end of 10000! ? If you can do this then you understand. Try it, when you get an answer of 2499 then you are there.

iamfriendly · 2010-02-15 08:54:43

I got it it's 2499! 10000! = [10000/3125]+[10000/625]+[10000/125]+[10000/25]+[10000/5]=a number with 2499 0s

Thank you

bobbym · 2010-02-15 18:45:53

Hi;

That's it. A number with 2499 zeros at the end of it.

Math Is Fun Forum

#1 2010-02-14 11:01:25

Anyone here has (or is) a speed calculator that can display 25 digits?

#2 2010-02-14 11:03:11

Re: Anyone here has (or is) a speed calculator that can display 25 digits?

#3 2010-02-14 11:04:25

Re: Anyone here has (or is) a speed calculator that can display 25 digits?

#4 2010-02-14 11:05:33

Re: Anyone here has (or is) a speed calculator that can display 25 digits?

#5 2010-02-14 11:38:46

Re: Anyone here has (or is) a speed calculator that can display 25 digits?

#6 2010-02-14 12:26:36

Re: Anyone here has (or is) a speed calculator that can display 25 digits?

#7 2010-02-14 13:23:37

Re: Anyone here has (or is) a speed calculator that can display 25 digits?

#8 2010-02-14 13:36:39

Re: Anyone here has (or is) a speed calculator that can display 25 digits?

#9 2010-02-14 18:08:26

Re: Anyone here has (or is) a speed calculator that can display 25 digits?

#10 2010-02-15 08:54:43

Re: Anyone here has (or is) a speed calculator that can display 25 digits?

#11 2010-02-15 18:45:53

Re: Anyone here has (or is) a speed calculator that can display 25 digits?

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