Math Is Fun Forum

moon79

Guest

Factorials

How many zeros will be there at the end of:
911! / 119! ?

You are asked to find the number of zeros at the end of the sum, not the total number of zeros.

Cheers,

Jai Ganesh

Administrator

Registered: 2005-06-28

Posts: 46,286

Re: Factorials

moon79,
every multiple of a factor of 2 and 5 gives an additional zero.
By calculating how many multiples of 2 and 5 are there in 911, and 119, the answer can be found.

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moon79

Guest

Re: Factorials

thanks, but:
there are 59 multiples of 2 and 23 of 5 in 119

there are 455 multiples of 2 and 182 of 5 in 919

the answer then would be 555 zeros but mathematica gives me 199 :-(

moon79

Guest

Re: Factorials

even if we look at the lower multiple - in this case 23 multiples of 5, we should have 23 zeros in 119, but I'm getting 28 in mathematica :-(

something is missing, but what?

cheers

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Factorials

You've done the first step by figuring out that it's the multiples of 5 that are important.
This is because each 0 requires a multiple of 2 and a multiple of 5 to make it. There are less multiples of 5, so they get used up first. There are lots of multiples of 2 left over but we don't care about those.

The reason your workings don't match up with Mathematica is that some multiples of 5 are hiding inside others.

Up to 119, the multiples of 5 are 5, 10, 15, 20, 25, ..., 105, 110, 115. That's 23 in total.
However, some of those are still multiples of 5 when you divide them by 5, and so must be counted again.

These are 25, 50, 75 and 100. That's 4 more multiples, and so the total is 27. I don't know why Mathematica has an extra one - maybe it's counting zero?

Working out the 0's on the end of 911 is done in much the same way.
You have 5, 10, 15, ..., 900, 905, 910, which is 182 multiples of 5.
Of those, 25, 50, 75, ..., 850, 875, 900 are hiding another multiple of 5, adding another 36 to the total.

Of those, 125, 250, 375, 500, 625, 750 and 875 have three multiples of 5, which adds on another 7.

Finally, 625 has yet another multiple, which means you add on a final one.

That means that the number of zeroes ending 911! is 182+36+7+1 = 226.

Take the previous answer away from this one, and we get that there are 226-27 = 199 zeroes ending 911!/119!

I like that answer. It's very appropriate.

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Why did the vector cross the road?
It wanted to be normal.