[Prime]

Just a bit more challenging factorial problem.

How many consecutive zeroes does the number 2008! end with?

For those not familiar with factorial operator "!", n! = 1*2*3* ... *(n-1)*n.

That is all numbers from 1 to n multiplied.

Edited October 5, 2008 by Prime

[unreality]

On calculators all across the Internet, 170! is the highest thing you can enter, 171! always says "infinity", or in Google's case, just does a normal search

I'll have to figure this out mathematically I guess

[Prime]

On calculators all across the Internet, 170! is the highest thing you can enter, 171! always says "infinity", or in Google's case, just does a normal search

I'll have to figure this out mathematically I guess

Even if your calculator could come up with exact decimal representation of 2008!, it would take you several minutes to count the zeroes at the end.

[Guest]

according to WinXP calculator (scientific form) 2008!= 8.64346418576710702052555578574491*10^5761, which would mean that there would be 5761-32= 5729 noughts on the end of 2008!?

[Prime]

according to WinXP calculator (scientific form) 2008!= 8.64346418576710702052555578574491*10^5761, which would mean that there would be 5761-32= 5729 noughts on the end of 2008!?

Sorry, that's incorrect. I did not mean for anyone to use calculator or computer program to count the actual number of zeroes at the end. I meant to find the answer analytically. Use calculator only for what fits entirely on the display without "E" notation.

As you have noticed:

Your calculator suggests that it is a 5762-digit number. Whereas, it only gives you the first 32 significant digits. (The last 33rd digit is a round off). Thus you lose the 5730 rightmost digits and have no way of telling how many trailing zeroes were there.

Edited October 5, 2008 by Prime

[Guest]

The number of 0s at the end of a number must be equal to the minimum of the number of times 2 is a factor and the number of times 5 is a factor of that number.

Now lets find the number of times 5 is a factor of 2008!

2008/ (5⁴) = 3. there are 3 numbers from 1 to 2008 that are divisible by 5⁴. So that adds up to 4*3 = 12 times that 5 is a factor

2008/ (5³) = 16. there are 16 numbers from 1 to 2008 that are divisible by 5³. Subtract from this the 3 that were already counted above. So we have 13 numbers that are divisible by 5³. So that adds to 3*13 = 39 times that 5 is a factor.

2008/ (5²) = 80. there are 80 numbers from 1 to 2008 that are divisible by 5². Subtract from this the 16 that were already counted above. So we have 64 numbers that are divisible by 5². So that adds to 2*64= 128 times that 5 is a factor.

2008/ (5) = 401 . there are 401 numbers from 1 to 2008 that are divisible by 5. Subtract from this the 80 that were already counted above. So we have 321 numbers that are divisible by 5. So that adds to 1*321= 321 times that 5 is a factor.

So total number of times 5 is a factor in 2008! should be 12 + 39 + 128 + 321 = 500.

The number of times 2 is a factor in 2008! is definitely more than 500. So there must be 500 trailing zeros in 2008!

[Guest]

It is 2008*2007*2006....

so the numbers that are multiplied include

10,20,.....90,100,110....200....1000,1010...1100...2000

Allthese will contribute to zeros at the end.

Till 100 we have 11 zeros

then for every hundred we have 11 more zeros

from 910 to 1000 we have 12 zeros

This adds up to 11*10+1

Same is for 1001 to 2000

so we should have 2*(11*10+1)=2*111=222

just saw it is not complete as it includes number of 2,5 then 12,15 and so on as well which will add zero so there should be 201 more zeros

that makes it 222+201

Edit: Found my answer incomplete and corected it

Edited October 5, 2008 by imran

[Guest]

On calculators all across the Internet, 170! is the highest thing you can enter, 171! always says "infinity", or in Google's case, just does a normal search

I'll have to figure this out mathematically I guess

my calculator can do like 345! without using E notation =)

but still too big i guess

will try to figure this one out =P

[Prime]

The number of 0s at the end of a number must be equal to the minimum of the number of times 2 is a factor and the number of times 5 is a factor of that number.

Now lets find the number of times 5 is a factor of 2008!

2008/ (5⁴) = 3. there are 3 numbers from 1 to 2008 that are divisible by 5⁴. So that adds up to 4*3 = 12 times that 5 is a factor

2008/ (5³) = 16. there are 16 numbers from 1 to 2008 that are divisible by 5³. Subtract from this the 3 that were already counted above. So we have 13 numbers that are divisible by 5³. So that adds to 3*13 = 39 times that 5 is a factor.

2008/ (5²) = 80. there are 80 numbers from 1 to 2008 that are divisible by 5². Subtract from this the 16 that were already counted above. So we have 64 numbers that are divisible by 5². So that adds to 2*64= 128 times that 5 is a factor.

2008/ (5) = 401 . there are 401 numbers from 1 to 2008 that are divisible by 5. Subtract from this the 80 that were already counted above. So we have 321 numbers that are divisible by 5. So that adds to 1*321= 321 times that 5 is a factor.

So total number of times 5 is a factor in 2008! should be 12 + 39 + 128 + 321 = 500.

The number of times 2 is a factor in 2008! is definitely more than 500. So there must be 500 trailing zeros in 2008!

Vimil has the exact number!

Imran came close, but missed the powers of 5.