which calculation is this number the result of? answer fits into

>>8169862
It could be anything

>>8169862
that number + 0 = that number

>>8169862

Judging from all the 0's at the end, I'd say it's something like "what is the smallest number divisible by every element in the set "blah" where blah is a certain subset of natural numbers. Actually, I think all the 0's at the end might be able to tell us that, but fuck counting.

>>8169866
>>8169868
whole calculation can be written down into one post with less than 20 digits without using 'that number'

It's probably a factorial that can be worked out to some extent by counting the number of 0s which correspondes to the number of factors of 5s

>>8169873
Or indeed using stirlings formula

2 prime numbers multiplied

File: choices.jpg (71KB, 543x640px) Image search: [Google]

71KB, 543x640px

>>8169862

Why not make a textfile first and let bored mathfags work it out?

>>8169862
Looks like some factorial

>>8169942
number is too large to fit into single post
solution can be written down in less than 20 digits

File: 1462227739667.png (170KB, 396x388px) Image search: [Google]

170KB, 396x388px

>>8169932
ayy it's 1243! for you autists
pic not related

>>8169958
>1243!
how did you solve that?

I recognized the number as a factorial from the string of 0s at the end, just checked all factorials up to 2000 with a small matching string

>>8169971

why is there a string of zeros at the end of a factorial?

>>8169862
Let me just head on to OIES real quick


shit I have to type all this out

>>8170001
You keep ending up with multiples of even numbers and 5s. Every even number and a five will result in another ten. And you can never go back.

>>8170001
Because everytime you multiply by a multiple of 10 you get a zero that does not go away

>>8169958
no its not

Can't you make an estimate of the interval that factorial is between based on the amount of zeroes it has? like, 10! has 2 zeroes because it multiplies by 10 and then by (2*5) again, 20! has 4 zeroes since it does that again, 30! has 7 for some reason, but still. Idk someone smart do the squigly lines and work it out pls

>>8170167
shit nvm I thought you said 1234!

>>8170182
Well you got it. Ever two zeroes is another multiple of ten. Better precision, ever 5 will have at least 2 even numbers that preceded it, so every 5 will have contributed another 0. You could get any factorial within +/- 2.5 of the right number.

>>8169962
OP said it was a short expression and the number of zeros at the end suggests it's some kind of repeated multiplication with a lot of terms 5 or 10. so probably a factorial.
If you count the zero you can find a small range of possible values. there's 307 trailing zeros, which mean it has been multiplied by 10 at least 307/2 times which means the number you're looking for is less than 1500! You can count that better if you consider that 25 contains 5 twice, and stuff like that but I can't be bothered with proper combinatorics so late in the night

>>8170213
yep, the number of zeros at the end of n! should equal the multiples of five below n

>>8170182
so, for example, 10! has two zeros, 20! has four zeros, and 30! has seven zeros. Multiples of 5 below 30:
5, 10, 15, 20, 25 = 5^2, 30
so 7 zeros

I also got 1243, counted the digits of the number (using GIMP) then used stirlings approximation and found the solution to nln n -n = ln c. Where is the given number and used it's first 4 digits times by 10^digit.

>>8169862
How would I know and why would I care

File: le go back trump.gif (349KB, 350x233px) Image search: [Google]

349KB, 350x233px

>>8170013
>And you can never go back.
Course you can Pedro

>>8169862
how many zeros at the end pls

>>8169872
Number-1 + 1 ?

>>8169862
the number is equal to 1250!

>>8171710
sorry rather 1241!