Which calculation is this number the result of?

which calculation is this number the result of?
answer fits into one post, no picture as answer allowed.

It could be anything

that number + 0 = that number

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.

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

Or indeed using stirlings formula

2 prime numbers multiplied

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

Looks like some factorial

>no picture as answer allowed.
>Question given as picture

What hypocrisy

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

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

>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

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

Let me just head on to OIES real quick


shit I have to type all this out

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.

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

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

shit nvm I thought you said 1234!

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.

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

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

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.

How would I know and why would I care

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

how many zeros at the end pls

Number-1 + 1 ?

the number is equal to 1250!

sorry rather 1241!