/sci/ - Science & Math » Thread #13914711

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Hard maths problem.

Anonymous Thu 02 Dec 14:22:09 2021 No.13914711 View Reply Original Report

Quoted By: >>13914769 >>13914917 >>13915605 >>13916776

Lets see what /sci/ is capable of.

Anonymous

Anonymous Thu 02 Dec 2021 14:37:39 No.13914769 Report

Quoted By:

>>13914711
You can just convert any number to a form where each digit has the base of it's position (save for 0th). It's trivial to see that the value of each corresponding digit equates it times successive factorials.

42069(base-10) -> 10222311(base-n)

Anonymous

Anonymous Thu 02 Dec 2021 15:11:54 No.13914907 Report

Quoted By:

where is the money to pay the grader?
in the real world, you have to pay graders to check math problems
on 4chan, there is no money
how are you going to get solutions graded, OP?!?!
Bottom line is you don't have the context to judge solutions, so it's a waste of time to generate candidate solutions.

Anonymous

Anonymous Thu 02 Dec 2021 15:13:07 No.13914917 Report

Quoted By:

>>13914711
More like, who's the most boring cunt.

Anonymous

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Anonymous Thu 02 Dec 2021 15:14:20 No.13914924 Report

Quoted By:

Anonymous

Anonymous Thu 02 Dec 2021 18:41:52 No.13915605 Report

Quoted By: >>13915632 >>13915830

>>13914711
Its easy to prove existence with induction. Uniqueness tho is problematic, since I can always add a_{k+1}=0 to any existing sequence, since a_i can be zero. Did you mean "unique disregarding the zeros"?

Anonymous

Anonymous Thu 02 Dec 2021 18:50:12 No.13915632 Report

Quoted By:

>>13915605
wait, nvm, you messed up, you probably meant (k+1)!<n

Anonymous

Anonymous Thu 02 Dec 2021 19:50:58 No.13915830 Report

Quoted By: >>13916161

>>13915605
>unique set
adding more zeros to a set with zeros preserves the set

Anonymous

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Anonymous Thu 02 Dec 2021 20:10:37 No.13915907 Report

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Existence: if k!<n< (k+1)! then n-k(k)! < (k+1)! - k (k!) =k! and we are done by induction.
Uniqueness: 1(1!) + 2(2!) + 3(3!) +... k(k!) = (k+1)! -1 a formula which I just now discovered by playing around with desmos. We prove it by induction: 1(1!) +... + k(k!) + (k+1)(k+1)! = (k+1)! - 1 + (k+1)(k+1)! = (k+2)(k+1)! - 1 = (k+2)!, and the base case is obvious QED.
Now how many combinations can we make of the form a_1(1!) + ... + a_k(k!)? It's exactly (k+1)!. And if a number is at least (k+1)! then by the formula above it must not be of this form, because it must use factors at least as large as (k+1)!. Thus we have a 1-1 pairing between expressions of the form a_1(1!) + ... + a_k(k!) and numbers in [0, (k+1)! - 1] and the uniqueness is proven.
This was amusing, albeit a bit easy. Thanks OP.

Anonymous

Anonymous Thu 02 Dec 2021 21:17:30 No.13916161 Report

Quoted By:

>>13915830
I guess you can view it that way.
If so, you can prove the uniqueness by assuming the opposite, lets say there are two representations a_i and b_i, and let k be the lowest term where they differ a_k != b_k. Watching the diference of n=sum a_i i! and n=sum b_i i! modulo k+1, we will get, a_k==b_k (mod k+1), and since they are both less or equal to k(and >0), they must be equal. Contradiction, it follows it must be unique.

Anonymous

Anonymous Thu 02 Dec 2021 23:26:09 No.13916776 Report

Quoted By:

>>13914711
$n = a+(2!)b+(3!)c...(k!)a{_k}$
1. if $x$ divides $k! \rightarrow x|(k+a)!$ with $a > 0$ .
Therefore, $k!\mod x = (k+a)!\mod x = 0$

From 1. $n = a+(2!)b+(3!)c...+(k!)a{_k}\mod k! = a+(2!)b+(3!)c....+(k-1)!a_{k-1}\mod k!$

If the equation above was not the case, then there would be an element $(k-n)!a_{k-n}$ with $n > 0$ such that
$(k-n)!a_{k-n}\mod k! = 0$ . Therefore, $(k-n)!a_{k-n}\mod k!$ can also be expressed as $(k)!b_{k-n}$ .
Now $n\mod k! = a+(2!)b+(3!)c+(k-1)!a_{k-1}\mod k! \rightarrow n = a_k + r$ .
Basic properties of the module guarantee that both variables $a_k, r$ are unique.
Repeat argument with $(k-1)!,(k-2)!...$ to get a set of unique numbers $a_k, a_{k-1}...a_2,a_1$ .

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