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384's Modding Queue [Closed]

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Total Posts
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Topic Starter
pw384
uguuuuuuuu

Finally a long summer vacation is coming! So I decided to reopen the queue.

RULES


1. See my last post.
2. If you have a ticket, you can pm me for mod when I am online or just leave a link here!
3. Nothing else.


More information

By the way, here's my quite simple modding exe: Baidu Mirror. I use it to generate mod posts.
Using delphi


Tickets
rezoons x 1

Reminder
https://osu.ppy.sh/s/174968
https://osu.ppy.sh/s/185346
awa
pasha
Topic Starter
pw384
open for 2 slots!
AnreFM
Reisen Udongein
Topic Starter
pw384
Oh well you guys are so fast!
closed.
[CSGA]Ar3sgice
when it is closed
Topic Starter
pw384

[CSGA]Ar3sgice wrote:

when it is closed
机智
Topic Starter
pw384
2 slots
Oriance
Hello owo

https://osu.ppy.sh/s/138210

thanks so much :3
Nymph
http://osu.ppy.sh/s/125644
手速不行了 老了
pishifat
https://osu.ppy.sh/s/148846

a marathon map, you'd be the first to mod it :P

if it's too long, just ignore. thanks :D
Topic Starter
pw384
closed~

EDIT:

pishifat wrote:

https://osu.ppy.sh/s/148846

a marathon map, you'd be the first to mod it :P

if it's too long, just ignore. thanks :D
oh sorry you're the third >.<

but I'll have a look at the map if I have time. Don't worry! :D


EDIT2:

香草妮姆芙
Topic Starter
pw384
Opened!


Solve the following question and you can get 2 free mods from me!

Suppose that p is a given prime. For all integer x that satisfies p|(x^n -1) ,we have (p^2)|(x^n -1).
Find out all the positive integer n. The answer should contain the prime p.
Proof is required.
rezoons
Hi :),

I'll try something:

(1+p)^n=1 modulo p
(1+p)^n=1+np modulo p^2

So if n is such that for all integer x, p|(x^n -1) implies (p^2)|(x^n -1) then we must have 1+np=1 modulo p^2 then we must have p|n.

Let's proove that n=p works:
If x is a multiple of p then x^p=0 modulo p.
If x is not a multiple of p then x^p=x modulo p.
So, x^p=1 modulo p if and only if x=1 modulo p.
Or, if x=1 modulo p then x=1+kp with k an integer.
So, x^p=1+kp^2 modulo p^2
-> x^p =1 modulo p^2

So, x^p=1 modulo p implies x^p=1 modulo p^2.

Now let's proove that the property is verified for every n multiple of p.
Let's take n=kp with k an integer. Then we have:
x^n=1 modulo p
-> (x^k)^p=1 modulo p
-> (x^k)^p=1 modulo p^2 (because we jsut proove that x^p=1 modulo p implies x^p=1 modulo p^2)
-> x^n=1 modulo p^2

So, i just prooved that if n verifies your property then n is a multiple of p and after that, i prooved that if n is a multiple of p then it verifies your property.
So, the answer is: all the n mulitples of p (including 0) verifies: for all integer x, p|(x^n -1) implies (p^2)|(x^n -1).
Topic Starter
pw384
I accept this solution so you can get 2 mods (or tickets) from me!
Feel free to post your map link :3


And also, the queue is closed for a while!
rezoons
Well, i only have one pending map so i will take only one mod :p

A+ Superstar (Nightcore Mix)

Thank you very much ^^
Topic Starter
pw384
2 normal slots.
here you go!
HanzeR
Abe Nana
Topic Starter
pw384
Closed!
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