| Account Details | |
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| SteamID64 | 76561198064711665 |
| SteamID3 | [U:1:104445937] |
| SteamID32 | STEAM_0:1:52222968 |
| Country | United States |
| Signed Up | November 10, 2014 |
| Last Posted | March 14, 2024 at 12:29 AM |
| Posts | 746 (0.2 per day) |
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v
I have a weird hand setup where pinky is on a, ring w, middle d so I just use my index for crouch
There haven't been any power rankings since preseason, so megaboy and I decided to write up our own: https://pastebin.com/wqNrt8sE
Both of us haven't really been following open at all this season, so this was mainly looking at matches, guesswork, and a few scrim results. There haven't really been many close matches so it was hard to rank some of the teams. Not sure if this will be a weekly thing or not.
oi made that meme can I get a nerd star for it
uhh bwelp sucks upfrag me
good group of dudes who want to improve at the game
pastebin 3:35 PM - mastrej: tbh, I want to do my friend fyg a favor
3:35 PM - mastrej: she's done so much for me
3:35 PM - mastrej: and she gets upset a lot when she carries and her team still can't capitalize
3:35 PM - mastrej: Consider this my little present to her
3:35 PM - mastrej: getting rid of dead weight
3:35 PM - mastrej: like u
what the fuck
if it didn't specify that they had to be positive integers, you don't get integers nearly as large: (11,9,-5) and (11,4,-1) are two examples
getting the answers are still difficult but it's cool how just by specifying that the ans has to be positive you get much bigger numbers
800 dpi
1.8 in game
11.4 in/360
I play any class
If it happens before school starts I'd definitely go, lots of friends that I want to meet
seen her around for a while, has good aim and nice from my experiences. very underrated pickup
Bob_Marleyidk if I misunderstand your notation, but 0 mod any number is 0. So
X^2 + 1 = 0 mod(p) = 0
x^2 = -1
so, x = +i, -i. Which aren't in Z.
0 mod p=/=0. any number that has a remainder of 0 when divided by p is equal to 0 mod p.
so if p=7 then 0, 7, 14, etc are all 0 mod p
suppose there is such a solution x in Z
since p = 3 (mod 4), let p = 4k + 3 for some k in Z
x^2 + 1 = 0 (mod p) => x^2 = -1 (mod p) => x^4 = 1 (mod p)
then x^(p-1) = x^(4k+2)= x^4k * x^2 = (x^4)^k * x^2 = 1 * -1 = -1 (mod p)
but Fermat’s Little Theorem states x^(p-1) = 1 (mod p), so we have a contradiction.
thus there are no solutions x in Z
https://www.youtube.com/watch?v=VKGL9frrRhI
been recording and editing nonstop since grand finals