Real Math by FirstSintax September 6, 2014 at 10:09 PM UTC
My old thread is full of spammers.here is new one
This thread was created because I saw some other people enjoy math (MilkCream)
Post anything mathy here so yeah.
no 1+1 stuff plz
Im serious
FirstSintax September 6, 2014 at 10:09 PM UTC
When
is divided by 5, what is the remainder?
FirstSintax September 6, 2014 at 10:09 PM UTC
Yes, cause there needs to be more posts of people bragging about how "smart" they are at a subject. On-Topic : 1+1 = 2
just stop
Infivix September 6, 2014 at 10:09 PM UTC
What is 4929x472-472+382?
TheMasterCow September 6, 2014 at 10:09 PM UTC
Thread cleaned up.ShiptheBattle September 6, 2014 at 10:09 PM UTC
FirstSintax, to answer your question, the remainder is 2.It's really simple if you just look for the pattern...
Took me less than half a minute .-.
2 ^ 100 divide 5 has a remainder of 1
3 ^ 100 has a remainder of 1
1 + 1 =2
FirstSintax September 6, 2014 at 10:09 PM UTC
FirstSintax, to answer your question, the remainder is 2.
It's really simple if you just look for the pattern...
Took me less than half a minute .-.
2 ^ 100 divide 5 has a remainder of 1
3 ^ 100 has a remainder of 1
1 + 1 =2
(a^x + b^x) is not equal to A+b^x
but what wuz the pattern? im curious
ShiptheBattle September 6, 2014 at 11:09 PM UTC
Sintax,2,4,8,6,2,4,8,6,etc.
3,9,7,1,3,9,7,1, etc.
Using Mod,
We have 6 and 1 as the remainders for 2 ^100 and 3 ^ 100 respectively.
Thus, 6 divide 5, has a remainder of 1.
In conclusion, 1 + 1 = 2
There's tons of these problems in Competitive Math, some easy point takers for me :D
JKawesome September 6, 2014 at 11:09 PM UTC
Stuck on a problem for days...Simplify log(2)*log(x)
It's not log(2x), I tried that
winterice123 September 6, 2014 at 11:09 PM UTC
Stuck on a problem for days...
Simplify log(2)*log(x)
It's not log(2x), I tried that
so right... it's log(x^(log 2))
aka log x to the log 2 power
LeonardoPlays September 7, 2014 at 3:09 AM UTC
What are u saying2!MasterProX September 7, 2014 at 6:09 AM UTC
erm wait no... lemme figure this out...
so right... it's log(x^(log 2))
aka log x to the log 2 power
"(Step A) (x – 3y)^4"
"(Step B) (4 choose 0)(x)^4 + (4 choose 1)(x)^3(–3y) + (4 choose 2)(x)^2(–3y)^2 + (4 choose 3)(x)(–3y)^3 + (4 choose 4)(–3y)^4"
"(Step C) (4! ÷ 0!4!)(x)^4 + (4! ÷ 1!3!)(x)^3(–3y) + (4! ÷ 2!2!)(x)^2(–3y)^2 + (4! ÷ 3!1!)(x)(–3y)^3 + (4! ÷ 4!0!)(–3y)^4"
"(Step D) x^4 – 12x^3y + 54x^2y^2 – 108xy^3 + 81y^4"
It may not look as pleasant when typed out rather than written by hand, but hopefully someone can decipher it.
JKawesome September 7, 2014 at 7:09 AM UTC
erm wait no... lemme figure this out...
so right... it's log(x^(log 2))
aka log x to the log 2 power
ParthDarthVader September 7, 2014 at 12:09 PM UTC
Sintax,
2,4,8,6,2,4,8,6,etc.
3,9,7,1,3,9,7,1, etc.
Using Mod,
We have 6 and 1 as the remainders for 2 ^100 and 3 ^ 100 respectively.
Thus, 6 divide 5, has a remainder of 1.
In conclusion, 1 + 1 = 2
There's tons of these problems in Competitive Math, some easy point takers for me :D
Milki September 7, 2014 at 12:09 PM UTC
I'm not bragging, Milkcream inspired me.
just stop
Remember to have fun! ^_^
Milki September 7, 2014 at 4:09 PM UTC
In case additional confirmation is needed, I can assure you that this is correct. Now... sorry to bother with a trifling, simple math problem — algebraic combination, with factorials and all that jazz — but can someone check if my work is correct? A couple of my classmates were thinking that I did a step or two incorrectly...
"(Step A) (x – 3y)^4"
"(Step B) (4 choose 0)(x)^4 + (4 choose 1)(x)^3(–3y) + (4 choose 2)(x)^2(–3y)^2 + (4 choose 3)(x)(–3y)^3 + (4 choose 4)(–3y)^4"
"(Step C) (4! ÷ 0!4!)(x)^4 + (4! ÷ 1!3!)(x)^3(–3y) + (4! ÷ 2!2!)(x)^2(–3y)^2 + (4! ÷ 3!1!)(x)(–3y)^3 + (4! ÷ 4!0!)(–3y)^4"
"(Step D) x^4 – 12x^3y + 54x^2y^2 – 108xy^3 + 81y^4"
It may not look as pleasant when typed out rather than written by hand, but hopefully someone can decipher it.
Oh, and the above problem was an original one.
If you're wondering what level the problem's around, I like to do mid-AIME to USAMO and beyond, so I'm guessing this might be late-AIME to early USAMO, but I'm not a judge of this.
What math competitions will you guys be attending this year?
JKawesome September 7, 2014 at 8:09 PM UTC
How can use the Haversine formula with longitude and latitude, to get points from a 50 mile radius from a database, and find the angle of the point in relationship to the origin?FirstSintax September 7, 2014 at 8:09 PM UTC
How can use the Haversine formula with longitude and latitude, to get points from a 50 mile radius from a database, and find the angle of the point in relationship to the origin?
Im only in 7th grade tho.
oh well have fun!
JKawesome September 7, 2014 at 9:09 PM UTC
wut?
Im only in 7th grade tho.
oh well have fun!
Also, how do you find the angle between the Y axis and a point from the origin?
FirstSintax September 7, 2014 at 9:09 PM UTC
um, kill yourself?jk(awesome)
how.... distance form?
JKawesome September 7, 2014 at 11:09 PM UTC
um, kill yourself?
jk(awesome)
how.... distance form?
Let's say I have 2 points on a coordinate plane. One is at the origin (0, 0), and the other one is at, say, (10, 7). If I were to draw a line between the 2 points, and calculate the angle between the positive side of the Y-axis and the line, clockwise, how would I do that?
Milki September 7, 2014 at 11:09 PM UTC
Given point (x,y):angle=arcsin(x/y)
'nuff said
Poll: What is your most hated topic in math?
For me, it's graph theory. It gets me almost every time.
This is why I'm a number theory/geometry/algebra person. Not a big fan of Combo.
JKawesome September 7, 2014 at 11:09 PM UTC
Given point (x,y):
angle=arcsin(x/y)
'nuff said
Poll: What is your most hated topic in math?
For me, it's graph theory. It gets me almost every time.
This is why I'm a number theory/geometry/algebra person. Not a big fan of Combo.
And the poll: I freaking hate trigonometry :3
FirstSintax September 8, 2014 at 12:09 AM UTC
Given point (x,y):
angle=arcsin(x/y)
'nuff said
Poll: What is your most hated topic in math?
For me, it's graph theory. It gets me almost every time.
This is why I'm a number theory/geometry/algebra person. Not a big fan of Combo.
freaking hate binomials
Infivix September 8, 2014 at 12:09 AM UTC
Personally, I am A LOT better at algebra stuff then anything Geometry related.JKawesome September 8, 2014 at 4:09 AM UTC
Personally, I am A LOT better at algebra stuff then anything Geometry related.
Infivix September 8, 2014 at 5:09 AM UTC
Lol, complete opposite of me. I can't do algebra at all.
FirstSintax September 13, 2014 at 11:09 PM UTC
Convert
to base 3.i hate bases!
garsdef September 13, 2014 at 11:09 PM UTC
69+420=1337steven5703 September 13, 2014 at 11:09 PM UTC
69+420=1337
FirstSintax September 13, 2014 at 11:09 PM UTC
69+420=1337
garsdef September 13, 2014 at 11:09 PM UTC
how did you get that with base 3?
Milki September 15, 2014 at 11:09 AM UTC
A good, extremely hard problem I worked on a few days ago:An
matrix whose entries come from the set 
is called a silver matrix if, for each 
, the 
-th row and the -th column together contain all elements of 
. Show that silver matrices exist for infinitely many values of 
.(1997 IMO, #4 part (b))
StewieFG September 15, 2014 at 2:09 PM UTC
45689x384+3895-345:(6-3+3674x4856830358689382)x0:D
PsychoStoner September 15, 2014 at 7:09 PM UTC
1 + 1 x 1 - 1=?ballzi September 15, 2014 at 9:09 PM UTC
oh god. why is there a math problem here?! i thought school wasnt incorporated here. well i will give it a go. how is it so hard?Zieno12345 September 15, 2014 at 10:09 PM UTC
Not really a math problem, but I think it fits here :D.In the night, four people need to cross a bridge. One person crosses in 1 minute, another in 2 minutes, another in 5 minutes, and the last in 10. At most, the bridge can only hold 2 people at a time, and when said people cross, they must go at the slower persons' pace. In addition, there must be a person that crosses back in order to hand the flashlight back to the other group. What is the order in which they cross for all of them to reach the other side in 17 minutes?
FirstSintax September 16, 2014 at 1:09 AM UTC
oh god these r so hardMrLavaCreeperFTW September 16, 2014 at 8:09 AM UTC
100+100......100100?XD NOO
sorry if I did dis.. ik you said no 1+1 stuff.. ^-^
(Joke, <3 you guys)
Milki September 16, 2014 at 10:09 AM UTC
Not really a math problem, but I think it fits here :D.
In the night, four people need to cross a bridge. One person crosses in 1 minute, another in 2 minutes, another in 5 minutes, and the last in 10. At most, the bridge can only hold 2 people at a time, and when said people cross, they must go at the slower persons' pace. In addition, there must be a person that crosses back in order to hand the flashlight back to the other group. What is the order in which they cross for all of them to reach the other side in 17 minutes?
1 and 2 cross.
1 goes back.
3 and 4 cross.
2 goes back.
1 and 2 cross.
This is a pretty famous problem. It's not exactly a math problem, but it's more of a logic puzzle, so also partially a math problem :D