We are all lazy in this time of the day, lets try this one: Find the number of quadratic polynomials \( ax^2 + bx + c \) such that: a) \( a, b, c \) are distinct. b) \(a, b, c \in \{1, 2, 3, ...2008\} \) c) \( (x + 1)\) divides \( ax^2 + bx + c \) PS:Just another problem uploaded, on the request of MR.Math in chat.
random guess. 2015028
Nopes :-(
It's the set of all solutions of \(c=b-a\) such that \(b\ne c \ne a\), \(a,b,c \in \{1,2,3,\dots, 2008\}\).
2014024? basically trying to count what Mr.Math posted.
Yes, joe you got that right, but your approach was very close too, you just needed to subtract few cases for b is even.
i was thinking of it like, a+c = b, so the question is really, have many pairs of distinct numbers can you pick from 1,...,2008, where their sum is also within 2008.
The answer you gave joe is the sum of 1 + 2 + 3 + 4 + ... + 2007
earlier*
yeah, i over counted in my first answer. I was counting pairs like (1004,1004), (1005,1005), etc.
yes, when b is even, there will be case of \(a = c\). We need to remove such cases as \( a \neq b \neq c\) Such cases = 2008/2 = 1004 so the answer becomes:2015028 - 1004 = 2014024
cool problem, now i can get back to cleaning lol. I saw the problem and was like, "its going to bug me if I dont try to asnwer it <.<" haha
lol, yeah just for fun :D
\[2(2)+2(4)+2(6)+\cdots+2(2006)=2\sum_{n=1}^{1003}2n=2014024.\]
I did have the answer before you posted it, but the page was freezing.
Let me explain Mr.Math's strategy, a + c = b now b = 3 gives (2, 1), (1, 2). b = 4 gives (3, 1), (1, 3) so again 2 solution like this .. so summing we get 2+2+4+4+· · ·+2006+2006 = 2·1003·1004 = 2014024
*n is odd*
Basically, what I did was this: b a c 1 - - 2 - - 3 1,2 2,1 4 1,3 3,1 5 1,2,3,4 4,3,2,1 6 1,2,4,5 5,4,2,1 We can see that for b=3 we have 2 solutions, same as b=4. For b=5 and b=6, we had 4 solutions. For b=7 and b=8, there will be 6 solutions (under the given constrains). We can use this pattern to formulate the formula I wrote above that gave us the number of quadratic polynomials that satisfy the given conditions.
Cases go from 3 to 2008. 3 = 1 case 4 = 1 + 3, 3 + 1 = 2 cases 5 = 2 + 3, 4 + 1, 3 + 2, 4 + 1 = 4 cases 6 = 4 + 2, 2 + 4, 5 + 1, 1 + 5, = 4 cases 7 = 4 + 3, 3 + 4, 5 + 2, 2 + 5, 1 + 6, 6 + 1 = 6 cases So odd numbers have n-1, even numbers have n-2 Series to use: 3 - 1 + 4-2 + 5 - 1 + 6 - 2 + ... + 2007 - 1 + 2008 - 2 2 + 2 + 4 + 4 + ... + 2006 + 2006 There are 2006 terms now there are 1003 2(2) + 2(4) + ... + 2006(2) Sum of this = \[\sum_{i=1}^{1003} 4i = 2014024\]
and I see I am terribly late. Damn.
lol
Sorry, made a typo, 3 has 2 cases.
We used the same approach @slaaibak!
Now, where is the problem you promised?
I posted it in chat, but I think the buggy chat lost it somehow. I will post it in a sec!
Not in the chat please! thanks :)
That's cheating across :P
Looks like I will ask across to help me with the programming I have to do. :P
My computer is my loyal slave *pets computer. computer barks.*
First problem to warm up: (Not that difficult) _, T, T, F, F, S, S, E, N, T, E, T, T, F, F, S, S, E, N, T Missing letter?
E
Nope
I can't find it. but if I have to make a guess, other than E, I would say N.
Nope. Think VERY basic.
it's F actually.
Not F either
That just leaves us with T...
wait don't answer now
kk
I asked this to someone I know, he said T. and when I asked why he replied "Why not? It's a finite thing, we can always create enough superficial demands for it to be anything." lol :D
hahaha. True, but that does not fit my answer :)
I'll give you a clue. English.
I'm not good at English! :(
yeah, same... I didn't get it either.
I want a problem in Zimbabwean language.
I don't think it gets any more basic than E, honestly.
Verstaan jy wat ek nou sê?
Hierdie is 'n taal wat in Zimbabwe gebruik word, so jy behoort dit te verstaan.
Verstaan \(\approx\) Verstehen = Understand ?\[\]
Yeah, I got that. So it's T! I knew it :D
Alright, what comes next in this sequence: 7, 4, 5, 4, 2, 4, [ ]
3
no, no
7,5,2,-2
Gosh I can't do simple addition and subtraction. -2 or 2
No across.
It's 4 !
No :-(
( Across, yes, verstaan = understand. Some of the german words are a bit similar to my language)
5 ?
Here is the answer : Alright|=7, |what|=4, |comes|=5, |next|=4, |in|=2, |this|=4 I am sure that this will bring a smile in your face :-)
Haha,unfortunately, the pattern isn't even mathematical... :D
4
you are so messed up
lol :D
lol,
lol. Another clue. My pattern isn't mathematical either..
8! :P
haha. It's a letter tho. Is everyone fine with me giving the answer?
You answer is O
yep! It's O. did you google, or figure it out?
One,two, three, four,...
Yep! Lame, I know... Well done!
You both suck!
Time for MrMath or across to give a puzzle.
lol :D
I have this link once to my German friend in Germany and he told me it was blocked! :0
Germany blocks most YouTube videos, especially music videos.
Really?!! Why?
Oh I got, copyrights.
Yes, why ?
My life without youtube would be incomplete.
I thought it had to do with the sensitivity about Nazism or something like that.
http://www.afterdawn.com/news/article.cfm/2009/04/02/youtube_blocks_music_videos_in_germany
No, it has nothing to do with the Hitler reference.
Hey my friend reaction to the answer of slaaibak's question: "Maybe we should rename "one" to begin with an "e" so as to preserve symmetry. I'm sure mathematicians will go for it with that rationale."
In my language, one begins with an E :D But yes, we should start a petition!
haha, :D
One also begins with an E auf Deutsch.
Poor German!
Germans*
Alright, time to sleep for me.
Gute Nacht, FFM. Träume schön und bis später!
haha ,across, the only German I know is Auf wiedersehen :P
Good night! Hoop jy slaap lekker.
Horra Borra Torra Fool!! (Good night in Zimbabwean)
lol :D Subho ratri (Good night in my mother tongue)
Join our real-time social learning platform and learn together with your friends!