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Mathematics 14 Online
OpenStudy (anonymous):

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.

OpenStudy (anonymous):

random guess. 2015028

OpenStudy (anonymous):

Nopes :-(

OpenStudy (mr.math):

It's the set of all solutions of \(c=b-a\) such that \(b\ne c \ne a\), \(a,b,c \in \{1,2,3,\dots, 2008\}\).

OpenStudy (anonymous):

2014024? basically trying to count what Mr.Math posted.

OpenStudy (anonymous):

Yes, joe you got that right, but your approach was very close too, you just needed to subtract few cases for b is even.

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

The answer you gave joe is the sum of 1 + 2 + 3 + 4 + ... + 2007

OpenStudy (anonymous):

earlier*

OpenStudy (anonymous):

yeah, i over counted in my first answer. I was counting pairs like (1004,1004), (1005,1005), etc.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

lol, yeah just for fun :D

OpenStudy (mr.math):

\[2(2)+2(4)+2(6)+\cdots+2(2006)=2\sum_{n=1}^{1003}2n=2014024.\]

OpenStudy (mr.math):

I did have the answer before you posted it, but the page was freezing.

OpenStudy (anonymous):

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

OpenStudy (mr.math):

*n is odd*

OpenStudy (mr.math):

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.

OpenStudy (slaaibak):

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 = \[\sum_{i=1}^{1003} 4i = 2014024\]

OpenStudy (slaaibak):

and I see I am terribly late. Damn.

OpenStudy (across):

lol

OpenStudy (slaaibak):

Sorry, made a typo, 3 has 2 cases.

OpenStudy (mr.math):

We used the same approach @slaaibak!

OpenStudy (mr.math):

Now, where is the problem you promised?

OpenStudy (slaaibak):

I posted it in chat, but I think the buggy chat lost it somehow. I will post it in a sec!

OpenStudy (mr.math):

Not in the chat please! thanks :)

OpenStudy (anonymous):

That's cheating across :P

OpenStudy (mr.math):

Looks like I will ask across to help me with the programming I have to do. :P

OpenStudy (across):

My computer is my loyal slave *pets computer. computer barks.*

OpenStudy (slaaibak):

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?

OpenStudy (across):

E

OpenStudy (slaaibak):

Nope

OpenStudy (mr.math):

I can't find it. but if I have to make a guess, other than E, I would say N.

OpenStudy (slaaibak):

Nope. Think VERY basic.

OpenStudy (anonymous):

it's F actually.

OpenStudy (slaaibak):

Not F either

OpenStudy (across):

That just leaves us with T...

OpenStudy (anonymous):

wait don't answer now

OpenStudy (slaaibak):

kk

OpenStudy (anonymous):

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

OpenStudy (slaaibak):

hahaha. True, but that does not fit my answer :)

OpenStudy (slaaibak):

I'll give you a clue. English.

OpenStudy (mr.math):

I'm not good at English! :(

OpenStudy (slaaibak):

yeah, same... I didn't get it either.

OpenStudy (mr.math):

I want a problem in Zimbabwean language.

OpenStudy (across):

I don't think it gets any more basic than E, honestly.

OpenStudy (slaaibak):

Verstaan jy wat ek nou sê?

OpenStudy (slaaibak):

Hierdie is 'n taal wat in Zimbabwe gebruik word, so jy behoort dit te verstaan.

OpenStudy (across):

Verstaan \(\approx\) Verstehen = Understand ?\[\]

OpenStudy (mr.math):

Yeah, I got that. So it's T! I knew it :D

OpenStudy (anonymous):

Alright, what comes next in this sequence: 7, 4, 5, 4, 2, 4, [ ]

OpenStudy (mr.math):

3

OpenStudy (anonymous):

no, no

OpenStudy (across):

7,5,2,-2

OpenStudy (slaaibak):

Gosh I can't do simple addition and subtraction. -2 or 2

OpenStudy (anonymous):

No across.

OpenStudy (slaaibak):

It's 4 !

OpenStudy (anonymous):

No :-(

OpenStudy (slaaibak):

( Across, yes, verstaan = understand. Some of the german words are a bit similar to my language)

OpenStudy (slaaibak):

5 ?

OpenStudy (anonymous):

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 :-)

OpenStudy (anonymous):

Haha,unfortunately, the pattern isn't even mathematical... :D

OpenStudy (across):

4

OpenStudy (across):

you are so messed up

OpenStudy (anonymous):

lol :D

OpenStudy (mr.math):

lol,

OpenStudy (slaaibak):

lol. Another clue. My pattern isn't mathematical either..

OpenStudy (mr.math):

8! :P

OpenStudy (slaaibak):

haha. It's a letter tho. Is everyone fine with me giving the answer?

OpenStudy (anonymous):

You answer is O

OpenStudy (slaaibak):

yep! It's O. did you google, or figure it out?

OpenStudy (anonymous):

One,two, three, four,...

OpenStudy (slaaibak):

Yep! Lame, I know... Well done!

OpenStudy (mr.math):

You both suck!

OpenStudy (slaaibak):

Time for MrMath or across to give a puzzle.

OpenStudy (mr.math):

http://www.youtube.com/watch?v=RCn7Xk_6sb8

OpenStudy (anonymous):

lol :D

OpenStudy (mr.math):

I have this link once to my German friend in Germany and he told me it was blocked! :0

OpenStudy (across):

Germany blocks most YouTube videos, especially music videos.

OpenStudy (mr.math):

Really?!! Why?

OpenStudy (mr.math):

Oh I got, copyrights.

OpenStudy (anonymous):

Yes, why ?

OpenStudy (slaaibak):

My life without youtube would be incomplete.

OpenStudy (mr.math):

I thought it had to do with the sensitivity about Nazism or something like that.

OpenStudy (across):

No, it has nothing to do with the Hitler reference.

OpenStudy (anonymous):

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."

OpenStudy (slaaibak):

In my language, one begins with an E :D But yes, we should start a petition!

OpenStudy (anonymous):

haha, :D

OpenStudy (across):

One also begins with an E auf Deutsch.

OpenStudy (mr.math):

Poor German!

OpenStudy (mr.math):

Germans*

OpenStudy (anonymous):

Alright, time to sleep for me.

OpenStudy (across):

Gute Nacht, FFM. Träume schön und bis später!

OpenStudy (anonymous):

haha ,across, the only German I know is Auf wiedersehen :P

OpenStudy (slaaibak):

Good night! Hoop jy slaap lekker.

OpenStudy (mr.math):

Horra Borra Torra Fool!! (Good night in Zimbabwean)

OpenStudy (anonymous):

lol :D Subho ratri (Good night in my mother tongue)

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