Posted: Sat Jun 28, 2003 6:15 pm Post subject: Math Solutions

ok, time to post some solvable math problems, I'll give bits to the person that comes up with these solutions. They don't have to be correct, but if there is something interesting that you did, I will also award bits. I just want to learn something new out of this.

*note: I haven't done any of these problems myself, so I will be participating at the same time as everyone else. Therefore, to prevent me from getting a head start, I will not do the problem until a day after.

Question #1:

show that there are no integers a, b, c for which a^2 + b^2 - 8c = 6

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krishon

Posted: Sat Jun 28, 2003 6:21 pm Post subject: (No subject)

welll, all i do is just substitute numbers in

i'd put c=1 and test that...and theres no solutions for that one
then i'd do c=2 and so on...but those dun't work either..so there's no solution.

AsianSensation

Posted: Sat Jun 28, 2003 6:27 pm Post subject: (No subject)

haha...funny....

I doubt that is a solution, because you cannot check for an infinite number of a, b and c.

anyways, to prevent spamming, or people fooling around, I will propose a negative bit system. The idea is I keep track of the participant, and if someone spams, then they receive minus something bits, and even if they send solutions later and receive bits, it will be added on to that negative value.

example:

-dodge_tomahawk spams
-dodge_tomahawk now have -20 bits(in this thread)
-dodge_tomahawk sends in a solution
-I award 15 bits
-now he has -5 bits(in this thread)

so as you can see, you will not receive bits if you continuously post stupid stuff.

krishon

Posted: Sat Jun 28, 2003 6:31 pm Post subject: (No subject)

well that's wuzn't really my solution...i wuz tellin u my method...

SilverSprite

Posted: Sat Jun 28, 2003 7:20 pm Post subject: (No subject)

where are you getting these from btw?

Crono

Posted: Sat Jun 28, 2003 7:28 pm Post subject: (No subject)

clearly i'll b cheatin if i did these problems, lol, it's just not fair fo the rest of u guyz...haha, nah, just j/k...anywho, want hints? ask me...

SilverSprite

Posted: Sat Jun 28, 2003 7:45 pm Post subject: (No subject)

yeee crono is your main math man... lol anyways i have the solution.. another 10 second one meng you gimp.. get harder ones than these or do you WANT to give free bits away.. well here it goes

when ever i do a2 or b2 it means a^2 or b^2

assume that there exists integer values for a,b,c such that
a2 + b2 - 8c = 6

a2 + b2 = 6+8c
look at this in mod4.. where a perfect square is either 1 or 0.. therefore three are two cases
CASE 1:a2=1 mod 4 and b2=1 mod 4 from this it follows that 6-8c = 2 mod 4 and c=0 mod 4
CASE 2: a2=0 mod 4 and b2=0 mod 4 from this it folllows that 4c=3 mod 4 and so this case is also extraneous since c is integral

we are left with CASE 1.. both a and b are odd and c is a multiple of 4..
from this it follows that m(m+1) + n(n+1) = 2c+1.. the RHS is obviously even and the LHS is obviously odd. Hence contradiction.. therefore no integer values of a,b,c satisfy the above condition

SilverSprite

Posted: Sat Jun 28, 2003 7:46 pm Post subject: (No subject)

and i'll need more than 15 bits for YOUR sorry excuse of a question making me do math this summer!!

SilverSprite

Posted: Sat Jun 28, 2003 7:55 pm Post subject: (No subject)

i used 6-8c instead of 6+8c.. however this does not change anything.. the result is the same.. my bad:S

SilverSprite

Posted: Sat Jun 28, 2003 7:58 pm Post subject: (No subject)

i also forgot to mention that i let a=2n+1 and b=2m+1... haha i'm wayy out of it today

AsianSensation

Posted: Sun Jun 29, 2003 10:04 am Post subject: (No subject)

yeah, it wouldn't be fair if Crono or Bugz start to do these questions, but there are still good solutions to some of the problems.

SilverSprite gave the correct solution, and that's the way I did it too. But I found a better solution, much cooler:

Alternate Solution:

every integer in the world is the forms: 4n, 4n+1, 4n+2, or 4n+3.

the modulus equivalent of each of the above cases squared is either 0, 1, or 4 (mod 8 ).

8c+6 is equivalent to 6(mod 8 ), and therefore, it is not possible to form something equal to 6(mod 8 ) from the sum of two perfect squares.

AsianSensation

Posted: Sun Jun 29, 2003 10:06 am Post subject: (No subject)

Score Board

I will tally up the scores for each participant in here.

So far...

SilverSprite: 0 bits
Bugzpodder: 0 bits
Crono(if he ever wish to participate): 0 bits

*Note, I will only pay out the bits at the end of each week, so it would be easier for me to "take away" bits. (I can't actually give you minus bits, Im not a mod, though I wish I could be, for this thread anyways )

AsianSensation

Posted: Sun Jun 29, 2003 10:14 am Post subject: (No subject)

Btw, I haven't done these problems myself, so I don't know how hard or ho easy it is, so if I post up a really hard question that looks easy, then it's not my fault.

The first question was easy, next few would get increasingly harder.

anyways, on to the next one

Question #2:

prove that for n = 1, 2, 3...

1 + 1/1! + 1/2! + 1/3! + ... + 1/n! < 3

krishon

Posted: Sun Jun 29, 2003 10:25 am Post subject: (No subject)

well here's how i did it

since 1 + 1/1! is already equal to two

1/2! + 1/3! ... 1/n! > 1

and then worked out the rest. when u add the other ones...the sum increases by a continually smaller amount so that it can never reach one. am i right?

SilverSprite

Posted: Sun Jun 29, 2003 11:59 am Post subject: (No subject)

asian. your way cooler solution is the same solution as mine..just presented differently:P.. so dont go saying its way cooler geek.. and 4 in modulus 4 is the same as 0 LOSER