Recent content by tongos

"find all integers n, k1,k2,k3...kn such that k1+k2...+kn=5n-4 and 1/k1+1/k2+...+1/kn=1" 1/k1+1/k2+...+1/kn is minimized when k1=k2=k3=k4...=kn= (5n-4)/n 1/k1=1/k2=...=n/(5n-4) 1/k1+1/k2+...+1/kn= (n^2)/(5n-4) this is the minimum value so for this to work, we need (n^2)/(5n-4)<1 when...

the putnam is very difficult exam, the problems are very motivating for the motivated math enthuisists though. All the years can be found at kalva demon as well as on american math competition site.

yeah, the art of problem solving is a good book, they have a forum for people who like to problem solving, its at artofproblemsolving.com, the problems are quite motivating. I was way into doing these problems last year (senior year) but now I've slowed down a bit and i think that I've actually...

well, you know the integral of sinx with limits. Now arcsin x will be the limits, and you can make a rectangle.

(2n-1)!/{(2^(n-1)(n-1)!(n!)}=<2^n {2^n(n!)}{2^(n-1)(n-1)!}>=(2n-1)! (2n-1)!/(n!)(n-1)!=<2^(2n-1) C(2n-1,n)=<2^(2n-1) True?

Some types of problems that i create and jog my brain are 1) pigeonhole principle problems 2) strategy problems 3) number theory problems (remainders, in-my-head multiplication) *algebra problems don't seem to stimulate (move this number to the other side isn't really thinking) I find...

if n^3-7n+3 is divisible by 3 then isn't n^3-7n is divisible by 3? n^3-7n= n(n^2-7) if n is not divisible by 3, then n^2-7 must. when 3 is in, n^2-7 is never divisible by 3, always a remainder of 2. n^2-7= n^2-4-3 (n-2)(n+2)-3 must be divisible by 3 when n is not divisible by 3...

thanks! I seem to know what to do now. The fun part about this problem is the pigeon hole principle. I love math problems like this one. one of my favorite math problems (though simple) is this: Prove that at any party, two friends at that party must have the same amount of friends present...

its for a different post. i wasnt talking about "is defined as". 1/3=0.333333... by division.

1/3 is too. simple division leads me to believe that. this is pointless.

how do we know that 1/3=0.33333... by this, we need to look at the geometric series to prove such.

i have 69 distinct positive whole numbers between 1 and 100. i pick out 4 integers a,b,c,d. prove that i can always pick out 4 integers such that a+b+c=d. can this always hold true with 68 positive integers?

yeah, i was to lazy to do the unfolding and folding of shapes, i got a 132 as well. Really though, I am not that smart.

I have an AP test coming up in a couple of weeks, and i just need to know what i should study for. I am hoping on getting a five btw, despite my grade in the class.

I know that there is a scholarship for scoring well on the AIME and making the USAMO.