Proof contest

chess64

I will post like one problem every week or something. PM me your proof. They will be graded on a scale from 0 to 7.

Scoreboard

Week 1

Let $f(x)=\frac{x}{3x-2}$ . Show that if $n$ has $d$ digits, then $f(n)=\frac{n}{10^d}$ if and only if $n$ is an integer and $n=\underbrace{33\cdots 33}_{(d-1) \text { threes}}4$ in base 10.

(Yes, the first problem is easy.)

chess64

2 days left...

**Posted:** Thu Apr 27, 2006 10:55 am

matt276eagles

Edit: Oops. I accidentally posted my proof. Blush

Don't worry, I deleted it immediately. I will PM it to you now.

**Posted:** Thu Apr 27, 2006 3:09 pm

chess64

I'm going to post the next problem now since I probably won't be able to tomorrow. This problem was really easy (look at the scores Smile

), but the next ones will be harder.

edit: Week 1 Solution

Week 2

Compute, with proof, all primes $p$ such that $2^p+p^2$ is also prime.

**Posted:** Fri Apr 28, 2006 11:16 am

chess64

Week 2 Solution:

**Posted:** Sat May 06, 2006 5:30 am

chess64

Week 3

You have three piles of stones containing 5, 49, and 51 stones. You can join any two piles together into one pile, and you can divide a pile with an even number of stones into two piles of equal size. Is it possible to eventually have 105 piles each with one stone?

**Posted:** Sat May 06, 2006 5:42 am

krassi_holmz

that's easy.

**Posted:** Sat May 06, 2006 11:54 am

chess64

Week 3 Solution

**Posted:** Sat May 13, 2006 11:36 am

chess64

Week 4

Let $\mathcal{S}$ be a subset of $[0,1]$ with total length greater than $\frac{1}{2}$ . Prove that there are two elements of $\mathcal{S}$ that differ by exactly $0.1$ .

**Posted:** Sat May 13, 2006 11:38 am

chess64

Regarding Presentation

- You don't need to use LaTeX for just a number, like 5. But if you have $5\cdot 5$ , you should use LaTeX instead of saying 5*5.
- The solution should be easy to read. You don't need to make the grammar perfect, just readable.
- If English isn't your first language, do your best. I won't take off points if I can tell that English isn't your mother tongue, as long as you make an effort to write a clear solution. (At least spell correctly Smile

)

EDIT. "Writing up Solutions" by the Canadian Mathematical Society.

darkdevil

what do you mean by total length? difference between biggest and smaller element? or what?

**Posted:** Sat May 13, 2006 11:58 am

_________________
$\mathfrak{Radvan Marius}$ $aka$ $\mathfrak{darkdevil}$

chess64

$\mathcal{S}$ can be a union of intervals. The length of an interval $[a,b]$ is $b-a$ . The total length is the sum of the lengths of all the intervals in $\mathcal{S}$ .

**Posted:** Sat May 13, 2006 12:23 pm

pkerichang

wait, but then if $S$ is a set of one interval, $[0, 0.6]$ , then the statement will not be true. (or the case that $S = \{[0, 0.5], [0.7, 0.71]\}$

**Posted:** Sat May 13, 2006 7:02 pm

_________________
For to me to live is Christ, to die is gain

matt276eagles

**Posted:** Sat May 13, 2006 7:11 pm

amcavoy · Yang-Mills Theory Offline Joined: 21 Mar 2005 Posts: 922

If $S_{k}=\left[a,b\right]$ , then $t\in S_{k}$ if and only if $a\leq t\leq b$ , meaning that the two elements that differ by $.1$ don't have to be in disjoint intervals.

chess64

**Posted:** Sun May 14, 2006 10:40 am

amcavoy · Yang-Mills Theory Offline Joined: 21 Mar 2005 Posts: 922

Edited.

One of the previous posters must have thought that they needed to be in different intervals (meaning that it wasn't realized that points inside an interval can differ by $.1$ ).

**Posted:** Sun May 14, 2006 10:46 am

chess64

Just a clarification: $\mathcal{S}$ is a union of intervals only.

**Posted:** Tue May 16, 2006 4:10 pm

krassi_holmz

I think I got it.

**Posted:** Fri May 19, 2006 5:50 am

chess64

Sorry for the delay guys, I'll try to get the solutions graded tomorrow.

Oh yeah and if somebody would like to take over the contest, please PM me. Otherwise, I don't think it will continue... Sad

**Posted:** Sat May 20, 2006 5:08 pm