induction
ceiling function
ratio
combinatorics proposed
combinatorics

Two Players A and B take turns removing chips from a pile that initially contians n chips.A cannot remove all chips on his first turn and must remove atleast one chip.Then onwards a player cannot remove more chips than how much his opponent removed the last turn.The player who removes the last chip is the winner.Find a winning strategy for A.

I will post my answer later.

An Alternate:
   What if a player is allowed to remove up to twice the number of chips removed by his opponent on the previous turn?

Obviously $n>1$. If $n$ is odd, A just takes one chip; then all the moves are forced one-chippers and A takes the last chip. If $n=2(2k+1)$, A takes two chips; if B takes two then A takes two again and so on, winning; if at some point B takes one chip, he leaves an odd number of chips and A can apply the strategy for $n$ odd to win.

In general, we claim that A wins iff $n$ is not a power of $2$. Let $n=2^x(2y+1)$ with $y \geq 1$. We'll prove that A wins by induction on $x$. He begins by taking $2^x$. As long as B takes $2^x$ too, A keeps taking $2^x$ and eventually wins, so suppose that at some point B takes $m<2^x$. The biggest power of $2$ that divides the resulting $n$ is at most $m$, so A can apply the strategy for smaller $x$ and win by induction.

Now to prove that B wins if $n=2^x$: Say A takes $m$ on his first move, leaving $2^x-m$ chips. If this is a power of two, then it is at most $2^{x-1}$, which means $m \geq n-m$; then B can take all the remaining chips and win. If $2^x-m$ is not a power of two, then, since $(2^x, 2^x-m)=(2^x, m) \leq m$ we know that B can take the biggest power of $2$ dividing $n-m$, and then apply A's original strategy.

Hey! That's the same solution I had got

What about the alternate problem?

For the alternate:

We can check that B wins for $2, 3, 5$ and A wins for $4, 6, 7$.

A lemma: suppose that $k$ is winning for B and let $n=k+r$ with $r<k/2$. Then $n$ is winning for A. Indeed, he can take $r$ on his first move: then B can't take the $k$ remaining chips since $k>2r$, and in any other case A wins with B's strategy for $k$. This means that if $k$ is winning for B then $k+1, \ldots, \lceil \frac{3k}{2} \rceil-1$ are winning for A.

Suppose that $2k$ is winning for B. Then $2k+1, \ldots, 3k-1$ are winning for A. We want to show that $3k$ is winning for B iff $k$ is (!). Indeed, in the beginning A can't take $k$ or more chips as B would then win immediately, so he will take some $r<k$ chips. From there on, if any player gets to $2k$ chips, he wins (playing B's strategy for $2k$ chips), so having an opportunity to get $2k$ chips or fewer is equivalent to winning. We can then imagine that there are just $k$ chips in the beginning, as the conditions are equivalent.

If $2k+1$ is winning for B, then $2k, \ldots, 3k+1$ are winning for A. We see that whoever wins for $3k+2$ wins for $k+1$, as above.

I don't really know how to go on here: inspection shows that B usually wins, if not for $3k$ or $3k+2$, then for the next number (the pattern: BBABAABAAAABAAAAAAAB...), but I wouldn't know if that's always true.

Edit: I have just noticed that B wins for $2, 3, 5, 8, 13, 21$. Coincidence? I think not!

I'll prove that B wins for all Fibonacci numbers. He wins for those up to $21$ by inspection. Now suppose that B wins for $F_3, F_4, \ldots, F_m$. Let $n=F_{m+1}$. If A takes $F_{m-1}$ or more chips, he obviously loses (since $\frac{F_{m+1}}{F_m}$ is bigger than $1.5$ in general). Whoever gets to $F_m$ wins, unless he gets there by a jump of at least $F_m/2$. However, if a jump that gets to $F_m$ at least $F_m/2$, then the previous one was at least $F_m/4$, so $F_{m-1} \geq 3F_m/4$, absurd. Then this doesn't happen (unless there is no previous move, but if A can't do it on his first move, so we're safe). Then the play is just as with $F_{m-1}$, so B wins.

This also implies that B wins for no other numbers, since any two B wins have a ratio of at least $1.5$, and $1.5^2=2.25>>1.618$.


calculus
integration
function
algebra
polynomial
rational function
calculus computations

I was hoping someone could help me out with 2 substitution problems that appear to run along a smilar vein.

The first is to eval. the integral of (x^2 / x+1 )dx and the second is to evaluate the integral of( x / 1+x^4) dx

I realize the second will yield an arctan with 1 + ( x^2)^2 but I don't know what to do with the x in the numerator.

Thinking back on my question, I just realized that (x^2)2 should be u and du should be 2x for the x/1+x^4 integral. That makes things real quick. However, I still am stumped on the other question, which Im sure is very similar.

For the other question: Divide through first. Don't try to integrate an improper rational function directly.
You should get a polynomial part and a fractional part, each of which will be routine.

for the firs one if you devide $x^2$ by $x+1$ you'll get $x-1$ and a remeinder of 1 so
$\int\frac {x^2} {x+1}dx = \int\(x-1dx+\int\frac {1} {x+1}dx$
which equals to
$\frac {(x-1)^2} {2} + ln (x+1) $


according to my math, this is incorrect--- BUT i did write this formula myself so there is a good chance my formula could be flawed,, but according to what i came up with today, you are incorrect

[quote="ckck"][quote="luke(mathnerd)"]A vein diagram is drawn consisting of two congruent circles of radius 17 units.  If each circle passes through the center of the other circle, what is the area of the football shaped region in the middle where the circles intersect?[/quote]
[hide]Draw from the center of the circle to the top arc where they intersect and the bottom arc where they intersect. Then connect the centers, to have two equilateral tiangles with a side length of 17. We need one more thing, the little circular region. That area is equal to the area of the circular sector minus the area of a triangle. Since its an equilateral triangle, the area of the circular sector is $\frac{289}{6}\pi$. The area of that equilateral triangle is $\frac{289\sqrt{3}}{4}$. So the little area is $\frac{289}{6}\pi-\frac{289\sqrt{3}}{4}$. But there are four of these sectors so we multiply by 4 to get $\frac{578}{3}\pi-289\sqrt{3}$. Then add in those two equilateral triangles to have our final answer which is $\frac{578}{3}\pi-\frac{289\sqrt{3}}{2}$.[/hide]
I hope I didn't do anything wrong, I did this all in my head, so theres bound to be something wrong.[/quote]
What do you mean there are four regions?, I thought there were only two of those regions, but I may have misunderstood...
edit: nvm, I drew the figure, and I see now. I got the same answer as aznwithabrain

[quote="Aznwithabrain"]dude
who are you?
i need to know
plz[/quote]

My name is Luke Wilson
Im from the northern ky MC chapter
got 1st in N. Ky chapter
I got 17th at state today
got like 25th in KAAC governors cup math state
i just started liking math this year
thats about it

8th(whoohooooo high school next year!)(sorry, ROE<Random Outburst of Excitement>)

this problem is pretty tough it took me a couple days to figure out how to do it


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number theory unsolved
number theory

Are there finitely or infinitely many pairs of positive integers $(m,n)$ such that \[2003m-2004n^2+2005(m^2+n)=1\]

Hi, Peng!
I just had one question. Where did you get the problem? It looks so familiar. I remember working on this problem very recently. Was this given at the seminar? :?

[quote="pengshi"]It's in the PC-PS4 that Dr. Recio gave us by email.[/quote]
Oh! :idea:  You are right! I forgot about them. I should keep working on them... :D

If this helps anybody, I think that the answer is that there are finitely many $(m,n)$'s. Indeed, I did a computer search on the first couple of thousand $m$'s and only found the solution $(0,1)$. Then again, I might be wrong.


i know what it is. the only use i know of it is when you have something like
how many sets of 3 single digit positive integers are there that have a sum of 8.

can give me a different type of problem involving this?

ive never heard of balls and urns could some one explain it.

Something like

David wants to eat either chocolate, glazed or rainbow-sprinkled donuts.  He can get 6 total and has an ample supply of each.  How many different assortments of donuts can he buy?

The solution would be:

Let there be 8 donuts: O O O O O O O O (mmm.... tastey  :D )

You want to turn 2 of these donuts into "dividers"  such that each divider starts a new category of donuts.  These categories would divide it into chocolate, glazed, or rainbow sprinkled.  For example,

O | O O O | O O  Would be 1 chocolate donut, 3 glazed donuts and 2 rainbow sprinkled donuts.

| O | O O O O O Would be 0 chocolate donuts, 1 glazed donut and 5 rainbow sprinkled donuts.

So in the end you just do $ \binom{8}{2}\equal{}28$ possible assortments.

That would be how to use balls and urns... some people call it stars and bars, and it's a VERY useful method of solving problems (eliminating a lot of annoying casework).

and if you had 4 different types of donuts it would be $ \binom{9} {3}$ right?

thanks.

yeah.  it would be like $ \binom{\text{total \plus{} number of different choices} \minus{} 1}{\text{number of different choices} \minus{} 1}$... i think


Catherine has six exterior doors on her dollhouse. In how many distinct ways can Catherine allow her doll to enter and then exit her dollhouse?

6 ways to enter and 6 ways to leave

answer : 36