I accidentally voted for "checking is a waste of time" lol.  But my belief is that checking your work right after is best.  My strategy is to be clear about how I solve a problem in the first place.  Then, I check right after to go over arithmetic while it's still fresh.  Sometimes I will go back and check again if I have time, and occasionally I can find flaws in my original thinking, but that usually arises because of casework.  In such cases I probably wasn't too clear on my thinking in the first place.

[quote="RC-7th"]I find that if I check my work right after I do the problem, I'm prone to repeating the same mistakes I made the first time.[/quote]

I agree completely.  Checking your work is an absolute must on the AIME, but it should be done effectively.  I think that going back  [i]after you've finished making progress on unsolved problems[/i], without looking at your old scrap work, is the best way to check.  In particular, you will almost certainly catch any computational errors. I find that, in working a problem twice in succession, I just go along with my previous calculations as opposed to really grinding through them again. Determining when you should go back to check is a judgement call; the amount of time depends on whether you are a risk-taker who would rather aim higher with greater chance for error or a conservative who would rather be assured of a slightly lower score. In the end though, I think it would be imprudent for anyone to completely neglect a thorough checking.

A brief (but hopefully instructive) story of my experience on last year's AIME I: After having answered the first 14 but made little headway on #15 with an hour to spare, I stopped to check my answers.  After obtaining the same answers on the first 8 problems, I decided to try to get #15, figuring that a fairly assured 14 was not really much better than a 13 or 12, but that a 15 would be a significant step better. As fortune would have it, I had overlooked a case on problem #9, and I did not get #15, so I wound up with a 13. In spite of the error, I was, and still am, satisfied with that result.

I agree too. Definitely check later on.

Mildorf, 13!? Geez. No wonder you can write AIME questions that are so difficult ;)

well, reading the question carefully again, and actually writing down the steps slowly but surely will weed out mistakes... write down the gist of your argument and pretend you don't know anything and understand it from what you have written down.

I agree with the checking at the end, but I voted for "right after solving each problem".  It seems that the strategy that works best for me is to do a quick "reasonability" check right after I get a problem, because I catch mathematical (non-stupid) mistakes then, when the problem is still fresh in my mind.  Then, in about the last 20-25 minutes, not 40, I check the rest of the problems.  I think 40 minutes is too much time to waste on checking- I have done practice AIME's in which I solved 2-4 problems in the last 40 minutes.  It usually takes me only about 10 minutes to check to my satisfaction, and then in the last 10 minutes I work on the problems I didn't get yet.  It worked last year for me- I solved one in the last 5 minutes and checked it with 30 seconds to go!

Generally, I'd check briefly when done, more thoroughly later on.  If you organize your work [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Mistakes.php]like this[/url], you'll both make less errors the first time around, and have a way easier job when you go back and proofread.  The AIME is the contest that's probably best structured for the approach above.

I always have thought that the best approach to checking is to try to do each question in 2 entirely different methods.  Obviously not always possible, but it helps avoid repeating the same mistakes.

[quote="Nosepuppy"]i check my answers after i figure out seven or eight of the problems.

then i go on.[/quote]
Good answer.  Organization also helps, especially with making diagrams and some of the simple stuff.  If you draw something innaccurately or make no room to write out followable steps to your base logic, you'll end up prone to make some ego-challenging errors that'll leave you spinning.  And I like this idea of 'pre-recorded' problem zones.  *Also, I can imagine having some random things like circles on grids or triangles drawn could come in handy--somehow, somewhere. [*That's a maybe.]

And I don't know about 'two different methods,' but looking back to see if there's anything that can be made simpler can improve your theory and do a lot of the checking for you when you find simple patterns.  (Making little counting mistakes is not an option when you realize [i]What[/i] you're counting to an exact understanding.

Personally I always check for about 45 minutes, and with indexes over 250 I think I can say I've had a good run. However, in spite of fairly solid checking skills, I have NEVER gotten #6 right on the AIME (8th, 9th or 10th grade) which I find to be quite amusing. This year I plan on doing #6 and then checking it about 10 times just to get that off my back.

P.S. I have known how to do #6 all 3 times, which make sit quite comical in my eyes.


Pentagon $ ABCDE$ is formed by connecting $ A$ and $ E$, the midpoints of two consecutive sides of a square. What is the sum of the lengths of the diagonals of the pentagon, expressed as a decimal rounded to the nearest tenth?
[asy]draw((0,0)--(1,0)--(1,1)--(.5,1)--(0,.5)--cycle);
draw((.5,1)--(0,1)--(0,.5),dashed);
label("4.00",(.5,0),N);
label("$A$",(.5,1),N);
label("$B$",(1,1),NE);
label("$C$",(1,0),SE);
label("$D$",(0,0),SW);
label("$E$",(0,.5),W);[/asy]

If you added correctly, you would get 23.5 :wink:

Yes i accidentally added 4.47 one too many times. Thank you for noticing.

[quote="Mewto55555"]WE have 5 diagonals:

DA=2rt5=4.47
DB=4rt2=5.66
EB=2rt5=4.47
EC=2rt5=4.47
AC=2rt5=4.47

28.0[/quote]

[color=red][size=150]I'm just wondering... Is there a faster way?[/size][/color] :read:


[url]http://www.ruletheseas.com/index.php?r=21282[/url]

Play this game  :)

STOMP

[color=white]Why do people always want to play MMORPG's???[/color]

Whoa, why are you stomping the thread???  :mad:   :P

Everyone should sign up and play this game.

MMORPG's are the best computer games ever!!!!!!!!!!!
I'll try this next week.

Im part of everybody, and I have school all year, if you didn't know.

How do you gain money, other than crime?

Once per day, at the "Things to do in the Loading Docks," you do things like Dig for Money, and Lucky Dip, which combined should give you several tens of thousands to work with (depends on the Lucky Dip). Slot machines are also good to earn money since they are slightly in your favor. And once you start leveling up, you can earn money by attacking/mugging people. What's your SN? Mine is the same as my AoPS one: Fraxia.

Excellent, finally I found a pirate MMORPG.

[quote="i_like_pie"]Did you really have to revive a month old topic with a post that has no value whatsoever to the community?[/quote]

Exactly what this post is.

Actually, it does have some value. I'm asking you not to revive old topics so the forum doesn't get boring with a bunch of old topics. The sole point of your last post, however, is to try to make me look bad. :roll:

Mods, please lock this.

If one person wants this thread to be locked but five other people don't want it to be locked, why should this thread be locked?

Anyway, this thread is about the Pirate MMORPG, and not whether my post is detailed or not.
I could start a brand new thread though, but one thread takes up more space than another post which saves some space which reduces lag.(Always use the search button before making a new thread)

So, a good way to start off making money on Rule The Seas is to vote for the game which gives you 24 points daily. Then sell the 24 points on the points market for 150k-200k which gives you around 3-4 million every day.
I suggest staying at the Loading Docks until level 5 because you can't be attacked or stolen from while you're there. I also think the success rate of crimes is increased while you're there.

Now if you had posted that the first time, I wouldn't have even said anything. And who said five people don't want this thread locked?

Just forget it...


geometry
circumcircle
Gauss
angle bisector
geometry unsolved

let triangle $ ABC$ be a triangle with orthocentre $ H$,and $ M$ and $ K$ denote the midpoints of $ AB$ and $ CH$.prove that the internal angle bisector of $ <CAH$ and$ <CBH$ meet at a point on the line $ MK$

Let $ D,E,F$ be respectively foots of $ A,B,C$ altitudes. Let $ o$ be the circumcircle of $ ABDE$ and $ P$ the intersection point of $ MK$ and $ o$. Since $ \angle EMK\equal{}\angle EFK\equal{}\angle DFK\equal{}\angle DMK$ and $ M$ is center of $ o$ we obtain $ \angle DBP\equal{}\angle PBE\equal{}\angle PAE\equal{}\angle HAP$ QED

Why QED? I am just learning. I see that you used the nine-point circle without stating it. Did you try to use some property of the circle of similitude, or Appolonius' circle?

Let's call $ { D } = AH \in BC$ and $ { E } = BH \in AC$ and let $ N$ be the midpoint of $ DE$. It is known that $ M, N$ and $ K$ are collinear ( Newton - Gauss line ).  $ \triangle AHC \sim  \triangle AED$ ( 1 ) and $ N$ and $ K$, being homologous points, they lie onto symmetrical lines about the angle bisector of $ \angle HAC$, i.e. the angles $ \angle HAC$ and $ \angle NAK$ have the same angle bisector. From (1) we shall get $ \frac {AN} {AK} = \frac {DE} {HC}$ ( 2 ).
In a similar way we can prove that N and K lie onto symmetrical lines about the angle bisector of the $ \angle HBC$ and $ \frac {BN} {BK} = \frac {DE} {CH}$ ( 3 ), from (2) and (3) getting that the subject angle bisectors divide the segment $ NK$ into the same ratio.

Best regards,
sunken rock

I think this problem is actually much easier.
Let $ D$ and $ E$ be the feet of the perpendiculars from $ A$ and $ B$. Then since $ \angle CEH \equal{} \angle CDH \equal{} 90$, $ CDEH$ is cyclic, and since $ K$ is the midpoint of $ CH$, $ K$ is the center of circle $ CDHE$. Similarly, $ M$ is the center of circle $ AEDB$. The bisectors of $ \angle EAD$ and $ \angle DBE$ will meet at the midpoint of arc $ DE$ of circle $ AEDB$, which obviously lies on the line connecting the two centers of the circles, which is $ KM$.

@dgreenb801: QED stands for Quod Erat Demonstrandum, which is latin for "which was to be demonstrated".

I know what QED is but I didn't know how showing $ \angle DBP \equal{} \angle HAP$ finished the proof, but I think I see now. You don't have to use the nine-point circle to prove it, though.


quadratics
modular arithmetic
number theory
prime factorization
number theory proposed

Let $ m,\ n \geq 3$ be positive odd integers. Prove that $ 2^{m}\minus{}1$ doesn't divide $ 3^{n}\minus{}1$.

@Rust: divides means "is a divisor of", not "is divisible by".

Nice problem :).

Assume, to the contrary, that there are such odd numbers $ m,n\ge3$ for which $ 2^m\minus{}1|3^n\minus{}1$.

Let $ p$ be a prime divisor of $ 2^m\minus{}1$ of the form $ 4k\plus{}3$. Because $ p|3^n\minus{}1$, we have that $ d\equal{}\text{ord}_3(p)$ is odd. Since $ d|p\minus{}1\equal{}4k\plus{}2$, we have $ d|2k\plus{}1$, hence $ 3^{\frac{p\minus{}1}2}\equal{}\binom{\underline3}p\equal{}1$. Then, by the Quadratic Reciprocity Law, we have $ \binom{\underline3}p\cdot\binom{\underline p}3\equal{}(\minus{}1)^{\frac{3\minus{}1}2\cdot\frac{p\minus{}1}2}\equal{}\minus{}1$ hence $ \binom{\underline p}3\equal{}\minus{}1$, so $ p\equal{}3t\plus{}2$.

Let now $ p$ be a prime divisor of $ 2^m\minus{}1$ of the form $ 4k\plus{}1$. A reasoning just as above and $ \binom{\underline3}p\cdot\binom{\underline p}3\equal{}(\minus{}1)^{\frac{3\minus{}1}2\cdot\frac{p\minus{}1}2}\equal{}1$ leads to $ \binom{\underline p}3\equal{}1$, hence $ p\equal{}3t\plus{}1$.

So let $ M$ be the multiset of prime divisors $ p$ of $ 2^m\minus{}1$ of the form $ 4k\plus{}3$, containing each prime with multiplicity equal to its exponent in the prime factorization of $ 2^m\minus{}1$. Because $ 2^m\minus{}1\equiv 3\pmod 4$, $ |M|$ is odd. But $ M$ contains precisely all prime divisors $ p$ of the form $ 3t\plus{}2$ of $ 2^m\minus{}1$. Then considering $ \text{mod}$ $ 3$, we have $ 2^m\minus{}1\equiv 2^{|M|}\equal{}2\pmod 3$, Contradiction.

Suppose that $ 2^{m} \minus{} 1$ divides $ 3^{n} \minus{} 1$.As we know $ n$ is odd,let $ n \equal{} 2k \minus{} 1$,where $ k\geq 2$.
It is easy to understand that $ 2^{m} \minus{} 1\equiv \minus{} 5$ mod $ 12$.
Since $ 2^{m} \minus{} 1$ is not divisible by $ 3$,we conclude that all prime divisors of $ 2^{m} \minus{} 1$ are congruent to either $ \plus{} 5, \minus{} 5$,or $ \plus{} 1, \minus{} 1$ mod $ 12$,but as we know  $ 2^{m} \minus{} 1\equiv \minus{} 5$,
it follows that, there exist $ p\equiv 5$ or $ p\equiv \minus{} 5$ mod $ 12$ and $ p$ divides $ 2^{m} \minus{} 1$,thus it divides $ 3^{n} \minus{} 1$,too,but then $ p$ divides $ 3(3^{n} \minus{} 1)$ as well,hence $ 3^{2k} \equal{} 3^{n \plus{} 1}\equiv 3$ mod $ p$,so $ 3$ is a quadratic residue mod $ p$,but this contradicts to the quadratic reciprocity theorem,because $ p\equiv 5$ or $ \minus{} 5$ mod $ 12$.

Other solution (same free mind but have a bit difference)
Call p is a odd  prime divisor of $ 3^n - 1$
 $ \Rightarrow p|3^{\frac {n + 1}{2}} - 3$
$ \Rightarrow \frac (\frac {3}{p}) = 1$
$ \Rightarrow p\equiv 1 (\mod 12)$ or $ p\equiv - 1 (\mod 12)$
So if $ k|3^n - 1$ then $ k\equiv 1 (\mod 12)$ or $ k\equiv - 1(\mod 12)$
But $ 2^n - 1\equiv 7 (\mod 12)$ ,it give contradiction .
More than $ rad(2^m - 1) \not |3^m - 1$

[quote="freemind"]
Let $ p$ be a prime divisor of $ 2^m \minus{} 1$ of the form $ 4k \plus{} 3$. Because $ p|3^n \minus{} 1$, we have that $ d \equal{} \text{ord}_3(p)$ is odd. Since $ d|p \minus{} 1 \equal{} 4k \plus{} 2$, we have $ d|2k \plus{} 1$, hence $ 3^{\frac {p \minus{} 1}2} \equal{} \binom{\underline3}p \equal{} 1$.[/quote]
 how you can say that if dl 4k+2 then dl 2k+1 for example p=7   7l 3^6-1 but  7l 3^3-1

$ d$ is odd because $ d|n$ and $ n$ is odd.
[quote="kim_ina_88"][quote="freemind"]
Let $ p$ be a prime divisor of $ 2^m \minus{} 1$ of the form $ 4k \plus{} 3$. Because $ p|3^n \minus{} 1$, [color=red]we have that $ d \equal{} \text{ord}_3(p)$ is odd[/color]. Since $ d|p \minus{} 1 \equal{} 4k \plus{} 2$, we have $ d|2k \plus{} 1$, hence $ 3^{\frac {p \minus{} 1}2} \equal{} \binom{\underline3}p \equal{} 1$.[/quote]
 how you can say that if dl 4k+2 then dl 2k+1 for example p=7   7l 3^6-1 but  7l 3^3-1[/quote]

Very simple, if we use Jacobi's symbol:
We note $3^{\dfrac{n-1}{2}}=a\Rightarrow 3a^2\equiv 1(mod\,\; 2^m-1)\Rightarrow (3a)^2\equiv 3(mod\,\; 2^m-1)\Rightarrow (\dfrac{3}{2^m-1})=1$. By use quadratic reciprocity, $(\dfrac{2^m-1}{3})*(-1)^{\dfrac{3-1}{2}*\dfrac{2^m-1-1}{2}}=1\Rightarrow (\dfrac{2^m-1}{3})=-1$, obviously false, because $2^m\equiv 2(mod\,\; 3)$.

After noticing that there exists a prime $p$ such that $p \mid 2^m-1$ and that $p \equiv 5,7 (mod 12)$ we conclude by quadratic reciprocity law.

I will solve the following alternative formulation of this problem:
[quote = Alternative Formulation]
Let $a$ and $b$ be positive integers such that $2^a - 1$ divides $3^b - 1$. Prove that either $a = 1$ or $b$ is even.
[/quote]

[b] Solution: [/b] Suppose $a > 1$ and $b$ is odd. Let $p \mid 2^a - 1$ be a prime. Then,
	\begin{align*}
		p &\mid 3^b - 1 \\
		p &\mid 3^{b + 1} - 3.
	\end{align*}
Hence $\left(\frac{3}{p}\right) = 1$. If $p \equiv 1 \pmod{4}$, we have by Quadratic Reciprocity that
	\begin{align*}
		\left(\frac{p}{3}\right) &= \left(\frac{3}{p}\right) \\
		&= 1,
	\end{align*}
so $p \equiv 1 \pmod{12}$. If $p \equiv -1 \pmod{4}$, we have by Quadratic Reciprocity that
	\begin{align*}
		\left(\frac{p}{3}\right) &= -\left(\frac{3}{p}\right) \\
		&= -1,
	\end{align*}
so $p \equiv -1 \pmod{12}$. Hence, $p$ must be of the form $12k \pm 1$. Since $a > 1$, we have
	\begin{align*}
		2^a - 1 &\equiv -1 \pmod{4}.
	\end{align*}
Also, note that
	\begin{align*}
		2^a - 1 \equiv 0, 1 \pmod{3}.
	\end{align*}
Hence
	\begin{align*}
		2^a - 1 \equiv 3, 7 \pmod{12}.
	\end{align*}
Hence there exists a prime $q \mid 2^a - 1$ that is not of the form $12k \pm 1$, contradiction. Thus, either $a = 1$ or $b$ is even, as desired. $\Box$

[hide=QR]

Assume for the sake of contradiction that $a \neq 1$, $b$ is odd. Then, we have $3^b \equiv 1 \mod (2^a-1)$, so $3^{b+1} \equiv 3 \mod (2^a-1)$. Because $b$ is odd, this implies that $3$ is a quadratic residue $\mod 2^a-1$. However, we have $$(\frac{3}{2^a-1})(\frac{2^a-1}{3})=(-1)$$ using the Jacobi Symbol. Because $(\frac{2^a-1}{3})=1$ for odd $a$ (and $a$ is odd otherwise $3|(2^a-1)$), we have $(\frac{3}{2^a-1})=-1$. Therefore, $3$ is a quadratic non-residue module some $p|2^a-1$, contradiction, so $b$ is even, as desired. $\blacksquare$

[/hide]

Suppose $b$ is odd and $2^a-1\mid 3^b-1$ for some $a.$ Consider some prime $p$ dividing $2^a-1.$ We have $$3^b-1\equiv 0\pmod{p}$$$$\implies 3^{b+1}\equiv 3\pmod{p}$$$$\implies \left(\frac{3}{p}\right)=1.$$ If $p\equiv 1\pmod{4},$ then $$p\equiv \left(\frac{p}{3}\right)\stackrel{QR}{=}\left(\frac{3}{p}\right)=1\pmod{3}.$$ If $p\equiv 3\pmod{4},$ then $$p\equiv \left(\frac{p}{3}\right)\stackrel{QR}{=}-\left(\frac{3}{p}\right)=-1\pmod{3}.$$ Therefore, $p\equiv\pm 1\pmod{12}.$ Since our choice of $p$ was arbitrary, we must have $$2^a-1\equiv\pm 1\pmod{12}.$$ This is true if and only if $a=1,$ so we are done.

We do the following stronger problem.
[quote = Stronger Statement]
Let $a$ and $b$ be positive integers such that $2^a-1\mid 3^b-1$. Prove that either $a=1$ or $b$ is even.
[/quote]
Suppose otherwise. As $2^a-1 > 1$, $2^a-1=N$ has some prime divisors $p$.

[color=#f00][b]Claim:[/b][/color] For $p\mid N$, we have that $p\equiv \pm 1\pmod{12}$.

[i]Solution[/i]: Let $o_p(3)$ be the order of $b$ modulo $p$. As $p\mid N\mid 3^b-1$, we have $o_p(3)\mid b$. Since $b$ is odd, we must have $o_p(3)\mid \frac{p-1}{2}$, since we can disregard powers of $2$ that divide $p-1$. This implies that $3$ is a quadratic residue modulo $p$, so $p\equiv \pm 1\pmod{12}$ as desired. $\fbox{}$

Now, note that this implies $N\equiv \pm 1\pmod{12}$. As $N\equiv -1\pmod{4}$, we get that $N\equiv -1\pmod{12}$. In particular, $2^a-1=N\equiv -1\pmod{3}$. But this would imply $3\mid 2^a$, absurd.

[quote=algebra_star1234]Since $b$ is odd, we have $3^{b+1} \equiv 3 \pmod{2^{a}-1}$ is a quadratic residue. By quadratic reciprocity,
\[ \left( \frac{3}{2^{a}-1}\right)=\left(\frac{2^{a}-1}{3}\right) (-1)^{2^{a-1}-1} = 1 \cdot -1 = -1 .\]
for $a > 1$. Therefore $b$ cannot be odd.[/quote]

This is close, but it doesn't work because it's missing the $(-1)^\frac{3-1}{2}=-1$ term, so
\[
  \left(\frac{3}{2^a-1}\right) = 1
\]

It would work if $3$ was changed in the original problem to a value that is 1 mod 4, such as $5$.

[quote=Rust]$ 3^n\minus{}1\not |2^m\minus{}1$, because $ 3^n\minus{}1$ even, $ 2^m\minus{}1$ odd.[/quote]

Odd number can divide an even number but even number can never divide an odd number.

We have that $3^{b+1} \equiv 3 \pmod{2^a-1}$ is a quadratic residue; by quadratic reciprocity we have $$\left( \frac{3}{2^a-1} \right) \left( \frac{2^a-1}{3} \right) = (-1)^{(2^a-2)(2)/4} = -1$$so $2^a-1$ is a quadratic nonresidue $\pmod{3}$. This is impossible.

The case where $a = 1$ is trivial. Otherwise, FTSoC assume $a > 1$ and $b$ is odd. Also ote that $a$ must also be odd, else $3 \mid 2^a - 1$.

Select a prime $p$ dividing $2^a - 1$; note that $2^{a+1} \equiv 2 \pmod p$ and $3^{b+1} \equiv 3 \pmod p$. Hence,\[\left(\frac{2}{p}\right) = \left(\frac{3}{p}\right) = 1\]must be true. From $\left(\tfrac{2}{p}\right) = 1$, we get that $\tfrac18(p^2 - 1)$ is even which yields $p \equiv \pm 1 \pmod 8$. [list]
    [*] If $p \equiv 1 \pmod 8$, then\[1 = \left(\frac{3}{p}\right) = \left(\frac{p}{3}\right)\]and clearly $p \neq 3$ since $a$ is odd, so $p \equiv 1 \pmod 3$. Thus $p \equiv 1 \pmod {24}$. \\
    [*] If $p \equiv -1 \pmod 8$, then\[1 = \left(\frac{3}{p}\right) = -\left(\frac{p}{3}\right)\]so $p \equiv -1 \pmod 3$. Thus $p \equiv -1 \pmod {24}$.
[/list]
Since all primes dividing $2^a - 1$ are $\pm 1$ modulo $24$, the number $2^a - 1$ itself must be $\pm 1$ mod $24$. This reduces to either $24 \mid 2^a$ or $24 \mid 2^{a} - 2$ which are both impossible, so we have our desired contradiction. $\blacksquare$

Suppose $2^m-1\mid 3^n-1$. Suppose $p\mid 2^m-1$, so $p\mid 3^n-1$. Hence
\[ \left(3^{\frac{n+1}{2}}\right)^2 \equiv 3^{n+1} \equiv 3 \pmod{p} \implies \left(\frac{3}{p}\right)=1. \]
Hence $p\equiv \pm1 \pmod{12}$. Hence all prime factors of $2^m-1$ are $\pm 1 \pmod{12}$, so $2^m-1 \equiv \pm 1\pmod{12}$. So $2^m\equiv 2 \pmod{12}$, which is a contradiction by mod 4.

Cool.

We will use the generalization of the Law of Quadratic Reciprocity for the Jacobi Symbol. \\
Assume FTSOC that $2^m-1 \mid 3^{2k+1}-1$, then notice that $3$ is a quadratic residue $\pmod{2^m-1}$. Moreover that $m$ must be odd as otherwise $3 \mid 2^m-1$ which is not possible and consequently, $$2^m-1 \equiv 1 \pmod{3}$$
We have that , $$1 = \left(\frac{2^m-1}{3}\right)= \left(\frac{3}{2^m-1}\right) \cdot \left(\frac{2^m-1}{3}\right) = (-1)^{\frac{2(m-1)}{4}} = -1$$ as $m$ is odd which is a contradiction. $\blacksquare$ \\

$a=1$ is trivial so assume $a>1.$

Consider some $p \mid 2^{a}-1.$ Then, $2^{a+1} \equiv 2 \pmod{p},$ meaning \[ \left( \frac{2}{p} \right) = (-1)^{1/8 (p^2-1)} =1 \iff p \equiv \pm 1 \pmod{8} \] as $a+1$ is even, meaning $2$ is a QR mod $p.$ Similarly, $p \mid 3^{b}-1$ and furthermore \[ \left( \frac{3}{p} \right) \left( \frac{p}{3} \right) = (-1)^{1/2(p-1)}. \] If $p \equiv 2 \pmod{3},$ then $ \left( \frac{p}{3} \right) = -1,$ which also gives $p \equiv -1 \pmod{3}.$ Combined, these give $p \equiv -1 \pmod{24}.$ The other case gives $p \equiv 1 \pmod{24},$ so $p \equiv \pm 1 \pmod{24}.$

Therefore, if $p \mid 2^{a}-1$ we also have $p \equiv \pm 1 \pmod{24}.$ Combined this implies that $2^{a} -1 \equiv \pm 1 \pmod{24}$ which are both contradictions due to divisibility reasons. Thus there are no solutions.

Can someone check this sol?

Obviously there exists a prime $p$ congruent to 3 mod 4 dividing $2^m-1$. Now, notice that as $m,n$ are odd, $m=2x+1$ and $n=2y+1$ thus $2^m\equiv 1 (p) \implies 2^{2x}.2\equiv 1 (p) \implies (\frac{1/2}{p})=1$, thus we have by legendre symbol properties that $(\frac{2.1/4}{p})=1=(\frac{1/4}{p}).(\frac{2}{p})=1$ which implies $2$ is a quadratic residue mod $p$, similarly for $3$. Respectively, this implies that $p \equiv \{1,-1\} (8), p \equiv \{1,-1\} (12)$, but as $p\equiv 3 (4)$ this implies $p\equiv -1 (mod 8)$ and $p \equiv -1 (mod 12)$. Quadratic recyprocity gives: $(\frac{3}{p}).(\frac{p}{3})=(-1)^{p-1/2}.(-1)^{1}=1$ and thus $p\equiv 1 (3)$ or $p=3$, which both contradict the fact that $p\equiv -1 (mod 12)$

[quote=yayups]Suppose $p$ is some prime factor of $3^n-1$. Then, we have that
\[\left(3^{\frac{n+1}{2}}\right)^2\equiv 3\pmod{p},\]
so $3$ is a quadratic residue mod $p$. Using quadratic reciprocity, this tells us that $p\equiv\pm 1\pmod{12}$.

Therefore, all the prime factors of $2^m-1$ are $\pm 1\pmod{12}$, so $2^m-1\equiv\pm 1\pmod{12}$, so in particular, $2^m\equiv 2\pmod{4}$. Thus, $m=1$, which is a contradiction since $m\ge 3$, so we're done.[/quote]

very good and correct solution :)

If $a=1$, clearly $2^a-1=1\mid 3^b-1$. Otherwise, if $a$ is even then $3\mid 2^a-1$, so $3\mid 2^a-1\mid 3^b-1$, a contradiction. Thus, we will only consider odd $a$.

[b]Claim 1:[/b]For $q\geq 5$, 3 is a quadratic residue mod $q$ if and only if $q\equiv 1,-1\pmod{12}$.
[b]Proof:[/b]If $q\equiv 1\pmod{4}$, then $q$ must be a quadratic residue mod 3, so $q\equiv 1\pmod{3}$ too, so $q\equiv 1$. Otherwise, if $q\equiv 3\pmod{4}$, then $q\equiv 2\pmod{3}$ by quadratic recirprocity, so $q\equiv -1$.$\square$

[b]Claim 2:[/b]For odd $a$, 3 is not a quadratic residue mod $2^a-1$
[b]Proof: [/b]Note that $2,3 \nmid 2^a-1$, and $2^a-1 \equiv 3 \pmod{4}$ and $2^a-1\equiv 1 \pmod{3}$, so $2^a-1\equiv 7\pmod{12}$. Thus, there exists some $q\not\equiv -1,1$ such that $q\mid 2^a-1$. Since by the previous claim, $a$ is not a quadratic residue mod $q$, $a$ is not a quadratic residue mod $2^a-1$.$\square$

Thus, if $b$ is odd, then $3^b=1$ is also not a quadratic residue mod $2^a-1$ a clear contradiction.

Ok how broken is quadratic reciprocity

We make use of the fact that if a prime $p$ divides $a^k-1$ for $k$ odd, then $a$ is a quadratic residue $\pmod{p}$, since $\mathrm{ord}_p(a)=\tfrac{p-1}{d}$ for some $d \mid k$.
Assume otherwise, and let $p$ be a prime dividing $2^m-1$. Then from the above fact we obtain $(\tfrac{2}{p})=1 \implies p \equiv 1,7 \pmod{8}$. Now, prime factorize
$$2^m-1=p_1^{e_1}p_2^{e_2}\ldots p_k^{e_k}.$$
Since $2^m-1 \equiv 7 \pmod{8}$, there are an odd number of indices $i$ such that $e_i$ is odd and $p_i \equiv 7 \pmod{8}$—call these indices [i]special[/i]. Note that the non-special indices $i$ either satisfy $p_i \equiv 1 \pmod{8}$ or $p_i \equiv 7 \pmod{8}$ and $e_i$ is even.
Now, by assumption, we must have $p_i \mid 3^n-1 \implies (\tfrac{3}{p_i})=1$. Consider the following cases:
[list]
[*] $p_i \equiv 1 \pmod{8}$. Then by quadratic reciprocity
$$\left(\frac{3}{p_i}\right)\left(\frac{p_i}{3}\right)=1 \implies \left(\frac{p_i}{3}\right)=1 \implies p_i \equiv 1 \pmod{3}$$
[*] $p_i \equiv 7 \pmod{8}$. Then by quadratic reciprocity
$$\left(\frac{3}{p_i}\right)\left(\frac{p_i}{3}\right)=1 \implies \left(\frac{p_i}{3}\right)=-1 \implies p_i \equiv 2 \pmod{3}$$
[/list]
Take both sides of $2^m-1=p_1^{e_1}\ldots p_k^{e_k}$ modulo $3$. The LHS is clearly $1 \pmod{3}$. However, since the only $i$ with $p_i^{e_i} \equiv 2 \pmod{3}$ are the special $i$, of which there are an odd number, and the rest are $1 \pmod{3}$—either because $p_i \equiv 1 \pmod{3}$ or $e_i$ is even—it follows that the RHS is $2 \pmod{3}$, which is a contradiction. Thus no such $(m,n)$ exist, as desired. $\blacksquare$

Suppose that $b$ is odd. If $a$ is even, then for mod $3$ yields a contradiction.

Thus assume $a$ is odd, Let $p \mid 2^a - 1$; thus $2^a \equiv 1 \pmod p$ for $a$ odd implies $\left(\frac 2p\right) = 1$. Similarly, $\left(\frac 3p\right) = 1$.

On the other hand, the first equality holds for $p \equiv \pm 1 \pmod 8$, and the other equality holds for $$(-1)^{(p-1)/2}\left(\frac p3\right) = 1 \iff p \equiv \pm 1 \pmod {12}.$$ So $p \equiv \pm 1 \pmod {24}$ and $2^a - 1 \equiv \pm 1 \pmod {24}$, contradiction.

Here is the phrasing I received the problem in.
[quote=Romania TST 2008 modified]
Let $a > 1$ and $b > 1$ be positive integers such that $2^a-1\mid 3^b-1$. Prove that $b$ is even.
[/quote]

The following below is my solution.

Firstly, if $a$ is even, then we get that $3 \mid 2^a-1\mid 3^b-1$ which gives a contradiction. So $a$ must be odd. Furthermore, FTSOC assume that $b$ is odd too. So we now use $a=2m+1$ and $b=2n+1$, where $m$ and $n$ are +ve integers.

Now pick any prime $p$ such that $p\mid 2^{2m+1}-1$. Clearly $p\not=3$. This gives us that $2^{2m+1}\equiv 1\pmod{p}\implies 2^{2(m+1)}\equiv 2\pmod{p}$, which means that $2$ is a quadratic residue $\pmod{p}$. Similarly, we get that $p\mid 2^{2m+1}\mid 3^{2n+1}$ which further gives that $3$ is a quadratic residue too. Now we proceed using Legendre's notation.

Firstly, from the fact that $\left(\dfrac{2}{p}\right)=1$, we get that $p\equiv \left\{+1,-1\right\}\pmod{8}$. Also, from $\left(\dfrac{3}{p}\right)=1$, we get that $\left(\dfrac{p}{3}\right)=(-1)^{\dfrac{3-1}{2}\cdot\dfrac{p-1}{2}}\cdot\left(\dfrac{3}{p}\right)=(-1)^{\dfrac{p-1}{2}}$.

Now if $p\equiv 1\pmod{8}$, then we get that $\left(\dfrac{p}{3}\right)=-1$ which further gives us that $p\equiv 1\pmod{3}$. Similarly if $p\equiv -1\pmod{8}$, then we get that $p\equiv -1\pmod{3}$. Now finally combining these two using C.R.T., we get that $p\equiv\left\{+1,-1\right\}\pmod{24}$.

Now as $2^{2m+1}-1$ is just a product of a bunch of such primes, we get that $2^{2m+1}-1\equiv \left\{+1, -1\right\}\pmod{24}$. Both of the cases give simple modular contradictions and we are done. :stretcher:


For any positive integer $n$,let $T(n)$ denote the number of positive divisor of $n$(including 1 and $n$ itself ).Determine all positive number $m$ such that $T(n^3)$ +$T(n)$=$mT(n^2)$ for some $n$.

Let $n = p_1^{a_1} p_2^{a_2} ... p_k^{a_k}$.  Then $T(n) = \prod_{i=1}^{k} (a_i + 1), T(n^p) = \prod_{i=1}^{k} (pa_i + 1)$

So we wish to find solutions to $\prod_{i=1}^{k} (3a_i + 1) + \prod_{i=1}^{k} (a_i + 1) = m \prod_{i=1}^{k} (2a_i + 1)$ for $a_i \in \mathbb{N}$.  First note that if the $a_i$ have a common factor $p$, we can take $\bmod p$ and get $1 + 1 \equiv 1 \bmod p$ - hence, they cannot all have a common factor, which means $n$ cannot itself be a power of another integer.

And, er, I'm stuck  :(  Who wants to post a solution?

[quote="othman"]$\prod_{i=1}^{k} (a_i + 1) \equiv0mod2$[/quote]
[b]othman[/b],can you write more clearly?


[quote="chien than"]Prove that:$ab+bc+cd+de \leq \frac{1}{4}$
where $a;b;c;d;e$ are non-negative and $a+b+c+d+e=1$[/quote]
$ab+bc+cd+de+ea \leq \frac{1}{4}$ is still true. :wink:

Maybe. But now I see my prove is not works with your problem.


calculus
integration
logarithms
real analysis
real analysis theorems

I know of a nice little trick to test convergence but I would like to know the name
I have only seen it once in a syllabus

let $u_{n}$ a sequence, with for all n $u_{n}\geq 0$ and it is a decreasing sequence

the sequence $\sum_{n=1}^{\infty}u_{n}$ converges if and only if

$\sum_{n=1}^{\infty}u_{2^n}$ converges

all generalizations or comments are welcome

And a very nice test it is.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=17772]An example of it being put to good use.[/url]

I see it as a natural generalization the argument given by Oresme in the middle of the 14th century that the harmonic series diverges - and example of how the best and most beautiful arguments can often be extended beyond their original application.

As a mundane test for the convergence of specific series, it is generably usable on the same series that the integral test is good for. Note that it shares a key hypothesis with the integral test: the terms are positive, nonincreasing, and tend to zero.

Try it on $\sum\frac1{n^p},$ $\sum\frac1{n(\ln n)^p}$ and $\sum\frac1{n(\ln n)(\ln\ln n)^p}$ to see it work.


function
abstract algebra
floor function
group theory
combinatorics unsolved
combinatorics

Let $ S\equal{}{1,2,...,2n}$ How many functions $ f: S\mapsto S$ such that
$ f(f(f(k)))\equal{}2n\minus{}k\plus{}1$ for all $ k\in S$ and $ f(f(k))$ is not equal to $ k$ for all $ k\in S$

No particular reason to restrict to the even case.  We want to know the number of elements $ \pi\in S_{n}$ such that $ \pi^{3}\equal{}\rho$, where $ \rho$ is the reversal permutation $ \rho(k) \equal{} n\plus{}1\minus{}k$.  Note that the map $ \sigma\mapsto\sigma^{3}$ is a group homomorphism from $ S_{n}$ to $ S_{n}$.  Thus, the size of the preimage of $ \rho$ is either 0 or it is equal to the size of the kernel of this homomorphism.  Since $ \rho^{3}\equal{}\rho$, the preimage is not empty, so the answer is the size of this kernel.  Any element of the kernel can be written in cycle notation; expressed this way, we see that the cycles must all either have length 3 or length 1.  If we wish to choose $ k\leq\frac{n}{3}$ cycles of size 3, we have $ \frac{n!}{k!(n\minus{}3k)!3^{k}}$ ways to do so, so our desired answer is $ \minus{}1\plus{}\sum_{0\leq k\leq\lfloor n/3\rfloor}\frac{n!}{k!(n\minus{}3k)!3^{k}}$ (where the $ \minus{}1$ accounts for the fact that we wish not to count $ \rho$ itself, the only member of the set which satisfies $ f(f(x))\equiv x$).  This sequence is in the [url=http://www.research.att.com/~njas/sequences/A001471]OEIS[/url], and this formula is the best given there.


Show that the arithmetic progression

8, 21, 34, 47, ...

contains infinitely integers, consisting only of 9's in decimal form.

[hide]you want us to show that there are infinitely many integers of the form 10^n-1 that are also of the form 13k-5, i.e.10^n:equiv:9 mod 13. Indeed, all n:equiv:2 mod 12 satisfy this condition. This is easy to see as 10^2=100:equiv:9 mod 13, and a^12:equiv:1 mod 13 (for a not a multiple of 13) by fermat. [/hide]


geometry
factorial
quadratics
algebra
polynomial
floor function
symmetry

1. Find positive integers A and B such that A^2 + B^2 = 2009.

2. Find the area of the region bounded by ! x^2+ y^2-2x+4y-6! < 8. (absolute value i mean, not factorial)

3. The roots of quadratic equations x^2 + ax+b =0 , x^2 + cx+d =0, x^2 + ex+f =0 are the same as the roots of 3x^3-2x^2+x-1=0, taken two at a time. Evaluate (b+d+f)/(ace)

4. Find all k such that  x^3+x^2-kx-k=0 and  x^3-x^2-3x+k=0 have a common root.

5. Find all integers n such that ! n^3- 4n^2-5! is prime. (absolute value i mean, not factorial)

6. Solve for x ! 3x-4!+!x-5! < !x! (absolute value i mean, not factorial)

7. Find all negative real roots of x^5-45x^4+168x^3-164x^2-49x+2009.

8. Find all ordered integer pairs (x,y) such that x+4y-4(sqrt x)-8(sqrt y)+4(sqrt xy) + 3 =0

9. How many terms are there in (x^3+1)^2007+(x^2+1)^2008+(x+1)^2009?

10. How many ordered integer triplets are there such that 2^a+3^b=5^c <2009.

[hide="number two"]
Factorise to
$ |(x \minus{} 1)^2 \plus{} (y \minus{} 2)^2 \minus{} 11| \le 8$

$ 3\le (x \minus{} 1)^2 \plus{} (y \minus{} 2)^2 \le 19$

$ A \equal{} \pi((\sqrt{19})^2 \minus{} (\sqrt{3})^2) \equal{} 16\pi$

[/hide]

[hide="number 4"]
Denote the two polynomials, $ P_1(x),P_2(x)$, if they both have a common factor $ (x \minus{} r)$
$ \gcd(P_1(x),P_2(x)) \equal{} \gcd(P_1(x) \plus{} P_2(x),P_1(x))$

Now $ P_1(x) \plus{} P_2(x) \equal{} 2x^3 \plus{} (3 \minus{} k)x \equal{} 0 \Longrightarrow x \equal{} 0, k \equal{} 2x^2 \minus{} 3$

Sub $ x\equal{}0 \longrightarrow \minus{}k\equal{}k \Rightarrow k\equal{}0$

Sub $ k \equal{} 2x^2 \minus{} 3$ into $ P_2(x)\longrightarrow x^3 \plus{} x^2 \minus{} 3x \minus{} 3 \equal{} (x \plus{} 1)(x^2 \minus{} 3) \equal{} 0$

Therefore $ k \equal{} 2(1) \minus{} 3 \equal{} \minus{} 1$ or $ 2(3) \minus{} 3 \equal{} 3$

$ k \equal{} 0,\minus{}1,3$  :) [/hide]

EDIT: Okay, fixed them up

#2 16pi

#4 0, -1, 3

others are correct..

[quote="ocha"][hide="number 1"]
$ 2009 \equiv 0 \mod 7$
Quadratic residues $ \mod 7$ are $ \{0,1,2,4\}\Rightarrow A,B \equiv 0 \mod 7$

$ \Rightarrow x^2 \plus{} y^2 \equal{} 41$
Again $ 41\equiv 2 \mod 3$ So $ x\equiv y \equiv 1\mod 3$ and since $ 0 < x,y < 7$
The only possibilities are $ x,y \equal{} \{4,5\}$
A quick check to see that they work and we have our answers

$ (A,B) \equal{} (28,35)$[/hide][/quote]
Is there a quick way to calculate the quadratic residues $ \bmod 7$ (without finding the remainder when $ a^2$ is divided by $ 7$ where $ 1\le a \le 9$?
EDITED TYPO

[quote="mathwizarddude"]
Is there a quick way to calculate the quadratic residues $ \bmod 7$ (without finding the remainder when $ a^7$ is divided by $ 7$ where $ 1\le a \le 9$?[/quote]

Yes, to find quad residues $ \mod n$ we just look that the remainder when $ a^2$ is divided by $ n$ for $ 0 \le a \le \left \lfloor \frac{n}{2} \right\rfloor$. Because of symetry we only need to calculate half of the set.

I didn't know It was the same as calculating $ a^n \mod n$...

[quote="ocha"]
 Because of symetry we only need to calculate half of the set. [/quote]
How do you prove there is symmetry?

The idea is showing that $ x^2\equiv (m\minus{}x)^2\pmod{m}$.  Expand out the right-hand side and you'll see why.

Number $ 5$ is very exacerbating, It doesn't factor or fall out to any obvious residue...
After giving up I puged in a buch of values and found primes at $ n\equal{}2,4,6,8,16,22,26,36,46...$
This doesn't seem to follow any pattern...
Do you have any hints orangefronted?
Btw, where did you get all these questions?

my shot at 5:

$ |n^3 \minus{} 4n^2 \minus{} 5| \equal{} p$
$ n^3 \minus{} 4n^2 \minus{} 5\minus{}p \equal{} 0$ or $ n^3 \minus{} 4n^2 \minus{} 5\plus{}p \equal{} 0$

By the rational root theorem, $ n|(\minus{}5\pm p)$.  So we could have solutions for... any $ p$ i guess.  not helpful.

Lets try analysis mod 6.

$ n \equiv 0$ implies $ p \equiv 1$ - could yield primes
$ n \equiv 1$ implies $ p \equiv 4$ - no primes (except possibly $ p\equal{}2$, if we take the opposite of the quantity)
$ n \equiv 2$ implies $ p \equiv 5$ - could yield primes
$ n \equiv 3$ implies $ p \equiv 4$ - no primes (except possibly $ p\equal{}2$, if we take the opposite of the quantity)
$ n \equiv 4$ implies $ p \equiv 1$ - could yield primes
$ n \equiv 5$ implies $ p \equiv 2$ - no primes (except possibly $ p\equal{}2$)

so lets test $ p\equal{}2$
$ n^3 \minus{} 4n^2 \minus{} 7\equal{}0$.  $ n|7$, the only possible integer roots are $ \pm 7, \pm 1$, none of which work.

$ n^3 \minus{} 4n^2 \minus{} 3\equal{}0$ so $ n|3$.  thus $ n \equal{} \pm1, \pm 3$, none of which are solutions.

if $ |n^3 \minus{} 4n^2 \minus{} 5| \equal{} p$, then $ p \ge 5$.

stuck here.

These are taken from some contest problems.

Sorry, I committed a mistake it should have been ! n^3- 4n^2+5! is prime.

Some of these problems I dont have an idea really. especially number 7 to 10. :(

Thank you for your help!

haha...well here is solution to $ 5$ then...

$ x^3 \minus{} 4x^2 \plus{} 5 \equal{} (x \plus{} 1)(x^2 \minus{} 5x \plus{} 5)$!!!!

So now if we want to find primes we can only have $ x \plus{} 1 \equal{} \pm 1$ or $ x^2 \minus{} 5x \plus{} 5 \equal{} \pm 1$
$ x \equal{} \minus{}2, 0, 1,2,3 ,4$

Checking we get; $ 5,2, 3, 4, 5$ respectively. Now discard $ 4$ because its not a prime

Answer: $ \boxed{\minus{}2,0,1,2,4}$

well that works out much better.  Factor:

$ (n \plus{} 1)(n^2 \minus{} 5n \plus{} 5) \equal{} 0$.  Now either $ n \equal{} \minus{} 2, 0$ ($ 0$ clearly works, $ \minus{} 2$ gives $ \minus{} 19$ which also works)  or $ n^2 \minus{} 5n \plus{} 5 \equal{} \pm 1$ implies either $ n^2 \minus{} 5n \plus{} 4 \equal{} 0$ or $ n \equal{} 1,4$ (both which work) or $ n^2 \minus{} 5n \plus{} 6 \equal{} 0$ implies $ n \equal{} 2,3$ and $ 2$ works while $ 3$ does not.  So our values are

$ n \equal{} \minus{} 2, 0, 1, 2, 4$.

EDIT: ocha, you beat me to it.  Almost correct, but you ignored the absolute value signs.
EDIT 2: seems you fixed it already, before I posted.

Oh! Factoring was the krux move in that problem! got it!

I hope you can help me with the other problems.. :)

[quote="facis"]
EDIT: ocha, you beat me to it.  Almost correct, but you ignored the absolute value signs.
EDIT 2: seems you fixed it already, before I posted.[/quote]

Yeah I noticed my mistake as I clicked the submit button, haha...

[hide="number 6 - solution outline"]
Draw the graphs $ y \equal{} \minus{} |x|, y \equal{} |x \minus{} 5|$ first, then do an addition-of-ordinates to get a graph that look sort of like a "Z". Then superimpose the graph $ y \equal{} |3x \minus{} 4|$ and check when it reaches the value $ y \equal{} 5$...
[/hide]

[hide="solution 7"]
Since $ 2009 \equal{} 41\cdot 7^2$ from the rational root theorem, rational roots could be $ 7, 41, 49, 287...$
This polynomial is very long and has large coefficients so synthetic division will help.([url=http://en.wikipedia.org/wiki/Polynomial_long_division]See Here[/url]--and scroll down)

Checking we find that $ 41$ is a root so
$ P(x) \equal{} (x \minus{} 41)(x^4 \minus{} 4x^3 \plus{} 4x^2 \minus{} 49)$

Factorising further;
$ P(x) \equal{} (x \minus{} 41)(x^2(x \minus{} 2)^2 \minus{} 49) \equal{} (x \minus{} 41)(x(x \minus{} 2) \plus{} 7)(x(x \minus{} 2) \minus{} 7)$

Now the quadratic formula yeilds

$ x \equal{} 41, \frac {2\pm \sqrt {4\pm 4\cdot 7}}{2}$

So our real roots are; $ 41, 1 \pm 2\sqrt {2}$

Negative roots are; $ \boxed{1 \minus{} 2\sqrt {2}}$
[/hide]

Yah!! Great great for number 7!! Few more! Thanks!

9) 1st, note from binomial theorem that $ (x\plus{}y)^n$ expands to $ n\plus{}1$ terms.  So initially it appears that we have a total of $ 2008\plus{}2009\plus{}2010 \equal{} 6027$ terms.

But some of the terms will combine together since they have the same exponents on $ x$.
$ (x^3\plus{}1)^{2007}$ gives exponents that are multiples of 3, from $ 0$ to $ 6021$
$ (x^2\plus{}1)^{2008}$ gives exponents that are multiples of 2, from $ 0$ to $ 4016$
$ (x\plus{}1)^{2009}$ gives all integer exponents from $ 0$ to $ 2009$

Our expansion clearly has $ x^0$ through $ x^{2009}$ terms ($ 2010$ terms here).
Our expansion will also include all even exponents from $ 2010$ to $ 4016$ ($ 1004$ terms)
and all exponents that are multiples of $ 3$ from $ 2010$ to $ 6021$ ($ 1338$ terms)

But these last two categories both include all exponents that are multiples of $ 6$ between $ 2010$ and $ 4016$, of which there are $ (4014\minus{}2010)/6\plus{}1 \equal{} 335$.  So there are $ 1338\minus{}335 \equal{} 1003$ terms with exponents that are multiples of $ 3$ and not already counted as multiples of $ 2$.

So there are a total of $ 2010\plus{}1004\plus{}1003 \equal{} 4017$ terms in the expansion.

As a side note, this problem was somewhat nice to us in that we did not have to expand anything being subtracted, so we knew we would not encounter the case that some of the terms in the expansion of one binomial cancel with terms from the expansion from another.  For example, in $ (x\plus{}1)^2 \plus{} (x\minus{}1)^2 \equal{} 2x^2 \plus{} 2$, each binomial has $ 3$ terms in its expansion but $ 1$ term cancels in the addition.

I got number 8 already after the proper factorization! Yahoo!!

Thanks facis for the explanation for number 9.

Additional Problem: How many terms are there in the simplified expansion of
    (x+y+z)^2008 +(x-y-z)^2008 ?


Now that I realize that this forum exists, I might as well post it here.

About a month ago, my font in AoPS has suddenly dramatically reduced in size, and is now only about half as large. This is usually not a problem, but is costing me quite a bit in For The Win, as many numbers are blurred there, making $ 8$ and $ 9$ look like each other. Does anyone know how to fix this?


abstract algebra
group theory
superior algebra
superior algebra unsolved

Let $ G$ be a finite generated abelian group,Let $ H$ be a subgroup of $ G$,Prove there exist $ x_1,x_2,\cdots,x_r \in G$ such that $ G \equal{} G_t\oplus \langle x_1 \rangle \oplus \langle x_1 \rangle \cdots \oplus \langle x_r \rangle$ and $ H \equal{} H_t \oplus [H \cap\left(\langle x_1 \rangle \oplus \langle x_1 \rangle \cdots \oplus \langle x_r \rangle \right)]$.

Where I $ G_t$ is the subgroup of $ G$ consist all of element of $ G$ which has finite order.I think the problem is nice if it is true.Enjoy it!

Hey,can somebody prove it or disprove it?

Ok, this is actually wrong.  :wacko:
Take $ G: \equal{}\mathbb{Z}\times (\mathbb{Z}/2\mathbb{Z})$ with free subgroup $ H\equal{}\mathbb{Z}(2,1\plus{}2\mathbb{Z})$.
Then $ H$ lies in no free complement of $ G_t\equal{}0\times (\mathbb{Z}/2\mathbb{Z})$ in $ G$.


$g(f(x)) = f'(x)$

What is the meaning of $g$ ?

Nothing. This is an autonomous first-order differential equation.

it need to have a meaning

if this does not. so it is unecessary?
its irrelevant?

i know differential equation

i just want a good asnwer. if you dont have. dont aswer

If you want a better answer- ask a better question. There isn't enough context to say anything meaningful here.

I remember [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=autonomous&t=67252]this problem[/url], but the conclusion is about $f$ - it is monotonic.

i want to fund $g$ in function only of $f(x)$  not of $f'(x)$

this is not a differential equation


It takes 24 minutes for Jana to walk one
mile. At that rate, how far will she walk in 10
minutes? Express your answer as a decimal to
the nearest tenth.

To solve this, we can set up a proportion.

$ \frac{24}{1} \equal{} \frac{10}{x}$

Cross-multiply:

$ 24x \equal{} 10$

$ x \equal{} \frac{5}{12}$

$ \frac{5}{12}$ is about $ \boxed{0.4}$, so that's our answer.


Let $A$ be a non-empty set of positive integers. Suppose that there are positive integers $b_{1}$, $\cdots$, $b_{n}$ and $c_{1}$, $\cdots$, $c_{n}$ such that [list] [*] for each $i$ the set $b_{i}A+c_{i}=\{b_{i}a+c_{i}\vert a \in A \}$ is a subset of $A$, [*] the sets $b_{i}A+c_{i}$ and $b_{j}A+c_{j}$ are disjoint whenever $i \neq j$.[/list] Prove that \[\frac{1}{b_{1}}+\cdots+\frac{1}{b_{n}}\le 1.\]

For any set $S$ of positive integers define $n_S(k) := | \{ s \in S | s \leq k \}|$.
For $B \subseteq A$ we define the density of $B$ in $A$ by $d_A(B) := \lim_{k \to \infty} \frac{n_B(k)}{n_A(k)}$ if this limit exists. Clearly $0 \leq d_A(B) \leq 1$.

We define $A_i := \{ b_i a + c_i \in A | a \in A\}$.
Some easy analysis gives $d_A(A_i)= \frac{1}{b_i}$.
Since the sets $A_i, A_j$ are disjont for $i \neq j$, we get that $n_A\left(\bigcup_{i=1}^n A_i\right) = \sum_{i=1}^n n_A(A_i)$, giving $d_A\left(\bigcup_{i=1}^n A_i\right) = \sum_{i=1}^n d_A(A_i)= \sum_{i=1}^n \frac{1}{b_i}$.
But now $1 \geq d_A\left(\bigcup_{i=1}^n A_i\right) = \sum_{i=1}^n \frac{1}{b_i} $.

[quote="ZetaX"]For any set $ S$ of positive integers define $ n_S(k) : \equal{} | \{ s \in S | s \leq k \}|$.
For $ B \subseteq A$ we define the density of $ B$ in $ A$ by $ d_A(B) : \equal{} \lim_{k \to \infty} \frac {n_B(k)}{n_A(k)}$ if this limit exists. Clearly $ 0 \leq d_A(B) \leq 1$.

We define $ A_i : \equal{} \{ b_i a \plus{} c_i \in A | a \in A\}$.
Some easy analysis gives $ d_A(A_i) \equal{} \frac {1}{b_i}$.
Since the sets $ A_i, A_j$ are disjont for $ i \neq j$, we get that $ n_A\left(\bigcup_{i \equal{} 1}^n A_i\right) \equal{} \sum_{i \equal{} 1}^n n_A(A_i)$, giving $ d_A\left(\bigcup_{i \equal{} 1}^n A_i\right) \equal{} \sum_{i \equal{} 1}^n d_A(A_i) \equal{} \sum_{i \equal{} 1}^n \frac {1}{b_i}$.
But now $ 1 \geq d_A\left(\bigcup_{i \equal{} 1}^n A_i\right) \equal{} \sum_{i \equal{} 1}^n \frac {1}{b_i}$.[/quote]

Hi,ZetaX.Could you please show why [b]Some easy analysis gives $ d_A(A_i) \equal{} \frac {1}{b_i}$.[/b]
I think this is the difficulty of this problem(maybe it can be solved by an advanced theorem?)Could you please help me with it? :)
You can also check http://www.mathlinks.ro/viewtopic.php?t=17335


AoPSwiki
geometry
3D geometry
pyramid
search
articles
function

Each week, we will select two entries in the [url=http://www.artofproblemsolving.com/Wiki/index.php]AoPSWiki[/url] for the community to spruce up -- these entries are the [url=http://www.artofproblemsolving.com/Wiki/index.php/AoPSWiki:WotW]AoPS Wiki Words of the Week[/url].  This week's words are [[Pyramid]] and [[iTest]].  You'll also find links to the Words of the Week atop the AoPS Forum, right under the time/date information.

Also, you can now easily link to terms in the AoPSWiki in the forum.  To do so, use double brackets, just like in the AoPSWiki.  For example, $ {[}[\text{Pyramid}{]}]$ produces [[Pyramid]].

Man I thought this was gonna be about World of Warcraft. To you know, give me another incentive to start playing again.

Disappointment, but pretty cool nonetheless.

[quote="AstroPhys"]The $ \left[\left[\right]\right]$ does not return the search page but, just the article with the name provided?
So the text between the brackets must be the exact title of the article?

Besides those questions, cool function for the forums :) .[/quote]

Correct; the name in the brackets is the name of the article in the Wiki.

Yeah, the only problem with $ [[]]$ is that there aren't that many articles...yet.

[[Newton's Sums]] --Okay, why do I have to click "search for page's title"?  Is it because of the apostrophe?

[[Octagon]] works.

It doesn't work because the title is [[Newton's sums]], not [[Newton's Sums]]. Shouldn't there be a redirect page for that?

Neither of the those links work. :maybe:

[[Newton's sums]] links to http://www.artofproblemsolving.com/Wiki/index.php/Newton%27s_sums for me.

[[Newton's sums]] links to http://www.mathlinks.ro/Wiki/index/Newton's%20sums for me. :maybe:

And who rated my post a two?

What browser are you using? And I fixed the redirect.

You're right, it does link to http://www.artofproblemsolving.com/Wiki/index/Newton's%20sums but it redirects to http://www.artofproblemsolving.com/Wiki/index.php/Newton%27s_Sums immediately. Unless you have meta redirects turned off?

It does not redirect, it says "HTTP 404 Not Found The webpage cannot be found".

What is in the address bar at the time when the error page shows up?

It has http://www.mathlinks.ro/Wiki/index/Newton's%20sums .

Check your internet options. Do you have HTTP 1.1 turned on? If so, then... I don't know. If not, turn it on.

One last thing: what version? If it's not IE7, you should upgrade. Or just switch to firefox.

It is Internet Explorer 7.
Tried it in Opera and it does not work either.

What happens if you just go to the Wiki and type in "Newton Sums"?

http://www.artofproblemsolving.com/Wiki/index.php/Newton_sums

Just tried Firefox also
"Not Found

The requested URL /Forum/redirect:/Forum/Wiki/index.php/Newton's sums/Newton's sums was not found on this server."

Could it be that I am using MathLinks and not AoPS? Since all the links I get with $ [[]]$ are with Mathlinks.

Anaxerzia that link works, just not the $ [[]]$ links.

I am using Mathlinks as well. Hm.... well, does it work with AoPS style?

I use the AoPS skin, and the bracket links work fine with both Firefox 2.0.0.6 and IE 6.

Works fine in the AoPS skin and all the lastest Windows versions of Firefox, Opera, Internet Explorer, and Safari.

[quote="Temperal"]The template looks quite ugly to me.
And how is the WoW chosen? Just the worst articles?
Also, when are you going to apply formatting? e.g. [[hi|hello]]

[[User:Temperal/testtemplate]]

I tried to make the WoW different.[/quote]

I agree; the color choice for the banner just doesn't appeal too much, or perhaps the large font size. It appears that our moderators get the decision for the WoWs (see A:WoW), as there isn't any place to put votes/suggestions/etc. I can't follow what the last question is asking.

As for the boxes that used to go on the bottom of the article for WoWs, I moved them to the talk pages because (1) the banner at the top suffices as a message and (2) its not directly related to the actual content of the article. But since the talk pages are all empty, the formatting doesn't work well there either. I think a design like the talk page templates on Wikipedia would be nice.

I disagree. The template with the preceding/following WoW should be moved to the side of the mainspace page, not the talk page, where it's confusing and few who read the article will see them.

EDIT: I tried float:right, but it didn't work.

I see what you’re saying, but the template is not related at all to the actual content of the article; hence it’s only bound to confuse a reader that is not so familiar with the AoPSWiki. Honestly, I don't think many people will be interested in knowing if an article once was a past WoW. On the other hand, a navigation box on the bottom connecting related topics (for example, a set of problems with a navigation box linking to all of the problems) keeps everything easily accessible.

I suppose a compromise is possible, like we created a category named Past WoWs or a minimalized template. It’s too bad the <hide/show> class doesn’t work (requires css which is disabled).

(by the way, could you use the "Show preview" more often? It's not really a big deal, but for example the history of the Template is now flooded with back-and-forth float:right experiments. I'm not sure, but maybe align:right might work.)

All right, sorry about all the testing edits.

And align:right results in a stretched box, like float:right.

[quote="1=2"]Why can't there be 5 WoWs? Because some pages really need sprucing up. For more details, look at my sig.[/quote]

For the same reason there can't be 364,000,000,000 WoWs, I guess. Maybe the admins will increase the number if WoW goes well at first.


Prove that:$(a+b+c+d)^{2}>8(ac+bd)$
where $a;b;c;d$ are real numbers satisfying $b>c>d$


ABC triangle. Prove that

3/2*(ab+bc+ca)>ma^2+mb^2+mc^2

my idea would be that you replace:
sum(ma^2) with 3/4 sum(a^2) and then you get an inequality in a,b and c and it should be easier !

cheers! hope it works cause I haven't tried it yet! I am out of town right now... :D

Yeah, that is how I thoght the problem too.

hey.. I forgot about this one.. sorry! :(

here is a full solution:

after doing what I said in the first post we get to the following ineq:

a^2+b^2+c^2 < 2(ab+ac+bc). this is true because:

a^2<ab+ac
b^2<bc+ba
c^2<ca+cb which are all true because in a triangle we have:
a<b+c
b<c+a
c<a+b

all done! a full solution!
cheers! :D :D


algebra
polynomial
search
algebra proposed

Prove that there exists a polynomial $ P(x, y)$ with real coefficients such that $ P(x, y) \ge  0$ for all real numbers $ x, y$, but $ P(x, y)$ cannot be written as the sum of squares of polynomials with real coefficients

Posted before, but I can't find the thread...

Edit:  Try http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=748721042&t=6131 .

Edit2:  See also http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=748721042&t=110254 .


I noticed that the usual list of cans/can'ts was not listed.  Instead, there was:

[code]
Rules_post_can
Rules_reply_can
Rules_edit_can
Rules_delete_can
Rules_vote_can
Rules_calendar_cannot
You can attach files in this forum
You can download files in this forum
You can moderate this topic

[/code]

The meaning is pretty clear, but is that supposed to happen (I'm pretty sure it used to be different)?

This is rather something for the support forum...

Well, I thought I'd post it here, because a lot of the things are moderator related, so I didn't know if it affected anyone else.

Peter is right. It's a small problem with the lang entries apparently. Fixed now.

Related to this, when splitting a topic, this comes:

[quote]The selected topic has been split successfully

Click_return_destination_topic

Click Here to return to the topic[/quote]

Again, understandable very well but could be said in human grammar. ;)

  darij


I've been looking at this for a while from a topic dated early August of this year.  How do you prove:

$\frac{ab}{a+b}+\frac{cd}{c+d}\leq \frac{(a+c)(b+d)}{a+b+c+d}$  for a,b,c,d greater than zero?

Dumba** it maybe? Clear the denominators, expand, and use Schur's and/or AM-GM and/or Cauchy.

Is there a more elegant method?... expanding seems to be quite a large task...

Expanding it isn't so bad, don't even need anytough inequalities probly... umm I see something more along the lines of $\frac{1}{\frac{1}{a}+\frac{1}{b}}$ yes I think I see a solution...

hmm won't let me repost
anyway use:
$\frac{1}{x}+\frac{1}{y}\geq \frac{4}{x+y}$ (easy proof lol)
also consider rhs as
[hide=" :D Peek when youre done with problem..."]$\frac{1}{\frac{1}{a+b}+\frac{1}{c+d}}$[/hide]

[quote="zeus_three"] How do you prove:

$\frac{ab}{a+b}+\frac{cd}{c+d}\leq \frac{(a+c)(b+d)}{a+b+c+d}$  for a,b,c,d greater than zero?[/quote]
$\frac{ab}{a+b}+\frac{cd}{c+d}\leq \frac{(a+c)(b+d)}{a+b+c+d}\Leftrightarrow(ad-bc)^{2}\geq0.$ :wink:

Nice, solution arqady, but I think he specifically said not to multiply out to solve :wink: I still think it is good... this is a tough problem. I noticed the following: \[\leftrightarrow \frac{a}{\frac{a}b+1}+\frac{c}{\frac{c}d+1}\leq  \frac{a+c}{\frac{a+c}{b+d}+1}\] which is a more interesting expression, anyone experienced with inequalities with fractions and combining denom.s etc. should know how to do this one...


Six bippitties equal three bops, and six bops equal two boos. How many boos are there in one bippitty, one bop and one boo? Express your answer as a common fraction.

[quote=gaussintraining]Equivalently, $ 12$ bippities=$ 6$ bops $ \implies$ $ 12$ bippities=$ 2$ boos $ \implies$ $ 1$ bippity=$ \frac{1}{6}$ boo. Also, $ 1$ bop=$ \frac{1}{3}$ boo.
So, the answer is $ \frac{1}{6}\plus{}\frac{1}{3}\plus{}1\equal{}\boxed{\frac{3}{2}}$.[/quote]

isn't it 2+1/6+1/3=2 1/2 or 5/2


calculus
integration
calculus computations

Intgrate with respect to z:

sin(x+y-z)^x + x^y/2 + 2y^z + xy(cos(2z^2))

this one is very hard if you can do it you are a master

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=26438

i suppose i could chain-rule bash my way through it but i'm too lazy to, lol


[quote="evarist"][color=purple] Let three positive real numbers $ x,y,z$. Prove or disprove that :
        $ x^2\frac {y}{z} \plus{} y^2\frac {z}{x} \plus{} z^2\frac {x}{y}\ge x^2 \plus{} y^2 \plus{} z^2$
  Please help me ! Thank alot :)[/color][/quote]
Try $ x\equal{}7,$ $ y\equal{}1$ and $ z\equal{}2.$ :wink:

The correct is $ x>\equal{}y>\equal{}z$ and is TST Vietnam. :wink:

All sums are cyclic. I will assume that $ x\ge y\ge z$.

Verify that $ \sum x^3y^2 - \sum x^2y^3 = (x - y)(y - z)(x - z)(xy + yz + zx)\ge 0$. Thus $ \sum x^3y^2\ge \sum x^2y^3$. Hence we have by AM-GM
\begin{align*}2\sum x^3y^2 & \ge \sum x^3y^2 + \sum z^2x^3 \\
& \ge 2\sum x^3yz \\
& = 2xyz(x^2 + y^2 + z^2)\end{align*}
Thus $ \sum x^3y^2\ge xyz(x^2 + y^2 + z^2)$ and dividing by $ xyz$ we get our desired inequality.


[quote="DonkeyKingKong"][img]http://i297.photobucket.com/albums/mm223/theeater820/State27.png[/img][/quote]
[hide]Let m be the number of members per row, and there are r rows. Therefore, with the given info, we have mr+2=(m+1)(r-2)=mr-2m+r-2, so r-2m=4. Since there must be less than 100 band members, the largest value of r we can use is 16, yielding m=6. So mr=96, and the total band members is 96+2=98 members.[/hide]


Does AP Physics C require the kind of problem solving skills that this site so strongly encourages?  Or will it just be like honors, where I had to just plug the numbers into the formulas?

it really depends on your teacher.

the ap test is somewhere in between, although less of the latter (especially since you can't use the calculator for the mult. choice)


The proof is attached to this reply.[/img]


[quote="andyciup"][hide]6[/hide][/quote]

I think you are wrong. It's obvious that $\ (x_1-x_2)^2+(x_2-x_3)^2+(x_3-x_4)^2+(x_4-x_5)^2+(x_5-x_6)^2+(x_6-x_1)^2\geq 6$, but equality never holds.  Let's supose that you're right, that is min of ${(x_1-x_2)^2+(x_2-x_3)^2+(x_3-x_4)^2+(x_4-x_5)^2+(x_5-x_6)^2+(x_6-x_1)^2}=6$. If for some $i, \mid x_i-x_{i+1}\mid \geq 2$, then for some $j<>i, \mid x_j-x_{j+1}\mid=0$, that is $x_j=x_{j+1}$, contradiction, since they are distinct. So for all i, $\mid x_i-x_{i+1}\mid =1$. It's obvious that for all i, $x_i-x_{i+1}$ have same value, that is either 1 or -1, which brings us that some $x_i$ are equal, contradiction.
Here is my solution:
let $x_1-x_2=a, x_2-x_3=b, x_3-x_4=c, x_4-x_5=d, x_5-x_6=e, x_6-x_1=f$, when we sum up these equations we get $a+b+c+d+e+f=0$. Let $f(a,b,c,d,e,f)=a^2+b^2+c^2+d^2+e^2+f^2$. It's obvious that $f(a,b,c,d,e,f)\geq f(\frac{a+b}{2},\frac{a+b}{2},c,d,e,f)$, and so on. So $min f(a,b,c,d,e,f)=f(t,t,t,t,t,-5t)=30t^2$ and since $t\geq1$, $min f(a,b,c,d,e,f)=f(1,1,1,1,1,-5)=30$. This is achieved for all $(x_1,x_2,x_3,x_4,x_5,x_6)=(6+k,5+k,4+k,3+k,2+k,1+k)$.

WRONG SOLUTION

No, Andrei, this isn't the correct answer.The best constant is bigger,bigger than $\ 6$.

[hide]
I think it is 18

WLOG, assume the numbers are 1...6
The sequence consists of increasing and decreasing subsequences.
Case 1: 1 increasing and 1 decreasing
WLOG, assume x1<x2<...<xn, xn>...>x6>x1
x1 = 1, xn = 6
|x1-x2|+|x2-x3|+|x3-x4|+|x4-x5|+|x5-x6|+|x6-x1| = 10
In order to minimize (x1-x2)^2+...+(x6-x1)^2, there should be four 2s and two 1s in |x1-x2|...|x6-x1|
the minimum is 18. (x1=1,x2=3,x3=5,x4=6,x5=4,x6=2)

Case 2: otherwise
let x1=1, xn=6
|x1-x2|+|x2-x3|+|x3-x4|+|x4-x5|+|x5-x6|+|x6-x1| >= 10
The minimum cannot be greater then 18

[/hide]

[quote="stancioiu sorin"]OK, modeler.That's the correct answer. ;) But I have a better solution than you.Do you want to post it?[/quote]

OK post your solution, please!

It's a long solution, but...OK
  Because the sum of the numbers $\ x_1-x_2,x_2-x_3,...,x_6-x_1$ is zero, we can say that between them we'll have an even number of odd numbers.Then, if the sum of that six squares from the problem isn't bigger than $\ 18$
then it is equal whith one of the numbers $\6,8,10,12,14 or 16$-even numbers.We can write this numbers as a sum of six squares in the following way:
$\ 6=1^2+1^2+1^2+1^2+1^2+1^2$
$\ 12=2^2+2^2+1^2+1^2+1^2+1^2$
$\ 14=3^2+1^2+1^2+1^2+1^2+1^2$
   Than rezults that the only posbilities for the numbers $\ x_1-x_2,...,x_6-x_1$ are
$\ (1,1,1,-1,-1,-1),\ (2,2,-1,-1,-1-1),\ (3,1,-1,-1,-1,-1).$   (1)
    But, the difference between the smallest and the greatest number from $\ {x_1,x_2,...,x_6}$ is at least $\ 5$ and this is equal whith the sum of some elements from$\ (x_1-x_2,...,x_6-x_1)$.That's impossible, if we look at (1).
So, the minimum value of that sum is $\ 18$.This it can be obtained if we have
$\ x_1=0,x_2=1,x_3=3,x_4=5,x_5=4,x_6=2$, q.e.d
    I hope you understood the solution.


solve this equation in $\mathbb{R}$ : $E(x^2)=(E(x))^2$
thank u
$E(x)$ is integer part of $x$
we can also write that :   $( \Large [x^2]=[x]^2)$

If it is what I think, I assume she's talking about the integer part.
I may be wrong though..

[quote="cats..."]sorry, what is the perfect part?[/quote]
sorry I mean  integer part

[quote="frt"]So... is the equation the same as $\lfloor x^2\rfloor =(\lfloor x\rfloor)^2$?[/quote]
yes of course
$\Large[x^2]=[x]^2 $

This is obviously true for all integers.
In order for this to be the case for all numbers in an interval, $(n, n+1)$ between integers, the squares of n and n+1 must differ by no more than 1.  We can see that for positive reals, the intervals is $(0, \sqrt{2})$, and for negative/zero numbers, the integer part function rounds down in the negatives, while when one squares the number, it becomes positive, so it's only true for $[-1, 0]$.  So the final interval is $\mathbb{Z}$ U $[-1, \sqrt{2})$

[quote="seamusoboyle"]Pick any interger say $n$

Now pick any r, with $0 \leq r < \sqrt{n^2+1}-n$,

Then $n+r$  is a solution.

You can extend this result for negative n, but i cant get the absolute value signs up.

Set $x=n+r$ and sub it in to solve.[/quote]
I have solve for positive n , but with negative n  :?  you can give me the solve with negative n plz

For negative n, change the $-n$ to $+n$. Thats what i needed the absolute value sign for.


On the free response section, there's always a reaction writing question.  It would be nice if we could have a thread that just lists general types of reactions which may come up on reaction writing.

I guess the classic list is as follows (but not limited to):

1. Synthesis
2. Decomposition
3. Oxidation/Reduction
4. Single Replacement
5. Double replacement
6. Disproportionation
7. Esterification
8. Mild Oxidation
9. Elimination

of course USNCO tests a lot of weird reactions as well, but Id say that these 9 reactions show up the most frequently.

Here are some practice reactions, maybe we can start a reaction marathon? (the person who answers a reaction right posts another reaction etc.)

1. Concentrated Nitric Acid is added to Iron II Sulfide
II. Bromine gas is added to propene (name major product)

1. Concentrated nitric acid is added to iron II sulfide
[hide="Reaction"]
$ 2\text{HNO}_{3(aq)} \plus{} \text{FeS}_{(s)} \to \text{H}_2\text{S}_{(g)}\plus{}\text{Fe}(\text{NO}_3)_{2(aq)}$
[/hide]

2. Bromine gas is added to propene
[hide="Reaction"]
$ 3\text{Br}_{2(g)}\plus{} \text{C}_3\text{H}_{6(g)} \to 3\text{CBr}_{2(g)}\plus{}3\text{H}_{2(g)}$
[/hide]

Sorry if there are mistakes..

Let's continue..
3. Barium acetate is added to hydrochloric acid.

4. Calcium hydroxide is added to ammonium phosphate.

For number 1, [hide]the only thing I would do different is leave NO3- as a spectator ion, since nitric acid is a strong acid, the nitrate ion plays no part in the reaction.[/hide]

For number 2, [hide]i'm pretty sure the alkene becomes an alkane so you get:
Br2 + C3H6 --> C3H6Br2.  It has the structure of propane, and the two bromines will be on opposites sides of the molecules because of steric hindrance.  Can someone confirm this for me?[/hide]

3. [hide]Ba2+ + CH3COO- + H+ + Cl- --> BaCl2 + CH3COOH[/hide]
4.) [hide]Ca2+ + OH- + NH4+ + PO43- --> Ca3(PO4)2 + NH3 + H2O[/hide]

One question about reaction 4, is the ammonia produced in the reaction aqueous or gaseous?  Thanks.

Okay new questions,
5. Flourine-18 undergoes positron emission.
6. Water is added to Magnesium Nitride.

yankeesrule, I agree with your #2.

[hide="5"]
You get oxygen-18.
[/hide]

[hide="6"]
I think you end up with [Mg(H2O)3(OH)]+ and H+?
[/hide]

7. 3-pentanone reacts with methylmagnesium in the presence of water.

No sorry 6 isn't right, where do the nitrogen atoms go then?

And for 7 i'm not sure but does it form
[hide]Magnesium hydroxide and 3-hexanol?[/hide]

EDIT: oops forgot to post another one,

Acidic Potassium dichromate and propanol are mixed and heated.
Bonus: What color change will accompany this reaction?

[quote="yankeesrule007"]
One question about reaction 4, is the ammonia produced in the reaction aqueous or gaseous?  Thanks.
[/quote]

Gaseous..  :D

[quote="yankeesrule007"]No sorry 6 isn't right, where to the nitrogen atoms go then?
[/quote]

[hide="Maybe.."]
$ \text{Mg}(\text{NO}{}_2){}_{2(aq)} + \text{H}_2\text{O}_{(l)} \to \text{MgO}_{(s)} + 2\text{HNO}_2_{(aq)}$
[/hide]

[hide]
I'm sorry if i'm wrong, 'cause I'm not good in chemistry.. T.T
[/hide]

[quote]6. Water is added to Magnesium Nitride[/quote]

you get Mg(OH)2 + NH3...

7.  Do you mean Methyl magnesium bromide?? i dont think methyl magnesium is even stable on its own.

(PLEASE STAY AWAY FROM ORGANIC REACTIONS... )

8. I believe you get Chromium III and proponoic acid. The solution turns green.

9. Sodium Hydroxide is added to Zinc nitrate

[hide]Mg2+ + OH- --> Mg(OH)2[/hide]

Ammonia is passed through a solution of silver nitrate.

Yeah sorry about #7 . . . I should have done methyllithium or something like that.  In any case, the reaction is interesting.

For the last one:
[hide]
I think it forms (NH4)+, AgOH, and (NO3)-?
[/hide]

11. Hydrogen carbonate ion is added to a solution of acetic acid and sodium acetate.

For the last one, that isn't correct.

11.[hide] HCO3- +CH3COO- ---> CO3- +CH3COOH?[/hide]

[quote="feynman_wannabe"]7.  Do you mean Methyl magnesium bromide?? i dont think methyl magnesium is even stable on its own.

(PLEASE STAY AWAY FROM ORGANIC REACTIONS... )

[/quote]

Excuse me for intruding for a moment. This is the Grignard Reagent and you are right it isn't particularly stable on its own. It is usually prepared in ether solvents and used on the spot. This is what made me wonder about the part where it says "react so and so in the presence of water". If it is indeed the Grignard Reagent then you want to stay away from water I'd imagine unless the poster means acidic work-up.

haha, yes I am aware of grignards reagents and their respective reactions.

BTW. if people could number the reactions that they are responding to than it would be much more organized.

btw.. for the reaction of ammonia passing through a solution of silver nitrate you get
[Ag(NH3)2] i think.

Don't forget about the positive charge: $ \ce{Ag(NH3)2}^{\plus{}}$. But the reaction would largely depend on the relative amount of ammonia you add to silver nitrate.

If you add a relatively modest quantity then you should expect the precipitation of $ \ce{Ag2O}$ (note I think silver hydroxide does not exist). If on the other hand you add ammonia in excess then yes the complex ion forms and you had observe the dissolution of the precipitate.

Hmm... interesting, then the ammonia wouldn't take any part in the reaction at all right?  Because the precipitation of silver oxide has nothing to do with ammonia right?

Remember ammonia is still itself a basic substance: $ \ce{NH3} \plus{} \ce{H2O} \rightleftharpoons \ce{NH4\plus{}} \plus{} \ce{OH\minus{}}$. Furthermore we have $ 2\ce{Ag\plus{}} \plus{} 2\ce{OH\minus{}} \rightarrow \ce{Ag2O} \plus{} \ce{H2O}$. It is because of ammonia that an appreciable amount of hydroxide is present to cause the second reaction.

Ah I see, well while we're asking questions, I have one more lol.  This is #59 2004 National Exam MC.

The question is what is the most common organic product in the reaction between
CH3CH=CH2 + HCl?
The answer choices are
A. CH3CHClCH3
B. CH3CH2CH2Cl
C. CH3CHCLCH2Cl
D. CH2ClCH=CH2

The answer is A.  My question is what makes A a better choice than B?  They both have one chlorine attached, the only difference is where its attached.

I read about reactions like that somewhere while looking for something else.  It is a consequence of Markovnikov's Rule.  See http://www.cem.msu.edu/~reusch/VirtualText/addene1.htm for an explanation.

By the way I think the answer to the Grignard question is wrong.

[hide=" Answer"] I think it should be $ \text{3\minus{}methyl\minus{}pentan\minus{}3\minus{}ol}$.[/hide]

[quote] The question is what is the most common organic product in the reaction between
CH3CH=CH2 + HCl?
The answer choices are
A. CH3CHClCH3
B. CH3CH2CH2Cl
C. CH3CHCLCH2Cl
D. CH2ClCH=CH2Cl[/quote]

basically in organic reactions you gotta think in terms of mechanisms. So first the HCl breaks up into H+ and Cl-, then the H bonds to the alkene to whichever carbon causes a more stable carbocation. In this case, it will bond to the terminal carbon in order to create a secondary carbocation which is stabler than a primary one. Then the Cl just adds to the carbocation, leaving us with CH3CHClCH3 - A.[/quote]

[quote]basically in organic reactions you gotta think in terms of mechanisms. So first the HCl breaks up into H+ and Cl-, then the H bonds to the alkene to whichever blah blah[/quote]

Or you could just remember [url=http://en.wikipedia.org/wiki/Markovnikov%27s_rule]this[/url].

Or you could play more StarCraft.

The choice is yours.

To continue with thread...

#13? A stream of chlorine gas is passed through a solution of potassium sulfate.

13. Cl2 + K2SO4 --> KClO + SO2

14. Barium peroxide is added to water

BaO2 + HOH ---> Ba(OH)2 + HO2-

Aqueous ammonia is added to a solution of Fe(NO3)3.

Fe[NH3]6 +3 ?

Ethene reacts with hydrochloric acid =) (too tired to think of a hard one)

Ethene reacts with HCl....

CH2CH2 + HCL -> CH2ClCH3

electrophilic addition?

Here is mine:

Ethene reacts with HBr in the presence of a peroxide.

Um...

$ \ce {C2H4} \plus{} \ce {HBr} \rightarrow \ce{C2H5Br}$

Is that right? I don't see how the peroxide would affect the overall reaction.

Assuming I'm correct, my question:

18. (Please continue numbering these!)
Sucrose is added to concentrated sulfuric acid. (Is it sulfuric or sulphuric?)

[quote="yanglunj"]Um...

$ \ce {C2H4} \plus{} \ce {HBr} \rightarrow \ce{C2H5Br}$

Is that right? I don't see how the peroxide would affect the overall reaction.
[/quote]

In this case peroxide would not affect the product as the ethene molecule is symmetric across the double bond. However, if the alkene molecule is not symmetric across the double bond the product will vary.

e.g. consider the same reaction but with propene ($ \ce{CH2\equal{}CHCH3}$). In the absence of peroxide we had get $ \ce{CH3CHBrCH3}$ but with peroxide present the product should be $ \ce{CH2BrCH2CH3}$.

The explanation of this lies in the mechanism of the two reactions. Without peroxide the mechanism involves carbocations (hence Markovnikov addition) while with peroxide the process involves free radicals (therefore anti-Markovnikov).


To calculate 76% of 25, John divides an integer n with 4. What is n?

[hide]76[/hide]

i dunno why i'm answering this

Tare, are you feeling bored right now? is that why you're posting all these problems? :?

Yup, however I'll have to rebuttal that "weird" comment since these problems are all taken from the Number Sense Stretch in the 2003-2004 MATHCOUNTS School Handbook (except I changed the numbers so people can't cheat)

I think I'm doing a good job of revitalizing the MATHCOUNTS Forum, don't you think? I mean over 5 people's been actively posting...

whats the point of these if they dont have solutions? and solving them first means nothing especially when you take into account that not everyone is on at the same time

hey, attacking me now, eh? (the eh was vocative... lol) j/k.

its alright... or all right... all of them work...

:?: I thought Neal was bashing me not Chinaboy, what did he ever do?

[quote="ml2"]hey, syntax, lets gang up  and attack chinaboi.[/quote]

lool.

and for tare, i was bashing tare because you might as well put 1+1 up there because theyre just sooo easy/weird. i liked some of the questions though, they did teach a concept like factoring the addition of cubes and stuff, but why make them COUNTDOWN ROUNDDDDDDDDSSSS

Hmmm.... im doing a project right now... so please don't attack me right now... lol.

Um, if it were hard it won't be Countdown Round. Also (and I'm saying this again) these problems will appear on MATHCOUNTS sooner or later no matter how easy they look, so meh to you.

Golden Rule of the Net: "If you don't like it, leave."
In this case, ignore my questions since nobody else thinks they're weird...in fact most people in the nation will not label this weird, especially because it came from such a trusted material (2003-2004 MATHCOUNTS School Handbook)

...I'm just trying :cry:

we all have the handbooks, and its not like were giving solutions. so meh to you


LaTeX
geometry
parallelogram
perpendicular bisector
geometry proposed

Two squares ACDE and ABFG are constructed out triangle ABC.
Perpendiculars from B and C to BC intersect GE at P and Q.
Prove :
         $BP+CQ \geq BC+GE$.
___________________________________________________________
No pain, no gain
[color=red][Moderator edit: corrected the LaTeX notation][/color]

There are some properties of the triangles $ABC$ and $AGE$ which I'll use without proof (which isn't hard, btw). First of all, the median of one of these two triangles corresponding to vertex $A$ is half of the side opposite to $A$ of the other triangle. Another property is that the median corresponding to $A$ in one of them is perpendicular to the side opposite to $A$ of the other one.

Let $M$ be the midpt of $BC$, $N$ the midpt of $GE$, $T$ the intersection of $GE$ and the perpendicular bisector of $BC$. According to the first property stated above, the conclusion is equivalent to $MT\geq AN+AM$. Let the parallel through $A$ to $GE$ cut $MT$ in $A'$. $AA'TN$ is a parallelogram because both $MT$ and $AN$ are perpendicular to $BC$ (according to the second property). We have $MT=A'T+A'M=AN+A'M\geq AN+AM$ because in the right triangle $A'AM$ $A'M$ is the hypothenuse and $AM$ is a leg, so it's shorter than $AM$.

Something I forgot to add: the equality takes place iff $AB=AC$.


what's the problem asking for, who's faster or their average speed?
[hide]20/3 = 6 and 1/3 =  John the Weak  30/5 = 6 = Davy the Strong
Davy's faster[/hide]

Their average speed.

[hide="big hint"]
John = 9 miles per hour
Davy = 10 miles per hour
[/hide]

Yup,

Treething. Was my problem not clear?

o, got a little confused at what I was multiplying. Sorry.  :D  :blush:

[quote="shinwoo"]John the Weak and Davy the Strong are preparing for the school race it's would be held the next day.

They tested their time..

John the Weak ran 3miles in 20 minutes

David the Strong ran 5miles in 30 minutes

average their speed

(this is a very simple question)[/quote]

We have $ V\equal{}\frac{S}{t}$

John$ \equal{}V\equal{}\frac{S}{t}\equal{}\frac{3}{\frac{1}{3}}\equal{}\frac{9}{1}\equal{}9$miles/h

Davy$ \equal{}V\equal{}\frac{S}{t}\equal{}\frac{5}{\frac{1}{2}}\equal{}\frac{10}{1}\equal{}10$miles/h


algorithm
modular arithmetic
number theory unsolved
number theory

Write down six consecutive natural numbers such that each of them is the multiple of the complete square of a certain number (except 1).  Please write down all those complete square numbers and show your calculation.


function
inequalities
inequalities proposed

hi,
If $ a_{1}, a_{2}, ..., a_{n}, b_{1} \leq b_{2} \leq ... \leq b_{n}$ positive numbers and
\[ a_{1} \leq b_{1} , a_{1} \plus{} a_{2} \leq b_{1} \plus{} b_{2}, ... , a_{1} \plus{} a_{2} \plus{} ... \plus{} a_{n} \leq b_{1} \plus{} b_{2} \plus{} ... \plus{} b_{n},\]
prove
\[ \sqrt{a_{1}} \plus{} \sqrt{a_{2}} \plus{} ... \plus{} \sqrt{a_{n}} \leq \sqrt{b_{1}} \plus{} \sqrt{b_{2}} \plus{} ... \plus{} \sqrt{b_{n}}.\]

PS.: i think there is a lot of theory similar to this problem. can somebody give ma a link or something like that with the theory, plz?

[quote="doicu"]hi,
If $ a_{1}, a_{2}, ..., a_{n}, b_{1} \leq b_{2} \leq ... \leq b_{n}$ positive numbers and
\[ a_{1} \leq b_{1} , a_{1} \plus{} a_{2} \leq b_{1} \plus{} b_{2}, ... , a_{1} \plus{} a_{2} \plus{} ... \plus{} a_{n} \leq b_{1} \plus{} b_{2} \plus{} ... \plus{} b_{n},
\]
prove
\[ \sqrt {a_{1}} \plus{} \sqrt {a_{2}} \plus{} ... \plus{} \sqrt {a_{n}} \leq \sqrt {b_{1}} \plus{} \sqrt {b_{2}} \plus{} ... \plus{} \sqrt {b_{n}}.
\]
PS.: i think there is a lot of theory similar to this problem. can somebody give ma a link or something like that with the theory, plz?[/quote]
See here: http://www.mathlinks.ro/Forum/viewtopic.php?t=147584

[quote="doicu"]hi,
If $ a_{1}, a_{2}, ..., a_{n}, b_{1} \leq b_{2} \leq ... \leq b_{n}$ positive numbers and
\[ a_{1} \leq b_{1} , a_{1} \plus{} a_{2} \leq b_{1} \plus{} b_{2}, ... , a_{1} \plus{} a_{2} \plus{} ... \plus{} a_{n} \leq b_{1} \plus{} b_{2} \plus{} ... \plus{} b_{n},
\]
prove
\[ \sqrt {a_{1}} \plus{} \sqrt {a_{2}} \plus{} ... \plus{} \sqrt {a_{n}} \leq \sqrt {b_{1}} \plus{} \sqrt {b_{2}} \plus{} ... \plus{} \sqrt {b_{n}}.
\]
PS.: i think there is a lot of theory similar to this problem. can somebody give ma a link or something like that with the theory, plz?[/quote]
It follows by Karamata's Inequality. Since $ (a_1,a_2,...,a_n\plus{}k) \prec (b_1,b_2,...,b_n)$ for some non-negative $ k$ and $ x \to \sqrt {x}$ is a concave function, we get that: $ \sqrt {a_{1}} \plus{} \sqrt {a_{2}} \plus{} ... \plus{} \sqrt {a_{n}\plus{}k} \leq \sqrt {b_{1}} \plus{} \sqrt {b_{2}} \plus{} ... \plus{} \sqrt {b_{n}}$. Now since $ k \ge 0$ and $ x \to \sqrt {x}$ is increasing we see that $ \sqrt {a_{n}} \le \sqrt {a_{n} \plus{} k}$ and we are done.

[quote="BG Yoda"][quote="Mathias_DK"][quote="doicu"]hi,
If $ a_{1}, a_{2}, ..., a_{n}, b_{1} \leq b_{2} \leq ... \leq b_{n}$ positive numbers and
\[ a_{1} \leq b_{1} , a_{1} \plus{} a_{2} \leq b_{1} \plus{} b_{2}, ... , a_{1} \plus{} a_{2} \plus{} ... \plus{} a_{n} \leq b_{1} \plus{} b_{2} \plus{} ... \plus{} b_{n},
\]
prove
\[ \sqrt {a_{1}} \plus{} \sqrt {a_{2}} \plus{} ... \plus{} \sqrt {a_{n}} \leq \sqrt {b_{1}} \plus{} \sqrt {b_{2}} \plus{} ... \plus{} \sqrt {b_{n}}.
\]
PS.: i think there is a lot of theory similar to this problem. can somebody give ma a link or something like that with the theory, plz?[/quote]
It follows by Karamata's Inequality. Since $ (a_1,a_2,...,a_n \plus{} k) \prec (b_1,b_2,...,b_n)$ for some non-negative $ k$ and $ x \to \sqrt {x}$ is a concave function, we get that: $ \sqrt {a_{1}} \plus{} \sqrt {a_{2}} \plus{} ... \plus{} \sqrt {a_{n} \plus{} k} \leq \sqrt {b_{1}} \plus{} \sqrt {b_{2}} \plus{} ... \plus{} \sqrt {b_{n}}$. Now since $ k \ge 0$ and $ x \to \sqrt {x}$ is increasing we see that $ \sqrt {a_{n}} \le \sqrt {a_{n} \plus{} k}$ and we are done.[/quote]

Mathias, that's wrong ...

Look it again and you will find the mistakes :roll:

First, notice that we don't know how is ordered the $ \{a_i\}_{i \equal{} 1}^{n}$ sequence, so we can't say with confidence that $ \{b_i\}_{i \equal{} 1}^{n}$ majorizes $ \{a_i\}_{i \equal{} 1}^{n}$ (still more when we have $ a_1\le \min{b_i}$ instead of $ a_1\le \max{b_i}$  :D ) and except this, when the function is concave, then the inequality is $ \sum f(a_i)\ge \sum f(b_i)$ in actually.[/quote]
Oh.. Thanks.. Don't know what i was thinking :S

If $ f$ is differentiable and concave, then $ f'$ is decreasing and nonpositive. Then

$ f(a_1) \plus{} f(a_2) \plus{} ... \plus{} f(a_n) \minus{} f(b_1) \minus{} f(b_2) \minus{} ... \minus{} f(b_n)\le$
$ (a_1 \minus{} b_1)f'(b_1) \plus{} (a_2 \minus{} b_2)f'(b_2) \plus{} ... \plus{} (a_n \minus{} b_n)f'(b_n)\le$
$ (a_1 \minus{} b_1)f'(b_2) \plus{} (a_2 \minus{} b_2)f'(b_2) \plus{} ... \plus{} (a_n \minus{} b_n)f'(b_n) \equal{}$
$ (a_1 \plus{} a_2 \minus{} b_1 \minus{} b_2)f'(b_2)) \plus{} ... \plus{} (a_n \minus{} b_n)f'(b_n)\le ... \le$
$ (a_1 \plus{} a_2 \plus{} ... \plus{} a_n \minus{} b_1 \minus{} b_2 \minus{} ... \minus{} b_n)f'(b_n)\le 0.$


Prove that $Card(N) = Card(N^2)$ where $N=$ the natural numbers.

Relatedly, or not-so-relatedly, construct a bijection between $R$x$R$ and $R$, $R$ being the reals.[/b]

Cantor-Bernstein theorem almost systematically solves those kinds of questions; for the example of $N$, you may find the typical construction on Google (numeroting the integers in an "array").

A more intresting question is: Prove that Card A=Card A^2 for any infinite set A.


geometry
trapezoid
ratio
quadratics
floor function
AMC
AIME

One base of a trapezoid is 100 units longer than the other base.  The segment that joins the midpoints of the legs divides the trapezoid into two regions whose areas are in the ratio $2: 3.$  Let $x$ be the length of the segment joining the legs of the trapezoid that is parallel to the bases and that divides the trapezoid into two regions of equal area.  Find the greatest integer that does not exceed $x^2/100.$

[hide="Answer"]We have trapezoid $ABCD$, with $DC=AB+100$. The segment joining the midpoints has length $AB+50$. Since the heights of the two smaller trapezoids are equal, we can reduce the ratio $\frac{2}{3}=\frac{AB+(AB+50)}{(AB+50)+(AB+100)}\Rightarrow AB=75$, $DC=175$. We label $h$ as the height of $ABCD$, $k$ as the height of the trapezoid with $DC$ as base, and the length of the parallel segment as $AB+y=75+y$. First, we notice the ratio $\frac{100-y}{100}=\frac{k}{h}$. From this, we deduce that $k=h\left(\frac{100-y}{100}\right)$ and $yh=100h-100k$. Now, since the areas of the two smaller trapezoids are equal, we have $\frac{75+(75+y)}{2}\cdot (h-k)=\frac{(75+y)+175}{2}\cdot k$. After some manipulations, we come up with $150h+yh=400k+2yk$. Substituting in our value of $yh$, we have $250h=500k+2yk=k(500+2y)$. Now, substituting in our value for $k$, we have $250h=h\left(\frac{100-y}{100}\right)(500+2y)\Rightarrow 250=\frac{100-y}{100}(500+2y)$, which gives the quadratic equation $y^2+150y-12500$. Solving, we have $y=-75+25\sqrt{29}$, so $x=75+y=25\sqrt{29}$. Therefore, $\frac{x^2}{100}=\frac{25(29)}{4}$, and $\left\lfloor \frac{25(29)}{4}\right\rfloor =\boxed{181}$.[/hide]

This is under the link for problem number fourteen for the alternate AIME 2000 on the AIME problem resources page. :?

Not sure how that happened exactly... It's fixed now.

Try doing the problem without using height as a factor in your algebra.  ;)

Ooh! Pick me! Pick me! :P

Anyway, as long as you know you can solve it with algebra, it would probably be smarter to pick a value for the height to speed up computations and give more time on other problems. ;)

[hide]Let's call the trapezoid $ABCD$, with $AB||CD$ and $AB+100=CD$. Let the median be $EF$, with $E$ and $F$ on $AB$ and $CD$, respectively. Some quick algebra yields $EF=AB+50$. We're given that $\frac{[ABFE]}{[CDFE]}=\frac{AB+25}{AB+75}=\frac23$, so $AB=75$ and $CD=175$.

Now, let's denote the height of the trapezoid $h$, such that the distance between the "solution line" and $AB$ is $kh$, where $k$ is some constant. It follows that the length of our solution line is $AB+100k$.

We can now set up an equation to solve for $k$. Because the area of newly formed trapezoid is half that of trapezoid $ABCD$, $\frac 12(kh)(2AB+10k)=\frac14(h)(2AB+100)$. Simplifying, $4k^2+6k-5=0$, and $k=\frac{\sqrt{29}-3}{4}$. The length of our line is then $75+25(\sqrt{29}-3)=25\sqrt{29}$, and our answer is $\lfloor \frac{(25\sqrt{29})^2}{100} \rfloor=\lfloor 181.25 \rfloor=\boxed{181}$.[/hide]

Phew... finally done.  :P

That problem 6 took like 20 minutes. :stretcher:

[hide]
After finding $AB=75$ and $CD=175$, extend $AD$ and $BC$ to meet at $E$.
Also, let $PQ$ be the segment that divides the trapezoid into equal areas. ($P$ is on $AD$ and $Q$ is on $BC$)

Since $AB: CD=3: 7$, $\triangle EAB: \triangle EDC=3^2: 7^2=9: 49$.

So we have $[EAB]: [ABQP]: [QPCD]=9: 20: 20$, which means $\frac{\triangle EPQ}{\triangle EDC}=\frac{29}{49}$.

Therefore, $x=PQ=\sqrt{\frac{29}{49}}CD =\frac{\sqrt{29}}{7}\cdot 175 =25\sqrt{29}$, which gives $\lfloor \frac{x^2}{100}\rfloor =181$.
[/hide]

Sorry to revive, but I keep getting lost at the point where people bring in the number $100$... where does that come from? i.e. in jesusfreak's solution, where did he get $\dfrac{100-y}{100} = \dfrac{k}{h}$? And in 4everwise's solution, how did he get $AB + 100k$? Help would be greatly appreciated, happiface.

Can someone post a solution, with assuming the height is 100?  Thanks!

Sorry for revive, but help would be exteremly helpful.

So for a generic trapezoid ABCD say I have AB=x and CD=y. The difference between them is then y-x.

I want to find the length of the line PQ if it lies parellel to AB and CD and P is on AD, Q is on BC.

Let the perpendicular distance from B down to our imaginary line PQ be equal to kh, where h is the height of ABCD and k is some constant, like in this problem with 4everwise's solution.

Can I always say that the length of PQ = x+k(y-x)?

Thanks! In other words, is the length of any kind of segment between AB and CD determined by the ratio of the heights for the made up segment and the longest one?

[quote=Blackhawk]/bump thanks![/quote]

yes.

I suppose this could be called the "state-of-the-art" solution?
[hide=Short]
Let $a$ and $a+100$ be the lengths of the bases. We have that the midsegment has length $a+50$, so the ratio of the areas of the two trapezoids is
\[\frac 23 = \frac{a+a+50}{a+100+a+50}\implies a = 75.\]
Now, it is well-known that the area of the segment dividing the trapezoid into equal segments has length $\sqrt{\frac{75^2+175^2}{2}}$ (proof: intersect the legs at a point and use similar triangles) so the answer is
\[\left\lfloor\frac{75^2+175^2}{200}\right\rfloor = \left\lfloor\frac{15^2+35^2}{8}\right\rfloor = \left\lfloor\frac{225+1225}{8}\right\rfloor = \left\lfloor\frac{1450}{8}\right\rfloor=181.\][/hide]

[quote=matt276eagles]This problem simplifies dramatically if you notice that the height of the trapezoid doesn't matter.  I set the height equal to 100.[/quote]

Just let it be $h$ and it cancels out  :)


invariant
inequalities proposed
inequalities

[quote="xzlbq"](3)
$ x_1,x_2,...,x_n \geq0,n\geq3,x_{n \plus{} 1} \equal{} x_1$,prove that:

$ \prod_{i \equal{} 1}^n{(x_i \plus{} x_{i \plus{} 1})} \geq 2^n\prod_{i \equal{} 1}^n{x_i} \plus{} \prod_{i \equal{} 1}^n{(x_i \minus{} x_{i \plus{} 1})}$

BQ[/quote]
Just AM-GM

$ \prod_{i \equal{} 1}^n\frac{(x_i \plus{} x_{i \plus{} 1})}2\minus{} \prod_{i \equal{} 1}^n\frac{(x_i \minus{} x_{i \plus{} 1})}2 \geq \prod_{i \equal{} 1}^n{x_i}$

 because the LHS is homogeneous and invariant w.r.t. $ x_1\to ...\to x_n\to x_1$, and if expanded, it would have only positive terms. :)


calculus
derivative
calculus computations

Ok I tried to do this problem and so far I cant proceed to the next step. The problem is the following:

Sketch the graph of $f(y)$ versus $y$ and determine the critical points, and also clasify them asymptotically stable or unstable.

$\frac{dy}{dt}= ay+by^{2}, a>0, b>0,-infinity < y_{0}< infinity$

So what I did was factor a $y$ and i was left with $y (a+by)$. Then I set $\frac{dy}{dt}= 0$ then I get two critical points which are $0$ and $-\frac{a}{b}$.

Now what I don't understand is how can I sketch it, and most of all how to determine if its stable or unstable.

[quote="Ottem"]This can either be done by solving the separable ODE and plotting the trajectories manually.. or noting the sign of the derivative for $y>0$ (positive), $0>y>-a/b$ (negative), and $y>-a/b$ (positive), hence $0$ is unstable, and $-a/b$ is stable..[/quote]

Ok so how do you state that its stable or unstable, can you explain a lil more please.

The values $y=0$ and $y=-a/b$ are of course constant solutions of the equation $y'=ay+by^{2}$, if $y_{0}>0$ is given sufficiently close to $0$, then since the derivative is increasing here, the solution curve corresponding to $y_{0}$ will just diverge from $y=0$ and similarily for $-a/b <y < 0$ (derivative is decreasing). Hence $0$ is an unstable critical point. Applying the same reasoning to $-a/b$ should convince you that it is indeed stable!


inequalities
geometry
calculus
calculus computations

Give an example (if possible!)of a region with finite surface and  infinite volume.(ie opposite of )
P.S Does'nt Gabriel's horn imply a deficiency in our definition of volume,...?I mean if we cannot fill a region ,
then how can we determine a volume to it?

There's no such region. This runs into the three-dimensional Isoperimetric Inequality: For a given volume, the region with the least surface area is the sphere. Or, equivalently, for a given surface area, the region with the greatest volume is the sphere.

Unfortunately, proofs of this inequality are not elementary.


probability
conditional probability
probability and stats

I understand that this probably a fairly basic probability problem  but I could use some help...

A bin contains 100 balls.  30 of them are green, 45 yellow and 25 blue.

a.  If two balls are selected from the bin without replacement, what is the probability that one will be green and one will be blue?

b. Two balls are drawn in succession without replacement.  What is the probability that first drawn is blue given that the second is green?

Thank you!

The second part is a conditional probability problem.  Given $ A$, the probability of $ B$ is $ P(B\mid A)\equal{}\frac{P(A\cap B)}{P(A)}$.  Here, $ A$ is the probability of first drawing a blue and $ B$ is the probability of then drawing a green.

So in regards to the first part of the problem, is it assuming that it doesn't matter what order that they were drawn in or are we assuming that a green was drawn first and then a blue?


calculus
derivative
geometry
real analysis
real analysis unsolved

Before when I was in high school, I always thought a differential, dx, was an infinitesimal. So I thought of it as an actual quantity that could be added or multiplied and what not. Now, that I am in a intro to analysis class, I know that infinitesimals are not rigorous (although I've read on wikipedia about something called non-standard analysis that rigorously defines them, but lets ignore that for now).

So my question is, what exactly is 'dx?' I've heard that it is just merely a tool to use to 'push symbols' to calculate things in elementary calculus, but as an engineering major, I've seen them treated as actual quantities (i.e. take an infinitesimal unit that has sides dx,dy,dz). Are they just abusing the notation by saying this to appeal to the students who don't care as much about the actual math? Is there an actual more rigirous way of stating what they are saying?

Again, I have also seen in my thermodynamics class that d(something) being treated as an actual quntity. Go here for example http://en.wikipedia.org/wiki/Second_law_of_thermodynamics in the section titled "Available useful work." It seems to be relating these differential quantities (or maybe derivatives with respect to nothing?) together. So is this incorrect, or is it actually rigiruous using something I have not learned yet (maybe that non-standard analysis? but the way people seem to talk about it, most people avoid it).

Thanks for all the help!

Wikipedia also has an article that explains several approaches to [url=http://en.wikipedia.org/wiki/Differential_%28infinitesimal%29]differentials[/url]. An engineering major can think of $ dx$ as a nonzero number that is sufficiently close to zero so that $ \frac{F(x\plus{}dx)\minus{}F(x)}{dx}\approx F'(x)$ with enough precision for the problem at hand. We had a lively [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=124472]discussion[/url] of this subject a while ago.

If you study differential geometry you'll get a pretty rigorous treatment of what "dx" means. The technical treatment is that it's a linear operator on the manifold's tangent space.

hi

in differential geometric yo can difine dx.
You may look here:
http://en.wikipedia.org/wiki/Differential_form

I don't know how you can proof now, that this object dx does what it shout do in thermodynamics etc...


analytic geometry
graphing lines
slope
function
calculus
calculus computations

Given $f(x)=\frac{2}{x-1}$, use the four step process to find a slope-predicator function $m(x)$. Then write an equation for the line tangent to the curve at the point $x = 0$.

Not knowing what the four-step process is, I can't answer your question. I wish I could talk to your teacher.


inequalities
real analysis
real analysis unsolved

Let $ k$ be a positive integer, $ z$ a complex number, and $ \varepsilon <\frac12$ a positive number. Prove that the following inequality holds for infinitely many positive integers $ n$: \[ \mid   \sum_{0\leq l \leq \frac{n}{k+1}} \binom{n-kl}{l}z^l  \mid \geq (\frac 12-\varepsilon)^n.\]

[i]P. Turan[/i]

use \left| instead of | it comes better. :)


It is given a directed graph. Every edge is a pipe with capacity c[i,j]. Through every pipe must flow at least l[i,j] liquids. (liquids that flow in every node is equal to the liquid that flow out from every node). Your task is to determine whether it is possible to make a graph satisfying the conditions above. c[i,j] and l[i,j] are known.

This is a known problem. A quick wiki search turned up[url=http://en.wikipedia.org/wiki/Edmonds-Karp_algorithm]Edmonds-Karp  algorithm[/url].[/url]


quadratics
modular arithmetic
arithmetic sequence
number theory unsolved
number theory

If I have n = 1 (mod 4), then we know by Dirichlet's theorem on primes in arithmetic progression,
that there is a prime such that p = 2n + 1 (mod 4n). Can someone prove that then (-n/p) = -1, where (n/p) denotes the Lagrange symbol?
Thanks.

Use the quadratic reciprocity theorem for the Jacobi symbol.

LOL. Have NEVER heard of something like that haha...but I'll find it on the internet somewhere...thanks anyway  :)

Since $p\equiv 3\pmod 4$, we have $\left(\frac{-1}p\right)=-1$, so we have to show $\left(\frac np\right)=1$. By the Quadratic Reciprocity Theorem (stated for Jacobi symbols), we have $\left(\frac np\right)=1\iff\left(\frac pn\right)=1$, which follows from $p\equiv (n+1)^2\pmod n$.


Hi to all !

I want to proove using calculus that $ f(x)=x^{2}-4x+4$ has one real root but;

I find that $ f$ is decreasing on $ A_{1}=(-\infty,2]$ and increasing on $ A_{2}=[2+\infty)$
and also $ f(A_{1})=[0,+\infty)$  and  $ f(A_{2})=[0,+\infty)$

so $ 0\in f(A_{1}) , f(A_{2})$  therefore $ f$ has $ 2$ real roots   :o

What have I make mistake and find 2 real roots?

The fact that $ 0$ is in both those images does not imply that there are two zeroes since their two preimages are not disjoint.

Rather, you should plug in the value at which there is a local minimum to find that the minimum is 0, so it only has one real root. The flaw in your logic is that 2 belongs to both A1 and A2, and it just so happened to be a root.


trigonometry
calculus
calculus computations

If [x] denotes the GIF and $ f(x) \equal{} [ n \plus{} p  \sin x ] , 0 < x < \pi  , n  \in  I ,$ and p is a prime number ,   then find the number of points where $ f(x)$ is not differentiable ..

$ f(x) \equal{} n \plus{} [p\sin x]$

Clearly f(x) fails to be non-differentiable when it becomes discontinuous i.e.$ p\sin x$ becomes an integer

For $ 0 < x < \pi$,$ 0 < \sin x\leq1$

Consider $ 0 < x < \frac {\pi}{2}$ where $ 0 < p\sin x < p$.

Thus $ p\sin x$ becomes an integer $ (p \minus{} 1)$ times.The same thing happens for $ 0 < x < \pi$.

Thus $ p\sin x$ becomes an integer $ 2(p \minus{} 1) \plus{} 1 \equal{} 2p \minus{} 1$ times(including the case when $ x \equal{} \frac {\pi}{2}$)

Answer should be $ 2p \minus{} 1$.

Somebody would please enlighten me what would happen if $ p$ is not prime.


integration
calculus
trigonometry
calculus computations

\[ Evaluate \int_{0}^{\frac{(pi)^{2}}{4}}(\frac{2}{csc\sqrt{x}} \plus{} \frac{\sqrt{x}}{sec\sqrt{x}}) dx\]

the integral is equivalent to $ \int 2\sin\sqrt {x} \plus{} \sqrt {x}\cos{\sqrt {x}} \,dx$

$ \equal{} 2x\sin\sqrt {x}$

Putting in the limits, we get $ \frac{\pi^{2}}{2}$.

hello, better would it be to write
$ \int2\sin(\sqrt{x})\plus{}\sqrt{x}\cos(\sqrt{x})\,dx\equal{}2x\sin(\sqrt{x})\plus{}C$.
Sonnhard.


Consider the sequence-based function defined by:

$a_{0}(x)=x$
$a_{n+1}(x)=\sqrt{1+a_{n}(x)}-a_{n}(x)$ for positive integers $n$

Suppose, for some positive integer $n$ greater than 1, $a_{n}(x)$ is graphed on the interval $[-1,3]$.

Find the $x$ value at which the graph reaches

a. its maximum point, and
b. its minimum point

It may or may not be helpful to consider the cases of odd and even values of $n$ separately.

Fact #1: if we know the extreme points of some function $g$, and if $h$ is strictly decreasing on the range of $g$, then $h\circ g$ has the same extreme points, only the roles of maxima and minima are interchanged.

Let $f(x)=\sqrt{1+x}-x$ for $x\in [-1,3]$. The function $f$ is increasing on the interval $[-1,-3/4]$ from $f(-1)=1$ to $f(-3/4)=5/4$. Then it decreases to $f(3)=-1$. This describes $a_{1}(x)=f(x)$.

Fact #2: the range of $f$ is $[-1,5/4]$.

Next, $a_{2}(x)=f(f(x))$. The critical points of this functions are $-3/4$ and $f^{-1}(-3/4)=\sqrt{2}+5/4$. The minimum is $a_{2}(-3/4)=1/4$; the maximum is $a_{2}(\sqrt{2}+5/4)=5/4$.

Fact #3: the range of $f\circ f$ is $[1/4,5/4]$, and $f$ is strictly decreasing on that interval. Since the minimum of $f$ on $[1/4,5/4]$ is $f(5/4)=1/4$, the range of $f\circ f\circ f$ is again contained in $[1/4,5/4]$, and so forth.

The rest is a repeated application of Fact #1: for any $n\ge 2$ the extreme points of $a_{n}$ are $-3/4$ and $\sqrt{2}+5/4$. When $n$ is even, $-3/4$ is the point of minimum and $\sqrt{2}+5/4$ the point of maximum. When $n$ is odd, it's the other way around.