induction
number theory
relatively prime
number theory unsolved

$ l \cdot 10^k \plus{} 2^x \equal{} 2^y$, where $ k$ is the number of digits of $ 2^x$, gives us

$ l\cdot 2^k \cdot 5^k \equal{} 2^x(2^{y\minus{}x} \minus{} 1)$.

Let $ a$ be the greatest power of $ 2$ dividing $ l$. Then $ x\equal{}a\plus{}k$. But $ l$ is a digit and therefore $ a\le 3$. So $ x\le k\plus{}3$.

For $ x\ge 7$, we have $ x > k\plus{}3$ (easy to prove by induction), so $ x\le 6$. We check all the cases and get that the only good pairs of powers $ (2^x, 2^y)$ are $ (2, 32)$ and $ (4, 64)$.

I will offer you a second solution.
Let $ m\equal{}2^x$ be the number without the first digit and $ n\equal{}2^y$ the number with the first digit. As djuro said, $ 2^y\equal{}2^x\plus{}l\cdot 10^k$. $ \frac{n}{m}\equal{}\frac{2^y}{2^x}$ is obviously a natural number. Since $ n$ only has one digit more than $ m$, obviously $ \frac{n}{m}<100$ (the smallest number with $ t$ digits is $ 10^t$ and the largest number with $ t\plus{}1$ digit is $ 10^{t\plus{}2}\minus{}1\equal{}100\cdot 10^t$. Considering that fact, we learn that we only have $ 6$ possible values for $ \frac{n}{m}$. These are:
I) $ 2^y\equal{}2\cdot 2^x$, meaning $ l\cdot 10^k\equal{}2^x$. The left side is obviously divisible by $ 5$, the right side isn't. Contradiction.
II) $ 2^y\equal{}4\cdot 2^x$, meaning $ l\cdot 10^k\equal{}3\cdot 2^x$. The left side is obviously divisible by $ 5$, the right side isn't. Contradiction.
III) $ 2^y\equal{}8\cdot 2^x$, meaning $ l\cdot 10^k\equal{}7\cdot 2^x$. The left side is obviously divisible by $ 5$, the right side isn't. Contradiction.
IV) $ 2^y\equal{}16\cdot 2^x$, meaning $ l\cdot 10^k\equal{}15\cdot 2^x$. However, than $ k\equal{}1$, because otherwise, the left side would be divisible by $ 25$ while the right wouldn't. So, we have $ l\cdot 10\equal{}15\cdot 2^x$. That means $ l$ is obviously $ 3$ or $ 6$(the left side must be divisible by $ 3$, but not by $ 9$ and $ l<10$). We have two cases:
 a) $ l\equal{}3$. Therefore, we have $ 2^x\equal{}2$. So, our numbers $ m$ and $ n$ are $ m\equal{}2^1\equal{}2$ and $ n\equal{}16\cdot 2\equal{}32$. We check and the solution is indeed correct.
 b) $ l\equal{}6$. Therefore, we have $ 2^x\equal{}4$. So, our numbers $ m$ and $ n$ are $ m\equal{}2^2\equal{}4$ and $ n\equal{}16\cdot 4\equal{}64$. We check and the solution is indeed correct.
V) $ 2^y\equal{}32\cdot 2^x$, meaning $ l\cdot 10^k\equal{}31\cdot 2^x$. The left side is obviously divisible by $ 5$, the right side isn't. Contradiction.
VI) $ 2^y\equal{}64\cdot 2^x$, meaning $ l\cdot 10^k\equal{}63\cdot 2^x$. The left side is obviously divisible by $ 5$, the right side isn't. Contradiction.

[quote="aakansh92"]can you show me the induction please.[/quote]

First, notice that when you multiply the number by $ 2$, the number of its digits remains the same or increases by $ 1$.

So we are proving $ x > k \plus{} 3 \iff x \minus{} k > 3$ for all $ x\ge 7$.

It holds for $ x \equal{} 7$.

Now when we increase $ x$ by $ 1$, which means we multiply $ 2^x \cdot 2$, then $ k$ (the number of digits) can increase only by $ 0$ or $ 1$. So the difference $ x \minus{} k$ can only increase but not decrease.

[quote="djuro"]Now when we increase $ x$ by $ 1$[/quote]

I think we cant use induction on x here! We can't increase x by 1 because it is not necessary that on increasing x by 1 a y will exist.

That doesn't matter!

I'm just proving that $ x > k \plus{} 3$, where $ k$ is the number of digits of $ 2^x$, for all natural numbers $ x\ge 7$.

When it is proved, I use this fact in the above solution.

[quote="djuro"]
$ k$ is the number of digits of $ 2^x$
[/quote]

Actually it is not necessary that $ 2^x$ has k digits; it may be that it has k-1 or k-2 or..... digits due to the presence of 0's.
for example $ 2^{10}$ = 1024 and 24 has 2 digits not 3.

[quote="mornik"] Since $ n$ only has one digit more than $ m$, obviously $ \frac {n}{m} < 100$ [/quote]

The difference between the number of digits in $ n$ and $ m$ can be more than 1 as I told Djuro.
In fact $ 2^{53}\equal{}9007199254740992$ and this difference is 2.  :lol:

Yes, quite, I see your problem. I'll look into it, but it does make it harder.

EDIT2: Please observe my second post - just after this one. Thanks.

OK, I've solved it! Hopefully. Be advised - the solution is rather long and pointless:

We have $ 2^y \equal{} l\cdot 10^k \plus{} 2^x$. From $ \frac {2^y}{2^x} \equal{} 2^t, t > 0$ we gather $ (2^t \minus{} 1)\cdot 2^x \equal{} l\cdot 10^k$. Now, let's make $ l \equal{} l_1\cdot 2^a$, $ a \equal{} 0$, $ a \equal{} 1$, $ a \equal{} 2$ or $ a \equal{} 3$. $ l_1$ is odd, of course. Obviously, then we have $ (2^t \minus{} 1)\cdot 2^x \equal{} l_1\cdot 2^{k \plus{} a}\cdot 5^k$. Now, it is obvious that $ x \equal{} k \plus{} a$ and we have $ 2^t \minus{} 1 \equal{} l_1\cdot 5^k$. We have five cases:
I) $ l_1 \equal{} 1$
 If $ t > 1$, the left side gives $ 3 (mod 4)$, while the right side gives $ 1 (mod 4)$. Contradiction.
 If $ t \equal{} 1$, we have $ k \equal{} 0$, but that's also a contradiction.
II) $ l_1 \equal{} 5$
 If $ t > 1$, the left side gives $ 3 (mod 4)$, while the right side gives $ 1 (mod 4)$. Contradiction.
 If $ t \equal{} 1$, we have $ k < 0$, but that's also a contradiction.
III) $ l_1 \equal{} 9$
 If $ t > 1$, the left side gives $ 3 (mod 4)$, while the right side gives $ 1 (mod 4)$. Contradiction.
 If $ t \equal{} 1$, we have $ k < 0$, but that's also a contradiction.
IV) $ l_1 \equal{} 7$
 Obviously $ t > 1$. The right side is divisible by $ 7$, so $ 2^t$ must give remainder $ 1 (mod 7)$. That means $ t \equal{} 3\cdot t_1$. Also, since $ k > 0$, the right side is divisible by $ 5$ and thus $ t \equal{} 12\cdot t_2$. (We can alternatively also check remainders of $ 2^j (mod 7)$, with $ j$ going from $ 1$ to $ 12$. We can even limit our checking to just a few numbers.). That means we have $ 4096^{t_2} \minus{} 1 \equal{} 7\cdot 5^k$. However, $ 4096^{t_2} \minus{} 1$ is also divisible by 3, and since the right side isn't, we reach a contradiction.
V) $ l_1 \equal{} 3$
 Obviously $ t > 1$.
  If $ k \equal{} 1$, obviously $ t \equal{} 4$ and we have $ l \equal{} 3$ or $ l \equal{} 6$.
   If $ l \equal{} 3$, we have $ 2^y \equal{} 30 \plus{} 2^x$. We can easily obtain that $ x \equal{} 1$, which means $ (1, 5)$ is the only solution to that, meaning we have the pair $ (2, 32)$.
   If $ l \equal{} 6$, we have $ 2^y \equal{} 60 \plus{} 2^x$. We can see that $ x < 3$. If $ x \equal{} 1$ we don't have a solution and if $ x \equal{} 2$, we reach a solution $ (2, 6)$, meaning we have the pair $ (4, 64)$.
  If $ k > 1$, the right side is divisible by $ 75$. Now, we easily come to the fact that $ t \equal{} 4\cdot t_1$ (from divisibility with $ 5$ and $ 3$). Now, let's see the remainders of $ t (mod 20)$. The possible remainders are $ 4$, $ 8$, $ 12$, $ 16$, $ 20$. Using simple calculations (or using orders, which we can also use if $ l_1 \equal{} 7$), we can establish that $ t$ is divisible by $ 20$. $ t \equal{} 20\cdot t_2$. That means we have $ ({2^{10}})^{2\cdot t_2} \minus{} 1 \equal{} 3\cdot 5^k$. However, using Fermat we have that $ 2^{10} \minus{} 1$ is divisible by $ 11$. Also, the left side equaling $ ({2^{10}})^{2\cdot t_2} \minus{} 1$ is divisible by $ 11$ while the right side isn't. Contradiction.

Since we have covered all the cases, we have found both solutions ($ (2, 32), (4, 64)$), there are no more solutions and the problem is solved.

Now, I might have missed something somewhere so... read carefully, I might be wrong somewhere, but I think that's it.

$ 1$) [quote="mornik"]The possible remainders are $ 4$, $ 8$, $ 12$, $ 16$, $ 20$. Using simple calculations (or using orders, which we can also use if $ l_1 \equal{} 7$), we can establish that $ t$ is divisible by $ 20$. $ t \equal{} 20\cdot t_2$.[/quote]
Can you show me those "simple calculations"?

$ 2$)
Many of your cases can be reduced into a few lines:

      In $ 2^y \equal{} l\cdot 10^k \plus{} 2^x$,$ x$ , $ y$ have the same parity(because $ 2^x$ and$ 2^x$ both end with same digits)
Taking the equation $ mod 3$ we get $ l \equal{} 0(mod 3)$
Hence l=3,6,9

1) The simplest way is to just try it and not think about how you can simplify it even more: $ 2^4 \minus{} 1 \equal{} 15$, $ 2^8 \minus{} 1 \equal{} 255$, $ 2^{12} \minus{} 1 \equal{} 4095$ and $ 2^{16} \minus{} 1 \equal{} 65535$ aren't divisible by 25, while $ 2^{20} \minus{} 1$ is (because of Euler's theorem). Therefore, $ t$ must be divisible by $ 20$.

2) That's great. I came up with a solution... you came up with a shorter one. Still, it's not much of a trouble no matter how you look at it. I just used brute force because it saves me a lot of thinking and the calculations are simple.

Nevertheless, a very nice problem. Where's it from?

[quote="aakansh92"]Find all powers of 2 such that whenever the first digit is removed the resulting number is still a power of 2.[/quote]
[hide="Solution"]
Let the number be $ 2^n \plus{} 10^a\cdot k$ with $ a\le 9$.  Now, when we remove the first digit, which is $ a$, then we are left with another power of $ 2$, which is $ 2^n$.  Now, $ 2^n$ must have the same units digit as $ 2^n \plus{} 10^a\cdot k$.  Notice that powers of $ 2$ cycle in blocks of $ 4$.  Hence, if we have a power of $ 2$, call it $ m$, then the lowest power of $ 2$ above $ m$ that has the same units digit as $ m$ is $ 2^4\cdot m \equal{} 16m$.  Hence, $ 2^n \plus{} 10^a\cdot k \equal{} 16\cdot 2^n$.  The next highest power of $ 2$ with the same units digit would be $ 256\cdot 2^n$.  Yet this clearly adds at least $ 2$ digits to $ 2^n$ and we only wish to add one.  It follows that $ 16\cdot 2^n \equal{} 2^n \plus{} 10^a\cdot k\implies 15\cdot 2^n \equal{} 10^a\cdot k$.  Since $ 3|15$ and $ 3$ is not a factor of $ 10$, we have that $ 3|k$.  Therefore, $ k \equal{} 3, 6, 9$.  Now, $ 5$ is the highest factor of $ 5$ that divides $ 15$ and no factor of $ 5$ divides $ 2$.  Thus, there only must be one factor of $ 5$ on the right hand side, but there are actually $ a$ factors of $ 5$.  Thus, $ a \equal{} 1$.  It follows that $ 15\cdot 2^a \equal{} 10k\implies 3\cdot 2^a \equal{} 2k$, so $ k \equal{} 3\cdot 1 \equal{} 3$ and $ a \equal{} 0$ or $ k \equal{} 3\cdot 2 \equal{} 6$ and $ a \equal{} 1$.  Hence, the solutions are $ (32, 2)$ and $ (64, 4)$.  Notice that there are no other solutions since $ k \equal{} 3, 6, 9$ and $ 9$ clearly does not work. [/hide]

[quote="The QuattoMaster 6000"]
Hence, $ 2^n \plus{} 10^a\cdot k \equal{} 16\cdot 2^n$.  The next highest power of $ 2$ with the same units digit would be $ 256\cdot 2^n$.  Yet this clearly adds at least $ 2$ digits to $ 2^n$ and we only wish to add one. [/quote]

I think, if I understood you correctly, you fell in the same trap as myself in the first "solution" (I think the second one is OK) and djuro above.
We may wish to add more than one digit since the power of 2 can have zeroes in the second (or not only second but third, fourth and so on) place from the beginning. For example, as aakansh92 posted above, $ 2^{10} \equal{} 1024$ and $ 2^{53} \equal{} 9007199...992$.

I apologize for my error.  I believe, however, that we can make a correction:
[hide="Correction"]
From the same argument as my previous "solution" we can see that $ 16^x\cdot 2^n \equal{} 2^n \plus{} 10^a\cdot r$ for an integers $ x$ and $ r$.  Thus, $ (16^x \minus{} 1)\cdot 2^n \equal{} 10^a\cdot r$.  Now, if $ x \equal{} 1$, then we get the same two solutions as before.  If $ x\ge 2$, then we will show that $ 16^x \minus{} 1$ has a prime factor greater than or equal to $ 11$.  Since $ k\le 9$ and $ 10 < 11$, we find that $ 16^x \minus{} 1$ has a prime factor that is greater than any of the prime factors of $ 10^a\cdot k$, and we can conclude that $ x \equal{} 1$.  Now, if $ x$ is even, then we find that $ 16^x \minus{} 1\equiv ( \minus{} 1)^x \minus{} 1\equiv 1 \minus{} 1\equiv 0\bmod 17$, so $ 17|(16^x \minus{} 1)$ if $ x$ is even.  Now, if $ x$ is odd, then let $ x \equal{} 2k \plus{} 1$.  It follows that
\[ 16^x \minus{} 1 \equal{} 2^{4x} \minus{} 1 \equal{} (2^{2x} \minus{} 1)(2^{2x} \plus{} 1)
\]
Yet this is equal to \[ (2^x \minus{} 1)(2^x \plus{} 1)(2^{4k \plus{} 2} \plus{} 2^{2k \plus{} 1}\cdot 2 \plus{} 1 \minus{} (2^{k \plus{} 1})^2)\] which becomes \[ (2^{2k\plus{}1} \minus{} 1)(2^{2k\plus{}1} \plus{} 1)(2^{2k \plus{} 1} \plus{} 1 \minus{} 2^{k \plus{} 1})(2^{2k \plus{} 1} \plus{} 1 \plus{} 2^{k \plus{} 1})\]
Now, we take pairs of factors and show that they are coprime.  $ \gcd (2^x\plus{}1, 2^x\plus{}1)\equal{}\gcd (2^x\minus{}1, (2^x\plus{}1)\minus{}(2^x\minus{}1))\equal{}\gcd(2, 2^x\minus{}1)$, but $ 2$ is not a factor of $ 2^x\minus{}1$, so $ 2^x\minus{}1$ and $ 2^x\plus{}1$ are coprime.  Also, \[ \gcd (2^x\plus{}1, 2^{2k\plus{}1}\minus{}2^{k\plus{}1}\plus{}1)\equal{}\gcd (2^{2k\plus{}1}\minus{}2^{k\plus{}1}\plus{}1, (2^x\plus{}1)\minus{}(2^{2k\plus{}1}\minus{}2^{k\plus{}1}\plus{}1))\equal{}\gcd (2^{2k\plus{}1}\minus{}2^{k\plus{}1}\plus{}1, 2^{k\plus{}1})|2^{k\plus{}1}\] but $ 2^{2k\plus{}1}\minus{}2^{k\plus{}1}\plus{}1$ is odd, so the only factor of $ 2^{k\plus{}1}$ which is odd is one, so $ 2^{2k\plus{}1}\minus{}2^{k\plus{}1}\plus{}1$ and $ 2^{2k\plus{}1}$ are coprime.  Similarly, $ 2^{2k\plus{}1}\plus{}2^{k\plus{}1}\plus{}1$ and $ 2^{2k\plus{}1}$ are coprime.  Furthermore, \[ \gcd (2^{2k\plus{}1}\minus{}2^{k\plus{}1}\plus{}1, 2^{2k\plus{}1}\plus{}2^{k\plus{}1}\plus{}1)\equal{}\gcd (2^{2k\plus{}1}\minus{}2^{k\plus{}1}\plus{}1, (2^{2k\plus{}1}\plus{}2^{k\plus{}1}\plus{}1)\minus{}(2^{2k\plus{}1}\minus{}2^{k\plus{}1}\plus{}1))\equal{}2^{2k\plus{}1}\minus{}2^{k\plus{}1}\plus{}1, 2^{k\plus{}2})|2^{k\plus{}1}\] and since $ 2^{2k\plus{}1}\minus{}2^{k\plus{}1}\plus{}1$ is odd, we see that by similar reasoning as before, $ 2^{2k\plus{}1}\plus{}2^{k\plus{}1}\plus{}1$ and $ 2^{2k\plus{}1}\minus{}2^{k\plus{}1}\plus{}1$ are coprime.  Finally, we see that $ \gcd (2^{2k\plus{}1}\minus{}2^{k\plus{}1}\plus{}1, 2^{2k\plus{}1}\minus{}1)|\gcd (2^{2k\plus{}1}\minus{}1, 2(2^k\minus{}1))$.  Yet $ (2^k\plus{}1)(2^k\minus{}1)\equal{}2^{2k}\minus{}1$, which is relatively prime with $ 2^{2k\plus{}1}\minus{}1$ since their difference is a power of $ 2$.  It follows that $ \gcd (2^{2k\plus{}1}\minus{}1, 2(2k\plus{}1))\equal{}1$, so $ 2^{2k\plus{}1}\minus{}2^{k\plus{}1}\plus{}1$ and $ 2^{2k\plus{}1}\minus{}1$ are coprime as are $ 2^{2k\plus{}1}\plus{}2^{k\plus{}1}\plus{}1$ and $ 2^{2k\plus{}1}\minus{}1$.  We  conclude that all four of the factors are pairwise coprime.

Notice that when $ x \equal{} 1$, then $ 2^x \minus{} 1 \equal{} 1$, so there are only three factors after we take out the one.  In all other cases, though, we see that all factors are greater than one. It follows that when $ x\ge 2$, there are four factors, all of which are relatively prime, but none of them are divisble by $ 2$.  Since they are relatively prime, none of them share the same prime factor.  Yet, there are only three prime factors other than $ 2$ that are less than $ 10$, so at least one of these factors has a prime factor greater than or equal to $ 11$, as desired.   [/hide]


linear algebra
matrix
algebra
polynomial
superior algebra
superior algebra solved

Let $A,B\in M_n(\mathbb C)$ s.t. $AB=BA$ and there exists $p\in\mathbb N$ s.t. $A^p=O_n$
Prove that $\det(A^2+AB+B^2)=detB^2$.

I think it can be shown fairly easily that $\det(xB-A)=\det B\cdot x^n$, and the result follows from $A^2+AB+B^2=(\varepsilon B-A)(\bar\varepsilon B-A)$.

It seems to me that the following is true:
From $A^p=O_n$ follows that all the eigenvalues of $A$ are equal to $0$. Then it is known fact that if $A,B$ commute then eigenvalues of $A+B$ could be obtained as a sum of the eigenvalues of $A$ and $B$ -the same is true for the eigenvalues of the product $AB$. Since $A^2$ and $AB$ commute and $A^2+AB$ and $B^2$ commute the eigenvalues of the matrix $A^2+AB+B^2$ are equal to the eigenvalues of $B^2$, since all the eigenvalues of $A^2$ and $AB$ equal to $0$. So taking into account the fact that the determinant of the matrix is equal to the product of its eigenvalues we obtain the requested result

$A$ is nilpotent its eigenvalues are $0$
$A$ is a matrix in $M_n(C)$ so it is trigonalisable.

$A,B$ commute there exist a common basis in wich
they are trigonalisable (for example triangular superior)

In this basis $A^2+AB+B^2$ is a triangular superior with diagonal
the eigenvalues of $B^2$ taking the determinant we have

$det(A^2+AB+B^2)=det(B^2)$

If $A,B \in M_n(C)$ s.t. $AB=BA$ if exist $p \in \mathbb N^*$ with $ A^p= O_n$

If $P(x,y)=xQ(x,y)+W(y)$ where $Q \in \mathbb R[X,Y], W \in \mathbb R[X]$

We have $det(P(A,B)) = det(W(B))$

Another application of a very classical result: if A is nilpotent and B commutes with A then $ det(A+B)=detB$. Just take for the place of A and B the matrices $ A^2+AB$ and $ B^2$.

I solved it in two cases
if $\det B\ne0$ then $\det(A^2+AB+B^2)=\det B^2\det(A^2B^{-2}+AB^{-1}+I_n)=\det B^2$ (since $C=A^2B^{-2}+AB^{-1}$ is nilpotent and has all its eigenvalues zero=> the characteristic polynomial is $P(X)=(-X)^n$).

if $\det B=0$ then let $C=B+\lambda I_n$ where $\lambda\in\mathbb C$ s.t. $C$ is invertible (there exist inifinitely many such $\lambda$ )
Now we aply the conclusion above and we get that $\det(A^2+AC+C^2)=\det C^2$.
If we consider these two determinants as polynomials in $\lambda$ (of degree $2n$) we get that they are equal in an infinitely many points, so they are equal in $\lambda=0$ too. Results $\det(A^2+AB+B^2)=\det B^2$...

the funny (sad for me) part was that they didn't understand my sol  :? .
I had to explain them an hour why if $A$ is nilpotent results that its eigenvalues are zero...  :(

[quote="Kuba"]

...they didn't understand my sol
I had to explain them an hour why if $A$ is nilpotent results that its eigenvalues are zero...  :([/quote]

Who did not understand your solution ?

I think Kuba was talking about the teachers grading the papers.


1) Find all palindromes that has the form :

         $\large{\underbrace{111\,\cdots\,11}_{n}\,\textrm{.}\,\underbrace{111\,\cdots\,11}_{m}}$

And can you tell me what is palyndrome? ;)

Oh Maybe palyndrom is symmetrical! :oops:


Hi, I was thinking of holding an informal contest of easy AIME problems over AIM, or MSN, or IRC, or what have you. I would voluteer to moderate, but I want to know if anyone else, expecially seasoned problem solvers , would like to moderate with me, so I can take a turn. Anyone is invited; I will throw in AMC level stuff if you guys want it. It will probably happen later, in a few hours, or it could happen very soon. Just post "Im in", your prefered chat service( but also a screenname for AIM just in case), and comments.

[quote="math92"]i'm in (MSN)[/quote]
Ok, but be ready to get an AIM name, a lot of people here use it.

I'd like to try it.  I'll use almost everything except for IRC.  I've tried to get that to work, but it never does for me.

We really need more participants...2 really isn't enough :?

ok, lets start in 50 minutes of this posting

My name is peachfuzzy007

Ok, everyone get on AIM and message me, from now until in about 30 minutes from this posting

[quote="Silverfalcon"]American Instant Messenger

http://www.aim.com[/quote]I thought the "A" was for "AOL"?  :?

Sorry, I couldn't find enough people. Until then, what are some times that work well for everyone throught the week?

I live in England so it can be anytime after school. :D
But I dont see how this is going to happen, hmm maybe ill ask some people. But 3 different time zones nearly 5 hours apart will be difficult to get a time!

I'd like to see you guys solve problems, and see how hard the problems are. I'm in highschool as a sophmore, so my level isn't that high, so I will just spectate.

AIM - megawerewolf EXE
(I hate AIM though)

Is this still going to take place? I might give it a try.

[quote="Treething"]AIM: Painkiller Abuse[/quote]
 :rotfl:
you are one funny person
 :rotfl:

:rotfl:  :rotfl: Painkiller Abuse! That's Hilarious!  :rotfl:  :rotfl:


[quote="RC-7th"][quote="Ani"]How can you know the nature of the roots of a trinomial without solving it?
Apart from figuring out the value of the discriminant, list all possible ways.[/quote]

When you say trinomial, do you mean something of the form ax^2+bx+c, or anything with three terms?[/quote]

I mean ax 2 +bx+c

I don't know, the fastest way I know of and use is through the dicriciminant.

Ok, because on my math test, it said to list all possible ways. So can you tell me if the following answer is right?:
You can turn the trinomial from standard form to vertex form. First look at the value of a. If it's negative you know that the parabola has a maximum. Then you look at the h and k value and if it's below the x axis, you know that it will never meet the x axis, so it has no real roots.
Is this an ok explanation?

Hmmm, I still think the discriminant is easier but don't you mean above the x-axis instead of below it?  If the vertex is  below the x-axis  there will be real solutions.

Ani was talking about the case where a was negative, meaning the parabola is upside down (so for no roots, the vertex is below the x axis). That works fine, but you still need to talk about what happens when a is positive.

And don't forget that any 'different' way of finding out if there are roots is fundamentally the same thing.


I've been using the gclc program to draw some pictures, but when I try to include them in a Latex document it won't draw any lines. All points are drawn exacly where they should be and there are no error messiges, just no lines.

I've tried not using the gclc package and typing in the definitions directly into the preamble of the document, tried various pictures including the working example provided and my own and it's all the same.

Has anyone had a similar problem and knows what to do?

Nevermind, I figured it out. But in case anyone else has the same problem:

The thing is you have to compile the tex file to postscript, not dvi, or no circles or lines will show up. And if you want a pdf file, then you have to compile it from a postscript file or the same problem will occur. Your file might not look so great after all those conversions, but it'll print just fine. In review:

either
tex -> dvi -> ps
or
tex -> dvi -> ps - > pdf

or nothing will show up.


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Support

Hello,

I would like to see what nice signatures you find in the forum. In my case, it is just a frase that I thought about a year ago when I finally realize I was way too overweighted. But I have found some nice ones. I appologize with the authors, the following are not mine and I don't remember who were the autors of each one of them. Nevertheless, authors, take your credit. I have found each of this in the signature of members of the forum. Some of the nicknames of the members are there.

I mainly like the first one.

[quote]There are two aspects of being good at competitive mathematics; the Art of Problem Solving and the Art of Lucky Guessing.[/quote]

Great, just great, and not only true for competitive mathematics, it is true in general in mathematics.

[quote]Nothing in life is impossible,... except: Slamming a revolving door; Clapping with one hand; Striking a match against wet toilet paper; Swallowing while you're talking; RC-7th
[/quote]

Wow, so enlightening!  :D

[quote]Words of Wisdom
1.When your dad is mad and asks you, "Do I look stupid?" don't answer him.
2.Never tell your mom her diet's not working.
3.If you want a kitten, start out by asking for a horse.
[/quote]

Those are definitelly wise words ;)

[quote]In the fall of 1972 President Nixon announced that the rate of increase of inflation was decreasing. This was the first time a sitting president used the third derivative to advance his case for reelection.
Hugo Rossi
[/quote]

Interesting!!

[quote]I wish you were $f(x)$ and I was $f(x)+|sin(\omega)|$, and $\omega$ a steadly increasing real number (sam3.14).[/quote]

 :blush:

Let's see what else can we find here!

I always liked this one:

[quote]I have ADD that stands for Attention Deficit hey look a birdie![/quote]

[quote="sam3.14"]I like the one that says "There are 11 types of people in this world, those that understand binary and those that don't." I'm not sure who the author is, though.[/quote]

Correction:

There are 10 types of people in this world, those that understand binary and those that don't.

Also, here's one.

Always remember that you are unique--just like everyone else.

One that makes people curious well at least i was so i adopted it. click on the hide thing  :lol:

edit: just keep clicking.

elvenchamp777

[quote="mathgeek2006"]Correction:

There are 10 types of people in this world, those that understand binary and those that don't.
[/quote]

I think part of the joke is that 11 is three, and not two.

[quote="djimenez"]Someone else?

BTW, those are good ones!

This one is from a friend, but in his email signature

[quote]
I am going to live forever... or die trying  :D
[/quote][/quote]

Haha, great!

Here's one that I saw somewhere and thought it was funny-

"Where there's a will, I want to be in it." :D

"When I feel blue, I start breathing again"
   -saw it somewhere.

 :D

[quote="sam3.14"]I like the one that says "There are 11 types of people in this world, those that understand binary and those that don't." I'm not sure who the author is, though.[/quote]
a variant ive seen somewhere
there are 5 types of people in the world: those who can count and those who cant

Here's a good one:

[quote="white_horse_king88's signature"]Life is a sexually transmitted disease with 100% mortality[/quote]

How true. *sighs*

 :P

Shows how pessimistic I am.  :D

[quote="dts"][quote="djimenez"]I mainly like the first one.

[quote]There are two aspects of being good at competitive mathematics; the Art of Problem Solving and the Art of Lucky Guessing.[/quote][/quote]

Hey, thanks!  Glad you like it. :)[/quote]

You're welcome. Its very true and very nice too!

It's all fun and games until somebody loses an eye - then it's just a game.

If at first you don't succeed, skydiving is not for you.

I read this:

"Hard work pays off later. Laziness pays of NOW!"  :D

When it feels like the whole worlds against you, remember: Switzerland's always neutral.

Can someone post a graph of that function, the innopropriate one of them.

Or better yet, pm me?

[quote="white_horse_king88"]Er, the function - why can't you just graph it yourself?

It's ...  ;)  :o[/quote]

Ive got no grapher.

Do it by hand. If you really have trouble, ask in the math forums about how to graph stuff. Or ask about free graphing calculator. Not that I support them, but Window's Calculator Powertoy has graphing capability.

I like this one I found on a calendar:

The bigger the risk, the more angels show off.

How do I get the steadily increasing thing in there?


Let $ \{a_n\}_{n \equal{} 1}^{\infty}$ be a sequence of natural numbers which haven't got a digit '$ 9$'.
Prove that:
$ \sum_{n \equal{} 1}^{\infty}\frac {1}{a_n} < 81$

The number of $ k$-digit terms of $ \{a_{n}\}$ equals $ 8\cdot 9^{k \minus{} 1}$. It is obvious that every $ k$-digit natural number is greater or equal $ 10^{k \minus{} 1}$.
Using this facts we see that:
$ \sum_{n \equal{} 1}^{\infty} \frac {1}{a_{n}}\leqslant 8\cdot\sum_{k \equal{} 1}^{\infty}\left(\frac {9}{10}\right)^{k \minus{} 1} \equal{} 80 < 81$

In fact, it is easy to prove the stronger result:
$ \sum_{n \equal{} 1}^{\infty} \frac {1}{a_{n}} \leqslant 10\cdot \sum_{n \equal{} 1}^{8}\frac {1}{n} < 25.4$
since every $ k$-digit natural number is greater or equal $ l\cdot 10^{k \minus{} 1}$, where $ l$ is the first (from the left) digit of this number.


Solve the equation:
$ 2-x^{3}=\sqrt{2-x}$

Do you want all three, or just the real one?


calculus
integration
trigonometry
calculus computations

Evaluate $ \int_0^{\frac{\pi}{4}} \frac{1}{\cos \theta}\sqrt{\frac{1\plus{}\sin \theta}{\cos \theta}}\ d\theta$.

hello
  [hide="solution"]
 $ \int_0^{\frac{\pi}{4}} \frac{1}{\cos \theta}\sqrt{\frac{1+\sin \theta}{\cos \theta}}\ d\theta$
=${ \int_0^{\frac{\pi}{4}} sec(x)\sqrt{tan(\frac{x}{2}+\frac{\pi}{4}})}dx$
=${ 2[\sqrt{tan(\frac{x}{2}+\frac{\pi}{4}})}]^\frac{\pi}{4}_{0}
=2[\sqrt{tan{\frac{3\pi}{8}}}-1]$
[/hide]
thank u

You have almost solved the problem.

$ \tan \frac{3}{8}\pi \equal{}\tan 67.5^\circ \equal{}\frac{1}{\tan 22.5^\circ}\equal{}\frac{1}{\sqrt{2}\minus{}1}\equal{}1\plus{}\sqrt{2}$.

$ \therefore \int_0^{\frac {\pi}{4}} \frac {1}{\cos \theta}\sqrt {\frac {1 \plus{} \sin \theta}{\cos \theta}}\ d\theta \equal{}2(\sqrt{1\plus{}\sqrt{2}}\minus{}1)$.

alternatively, the problem can be written as

${ \int_{0}^{\frac{\pi}{4}}\sec\theta}\sqrt{\sec\theta + \tan\theta}\ d\theta$

Now let $ u = \sec\theta + \tan\theta$.   The rest is easy.   :)


[quote="confuted"]I know how to draw horizontal lines with \hrule, but I would like to also be able to make a similar line that goes from the current indentation to the right margin. Does anyone know how to do this?[/quote]

I know of three possibilities.

First, it is possible to control the dimensions of an \hrule with the parameters height, depth, and width. For example, "\hrule width 1in" gives
$\hrule width 1in$
There is a vertical line, \vrule, that works the same way.

Another possibility is to use leaders. A leader is a line, usually of dots, used to  connect text across a distance, such as connecting names to numbers in a table of contents. The line of dots is \dotfill, but a solid line is \hrulefill. The \leaders command can be used to create a custom leader.
For example, "Left\hrulefill Right"
$\text{Left}\hrulefill \text{Right}$

Third, you could temporarily redefine the width of a page with a parbox or a minipage environment.  Then the \hrule that stretches across the page will only stretch the current page width. For example, "\parbox{1in}{\hrule}" should give a 1-inch line, but the example does not work in this forum.

Erin Schram


probability
expected value
probability and stats

It is said that different states of a Markov Chain are symmetric if:

     $ P_i[T_j<T_i]\equal{}P_j[T_i<T_j]$.

  Let i and j be two symmetric states. Prove that given $ X_0\equal{}i$ , the expected number of visits to state j
   before the chain goes back to i is one.

These are just rough ideas.

$ \xi$ - number of visits of $ j$ before going to $ i$. We write $ i\to i$ if chain starting form $ i$ visits $ i$ before $ j$, $ i\to j$ if it visits $ j$ before $ i$, etc.
\[ E_i[\xi] \equal{} E_i[\xi 1_{i\to i}] \plus{} E_i[\xi 1_{i\to j}].\]
First term is just $ 0$.

Further,
\[ E_i[\xi 1_{i\to j}] \equal{} E_i[ E[\xi 1_{i\to j} / F_{T_j}]] \equal{} E_i[1_{i\to j} E[\xi/F_{T_j}]] \equal{} E_i[1_{i\to j}E_j[(1\plus{}\xi)]].\]
The last equality is true because of Markov property.

Now (*)
 \[ E_i[1_{i\to j}E_j[(1\plus{}\xi)]]\equal{} P_i(i\to j) (1\plus{} E_j[\xi]) \equal{} P_j(j\to i) (1\plus{} E_j[\xi])\]
Finally,
\[ E_j[\xi] \equal{} E_j[\xi 1_{j\to i} ] \plus{} E_j[\xi 1_{j\to j}] \equal{} E_j[\xi 1_{j\to j}] \equal{} P_j(j\to j)(1\plus{} E_j[\xi]),\]
where the last equality is deduced as above.
Hence $ E_j[\xi]P_j(j\to i) \equal{} P_j(j\to j)$. Plugging this in (*), we have
\[ E_i[\xi] \equal{} P_j(j\to i) \plus{} P_j(j\to j) \equal{} 1.\]

[quote="marko avila"]What  is $ F_{T_j}$?[/quote]
Sorry, forgot to write. This is $ \sigma$-algebra corresponding to stopping time $ T_j$, i.e. $ \sigma$-algebra
generated by events of the form
\[ \{X_1\equal{} i_1,\dots, X_2 \equal{} i_2,\dots, X_n\equal{}i_n, T_j\equal{}n\}\]

[quote="limsup"][b]marko avila[/b]: What do you mean by symmetric states?[/quote]
Read teh formulation:
[quote="marco avila"]It is said that different states of a Markov Chain are symmetric if...[/quote]

Thanks zhoraster. Sorry I didn't read it properly :blush:


ceiling function
floor function
number theory proposed
number theory

Let $N$ be a positive integer.
Determine the number of positive integers $n\leq N$, which have a multiple which uses (in decimal system) exactly the digits 2 and 6 (not neccessarily equally often).

Clearly $2|n$ and $4$, $5$ do not divide $n$. Let's show that all such $n\le N$ satisfy the conditions of the problem. Denoting $m=\frac{n}{2}$ we have to construct a number divisible by $m$ using digits $1$ and $3$ only. Since $(m, 10)=1$ $\exists d\colon 10^d\equiv 1\mod m$. Now let $A=31\ldots 1$ so that $A$ has $d-1$ 1's. Then $B=\sum_{i=0}^{m-1} 10^{di}A \equiv m\cdot A \equiv 0 \mod m$ satisfies the conditions. Counting all such $n$ is straight forward and if I haven't mistaken the answer equals $\left\lceil\frac{1}{2}\left\lfloor\frac{n}{2}\right\rfloor\right\rceil-\left\lfloor\frac{n}{10}\right\rfloor$.


In general, what is the largest coefficient of $ (ax \plus{} b)^n$ for positive real numbers a and b, and positive integer n?

[hide]
The largest coefficient is $ \binom{n}{p}a^{p}b^{n\minus{}p}$  where $ p\equal{}\left\lceil\frac{bn\minus{}a}{a\plus{}b}\right\rceil$

I used a general formula for a ratio between coefficients:

$ \frac{\binom{n}{p\plus{}1}b}{\binom{n}{p}a}$

Now I set it equal to 1, simplified, solved for p, then used the ceiling function to make the answer an integer.
[/hide]


[hide="Hint"]Can you mimic the procedure of constructing the limit inferior over the reals?[/hide]


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $H$ and $K$ be subgroups of $G$. Prove that $HK=\{x \in G|$ there exist elements $h \in H$ and $k \in K$ such that $x=hk \}$ is a subgroup of $G$ if $H \cap K =\{e\}$, where $e$ denotes the identity element of $G$.

Counterexample: let $H,K$ be any two 2-element subgroups of $\mathbb{S}^{3}$.

You can show that $HK$ is a group provided that $H \subseteq N(K)$ or $K \subseteq N(H)$.

[quote="ZetaX"]There is a result for such things that normally suffices, but is notgenerally true for your problem.

The following are equivalent for two subgroups $A,B$:
a) $AB$ is a subgroup
b) $AB=BA$
c) $AB \subset BA$
d) $BA \subset AB$[/quote]

Although it is easily proven if the group is Abelian.
(Mainly because then $AB = BA$.)

that's strange. i was having difficulty with the problem, mainly because the group is not abelian, like you guys would suggest. but the professor says that the group needn't be abelian.

Maybe you misunderstood the wording of the problem, because I cannot image a professor to commit such a mistake ...

The professor did indeed make a mistake. I have now fixed the problem. The subgroups are in fact normal subgroups of $G$.

By the criterion I gave above, the trivial intersection is not necessary then  :maybe:
And we would onle need one of them to be normal.

You only need to show closure and existence of inverses.
[hide]Let $x_{1}=h_{1}k_{1}$ and $x_{2}=h_{2}k_{2}$ be elements of $HK$. Then $x_{1}x_{2}=h_{1}k_{1}h_{2}k_{2}$. Use the given that $H$ is normal ($\forall x\in G,\forall h\in H,\exists h'\in H,\ s.t.\ xh=h'x$). Then $x_{1}x_{2}=h_{1}(k_{1}h_{2})k_{2}=h_{1}(h_{3}k_{1})k_{2}=hk$. The last equality is true since $H$ and $K$ are closed.
Similarly, you can show $k^{-1}h^{-1}$ is an element of $HK$ using the property that $H$ is normal.
Like ZetaX said, I think the restriction on their intersection is unnecessary.[/hide]


How many integers with four different digits are there between $1,000$ and $9,999$ such that the absolute value of the difference between the first digit and the last digit is $2$?

$\textbf {(A) } 672 \qquad \textbf {(B) } 784 \qquad \textbf {(C) } 840 \qquad \textbf {(D) } 896 \qquad \textbf {(E) } 1008$

[hide="Solution"]The types of first and last digit combinations may be $(0,2)$, $(1,3)$, and etc, all the way to $(7,9)$.

These can also be interchanged, except for $(0,2)$. By interchanged, I mean $2xx4$ and $4xx2$ both work.

Now, for each combination of first to last digits, there are $8\cdot 7=56$ ways to fill in the remaining two numbers. Our answer is thus $15\cdot 56=840$ numbers.[/hide]

Cool problem.
[hide]
Ordered pairs that work for first and last digit:
(0,2), (1,3),(2,4)...etc

they're interchangeable except for (0,2) cause of the 0 first digit.

there are 15 ordered pairs.

8*7 to arrange the two numbers in the middle because of the different digits clause.

15*56=\boxed{840}
\boxed{C}[/hide]

I know this is an old problem, but it is linked to as an AMC problem so it should probably be corrected:

it's 14 pairs of first and last because (0,2) creates a 3 digit number.

So 14 * 56 = 784

then what about a number such as 2000? that isn't included in the pairs. there are 15 pairs.

[quote="4everwise"]How many integers with four different digits are there between $ 1,000$ and $ 9,999$ such that the absolute value of the difference between the first digit and the last digit is $ 2$?

$ \text{(A)}672 \qquad \text{(B)}784 \qquad \text{(C)}840 \qquad \text{(D)}896 \qquad \text{(E)}1008$[/quote]

[hide="Solution"] First and last digit can only be $ (0,2), (1,3), (2,4), (3,5), (4,6), (5,7), (6,8), (7,9)$. Except for the first pair, all the other ones can be interchanged as first and last digits giving $ 15$ sets of first and last digit possibilities. Now suppose we have chose the first and last digits. $ 8$ numbers are left and we have $ P^{8}_{2}= 56$ permutations of them. So: $ 15 \times 56 = 840$ numbers. $ C$ is the answer.[/hide]


invariant
function
absolute value
combinatorics proposed
combinatorics

Kosovo National Olympiad

In a circle we have fixed four distinct points. Let in each point be a real number : $ x_1,x_2,x_3,x_4$,such that $ x_1 \plus{} x_2 \plus{} x_3 \plus{} x_4 > 0$.
Let's make a game with these numbers. If there exists a negative number, say $ x_i$, then the player makes a step taking that number, adding to $x_i$ its nearby numbers, and changes the sign of selected number. The game is finished when all numbers are positive. Prove that this game will end after a finite number of steps!

This is a variation on a classical {\it invariant} motif (see IMO 1986 -- Problem 3). Denote, for simplicity, the four numbers by $ a, b, c, d$, and have $ \sigma : \equal{} a\plus{}b\plus{}c\plus{}d > 0$. It is clear that $ \sigma$ is invariant to the steps allowed. Let us build a semi-invariant $ \Sigma(a,b,c,d) : \equal{} a^2\plus{}b^2\plus{}c^2\plus{}d^2 \plus{} \frac {1} {2} (a\minus{}c)^2 \plus{} \frac {1} {2} (b\minus{}d)^2$; clearly this is invariated by a cyclic permutation of its arguments. Then, e.g. for $ a<0$, simple computations yields $ \Sigma(a',b',c',d') \equal{} \Sigma(\minus{}a,b\plus{}a,c,d\plus{}a) \equal{} \Sigma(a,b,c,d) \plus{}2a\sigma < \Sigma(a,b,c,d)$. All would be simple if the four numbers were integers, since then $ 0 < \Sigma(a',b',c',d') \leq \Sigma(a,b,c,d) \minus{} 2\sigma$, and the semi-invariant decreases by at least $ 2\sigma$ with each step, absurd, since it must stay positive.

Given the four numbers just reals, one needs know there exists a lower bound $ m>0$ for the absolute value of any negative number obtained through steps; then $ 0 < \Sigma(a',b',c',d') \leq \Sigma(a,b,c,d) \minus{} 2m\sigma$, and the same argument applies. This is obtained by noticing that any number obtained through steps is of structure $ x \plus{} k\sigma$, $ (x\plus{}y) \plus{} k\sigma$, or $ (x\plus{}y\plus{}z) \plus{} k\sigma$, for some $ \{ x,y,z \} \subset \{ a,b,c,d \}$ and $ k \in \mathbb{Z}$. Then clearly the absolute values of negative numbers obtained cannot be lower than some bound.

A simpler example is with just three numbers (the IMO problem is on five numbers). Denote $ \sigma$ as above, and $ \Sigma(a,b,c) : \equal{} a^2\plus{}b^2\plus{}c^2$. Then, e.g. for $ a<0$, simple computations yields $ \Sigma(a',b',c') \equal{} \Sigma(\minus{}a,b\plus{}a,c\plus{}a) \equal{} \Sigma(a,b,c) \plus{}2a\sigma$, with similar reasoning. Now it is easier to see that any number obtained through steps is of structure $ \pm x \plus{}k\sigma$, for some $ x \in \{ a,b,c \}$ and $ k \in \mathbb{Z}$.

Thank you for replying , what you think for the official solution ?

$ R(x_1,x_2,x_3,x_4) \equal{}|x_1| \plus{}|x_2 | \plus{}|x_3| \plus{}|x_4| \plus{}|x_1 \plus{}x_2 | \plus{}|x_2 \plus{}x_3| \plus{}|x_3 \plus{}x_4| \plus{}|x_4 \plus{}x_1| \plus{}|x_1 \plus{}x_2 \plus{}x_3| \plus{}|x_2 \plus{}x_3 \plus{}x_4 | \plus{}|x_3 \plus{}x_4 \plus{}x_1| \plus{}|x_4 \plus{}x_2 \plus{}x_1|$

We can see that this function ,which is strictly positive ,and it decrease each time the player make a step ,thus the game will end after finitely steps.

The best general semi-invariant, working for any $ n \geq 3$ integer numbers $ x_0, x_1, \ldots, x_{n\minus{}1}$ (indexed by $ \mathbb{Z}_n$) with $ \displaystyle \sigma : \equal{} \sum_{i\equal{}0}^{n\minus{}1} x_i > 0$, is given by the expression $ \displaystyle \Sigma(x_0, x_1, \ldots, x_{n\minus{}1}) : \equal{} \sum_{k\equal{}1}^{n\minus{}1} \sum_{i\equal{}0}^{n\minus{}1} | \sigma(i, k) |$, where $ \displaystyle \sigma(i, k) : \equal{} \sum_{j\equal{}0}^{k\minus{}1} x_{i\plus{}j\minus{}1}$. This is exactly the one you refer to for $ n\equal{}4$.
Then, e.g. for $ x_i<0$, simple computations yield
$ \Sigma(x_0',x_1', \ldots, x_{n\minus{}1}') \equal{} \Sigma(x_0, \ldots, x_{i\minus{}1} \plus{} x_i, \minus{}x_i, x_{i\plus{}1} \plus{} x_i, \ldots,x_{n\minus{}1}) \equal{} \Sigma(x_0,x_1,\ldots,x_{n\minus{}1})\minus{}| \sigma \minus{} x_i |\plus{}| \sigma \plus{} x_i |$.
But $ | \sigma \minus{} x_i | \minus{} | \sigma \plus{} x_i | \equal{} 2\min(\sigma, \minus{}x_i) \geq 2$, and similar considerations with the ones exposed before apply.
Clearly though, the issue is when the numbers are real. Then the argument as such is insufficient, since a positive quantity may decrease forever and still stay positive.
One needs a lower bound, like explained before, and in fact the method of extending this result to reals works, as above.

The problem in olympiad was first for real numbers ,but 5 minutes latter they  came and said :
solve the problem in integers.

Is the solution correct for integers ?

Yes it is.
For the integer case you can also give a simple argument. Consider the maximum element and note that an operation can either decrease it or leave it unchanged.
since the sum is positive then the maximum has to be positive. Because we can not have an infinite decreasing sequence of naturals we conclude that if the game would not finish in a finite number of steps then after somepoint the maximum is left unchanged.
Since the maximum is unchanged that means that no operations are done on its neighbors and therefore an operation is done infinitely often at the same position. This is impossible since we can not apply the operation consecutively to the same position.


Stanford
college
inequalities proposed
inequalities

Let $ a_{1},a_{2},...,a_{2n}$ be non-negative with sum $ 1$ such that $ a_{1}\ge a_{2}\ge ...\ge a_{2n}$. Find the maximum value of

\[ (a_{1}^{2}\plus{}a_{2}^{2})(a_{3}^{2}\plus{}a_{4}^{2})...(a_{2n\minus{}1}^{2}\plus{}a_{2n}^{2}).\]

This is a problem in Secrets in Inequalities. I hope someone can give a solution in a few weeks. Waiting you!

Does anyone get any idea for this one?  :blush:

[quote="zaya_yc"]$ a_{1}\equal{} a_{3}\equal{} ... \equal{} a_{2n\minus{}1}\equal{}\frac{1}{n}$ and $ a_{2}\equal{} a_{4}\equal{} ... \equal{} a_{2n}\equal{} 0$[/quote]
This doesn't satisfy the condition of the problem :
$ a_{1}\ge a_{2}\ge ...\ge a_{2n}$.  :wink:

Hungkhtn          $ max\equal{}\frac{2^{n\minus{}1}}{(2n\minus{}1)^{2n}}$ ?

Problem 99 in my book is the following:

Let $ a_{1},a_{2},...,a_{2n}$ be non-negative real numbers such that $ a_{1}\ge a_{2}\ge...\ge a_{2n}$ and $ a_{1}\plus{}a_{2}\plus{}...\plus{}a_{2n}\equal{} 2n\minus{}1$. Prove that
\[ (a_{1}^{2}\plus{}a_{2}^{2})(a_{3}^{2}\plus{}a_{4}^{2})...(a_{2n\minus{}1}^{2}\plus{}a_{2n}^{2})\le 2^{n\minus{}1}.\]
Base on this result, the anticipation of Zaya is true.

I did come to Stanford now, and at this time, can anyone solve this problem? I know that the strictly increasing condition is not so familiar, but just make a try again  :blush:


vector
linear algebra
linear algebra unsolved

Let A and B be matrices such that
A :V --->W    B:Z--->V   Where V,W,Z are vector Spaces

[b]1-Prove Rank AB<= Rank A [/b]
2- What is the condition the equality is satified [b]Rank AB = Rank A[/b]

Hint: Treat the matrices as linear transformations, and use the definition that the rank of a linear transformation is the dimension of its image.


After her first roll, she must roll a bad number with probability $ \frac56$ eight times, and then get it with probability $ \frac16$ the last time, for a probability of $ \left(\frac56\right)^8\cdot\frac16\approx\boxed{0.039}$.

I'm stupid... I missed the last part of the question. [quote=GameBot]Express your answer as a decimal to the nearest thousandth.[/quote]

I'm a little unsure. How come you don't subtract $5^9/6^{10}$ by $1$?

[quote=GameBot]If Ella rolls a standard six-sided die until she rolls the same number on consecutive rolls, what is the probability that her 10th roll is her last roll? Express your answer as a decimal to the nearest thousandth.[/quote]

We can use constructive counting to count the number of successful outcomes.

For example, we have $6$ choices for the first outcome. Then, we cannot have the second roll equal the first, so we must have $5$ outcomes for the 2nd roll.

We continue in this fashion until we find that the total number of successful outcomes is $6 \cdot 5^8$. We divide this by the total number of possibilities, which is $6^6$.

omg i accidently entered $\frac{390625}{10077696}$ lel


inequalities
combinatorics unsolved
combinatorics

We chose some non-empty subset of S={1;2...n}.For each chosen subset A,there exists an element a such that:
 $a\notin A$ andif B is other chosen subset such that $a\notin B$ ,then A and B are disjoint.Find the maximum number of chosen subset

I think the answer is $n$.

One direction is easy: we can always find $n$ such sets by considering the $n$ singletons in $\overline{1,n}$.

Conversely, we have to prove that such a system of sets has at most $n$ members. Let $A_{1},\ldots,A_{k}$ be our sets, and let $x_{i}$ be the point corresponding to $A_{i}$.

Suppose $A_{1},\ldots,A_{t},\ t\ge 2$ are sets in our collection such that every $2$ of them have non-empty intersection, and that $A_{t+1}$ intersects $A_{1}$. Then, on the one hand, $x_{1}$ belongs to $A_{2},\ldots,A_{t}$, and on the other hand, it belongs to $A_{t+1}$. This means that $A_{t+1}$ intersects $A_{2},\ldots,A_{t}$, and after replacing $A_{1}$ with $A_{2}$ we conclude that it intersects $A_{1}$ as well. This means that our collection $A_{1},\ldots,A_{k}$ of sets can be partitioned into subcollections with disjoint unions such that every two members of every subcollection intersect.

In order to get $k\le n$, it suffices to prove that in a subcollection such as the ones described above there are at most as many sets as there are elements in the union of the subcollection. If the collection consists of one set only, this is clear, so assume we’re dealing with the subcollection $A_{1},\ldots,A_{t},\ t\ge 2$. What we want to prove is that $t\le\left|\bigcup_{i=1}^{t}A_{i}\right|\ (*)$. Of course, all the $x_{i},\ i=\overline{1,t}$ are different, and since $x_{1},\ldots,x_{t-1}\in A_{t}$, we have $t-1\le|A_{t}|$. This means that $t\le|A_{t}|+1$, and because $A_{t}$ cannot contain another $A_{i}$, the RHS in this inequality is not larger that the RHS in $(*)$. This proves $(*)$, and we are done.

Suppose $A$ and $B$ are disjoint chosen subsets.  Suppose there is some other chosen subset $C$ such that $A \cap C \neq \{\}$.  Then $C$ must contain all of $S-A$.  But then $B \cap C \neq \{\}$, so $C$ contains $S-B$.  So $C = S$, but then $C$ does not satify the chosen condition.  So it follows that all sets must be pairwise disjoint, so there can be no more than $n$ of them.

Otherwise, there can be no pair of disjoint subsets, which means for all $A,B$, $A \cup B = S$.  But then the complements of the chosen sets would be pairwise disjoint, so again there can be no more than $n$ of them.

You seem to believe that whenever two chosen subsets intersect, their union has to be the entire $S$. This is certainly not true :?. Take $n$ large, for example, and consider only two chosen subsets, $A=\{1,2\},B=\{1,3\}$, with $3$ being the point corresponding to $A$ and $2$ the point corresponding to $B$.

Yup, yet another misread problem.  :oops:


integration
calculus
real analysis
real analysis unsolved

This is a very strange problem as written: if $x(t)=-2+t$, the integral happily diverges to $-\infty$, so there is no minimum. Seems like some restrictions on $x(t)$ are missing...

Yes your comment is correct thanks fedja.
I rewrite problem by following :
$\int_{0}^{1}\frac{4\sinh^{2}(x(t))}{t}(x'(t))^{2}dt+5\cosh^{2}(x(0))-4\cosh(x(0))\cosh(x(1))-2\cosh(x(0))\to min$
find min and $x(t)$


Rebecca remembered the first six digits of a friend’s 7-digit phone number. If she randomly selects a digit to replace the last digit she can’t remember, what is the probability she will dial the correct number? Express your answer as a common fraction.

1 digit works in 10 possibilities.
1/10.


Charles the greengrocer got a shipment of watermelons (yum yum!). He spread them out on the floor 1 deep; he did not stack them. What is the sum of the possible lengths of his stack of watermelons if he had 900 watermelons?

There is a trick to this :) I'll post it Saturday.

Billy

EDIT: i got it right this time[hide="answer"]We are looking at the sum of all of the factors of $900$, because the watermelons must be in the shape of rectangles and there are $900$ watermelons. That sum is $\boxed{2821}$[/hide]

hi solafidefarms:

  I'm not quite clear on what you're asking for. In the first sentence, you state that you don't have the watermelons stacked. Then in the next sentence we are asked to
find the sum of the lengths of his "stack". When you say the sum of the lengths that leads me to believe there is more than one. How would you describe this "stack"?
Could you please clarify?

Basically the problem is asking what is the sum of all the possible widths of a rectangle with area 900.


Find all a function $f: \mathbb R^{\geq 0} \to \mathbb R^{\geq 0}$ satisfying the following conditions:
i) For any $x \geq 0$, \[ f(f(x))\equal{}2009\cdot 2010x\minus{}f(x),\]
ii) For any $x>0$, we have $f(x)>0$.

[quote="Pirkuliyev Rovsen"]Find all a function  $ f: [0; \plus{} \infty)\rightarrow [0; \plus{} \infty)$ satisfying the following conditions
i)$ f(f(x)) \equal{} 2009\cdot 2010x \minus{} f(x)$ ,
ii)$ f(x) > 0$ $ x > 0$
$ x \in [0; \plus{} \infty)$[/quote]
For a given $ x>0$, Consider the sequence :
$ u_0\equal{}x$
$ u_1\equal{}f(x)$
$ u_{n\plus{}2}\equal{}f(u_{n\plus{}1})\equal{}\minus{}u_{n\plus{}1}\plus{}2009\cdot 2010 u_n$

It's easy to solve this sequence : $ u_n\equal{}\frac{2010x\plus{}f(x)}{4019}2009^n\plus{}\frac{2009x\minus{}f(x)}{4019}(\minus{}2010)^n$

Which may be written : $ u_n\equal{}\frac{2009^n}{4019}(2010x\plus{}f(x)\plus{}(2009x\minus{}f(x))(\minus{}\frac{2010}{2009})^n)$

We know that $ u_n>0$ and so $ f(x)\equal{}2009x$, else $ u_n$ would be negative for any odd or even (depending on sign of $ f(x)\minus{}2009x$) n great enough.

And it is easy to check back that this indeed is a solution.

Hence the unique solution : $ \boxed{f(x)\equal{}2009x}$ $ \forall x\ge 0$


algorithm
induction
pigeonhole principle

2n ambassadors are invited to a banquet. Each of them has at most n - 1 enemies among the other guests. Prove that they can be seated at a round table so that no one sits next to an enemy.  (Assume that being an enemy is symmetric, in that A is an enemy of B if and only if B is an enemy of A).

Cute problem !

[hide]I think we can build a short algorithm to prove it.

Let's start we a random distribution (ambassadors seated randomly).

<<loop>>

Take a ambassador A

. if he has no enemy around him : do nothing

. if he has one enemy E to his right : look clockwise for the first ambassador B who doesn't not have E as an enemy AND such as the ambassador to the right of B (E') is a friend of A (possible because there are only n-1 enemies). Having found E and E' switch them.

So in this case the number of ambassadors with an enemy to their right decreases by 1 or 2.

Take the ambassador to the right of A and repeat <<loop>>.

In at most n steps, every ambassador will have a friend to his right, since they are on a circle, it' done.

Does it make sense ??[/hide]

Would [hide]   induction   [/hide] offer a more rigorous proof than belenos' explanation?

Belenos - nice work, but to make it a bit more rigorous you may want to start out as (in spoiler) [hide]seat the ambassadors as to minimize the number of clockwise adjacencies, and then assume that there exists A and B sitting next to one another, and then use your method to decrease the number of adjacencies.[/hide]

Rlee - yes this problem can be done by induction, anyone want to find the inductive solution?

What about  [hide]the pigeonhole principle?[/hide]

I don't have much experience with proofs, but I'll give it a go.  However, it seems like I can't find a nice clean way to write what I am thinking.

[hide]

Take 2k people sitting at the table (each with at most k-1 enemies) and add 2 - persons A and B.

Case 1 - they are enemies

Each has at most k-1 enemies seated at the table.  Taking all 2k pairs of adjacent people (with each person in 2 groups), there are a total of at most 2(k-1) enemies of person A, so there is at least 2 adjacent pair of friends, you can seat A between them.  Same with person B - there will be at least 1 pair of adjacent friends of B, and you can sit him there.

Case 2 - they are friends

Each will then have at most k enemies seated.  With 2k people seated, for every person that is an enemy of both A and B, there must be at least a person that is friendly to both.

(If a friendly person exists) So since the person who is friendly cannot be surrounded by enemies of both, you can sit A and B between an enemy of one and the person friendly to both.

(this is where I wonder if I can shorten it up)

However, there is is the problem if there are k friendly and k hostile to both A and B, where each hostile one is friendly only with people that are friendly with A and B, and thus two people who are friendly to A and B are not sitting together.  However, with would give each person hostile to both A and B k+1 enemies.

Now if there are no enemies of both A and B, its easy, you can put A and B between an enemy of B and an enemy of A or between an enemy of one and a person who is not an enemy of either (or 2 people friendly to both).  Since at least one of these pairs must exist, or everyone must be an enemy of one person.

And with an arrangement of 2k, we can make a valid arrangement of 2(k+1). Base case of 2 works, so I think I'm done.

Now some other thoughts:

For friendly A and B -

In order to be unable to put A and B in the circle, every adjacent pair must be each hostile to the same person (A or B).  Since every pair can't be hostile to just A or just B, the only way to do this would be to have a string of people hostile to A, then a person hostile to both A and B, then a string of people hostile to B and finally another person hostile to both.  And this would result in too many enemies.

[/hide]

Any problems? Any way to make it sound better? Any way to make it easier?

[quote="philosophking"]What about the pigeonhole principle?[/quote]

A good thought, but the pigeonhole principle is really the exact opposite of what you want here.  The pigeonhole principle tells you that "if you have at least blah1, then blah2 must happen."  What we need here is a statement that says, "if you have at most blah1, then blah2 could happen."

Geomar there is a very easy way to make your method simpler: get rid of the fact that there are 2n people there.  Surprisingly, you can prove that if n people are at a party and each person is friends with at least half the people there, then they can be seated in a circle s.t. each person sits next to two friends.  Notice that this lemma would prove the original statement, since if each person has at most n-1 enemies, they have at least n friends (assuming A and B are friends iff A and B are not enemies).

Well with at least n friends,  wouldn't that be at most n-1 enemies... since 1 person is himself. (but that is irrelevant)
So prove that for n people, with each person having [(n+1)/2] friends, he can be seated with friends on both sides.  And it would be sufficient since the problem is the same for even n's?

yea, and it makes the induction really easy

and I forgot to mention, that this is really a famous theorem of Dirac in disguise: http://mathworld.wolfram.com/DiracsTheorem.html


Determine all polynominals $ W\left(x\right)$ with real coefficients such that for every real number $ x$, the equation $ W\left(x^{2}\right)\cdot W\left(x^{3}\right) \equal{} \left(W\left(x\right)\right)^{5}$ holds.

Clearly any polynomial of type $ W(y)\equal{}Ay^{n}$ fulfills the relationship iff $ A\equal{}0,1$ holds.

So from now on assume $ W(y)\equal{}Ay^{n}\plus{}By^{m}\plus{}\mbox{lower terms}$ with $ 0\leq m<n$ and $ AB\not\equal{} 0$. Then

$ (W(x))^{5}\equal{}A^{5}x^{5n}\plus{}5A^{4}x^{4n\plus{}m}\plus{}\mbox{lower terms}$ and

$ W(x^{2})W(x^{3})\equal{}A^{2}x^{5n}\plus{}ABx^{3n\plus{}2m}\plus{}\mbox{lower terms}$, a contradiction to the required identity.

Yep, but it is not a complete proof, I cannot understand it...


When the polynomial $x^3-2$ is divided by the polynomial $x^2-2$, the remainder is

$\textbf{(A)} \ 2 \qquad \textbf{(B)} \ -2 \qquad \textbf{(C)} \ -2x-2 \qquad \textbf{(D)} \ 2x+2 \qquad \textbf{(E)} \ 2x-2$

[hide]$x^3-2=x(x^2-2)+2x-2$ so the remainder is $2x-2$ $\boxed{E}$[/hide]

[hide="Solution"]

I just did long division. I rewrote $x^3 - 2$ as $x^3 - 0x^2 - 0x - 2$.

$(x^2 - 2) \cdot x = x^3 - 2x$.

If you add the remainder, $2x - 2$, to $x^3 - 2x$, you get $x^3 - 2$.

[/hide]

[quote="boxedexe"][hide="Solution"]
Through long division, we see that teh remainder is 2 $\boxed{A}$
[/hide]
Masoud Zargar[/quote]
Your claim: $p(x)=x^3-2=q(x)(x^2-2)+2$. If you double-checked,
$p(\sqrt2)=(\sqrt2)^3-2\ne 0+2$  ;)


Does anyone have suggestions for a Brahms or Liszt piece for piano? I'm fairly advanced.

You might like to try these:
Liszt
1.Les jeux d'eaux a la Villa d'Este from Years of Pilgrimage-Third Year

2.Paraphrase on themes from Rigoletto

Brahms
1.Variations on a theme of Robert Schumann

2.Hungarian Dances in the solo piano version.

Possibly you are coming to these composers for the first time.Possibly you are trying  to select suitable material for a music exam or solo recital.?In each case I think there would be a different selection.In my opinion anything in the Third year of Pilgrimage  is worthy of study and performance but as most of it is introspective almost funereal music quite different from the Hungarian rhapsodies you would need to take this into account if you intended to include them in a recital.

[quote="HilbertThm90"]Brahms - op. 118, no. 2, magnificent![/quote]

One of my favorites. But definitely very demanding musically.

Really fun

Liszt-Sonnetto del Petrarca no. 104.

Anything from the Op. 116 Fantasien, or the 118 intermezzi.  Both rhapsodies are also good, but many prefer G-minor.

Liszt's Mephisto Waltz is a good piece if you're looking for a technique challenge.

Thanks everyone!

I've decided on Brahms' Rhapsody in G minor and Liszt's Liebestraume.

for Lizst, try the transcendental studies or the liebestraume.

[quote="DiscovererOfGUT"]for Lizst, try the transcendental studies [/quote]

I wouldn't think about that just yet. Seriously, some of the Liszt pieces are really technically demanding. Sometimes (this has happened before) people are so obsessed with trying to cram all the notes in that they neglect a lot of musical elements (e.g. expressions, structure etc.).


Please post the theorem of Ava Celin and his proof. thanks

I've never heard of this theorem..
Does anyone actually know what it is?

Possibly some typo, but google doesn't know it either...
Can you give the context or something similar¿


analytic geometry
calculus
derivative
graphing lines
slope

A curve has the equation

$ y^{3} \equal{} x^{3} \plus{} a^{3} \plus{} b^{3}$,

where $ a$ and $ b$ are positive constants. Show that the tangent to the curve at the point $ (\minus{}a, b)$ is

$ b^{2}y \minus{} a^{2}x\equal{} a^{3} \plus{} b^{3}$.

In the case $ a\equal{}1$ and $ b\equal{}2$, show that the $ x\minus{}$coordinates of the points where the tangent meets the curve satisfy

$ 7x^{3} \minus{}3x^{2} \minus{}27x \minus{} 17 \equal{} 0$.

Hence find positive integers $ p, q, r$ and $ s$ such that

$ p^{3} \equal{} q^{3}\plus{}r^{3}\plus{}s^{3}$.

Part 1: By implicit differentiation:

$ 3y^2\frac{dy}{dx}\equal{}3x^2$

$ \frac{dy}{dx}\equal{}\frac{x^2}{y^2}$

When $ x\equal{}\minus{}a, y\equal{}b$, then $ \frac{dy}{dx}\equal{}\frac{a^2}{b^2}$

Now we use the point-slope equation of a line:

$ (y\minus{}b)\equal{}\frac{a^2}{b^2}(x\plus{}a)$

$ b^2(y\minus{}b)\equal{}a^2(x\plus{}a)$

$ b^2y\minus{}a^2x\equal{}a^3\plus{}b^3$.

Part 2: If $ a\equal{}1$ and $ b\equal{}2$, then the equation of the tangent line becomes:

$ 4y\minus{}x\equal{}9\Rightarrow y\equal{}\frac{x\plus{}9}{4}$

Substituting for $ y$ in the cubic equation yields:

$ \left(\frac{x\plus{}9}{4}\right)^3\equal{}x^3\plus{}9$

$ \frac{x^3\plus{}27x^2\plus{}243x\plus{}729}{64}\equal{}x^3\plus{}9$

$ x^3\plus{}27x^2\plus{}243x\plus{}729\equal{}64x^3\plus{}576$

$ 63x^3\minus{}27x^2\minus{}243x\minus{}153\equal{}0$

$ 7x^3\minus{}3x^2\minus{}27x\minus{}17\equal{}0$

Part 3: The above equation can be factored.

$ 7x^3\minus{}3x^2\minus{}27x\minus{}17\equal{}(x\plus{}1)(7x^2\minus{}10x\minus{}17)$

$ 7x^3\minus{}3x^2\minus{}27x\minus{}17\equal{}(x\plus{}1)^2(7x\minus{}17)$

Unfortunately, I'm at a brick wall. I'll come back to this problem after some sleep. By the way, where did you get this problem?

Oh, I see. I was trying to figure out the meaning of the tangent line and of the equation $ 7x^{3}\minus{}3x^{2}\minus{}27x\minus{}17\equal{}0$. I guess all I had to do was factor it and plug it in the equation with the cubed terms.


geometry
USAMO
trigonometry
angle bisector
Law of Cosines

In triangle $\, ABC, \,$ angle $\,A\,$ is twice angle $\,B,\,$ angle $\,C\,$ is obtuse, and the three side lengths $\,a,b,c\,$ are integers.  Determine, with proof, the minimum possible perimeter.

i think that is 77.
I am certain?

let (a,b,c) the sides of the triangle with minimum perimeter.
suppose that gcd(a,b,c) = 1, (if exist p>1 such that a = px, b = py, c = pz, we have that (x,y,z) is other solution with x+y+z < a+b+c).

we have that a² = b²(b+c), because m(A) = 2.m(B).
if gcd(b,c)>1, we have gcd(a,b,c)>1. this implies that gcd(b,c) = 1.
so, gcd(b,b+c) = 1 and b= m², b+c = n² and a = mn.
by the sine law, we have n/m = a/b = 2cos(B).
C = 180 - 3B > 90, that implies B < 30°.
so, sqrt(3) < n/m < 2. is easy to see that for m = 1,2,3 this inequality no have integer solutions.
so, m>=4, n>=7.

the pair (m,n) = (4,7) get the solution (a,b,c) = (28,16,33).  :)

I know this is reviving an old thread, but I don't think that answer's right, and AoPS wiki doesn't have the solution, so here's mine:

After drawing the triangle, also draw the angle bisector of $ \angle A$, and let it intersect $ \overline{BC}$ at $ D$.  Notice that $ \triangle ABC \sim \triangle DAC$, and let $ AD \equal{} x$.  Now from similarity,
\[ \frac {AC}{AD} \equal{} \frac {BC}{AB} \Longleftrightarrow \frac {b}{x} \equal{} \frac {a}{c}\Longrightarrow x \equal{} \frac {bc}{a}
\]
However, from the angle bisector theorem, we have
\[ \frac {AB}{BD} \equal{} \frac {AC}{CD}\Longleftrightarrow BD \equal{} \frac {ac}{b \plus{} c}\text{ and }CD \equal{} \frac {ab}{b \plus{} c}
\]
A few steps were skipped to acheive this result.  We see that $ \triangle DAB$ is isosceles, so $ BD \equal{} AD \equal{} x$, which means
\[ \frac {ac}{b \plus{} c} \equal{} \frac {bc}{a}\Longrightarrow \boxed{a^2 \equal{} b^2 \plus{} bc}
\]
so all side lengths which satisfy the conditions also meet the boxed condition.  We can make a list of all possible values for $ a, b, c$ and impose the condition that $ c$ is the longest side since $ \angle C$ is obtuse.  The first set of values that works is $ (a, b, c) \equal{} (15, 9, 16)$ which yields a perimeter of $ \boxed{40}$.  A thorough check shows that this is the least perimeter, and we can verify that it satisfies the original conditions by using the law of cosines and the fact that $ \cos(2\theta) \equal{} 2\cos^2(\theta) \minus{} 1$ to show that $ \angle A \equal{} 2\angle B$.

EDIT: Somehow kalva says it is 77.  Dang it! :mad:   I forgot to show that <C is obtuse by a^2+b^2<c^2.  Oops :oops:

Here is a solution, but...
[url]http://www.animath.fr/old/tutorat/dossier_02033sol.html[/url]

We can also use calculus ...(maxima and minima) and the answer is 77

[quote=dgreenb801][quote="ZzZzZzZzZzZz"]$ \boxed{a^2 \equal{} b^2 \plus{} bc}$
[/quote]
You can obtain this with the law of sines also.
$ \frac{sin2x}{a}\equal{}\frac{sin x}{b}\equal{}\frac{sin 3x}{c}$
$ \frac{2cosx}{a}\equal{}\frac{1}{b}\equal{}\frac{\minus{}4sin^2x\plus{}3}{c}$
So $ cos x\equal{}\frac{a}{2b}$, so $ sin^2x\equal{}1\minus{}cos^2x\equal{}\frac{4b^2\minus{}a^2}{4b^2}$
Substituting,
$ \frac{\frac{a^2\minus{}4b^2}{b^2}\plus{}3}{c}\equal{}\frac{1}{b}$, so $ a^2\equal{}b^2\plus{}bc$.[/quote]

No one said that $C = 3A$

Note that the obtuse condition reduces to $B < 30^{\circ}$.

[b][color=#f00]Claim:[/color][/b] $a^2=b^2+bc$.
[i]Proof.[/i] By angle bisector theorem and isosceles triangle $\triangle ABD$ we have $AD=BD=\frac{ac}{b+c}$ and since $\triangle ABC \sim \triangle DAC$ setting ratios gives the desired. $\square$

Now note that $\gcd(a,b,c) = 1$, so $b$ and $b+c$ are perfect squares. Let $a=xy, b=x^2, b+c=y^2$. By LoC on angle $B$ we get $$\sqrt2 < \frac{y}{x} < 2 \rightarrow x \geq 4,$$ and checking gives $(x,y)=(4,7)$ as a valid pair, so the minimum perimeter is $77$, achieved at $(a,b,c)=(28,16,33)$. $\blacksquare$

Let the side lengths be $a,b,c$ and let the angle opposite $b$ be $x$ so that the other two angles are $2x$ and $180-3x.$ By Law of Sines, $$a:b:c=\sin(2x):\sin(x):\sin(3x)=2\cos(x):1:3-4\sin^2(x)=2\cos(x):1:4\cos^2(x)-1.$$ Let $p=\cos(x)$. Note that by the obtuse condition, $\sqrt{3}/2<p<1.$ Therefore, $$a:b:c=2p:1:4p^2-1.$$ Let $p=\frac{m}{n}$ for relatively prime positive integers. This means $$a:b:c=\frac{2m}{n}:1:\frac{4m^2-n^2}{n^2}=2mn:n^2:4m^2-n^2.$$

Case 1: $n$ is even. Then, this can be reduced by a factor of 4 (but no further due to relatively prime $m$ and $n$), so we want to minimize $$\frac{1}{4}(2mn+4m^2)=(1/2)m(n+2m).$$ Note that the constraint $\sqrt{3}/2<p<1$ makes it so that $m\geq7, n\geq 8$, so the minimum such value occurs at $m=7,n=8$, which gives $$(a,b,c)=(28,16,33)$$ as the smallest solution giving a perimeter of 77.\\

Case 2: $n$ is odd. Clearly, this is not optimal, since we won't be able to reduce by the factor of 4, and $n=9,m=8$ gives a worse value, so we are done.

"It's very easy to $C$ that bashing is not the *right* way of solving this problem"

Wow that was a horrible joke but whatever...
So we let $B = \alpha, A = 2\alpha, C=180-3\alpha$. By law of sines we have,
$$\frac{b}{\sin \alpha} = \frac{a}{\sin 2\alpha} = \frac{a}{2\sin \alpha \cos \alpha} \Rightarrow b = a(2\cos \alpha)$$
$$\text{and}$$
$$\frac{b}{\sin \alpha} = \frac{c}{\sin(180-3\alpha)} = \frac{c}{\sin 3\alpha} = \frac{c}{\sin \alpha(4\cos^2 \alpha - 1)} \Rightarrow c=a(4\cos^2 \alpha - 1)$$

As $0 < \alpha < 30$, we have that $\sqrt{3}a < b < 2a$. By inspection, a smaller denominator (when reduced) will be most optimal[color=#f00]*[/color]. Hence, we have $b = \frac{7}{4}a$, and, therefore, $c = \frac{33}{16}$. This gives $(a, b, c) = (16, 28, 33)$ and a minimal perimeter of $\boxed{77}$.

[size=75][color=#f00]*Would I have to prove this in-contest?[/color][/size]


Let's say that I have a measurement of 32.5 $ \pm$ 0.05 cm.  I want to divide this number by $ \pi$.  What is the result?  How do I write the plus-minus statement?

Thanks in advance!

Divide both the estimated value and the error by $ \pi$. The $ \pm$ notation merely means a 2-sided inequality and should be treated as such.


Write the numbers $1$ to $15$ in a row such that the sum of every two numbers beside each other is a perfect square.
How many ways there are so we can do such a thing?

[hide] the way i see it, 9 can only be next to 7 as 25-9 = 16, so we have that at one end.  then step by step we can get that the order is  9-7-2-14-11-5-4-12-13-3-6-10-15-1-8  and of course that can be reversed, so I get 2 ways total[/hide]

[hide]both are correct . there is only two ways (one is 8 1 15 10 6 3 13 12 4 5 11 14 2 7 9 and the other the reversion)  . Well the one who wants to solve this problem must know that the only perfect squares that can be seen in the sum of any two numbers ( from 1 to 15) beside each other can be 4,9,16,25 :) [/hide]
Now Can anyone solve this again For $1$ to $35$ ? :D


Can someone provide me a proof for synthetic division? Thanks.

http://en.wikipedia.org/wiki/Synthetic_division

Not a proof but has some good links and info.


The perimeter of a particular square and the circumference of a particular circle are equal. What is the ratio of the area of the square to the area of the circle? Express your answer as a common fraction in terms of π.

[hide] We have that $ 2\pi r\equal{}4s \implies \frac{\pi r}{2}\equal{}s \implies\frac{\pi^2 r^2}{4}\equal{}s^2 \implies \frac{s^2}{\pi r^2}\equal{}\boxed{\frac{\pi}{4}}$.[/hide]


geometry
3D geometry
pyramid
prism
ratio

what make me feel really lame is the fact that the 1st one is for 8th graders.

[b]problem 1[/b]
[img]http://www.math.utah.edu/~alfeld/1050/bb200.gif[/img]

[b]problem 2[/b]
In your summer home you have a fuel tank that is in the shape of a pyramid sitting on its base. Its height is H. The depth of the fuel when the tank is half full equals_________

[hide="HINT"]
Make two equilateral triangular prisms and pyramids, one of the entire shape and one of the empty.  Subtract.
[/hide]

[hide="similar hint"]calculate the volume of the entire prism first..then figure out the height of a smaller equilateral prism within that bigger one such that its volume is half that of the bigger one.. it should be easy because u know the lenght already. [/hide]

[hide]the ratio of the volumes is the cube of the ratio of the sides for similar polyhedrons

actually, the ratio of the areas is the square ofthe ratio of the sides is more useful for this problem.[/hide]

[quote="mihail911"]yeah, jli, i agree, i think i know where you were trying to go with that ratio of volumes thing but u must not forget that the smaller equilateral prism is not similar to the bigger one.  :)[/quote]

i know...thats y i included the area one

so i figured out the answer for the 1st one but i cant get the 2nd problem correct, Can anyone supply me with the answer. Thanks

by the way the answer will depend on h

[hide="hint"]

The volume of a pyramid is (surface area times height) / 3

[/hide]

thanks for the hint but i already know that info, im doing something else wrong. I understand its similar to the other problem when it comes down to how to go about solving it. The answer should end up being all in variables and thats where im getting things wrong. somewhere in combining equations and what not im getting the answer wrong

[hide]say your answer for the first problem is x

then $ \frac{x}{8}\equal{}\frac{A}{H}$

where A is the answer to this problem

so $ A\equal{}\frac{x}{8}H$[/hide]

1. General intro:

For an equilateral triangle we know $ h ={a\sqrt{3}\over 2}$ and $ A ={a^{2}\sqrt{3}\over 4}$. Expressing $ a$ in terms of $ h$, we get $ a ={2h\over\sqrt{3}}\implies A ={h^{2}\over\sqrt{3}}$

The main problem:

If the full height is $ H$, then the volume of the tank is $ {H^{2}\over\sqrt{3}}L$, where $ L$ is length.

Now, let $ x$ be the height of the EMPTY SPACE above the contents when the tank is half full. Since THAT is a triangular prism too, its volume is $ {x^{2}\over\sqrt{3}}L$. But we know this is exactly the half of the total volume, hence

$ {x^{2}\over\sqrt{3}}L ={1\over 2}\cdot{H^{2}\over\sqrt{3}}L\implies x ={H\over\sqrt{2}}$

Then the desired depth is $ h = H-x = H\left(1-{\sqrt{2}\over 2}\right) = 8-4\sqrt{2}$ feet

2. If $ x$ is the height of the "leftover" pyramid (the small pyramid above the contents), then its base has $ \left({x\over H}\right)^{2}$ ratio over the big base. Therefore

${ {1\over 2}\cdot{1\over 3}BH ={1\over 3}\left({x\over H}\right)^{2}Bx\implies x^{3}={H^{3}\over 2}\implies x ={H\over\sqrt [3]{2}}={H\sqrt [3]{4}\over 2}}$

Hence $ h = H-x = H\left(1-{\sqrt [3]{4}\over 2}\right)$ (approximately $ 0.2H$, which might appear unexpectedly low)


Find the area of a triangle with vertices consisting of the centroid, orthocenter, and incenter of a triangle with sides 10,20 and 30

A triangle with sides 10, 20 and 30 isn't a triangle.

[quote="jasonwingofcarmelin"][color=darkred]Find the area of a triangle $GHI$ with vertices  $G$- the centroid, $H$- orthocenter, and $I$- incenter of a triangle $ABC$ with the sidelengths $\{a,b,c\}$.[/color][/quote][hide="Hint."]
$\{a,b,c\}\implies \{p,R,r\}\ ;$

$S=[ABC]\implies 16S^{2}=2\sum b^{2}c^{2}-\sum a^{4}\ ;$

$G\in (HO)\cap (NI)\ : \ HG=2\cdot GO\ ,\ NG=2\cdot GI\ ;$

$[GHI]=\frac{2}{3}\cdot [ION]\ : \ IO^{2}=R^{2}-2Rr\mathrm{\ ;\ }$ $ON=R-2r\ ;\ IN^{2}=p^{2}+5r^{2}-16Rr\ .$[/hide]


calculus
integration
trigonometry
function
algorithm
derivative
calculus computations

Sometimes, you have to accept your function is non-elementary as this one is.

Do you need the indefinite integral, or a definite integral instead (say, may be, the integral from $-\infty$ to zero or something like that?)

djiminez has a point.  If your looking for the value of a definite integral, you can always use Simpsons or Trapezoidal Rule (or of course a calculator) to approximate the solution. Does your problem suggest any bounds?

[quote="page275"]No, there is no bounds at all. I found this on my dad's notebook (when he was a student) in a section noted as "to be solved" :)[/quote]Did he ever try it?

[quote="page275"]$\int{e^x \sqrt{5-\cos{x}}} dx$

I tried integration by parts, but it seems to get more complicated :([/quote]
a very interesting question....
i worked on it, during my english class...yet i got no where...
you're dad is indeed a brave man.... :lol:

[quote="Jimmy"]Sorry, that function doesn't have an elementary antiderivative.[/quote]Can you prove it?

[quote="Jimmy"]Not exactly, I tried the function on Mathematica's Integrator, which uses the Risch Algorithm, which guarantees a solution if it is expressable in elementary functions.  Since it just returned what I typed in, there is no elementary anti-derivative.[/quote]You may want to give a look to this,
[quote="MathWorld"]...The case of algebraic extensions is quite complicated and is therefore not completely implemented in any computer algebra system...[/quote]And it is said by the same people who write Mathematica. Then it is possible it does have an elementary function. If you want the link to the complete paragraph, [url=http://mathworld.wolfram.com/RischAlgorithm.html]here it is[/url]. Either way, thanks for the data, I didn't know that such an algorithm existed.

How could you prove that it doesn't ahve an elementary anti-derivative?

Honestly I don't know, but it just says that, even when the Risch Algorithm is shown to give you an answer, it is too complicated and hasn't been fully implemented.

Does anybody knows how could you prove such a thing?

This link that fedja gave me might help.

http://math.hunter.cuny.edu/ksda/papers/churchill.pdf

has anyone tried,  using infinite series?

$e^x \sqrt{5-\cos{x}}=\sqrt{5}e^x\sqrt{1-\frac{\cos x}{5}}$

Use power serie expansion of $\sqrt{1-u}$

you have to compute  $K_n=\int{e^x(\cos x)^{n}dx}$ with a recurrence  relation

Well, going on the series expansion suggested earlier:

$e^x\, \sqrt{5 - \cos\left(x\right)}$
can be written as this infinite series

$2 + 2\, x + \frac{9\, x^2}{8} + ... $

Hence the integral
$\int\mathrm{e}^x\, \sqrt{5 - \cos\left(x\right)}\;\mathrm{d}x$ can be written as

$2\, x + x^2 + \frac{3\, x^3}{8} + ... $

(Generally if it were $e^x\, \sqrt{a - \cos\left(x\right)}$) the expansion (as correctly suggested earlier) would be

$\sqrt{a - 1} + x\, \sqrt{a - 1} + x^2\, \left(\frac{1}{4\, \sqrt{a - 1}} +\
 \frac{\sqrt{a - 1}}{2}\right) + ... $
hence the integral would be

$x\, \sqrt{a - 1} + \frac{x^2\, \sqrt{a - 1}}{2} + x^3\, \left(\frac{1}{1\
2\, \sqrt{a - 1}} + \frac{\sqrt{a - 1}}{6}\right) + ...$

Can anyone verify this, please?

(Sorry I only did 3 terms, too tedious, but you should get the picture :P)

-Ixeus


[quote="bos1234"]I done the maths comp. for Australia today. This was one of the qns that I can remember!

$ (x^{2}+3x+1)^{x+1}=1$ How many possible soln. are there for x??

1
2
3
4
5

Please HIDE YOUR ANSWER!!!!![/quote]
[hide]I found $ 3$ solutions: $ -3$, $ -1$, and $ 0$. There might be more, but for now, I'll say $ \boxed{3}$.[/hide]

[hide="Answer"]Apply natural logarithms to both sides of the equation to get

$ \ln\left[(x^{2}+3x+1)^{x+1}\right] = 0 \Rightarrow  (x+1) \ln(x^{2}+3x+1) = 0$
$ \Rightarrow  x+1 = 0$ or $ x^{2}+3x+1 = 1$
$ \Rightarrow  x =-1$ or $ x(x+3) = 0$
$ \Rightarrow  x =-1$ or $ x = 0$ or $ x =-3$.

[/hide]

alternate soln.

[hide]$ 1^{n}$ = 1

therefore $ x^{2}+3x+1 = 1$
solving, x = 0 or 3

$ n^{0}= 0$

x+1 = 0
$ x=-1$[/hide]

How about if the question was $ x^{2}-3x+1 = 1$
How many soln. would there for for x now?


Determine all ordered pairs $ (a,b)$ such that $ ab\plus{}a\plus{}b\equal{}a^b$ if $ a,b\in\mathbb{Z}$.

[hide="Hint"] Rewrite as $ (a \plus{} 1)(b \plus{} 1) \equal{} a^b \plus{} 1$.  The rest is casework. [/hide]

[quote="CircleSquared"][quote="Fraenkel"]Im kinda stuck.
[hide="what i know"]
I have been able to deduce three things.

1) $ gcd(a,b) \equal{} a$
2) $ (1,0)$ is a possible solution
3) $ ( \minus{} 1, \minus{} 1)$ is a possible solution.

I dont know what else to do  :([/hide][/quote]

[hide]
How did you get that first one?  It looks important.
[/hide][/quote]

[hide]I looked at $ ab \plus{} a \plus{} b \equal{} a^b$
I noticed it can be written as this-  $ ab \plus{} a \minus{} a^b \equal{} \minus{}b$
Since $ a$ divides the left hand side $ a$ divides the right hand side. [/hide]

[quote="Fraenkel"]$ a$ divides the left hand side[/quote]

Unless $ b \equal{} 0$.  Careful!

[hide="Let me clarify my hint"] Casework on $ a$ will solve the problem easily.  I would be very surprised if there were any solutions for $ a > 2, b > 1$.  The negative solutions are messier but can be deduced similarly. [/hide]

[quote="i_like_pie"]if $ a,b\in\mathbb{Z}$.[/quote]

Some of my math have notation $ \in$ too, so for my practice, yours read:

If $ a,b$ is in or an element of all real numbers. (?)

$ \mathbb{Z}$ refers to the integers, not the real numbers.

I have been able to find 3 solutions so far:

$ (1,0)$

$ (\minus{}1,1)$

$ (\minus{}1,\minus{}1)$

I'm not sure if there are any others.

[quote="Identity"]Please tell me if I'm right[hide]

$ (a \plus{} 1)(b \plus{} 1) \equal{} a^b \plus{} 1$

$ (a \plus{} 1)(b \plus{} 1) \equal{} (a \plus{} 1)(1 \plus{} a \plus{} a^2 \plus{} ... \plus{} a^{b \minus{} 1})$

Then

$ b \plus{} 1 \equal{} 1 \plus{} a \plus{} a^2 \plus{} ... \plus{} a^{b \minus{} 1}$

$ b \equal{} a \plus{} a^2 \plus{} ... \plus{} a^{b \minus{} 1}$

$ ab \equal{} a^2 \plus{} a^3 \plus{} ... \plus{} a^b$

$ ab \minus{} b \equal{} a^b \minus{} a$

$ a \plus{} ab \minus{} b \equal{} a^b$

Hence, by long transitivity, $ a \plus{} ab \minus{} b \equal{} ab \plus{} a \plus{} b$

$ \minus{} b \equal{} b$

$ 2b \equal{} 0$

$ b \equal{} 0$

Giving (1,0)

[/hide]

I don't know how to get the other solution(s) though[/quote]

Very nice! The only thing is, when you divide both sides by a+1, you have to make sure a+1 is not zero, that is, a=-1. This is a seperate case, and gives $ ( \minus{} 1,2k \plus{} 1)$ for all integers k.

[edit]Oh, whoops, I wasn't reading carefully enough I guess. $ a^{b}\plus{}1\neq (a\plus{}1)(a^{b\minus{}1}\plus{}a^{b\minus{}2}\plus{}\ldots\plus{}1)$, the signs in the second factor alternate.

[quote="13375P34K43V312"]

[edit]Oh, whoops, I wasn't reading carefully enough I guess. $ a^{b} \plus{} 1\neq (a \plus{} 1)(a^{b \minus{} 1} \plus{} a^{b \minus{} 2} \plus{} \ldots \plus{} 1)$, the signs in the second factor alternate.[/quote]

Sweet. I did not know you can factor that. That is all I need to solve the problem.

[hide="Solution"]
If $ a \ne 0$.
$ a \mid b$ as I have already shown.

since $ (a\plus{}1)(b\plus{}1)\equal{}a^{b}\plus{}1$
        $ (a\plus{}1)(b\plus{}1)\equal{}(a\plus{}1)(a^{b\minus{}1}\minus{}a^{b\minus{}2}\plus{}.... \plus{}1)$
        $ (b\plus{}1)\equal{}(a^{b\minus{}1}\minus{}a^{b\minus{}2}\plus{}.... \plus{}1)$
        $ b\equal{}a^{b\minus{}1}\minus{}a^{b\minus{}2}\plus{}.... \minus{}a$

Thus $ b \mid a$ iff $ a \ne \minus{}1$ & $ b \ne 0$.

Thus $ (\forall a \ne 0,\minus{}1)$ & $ a\equal{}b$

Since $ a\equal{}b$
$ a^2\plus{}2a\equal{}a^{a}$
$ a^{a\minus{}1}\minus{}a\equal{}2$
the only integer solution is $ \minus{}1$. (is there a way to find this without trial and error?)

The above statement is not true if $ a\equal{}0,a\equal{}\minus{}1$ or$ b\equal{}0, b\equal{}\minus{}1$

Now we simply have to test these points only.

Our solution
$ (\minus{}1,2k\minus{}1)$
$ (1,0)$

Let me know if I made a mistake.
Wooh 3 days on this problem.  :lol:
[/hide]

The problem is that you can factor $ a^b\plus{}1$ only if $ b$ is odd.

[quote="i_like_pie"]@ Fraenkel - Nice solution; I'm pretty sure your answers are correct. :)[/quote]

Uh, did you even read Farenhajt's post?

Uh, notice that $ 2k\minus{}1$ is always odd?

:huh:

$ a \plus{} 1|a^b \plus{} 1$ only if $ b$ is odd. What part of that is unclear? The method doesn't work for even $ b$.

By writing $ a^b\plus{}1\equal{}(a\plus{}1)(\mathrm{the\ rest\ of\ it})$ you implicitly assumed $ b$ is odd. And after you got your initial assumption as a result, you consider that a finished proof?

[quote="Farenhajt"]:huh:

$ a \plus{} 1|a^b \plus{} 1$ only if $ b$ is odd. What part of that is unclear? The method doesn't work for even $ b$.

By writing $ a^b \plus{} 1 \equal{} (a \plus{} 1)(\mathrm{the\ rest\ of\ it})$ you implicitly assumed $ b$ is odd. And after you got your initial assumption as a result, you consider that a finished proof?[/quote]

lol. I just learn you can factor that thing today and also learn you cant. hahaha. Maybe someone can teach me this properly.

Well I guess if b is even then $ a^{b}\plus{}1$ is irreducible in the reals. (Im assuming cause I dont know much,(Im only in the $ 4^{th}$ grade!))

Then I guess that would Imply that $ (a\plus{}1)\equal{}1$ or $ (b\plus{}1)\equal{}1$. And solve that.

No, that just means you must find a whole new method for $ b$ even :)

The fact that $ a \plus{} 1 | a^b \plus{} 1$ proves that $ b$ is odd (or that $ a$ is $ 1$).  Unfortunately, a complete solution has still not been posted in this thread.

[quote="Fraenkel"]$ b \equal{} a^{b \minus{} 1} \minus{} a^{b \minus{} 2} \plus{} .... \minus{} a$

Thus $ b \mid a$[/quote]

You mean $ a | b$.  The way you conclude that $ a \equal{} b$ is a mystery to me.

[hide="Solution"] The original problem can be rewritten

$ (a \plus{} 1)(b \plus{} 1) \equal{} a^b \plus{} 1$

[b]Case:[/b]  $ a \equal{} \minus{} 1$.  We then have a solution whenever $ b$ is odd, so we have the solutions $ \boxed{ ( \minus{} 1, 2k \minus{} 1) }$.

[b]Case:[/b]  $ a \equal{} 1$.  Then $ b \plus{} 1 \equal{} 1$, so we have the solution $ \boxed{ (1, 0) }$.

In every other case, $ a \plus{} 1 | a^b \plus{} 1$, so $ b$ is odd and

$ b \equal{} a^{b \minus{} 1} \minus{} a^{b \minus{} 2} \plus{} ... \minus{} a$

[b]Case:[/b]  $ a \equal{} 0$.  It would appear that $ b \equal{} 0$, but $ 0^0$ is undefined, so there are no solutions in this case.

[b]Case:[/b]  $ a \le \minus{} 2$.  Clearly we cannot have $ b \le 1$.  All of the terms on the RHS of the modified equation are now positive; for $ b \ge 3$, in fact, we have $ b < a^{b \minus{} 1} \minus{} a^{b \minus{} 2} \plus{} \minus{} ... \minus{} a$ by induction, hence there are no solutions in this case.

[b]Case:[/b]  $ a \ge 2$.  Again we cannot have $ b \le 1$.  For $ b \equal{} 3$ there is no solution, and for $ b \ge 5$ we have $ b < a^{b \minus{} 1} \minus{} a^{b \minus{} 2} \plus{} ... \minus{} a$ by induction, hence there are no solutions in this case. [/hide]

I did make a mistake assuming $ b \mid a$.

I have two questions about your solution.

1) Is it possible that $ b \plus{} 1 \mid a^b \plus{} 1$  and  $ a \plus{} 1 \nmid a^b \plus{} 1$ ?

2) What is induction?

[quote="Fraenkel"]1) Is it possible that $ b \plus{} 1 \mid a^b \plus{} 1$  and  $ a \plus{} 1 \nmid a^b \plus{} 1$ ?[/quote]

$ a^b \plus{} 1 \equal{} (a \plus{} 1)(b \plus{} 1)$.  Both $ a \plus{} 1$ and $ b \plus{} 1$ divide $ a^b \plus{} 1$ by definition.

[quote="Fraenkel"]2) What is induction?[/quote]

Induction is a general method for proving that some statement is true for any positive integer $ n$ by proving that it is true for $ 1$, and then proving that [b]if[/b] it is true for $ k$, [b]then[/b] it is true for $ k \plus{} 1$.  We can then conclude that if it is true for $ 1$, it must be true for $ 2$, and $ 3$, and $ 4$, and so on.

A common exercise:  Prove that $ 1 \plus{} 2 \plus{} 3 \plus{} ... \plus{} n \equal{} \frac{n(n\plus{}1)}{2}$.

In this particular case, I am inducting on $ b$.

[hide="Explanation"] When $ a \le \minus{}2$ we wish to prove that for $ b \ge 3$ we have

$ b < a^{b\minus{}1} \minus{} a^{b\minus{}2} \plus{}\minus{}... \minus{} a$

Well, since every term on the RHS is positive we simply have to prove that

$ b < 2^{b\minus{}1} < a^{b\minus{}1}$

For $ b \equal{} 3$ we have $ 3 < 2^2 \le a^2$, which is true.  If the inequality is true for $ b \equal{} 2k\minus{}1$ then

$ 2k\minus{}1 < 2^{2k\minus{}2} \implies$
$ 2k\plus{}1 < 2^{2k\minus{}2} \plus{} 2 < 2^{2k}$

hence the inequality is true for $ b \equal{} 2k\plus{}1$.  We conclude by induction that the inequality is true for all odd $ b$.

Similarly, when $ a \ge 2$ we wish to prove the same statement.  Since the terms actually alternate signs, we now wish to show that

$ b < 2^{b\minus{}2} \le a^{b\minus{}2} < a^{b\minus{}1} \minus{} a^{b\minus{}2} < a^{b\minus{}1} \minus{} a^{b\minus{}2} \plus{}\minus{}... \minus{} a$

When $ b \equal{} 5$ we have $ 5 < 2^3 \le a^{b\minus{}2}$ as desired.  If the inequality is true for $ b \equal{} 2k\minus{}1$ then

$ 2k\minus{}1 < 2^{2k\minus{}3} \implies$
$ 2k\plus{}1 < 2^{2k\minus{}3} \plus{} 2 < 2^{2k\minus{}1}$

hence the inequality is true for $ b \equal{} 2k\plus{}1$.  We conclude by induction that the inequality is true for all odd $ b$.  (It is true for all even $ b$ in this case too, but this is an unnecessary detail.) [/hide]

[quote="Farenhajt"][quote="t0rajir0u"]The fact that $ a \plus{} 1 | a^b \plus{} 1$ proves that $ b$ is odd (or that $ a$ is $ 1$).[/quote]

I would like to see the proof of this (over INTEGERS, not over polynomials).[/quote]

Sorry, the statement is not quite correct.  :oops:

[hide="Corrected statement and finish"] $ a^b \plus{} 1 \equiv (\minus{}1)^b \plus{} 1 \bmod (a \plus{} 1)$

Hence either $ b$ is odd or $ a \plus{} 1 | 2 \implies a \plus{} 1 \equal{} 2, 1, \minus{}1, \minus{}2 \implies a \equal{} 1, 0, \minus{}2, \minus{}3$.

I've handled two of these cases specially; the other two:

[b]Case:[/b]  $ a \equal{} \minus{}2$.  Then

$ \minus{}(b \plus{} 1) \equal{} (\minus{}2)^b \plus{} 1$

Clearly for $ b \le 1$ there is no solution by inspection.  For $ b > 1$ we require that $ b$ is odd and the previous argument holds.

[b]Case:[/b]  $ a \equal{} \minus{}3$.  Then

$ \minus{}2(b \plus{} 1) \equal{} (\minus{}3)^b \plus{} 1$

Again, for $ b \le 1$ there is no solution by inspection and for $ b > 1$ we require that $ b$ is odd and the previous argument holds. [/hide]


function
search
SuMAC
combinatorics unsolved
combinatorics

Along a one-way street there are n parking lots, and n cars numbered 1 to n enter the street in that order. Each driver of the i-th car heads to his favorite parking lot Ai (Ai is one from 1, 2, ..., n), and, if it is free, he occupies it. Otherwise, he continues to the next free lot and occupies it. But if all succeeding lots are occupied, he leaves for good. How many sequences (A1; A2; : : : ; An) are there such that every driver can park?

These are called parking functions (for obvious reasons) -- counting them is nontrivial, and it doesn't look like the full solution has been presented on the forum before (at least not with the word "parking" appearing anywhere in the thread, which is what I searched for).  Here are some threads where it has been discussed:

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1790615272&t=259129
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1790615272&t=221914
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1790615272&t=172305

Thanks for your help.

I actually knew that the answer was $ [n \plus{} 1]^[n \minus{} 1]$
and I was tring to prove that.
I found a recurrent relation:
f(n)=$ suma(C(n \minus{} 1;i)(i \plus{} 1)f(i)f(n \minus{} 1 \minus{} i))$

Can it be solved using generation functions?


At what time do the hour needle ,the minute needle and the secon needle of an analog clock meet in a line?

At 12:00:00 they do. There might be more, though.

Yes, there is more than one solution. 12:00:00 is one.
Second, 1:05:10 etc.

Actually, for the time 1:05:10, all three hands will not be lined up because the if the minute hand is at 5, the hour hand will have moved slightly past the number 1 because of the gears. I think that the only 2 times that work are 12:00:00 AM and 12:00:00 PM.

At twelve o' clock, so once. Otherwise the hands are between the hours hands and don't overlap each other. Glad to help you!  :lol:

there are actually 22, 11 in am, 11 in pm, one example of such is 1:5:12, 2:10:24, and so on, i remember this problem in MC,

The second hand means that less of those solutions work.

The other factor is that the hands being in a line doesn't meet they all point in the same direction.

[hide="hint"]
Using the fact that the angle between the hour and minute hands at $ h$ hours $ m$ minutes is $ |30h\minus{}5.5m|$, set this such that those two hands are on the same line, then just check the second hand.
[/hide]

i remember a similar question but involving only hour and minute hands. i agree with the  last post; probably only 12oclock am and pm work in my opinion

I went to a clock just for fun, it works 22 times in a day, ( I checked the clock every hour)


Ring Theory
calculus
integration
quadratics
algebra
function
domain

How can we check if the ring $\mathbb{Z}(\sqrt{D})$ is UFD?

Is D positive? or negative? or both?

Anyway, for D negative and smaller than $-2$ , it is easy to see that it is not UFD, try to an irreducible element not prime

Both, $D$ is square-free also.

Hmm, but I was sure that $\mathbb{Z}(\sqrt{-3})$ is UFD? For example, the proof of the case $n=3$ of Fermat's Last Theorem which I know uses this fact.

No, the integral clousure of $\mathbb{Z}[\sqrt{-3}]$ is UFD, in your ring you have $4=2 \cdot 2 = (1+\sqrt{-3})(1-\sqrt{-3})$ which are different factorisations.

Sorry, generally I'm not familiar with Superior Algebra but I'm trying to educate :).
When I say $\mathbb{Z}[\sqrt{D}]$ I mean the ring defined on the set:
$a+b\sqrt{D}$ if $D \equiv 2 \mod{4}$ or $D \equiv 3 \mod{4}$
and
$a + b\frac{1+\sqrt{D}}{2}$ if $D \equiv 1 \mod{4}$

I suppose that this is the integral clousure of $\mathbb{Z}[\sqrt{D}]$ you've mentioned. So, what in this case? I mean, when the integral clousure of $\mathbb{Z}[\sqrt{D}]$ is UFD?

Yes thats the integral closure then.
Being UFD should be equivalent to be PID, so to having class number $1$. Thus one could use the quadratic class number formulas... (well, not a very nice way)
For special cases, this could also be done by showing that the ring is euclidean, but that works only for small $D$.

Oh, thanks :)! I think that the following lemma can be helpful:
If $|a| \leq \frac{1}{2}$ and $|b| \leq \frac{1}{2}$ always implies that |$N(a+bK)| < 1$ then integral clousure of $\mathbb{Z}[\sqrt{D}]$ is PID. ($K$ depends on $D \mod {4}$).

I think you are hinting at Euclidean domain
all euclidean domains are pid and thus ufd, but not the converse

if you take D=-19 you will encounter this

For negative squarefree D, you must have D=-1,-2,-3,-7,-11,-19,-43,-67,-163
those are the only ones


Identify compounds A to H  :maybe:  :maybe:

Are you sure the initial compound is correct?

yes.....it is something like benzyl chloride!(?) :maybe:  :maybe:

In your opinion what might be compound A?

And how could that be possible? We have an alkyl halide reacting with a strong, crowded, base. Beta-elimination is impossible here. What other options are there?

Why the difference in acetone and water?

[quote="Carcul"]Solvation in water is far greater.[/quote]

And solvation favours SN1, correct?(I'm not talking abt this rn....in general)

And does solvation(i.e solvent) have any role to play in the competetion between substitution and elimination? :maybe:

And can you ans this ques. (:P)
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=220236

Yes, as far as i know, increase in polarity favours sn1 due to solvation of the nucleophile and hence, decrease in nucleophilicity.

wrt sn and e, i think increase in polarity should favour substitution (like the diff in alc. KOH and aq.KOH)


1. rezolvati ecuatia: ctg[x] * tg{x} +1=0

2. sa se arate ca pt orice n natural, n mai mare sau egal cu 6, un patrat poate fi impartit in n patrate.

3. sa se determine functiile monotone f:R->R a.i.  f(a^x) + 2*f(x)=1, oricare ar fi x din R, a mai mare sau egal cu 2.

[quote="pohoatza"]ca o chestie de divertisment asa, problema 2, este rezolvata pe [b]coperta[/b] cartii de la gil , "inductie matematica".[/quote]

care carte de la gil? ca ie mai multe :D

A zis acolo, "Inductia Matematica" (n-am auzit de ea).

Oricum, ideea e ca putem face dintr-un patrat mai mare patru mai mici, deci e suficient sa demonstram pentru cativa $n$ mici.

Oricum, ideea e ca putem face dintr-un patrat mai mare patru mai mici, deci e suficient sa demonstram pentru cativa $n$ mici.[/quote]

sa zicem ca ai demonstrat pt n=6....sa zicem. dup-aia pasul inductiv cum il faci?

ziceam de "Inductia Matematica" , autori : L & M.E Panaitopol si Mircea Lascu (e una galbena), adevarat $P(n) \rightarrow P(n+3)$, doar pt $n=6,7,8$ e `problema`  :P

am inteles ideea cu inductia si am reusit sa impart patratul in 6 si in 7, insa nu reusesc in 8 :?:  la problmea 1 am obtinut solutia k*pi, k intreg nenul

imparte 2 laturi alaturate in cate 4 parti egale, si du conturu ala de 7 patratele egale, avand unul comun in varful comun laturilor.


inequalities
trigonometry
function
inequalities unsolved

Prove:
$x+y+z\geq x(\frac{\sin y}{\sin x})+y(\frac{\sin z}{\sin y})+z(\frac{\sin x}{\sin z})$
for any $x,y,z$ between $0$ and $\frac{\pi}{2}$

$\left(\frac{x}{sinx}\right)'=\frac{cosx(tanx-x)}{sin^2x}>0,$ where $x\in(0,\frac{\pi}{2})$
Hence, if $x\geq y\geq z$ then $sinx\geq siny\geq sinz$ and $\frac{x}{sinx}\geq\frac{y}{siny}\geq\frac{z}{sinz}.$
Hence, $x+y+z=\frac{x}{sinx}\cdot sinx+\frac{y}{siny}\cdot siny+\frac{z}{sinz}\cdot sinz\geq$
$\geq\frac{x}{sinx}\cdot siny+\frac{y}{siny}\cdot sinz+\frac{z}{sinz}\cdot sinx=x(\frac{\sin y}{\sin x})+y(\frac{\sin z}{\sin y})+z(\frac{\sin x}{\sin z}).$ :)

What about
$x+y+z\geq x(\frac{\sin y+\sin z}{2\sin x})+y(\frac{\sin z+\sin x}{2\sin y})+z(\frac{\sin x+\sin y}{2\sin z})$

Your solution looks very messy, arqady.
This one maybe simpler.

[quote="nguyentrang"]Prove:
$x+y+z\geq x(\frac{\sin y}{\sin x})+y(\frac{\sin z}{\sin y})+z(\frac{\sin x}{\sin z})\ \ (*)$
for any $x,y,z$ between $0$ and $\frac{\pi}{2}$[/quote]
Assume that $x \ge y \ge z$, then:
\[ (*) \iff \left(\frac{x}{\sin x}-\frac{y}{\sin y}\right)(\sin x-\sin y)+\left(\frac{x}{\sin x}-\frac{z}{\sin z}\right)(\sin x-\sin z) \ge 0  \]
And $f(k)=\frac{k}{\sin k}$ is increasing function, isn't it?


dunno if its in the right place but newayz..the problem is
$3!=n!(3-n)!$

find the value of n.

Quite simple...
[hide]
Since we take $(3-n)!$, and the number being factorialed must be nonnegative, we know $(3-n) \geq 0$ so $0 \leq n \leq 4$.
We can just try the values, and knowing $0!=1$, we see that 3 works.[/hide]
I doubt this is intermediate level.

true, thnx , but cant we do it the other way
i mean make n the subject, n=somthing

Well if you want to get really hardcore, you can use the gamma interpolation of the factorial and break it up case-wise, from where you could probably beat it to death with concavity... which is an overkill.

[quote="Klebian"]
Since we take $(3-n)!$, and the number being factorialed must be nonnegative, we know $(3-n) \geq 0$ so $0 \leq n \leq 4$.
I doubt this is intermediate level.[/quote]

Well, first, $3-n\ge 0$ implies that $n\le 3$, not $n\le 4$.

But no, it isn't really the right forum -- you just check a few numbers, and you're done.

[hide]Divide both sides by $n!(3-n)!$
$1=\binom{3}{n}$
Clearly n=0 or 3[/hide]

all the answers are right but i mean something else

$\frac{n!}{b!(n-b)!}=x$

express n in terms of b and x.


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hi i am a new comer .....i want to learn how to solve functional equations..can anyone give me  a good note for solving functional equation??? :maybe:

Please use Search , you can find more problems for your study  :wink: