induction
algebra
polynomial
number theory unsolved
number theory

the sequence of numbers $ u_n$ defined by:
       $  u_n=2^n+3^n$
Prove the existence of $ n_0$ so that $ u_{n_0}$ has exactly 22 distinct prime divisors.

PNT says that this problem is very hard, and we can replace the number 22 by a random positive integer.
If no one have a proof, I will ask PNT.

Well, I give up. Could you please show us PNT's solution? This problem is far too interesting!

[quote="harazi"]Well, I give up. Could you please show us PNT's solution?[/quote]
No, no!
I see this problem first time today!

Ok, stop! Do not post any solution! We will wait, even if I am already tired of trying it. Myth, find a solution, please.

Consider $f(x)$ the number of prime factors of x. Then, look at the sequence $ x_n=3^{3^n}+2^{3^n}$. We have $ x_{n-1}$ divides $x_n$. Moreover, I think it is easy to prove that in fact $f(x_n)>f(x_{n-1})$. But $f(x)$ behaves like $ln ln x$ or something like this, thus $f(x_n)$ varies like $n ln3$. Oh, no, this is just a stupid idea, it won't work. I have to go and sleep and wait for a beautiful solution of this beautiful problem. Sorry for this non-sense. Good night!

well i prove u_n can have as much primes as one wants, but i didnt prove that it could have exsactly k primes for every k.

Lemma: let p be an odd prime. and suppose p^k is the max power of p that divides a-b, k>=1 and if p^r is the max power of p that divides n then p^(r+k) is the max power of p that divides a^n-b^n.(also a and b are coprimes with p).

To prove this lemma we prove by induction that
if a=b(modp^t) then a^p=b^p(modp^(t+1)). (exercise)

and finally we prove that if a=b(modp^r) and (n,p)=1 then a^n=b^n(modp^r) but not mod p^(r+1).

This two simple facts imply the lemma.

Now, Let F(u_n) be the number of primes dividing u_n.
Let p be an odd prime that does not divides u_n, then u_pn is bigger than u_n, but the primes dividing u_n stay being the same with the same exponents in u_pn, so u_pn has new primes....

I just saw this part was very obvious...but maybe the lemma can help somebody to find a solution...

Also Harazi's statement is easily verified...

It seems I have too small chances to solve this problem. So I will glad to see PNT's solution.

Vite, vite, show us PNT's solution! Please... :maybe:  :maybe:

I am very sorry for my poor knowledge but :
Who is PNT? Am I missing something?

I really do not know who is PNT, but he should be damn good to create such a problem.

Well,I am sorry but I can't post the solution now. Please wait for some days, after I meet PNT I will post the solution (maybe it's Monday).
   I think 2 days is a good time for us to find a nice and short solution, let shoot this wolf and finish the hunt!  ;)

I think both PNT and darkmaster can't solve this problem ! PNT's solution is wrong !

I'm very sorry but Nttu is right!
After checking, PNT found a big hole in the middle of his sol. He tried hard to   correct it but it's not sucessful.
Now, the only thing I can do is to say "sorry" because of wasting your time.Please forgive me!

Aww, and we were so anxious to see PNT's solution! :(
Anyways, Myth or harazi or grobber or pbornsztein or Moubi, please solve it! :blush:
It's getting on my nerves this problem grrrrr  :P

[quote="darkmaster"]Again I'm sorry. But for now it is just an "open question", I'm not sure there can be a proof.So trying solving it may be a waste of time. To me, I give up. :cry:[/quote]
hi darkmaster, i don't think PNT'solution is wrong
he created this problem from another problem, so why don't we see his first problem? ;)

We see Master and darkmaster. Is there lightmaster?  :D

[quote="darkmaster"][quote="Master"]hi darkmaster, i don't think PNT'solution is wrong[/quote]
Why do you think that?
He himself told me that his proof was wrong.I believe what he said.[/quote]
Maybe he means that PNT's solution is correct for initial problem which Master called "first problem".

Clearly $3^n+2^n=\prod_{t|n}R_t(-3/2)2^{\varphi(t)}$ where $R_n(x)$ is the n-th cyclotomic polynomial, then I suppose to solve is important to know th behavior in quantity of primes of $R_t(-3/2)2^{\varphi(t)}$ for differents $t$'s, but is a REALLY hard problem.

But I am going to think in this.

Best regards.


Let $ a, b, c$ be positive numbers and $ abc\equal{}1$. Prove that

$ \frac{2}{a\plus{}b\plus{}c}\plus{}\frac{4}{3}\geq1\plus{}\frac{3}{ab\plus{}bc\plus{}ca}\geq\frac{6}{a\plus{}b\plus{}c}$

---------------------
 :wink: Panagiote LIGOURAS

[hide="Left inequality"]
Let $a = x/y$, $b = y/z$, $c = z/x$, $x = \min\{x,y,z\}$, $y = x + u$, $z = x + v$, $u,v \geq 0$ then we have to prove \[u^3 v^3 + (u^4 v + 9 u^3 v^2 - 6 u^2 v^3 + u v^4) x + (u^4 +
    10 u^3 v + 3 u^2 v^2 - 5 u v^3 + v^4) x^2 + (3 u^3 + 9 u^2 v -
    6 u v^2 + 3 v^3) x^3 + (3 u^2 - 3 u v + 3 v^2) x^4 \geq 0\] which is easily checked to be true by checking the non-negativity of the coefficients, which is easy, equality if and only if $u = v = 0$, i.e. $a = b = c = 1$
[/hide]
[hide="right inequality"]
Let $a = x/y$, $b = y/z$, $c = z/x$, $x = \min\{x,y,z\}$, $y = x + u$, $z = x + v$, $u,v \geq 0$ then we have to prove \[u^3 v^3 + (u^4 v - 3 u^3 v^2 + 6 u^2 v^3 + u v^4) x + (u^4 -
    2 u^3 v + 3 u^2 v^2 + 7 u v^3 + v^4) x^2 + (3 u^3 - 3 u^2 v +
    6 u v^2 + 3 v^3) x^3 + (3 u^2 - 3 u v + 3 v^2) x^4 \geq 0\] which is again obvious, equality if and only if $u = v = 0$, i.e. $a = b = c = 1$.
[/hide]


I was just wondering if someone could figure out the number of USAMO qualifiers from last year that didn't qualify this year (not including seniors).  I can't think of any easy way to do it, but if someone could, that would be great (I know in our state, that's only 1 person, and we have like 5 people from last year that could do it this year).

if you're that curious you could just look at the qualifier lists...

I see 116 from my grep-fu.

Edit: Wow. That's pretty large.

Edit 2:
06-07: 116 (310 total non-senior qualifiers)
05-06: 52 (180)
04-05: 76 (176)

I think theres one person from Indiana that didnt..

[quote="stupidkid"][quote="worthawholebean"]I see 116 from my grep-fu.

Edit: Wow. That's pretty large.[/quote]
o.O, Linux user? What distro?[/quote]
Mac, not Linux.

[quote="worthawholebean"]I see 116 from my grep-fu.

Edit: Wow. That's pretty large.[/quote]

Do you think you could do this for 2005 to 2006, and 2004 to 2005 too?

Thanks.

Sure, when I get back from dinner. I'll set up some sort of automation.

[quote="kstan013"]I'm sure many of them were sophomores last year and juniors this year[/quote]Heh, that's me! 9 on AIME last year, 7 this year.

[quote="Teki-Teki"]this is kinda sad, considering the index fell about 20 points.  i mean, my index fell by 3.5 but that's b/c i took the 12 as a sophomore.  but seriously, with indices this low i coulda almost qualified in 8th grade with a 192.  strange year.[/quote]

Is it mostly the expansion, or was this one supposedly harder?

It annoys me to think I missed it in 9th grade with 218/8...

1.  I think this can be attributed to the expansion, most likely.
2. I agree with kstan013 that there are consistently over 20% of non-seniors that do not qualify the following year because many of them go from 10th to 11th.


why is the font so small and why won't it let me type words with latex, I tried a semicolon...

how do I change the size of the font because I could only make the title big, not the body of the paper.

How do I type words with LaTeX in TeXsHoP? I typed a semicolon but in the typeset, it showed the semicolon and the words were the same. how do I type words in latex, which are bigger and blacker??

Thanks!  :blush:

[b]Complete[/b] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157945]minimal example[/url] is required to explain what you mean.

As he said, it is hard to tell what you want.  Some Latex things maybe related to what you want...

\textbf{blacker text}

\section{Larger Size Section Heading}

\$\big(x+(y+z)\big)\$ bigger parentheses

\documentclass[12pt]{article} specify 12 point font


Okay hurt=-2 and heal=+1
The only reason why hurt is minus 2 instead of one, is because otherwise it could last forever.

Math - 21
Reading - 18
English - 20
Science - 20
Technology/Computers - 20
Foreign Language - 20
P.E. - 20

There's a freakin poll on this - go use the search button and stop hurt n' heal everything.


calculus
integration
function
number theory proposed
number theory

Prove that the number of nonnegative integral solutions of the equation x + 2y + 3z = n is the nearest integer to [(n + 3)^2]/12.

A quite boring but efficient way : Just look at the coefficient of $x^n$ in the generating function
$f(x) = (1+x+x^2 + \cdots )(1+x^2+x^4 + \cdots )(1+x^3+x^6 + \cdots) = \frac {Q(x)} {(1-x^6)^3},$
where $Q(x) = x^{12}+x^{11}+2x^{10}+3x^9+4x^8+5x^7+4x^6+5x^5+4x^4+3x^3+2x^2+x+1.$

Pierre.


I am looking for a good book in Enumerative Combinatorics.
Any suggestion?  :)

Enumerative Combinatorics, Vols I,II by Stanley.  They are cheap, too! ($ 20 and$35 on amazon, respectively)

The price of the second volume increased a lot in just ONE DAY!


A circle of radius $ 10$ inches has its center at the vertex $ C$ of an equilateral triangle $ ABC$ and passes through the other two vertices. The side $ AC$ extended through $ C$ intersects the circle at $ D$. The number of degrees of angle $ ADB$ is:

$ \textbf{(A)}\ 15 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 60 \qquad\textbf{(D)}\ 90 \qquad\textbf{(E)}\ 120$

Since ABC is an equilateral triangle, we know that angle DAB is 60. We also know that AD is a diameter since line CA, which contained the center, was extended to another point on the circle. Therefore triangle ABD must be an inscribed rght triangle with angle ABD is 90 degrees. Since the sum of all angles in a triangle is 180, angle ADB is 180 - 60 - 90 = 30 (B)


suppose p is prime
1) show that if p divides a^3 then p divides a
2) show that if p divides b and p divides a 2 + b 2 , then p divides a

1) Proof by contradiction. Assume temporarily p does not divide a, or in other words, p is not a factor of a. Since a^3 has the same factors as a, adding no new ones, p can't be a factor of a^3. Since we know p must be a factor of a^3, we have a contradiction and p must divide a.
2) If p divides b, p divides b^2. So, if p divides a^2 + b^2, since it divides b^2, it must also divide a^2. Since p is prime, if it is a factor of a^2, it must be a factor of a.


John has an ample supply of 10-cent and 15-cent stamps. He will buy an ample supply of stamps of one more denomination, which is a whole number. He needs to be able to combine the stamps to make any value of 30 cents or more. In cents, what is the greatest denomination of stamps he can buy which will enable him to do this?

I know how to do this problem after some work, but I'm not sure if there was a formula for this type of problem. Is there?

This isn't a direct answer to your question, but if you have stamps of denomination a and stamps of denomination b where a and b are relatively prime, then the largest value you cannot make with those stamps is [hide]ab-a-b[/hide]

thanks for refreshing my memory. knowing me, i'll probably forget it again before nationals


group theory
abstract algebra
superior algebra
superior algebra unsolved

If $ G$ is a commutative group prove that $ G$ is cyclic if and only if for any prime divisor $ p$ of $ |G|$ the equation $ x^{p}\equal{}1$ has at most $ p$ solutions in $ G$.

It immediately follows $ x^p\equal{}a$ has at most $ p$ solutions (consider the quotients of the solutions) and $ x^q\equal{}1$ has at most $ q$ soltions for any prime power $ q$. Thus all cyclic subgroups of order a prime power are normal in $ G$ and hence, $ G$ is abelian.


We can simplify the right side as $ 6(6^x) \implies 6^{x \plus{} 1}$. The left side is $ 6^3 \times 6^6 \implies 6^9$, so we have $ 6^9 \equal{} 6^{x \plus{} 1}$. Therefore, $ x \plus{} 1 \equal{} 9 \implies x \equal{} 8$.


prove that there are infinitely many positive integers n such that n(n+1) can be expressed as a sum of two positive squares in atleast two different ways (a^2+b^2 and b^2 + a^2 is considered as same way)


Let $H$ be the orthocenter of an acute triangle $ABC$.  The tangents from $A$ to the circle with diameter $BC$ touch the circle at $P$ and $Q$.  Prove that the points $P$, $Q$, and $H$ are collinear.

[color=red][Moderator edit: This problem has also been discussed on http://www.mathlinks.ro/Forum/viewtopic.php?t=5850 .][/color]

This is a classic one-liner with polar geometry.

Let $M$ and $N$ be the feet of altitude from $B$ and $C$ respectively. Since $A$ and $H$ are two diagonal points respect to the inscribed quad $BCMN$, they are conjugate points, it follows that $H$ lies on the pole of $A$, which is $PQ$.

Call O, I are the midpoint of BC, PQ, respectively; D is the foot of the altitude . We have:
$AQ^2 = AN\cdot AB = AH\cdot AD$
Moreover $AQ^2 = AI\cdot AO$
=> $AI\cdot AO = AH\cdot AD$
=> the quadrilateral HIOD is inscribed.
But <HDO = 90°
=> HIO = 90°
=> HI // PQ
=> H lies on PQ.

I have another solution for this problem
We call O as the midpoint of BC, T as the projection of H on AO.
We can easily prove OT.OA = 1/4.BC^2
vector(OH).vector(OA) = 1/4 BC^2
=> OA.OT = 1/4 BC^2
PQ cuts OA at E
=> OE.OA = OP^2 = 1/4 BC^2
=> OE = OT => P, H, T are collinear.
Similarly, we can prove Q, H, T are collinear.
So that H, P, Q are collinear.

Sorry about reviving such an old thread, but does this synthetic solution work? (I don't understand Huynh Anh Hao's solution)

Let $X$ and $Y$ be where $AB$ and $AC$ meet the circle; let $D$ be the foot of the altitude from $A$ to $BC$; let $O$ be the midpoint of $BC$.  Clearly, $APOQ$ is cyclic, and since $\angle ADO$ is right, the points $A,P,D,O,Q$ are concyclic.  This implies $\angle ADP = \angle AQP$.  Now, note that $AP^2=AX\cdot AB = AH\cdot AD$, since $XHDB$ is cyclic.  It follows that $\triangle APH \sim \triangle ADP$, so $\angle APH = \angle ADP = \angle AQP = \angle APQ$, since $AP=AQ$, and this implies the result.

[edited]

[quote="ThAzN1"](I don't understand Huynh Anh Hao's solution)[/quote]

What exactly don't you understand about Huynh Anh Hao's solution? Maybe you don't understand why $AQ^2=AI\cdot AO$ ? Well, this can be explained as follows:

The point O is the midpoint of the segment BC. Thus, the point O is the center of the circle with diameter BC. The points P and Q are the points where the two tangents from the point A to this circle touch this circle. Thus, the points P and Q are symmetric to each other with respect to the line AO. Consequently, we have $PQ\perp AO$, and the midpoint I of the segment PQ lies on the line AO.

Now, since $PQ\perp AO$, we have < AIQ = 90°; on the other hand, since a tangent to a circle at a point is perpendicular to the radius through this point, we have $AQ\perp OQ$, and thus, < AQO = 90°. Thus, < AIQ = < AQO. Also, trivially, < IAQ = < QAO. Thus, the two triangles AIQ and AQO are similar; it follows that $\frac{AQ}{AI}=\frac{AO}{AQ}$; in other words, $AQ^2=AI\cdot AO$. And we are done.

As for your solution, it is correct and I find it very nice. Only at one point I have an objection:

[quote="ThAzN1"]Clearly, $APDQ$ is cyclic, and since $\angle ADO$ is right, the points $A,P,D,O,Q$ are concyclic.[/quote]

This reasoning is incorrectly formulated. I guess you wanted to say the following:

Since < ADO = 90° (this is because the line AD is an altitude in triangle ABC) and < APO = 90° and < AQO = 90° (this is because the lines AP and AQ are tangents to the circle with diameter BC, and O is the center of this circle), the points D, P and Q lie on the circle with diameter AO. Thus, the points A, P, D, O, Q are concyclic.

[b]PS.[/b] Sorry for the stupid question, but I remember having met, on the way to the IMO 2004, in the Frankfurt airport, an US participant with a very strange ML nickname which I couldn't force myself to memorize. This was in July, just a few days after the ML-AoPS merge, and the two communities were still quite separated from each other, so that name didn't say anything to me. Now, is it possible that this participant was you? Excuse me again for the question and for my forgetfulness  :) .

  darij

[quote="darij grinberg"]
[b]PS.[/b] Sorry for the stupid question, but I remember having met, on the way to the IMO 2004, in the Frankfurt airport, an US participant with a very strange ML nickname which I couldn't force myself to memorize. This was in July, just a few days after the ML-AoPS merge, and the two communities were still quite separated from each other, so that name didn't say anything to me. Now, is it possible that this participant was you? Excuse me again for the question and for my forgetfulness  :) .

  darij[/quote]

Heh, I wish I was on the IMO team...  :)   Yeah, it wasn't me.  Anyways, thanks for your feedback!

[edit] Oh yeah, I think I meant that $APOQ$ is cyclic, and $\angle APO = \angle ADO = 90^{\circ}$.


I'm not sure if this works (wasn't sure about the meaning of "pairwise different" here), but here's my solution.

[hide="Solution"]Lemma: $\sum_{i=1}^{k}i^{i}<(k+1)^{k+1}$ for all natural numbers $k$.

Proof: We will use induction.  For $k=1$, the inequality gives $1^{1}<2^{2}$, which is obviously true.  Now assume the inequality holds for $k$.  Notice that $2(k+1)^{k+1}<(k+2)^{k+2}$.  Rearranging gives $(k+1)^{k+1}<(k+2)^{k+2}-(k+1)^{k+1}$.  Adding this to the inequality we assumed to be true for $k$, we get $\sum_{i=1}^{k+1}i^{i}<(k+2)^{k+2}$, so the inequality is true for $k+1$.  The induction is complete, and the lemma is proved.

We will use induction on $n$ to prove that $\{x_{1},x_{2},\ldots ,x_{n}\}=\{y_{1},y_{2},\ldots ,y_{n}\}$.  For $n=1$, we have $x_{1}^{x_{1}}=y_{1}^{y_{1}}$.  Since the function $f(x)=x^{x}$ is monotonically increasing for positive $x$, we have $x_{1}=y_{1}$ as desired.  Now suppose that the problem statement is true for $n=k-1$, and consider $n=k$.

Assume WLOG that the largest of the $2k$ numbers is $x_{k}$, so that $x_{k}$ is greater than or equal to every other element.  Suppose none of the $y_{i}$ are equal to $x_{k}$.  Then the left side of the equation is greater than $x_{k}^{x_{k}}$, and the right side of the equation is at most $\sum_{i=1}^{x_{k}-1}i^{i}$, which is less than $(x_{k})^{x_{k}}$ by our lemma.  Therefore, one of the $y_{i}$ is equal to $x_{k}$.  WLOG, let $x_{k}=y_{k}$.  Then the $x_{k}^{x_{k}}$ and $y_{k}^{y_{k}}$ terms cancel, leaving us with the problem statement for $n=k-1$, which we assumed to be true in our inductive hypothesis.  The induction is complete, so for all $n$, we have $\{x_{1},x_{2},\ldots ,x_{n}\}=\{y_{1},y_{2},\ldots ,y_{n}\}$ in the given equation.[/hide]

[quote="N.T.TUAN"]i.e $x_{i}\not = x_{j}\forall i\not = j$ and $y_{i}\not =y_{j}\forall i\not = j$.[/quote]

OK, that's what I assumed it meant.  :)


Prove that for any whole number $k$, numbers $k$ and $k^{5}$ always finish with the same number (number them units).

[quote="mathmanman"]http://www.mathlinks.ro/Forum/viewtopic.php?t=48725[/quote]
sorry and thank you!

I have another proof  :lol: ,different from the others.
If $(k,5)=1$ then by Fermat's Little Theorem $\rightarrow$ $k^{5}-k \equiv 0 (mod 5) (1)$  .
But $k^{5}-k=k(k+1)(k-1)(k^{2}+1)\rightarrow k^{5}-k\equiv 0 (mod 2) (2)$ .

From (1) and (2) $\rightarrow$ $k^{5}-k \equiv 0 (mod 10)$
If $(k,5)\neq 1$ then the conclusion it's obvious.


function
topology
advanced fields
advanced fields unsolved

If the statement is true then provide a proof.  If the statement is false then provide a counterexample.

A function f: R ---->R is continous if the preimage of f(X) is closed for every closed set X in R.

(R = real numbers)

The following are equivalent: $ f: (M,d)\rightarrow (N,\rho)$ is continuous on $ M$ and if $ E$ is closed in $ N$, then $ f^{\minus{}1}(E)$ is closed in $ M$.

Given $ x\in M$ and $ \varepsilon>0$, the set $ B_{\varepsilon}^{\rho}(f(x))$ is open in $ (N,\rho)$ and so the set $ f^{\minus{}1}(B_{\varepsilon}^{\rho}(f(x)))$ is open in $ (M,d)$.  But then $ B_{\delta}^{d}(x)\subset f^{\minus{}1}(B_{\varepsilon}^{\rho}(f(x)))$, for some $ \delta>0$, because $ x\in f^{\minus{}1}(B_{\varepsilon}^{\rho}(f(x)))$.  Now just apply complements to $ f^{\minus{}1}$.


Prove that $ \mathbb{Q}^{*}\equal{}\left\{\frac{p}{2q}: p,q\in\mathbb{Z},p\hbox{ is odd, }\gcd\left(p,q\right)\equal{}1\right\}$ is a dense subset of $ \mathbb{R}$.

Take $ s_n\equal{}\frac{2^na_n\plus{}1}{2^nb_n}\equal{} \frac {a_n}{b_n}\plus{}\frac 1{2^nb_n}$ with $ a_n,b_n$ integers . Manifestly $ s_n \in Q^*$
$ Q$(rational numbers set) is a dense subset of $ R$ , so for all $ x \in R$  there is a rational sequence $ \frac {a_n}{b_n}$ convergent to $ x$.
But $ \frac 1{2^nb_n}$ is convergent to $ 0$, so there is a sequence $ s_n$ of $ Q^*$  convergent to $ x$ , And so on ..
 :cool:


My school does not participate or even know of the AMC/AIME competitions.  :dry:
 :)
Congrats though.

azjps, Go Team Dave!   Good times.

I'm from South Jersey, I think that identifies me.

that person Andy Zhu seems really smart, does anyone here know him or know who he is?? :roll:  [size=59]i am kidding for all of you gullible ppl :rotfl: [/size]

This is 6 months late, but I just noticed this topic :) I'm on AoPS, and I was a USAMO qualifier in April :)

[quote="azjps"][url=http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2008-ua/2008usamoqual.shtml#nj]List of qualifies in state[/url].

Congratulations to all who made it!  :)  Anyone here on AoPS? (I recognized a couple of names) By the way, I am the 11th grader from Edison.[/quote]
Edison? I used to live there, do you go to Edison High School or J.P.Stevens?

1/2 year bump. Congratulations.

Please stop reviving such incredibly old threads. As you probably are/should be aware, AoPSers are only allowed to post on threads in which the most recent post has been with in the last month. This is almost your 10th offense. Please mind the rules, or you might be banned.

[quote="filetmignon821"]its not andy zhu its Alex Zhu.[/quote]
Perhaps I did not make USAMO in 2008.


[quote="weyming001"]The square of the number 12345678 is an [b][i]m[/i][/b]-digit number. The value of [b][i]m[/i][/b] is[/quote]

I prefer this solution...

$ 10^7<12345678<13\cdot10^6$
$ 10^{14}<12345678^2<169\cdot10^{12}$
Notice that the LHS has 15 digits, as does the RHS, so $ m\equal{}15$. This problem didn't really need bounds, but this works for more complicated problems of this sort.

[quote="weyming001"]The square of the number 12345678 is an [b][i]m[/i][/b]-digit number. The value of [b][i]m[/i][/b] is[/quote]

My way is kinda tricky, but I think its still okay to understand.  What I do is look at it the number of digits this number has--which is 8.  The square of a number will, with the number of digits it has $ n$, be $ 2n$ or $ 2n\minus{}1$.  So $ m$ has to be 16 or 15.  It'll be the higher one of the number was around 30000000, but below that would be 15 (there are 8 digits, the if the first one is less than 3, than its the lower one, of it's more than three, then its the higher one.  If its on three, then this way is not very accurate).  So $ m\equal{}15$.  I don't know if this way works every time, but in this situation, it works.

It does, because 3^2=9.  If you get something where the first digit is 3 you shoudl check.  4-9 will be 2n.  1-2 will be 2n-1

[quote="Poincare"]i don't understand the wavy symbol :([/quote]

Here, the wavy symbol stands for hand-waving approximations ;).

It basically means that when we have $ 7.\sim $, we don't care about anything after the decimal point, whatever it may be.

thank you yongyi that makes matters sooo much more clear :)


im having troubles with my probability math homework. particularly one question. maybe you can help me out?

a high school is considering forming a band to play at the next city's parade. a total of 33 boys and 22 girls show up at the music room of the high school to audition. the high school band, however, will be made up of just 10 students and of these 10 students, 5 must be boys and 5 must be girls.

now the question to this is - if you are a girl and attended the audition, what is the probability you will be selected to play in the school band?

[hide="Tentative solution"]
The number of ways to choose 5 boys and 5 girls is $ \binom{33}{5}\binom{22}{5}$

The number of ways to choose them assuming you are already chosen would be $ \binom{33}{5}\binom{21}{4}$.

Therefore, the probability is $ \frac{21!17!5!}{22!17!4!}\equal{}\boxed{\frac{5}{22}}$

I can't see anything wrong with the logic here, but the information involving the boys is not used in the solution, which is weird. However, I don't see why the number of boys or the proportion of boys would affect the girls' probability.
[/hide]

That sounds correct.
It really shouldn't matter how many boys there are.

The girls are who you are competing against for 5 positions, so you don't have to worry anything at all about the boys competing for their 5 spots.

Yes, that's what I figured, but it's always disconcerting to be given information that isn't used in the problem.

The quickest way is to note that there are 22 girls and 5 slots, therefore a $ \frac{5}{22}$ chance of being selected.


What is the least integer such that the sum of the integer and its reciprocal is greater than $ \minus{}6$?

[quote="GameBot"]What is the least integer such that the sum of the integer and its reciprocal is greater than $ \minus{} 6$?[/quote]

$ x\plus{}\frac{1}{x} > \minus{}6$
$ x^2\plus{}6x\plus{}1 < 0$ (since x is negative)

$ x\equal{}\minus{}5$, since $ x\equal{}\minus{}6$ doesn't work.


Given $1962$ -digit number. It is divisible by $9$. Let $x$ be the sum of its digits. Let the sum of the digits of $x$ be $y$. Let the sum of the digits of $y$ be $z$.
Find $z$.

$z=9$

we know that $x\leq 1962 \times 9 <1999$ so that $y\leq 1+4\times 9=37$ then $x\leq 9$  :D


geometry
3D geometry
tetrahedron
algebra
function
domain

In $ xyz$ space, 4 points $ A(1,\ 1,\ 2),\ B(1,\ 0,\ 4),\ C(0,\ 1,\ 7), D(2,\ 2,\ 5)$. Draw the perpendicular from $ A$ to the plane determined by 3 points $ B,\ C,\ D$, denote the foot of perpendicular by $ H$.

(1) Find the area of $ \triangle{BCD}$.

(2) Find the area $ S$ of the domain in which the point $ P$ moves , expressed by $ \overrightarrow{BP} \equal{} \alpha \overrightarrow {BC} \plus{} \beta \overrightarrow{BD},\ 0\leq \alpha ,\ 0\leq \beta ,\ 1\leq \alpha \plus{} \beta \leq 2$.

(3) Find the value of $ s,\ t$ such that $ \overrightarrow{BH} \equal{} s\overrightarrow {BC} \plus{} t\overrightarrow{BD}$.

(4) Find the volume of tetrahedron $ ABCD$.

[quote="kunny"]In $ xyz$ space, 4 points $ A(1,\ 1,\ 2),\ B(1,\ 0,\ 4),\ C(0,\ 1,\ 7), D(2,\ 2,\ 5)$. Draw the perpendicular from $ A$ to the plane determined by 3 points $ B,\ C,\ D$, denote the foot of perpendicular by $ H$.

(1) Find the area of $ \triangle{BCD}$.

(2) Find the area $ S$ of the domain in which the point $ P$ moves , expressed by $ \overrightarrow{BP} = \alpha \overrightarrow {BC} + \beta \overrightarrow{BD},\ 0\leq \alpha ,\ 0\leq \beta ,\ 1\leq \alpha + \beta \leq 2$.

(3) Find the value of $ s,\ t$ such that $ \overrightarrow{BH} = s\overrightarrow {BC} + t\overrightarrow{BD}$.

(4) Find the volume of tetrahedron $ ABCD$.[/quote]

(1)
$ \overrightarrow{BC} = ( - 1,1,3)$
$ \overrightarrow{BD} = (1,2,1)$
$ \overrightarrow{BC}・\overrightarrow{BD} = 4$
より
$ \triangle{BCD} = \frac {1}{2}\sqrt {|\overrightarrow{BC}|^2|\overrightarrow{BD}|^2 - (\overrightarrow{BC}\cdot \overrightarrow{BD})^2 } = \frac {5}{2}\sqrt {2}$

(2)
点PはBC,BDを２：１に外分するC’、B'によってできる台形BB'C'Cに存在するのでその面積Sは
$ S = 3\triangle{BCD}$
 $ = \frac {15}{2}\sqrt {2}$
(3)
$ \overrightarrow{OH} = (a,b,c)$
と置くと
$ \overrightarrow{BH} = (a - 1,b,c - 4) = ( - s + t,s + 2t,3s + t)$
$ \overrightarrow{AH} = (a - 1,b - 1,c - 2)$
$ \overrightarrow{AH}\cdot\overrightarrow{BC} = 0$
$ \overrightarrow{AH}\cdot\overrightarrow{BD} = 0$
よりこれらを整理すると
$ s = - \frac{18}{25}$
$ t = \frac{12}{25}$

(4)
$ |\overrightarrow{AH}| = \frac {\sqrt61}{5}$
より
ABCDの体積Vは
$ V = \frac {|\overrightarrow{AH}|S}{3}$
 $ = \frac {\sqrt122}{6}$

Your answer to (3), (4) is incorrect. :(


function
probability
real analysis
real analysis solved

Given $E\in{\mathbb{R}}$ countable set.Give an example of an increasing function $f:{\mathbb{R}}{\rightarrow}{\mathbb{R}}$ s.t. the set of points of discontinuity of $f$ is $E$.

let $u_n > 0$ be a seires such that $\sum_{1}^{\infty} u_n < +\infty$
suppose that $E=(e_1,e_2,..,e_k,..)$
$f(x)=\sum_{i | e_i<x} u_i$ seems to work ?

It may be just me and the late hour at which I'm writing this (:)), alekk, but it looks to me like that function is continuous everywhere if $(e_n)_n$ is dense.

I was thinking we could do this: for each $n$, find an increasing function $f_n$ which is discontinuous only in $e_n$ ($e_n$ being, just like in the post above, an enumeration of $E$), and s.t. the series $\sum f_n(x)$ converges uniformly.

Does it work?

alekk's answer is of that form, grobber. His $f_n$ are step functions.

Yes, I can see where I went wrong: his function is continuous to the left in all points, and when I checked it I only looked at that, without realizing that it's not continuous to the righ in any of the $e_i$. Sorry, everybody.. :)

the example alekk gave induce us to assert the same fact with distibution function(because for descrete probability variable it's very close to the alekk example):
For every countable set $E\in\mathbb{R}$ there exists descrete probability variable, which has the distibution function with $E$ as it's discontinuities points


Mr and Mrs Jones went to a party with four other couples. Some of the people shook hands when they met, but no partners shook hands with each other, no two people shook hands more than once and no-one shook hands with himself or herself. Later, Mr Jones asked everyone else how many times they had shaken hands and received a different answer from each person. How many times did Mrs Jones shake hands?

Let me try
All 6 persons = C(10,2)
minus 5 times( between couples)
=45-5= 40 times
correct me if it is wrong!

No two shook hands more than once.

Since Mr Jones got different answers from $ 9$ people, those different answers must have been $ 0, 1, 2, 3, 4, 5, 6, 7$ and $ 8$.

How can you determine how many times Mrs. Jones shook hands?  Couldn't it be any of the answers AndrewTom listed?

How many times did Mrs Jones shake hands?
that means they do not have to shake hand to each other.
0,1,2,3,4,5,7,8,
Mrs Jones must be the 8 times

Start with the person who shook hands $ 8$ times, who must be married to the person who shook hands $ 0$ times. Then consider the person who shook hands $ 7$ times. ...


geometry
3D geometry
calculus
integration

Consider any three consecutive positive integers. Prove that the cube of the largest cannot be the sum of the cubes of the other two.

Fermat's Last Theorem, done.

Where do you get all these problems?

EDIT: Real solution:

[hide]

Let the three positive integers be x-1, x, and x+1; we wish to show there is no such integer x for which $ (x-1)^{3}+x^{3}=(x+1)^{3}$.

That equation implies $ x^{3}-6x^{2}-2=0=F(x)$. Clearly F(x)>0 for all integral x>6. We can also immediately eliminate odd x because $ x^{3}$ is the only odd term, and a quick check of x=2,4, or 6 proves there is no solution.

[/hide]

I get the problems from all over the place, and some(very few) I make up myself.

[quote="bthings_2000"]Fermat's Last Theorem, done.[/quote]

Oh, I at least expect to hear "the elliptic curve $ y^{2}= x(x-a^{3})(x+(a+1)^{3})$ is not modular by Ribet's theorem, contradiction" from a joke solution :P


Take the number $111111111...1$ where the number has $1998$
digits, all of them being $1$ (The number is denoted in decimal notation).

Now consider the square root of this number.

What is the thousandth digit after the decimal point in this number?

[quote="lordWings"][hide="Hint"][hide="Taylor series"]How about this?
With $n$ digits,
$f(n)=\sqrt{\underbrace{1\dots 1}_{n}}=\sqrt{\frac{10^{n}-1}{10-1}}=\frac{10^{n/2}}{3}\sqrt{1-10^{-n}}$
Using Taylor series,
$f(n)=\frac{10^{n/2}}{3}(1-\frac{1}{2}10^{-n}-\frac{1}{8}10^{-2n}+\dots)$
Therefore
$3f(2m)=\underbrace{9\dots 9}_{m}.\underbrace{9\dots 9}_{m}4\underbrace{9\dots9}_{2m}\dots$
and
$f(2m)=\underbrace{3\dots3}_{m}.\underbrace{3\dots3}_{m}1\underbrace{6\dots6}_{2m}\dots$
Then the $1000$-th digit of $f(1998)$ would be a $1$.[/hide][/hide][/quote]

You can also estimate $1-10^{-n}$ to be $1-2\frac{1}{2*10^{n}}+\frac{1}{4*10^{2n}}=(1-\frac{1}{2*10^{n}})^{2}$  to avoid using Taylor series.

This works in this case if because the inaccuracy comes in far beyond the 1000th decimal digit. How can you solve this problem rigorously without Taylor series?

[hide="Well, this is kinda like"] Taking the square root of $\frac{10^{2n}}{9}$ [/hide] [hide="Which is"] $\frac{10^{n}}{3}$ [/hide] [hide="Which suggests that"] The first $1998$ digits of the result are just $3$s.  Now, there are going to be $999$ digits before the decimal point, so the first $999$ digits after the decimal point are going to be $3$s.  Because $\sqrt{10}\approx 3.16$ we're tempted to conclude that the digit after the threes is probably not $2$, since it would probably make the square too large (very rigorous, I know :P), so it's [b]probably[/b] $1$.  :D [/hide]

Nothing new, (Taylor's series is easy enought ) ...  but one way to think

You want to find out $\frac{\sqrt{10^{1998}-1}}{ 3}$

Assume it is $\frac{10^{999}-a 10^{-999}}{3}$
Easy to see  that  $a>0$  [hide="because"](Because $10^{1998}>10^{1998}-1$ [/hide]
and also  $a<1$     [hide="because"]${10^{1998}-2+10^{-1998}a^{2}< 10^{1998}-1}$ [/hide]
In fact , if you realize that you are looking  in the neighborhood of $2a=1$  and notice that it ($a$)    has to be between 0.4 and 0.6 is enough to deduce the decimal digit you are looking for is 1.[hide] (easy to see  when when a=1, the digit is 0, a=.7 digit is 1  and when a=.4 it is 2 etc ..[/hide]


number theory
greatest common divisor
algebra
polynomial
prime factorization

How do you find the gcd of the two polynomials in $ \mathbb{Q}[x]$: $ (x^2 \plus{} 3x \plus{} 6)^2(x \plus{} 1)^3(x \minus{} 3)^2$ and $ (x^2 \plus{} 3x \plus{} 6)(x \plus{} 1)^4(x \minus{} 2)^2$  :|

This is not pre-olympiad level.

Polynomials over $ \bf Q$ form a unique factorization domain.  You have two polynomials factored into irreducibles.  You do exactly the same thing you would do if presented with the prime factorization of two integers.


Show that for all positive a,b,c,
$ \frac{a(b\plus{}c)}{a^2\plus{}(b\plus{}c)^2}\plus{}\frac{b(c\plus{}a)}{b^2\plus{}(c\plus{}a)^2}\plus{}\frac{c(a\plus{}b)}{c^2\plus{}(a\plus{}b)^2} \le \frac{6}{5}$

[quote="dgreenb801"]Show that for all positive a,b,c,
$ \frac {a(b \plus{} c)}{a^2 \plus{} (b \plus{} c)^2} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} (c \plus{} a)^2} \plus{} \frac {c(a \plus{} b)}{c^2 \plus{} (a \plus{} b)^2} \le \frac {6}{5}$[/quote]

$ \sum \frac {a(b \plus{} c)}{a^2 \plus{} (b \plus{} c)^2} \leq \sum \frac { a(b \plus{} c) }{ a(b \plus{} c) \plus{} \frac {3(b \plus{} c)^2}{4} } \equal{} \sum \frac {4a}{4a \plus{} 3b \plus{} 3c}$

It suffices to prove that :

$ \sum \frac {a}{4a \plus{} 3b \plus{} 3c} \leq \frac {3}{10}$

Or,equivalently :

$ \sum \frac {1}{4a \plus{} 3b \plus{} 3c} \geq \frac {9}{10(a \plus{} b \plus{} c)}$

Which is obvious using Caushy-Schwarz inequality.

Done !

[quote="earldbest"][quote="rachid"]

$ \sum \frac {a(b \plus{} c)}{a^2 \plus{} (b \plus{} c)^2} \leq \sum \frac { a(b \plus{} c) }{ a(b \plus{} c) \plus{} \frac {3(b \plus{} c)^2}{4} } \equal{} \sum \frac {4a}{4a \plus{} 3b \plus{} 3c}$

[/quote]

why?  :([/quote]

$ a^2\plus{}(b\plus{}c)^2 \ge  a(b \plus{} c) \plus{} \frac {3(b \plus{} c)^2}{4} \iff 4a^2\plus{}(b\plus{}c)^2 \ge 4a(b\plus{}c)$ which is true by AM-GM. Nice solution!

[quote="dgreenb801"]Show that for all positive a,b,c,
$ \frac {a(b \plus{} c)}{a^2 \plus{} (b \plus{} c)^2} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} (c \plus{} a)^2} \plus{} \frac {c(a \plus{} b)}{c^2 \plus{} (a \plus{} b)^2} \le \frac {6}{5}$[/quote]
Let $ a\plus{}b\plus{}c\equal{}3.$
Hence, $ \sum_{cyc}\frac {a(b \plus{} c)}{a^2 \plus{} (b \plus{} c)^2}\leq\frac{6}{5}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(\frac{2}{5}\minus{}\frac{a(3\minus{}a)}{2a^2\minus{}6a\plus{}9}\plus{}\frac{9(a\minus{}1)}{25}\right)\geq0\Leftrightarrow\sum_{cyc}\frac{(a\minus{}1)^2(2a\plus{}1)}{2a^2\minus{}6a\plus{}9}\geq0.$

[quote="dgreenb801"]Show that for all positive a,b,c,
$ \frac {a(b \plus{} c)}{a^2 \plus{} (b \plus{} c)^2} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} (c \plus{} a)^2} \plus{} \frac {c(a \plus{} b)}{c^2 \plus{} (a \plus{} b)^2} \le \frac {6}{5}$[/quote]

$ \frac {a(b \plus{} c)}{a^2 \plus{} (b \plus{} c)^2} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} (c \plus{} a)^2} \plus{} \frac {c(a \plus{} b)}{c^2 \plus{} (a \plus{} b)^2} \le \frac {6}{5}$
$ \Leftrightarrow \sum{S_c (a\minus{}b)^2} \ge 0$
where $ S_c\equal{}(3 a^2 \plus{} 3 b^2 \plus{} 2 a c \plus{} 2 b c \minus{} c^2)(a^2\plus{}b^2\plus{}c^2\plus{}2ab)$.
WLOG, assume that $ a\ge b\ge c$, easy to prove that $ S_b,S_b\plus{}S_a,S_b\plus{}S_c\ge0$, by S.O.S theorem, ineq holds...

[quote="kuing"][quote="rachid"]
$ \sum \frac {a}{4a \plus{} 3b \plus{} 3c} \leq \frac {3}{10}$

Or,equivalently :

$ \sum \frac {1}{4a \plus{} 3b \plus{} 3c} \geq \frac {9}{10(a \plus{} b \plus{} c)}$
[/quote]

why equivalently?
 :o[/quote]

Because $ \sum\frac{10(a\plus{}b\plus{}c)}{4a\plus{}3b\plus{}3c} \ge 9 \Leftrightarrow \sum\left(\frac{\frac{10}{3}(4a\plus{}3b\plus{}3c)\minus{}\frac{10}{3}a}{4a\plus{}3b\plus{}3c}\right)\ge9 \Leftrightarrow \frac{3}{10}\ge \text{LHS}$

I think that rachid's solution is nice and clear  :)

[hide="@rachid"] amalhoum hadou mayfehmouch  :D  [/hide]

[quote="Evariste-Galois"][quote="kuing"][quote="rachid"]
$ \sum \frac {a}{4a \plus{} 3b \plus{} 3c} \leq \frac {3}{10}$

Or,equivalently :

$ \sum \frac {1}{4a \plus{} 3b \plus{} 3c} \geq \frac {9}{10(a \plus{} b \plus{} c)}$
[/quote]

why equivalently?
 :o[/quote]

Because $ \sum\frac {10(a \plus{} b \plus{} c)}{4a \plus{} 3b \plus{} 3c} \ge 9 \Leftrightarrow \sum\left(\frac {\frac {10}{3}(4a \plus{} 3b \plus{} 3c) \minus{} \frac {10}{3}a}{4a \plus{} 3b \plus{} 3c}\right)\ge9 \Leftrightarrow \frac {3}{10}\ge \text{LHS}$

I think that rachid's solution is nice and clear  :)

[hide="@rachid"] amalhoum hadou mayfehmouch  :D  [/hide][/quote]

I got it, thanks...
 :lol:


[b] How many distinct solutions $ (a,b,c)$ are there to the equation $ a^{2}\plus{}bc\equal{}b^{2}\plus{}ac$ if $ a, b, c$ are integers between 1 and 5 inclusive.[/b]

[quote="mihail911"][b] How many distinct solutions $ (a,b,c)$ are there to the equation $ a^{2}\plus{}bc \equal{} b^{2}\plus{}ac$ if $ a, b, c$ are integers between 1 and 5 inclusive.[/b][/quote]

[hide="eh."]
$ a^{2}\plus{}bc \equal{} b^{2}\plus{}ac\Rightarrow{}a^{2}\minus{}b^{2}\equal{}ac\minus{}bc\Rightarrow(a\minus{}b)(a\plus{}b)\equal{}c(a\minus{}b)$
Since $ a$ does not eqaul $ b$, we can divide:
$ a\plus{}b\equal{}c$

So the solutions are for (a,b.c): (1,2,3), (1,3,4), (1,4,5), and (2,3,5), so there are $ \boxed{4}$
[/hide]

I got $ 8$.

By meaning different solutions you don't include permutations ?

I just assumed $ a\neq b\neq c$, so, for example, $ (5,4,1)$ and $ (5,1,4)$ are still different solutions.

[quote="mihail911"][b] How many distinct solutions $ (a,b,c)$ are there to the equation $ a^{2}\plus{}bc \equal{} b^{2}\plus{}ac$ if $ a, b, c$ are integers between 1 and 5 inclusive.[/b][/quote]

[hide="solution"]Rewrite the equation:

$ (a\minus{}b)(a\plus{}b)\equal{}c(a\minus{}b)$.

Case 1: if $ a\equal{}b$.

We have $ c\equal{}c$ so if $ a\equal{}b$ and $ c$ are any integers between $ 1$ and $ 5$ then there are $ 25$ triplets.

Case 2: if $ a\ne b$.

Then we can divide by $ a\minus{}b$ to get $ a\plus{}b\equal{}c$.  As shown by the posters above, there are $ 8$ such triplets.

Therefore $ 25\plus{}8\equal{}\boxed{33}$.[/hide]

[quote="mihail911"]jli, ur answer is correct. good job! btw, how did u get a=b for the first case? i missed that in my answer.

anaxerzia, i got a similar answer, using that same factorization, but i missed the second limit. a=b/c=c. i only got 8.[/quote]
[hide]
If $ a\equal{}b$, we have $ c\equal{}c$ so any $ c$ will work for any $ a\equal{}b$.

There are 5 choices for $ c$ and 5 for $ a\equal{}b$.[/hide]

i understand you got a=b, but how did u get it? your factorization was $ (a\minus{}b)(a\plus{}b)\equal{}c(a\minus{}b)$ but how did u know to consider a=b as a case?

Because if $ a\equal{}b$ then $ a\minus{}b\equal{}0$ and then we would be dividing by 0 which is bad. We have to consider both cases separately. Always make sure you are not dividing by zero in algebraic division... this leads to bad things.


linear algebra
matrix
function
logarithms
calculus
derivative
probability

Show that the binomial coefficient:  $\begin{pmatrix}n \\ m \\ \end{pmatrix}$ has a maximum for $m=n/2$, if $n$ and $m$ are both positive integers, $m<n$ and $n$ is even.
[hide]
Set $m = xn$, with $0<x<1$. Introduce gamma functions to be able to differentiate with respect to $x$:
\[\begin{pmatrix}n \\ m \\ \end{pmatrix}= \frac{\Gamma (n+1)}{\Gamma(n-x n+1) \Gamma(x n+1)}= \Gamma (n+1) e^{-\log ( \Gamma (n-x n+1))-\log(\Gamma(x n+1))}\]
Differentiation with respect to $x$ gives:
\[\Gamma (n+1) e^{-\log(\Gamma(n-x n+1))-\log(\Gamma(x n+1))}\left(n \frac{d}{dx}\log(\Gamma(n-x n+1))-n \frac{d}{dx}\log(\Gamma(x n+1)) \right) \]
Dropping the overall factors, which are nonzero for all $n$ and $x$, gives the equation for the maximum:
\[n \frac{d}{dx}\left( \log(\Gamma(n-x n+1))-\log(\Gamma(x n+1)) \right)=0 \]
The term between parentheses  is zero if $\Gamma(n-x n+1)= \Gamma(x n+1)$. This gives $n-x n = x n$ and hence $x=1/2$. We can argue that this is the only maximum by using the fact that the binomial coefficient is symmetric around $m = n/2$. $\begin{pmatrix}n \\ m \\ \end{pmatrix}= \begin{pmatrix}n \\ n-m \\ \end{pmatrix}$[/hide] My answer is probably a little cumbersome.

Yes; I would stick with the discrete here. For any $m<n$, $\binom{n}{m+1}=\frac{n-m}{m+1}\binom{n}{m}$, so $\binom{n}{m+1}<\binom{n}{m}$ if $m<\frac{n-1}2$ and $\binom{n}{m+1}>\binom{n}{m}$ if $m>\frac{n-1}2$.

OK, let's raise the stakes a little. Fix a number $p$ with $0<p<1,$ and fix $n.$ For what integer $m$ is the quantity $\binom{n}{m}p^{m}(1-p)^{n-m}$ maximized?

[hide="Vague and not very useful hint."]Of course, this is the probability function of a binomial distribution with $n$ trials and probability of success $p$ on each trial. The mean of that distribution is $np.$ The question asks what the mode is. The plausible guess is that the mode ought to be as close to the mean as you can get it.[/hide]

[hide="A more useful hint."]Read jmerry's post. What would you have to do to apply his methods to this case?[/hide]


probability
geometry
3D geometry
prism
pyramid
graph theory

chellie bob designed a 7-faced die. in three seperate rolls of this die, what is the probability that she will roll two 1's and one 6 in no particular order?

how would u do this?

[hide]For any ordered triple of numbers the probability of getting them is $\frac{1}{7}\cdot\frac{1}{7}\cdot\frac{1}{7}=\frac{1}{243}$.  Since there are three ordered triples here {(6,1,1),(1,6,1),(1,1,6)} the answer is $\frac{3}{343}$[/hide]

[hide]the answer is $\frac{1}{7^{3}}\cdot\frac{3!}{2!\cdot1!}=\boxed{\frac{3}{343}}$[/hide]

[quote="meenamathgirl"]I think your answer is right but your method is off.  $\frac{3!}{2!\cdot1!}$ isn't 3, it is 1.5.[/quote]
Why is it 1.5? That's $\frac{6}{2 \cdot 1}$, which is 3.

Here's the general method for solving a problem like this (called overcounting, very useful method):
1) First, let all of the numbers be different even if you have [i]2 or more of the same number you want[/i]. Calculate the number of ways you can do this if they were different. [Example: In your case there were 3 numbers, so there would be $3\cdot 2\cdot 1$, or $6$.

2) Now, divide by the number of repeats you have. [Example: In your case, there were 2 repeats, $\frac{6}{2!}= 3$].

3) Take the result in 2, and divide it by the number of sides to the power of the number of numbers. Numerator=Result in #2. Denominator=(Number of sides)^(number of numbers) [Example: There were $7$ sides, and $3$ numbers. Denominator=$7^{3}=343$. Numerator=$3$].

That's it.

it can actually be proven that there is no seven-sided regular polyhedron.  I don't personally know the proof, but I know it is covered in AoPS vol. 2.

The Platonic solids are the only "regular" polyhedra, and none of them have seven sides.

Well you can probably design one that is fair, but has weird polygons as faces...I don't know much about 3-D and graph theory, so who am I to say.

I dunno, I suppose one could make a heptagonal prism and roll that, assuming there was no chance it could land on the ends.

or cut a pyramid off one corner of a die and weight it accordingly

btw what kind of name is chellie bob

I know there is a way to make two different dies give same results/probability as 2 sixes(generating functions).  Maybe there is a way to do that with sevens?


If a square has side length $ s$, then what is the largest equilateral triangle that can fit inside the square?  Prove your answer.

Thanks.

Proofs, are [b]not[/b] or this forum. Probably in HSB.

hmmm, wouldnt the answer be obvious-make the equilateral triangle side length the length of the side of the square?

Yeah, that sounds right, right?

[geogebra]3e25face515c10268917ff1cf2db6cbf8c3f9405[/geogebra]

Well, that has a larger area ;)


Alguien tiene los e-books (libros en pdf) de los siguientes libros???

Estrategias de resolución de problemas (Engel)
Fundamentos de teoría de Números (Vinogradov)

Gracias.....

saludos jose me parece que ese libro d engels lo puedes bajar por emule
 carlos bravo sabe un poco mas de ese libro

Ola Jose,
respecto a los ebooks que podes encontrar en la web,
podes instalar un P2P downloader...
asi como emule...

una vez instalada, en la busqueda haga una consulta:
Math Olympiad....

Saludos,

Carlos Bravo ;)
Lima- PERU


[quote="21 7 15"]1/a=0 what is a(dont't tell me a=infinity)[/quote]

There have been two threads recently on the topic.

[url]http://www.mathlinks.ro/viewtopic.php?t=180690[/url]

Here is a proof that no real value of $ a$ exists assuming virtually nothing, because I'm a little tired of discussing this.

[hide]By the definition of division, $ 1/a \equal{} 0 \Leftrightarrow 1 \equal{} a \cdot 0$.
[b]Lemma[/b]: $ \forall a \in \mathbb{R}, 0 \equal{} a \cdot 0$.
[hide="Proof"]By additive inverses, $ 0 \equal{} a \plus{} (\minus{}a)$
By a property of negative numbers, $ 0 \equal{} a \plus{} (\minus{}1 \cdot a)$
By multiplicative identity, $ 0 \equal{} (1 \cdot a) \plus{} (\minus{}1 \cdot a)$
By distributivity, $ 0 \equal{} (1 \plus{} (\minus{}1)) \cdot a$
By additive inverses $ 0 \equal{} 0 \cdot a$.
By commutativity of multiplication, $ 0 \equal{} a \cdot 0$.[/hide]
By Lemma 2 and the transitive property of equality, the existence of $ a$ would imply that $ 1 \equal{} 0$. Contradiction. QED.[/hide]


Evaluate the product $ \prod^{n}_{k\equal{}1}(1\minus{}\tan^{2}\frac{2^{k}\pi}{2^{n}\plus{}1})$


Rank the following five bases in order of increasing basicity:

$ CH_3CH_2^\minus{}$, $ HO^\minus{}$, $ RC \equiv C^\minus{}$, $ NH_2^\minus{}$, and $ CH_2\equal{}CH^\minus{}$.

increasing order
$ \text{OH}^{ \minus{} } < \text{RC}\equiv \text{C}^{\minus{}} < \text{CH}_{2} \equal{} \text{CH}^ \minus{} < \text{MeCH}_2^{ \minus{} } < \text{NH}_{2}^{ \minus{} }$

is there a mistake in any particular compound

I think it would be

$ \text{OH}^\minus{}<\text{NH}_2^\minus{}<\text{RC}\equiv\text{C}^\minus{}<
\text{CH}_2\equal{}\text{CH}^\minus{}<\text{CH}_3\text{CH}_2^\minus{}$

Nearly all C-H bonds are much harder to ionize than any O-H or N-H bond; the easiest C-H bonds to ionize would be those adjacent to a carbonyl group (as in acetone) or is a similarly electron-withdrawing $ \pi$ environment. On the right end of that chain of inequalities - what would it take to demonstrate that an alkane had any acidity at all? Anything that could ionize that might very well destroy multiple other bonds.

We know that terminal alkynes are slight acidic:

Acidity: $ RC \equiv CH > CH_2 \equal{} CH_2 > CH_3CH_3$ so I would guess that

Basicity: $ RC \equiv C^{ \minus{} } < CH_2 \equal{} CH^{ \minus{} } < CH_3CH_2 ^{ \minus{} }$

So my order would be the same as Kent Merryfield's if I was to guess.

but is $ \text{NH}_3$ more acidic than say $ \text{RC}\equiv \text{CH}$

Kent Merryfield's ranking is also incorrect.

[quote="Kent Merryfield"]Nearly all C-H bonds are much harder to ionize than any O-H or N-H bond[/quote]

That's also incorrect.

[quote="Kent Merryfield"]the easiest C-H bonds to ionize would be those adjacent to a carbonyl group (as in acetone) or is a similarly electron-withdrawing pi environment.[/quote]

That's not also strictly true: there are several counter examples.

OK, the following things have to be true:

$ \text{OH}^\minus{}<\text{RC}\equiv\text{C}^\minus{}< \text{CH}_2\equal{}\text{CH}^\minus{}<\text{CH}_3\text{CH}_2^\minus{}$ and

$ \text{OH}^\minus{}<\text{NH}_2^\minus{}<\text{CH}_3\text{CH}_2^\minus{}$

The only issue then is the placement of $ \text{NH}_2^\minus{}$ with respect to the alkyne and the alkene. That leaves two possibilities, with the next most likely being to put ammonia in between the alkyne and the alkene.

Yes, that's right. As such, sodium amide is often used to trigger eliminations that ultimately lead to the formation of a sodium alkynide. More strikingly, simple derivatives of $ NH_2^ \minus{}$ are also used in the acetylene zipper reaction.

[quote="Carcul"]That's not also strictly true: there are several counter examples.[/quote]

Exercise: can someone provide an example?


I have a bag with yellow and blue marbles in it. At the moment, the ratio of blue marbles to yellow marbles is 4:3. If I add 5 blue marbles and remove 3 yellow marbles, the ratio will be 7:3. How many blue marbles were in the bag before I added more?

[quote="isabella2296"]We can say \[ \frac{4x \plus{} 5}{3x \minus{} 3} \equal{} \frac{7x}{3x}\] Cross multiplying gives $ 21x^2 \minus{} 21x \equal{} 12x^2 \plus{} 15x$ or $ 9x^2 \minus{} 36x \equal{} 0$, which means $ x \equal{} 4$. Thus, there are $ 4 \cdot 4 \equal{} \boxed{16}$ blue marbles.[/quote]

is theren't a faster way?


quadratics
geometry
3D geometry
number theory
Diophantine equation
prime factorization
number theory open

Find the solutions in positive integers of the following diophantine equation

$ 2y^{2}\equal{}x^{4}\plus{}x$.

Note that $ (1,1)$ and $ (2,3)$ are solutions of this equation.  Are there any other?

Let $ p$ be the any prime factor dividing $ x$ except $ 2$; let $ k$ be the largest integer such that $ p^{k}|x$; then $ k\ge 1$. Now, $ x^4\plus{}x\equal{}x(x^3\plus{}1)$ hence as $ \gcd(x,x^3\plus{}1)\equal{}1$ have $ p^k$ is the largest power of $ p$ dividing $ x^4\plus{}x$, and so is the largest power of $ p$ dividing $ 2y^2$. But as $ gcd(2,p)\equal{}1$ then the largest power of $ p$ dividing $ 2y^2$ is $ p^l$ where $ l$ is even. Hence $ k$ is even.
Now, if $ x$ is even, let $ m$ be the largest integer such that $ 2*2^m$ divides $ x$; then again, as $ gcd(x,x^3\plus{}1)$ is $ 1$ have $ 2*2^m$ is the largest power of $ 2$ dividing $ x^4\plus{}x$ and so $ 2y^2$. But now, the largest power of $ 2$ dividing $ y^2$ is $ 2^n$ where $ n$ is even so the largest power of $ 2$ dividing $ 2y^2$ is $ 2*2^n$ where $ n$ is even. So, $ m$ is even.
Therefore, if $ x$ is odd, if $ x\equal{}p_{1}^{e_1}p_{2}^{e_2}...p_{t}^{e_t}$ is the prime factorization of $ x$ then $ e_1,e_2,...,e_t$ are even so $ x$ is a perfect square, i.e. $ x\equal{}z^2$. But because $ x\equal{}z^2|y^2$, $ z|y$, so $ y\equal{}zc$.
Then get: $ 2y^2\equal{}x^4\plus{}x\Leftrightarrow 2z^2c^2\equal{}z^2\plus{}z^8$
$ \Leftrightarrow 2c^2\equal{}1\plus{}z^6$ which can be rewritten in the form $ 2c^2\minus{}d^2\equal{}1$ where $ d\equal{}z^3$ so it just becomes a Pell's Equation and the problem just becomes to determine which solutions this equation has where $ d$ is a perfect cube, because once you have $ c,d$ you can build $ x,y$ and vice versa.
Now, if $ x$ is even, $ x\equal{}2z^2$ and $ z^2|y^2\Rightarrow z|y$ so $ y^2\equal{}z^2x^2$ so:
$ 2y^2\equal{}x^4\plus{}x\Leftrightarrow 2z^2c^2\equal{}2z^2\plus{}16z^8$
$ \Leftrightarrow c^2\equal{}1\plus{}8z^6$ which can be rewritten in the form $ c^2\minus{}8d^2\equal{}1$ where $ d\equal{}z^3$ so it just becomes a Pell's Equation and the problem just becomes to determine which solutions this equation has where $ d$ is a perfect cube, because once you have $ c,d$ you can build $ x,y$ and vice versa.
I am not sure how to determine if a Pell's equation has solutions where one of the variables is a perfect cube. It seems really difficult. :maybe:

Substituting $ s \equal{} 2z^2$, you end up with $ c^2 \minus{} s^3 \equal{} 1$, having only the solutions $ (c,s) \in \{(1,0),( \minus{} 1,0),(3,2),( \minus{} 3,2)\}$ (posted before, but not hard; a very special case of Catalan's conjecture).


number theory
prime numbers
number theory open

In one competition in Slovakia a mistake was made and instead one, maybe well-known problem, this was written: Find all prime numbers p,q for which
$
(5^p-2^q)(5^q-2^p)/{pq}
$
Can you solve it ? We also found some (non-trivial) solutions: (3,3),(41,3),(467,41) and (79631,751) so it can be quite hard to solve this problem (maybe its open). The original formula was:
$
(5^p-2^p)(5^q-2^q)/{pq}
$
But dont post solution to the second one, becouse the competition is still running.

Thanks

[quote="ondrob"]In one competition in Slovakia a mistake was made and instead one, maybe well-known problem, this was written: Find all prime numbers p,q for which. The original formula was:
$ (5^p - 2^p)(5^q - 2^q)/{pq}$
[/quote]
With this problem we can solve as follow
[u]Case1[/u] $ p=q$
Easy to check $ p=q=3$
Now consider $ p\geq q$
$ \gcd(5^p-2^p,5^q-2^q)=3$
If $ q=3$
then $ p=13$
Now consider $ p>3$
$ \gcd(q,5^q-3^q)=1$ so
$ q|5^p-2^p$ then
${ q|\gcd(5^{q-1}-2^{q-1},5^p-2^p}=3$ tradition.
So we has solution:
$ (p,q)=(3,13),(3,3)$
More
$ \gcd(a,b)=1$
Find all $ (p,q)\in P$ such that
$ \frac{(a^p-b^p)(a^q-b^q)}{ab}$ is an integer.

Your problem is very similar to Bulgaria 1996 - IV round. You may take a look at this: http://www.math.bas.bg/bcmi/.


calculus
integration
LaTeX
trigonometry
calculus computations

1. the integral from 0 to 3 of 10pisin(pi/3 * t) + 30
2. if the integral of f(u) du from 1 to 3 = 7, find the value of the integral from 1 to 2 of f(5-2x)dx

Try to use $\LaTeX$ to get well what you need.

Sorry i edited it.
But are you sure about the first one?

Sorry but can't se my mistake  :oops:
[color=red]Could you please point it out? [/color]

You can be playing bad with a sign, or something.

$\int{10\pi \sin \left({\frac{{\pi t}}{3}}\right)dt}$

Now set $u={\frac{{\pi t}}{3}}$

[quote="goldbach"]for  the first one the integral is :  90
for the second you cas pose that
u=5-2x
==>dx=-1/2du

u=1 ==>x=2
u=3 ==>x=1

then the value of the integral from 1 to 2 of f(5-2x)dx=7/2[/quote]

so then I got -1/2 * the integral from 1 to 3 of f(u) du.

shouldn't it be -7/2?? where did the -2 disappear to??

[b]EquinoX[/b]: Yes it should be $-\frac{7}{2}$ and concernings your question on
why I got $-\cos{u}\ldots$ : It's because the antiderivative of $\sin{u}$ is $-\cos{u}$
Just differentiate and you will se. ($u={\frac{{\pi t}}{3}}$)

this is weird cause the answer question gives me 7/2??

how about for number 2, the answer key is 7/2 without the minus sign

The limits are not wrong- your application of the transformation is. What is $x$ when $u=3,1$?

Edited.  Thanks.. I guess plugging the bounds into the variable on the right becomes so mechanical with u-substitutions that I didn't even think.  My bad.

[quote="skimnc"]If $\int_{1}^{3}f(u) du=7$, find $\int_{1}^{2}f(5-2x)dx$.

[hide]$\int_{1}^{3}f(u) du=7$

Let:
$u=5-2x \Rightarrow x=\frac{5-u}{2}$
$du=-2 dx$

Bounds:
$a=\frac{5-1}{2}=2$
$b=\frac{5-3}{2}=1$

$\int_{2}^{1}f(5-2x) (-2 dx)=7$
$\boxed{\int_{1}^{2}f(5-2x) dx=\frac{7}{2}}$[/hide][/quote]

so the - sign comes because you reverse the interval of the upper bound to the lower bound and vice versa??

$\boxed{\int_{a}^{b}f(x) dx =-\int_{b}^{a}f(x) dx}$

This is easily proven by:
$\int_{a}^{b}f'(x) dx=f(x)|_{a}^{b}=f(b)-f(a)=-\left[ f(a)-f(b) \right] =-\left[ f(x) \right]_{b}^{a}=-\int_{b}^{a}f'(x) dx$


calculus
integration
calculus computations


group theory
abstract algebra
search
superior algebra
superior algebra unsolved

Let $G$ and $H$ be two groups. $G_{1}$ is a subgroup of $G$ and is isomorphic to $H$, $H_{1}$ is a subgroup of $H$ and is isomorphic to $G$.
Show that $G$ and $H$ are not necessarily isomorphic.

Note: It's not difficult, and probably classical but I'm interested to see several exemples.

We already had this a couple of times, use search function.


function
logarithms
calculus
calculus computations

I have no idea how to do this diffential equation question  :( . Can anyone help ?

A tank contains $40$ litres of water with $10$ kg of salt dissolved in it. Brine with $1 $ kg of salt per litre is poured into the tank at the rate $5$ litres per second and the solution flows out at the same rate. Assume that the solutions are perfectly mixed at all times.

(a) Find a formula for the amount of salt in the tank at time t.
(b) Using your solution in part (a), find the limiting amount of salt (i.e. the amount of salt as t approaches infinity).

There are some things you can see before you even derive the differential equation. One is the answer to (b). In the long run (the limiting case), the original contents of the tank will be replaced by brine of the composition being poured into the tank, so the limiting amount of salt will be $40\text{ l}\cdot\1\,\frac{\text{kg}}{\text{l}}=40\text{ kg}.$

Now imagine that a membrane is placed in the tank between the original contents and the inflow that prevents mixing. Since we're replacing the $40\text{ l}$ contents of the tank at a rate of $5\text{ l}/\text{s}$, that membrane moves through the tank in 8 seconds. With mixing, that's not exactly what happens. However, do expect 8 seconds to show up as a kind of fundamental time scale for the solution of the differential equation.

Now, as for that differential equation itself. Let $y=y(t)$ be the amount of salt in the tank, in kg. We're also interested in the concentration of salt. In kg pre liter, that concentration will be $\frac{y}{40}.$

Consider a time period $dt.$ (Note: $dt$ is measured in the same units as $t,$ namely seconds.) What happens during that time period? A volume of solution $5dt$ leaves the tank by the outflow pipe and the same amount of solution $5dt$ enters by the inflow pipe. The volume of solution in the tank stays constant at $40\text{ l}.$ The amount of salt that enters the tank is $1\cdot5dt.$ Let's check the unit of measurement on that:

$1\frac{\text{kg}}{\text{l}}\cdot5\frac{\text{l}}{\text{s}}\cdot dt(\text{s}) =(5dt)\cdot\frac{\text{kg}}{\text{s}}$

To find the amout of salt that leaves the tank during time period $dt$, the calculation is the same except that instead of multiply by concentration $1,$ we multiply by the actual concentration in the tank, which is $\frac{y}{40}.$ Thus the amount of salt that leaves during time period $dt$ is $\frac{y}{40}\cdot 5dt=\frac{y}8\,dt.$

So our total budget of salt looks like this:

$dy=5dt-\frac{y}8\,dt$ or, if you prefer, $\frac{dy}{dt}=5-\frac{y}8.$

We have the initial condition $y(0)=10.$

So that's your differential equation. I'll let you solve it - and see if you can verify what I said at the top.

I said "However, do expect 8 seconds to show up as a kind of fundamental time scale for the solution of the differential equation."

Where it shows up is that the only time-dependent term should contain $e^{-t/8}.$ That 8 is actually 8 seconds. If we're trying to use units of measurement ("dimensional analysis") to help us check our solutions, one thing to look for is that the arguments of transcendental functions like exp, sin, or cos should be pure unit-less numbers. That works here, because with $t$ measured in seconds, $\frac t8$ is a pure unit-less number.

The half-life of this process - that is, the time it takes for the salt concentration to get halfway from where it started to where it's going - would be $8\ln 2$ seconds.


[quote="nutz_for2.718281828"]Matilda just made x out of y basketball shots, where x is less than y/2. What is the number of shots she must make in a row to increase the ratio of shots made to total shots to 1/2?[/quote]
[hide]
she must make at least $y$ shots to garentee it is one half, and at most $1$ shot to satisfy one half in the best circumstances.

best circumstances (need $1$ shot):
$\frac{1}{3}$, making $1$ more shot gets her $\frac{2}{4}$;
$\frac{2}{5}$, making $1$ more shot gets her $\frac{3}{6}$;
etc

worst circumstances (need $y$ shots):
$\frac{0}{5}$, making $y$ more shots gets her $\frac{5}{10}$;
$\frac{0}{3}$, making $y$ more shots gets her $\frac{3}{6}$;
etc

[/hide]

[hide]Set up an equation, where $a$ is the number of shots:

$\frac{x+a}{y+a}=\frac12$

$2x+2a=y+a$
$a=y-2x$[/hide]


geometry trigonometry circumcircle ratio

Prove that the ortocentre of a triangle whose vertices are $ (a \cos \alpha, a \sin \alpha)$, $ (a \cos \beta, a \sin \beta)$, $ (a \cos \gamma, a \sin \gamma)$ is
$ (a \cos \alpha \plus{} a \cos \beta \plus{} a \cos \gamma, a \sin \alpha \plus{} a \sin \beta \plus{} a \sin \gamma)$

Observe that Circumcenter S of the given triangle is (0,0) (each vertex is at a distance a from the origin). centroid G of the triangle is

 $ (\frac{a \cos \alpha \plus{} a \cos \beta \plus{} a \cos \gamma}{3}, \frac{a \sin \alpha \plus{} a \sin \beta \plus{} a \sin \gamma}{3})$
Now G divides SH in 2:1 ratio internally.
So we get orthocenter as

$ (a \cos \alpha \plus{} a \cos \beta \plus{} a \cos \gamma, a \sin \alpha \plus{} a \sin \beta \plus{} a \sin \gamma)$


Let $P$ be an arbitrary point on the side $BC$ of triangle $ABC$ and let $D$ be the tangency point between the incircle of the triangle $ABC$ and the side $BC$. If $Q$ and $R$ are respectively the incenters in the triangles $ABP$ and $ACP$, prove that  $\angle QDR$ is a right angle.

Prove that the triangle $QDR$ is isosceles if and only if $P$ is the foot of the altitude from $A$ in the triangle $ABC$.

see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=55388]this[/url]. It's basically the same problem.


geometry
calculus
function
algebra
domain
integration
calculus computations

Sketch the region bounded by the graphs of the algebraic
functions and find the area of the region.

y = x^3− 6x, y = −2x

I know that it intersects at three points

Well if they intersect 3 times, you have to compute 2 integrals.  I don't have the picture in front of me but you have to do area between two curves twice on whatever the domains are (I assume you know how to do that).

$ \left\|\begin{array}{c}
f(x)=x^3-6x\\\
g(x)=-2x\end{array}\right\|\ .$ Thus, $ f(x)=g(x)$ $ \Longleftrightarrow$ $ x^3-6x=-2x\Longleftrightarrow x\in\{-2,0,2\}\ .$

${ A\equiv\int_{-2}^2}\left|f(x)-g(x)\right|\ \mathrm {dx}=$ $ \int_{-2}^2\left|\left(x^3-6x\right)-(-2x)\right|\ \mathrm {dx}\Longrightarrow$

$ A=\int_{-2}^2\left|x^3-4x\right|\ \mathrm {dx}=$ $ 2\cdot\int_0^2\left|x^3-4x\right|\ \mathrm {dx}=$ $ 2\cdot\int_0^2\left(4x-x^3\right)\ \mathrm {dx}$ a.s.o.


Let $ n \geq 2$ be an integer.Let $ S\equal{}\{1,2,...,n\}$.Show that there exists a subset $ A \subset S$ such that:
(i)$ |A|$≤$ 2[\sqrt{n}]\plus{}1$
(ii)$ \{x\minus{}y|x,y \in A,x>y\}\equal{}S$

Note:
$ |A|$:the number of elements of set $ A$
$ [x]$:the floor function.

Is there a typo, because $ n \not\in \{x\minus{}y|x,y\in A,x>y\}\equal{}S$?


Hello everyone,

Please check to see if I have computed the right answer. Thanks in advance.

[u]1. Find the amount of water, in litres, discharged in 10 minutes from a pipe, 4 cm in diameter measured internally, if the water flows at a speed of 50 cm per second. (pi = 3.14)[/u]

[u]2. An open box is constructed by gluing a number of 1cm cubes together to form the bottom and the sides. The outside dimensions of the finished box are 10 cm by 10 cm by 10 cm, and the sides and the bottoms are all 1 cm thick. Determine the number of cubes required to construct the box.[/u]

-----
[size=134][u]
My Work for #1:[/u][/size]

If the water flows at 50 cm/s and the pipe's diameter is 2 cm, the volume of water being discharged = the volume of water that would fit in a cylinder with a height of 50 cm and radius 2.

Volume of discharging water = (pi)(4)(50)
= 628.32 cm^3

Therefore, 628.32 cm^3/s is being discharged --> 37699.2 cm^3/min --> 376992 cm^3/10 mins.

Since 1 L = 1000 cm^3: 376992 cm^3/10 mins = [b]376.992 L/10 mins[/b]

[color=indigo]• In 10 minutes, 376.992 liters of water will be discharged from this pipe.[/color]

-------------

[u][size=134]My Work for #2:[/size][/u]

Since the side and bottom is 1 cm thick, I am imaging them as a rectangular solid which is 10 cm by 10 cm by 1 cm.

Therefore, the area of 1 of these sides is 100 cm^3 and since this is an open box, there are 5 of these sides = 500 cm^3.

500cm^3 / 1 cm^3 cubes = [b]500 cubes[/b]

For Number 2
[hide]
Front $10 \cdot 10 = 100$
Back $10 \cdot 10 = 100$
Sides $2 \cdot 10 \cdot 8 = 160$
Bottom $8 \cdot 8 = 64$
Add $424$.
[/hide]

[hide="a hopefully better way for 2"]
Just take the whole 10x10x10 box.  Since it is hollow, we have to remove a 8x8x9 box from it.  After the subtraction the answer is 424.[/hide]

That is a better way CircleSquared,  :)  think inside the box instead of outside.


A right triangle has perimeter of length $ 7$ and hypotenuse of length $ 3$. If $ \theta$ is the largest non-right angle in the triangle, what is the value of ${ \cos{\theta}}$?

Intitution: The trinagle has 3 sides of lengths $ 7, 5, 3$, yields $ \cos \theta \equal{}\minus{}\frac{1}{2}$.

^?
I don't think I understand what you're trying to say, a 7,5,3 triangle is not a right triangle.

Sorry, I missed to look the condition a right triangle. :blush:

[hide]
[asy]pair A,B,C;
C=(0,0);
A=((4-sqrt(2))/2,0);
B=(0,(4+sqrt(2))/2);
draw(A--B--C--A);
label("$y$",(C+B)/2,W);
label("$x$",(C+A)/2,S);
label("3",(A+B)/2,NE);[/asy]
Let the legs be $ x$ and $ y$, and WLOG $ x<y$.  Then we have $ x\plus{}y\equal{}4$ and $ x^2\plus{}y^2\equal{}9$.  Squaring the first equation and subtracting the second equation from this yields $ 2xy\equal{}7\Rightarrow xy\equal{}\frac{7}{2}$.  Note that $ x$ and $ y$ are the roots of the equation $ 2x^2\minus{}8x\plus{}7\equal{}0$ (think Vieta's formulas).  Solving yields $ x\equal{}\frac{4\minus{}\sqrt{2}}{2}$ and $ y\equal{}\frac{4\plus{}\sqrt{2}}{2}$.  Then $ \theta\equal{}\cos^{\minus{}1}\frac{y}{3}$, so $ \cos\theta\equal{}\frac{\frac{4\plus{}\sqrt{2}}{2}}{3}\equal{}\boxed{\frac{4\plus{}\sqrt{2}}{6}}$.
[/hide]

[hide="alternately"]$ 7 \equal{} 3\sin{\theta} \plus{} 3\cos{\theta} \plus{}3$

$ \sin{\theta} \plus{} \cos{\theta} \equal{} \frac{4}{3}$

Since $ (\sin{\theta} \plus{} \cos{\theta})^2 \plus{} (\sin{\theta} \minus{} \cos{\theta})^2 \equal{} 2$

we have $ (\sin{\theta} \minus{} \cos{\theta})^2 \equal{} 2 \minus{} (\frac{4}{3})^2$

$ \sin{\theta} \minus{} \cos{\theta} \equal{} \frac{\sqrt{2}}{3}$

Subtracting from our second equation we have $ \cos{\theta} \equal{} \frac{4\minus{}\sqrt{2}}{6}$

[/hide]


geometry
incenter
inradius
trigonometry
circumcircle
trig identities
Law of Sines

Let $ ABC$ be a triangle with incenter $ I$ and let $ \Gamma$ be a circle centered at $ I$, whose radius is greater than the inradius and does not pass through any vertex. Let $ X_{1}$ be the intersection point of $ \Gamma$ and line $ AB$, closer to $ B$; $ X_{2}$, $ X_{3}$ the points of intersection of $ \Gamma$ and line $ BC$, with $ X_{2}$ closer to $ B$; and let $ X_{4}$ be the point of intersection of $ \Gamma$ with line $ CA$ closer to $ C$. Let $ K$ be the intersection point of lines $ X_{1}X_{2}$ and $ X_{3}X_{4}$. Prove that $ AK$ bisects segment $ X_{2}X_{3}$.

It is very easy problem.
Solution:
Denote $ P \equal{} AK\cap BC$.Using Menelaus theorem for $ \triangle ACP,\triangle ABP$ and transversals $ \overline{X_{4}X_{3}K}$ and $ \overline{X_{1}X_{2}K}$:
$ PX_{3}\equal{} PX_{2}\Longleftrightarrow\frac{BX_{2}}{BX_{1}}\cdot\frac{CX_{4}}{CX_{3}}\cdot\frac{AX_{1}}{AX_{4}}\equal{} 1$.The last one is obvious,because $ BX_{1}\equal{} BX_{2},CX_{3}\equal{} CX_{4},AX_{4}\equal{} AX_{1}$.(use that the center of $ \Gamma$ is incenter of $ \triangle ABC$.)Thus we are done. :)

Let $ D$, $ E$,$ F$ be the points of contact of the incircle with the sides $ BC$, $ AC$ and $ AB$ and let $ K_{1}$ and $ K_{2}$ be the points where $ X_{1}X_{2}$ and $ X_{3}X_{4}$ cut $ AD$. Notice that $ D$ is the midpoint of $ X_{2}X_{3}$ and that $ FX_{1}\equal{} EX_{4}$. It is easy to prove that $ X_{1}X_{2}\parallel DF$ and that $ X_{3}X_{4}\parallel DE$ so $ \Delta AFD\sim AX_{1}K_{1}$ and $ AED\sim AX_{4}K_{2}$, then $ AK_{1}\equal{} AD\frac{AF\plus{}FX_{1}}{AF}\equal{} AD\frac{AE\plus{}EX_{4}}{AE}\equal{} AK_{2}$ so $ K_{1}\equal{} K_{2}\equal{} K$, which completes the proof.

To dM@gGo+:
Did you take participant in this olympiad?

No. David is Panama's deputy leader, not a contestant.  :) By the way, nice solution, David! :D

It's easy to note that $ CX_{3}\equal{}CX_{4}$, $ BX_{2}\equal{}BX_{1}$ and $ AX_{4}\equal{}AX_{1}$.
Let $ L$ the line parallel to $ BC$ through A, let $ P\equal{}X_{3}X_{4}\cap L$ and $ Q\equal{}X_{2}X_{1}\cap L$ then, $ AP\equal{}AX_{4}\equal{}AX_{1}\equal{}AQ$ so $ A$ is the midpoint of $ PQ$, since $ X_{2}X_{3}$ is parallel to $ PQ$, we conclude that $ AK$ passes through the midpoint of $ X_{2}X_{3}$.

$ Tipe$

i have a trigonometric solution (as ussual  :lol: ), i'll post it later when i have a little more time... it uses that $ BX_{1}\equal{}BX_{2}$, $ CX_{3}\equal{}CX_{4}$, some power or a point and some law of sines for the gergonne cevian (that is what we want to prove) combined with briggs formulas...

to david: how did the contestants do at this problem? it seems much more easier than last year's q5..

[quote="campos"]i have a trigonometric solution (as ussual  :lol: ), i'll post it later when i have a little more time...[/quote]

There's no solution if we can't see it...  :P  Just kidding! Well, I have to say it was a 50/50: near 50% of the contestants solved it throughly. Problems 3 and 6 of this Iberoamerican Math Olympiad were more interesting cases: nobody solved them.

[quote="Erken"]It is very easy problem.
last one is obvious,because $ BX_{1}\equal{} BX_{2},CX_{3}\equal{} CX_{4},AX_{4}\equal{} AX_{1}$.(use that the center of $ \Gamma$ is incenter of $ \triangle ABC$.)Thus we are done. :)[/quote]
proof?

This is probably much more roundabout and complicated than needed but here's an alternative method.

Continue labeling the points counterclockwise around the circle as $ X_{5}, X_{6}$.  Clearly $ X_{2}X_{3}\equal{}X_{4}X_{5}\equal{}X_{6}X_{1}$, because $ I$ is equidistant from the sides of the triangle.  Then those three arcs have equal measure in this circle.  From angle chasing, we get $ m\angle{AX_{1}X_{2}}\plus{}m\angle{KX_{4}X_{1}}\equal{}180^{\circ}$, and so the circumcircle of $ \triangle{KX_{1}X_{4}}$ is tangent to $ AB$ and $ AC$.  Then $ KA$ is the symmedian of $ \triangle{KX_{1}X_{4}}$, and so it's the median in $ \triangle{KX_{2}X_{3}}$.  The conclusion follows.

Is easy to see that $AX_1X_4,  BX_2X_3,  CX_2X_3$ are isosceles triangles, thus $\angle{X_1KX_4}=\angle{AX_1X_4}=\angle{AX_4X_1}=\frac{\pi-A}{2}$ so $AX_1$ and $AX_4$ are tangents to the circumcirle of $X_1X_4K$. Finally $KA$ is symmedian and $X_2X_3$ is antiparallel to $X_1X_4$; hence done.

Let the incircle touch $BC,CA,AB$ at $D,E,F$,respectively. Then $DEF$ and $X_1X_4K$ are homothetic with homothety center $A$. So $A,D,K$ are collinear since $D$ is the midpoint of $X_2X_3$ we get the desired result.


Suppose $ \Delta ABC$ is isosceles with $ |AC| \equal{} |BC|$ and that its incircle touches $ AB$ in $ D$ and $ BC$ in $ E$. Let $ G$ be a point on the incircle different from $ E$ and suppose that $ AG$ intersects the incircle in $ F$. Let $ K$ be the intersection of $ AB$ and $ EF$ and $ L$ be the intersection of $ AB$ and $ EG$.

Prove that $ |KD| \equal{} |LD|$.

[hide]let EM a line parallel to AB that intersects the circle in M. so M is the point where the incircle touches AC.
by the equality of the AMK triangle and the LBE we get AK=LB and
Finally in ABC(isosceles) we know $ AB\equal{}AC\equal{}>AD\equal{}DB\equal{}>AD\minus{}AK\equal{}DB\minus{}LB\equal{}>KD\equal{}LD$[/hide]

[quote="Athinaios"][hide]by the equality of the AMK triangle and the LBE[/hide][/quote]

Could you explain this further, please? :)

[quote="Athinaios"][hide]let EM a line parallel to AB that intersects the circle in M. so M is the point where the incircle touches AC.
by the equality of the AMK triangle and the LBE we get AK=LB and
Finally in ABC(isosceles) we know $ AB \equal{} AC \equal{} > AD \equal{} DB \equal{} > AD \minus{} AK \equal{} DB \minus{} LB \equal{} > KD \equal{} LD$[/hide][/quote]You haven't proof that $ AMK$ and $ LBE$ is equal (maybe you mean congruent?).

Let $ N$ be the tangent point on $ AC$ and $ O$ the intersection of $ NL$ and $ KE$. Clearly, the distances between $ E$ to $ AB$ and $ N$ to $ AB$ are equal. Hence, the areas of $ KLN$ and $ KLE$ are equal, and the areas of $ NKO$ and $ LOE$ are also equal. But $ NO\parallel{}OL,KO\parallel{}OE$, which implies that $ NKO\cong LOE$. So $ NK=EL$. Since $ AN=EB,NK=EL,\angle NAK=\angle EBL$, we have $ NAK\cong ELB$ and $ AK=LB$. So $ KD=AD-AK=DB-BL=DL$, as desired.

I'm not sure about my proof that $ NKO\cong LOE$ by the way. :P

Denote the incircle touches $AC$ at $M$, then $MFDG$ is harmonic, so $E(MD;KL) = E(MD;FG) =  - 1$. Because $EM\parallel KL$, we have $KD=DL$.


In a shooting match, 8 clay targets are arranged in two hanging columns of

three each and one column of 2. A marksman must break all 8 targets according

 to the following rules:

#1 The marksman first chooses a column.

#2 the marksman must then break the lowest unbroken target in the chosen column.

If these rules are followed and then marksman never misses, in how many different orders can the targets be broken?

Im not really looking for a solution im just wondering if back in the 1990 the AIME was easier?

This question seems fairly easier then most modern AIME counting questions

So in the new AIMEs these would be like a 2...or maybe like a 20 on the amc 12? :huh:

Asked and answered -- [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=439577]see this thread[/url]

When you are in the forum, click "Contests" in the light bluish horizontal navigation bar near the top (the one with Profile and Private Messages) to access problems and solutions to many major contests.

All (or almost all) AMC and AIME problems are posted in the USA section of the Contests page  :)

I think you mis-read the intention of the above user's post, TZF.  :wink: (I would also add the [[AoPSWiki]] to that list, which is gradually getting better.)

I think that in general as a trend, the difficulty of the AIME problems has been increasing since the earlier tests like this one. Honestly though, I thought that #9 on this year's AIME I required the same formula and wasn't much more difficult; it required a minor bit of number theory but the way to approach the problem was even more obvious.


Derive an algorithm that can be used to find the square root of any positive real number.

The Bablyonian algorithm, if I remember right, does just that.

You can look it up on wikipedia.

A whole bunch of them can be found [url=http://mathworld.wolfram.com/SquareRootAlgorithms.html]here[/url]; I think [url=http://mathworld.wolfram.com/NewtonsIteration.html]this one[/url] is the most common.

I couldn't find the one DFT mentioned in either Wikipedia or Mathworld (even after correcting the spelling to Babylonian).

[quote="DiscreetFourierTransform"]http://www.mathpath.org/Algor/squareroot/algor.babylon.htm

is the first link on google.

Search Harder?

jk jk  :lol:[/quote]
That's just another name for [url=http://mathworld.wolfram.com/NewtonsIteration.html]Newton's Algorithm[/url]...

Okay then... how about calculating a cube root or an nth root using an algorithm.

Well, you can still use Newton's algorithm. Say you want to find the $ n$th root of $ k$. Start by doing $ \frac{k}{x^{n-1}}$. Whatever the result is, average it with $ x$ and repeat until your average is very close to the original value of $ x$.

Here's the algorithm I'm familiar with:

http://www.geocities.com/cnowlen/Cathy/Emat4680/Squareroot.htm


geometry
geometric transformation
rotation
analytic geometry
combinatorics unsolved
combinatorics

On a $ 10 \times 10$ chessboard, some $ 4n$ unit squares are chosen to form a region $ \mathcal{R}.$ This region $ \mathcal{R}$ can be tiled by $ n$ $ 2 \times 2$ squares. This region $ \mathcal{R}$ can also be tiled by a combination of $ n$ pieces of the following types of shapes ([i]see below[/i], with rotations allowed).

Determine the value of $ n.$

Firstly I found that n must be even, but after some time I supposed that such region $ \mathcal{R}$ doesn't exist. Am I correct?  :huh:

Maybe like this...
Color the 4n squares red. Suppose they can be covered by 2x2 squares. Look at the lower left red square, suppose its coordinates are (0,0). Then (0,0) must be covered by S or Z. But in both cases (2,0) must be also red, and (4,0), and (6,0),... or (0,2), and (0,4), and (0,6),...

I guess the conclusion of this problem should be $n=4,6,8,12,14,16$ though I haven't got a proof yet.
On the other hand the actual problem is "determine the minimum of n", which is much easier.

Could you give an example with n=4?  :o  :o

123eeee is right and I was wrong. Nice solution!


calculus
derivative
vector
calculus computations

Let $ f(x,y)\equal{}x^{3}\minus{}3xy\plus{}4y^{2}$ and $ u$ is the unit vector given by the angle $ \theta \equal{}\frac{\pi }{6}$. Find the directional derivative $ D_{u}f(x,y)$. Calculate $ D_{u}f(1,2)$.

We know that $ D_{u}f(x,y) \equal{} \bigtriangledown f(x,y) \cdot u$

From the given information, we have

$ u \equal{} \frac {\sqrt {3}}{2}\vec{\imath} \plus{} \frac {1}{2}\vec{\jmath}$ and $ \bigtriangledown f(x,y) \equal{} (3x^2 \minus{} 3y)\vec{\imath} \plus{} ( \minus{} 3x \plus{} 8y)\vec{\jmath}$.

Thus, the directional derivative is given by

$ D_{u}f(x,y) \equal{} \left(\frac {\sqrt {3}}{2}\right)(3x^2 \minus{} 3y) \plus{} \left(\frac {1}{2}\right)( \minus{} 3x \plus{} 8y).$

At a particular point (1,2),

$ D_{u}f(1,2) \equal{} \minus{} 3\left(\frac {\sqrt {3}}{2}\right) \plus{} 13\left(\frac {1}{2}\right) \equal{} \dfrac{13 \minus{} 3\sqrt {3}}{2}$.


algebra
polynomial
complex analysis
complex analysis unsolved

Why is a nonsingular homogeneous polynomial irreducible? (all in the complex numbers)

Is it?  What about something like $ \left(x^{2}\plus{}y^{2}\right)^{2}$?


geometry
geometric transformation
rotation
Duke
college

Well, Opening Day is 2 days away. Any predictions?

Here's mine:(although I am rooting for the Red Sox to win it all :D )
[hide="American League"]
Red Sox win AL East
Tigers win AL Central
Angels win AL West
Athletics win wild Card
[/hide]

[hide="National League"]
NY Mets win NL East
Cardinals win NL Central
Dodgers win NL West
Cubs win wild Card
[/hide]

[hide="Playoffs"]
AL:
Red Sox over Athletics in 4 games
Tigers over Angels in 5 games
Tigers over Red Sox in 6 games

NL:
Dodgers over Cubs in 5 games
Cardinals over Mets in 5 games
Dodgers over Cardinals in 7 games
[/hide]

[hide="World Series"]
Tigers over Dodgers in 6 games[/hide]

I don't come here to talk baseball - I get that at http://www.baseballthinkfactory.org .

um...
go A's?
nick swisher and rich harden will pwn all :)
hopefully

I can only speak for the Twins as they're the only team I actually know anything about.  They haven't changed too much from last year--really shaky starting pitching, but good relief, great closing, and a solid if not spectacular offense.  And this year, they'll have to deal with the shaky rotation all year with Francisco Liriano out.

A's!
starting lineup: above average

starting pitching: no clear-cut ace, but the top 4 are somewhat solid. not too sure about joe kennedy though.

relievers: among the best in the league - Duke, Street

bench: pretty good, especially w/ "super-sub" Scutaro

defense: among best in the league

So, overall, I think we'll have a very good chance at winning the AL West. again.


function
calculus
integration
inequalities
algebra
binomial theorem
real analysis

[hide="Solution"]
It is easy to show that the function $ \Gamma(x + 1)^{\frac {1}{x}}$ is monotonically increasing on $ [0,\infty)$.  Thus we have the integral estimate
\[ \sum_{k = 1}^n{(k!)^{\frac {1}{k}}} > 1 + \int_1^n{\Gamma(x + 1)^{\frac {1}{x}}dx}
\]
Next we establish $ \Gamma(x + 1)^{\frac {1}{x}} > \frac {x}{e} + \frac {1}{2e}$ for $ x \geq1$.  Splitting the integral we have
\[ \int_0^\infty{t^xe^{ - t}dt} = \int_0^x{t^xe^{ - t}dt} + \int_x^\infty{t^xe^{ - t}dt}
\]
Since $ t > x$ in the second integral and $ e^{ - t} > e^{ - x}$ in the first integral we have the inequality
\[ \Gamma(x + 1) = \int_0^\infty{t^xe^{ - t}dt} > \int_0^x{t^xe^{ - x}dt} + \int_x^\infty{x^xe^{ - t}dt} = \left(\frac {x}{1 + x} + 1\right)\left(\frac {x}{e}\right)^x
\]
Thus to prove the desired inequality it suffices to show $ \left(\frac {x}{1 + x} + 1\right)^\frac {1}{x} > 1 + \frac {1}{2x}$.  Indeed, by the binomial theorem we have
\[{ \left(\frac {x}{1 + x} + 1\right)^\frac {1}{x} > 1 + \frac {1}{x}\left(\frac {x}{x + 1}\right) - \frac {1}{2x}\left(\frac {x - 1}{x}\right) \left(\frac {x}{x + 1}\right)^2 = 1 + \frac {x + 3}{2(x + 1)^2}}
\]
For $ x\geq1$ we have ${ \frac {x + 3}{2(x + 1)^2}}\geq\frac {1}{2x}$ so we have established the inequality $ \Gamma(x + 1)^{\frac {1}{x}} > \frac {x}{e} + \frac {1}{2e}$ for $ x\geq1$.  Using this and the previous estimate gives
\[ \sum_{k = 1}^n{(k!)^{\frac {1}{k}}} > 1 + \int_1^n{\Gamma(x + 1)^{\frac {1}{x}}dx} > \frac {n(n + 1)}{2e} + 1 - \frac {1}{e}
\]
The upper bound can be done similarly, but it involves some messier estimates.
[/hide]


Without stoppage a train travels a certain distance with an average speed of 60km/hrand with stoppage, it covers the same distance with an average speed of 40 km/hr..on an average how many minutes per hr does the train stop during the journey?

It travels 60mph usually and 40mph with stoppage. So he must stop 20 minutes an hour right. If I am reading this problem rigt then i would say this is a basic mathcounts question!

$ \frac {60\,\mathrm{km}}{(1 \plus{} x)\,\mathrm{hr}} \equal{} 40\frac {\mathrm{km}}{\mathrm{hr}} \implies 1 \plus{} x \equal{} \frac 32\implies x \equal{} \frac 12$.

Here, $ x$ represents the [i]additional[/i] time after one hour that the train stops.

"Minutes per hour" does not quite make sense here; they are not linearly independent.

If you meant: "In one hour, how much of that hour is taken up by stoppage?", then the answer can be found by dividing $ \frac 12$ by $ \frac 32$ to get $ \frac 13$, or 20 minutes.

Equation for second interpretation:
$ 60\frac {\mathrm{km}}{\mathrm{hr}}\cdot(1 \minus{} x)\,\mathrm{hr} \equal{} 40\,\mathrm{km} \implies 1 \minus{} x \equal{} \frac 23 \implies x \equal{} \frac 13$, or 20 minutes.

[b]Note:[/b] The fact that this answer can be "obtained" by subtracting 60 by 40 is just pure coincidence.


calculus
integration
function
derivative
trigonometry
limit
calculus computations

[b]given:[/b] $ \textbf f\,(x)$ is a continuous function (having a continuous derivative) for values of $ x\in\left[x_0,\,x_1\right]$ where $ \{x_0,\,x_1\}\in\mathbb R$. it is also given that $ \int_{x_0}^{x_1}\;\;\textbf f\,(x)\;\;\textbf dx\;\equal{}\;\pi$

[b]assertion:[/b] integrals such as these, after a suitable substitution $ y\equal{}\textbf g\,(x)$ where $ \textbf g\,(x)$ is a continuous function for values of $ x\in\left[x_0,\,x_1\right]$, can [u][i]always[/i][/u] be transformed into say $ \int_{\minus{}\infty}^\infty\;\;\dfrac{\textbf dy}{1\plus{}y^2}$, which then readily yields the answer.

[b]question:[/b] prove or disprove the assertion.

I'm not very sure about this, but, just a thought  :blush:

For solve this problem I Will divide it onto 2 parts.

$ f(x) dx \equal{} \frac {dy}{y^2 \plus{} 1}$ Rearranging $ \left(y^2 \plus{} 1\right) f(x) \equal{} \frac {dy}{dx}$

$ f(x) \left(y(x)^2 \plus{} 1\right) \equal{} y'(x)$

This ODE is very simply $ y(x) \equal{} \tan \left(\int_{c_1}^x f(t) \, dt\right)$

Now the second part

$ \lim_{t\to x_1} \, y(t) \equal{} \infty$ , $ \lim_{t\to x_0} \, y(t) \equal{} \minus{} \infty$

This condiction need to be true. Otherwise, the limits transformations will not match.

Is this possible?


$F_n=F_{n-1}+3F_{n-2}$ and $F_0=0,F_1=1$

Find a closed form for $F_n$.

[quote="boxedexe"]The conditional polynomial of $F_{n+2}-F_{n+1}-3F_n=0$ is $x^2-x-3$[/quote]

wait what?


(1/3)<a<b<12 then find the integer values of 2/a+3/b
[hide]
I am wondering if 2/a, 3/b can only be integers.
If so, 2/a=1, 2, 3, 4, ...
a=2, 1, 2/3, 1/2...
but (1/3)<a<12
so a=...
and 3/b=1, 2, 3, ...
b=3,....
I give up
 [/hide]

[quote="KBabe"]i know the definition of integers.....but i never understood integers..not one bit.... :(  so can someone explain it to me???[/quote]
INTEGERS AT THE INTUITIVE LEVEL:

$\mathbb{Z}=\{\cdots,-3,-2,-1,0,1,2,3,\cdots\}$ is the set of integers. They are whole numbers with positive/negative signs. Any questions?

[quote="KBabe"]yea... one more question. Can u give an easy example?[/quote]

Of what? A question? For how many integer $x$s where $1<x<11$ is $\frac x2$ an integer?

[quote="paganinio"]I am wondering if 2/a, 3/b can only be integers. [/quote]

$\frac 2a + \frac 3b = \frac{3a+2b}{ab}$

So maybe not...

[quote="KBabe"]yea... one more question. Can u give an easy example?[/quote]
EXAMPLES???
1 is an integer
2 is an integer
0 is an integer
-100 is an integer
10000000000000 is an integer
9287493874 is an integer
-28938493849084 is an integer
$\texttt{googol}=10^{100}$ is an integer
Fractions are [b]not[/b] integers
1/2 is not an integer
$\sqrt{2}$ is not an integer
$\pi$ is not an integer
INTEGERS = WHOLE NUMBERS WITH SIGNS.
What is a 'sign' of a number? A 'sign' of a number determines whether the number is bigger or smaller than zero.
Positive number means bigger than zero. Negative number means smaller than zero.
Those are the most intuitive examples I can give you. However, if you are looking for an inductive definition... :D  :D  :D

[quote="RC-7th"][quote="3X.lich"]INTEGERS = WHOLE NUMBERS WITH SIGNS.
What is a 'sign' of a number? A 'sign' of a number determines whether the number is bigger or smaller than zero.
Positive number means bigger than zero. Negative number means smaller than zero.
Those are the most intuitive examples I can give you. However, if you are looking for an inductive definition... :D  :D  :D[/quote]

I must disagree. What about zero? Last time I checked zero was an integer. *mutters...* They should tell people when they change stuff like this *Is Unhappy About Not Knowing Something*[/quote]
OK, boy, you are so picky, here you go the correction
INTEGERS = WHOLE NUMBERS WITH SIGNS+ZERO

ok..... i think i know what integers are now.....thx.....

[quote="mathmaniack"](1/3)<a<b<12  then find the integer values of  2/a+3/b[/quote]

This might make it a bit easier for you:

[hide="What the Question is Asking"]
The first thing that you need to understand is what an INTEGER is:

in·te·ger
n. Mathematics

   1. A member of the set of positive whole numbers 1, 2, 3,... , negative whole numbers -1, -2, -3,... , and zero {0}.
   2. A complete unit or entity.

So now that we have established that lets look at what the question is asking us:

If $\frac{1}{3}<a<b<12$, find the integer values of $\frac{2}{a}+\frac{3}{b}$.

Basically it is asking for all the integers that can come out of $\frac{2}{a}+\frac{3}{b}$ when $a$ and $b$ satisfy $\frac{1}{3}<a<b<12$.

[/hide]
[hide="How to Solve"]
So lets break this down even further.

Don't attempt to plug in numbers and solve.  Its a waste of time.

$\frac{2}{a}+\frac{3}{b}=x$

What is the least possible integeral value for $x$ that can be made when  $a$ and $b$ satisfies the inequality $\frac{1}{3}<a<b<12$?

Well we know that the least possible integer can't be 0, because $a$ and $b$ have to be greater then 0.  The next integer is 1, and I am sure that there is someway that $a$ and $b$ can make 1 (considering that they can be any integer between 1/3 and 1).

So what is the largest number we can get for $x$?

Here is where you are going to have to try values...

You know it definately has to be lower then 12 (considering it has to be greater then 1/3 and lower then 12).

In order to find the largest number, you need to divide by the smallest... So lets try to think of two easy small numbers.

How about $a=\frac{1}{2}$ and $b=1$?

If you plug that in you get that $x=7$

Thats a nice number to start around.

So could x=8?

Pick a value for either $a$ or $b$, and plug it in and set $x$ equal to 8.  See you can come up with any fractions that satisfy $\frac{1}{3}<a<b<12$ and allow $x$ to be 8.

If you try $a=\frac{34}{100}$ you'll see that $b$ doesn't come even close to satisfying $\frac{1}{3}<a<b<12$.

So your range is between 1 and 7 inclusive,  so the possible integeral values that $\frac{2}{a}+\frac{3}{b}$ can output are 1, 2, 3, 4, 5, 6, and 7.

Answer: 1, 2, 3, 4, 5, 6, and 7

:D
[/hide]
:P

Farazy: [hide]that was a good idea except you made a couple of mistakes and missed some possible values.  For example, what is 2/a + 3/b when a = 1/2 and b = 3/4 ?  Also, be careful -- you acted as if x had to be in the range between 1/3 and 12, but it doesn't: only a and b have to be.[/hide]

Let me re-state what the question is asking, just in case anyone is still confused.  I have two numbers, a and b, that satisfy the given inequalities.  I want to know what possible values the expression $\frac{2}{a} + \frac{3}{b}$ can take on.  For example, I could let a = 2, b = 3.  Then I get $\frac{2}{a} + \frac{3}{b} = \frac{2}{2} + \frac{3}{3} = 2$.  So I know that 2 is one possible value for that expression.  Also, I could let a = 4, b = 12.  Then $\frac{2}{a} + \frac{3}{b} = \frac{2}{4} + \frac{3}{12} = \frac{3}{4}$, so $\frac{3}{4}$ is another possible value of the expression.  What the question asks you is which integers are possible values.  This is not an easy question at all, so you should think about it quite a while.  If you're interested in solving it, you should try answering the following 4 pairs of questions.  They'll help break the question down for you.
First:
[hide](1) If I increase a, what happens to the value of $\frac{2}{a} + \frac{3}{b}$?
(2) What if I increase b?[/hide]
Then:
[hide](3) As a result, of the first two questions, if I want to make $\frac{2}{a} + \frac{3}{b}$ as small as possible, what should I do to a and b?
(4) What if I want to make $\frac{2}{a} + \frac{3}{b}$ as large as possible?[/hide]
After that:
[hide](5) So, using the answers to (3) and (4) and the fact that a and b are between 1/3 and 12, what is an integer that is larger than all possible values of $\frac{2}{a} + \frac{3}{b}$?
(6) What a number that is smaller than all possible values of the expression?[/hide]
And finally:
[hide](7) Using part (6), what are all of the possible integer values of $\frac{2}{a} + \frac{3}{b}$?
(8) Can you find values for a and b that make each of those values work?[/hide]

[quote="JBL"]Farazy: [hide]that was a good idea except you made a couple of mistakes and missed some possible values.  For example, what is 2/a + 3/b when a = 1/2 and b = 3/4 ?  Also, be careful -- you acted as if x had to be in the range between 1/3 and 12, but it doesn't: only a and b have to be.[/hide]
[/quote]

Sorry I was a little bit tired when I did that... Just took a nap

I just redid my work...  Replace this ending with what I had before...And it should be correct.  Unless I did something wrong again...

[hide]
So what is the largest number we can get for $x$?

Here is where you are going to have to try values...

In order to find the largest number, you need to divide by the smallest... So lets try to think of he smallest possible numbers.

How about $a=\frac{34}{100}$ and $b=\frac{341}{1000}$?

If you plug that in you get that $x=14.5$

Just to be sure lets go up one more and see if 15 is at all possible...

If you try $a=\frac{3334}{10000}$ you'll see that $b$ doesn't come even close to satisfying $\frac{1}{3}<a<b<12$.

So your range is between 1 and 14 inclusive, so the possible integeral values that $\frac{2}{a}+\frac{3}{b}$ can output are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, and 14.

Answer: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, and 14[/hide]


Let $ A$ be a $ n\times n$ matrix. Does it exist a matrix $ B$ for every matrix $ A$ such that $ B^{2}\equal{}A$?

No matter what the field is, as long as $ n>1$, the answer is no. Look at nilpotent matrices that aren't zero, such as $ \begin{bmatrix}0&1\\0&0\end{bmatrix}$.

What properties should $ A$ satisfy if A exists "square root"?


Κατ'αρχάς καλησπέρα σε όλους!! Είμαι καινούριος στο forum και έχω ένα μεγάλο πάθος με τα μαθηματικά!! Είμαι στη Β λυκείου και σκέφτομαι να πάρω μέρος στο διαγωνισμό της ΕΜΕ. Η 1η φάση είναι περίπου σε 2 μήνες και θέλω να με βοηθήσετε στην προετοιμασία  :lol:  Αρχικά ποια ύλη πρέπει να καλύψω απο την Αλγεβρα της Α' Λυκείου και τη Γεωμετρία?? Επίσης ποσο πιο δύσκολα είναι τα θέματα σε σχέση με εκείνα του σχολικού (στην 1η φάση εννοείται)? Τέλος ως μια επιπρόσθετη βοήθεια έχω ένα βοήθημα άλγεβρας Α' Λυκείου, αλλα κυρίως θα ήθελα να δώσω έμαφαση σε ένα άλλο που έχω και λεγεται <<Ολυμπιαδες Μαθηματικών>> Α' Λυκειου του κ.Μπάμπη Στεργίου, το οποίο είναι εξαιρετικό και συγχαρητ'ηρια στον κ.Στεργίου που βοηθάει πολλά παιδιά που ενδιαφέρονται να αναπτύξουν τις γνώσεις τους στα μαθηματικά, απο το οποίο δεν ξέρω ποια κεφάλαια πρέπει να ξέρω οπωσδήποτε και ποια να παραλείψω... Ψάχνω απεγνωσμένα και το Problem Solving Strategies του Engels γιατι ξέρω οτι είναι κορυφαίο αλλα δεν όπως πάει δεν υπάρχει περίπτωση να το βρω κάπου για ελεύθερο κατέβασμα (παράνομο μεν αλλα...)

Αυτά τα λίγα  :D

Καλως ορισες.
Γενικα το επιπεδο του θαλη ειναι ελαφρως δυσκολοτερο απο το σχολικο επιπεδο των εξετασεων και οι απαιτουμενες γνωσεις ειναι απο την σχολικη υλη.
Κυριως μπαινει πολλη αλγεβρα και σχετικα ευκολη γεωμετρια . Χρειαζεσαι ομως και στοιχειωδεις γνωσεις θεωριας αριθμων.
Επειδη η ιστοσελιδα της ΕΜΕ δεν εχει πλεον παλια θεματα θα σου προτεινα να προμηθευτης το βιβλιο της ΕΜΕ με τα παλια θεματα το οποιο μπορεις να βρεις στα γραφεια της ΕΜΕ η στον Σαββαλα. Αν ξεκινας τωρα δεν νομιζω οτι πρεπει να ξεκινησεις με τον Engel.

Κατ' αρχάς καλώς ήρθες στο forum,και καλά posts :welcomeani:

Όντως ο Engel για αρχή είναι πολύ δύσκολος και ίσως απογοητευτείς.Θα σου πρότεινα να δουλέψεις πολύ καλά το βιβλίο του κ.Στεργίου για τη α'λυκείου και στη συνεχεια τα άλλα του βιβλία στις ανισότητες.Από εκεί και πέρα πρέπει να δουλέψεις πολύ με τους αριθμούς και τη γεωμετρία. :wink:

Δημήτρης

Ευχαριστώ για τις απαντήσεις παιδια!!!!  :lol:  Τελικά θα κάτσω να δουλέψω κυρίως  το βιβλίο του Στεργιου το οποίο είναι υπερπλήρες!! Αν δεν βρω θεματα στο νετ θα πάρω κανενα βιβλιαράκι της ΕΜΕ  :)


what  does  a circle  tangent  to  the circle  of inversion  that  doesnt  pass  through  the center  of  the circle of inversion  invert  to??

It inverts to another circle tangent at the same point. If the radius of the circle you are inverting about is $R$, then there are two cases. It depends whether the circle is internally or externally tangent. Either one of them inverts to the other. If the radius of the internal circle is $r_a$, and the radius of the external circle is $r_b$, than to find one or the other, use the relation:

$(R-r_a)\cdot(R+r_b)=R^2$ (By the definition of inversion)

At least I think its the definition. Anyway, its how inversion works.


Hello,

As some time ago someone asked about how many languages people here speak, and everyone was talking about natural languages, I wonder about artificial languages (many people don't even consider them languages).

Personally, I started learning Esperanto. For those speaking romance languages (Spanish, French, Italian, Portuguese, Romanian, Sardinian) it should be very easy. I thought that it could be interesting to create one of myown, but I havent finish (I barely started!)

Best,

Do Pig-Latin Bork, Bork Bork, and Elmer Fudd count?

I know some Goa'uld and some Klingon :P :D I'm planning to learn some Ancient too in the future :)


Let $ a \plus{} b \plus{} c \equal{} p,\ ab \plus{} bc \plus{} ca \equal{} q,\ abc \equal{} r$,

$ a^3 \plus{} b^3 \plus{} c^3 \equal{} p(p^2 \minus{} 3q) \plus{} 3r,\ a^3b^3 \plus{} b^3c^3 \plus{} c^3a^3 \equal{} q(q^2 \minus{} 3rp)\plus{}3r^2$,

$ 27(a^3 \plus{} 3)(b^3 \plus{} 3)(c^3 \plus{} 3)\geq 4(3ab \plus{} 3bc \plus{} 3ca \plus{} abc)^2$

$ \Longleftrightarrow 27\{27 \plus{} 9(p^3 \minus{} 3pq \plus{} 3r) \plus{} 3(q^3 \minus{} 3pqr \plus{} 3qr^2) \plus{} r^3\}\geq 4(3q \plus{} r)^2$

To be continued....


x y and z are positive integers, x>y and y>z  when x,y,and z are entered into a machine, the result is $(x-y)(y-z)$ what is the probability that the result is not odd?

[hide]hmmmmm, im guessing $\frac{3}{4}$[/hide]

I think you are right, because there are 4 possibilities (odd, odd; odd, even; even, odd; and even, even.) Only one of these are odd when multiplied together, so when you want the probability against it, it's 1 - 1/4 or $\frac{3}{4}$

more concrete?

[hide]

say y is even.

the only way the product will be odd is if both x and z are odd (1/4 probability).

say y is odd.

the only way the product will be odd is if both x and z are even (1/4 probability).

thus P(odd) = 1/2*1/4 + 1/2*1/4 = 1/4, and [b]P(even) = 3/4[/b][/hide]


[quote="thanhnam2902"][color=darkblue]Let $ (O)$ is a circle, let $ M$ is a point lie out site of circle. Let $ MA$ be tangent to $ (O)$ at point $ A$, let $ (d)$ is a change point and $ (d)$ pass point $ M$ and $ (d)$ meet circle $ (O)$ at $ B$ and $ C$, $ (MB < MC)$. Let $ D$ is midpoint of $ BC$, let $ AD$ meet circle $ (O)$ at point $ E$. Find place of $ (d)$ for area of $ MEC$ triangle is maximum.[/color][/quote]
Let $ MI$ be the tangency of $ (O)$. We have quadrilateral $ BICA$ is harmonic and this result is well-known: $ IE//BC$
So $ S_{MEC}\equal{}S_{MIC}\equal{}\frac{1}{2}MI.CI.sin \angle MIC\leq \frac{1}{2} MI.R.sin 90^o$
Therefore $ Max S_{MEC}\equal{}1/2MI.R.sin 90^o$ If and only if BE is the diameter of (O)