I've been trying to solve this problem using the Chinese Remainder Theorem, but it's not coming out right.

What is the smallest integer that leaves a remainder of 4 when divided by 6, and leaves a remainder of 2 when divided by 9?

Here's my work:

$N\equiv4(\bmod6), N\equiv2(\bmod9)$
$N=6i+4, N=9j+2$
$6i+4=9j+2$
$9j=6i+2$
$9j\equiv2(\bmod6)$

Here, I couldn't come up with any $x(\bmod6)$ where $\frac{x}{9}$ was an integer, so I just left it.  I don't know if this is the problem or not, or what to do if it is...?

$j\equiv\frac{2}{9}(\bmod6)$
$j=6d+\frac{2}{9}$
$N=9(6d+\frac{2}{9})+2$
$N=54d+4$
$N\equiv4(\bmod54)$

This gives an answer of 4, but then you have $\frac{4}{9}$, which leaves a remainder of 4, not 2...

Could someone please tell me why this isn't working?

$9j = 6i + 2$ has no solutions in integers; taking $\bmod 3$, we have $0 \equiv 2 \bmod 3$, obviously absurd.  Are you sure the problem is correct?

(Edit and Note:  The Chinese Remainder Theorem is only applicable of all of the $\bmod$s that you're using are pairwise coprime.  In this case, they aren't, and so a solution is not guaranteed.)

[quote="DanK"]Thank you for clarifying that.  I have to be more careful when I insert random numbers into a problem to practice a new theorem...

And just because I am severely sleep deprived right now, could you refresh my memory on what pairwise coprime means?[/quote]

None of the mods can have a common factor.  That is, given the mods $n_1, n_2, n_3, ... n_k$, then $i \neq j \implies gcd(n_i, n_j) = 1$.

[quote="DanK"]Ok, I get it now.  Thank you.

So is there a way to solve problems where they aren't pairwise coprime, or do those just never have any real solutions?[/quote]

They're not guaranteed to have solutions, but they still can.  For example, $x \equiv 2 \bmod 6$ and $x \equiv 2 \bmod 9$ obviously has an infinite number of solutions.  :)


calculus
integration
number theory proposed
number theory

Find the pair of integer $(a,b)$ such that $0\leq a,\ 0\leq b\leq 5,$ for which the following system of equation with respect to $x,y$ has integral solution.

\[ \begin{cases} ax+3by=1 \\ bx+ay=0 \end{cases}  \]

is there any other way than just case work one by one?

[hide]
well , i just case work from $b=0$ to $b=5$ and get

$(a,b)=(1,0),(2,1),(7,4)$
[/hide]

editted mistake

shyong, $a$ isn't negative, anyway there is another method.

Can anyone solve the problem in another method? :)

the first equation implies that (x,y)=(a,b)=1. from the second one we get that b/y=-a/x=(+/-) 1. Hence, (a,b)=(x,-y) or (-x,y)
the second one leads to $ 3b^2\equal{}a^2\plus{}1$ which is impossible mod 3
The first one gives $ a^2\equal{}3b^2\plus{}1$ and then plugging in values for b from 0 to 5, we get the possibilities


geometry
rectangle
number theory
least common multiple

[hide]I think it would be the same as finding the LCM of $10$ and $15$, which is $= 30$, and so it would be $(30/10)*(30/15) = 6$[/hide]

[quote="MCrawford"]What is the smallest number of 10" by 15" rectangular tiles required to form a square with no cutting or gaps?[/quote]
[hide="solution"]
30" by 30".

30"/10" = 3
30"/15" = 2

3 x 2 = [b]6[/b]
[/hide]

Hm...

we could fold it! :D

or, if it's a colored paper,

put it on a white desk and hide a portion with a white paper :D

[hide]You have to find the least common multiple of 10 and 15, which is 30.  You have to times 10 by 3 and 15 by 2.  2 x 3 =[b]6[/b][/hide]


inequalities
trigonometry
quadratics
inequalities theorems

I want to open a topics, in wich we found the substitution which keeps  conditioned inequalities.Please anyone to contribute at this topics!It is for the good of this community!Have a nice day! :lol:

1. If $ abc\equal{}1$ where $ a,b,c \in \mathbb{R}^{\plus{}}$ then:

$ a\equal{} \frac{x}{y}, b\equal{}\frac{y}{z}, c\equal{}\frac{z}{x}$ where $ x,y,z \in \mathbb{R}^{\plus{}}$

2. If $ a,b,c$ are sides of a triangle then:

$ a\equal{}x\plus{}y, b\equal{}y\plus{}z, c\equal{}z\plus{}x$ where $ x,y,z \in \mathbb{R}^{\plus{}}$.

Trigonometric substitution:
1.If $ x^2\plus{}y^2\plus{}z^2\plus{}2xyz\equal{}1$,then $ x\equal{}\cos{\alpha},y\equal{}\cos{\beta},z\equal{}\cos{\gamma}$.
2.If $ x\plus{}y\plus{}z\equal{}xyz$,then $ x\equal{}\tan{\alpha},y\equal{}\tan{\beta},z\equal{}\tan{\gamma}$.
3.If $ xy\plus{}yz\plus{}zx\plus{}2xyz\equal{}1$,then $ xy\equal{}\cos^2{\alpha},xz\equal{}\cos^2{\beta},yz\equal{}\cos^2{\gamma}$.
4.If $ xyz\equal{}x\plus{}y\plus{}z\plus{}2$,then $ \frac{1}{xy}\equal{}\cos^2{\alpha},\frac{1}{xz}\equal{}\cos^2{\beta},\frac{1}{yz}\equal{}\cos^2{\gamma}$
5.If $ xy\plus{}yz\plus{}zx\equal{}1$,then $ x\equal{}\cot{\alpha},y\equal{}\cot{\beta},z\equal{}\cot{\gamma}$.

If $ xy\plus{}yz\plus{}zx\plus{}xyz\equal{}4$ then,there exist $ a,b,c\in\mathbb{R}^{\plus{}}$,such that
$ x\equal{}\frac{2a}{b\plus{}c},y\equal{}\frac{2b}{a\plus{}c},z\equal{}\frac{2c}{a\plus{}b}$.

[quote="quangpbc"][quote="Erken"]If $ xy \plus{} yz \plus{} zx \plus{} xyz \equal{} 4$ then,there exist $ a,b,c\in\mathbb{R}^{ \plus{} }$,such that
$ x \equal{} \frac {2a}{b \plus{} c},y \equal{} \frac {2b}{a \plus{} c},z \equal{} \frac {2c}{a \plus{} b}$.[/quote]

You also see this substitution in : http://trungtuan.wordpress.com/2007/08/11/cac-phep-dat-thuong-dung-trong-chung-minh-bat-dang-thuc but not English  :P[/quote]
it has been posted on this forum too,look at:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=169195[/url]

You also see some problems for application this substitution  :wink:

This is not really a condition but I have seen questions involving the expression:

$ \frac{x\plus{}y}{1\minus{}xy}$ and the substitution is $ x\equal{} \tan{\alpha}, y\equal{} \tan{\beta}$

anything involving nasty quadratic terms under roots -> trigonometric substitution :) (just be sure to know many addition theorems)

for $ abc \equal{} 1$ the substitution $ x\equal{}\frac{1}{a}$, $ y\equal{}\frac{1}{b}$, $ z\equal{}\frac{1}{c}$ can also help.

[quote="bertram"]
for $ abc \equal{} 1$ the substitution $ x \equal{} \frac {1}{a}$, $ y \equal{} \frac {1}{b}$, $ z \equal{} \frac {1}{c}$ can also help.[/quote]
Could you show some examples,where you can use this substitution?

I only know IMO 1995.2:

$ abc \equal{} 1\Rightarrow \sum_{cyc}\frac{1}{a^3(b \plus{} c)}\ge\frac {3}{2}$

It can help bringing variables from the denominator to the numerator as you have $ xyz \equal{} abc \equal{} 1$ too.


i must admit that i got the last prob on the AMC 10 B wrong due to the fact that i thought it said it added up to 120 instead of 125.  For the 120 igot 68 which was an answer and i didn't reread and discover that i made a mistake.  READ CAREFULLY!!!  It can save you a lot of frusteration, time and points.

I'd have to agree with that, last year I missed AIME qualification by one problem, which i would have gotten correctly had I read correctly.  I've seen friends improve math scores without any studying, just taking a bit longer to read more carefully.

The Art of Reading Carefully... someone needs to write a book on that.  :P

yes yes...read carefully...and know that 1+1=2 not 1  >.<

I got number 13 wrong, I read the problem as "How many number between 1 and 2005 are integer multiples of 3 or 4?". And that was the only one I got wrong. :(

Reading carefully also applies to your own scratchwork..
(make sure you don't read crossed-out 2's as x's, etc)

I agree with the reading carefully thing. Especially on last year's AIME. Uchh. Remember the worker problem. Everyone who read/interpreted one way got 500, while the other reading/interpretation (that the problem author intended) yielded 766. I was one of the 500 putters, and it cost me the USAMO. Be wary of such nice answers.


Let ABC be a triangle such that $\angle ACB = 2\angle ABC$ and let D be ap oint in the interior of ABC satisfying $AD=AC$ and $DB=DC$. Prove that $\angle BAC = 3\angle BAD$


In the first equation we obtain $x_{n}=0$ or $x_{n}=\frac{-1}{4n^{2}}$
so we can get $ab \leq e^{3}$

Hello N.T.TUAN
[quote="N.T.TUAN"]Find all pairs of positive real numbers $(a, b)$ such that for every $n \in\mathbb{N}$ and every real root $x_{n}$ of the equation $4n^{2}x = \log_{2}(2n^{2}x+1)$ we have $a^{x_{n}}+b^{x_{n}}\geq 2+3x_{n}.$[/quote]

For $n=0$, every real x is root of the equation, and so, no matter what are the $x_{i}$ for $i>0$, the requirement is

What are the pairs of positive real numbers $(a, b)$ such that for every $x \in \mathbb{R}$ $a^{x}+b^{x}\geq 2+3x.$

Let$f(x)=a^{x}+b^{x}$ and $g(x)=3x+2$
$f(x)$ is convex and  $f(0) = g(0)$
So the requirement is $f'(0)=g'(0)$ $\Rightarrow $ $ln(a)+ln(b)=3$

Hence $ab=e^{3}$

--
Patrick

[quote="N.T.TUAN"]But here $\mathbb{N}$ is set of all positive integer numbers.[/quote]

You're right. I Make always the same error. Here is the correction.

We have $f(x)=a^{x}+b^{x}$ and $g(x)=3x+2$ and we want $f(x_{n})\geq g(x_{n})$
For $x_{n}=0$, the requirement is fulfilled.

So we want only $f(-\frac{1}{4n^{2}})\geq g(-\frac{1}{4n^{2}})$ $\forall n\in \mathbb{N}$

Since $f(0)=g(0)$ and $f(x)$ is convex, it's immediate to see that :
If $f(-\frac{1}{4n1^{2}})\geq g(-\frac{1}{4n1^{2}})$ and $n2 < n1$ then $f(-\frac{1}{4n2^{2}})\geq g(-\frac{1}{4n2^{2}})$

So the requirement is for the limit (when $n \rightarrow+\infty$) and so is $f'(0)\leq g'(0)$, so $ln(a)+ln(b) \leq 3$, so $ab \leq e^{3}$ as kyoshiro_hp said (shortly  :)  ).

--
Patrick

Sorry, I don't understand here.
[quote="pco"]

Since $f(0)=g(0)$ and $f(x)$ is convex, it's immediate to see that :
If $f(-\frac{1}{4n1^{2}})\geq g(-\frac{1}{4n1^{2}})$ and $n2 < n1$ then $f(-\frac{1}{4n2^{2}})\geq g(-\frac{1}{4n2^{2}})$
[/quote]
But I can have $ab\leq e^{3}$, by $f'(0)\leq g'(0)$. ( Not use above text) . In fact, It is from
\[\frac{f(\frac{-1}{4n^{2}})-f(0)}{\frac{-1}{4n^{2}}-0}\leq \frac{g(\frac{-1}{4n^{2}})-g(0)}{\frac{-1}{4n^{2}}-0}, \]
now let $n\to\infty$. Maybe I know little on convex functions.
Final, if $(a,b)$ is answer then $ab\leq e^{3}$, but if $ab\leq e^{3}$ then is it answer?  :maybe:

[quote="N.T.TUAN"]Final, if $(a,b)$ is answer then $ab\leq e^{3}$, but if $ab\leq e^{3}$ then is it answer?  :maybe:[/quote]

Ok, I'll try to be more precise :

$f(x)=a^{x}+b^{x}$  with $a>0$ and $b>0$
$g(x)=3x+2$

Let $x_{n}=\frac{-1}{4n^{2}}$

We have $f(x_{n})=f(0)+f'(0)x_{n}+f''(h_{n})\frac{x_{n}^{2}}{2}$ for some $h_{n}\in [x_{n},0]$ and $g(x_{n})=g(0)+x_{n}g'(0)$
$f(x_{n})\geq g(x_{n})$ $\Leftrightarrow$ $f(0)+f'(0)x_{n}+f''(h_{n})\frac{x_{n}^{2}}{2}\geq g(0)+x_{n}g'(0)$ and, since $f(0)=g(0)$ and $x_{n}<0$:
$f(x_{n})\geq g(x_{n})$ $\Leftrightarrow$ $f'(0)+f''(h_{n})\frac{x_{n}}{2}\leq g'(0)$ and :
$f(x_{n})\geq g(x_{n})$ $\Leftrightarrow$ $f'(0)-g'(0)\leq f''(h_{n})\frac{-x_{n}}{2}$

Then, and since $f''(h_{n})>0$ (f is convex) :
1) $f(x_{n})\geq g(x_{n})$ $\Rightarrow $ $f'(0)-g'(0)\leq f''(h_{n})\frac{-x_{n}}{2}$ $\forall n\in\mathbb{N}$ $\Rightarrow $ $f'(0)-g'(0)\leq 0$ and  $f'(0)-g'(0)\leq 0$ is a necessary condition
2)  $f'(0)-g'(0)\leq 0$ $\Rightarrow $ $f'(0)-g'(0)\leq 0\leq f''(h_{n})\frac{-x_{n}}{2}$ $\forall n\in\mathbb{N}$ $\Rightarrow $ $f(x_{n})\geq g(x_{n})$ and  $f'(0)-g'(0)\leq 0$ is a sufficient condition

So $ab\leq e^{3}$ is a necessary condition AND a sufficient condition.
I hope this would be clear enough (even with my poor english language)  :)

--
Patrick

[quote="N.T.TUAN"]Thank you very much! But this
[quote="pco"] $f'(0)-g'(0)\leq f''(h_{n})\frac{-x_{n}}{2}$ $\forall n\in\mathbb{N}$ $\Rightarrow$ $f'(0)-g'(0)\leq 0$ [/quote]
Why?[/quote]

You just have to do $n\rightarrow\infty$ in the inequality. Then $x_{n}\rightarrow 0$, $h_{n}\rightarrow 0$ and $f''(h_{n})\frac{-x_{n}}{2}\rightarrow 0$.

--
Patrick

Sorry  :blush: Thanks! Now this topic has got the two solutions  :wink:


group theory
abstract algebra
induction
AMC

n people are attending Fierytycoon's birthday party.  Every subgroup of 4 people contains at least one person who knows the other three people.  What is the minimum number of people that know everybody?

Proof: <by induction>

[hide]For n people, we need at least n-3 to know everybody.

Obviously, n = 4 works.

First, we must show that there must be at least 1 who knows everybody. We can show this by contradiction...if there is no one that knows everybody, we can select two pairwise individuals who do not know each other, and thus our condition is broken.

Lets say n = k works. Now look at n = k+1. Since there must be at least one person who knows everyone, let us take that person out. In the remaining group of k, we know that there are k-3 that know everyone in that group. Since the person we took out knew everyone, these k-3 people must also know him, and thus they know everyone. We then have (k+1) - 3 people who know everyone.

[/hide]


Let $ a,b,c$ be positive real numbers,then prove that
${ \dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a} \geq 3\sqrt{\dfrac{a^2+b^2+c^2}{ab+bc+ca}}}$

[quote="Phun_TZ"]Let $ a,b,c$ be positive real numbers,then prove that
${ \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a} \geq 3\sqrt {\dfrac{a^2 + b^2 + c^2}{ab + bc + ca}}}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=179719
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=217810

You can also try to prove this easier inequality (it's got an easier proof)
$ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\geq 3 \sqrt[3]{\frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca}}$ for a,b,c>0 ;)

[quote="limes123"]You can also try to prove this easier inequality (it's got an easier proof)
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\geq 3 \sqrt [3]{\frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}}$ for a,b,c>0 ;)[/quote]
Ofcourse my friend, since we have
\[ \frac{a}{b} \plus{} \frac{b}{c} \plus{} \frac{c}{a} \ge \frac{{\left( {a \plus{} b \plus{} c} \right)^2 }}{{ab \plus{} bc \plus{} ca}} \equal{} \frac{{a^2  \plus{} b^2  \plus{} c^2 }}{{ab \plus{} bc \plus{} ca}} \plus{} 2
\]
Denote $ t \equal{} \sqrt[3]{{\frac{{a^2  \plus{} b^2  \plus{} c^2 }}{{ab \plus{} bc \plus{} ca}}}}$. Hence it suffices us to show that
\[ t^3  \minus{} 3t \plus{} 2 \ge 0
\]
\[ \Leftrightarrow \left( {t \minus{} 1} \right)^2 \left( {t \plus{} 2} \right) \ge 0
\]
which is true. :)

Or
$ \frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca}\plus{}1\plus{}1\geq3\sqrt[3]{\frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca}}$ by AM-GM.


Find the values of $ m\in \mathbb{R}$ such that $ E\equal{}(2\minus{}m)x^2\plus{}mx\minus{}3<0\ ,\ (\forall)x< 0$ .

Its a quadratic equation. Use the quadratic formula.

[hide="How to solve this"]
First solve the equality case, you get 2 solutions(probably, if less than it's easier)

Then, divide the set of real numbers into 3 intervals, with boundaries on solutions, and try solving for any number in those intervals, and if it works for these numbers, then it works for these intervals...
[/hide]


hi guys,

can anyone provide me with a detailed solution of problem 2 :?:

rajatgauss
______________________________
genius is in everyone

Go here for the discussion on problem 2

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1380236#1380236[/url]

Befuddlers


function
real analysis
real analysis unsolved

Let $ f: (0,1)\to\mathbb R$ be a continuous function such that $ f\left(\frac12\right)\ge f(x)$ for all $ x\in(0,1).$ Show that $ f$ is not an injection. (Don't assume that $ f$ is differentiable.)

Let  $ m\equal{}\max \{f(1/4),f(3/4)\}$, and $ y_0\in (m,f(1/2))$. Since $ f$ is continuous there exists $ x_1\in[1/4;2/4]$ such that $ f(x_1)\equal{}y_0$, and there exists $ x_2\in[2/4;3/4]$ such that $ f(x_2)\equal{}y_0$.
So $ f(x_1) \equal{} f(x_2)$ and $ x_1\neq x_2$

[quote="anik_andrew"]Let  $ m \equal{} \max \{f(1/4),f(3/4)\}$, and $ y_0\in (m,f(1/2))$. Since $ f$ is continuous there exists $ x_1\in[1/4;2/4]$ such that $ f(x_1) \equal{} y_0$, and there exists $ x_2\in[2/4;3/4]$ such that $ f(x_2) \equal{} y_0$.
So $ f(x_1) \equal{} f(x_2)$ and $ x_1\neq x_2$[/quote]

That's certainly the right idea, but you don't know $ m < f(1/2)$, so you have to add a bit. Me, I would argue that $ f([1/2,1))$ is an interval and so is $ f((0,1/2])$. If either of these intervals is a single point, then f is not injective. Otherwise there is $ \epsilon > 0$ such that each of these intervals contains $ f(1/2)\minus{}\epsilon$, which  again implies $ f$ is not injective.

i think that it's obvious that if $ f(1/2)\equal{}f(1/4)$ or $ f(1/2)\equal{}f(3/4)$ then f is not injective...

Which brings up the question of how detailed our solutions should be. I usually just sketch the main ideas and go over the really difficult parts; Kent's posts, on the other hand, are of exemplary meticulousness. In school level fora it is important to write all solutions in full (learning correct mathematical writing is a part of the process there) but in CP one may, probably, be more relaxed. If I were grading an exam, I would definitely reduce some points for the missing cases in anik_andrew's solution. On the other hand, if I were explaining the problem to a student, I would say even less than he did leaving more details to the student to figure out.

Anyway, Happy New Year!  :D


geometry
3D geometry
inequalities
symmetry

Find the least real number, $ r$, such that for each triangle with sidelength $ a$, $ b$, and $ c$, we have \[ \frac{\text{max}\, \{a,b,c\} }{\sqrt[3]{a^3\plus{}b^3\plus{}c^3\plus{}3abc}}<r\]

[hide]Without Loss of Generality, assume that c = max{a,b,c}. Since a,b,c form a (I assume nondegenerate, because if degenerate triangles are allowed, there will be equality) triangle, a+b>c. Now
\[ 2 \equal{} \frac {2c^3}{c^3} < \frac {(a \plus{} b)^3 \plus{} c^3}{c^3} \equal{} \frac {a^3 \plus{} b^3 \plus{} 3ab(a \plus{} b) \plus{} c^3}{c^3}
\]

\[ < \frac {a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc}{c^3}
\]
Therefore, 2<$ \frac {a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc}{c^3}$. Recipricating (which changes < into >) and cube rooting gets $ \frac {1}{\sqrt [3]{2}} > \frac {c}{\sqrt [3]{a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc}}$. Let a=b and let c approach 2a. Then the expression approaches $ \frac {2a}{\sqrt[3]{a^3 \plus{} a^3 \plus{} 8a^3 \plus{} 6a^3}} \equal{} \frac {2a}{\sqrt [3]{16a^3}} \equal{} \frac {1}{\sqrt [3]{2}}$. Therefore, r = $ \frac {1}{\sqrt [3]{2}}$[/hide]

[quote="pythag011"][hide]Without Loss of Generality, assume that c = max{a,b,c}. Since a,b,c form a (I assume nondegenerate, because if degenerate triangles are allowed, there will be equality) triangle, a+b>c. Now
\[ 2 \equal{} \frac {2c^3}{c^3} < \frac {(a \plus{} b)^3 \plus{} c^3}{c^3} \equal{} \frac {a^3 \plus{} b^3 \plus{} 3ab(a \plus{} b) \plus{} c^3}{c^3}
\]

\[ < \frac {a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc}{c^3}
\]
[/hide][/quote]
I think this inequality is not true since (a+b)>c, so the sign should be flipped from < to >, which means: \frac {a^3 + b^3 + 3ab(a + b) + c^3}{c^3}
\]

\[
> \frac {a^3 + b^3 + c^3 + 3abc}{c^3}.
My approach:
Depending on the "natural experience and guessing",I guess the equality happens in the special case: the triangle is degenerated with two sides are equal, which means I will prove that r=1/third root(2) will be the least real number.
Now, we need to prove two steps, 1st is to show that the inequality satisfied by r=1/third root(2) and 2nd is to show that this is the least number. The 2nd step, as we can see that it easily shows by pointing out c=max(a,b,c) and c=2a=2b. So, we just need to consider the 1st step, which will be proved as following:
Clearing all denominator and cube both sides, we will need to prove:
a^3+b^3+3abc>c^3, which is equivalent to: (a+b-c)[(a+b)^2+(a+b)c+c^2-3ab]>0(1)
W.L.O.G, assume that c=max(a,b,c), we will have: ac>ab
                                                                     bc>ab
                                                                     c^2>ab
Adding sides by sides, we proved that (a+b)^2+(a+b)c+c^2-3ab>0.Therefore, (1) is true an the inequality above is proved.The equality should happen since I think this problem is little ambiguous(we can accept the special case of the triangle).In conclusion, [b]r=1/third root(2) [/b]is the sufficient and necessary value to satisfy the inequality above(Q.E.D)

@pythag011: ghjk's comment is correct...I didn't catch that mistake when I read your proof the second time, but the second use of $ a \plus{} b > c$ is the wrong way.

@ghjk: There is a slightly nicer way to finish the problem from here:

$ a^3 \plus{} b^3 \plus{} ( \minus{} c)^3 \minus{} 3(a)(b)( \minus{} c)$

Note that this is in the form of the common factorization... now it is just

$ (a \plus{} b \minus{} c)\frac {1}{2}\cdot ((a \minus{} b)^2 \plus{} (a \plus{} c)^2 \plus{} (b \plus{} c)^2) > 0$

---

I agree that a good way to motivate a solution is to check equality cases. They generally occur when two of the variables are equal (especially when there is symmetry) or if some of the variables are equal to zero.

The above can be used to prove AM-GM in three variables.


trigonometry
calculus
integration
function
inequalities
calculus computations

[quote="Jimmy"]Prove that
\[ \sum \frac{|\sin n|}{n}  \]
diverges.[/quote]

using integral i thinks

use comparation test with harmonic series. The rest is yours. You can do it yourself

[quote="Handhi"]use comparation test with harmonic series. The rest is You can do it yourself[/quote]
no you can't. the comparation test give nothing.
My idea silece that $\int_{0}^{\infty} \frac{|\sin x|}{x} \, dx$ diverges.

Anyway again somepeople from indonesia register. it has been three people within 2 days.

[quote="blahblahblah"]Neither of those approaches works, unless you are referring to something far more involved than what you hinted at.[/quote]

blahblahblah you thinks that works  :? . How it can be?

things can be done far more easy.

first of all, integral test doesn't work: sequence must be decreasing, and the function you consider has to be decreasing, too.
for example, just consider a slightly different series: $\displaystyle \sum \frac{|\sin(\pi n)|}{n}$.
this series of course converges (being constantly $0$), but the integral test fails.

again, the comparation test (with harmonic series) doesn't work, since the inequality comes out to be reversed...

we can now play with numbers: consider for any integer $n$, consider $\left[n\right]$ as the only real number $x \in \left[-\frac{\pi}2, \frac{\pi}2\right)$ such that there exist $k$ integer such that $n = x + k\pi$.
now, consider the the interval $I = (-\frac12,\frac12) \subset \left[-\frac{\pi}2, \frac{\pi}2\right)$. if $\left[n\right] \in I$ then $\left[n\right] \not\in I$, for the choice we've made.
of course, $|\sin n| = |\sin\left[n\right]|$. call now $\alpha = \sin 1> 0$.
the term has all positive terms, so we can write \[ \displaystyle 2\sum \frac{|\sin n|}{n} = \sum \frac{|\sin n|}{n} + \frac{|\sin(n+1)|}{n+1} > \sum \frac{\alpha}{n} \]
but the last one diverges, so does the first one.

[quote="Jimmy"]Prove that
\[ \sum \frac{|\sin n|}{n}  \]
diverges.[/quote]

$\sum \frac{|\sin n|}{n} \geq \sum \frac{\sin^2 n}{n} = \sum \frac{1-\cos 2n}{2n}$.

Since $\sum \frac{1}{n}$ diverges, and $\sum \frac{\cos n}{n}$ converges (by Dirichlet's Test), we know that $\sum \frac{1-\cos 2n}{2n}$ diverges.

[quote="ma_go"]things can be done far more easy.

first of all, integral test doesn't work: sequence must be decreasing, and the function you consider has to be decreasing, too.
for example, just consider a slightly different series: $\displaystyle \sum \frac{|\sin(\pi n)|}{n}$.
this series of course converges (being constantly $0$), but the integral test fails.

again, the comparation test (with harmonic series) doesn't work, since the inequality comes out to be reversed...

[/quote]
 I'm excactly agree with you, That's what i realized after a few minutes posting it. Thanks for correction  :P .


hi,
I am wondering if some one can help me on this problem...I checked out the one that is already posted but my problem is slightly a little different.

Using lemma 1 that I posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=168879, we have that the number of positive integer solutions $ (a_1,a_2,...,a_n)$ to the equation $ a_1 \plus{} a_2 \plus{} ... \plus{} a_n \equal{} k$ where $ k\ge n$ is $ \binom{k \minus{} 1}{n \minus{} 1}$, then the number of integer solutions $ (x,y,z)$ to the equation $ x \plus{} y \plus{} z \equal{} k$ with $ k\ge 3$ is $ \binom{k \minus{} 1}{2}$
Then the number of positive integers $ x,y,z$ such that $ x \plus{} y \plus{} z\le 20$ is:
$ \sum_{i \equal{} 3}^{20} \binom{i \minus{} 1}{2} \equal{} \sum_{i \equal{} 2}^{19} \binom{i}{2} \equal{} \sum_{i \equal{} 2}^{19} \binom{i}{i \minus{} 2} \equal{} \sum_{i \equal{} 0}^{18 \minus{} 1} \binom{2 \plus{} i}{i} \equal{} \binom{18 \plus{} 2}{18 \minus{} 1}$ (by the hockey stick theorem - http://mathworld.wolfram.com/ChristmasStockingTheorem.html )
$ \equal{} \binom{20}{17} \equal{} \binom{20}{3} \equal{} 1140$
[b]Answer: [/b]$ 1140$

[hide="Instead of those summations"]
We introduce another variable. If $ x \plus{} y \plus{} z\le 20$, then there exists a positive integer $ a$ such that $ x \plus{} y \plus{} z \plus{} a \equal{} 21$. So we just need to find the number of positive integer solutions to that. Thus the answer is

$ \binom{21\minus{}1}{4\minus{}1}\equal{}\binom{20}{3} \equal{} 1140$
[/hide]

this is wonderful, thanks a lot.

In meanwhile I was doing it in a long way buy counting all the possible solution for cases when x=1 and y takes all the values from 1 to 18 and and z takes all the values from 18 to 1, then when x=2 and so on ... and finally I got 1140.

well,  I am curioz to know from where is the lema taken and how do you know all  of this so well....

The lemma is available [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=168879]here[/url] if you want it like he said in his post, but tjhance's solution doesn't explicitly involve any lemmas (not really) and is faster. The lemma itself is just proven by rem.

In fact, I think you said you did exactly the same thing as has been done here - counted all the combinations - but he took a shortcut by using [url=http://en.wikipedia.org/wiki/Binomial_coefficient]binomial coefficients[/url], or, in other words, this shortcut formula to find the number of combinations, with $ n$ to choose from and $ r$ being chosen:

$ \frac{n!}{r! \cdot (n-r)!}$

So you had to find the number of combinations of $ 20$ objects (in this case numbers), picking $ 3$ of them each time, so $ n = 20$ and $ r = 3$. Instead of counting them all out which takes bloody forever:

\begin{align*}
\binom{20}{3} &= \frac{20!}{3! \cdot (20-3)!}\\
&= \frac{18 \times 19 \times 20}{3 \times 2}\\
&= \frac{380 \times 18}{6}\\
&= \frac{6840}{6}\\
&= \boxed{1140}
\end{align*}

And most calculators now have an $ ^{n}C_{r}$ button which you can use to type in $ n = 20$, $ r = 3$ and have it work out the $ \frac{n!}{r! \cdot (n-r)!}$ formula for you to get straight to the answer.

It's a pretty common formula and method, especially with questions involving the field of [url=http://en.wikipedia.org/wiki/Combinatorics]combinatorics[/url], like this one.


How does one label and refer to a phrase or a sentence that is neither a theorem nor an equation?

Use a new counter which is linked to label by \refstepcounter:
[code]\newcounter{statement}
\newcommand{\mylabel}[1]{\refstepcounter{statement} (\thestatement)\label{#1}}
This is a statement \mylabel{important} \\
This is another statement \mylabel{rubbish}\\
This statement \ref{rubbish} is nonsense but you must remember this one \ref{important}[/code]
You could also use optionally use \renewcommand{\thestatement}{\fnsymbol{statement}} after \newcounter{statement} which will change the numbering to a sequence of no more than 9 symbols.


[quote="Altheman"][url]http://mathworld.wolfram.com/Derangement.html[/url][/quote]

I am not able to relate these two problems i.e. derangemnet and my problem

oh woops, what was i thinking, nevermind that

[quote="RohanSingh"]Suppose I want to put the natural numbers from 1 to n in such a way that no number is followed by its immediate sucessor. Example: 14235 is not valid where 13254 is valid. Find a recuurence relation for this.[/quote]

Let f(n) be the number of valid arrangements for n numbers.

f(1) = 1
f(2) = 1. The only sequence is 2, 1.

Now, each sequence with n = 3 can be formed by adding a 3 to a n = 2 sequence. There are two ways to do this: 3, 2, 1 and 2, 1, 3. So f(3) = 2

Similarly, each sequence with n = 4 can be formed by adding a 4 in any of the three appropriate spots for any of the two sequences. So f(4) = 3f(3).

Continuing in this manner, we find that f(n+1) = n*f(n).

Since f(n) is actually equal to (n-1)!, we can provide the explicit formula, f(n+1) = n!, or f(n) = (n-1)!.

Denote by $a_n$ number of proper arrangments i.e. arrangments where no number is followed by its immediate sucessor.
If $n>2$ then the recurrence relation is: $a_n=(n-1)a_{n-1}+(n-2)a_{n-2}$.

Proof.
Proper arrangments can be made in exactly two distinct manners.

[b]1.[/b] If numbers $1,2,\ldots,n-1$ are properly arranged, then there are only $n-1$ places where we can put $n$ (place that succeeds number $n-1$ is forbidden).
There are $(n-1)a_{n-1}$ ways to do that.

[b]2.[/b] If numbers $1,2,\ldots,n-1$ are arranged in a manner that [b]exactly[/b] one pair of numbers is in forbidden order, then we can consider that pair as an "object" and consider given arrangment as a proper arrangment of $n-2$ objects. Now, we can insert number $n$ in between those two improperly ordered numbers and make a proper arrangment. There are $(n-2)a_{n-2}$ ways to do that.

If we have proper arrangment and remove number $n$ from it, we will get either proper or improper arrangment. So there is no proper arrangment obtainable with both methods described.

We conclude $a_n=(n-1)a_{n-1}+(n-2)a_{n-2}$.


http://www.flickr.com/photos/30014811@N06/3080097994/

What do those triangles represent?(and the 'a', '2a', '0' etc written in the triangle?)
And whats the solution?

LOL I didn't understand the question at first I say D is the answer but when I see =0 I get mixed up  :(

[quote="TZF"]I think it was supposed to provide a definition of what that triangle means where it says, "for questions 13 and 14, use the following definition".

If that definition is not there, it is a printing error, and you cannot do the problem.[/quote]

Yes, it's definitely a printing error. "For any numbers $ a$, $ b$, and $ c$..." It's clearly not complete!


If that's all that you can post then nobody is going to be able to help you.

sorry didn't mean to press send before...
i went to the downloads page and clicked on the link next to "TeXnicCenter Setup, Version 1 Beta 7.01 ('Greengrass')" and after i clicked "install a link onto desktop" it said there was an error.
Hope i made that clearer...

[quote="myyellowducky82"]Hope i made that clearer...[/quote]Sigh! What downloads page? What link? What error? Why are you 'installing a link'? A link to what? TexnicCenter is not LaTeX but an editor for latex files so why have you called this thread "can't install latex"?

I know there are some very bright people on this forum but I doubt that any of them can read your mind or see what is happening on your computer.

So start from the very beginning explaining [i]exactly[/i] what you are trying to achieve and [i]exactly[/i] what is happening and [i]exactly[/i] what is going wrong. But first make sure you have read and followed the instructions at [[LaTeX:Downloads]].

You don't need to download LaTeX for the forum, you know. Just  enclose equations in dollar signs and it'll look pretty right there. You only need to download it if you want to [u]make custom documents with LaTeX.[/u] For posting on AoPS, you don't need to download anything :)

Example:
[code]$x^2-4=0$[/code]

Gives $ x^2 \minus{} 4 \equal{} 0$.

You can test LaTeX in the test forum.


calculus
integration
calculus computations

Let $ \alpha ,\ \beta$ be real numbers with $ 0\leq \alpha \leq \beta$ and $ f(x)\equal{}x^2\minus{}(\alpha \plus{}\beta)x\plus{}\alpha \beta$ such that $ \int_{\minus{}1}^1 f(x)\ dx\equal{}1$. Find the maximum value of $ \int_0^{\alpha} f(x)\ dx$.

Let $ g(\alpha) \equal{} \int_0^{\alpha} f(x)\ dx$.

Evaluating the integral shows that:

$ g(\alpha) \equal{} \minus{} \frac {1}{6}{\alpha}^3 \plus{} \frac {1}{2}{\alpha}^2 \beta$

$ \int_{ \minus{} 1}^1 f(x)\ dx \equal{} 1$  can be used to find that $ \alpha \beta \equal{} \frac {1}{6}$

Using that,

$ g(\alpha) \equal{} \minus{} \frac {1}{6}{\alpha}^3 \plus{} \frac {1}{12}\alpha$

From here, one can use his or her preferred technique for maximizing functions.

$ g'(\alpha) \equal{} \minus{} \frac {1}{2}{\alpha}^2 \plus{} \frac {1}{12}$

The zero of $ g'(\alpha)$ is $ \alpha \equal{} \frac {1}{\sqrt {6}}$ ($ 0\leq \alpha \leq \beta$).  The maximum of $ g(\alpha)$ must occur at this value.

$ g(\frac {1}{\sqrt {6}}) \equal{} \frac {\sqrt {6}}{108}$

So, the maximum value of $ \int_0^{\alpha} f(x)\ dx$ is

$ \frac {\sqrt {6}}{108}$

That's the correct answer. :)

But, what's the range of $ \alpha$?


My teacher gives me some question but I am not sure how to answer it .

(i) A particle , travelling along a straight path with velocity $ v $ , explodes into two equal parts . The force due to the explosion act along the original direction of travel . The explosion causes the kinetic energy of the system to be doubled . Determine the subsequent velocities of the segments .

(ii) Some water is placed in a vacuum flask , and is shaken up . Has the internal energy of the water been altered ?

is the first one $ \sqrt{2}v $of each particle at an angle of $ 45' $ from the original direction of the particle???

as for the second one , i aint too sure but what stupid thing i thought is it wont change because internal energy means the kinetic energyof the molecules which in turn is .......the heat energy! hence as temperature isnt changing , there is no change in internal energy.

DUNN know if thats right :oops:

Answer to 1st one,
1 particle will have velocity $v-v\sqrt{2}$ and other $v+v\sqrt{2}$

as for the 2nd one I agree with both eminem and bus

[quote="bus"][quote]$\text {Life is complex: it has both real and imaginary components.}$[/quote]Wow, cool :P .

Anyway, try to concentrate on the word "along". It means that after the explosion the particles will be still moving along the same line as before the explosion - one of them faster and the other one slower (or even in opposite direction). If you properly understand this part, the rest is easy. (It also means that eminem's 45' is wrong :( - it is alway either 0' or 180'.)

As for the flask question, if you shake the water, you [i]could[/i] give the molecules some extra kinetic energy and this energy would in turn increase the water's temperature! Also there is friction and probably a lot of other things we haven't thought of so the energy definitely was [i]altered[/i]. (And likely increased - but that's just my opinion.)[/quote]

Ummm,bus,it's a vaccum flask,so i dont think there would be any air friction,i guess.What is the answer to this question,shyong?And by the way,welcome to the physics forum :lol:
I think it would alter(bridge :D )....

[quote="shyong"]Actually I am confused also with the question saying that

[quote]
The force due to the explosion act along the original direction of travel [/quote]

which I think that both the exploded particle actually move in same direction  :huh:

As for the second one , Im wondering could the kinetic energy for which you shake it can change to internal enrgy of the water :roll:  .... though you maybe correct that " heat " is more important factor ....[/quote]

aaaarrrrggggghhhh, :wallbash_red:  i forgot about the same direction stuff.... i guess riddler must be right now??? i dont know i 'll solve it again ......

hey bus , when you say that you shake the wwater so the energy (internal) would rise , ummmmm  then what if  suppose you heat the water then the internal energy will obviusly rise , ok. and you say while shaking it would rise too !  means heating and shaking it is the same?????

what i think is when you shake it the matter as a whole moves around temporarily whereas when you say heat energy or internal kinetic energy of the molecules i think it is calculated when MACROSCOPICALLy  the water is stable, thast is the kinetic energy of the molecules WHEN the water is stable !,

or it could be that there is a very very , $very^{very}$ small change in temp , due to shakin it???

[quote="riddler"]Answer to 1st one,
1 particle will have velocity $v-v\sqrt{2}$ and other $v+v\sqrt{2}$
[/quote]

I dont get this , if the velocity is like what you said , then the kintetic energy after explosion would be

$\dfrac{1}{2}\dfrac{m}{2}(6v^2)=1.5mv^2$

But the original kinatic energy is $0.5mv^2$ which doesnt fullfill the question saying that the KE is doubled  :?

Well, whatever happens the CM continues to move with same velocity as the force of explosion is an internal force.

so

{m[b]v[/b] + M[b]V[/b]}/{m + M}=[b]V[size=59]in[/size][/b]

now write energy eqn. Since the force acts along the direction of motion [b]v[/b] & [b]V[/b] should be of same sign. So you should get [b]v[/b]=2V and [b]V[/b]=0

Is this correct?

[quote="shadysaysurspammed"]Ummm,bus,it's a vaccum flask,so i dont think there would be any air friction,i guess.[/quote]That's right. But there will still be friction between the water and the flask and also between the water particles themselves (this is called viscosity).

As to eminem: when you shake the water, the water won't stop moving after you stop shaking the water! Well, it will but only because the "macroscopic" kinetic energy will change into "microscopic" kinetic energy due to friction, hydrogen bonds and so on. It's in fact still the same kinetic energy but the orderly motion of molecules of water changes into chaotic motion. So yes - shaking and heating is (somewhat) the same thing! And again yes, the temperature change would be probably very very small.

So Sound and heating and shaking are just the same thing?
 :huh:

[quote="riddler"][quote="shyong"][quote="riddler"]Wait I think Im worng an it should be 1 particle stops and the other goes with $2v$ velocity :D[/quote]

Well , how do you explain $1$ particle stops ? The energy during the explosion should be enough to give both the particles energy to move  :huh:[/quote]
No, you are wrong
the force is applied to 1 particle opposite its direction of motion and 1 particle same direction of motion.[/quote]

Are you saying that the particle is exploded at the center causing the behind part of the particle to move backward while front part to move forward ? But then I still dont know how you get one particle stops and other keeps moving ?  :( Are there any chance you are using Principle of conservation Momentum  :huh:

[quote="eminem"][quote="bus"]On the lowest level - yes. You can make things move / shake / break using only sound, you can make things move using heat (steam engine) and therefore it also works the other way around. (You can heat things by moving / shaking them.)[/quote]

maybe thats true bus, but then wont there be a neglegible increase in temp??[/quote]
That is what conufuses me, cos if heat=vibration=sound then what is definition of heat ????? they should be all the same,
I mean people should only say things vibrate not things get heated up or somethin else. :huh:
Our physics teacher says heat is vibration of particles then where does the temp come from????
and then sound should also increase the same amount of temp change. :?:

[quote="shyong"][/quote][quote="riddler"][quote="shyong"][quote="riddler"]Wait I think Im worng an it should be 1 particle stops and the other goes with $2v$ velocity :D[/quote]

Well , how do you explain $1$ particle stops ? The energy during the explosion should be enough to give both the particles energy to move  :huh:[/quote]
No, you are wrong
the force is applied to 1 particle opposite its direction of motion and 1 particle same direction of motion.[/quote][quote="shyong"]

Are you saying that the particle is exploded at the center causing the behind part of the particle to move backward while front part to move forward ? But then I still dont know how you get one particle stops and other keeps moving ?  :( Are there any chance you are using Principle of conservation Momentum  :huh:[/quote]
See take that there were 2 particles from the start before the explosion.
so when the force is applied (obviously from the centre) then one particle will experience force against its direction and one with same direction.

Temperature is a tricky property. It's definition in physics is really complicated (I don't understand it myself: :huh: ) but it's not any "new" property of a given substance - it is only a measure of how fast the particles in the substance move / vibrate. The only reason why we don't say that things vibrate instead of things are hot is that using temperature is much more comprehendable / natural / intuitive, people are just used to it. We percieve things as hot or cold but what they really are is vibrating. Heat [i]is[/i] the exact same thing as kinetic energy of particles in the substance (plus their internal potential energy).

And if guys were asking about something else but I didn't get your questions right then please ask again :roll: ...

so I say sound is less effective than heat cos it only vibrate slowly? and the thing vibrating doesnt get heated up? :huh:

Hm, I'm not sure. You could be right - people usually don't heat things up by yelling at them so it really might not work.... but the rate of vibration doesn't necessarily have to be that important - particles of the thing you want to heat up may still gain some energy even after a collision with slower moving particles. And sound is not really vibration of molecules of air - molecules of air don't vibrate, air is a gas so they move randomly all around the place. Sound is merely a pressure wave - a "vibration" of pressure, a change of air pressure propagating as a wave. I hope you know what I'm talking about :). But as I have said, I'm not sure. And the temp change would be really small anyway. Unless we're talking about some really loud sounds...


In a certain big city, all the streets go in one of two perpendicular directions. During a drive
in the city, a car does not pass through any place twice, and returns to the parking place
along a street from which it started. If it has made 100 left turns, how many right turns must
it have made?
.


Hello again,

The question is:

1. The parabola y = x^2 + bx + c has the vertex at V(2,-3). Find b and c. (Hint: you can use the equation of the parabola when the vertex is given)

Would there be any other ways to solve this equation other than what I did below?

---------

I did the following:

a = 1 since x^2 = 1x^2

h = -b/2a

Substitute a & h value:

2 = -b/2(1)

-b = 4

b = -4

Substitute x, y, and b values into quadratic equation:

2^2 + 2(-4) + c = -3
4 - 8 + c = -3
-4 + c = -3

c = 1

Answer : b = -4   c = 1

[hide]
Since the vertex is at $(2,-3)$ and we know that the graph of the parabola hasnt been stretched or compressed so we can create an equatiopn for the parabola in vertex form.

the graph of $x^{2}$ has been shifted tot he right 2 units and down 3 units so it follows that

$y = (x-2)^{2}-3$

expand

$y = x^{2}-4x+1$

Therefor b= = -4 and c = 1
[/hide]

[quote="7h3.D3m0n.117"]I'm very sleepy right now. Use the form of equation of a parabola:

$y=a(x-h)^{2}+k$ where the vertex is $(h,k)$.

I think it works.  :maybe:  Im on a laptop right now next to a sleeping person so i have to be quiet =).[/quote]
dude your way works
it just amuses me but what...do you bang on your keyboard or scream "OMG HOW CULE I DID IT! I FINISHED A MATH PROBLEM!!!!!1" every time you finish a problem on AoPS or something?
sorry I truly read this and apologise if I seem to spam but I agree with 7h3.D3m0n.117 on this one.

Thank you for your assistance in replying.

I smacked myself after seeing that the equation of the parabola simply meant y = a(x - h)^2 + k; I forgot what the equation of the parabola was!

From then on, this was a simple question!

Thanks again!


ARML
AMC
AIME
geometry
geometric transformation
reflection
HMMT

Hi, I'm really interested in trying out for the Chicago ARML next year... do you have any suggestions on how I can prep for it over the summer? Thanks a lot!!!

another thing to keep in mind is that "problem-solving" problems are relatively contrived; there is a set of tricks that you have to learn; i find that i have seen a substantial number of problems when we do arml practices/hw. i think the best practice is to do lots and lots of problems.

[quote="Altheman"]another thing to keep in mind is that "problem-solving" problems are relatively contrived;[/quote]

 :huh: I don't know what you mean  :D

well ok when i go into a problem, i assume that there is a nice solution, in real life contexts, either mathematical research or applied mathematics, there is not usually a solution behind each problem, when you do an aime problem, you know there is a simple solution

when you see a infinite series problem, you can be pretty confident that there is some telescoping/taylor series/geometric/aritmetic/polynomial/recurrence/difference equation way of doing it...arml is a matter of mastering these techniques and developing an ability to recognize them and compute quickly

one thing though, this is not necessarily easy, i suppose it reflects some sense of mastery if you are able to say: "here is a type of problem, i know all the techniques that will come up," it takes a lot of work to get to that point in any one subject, one of the reasons that i placed 3rd on the hmmt algebra individual is that i know most of the techniques available in the "problem-solving" algebra topic,

of course, there are those rare problems that come up where none of the standard techniques help, i guess you just need ingenuity for these problems, but i really feel that you only see problems of that difficulty on difficult olympiad problems

thanks for the explanations! This summer is going to be [i]the[/i] summer for me and I can feel it  :lol: I always picture myself doing a lot of math but somehow, there's this "resistance" going on mentally (or "laziness"?) that is a bit tough to control. I will smash that :D

Just judging by how much you post on these boards, I doubt you will have to worry about making ARML next year (just don't bomb the tryouts).

Also if you are looking for something to do this summer, have you considered applying to a math summer program?  There are some great ones out there, and applying is usually free. I can say from personal experience that when I was at HCSSiM last year, I improved much faster than I ever have by studying on my own.

he has a good point; if you post a lot on aops, you probably will do well...post as in try and solve problems [obviously]

[quote="nateharman1234"]I can say from personal experience that when I was at HCSSiM last year, I improved much faster than I ever have by studying on my own.[/quote]

really! I guess self-studying can be too much sometimes. I work during the summer and squeeze math in here and there. btw- do you know of any place around Chicago that coach competition math? and just require "enthusiasm"? Although my school had a pretty strong math team, there wasn't much training available for the less experienced...  I took online aops courses, but I don't have enough discipline to follow through this type of class and make the best out of it...

thanks!

See if you can organize a group of people to practice with.  Have people who are good at all types of problems so that if you miss one, they can help you understand the solution... Other than that, practice by yourself, but remember to always try and find alternate or better solutions to some problems.

i think somebody wanted to start a math circle...i will get back to you on that when i don't have tons of english homework

thank you! You guys are so supportive  :D


Does anyone know if there were any perfects?  If so, who were they?

Haven't found any. I know of 3 14's so far. (Or at least 3 ppl claiming 14's. Two are confirmed.)

I know of 3 14's as well. However, I still have not heard of any 15's.

thus I think it wasn't really ALL that easy...

Well of the 4 perfect scorers last year, 3 are in college. Fergie probably had the capability to snatch a 15 on this year's, but everyone makes mistakes once in a while. :-)

Besides, the number of perfect scores is hardly a good indication of the difficulty level of the test overall.

Two more 14s (assuming you haven't counted them yet): me and Nathan Pfleuger.

-Adam Hesterberg

Ah...nice...gj Adam. Go poll if you haven't already.

There was a 14 from my school, in the form of Abhi Gulati.

[quote="joml88"]I heard Zach Abel got a 14.[/quote]

You're right. :-) Although, there were plenty of students who probably had the ability to achieve a perfect score but made a mistake of two due to calculation/interpretation/etc errors.

Exeter most likely does not have 3 perfect scores. mathfanatic is probably very well informed of Exeter's situation. Ask him. ;-)

It was easy, but some of the last problems were hard.... and the general consensus is that 13 (the trap pattern) and 14 (for most) were the most difficult. By easy, everyone means the first 10 were easy to anyone with high school math, and number two was Mathcounts level.

Nobody at Exeter got a perfect. Top score there was 13 in the form of two people.

hehehe "in the form of two people". I like it :-)

[quote="hockeyadi23"]Wow... and Exeter is the best in New England![/quote]

Some schools in New England haven't taken the AIME yet  ;)

Haha, not like we're going to have a perfect anyways.

[quote="guangweicao"]there was a rumor going around that 3 people in exeter had perfect scores.  :huh:  but i highly doubt that.[/quote]

Yes, no perfect scores here.  You might be confused with AMC10?

And Albert is right.  We had two 13s, three 12s, two 11s, and a 10.  Lots of silly mistakes.

[quote]there was a rumor going around that 3 people in exeter had perfect scores. Huh? but i highly doubt that.[/quote]

More than 3 people there were more than capable of it.

[quote="darktreb"]More than 3 people there were more than capable of it.[/quote]

Plenty of people are capable of it this year. It just seems like everyone made a stupid mistake. (And some of us made more than 1 or 2 stupid mistakes.)

When'll the official scores come back? Do they wait for the alternates or something? Anyways good luck to anyone taking the alternates... funny that in such an "easy" test, there are no reported (so far) perfects, even from the great private schools... of course there always are the alternates, but still..


The area of the shaded region is 78 square inches. All angles are right angles and all measurements are given in inches. What is the perimeter of the non-shaded region?

[asy]size(101);
filldraw(((0,0)--(0,8)--(10,8)--(10,-2)--(6,-2)--(6,0)--cycle^^(2.5,3)--(2.5,5)--(7.5,5)--(7.5,3)--cycle),gray(.6)+fillrule(1),linewidth(1));
label("$2''$",(5.3,-1),fontsize(10pt));
label("$4''$",(8,-2.7),fontsize(10pt));
label("$2''$",(3.3,4),fontsize(10pt));
label("$10''$",(5,8.7),fontsize(10pt));
label("$10''$",(11,3),fontsize(10pt));[/asy]

If there were no white section removed, the area of the shaded region would be $ 10*10\minus{}2*6\equal{}88$.  Therefore, the area of the white section is $ 88\minus{}78\equal{}10$.  Since the height is 2, the base is 6.  The perimeter of a $ 2x5$ rectangle is $ \boxed{14}$.


The sequence of positive integers (1, 1, 1, 1, 2) has the property that the sum of any three numbers in the sequence is divisible by the sum of the other two. Is there a sequence of five distinct positive integes with this property?

NZIMO? Since when did we have an NZIMO. Ah, I suppose you mean team training?


Stolen from Western Mass. ARML 2002

A set of distinct, nonzero real numbers is placed along the circumference of a circle. Each of the numbers is equal to the product of the two numbers adjacent to it. What is the least possible number of numbers in the set?

where did you get this?  i dont remember seeing this...

The problem seems familiar. I think it might have been the first homework assignment here (though I am too lazy to check right now ;) ) :

http://arml.yanco.com/summer2002/index.html

[color=cyan]I like this one.

We ignore the case where there are only 1 or 2 numbers, although I might argue that since the null product is 1, 1 number could work . . . ignore that, please.  Anyhow, we know that if there are n numbers, and we call them a1, a2, . . . an, we have
a1 = a2 * an
a2 = a1 * a3

Thus
a1 = a1* a3 * an
Thus
a3 * an = 1.
By symmetry, ak * a(k+3) = 1, where the +3ing is mod n.  (In other terms, this means that the product of any two numbers on the circle which are seperated by a length of 2 is 1).
So, for three numbers this means that a1*a1 = 1, a2*a2 = 1, and a3*a3 = 1, which will not give us three distinct real numbers.  For four numbers, it gives us a1*a4 = 1 = a2*a1, which makes a2 = a4 which is bad.  For five numbers, it gives us a1*a4 = 1 = a4*a2 which makes a4 = a2.  For six numbers, however, it tells us nothing new.  In fact, we can choose 6 numbers that will work:
2, 3, 3/2, 1/2, 1/3, 2/3 is a nice-looking set.[/color]

[color=cyan]In fact, 6 is the [i]only[/i] number of numbers for which all the conditions can be satisfied, I believe.[/color]

Ah, my interpretation of the problem was entirely wrong... I thought it meant "Of all numbers in the set, what is the least that any of them could be?" Then I spent a long time wondering where they'd pulled the 6 from and why I couldn't get an answer...

Yes, this was taken from the homework assignments (problem 28).

Ah, you weren't reading carefully enough :)


geometry
perimeter
ratio
geometric sequence
arithmetic sequence

Hi, I'm kinda new here (on the forums at least) and I'm preparing for the 2008 State competitions.  I recently did the 2003 National Sprint round and I don't think that there are solutions for National Rounds.  I have the answers but I don't know how to do some of them.  I'd appreciate any help that you guys could give me with them.

Here they are:

[b]Problem 8:[/b]
What is the greatest three-digit number "[i]abc[/i]" such that 4, [i]a[/i], [i]b[/i] forms a geometric sequence and [i]b[/i], [i]c[/i], 5 forms an arithmetic sequence?

Answer:
[hide]697[/hide]

[b]Problem 16:[/b]
A, B, C and D are distinct positive integers such that the product AB=60, the product CD=60 and A-B=C+D.  What is the value of A?

Answer:
[hide]20[/hide]

[b]Problem 17:[/b]
When a positive three-digit dividend with a units digit of 2 is divided by a positive one-digit divisor, the result is a whole-number quotient with a remainder of 1.  How many distinct values possible for this three-digit dividend?

Answer:
[hide]38[/hide]

[b]Problem 19:[/b]
In how many different ways can the digits 1, 2, 3, 4, 5, 6 be placed in the boxes below, one per box and without repetition, so that the numbers in each row decrease from left to right and the numbers in each column decrease from top to bottom?
[img]http://farm4.static.flickr.com/3029/2287005764_7026f60ca4.jpg[/img]

Answer:
[hide]5[/hide]

[b]Problem 23:[/b]
Brad bicycles from home at an average speed of 9 miles per hour until he gets a flat tire.  With no way to fix the tire, Brad walks his bike back home by the same route, averaging 3 miles per hour.  If the entire round trip of biking and walking took a total of 6 hours, what was Brad's average speed in miles per hour for the entire round trip?  Express your answer as a decimal to the nearest tenth.

Answer:
[hide]4.5[/hide]

and finally,

[b]Problem 25:[/b]
The square with vertices (-[i]a[/i], -[i]a[/i]), ([i]a[/i], -[i]a[/i]), (-[i]a[/i], [i]a[/i]) and ([i]a[/i], [i]a[/i]) is cut by the line [i]y[/i]=[i]x[/i]/2 into congruent quadrilaterals.  What is the number of units in the perimeter of each quadrilateral? Express your answer in simplified radical form in terms of [i]a[/i]?

Answer:
[hide]4[i]a[/i]+[i]a[/i]√5[/hide]

Once again, I'd appreciate any help that I could get for these problems.

Thanks a bunch,

~~amgupt01

[hide="8"]
go through possible values for a and b first. since a must be as large as possible, start with 9 and work down.
so, if a was 9, then the geometric sequence would have a common ratio of 9/4, which would obviously make b>9 (and not a digit)
if a was 8, common ratio would be 2 and b would be 16 (not a digit)
if a was 7, b would be some weird fraction (not a digit)
if a was 6, common ratio would be 1.5, and b would be 9.
Since there is only one possible case per hundreds digit and 6 is the largest hundreds digit possible, these must be the ideal values for a and b. so a=6, b=9, and putting 9 for b in the arithmetic sequence, the common difference would be -2 and c would be 7. Thus we have a=6,b=9,c=7, the answer is the number $ abc$, which is obviously 697.
[/hide]

I bet there is an easier way to do this, but:[hide="Problem 23 Solution"]Call the total distance $ D$. Then, we set up an equation to find $ D$:\[ \frac {\frac {D}{2}}{9} \plus{} \frac {\frac {D}{2}}{3} \equal{} 6
\]Solving, we see that $ D \equal{} 27$.
Since average speed equals total distance over total time:\[ S \equal{} \frac {27}{6} \equal{} \boxed{4.5}
\][/hide]

[hide="Problem 25 Solution"]The square that is described has a side length of $ 2a$.
The line $ y\equal{}\frac{x}{2}$ cuts the square into two trapezoids. Three of the sides have lengths $ \frac{a}{2}$, $ 2a$, and $ \frac{3a}{2}$, respectively. The last side (the side that is $ y\equal{}\frac{x}{2}$), by the Pythaogrean Theorem, has a length of $ \sqrt{a^2\plus{}{\left(2a\right)}^2}\equal{}a\sqrt{5}$.
Thus, the perimeter is $ \frac{a}{2}\plus{}2a\plus{}\frac{3a}{2}\plus{}a\sqrt{5}\equal{}\boxed{4a\plus{}a\sqrt{5}}$[/hide]

Thanks for your help guys.

[quote]verizonwires:
did u get all the other ones rite??
[/quote]

Yeah. I got the others right.

For
Problem 16:
A, B, C and D are distinct positive integers such that the product AB=60, the product CD=60 and A-B=C+D. What is the value of A?

I think trial and error would be the quickest way in the competition.
First, find all the pairs that have a product of 60.
1 60
2 30
3 20
4 15
5 12
6 10

Then look for pairs that satisfy the equation above.
which are
20-3=12+5
A=20 and you get your answer

Hope I helped! :)

[hide="17"]
Since the dividend with units digit 2 leaves a remainder of 1 when divided by a one-digit divisor, the dividend with units digit 1 is divisible by that one-digit divisor. It can readily be seen that that divisor [b]cannot[/b] be 2, 4, 5, 6, or 8. That leaves divisors of 3, 7, and 9. However, any number divisible by 9 is automatically divisible by 3, so we don't have to check 9.

There are 90 numbers to check. Out of those 90, 30 of them are divisible by 3, 12 of them divisible by 7, and 4 of them are divisible by 21. Then the answer is $ 30 \plus{} 12 \minus{} 4 \equal{} \boxed{38}$.[/hide]

[hide="19"]
I find it easier to reverse the situation, i.e. to make all the rows and columns [i]increasing[/i]. That doesn't change the answer, because of symmetry. Then we can see that there are not that many combinations, so it's not that hard to list them out. Note that the 1 must be in the top-left and the 6 must be in the bottom-right.

1 2 3
4 5 6

1 2 5
3 4 6

1 3 5
2 4 6

1 2 4
3 5 6

1 3 4
2 5 6[/hide]

how did you solve problem 23
i need the procedures. Thank you


abstract algebra
algebra
polynomial
Ring Theory
calculus
integration
function

Let $ k$ be a field, and $ X$ a free polynomial variable over $ k$. Let $ 0\to A\to B\to C\to 0$ be a short exact sequence of $ k\left[X\right]/\left(X^2\right)$-modules. Prove that if two of the three modules $ A$, $ B$, $ C$ are free, then the third one is free as well.

Free modules are precisely those that lack torsion.  Torsion submodules are carried backwards nicely by morphisms.  That's enough to hack out a proof for all but the case where B and C are free.  For that last case, note that exactness stipulates that the map A -> B will be injective.  So A is isomorphic to its image, which is a submodule of a free module.

[quote="darij grinberg"][quote="Functormorphism"]Free modules are precisely those that lack torsion.[/quote]

Sorry for answering in a rush, but isn't every $ k\left[X\right]/\left(X^2\right)$-module torsion-free? And are you sure that a submodule of a free module is always free? We are working over $ k\left[X\right]/\left(X^2\right)$ which is not quite a PID for this all to hold.

  darij[/quote]

Why isn't it a PID? $ k[X]$ is a PID, and I think a quotient of a PID is a PID.

[quote="Nukular"]Why isn't it a PID? $ k[X]$ is a PID, and I think a quotient of a PID is a PID.[/quote]

Depends on what you call PID - I am talking about PIDs which are integral domains. For non-integral domains, even if they satisfy the rest of the PID axioms, we can hardly say that submodules of free modules are free (think of $ k\left[X\right]\slash\left(X^2\right)$ as a $ k\left[X\right]\slash\left(X^2\right)$-module, and of $ Xk\left[X\right]\slash\left(X^2\right)$ as the submodule), unless you seriously redefine "free".

  darij

In which case it also matters how you define torsion. That module is $ \mathbb{Z}$ torsion-free, but not $ k[X]/(X^2)$ torsion-free, and we can see its not a free module.


modular arithmetic
number theory unsolved
number theory

Prove that for all positive integer $ m,n$ there exist a positive integer $ x$ such that :
$ 2^x \equiv 1997(\mod 3^n)$ and $ 2^x \equiv 2007(\mod 5^n)$

note that $ 2$ is a primitive root modulo $ 5^n$ and also $ 3^n$.so there exists
two natural numbers like $ a,b$ such that $ 2^{a}\equiv 2007(\mod 5^{n})$ and $ 2^{b}\equiv 1997(\mod 3^{n})$
one can easily check that $ 4 | a \minus{} 1$(by considering the fact that the first congruency holds also modulo $ 5$) and similarly $ 2 |b \minus{} 1$
then the rest will be easy since by crt the congruencies
$ x \equiv a \pmod {5^{n \minus{} 1}}$
$ x \equiv b \pmod {3^{n \minus{} 1}}$
$ x \equiv 1 \pmod {4}$
has a solution.
then we have $ x \equal{}5^{n \minus{} 1}.4k \plus{} a \equal{} 3^{n \minus{} 1}.2t \plus{} b$
we can easily check that $ x$ satisfies the required conditions.
we r done!


[b]Who can give me some examples which proved by S.O.S ?
or Some equation can be used in S.O.S ?I know what is SOS ,but I can not find more examples or ..equations . :(
Thank you! [/b]

http://www.mathlinks.ro/viewtopic.php?t=80127

In this topic I posted a remarkable number of examples which can solved by SOS. A lot of equations are involved in this topic, too.


Let $ x,y,z \in Z^ \plus{}$ and $ x^n \plus{} y^n \equal{} z^n$, with $ n \in N$. Prove that $ min(x,y) \ge n$.

the only $ n$'s that will work is $ 1$ and $ 2$.
if $ n\equal{}2$ smallest Pytagorean triple is $ 3,4,5$
if $ n\equal{}1$ then $ x\equal{}y\equal{}1$ works, then $ min (1,1)\equal{}1$

Thank. But you used Fermat theorem for your slution. I don't want so. Because we prove Fermat theorem are not easy. So we prove this problem not use Fermat theorem. Can You?


All five squares in the figure below are congruent. How many lines of symmetry does the figure have?

[asy]path A = (1,3)--(-1,3)--(-1,-3)--(1,-3)--cycle;
draw(A); draw(rotate(90)*A);[/asy]

It's just horizontal and vertical, so $ \boxed{2}$.

The two diagonals of the middle square also create two more lines of symmetry, so therefore the answer is $ \boxed{4}$.


We have $\sum x_1 - y_1 = 0$ and $\sum (x_1 + y_1)(x_1 - y_1) = 0$.  Write $a_n = \frac{x_n - y_n}{2}$ and $b_n = \frac{x_n + y_n}{2}$ and we wish for $b_n + a_n, b_n - a_n$ to be disjoint.  Set $b_n$ to a large constant $\gg n$ and give $a_n$ small values such that for no $i, j$ is it true that $|a_i| = |a_j|$ and we are done.

(Edit:  And, of course, $\sum a_n = 0$.  For example, for $n = 3$ you could use $b_n = 7, a_1 = 1, a_2 = 2, a_3 = -3$ and this would produce $x_1 = 8, x_2 = 9, x_3 = 4$ and $y_1 = 6, y_2 = 5, y_3 = 10$ which works.)


[i]In triangle $ABC$ we have that $\frac{a}{b}=2+\sqrt{3}$.  If $\angle C=60^{\circ}$, find the other two angles.[/i]

Masoud Zargar

Let $P$ be a point on $AC$ such that $PB \perp CB$, then $wlog$ $AC=2+\sqrt{3}$, $CB=1$

$CP=2$, $PA=\sqrt{3}$, $PB=\sqrt{3}$ $\Rightarrow$ $\measuredangle PAB= \measuredangle PBA=15^\circ$

$\Rightarrow$ $\boxed{\measuredangle A= 15^\circ, \measuredangle B=105^\circ}$


There are an infinite number of[ Pytogorean Triples.

Thanks

any odd number can be written as the difference of two consecutive squares.
(x+1)^2-x^2=2x+1. For them to be triplets it must be an odd square. There are infinite number of odd squares there for there are infinite number of triplets.

hello, see also here
http://www.mathsisfun.com/numbers/pythagorean-triples.html
Sonnhard.

Primitive Pythagorean triples are of the form $ r^2\minus{}s^2$, $ 2rs$, and $ r^2\plus{}s^2$.

In any case, take the primitive triple $ (3,4,5)$, multiply it by any integer, and you have a new triple, so there are obviously infinitely many.

[quote="grn_trtle"]Primitive Pythagorean triples are of the form $ r^2 \minus{} s^2$, $ 2rs$, and $ r^2 \plus{} s^2$.

In any case, take the primitive triple $ (3,4,5)$, multiply it by any integer, and you have a new triple, so there are obviously infinitely many.[/quote]

yes, but I think the initial problem meant that there are an infinite number of primitive triples. Because otherwise, the problem is very trivial. :P

[quote="limac"]yes, but I think the initial problem meant triples that are at their simplest forms. Because otherwise, the problem is very trivial. :P[/quote]

It didn't say primitive triples, but yeah, probably :)

In that case, you'd use the fact that all primitive triples are $ r^2\minus{}s^2$, $ 2rs$, and $ r^2\plus{}s^2$, as mentioned

[url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=285286]Here[/url] his a somewhat different kind of proof.


trigonometry
geometry
circumcircle
trig identities
Law of Cosines
power of a point

[quote="13375P34K43V312"][quote="mihail911"]I was reading the chapter on the law of cosines in AoPS vol 2 and I was a little confused on their explanation of how to prove that the law of cosines works when B is on the circle and outside of it (for those of you that have vol 2)

other than that, does anybody else have different proofs of the law of cosines? im always not sure where to put the circumcircle or an additional line.

thanks![/quote]

What do you mean, "where to put the circumcircle?" You just draw it.[/quote]

Well yeah there can only be one circle that passes through all three verticies.

ok...maybe that was a stupid way to ask it, but like, how do u know when to put the circumcircle or when to draw that extra line (as in the proof in aops vol 2)

[quote="xpmath"]Try [url=http://www.themathpage.com/aTrig/law-of-cosines.htm]this[/url]
Near the bottom[/quote]

thanks! that proof was a lot simpler, and i understand it now. :)

[quote="archimedes1"]There are many, many ways to prove the Law of Cosines, and I think the one AoPS uses is really cool, but hard to come up with on your own.[/quote]

i agree it's genius, but i would never have come even close to coming up with it on my own. :D

[quote="mihail911"]how do u know when to put the circumcircle or when to draw that extra line (as in the proof in aops vol 2)[/quote]

This is just a part of the difficulty of geometry.  There are no hard and fast rules here; just do a lot of problems and/or read a lot of solutions and you'll gradually develop an intuition and one or two rules of thumb.  I myself am terrible at this, so I can't give you any of those :oops:

Edit:  Well, circumcircles are generally used when one wishes to find cyclic quadrilaterals (which have a lot of useful properties) or use chord angle relations.  If you're dealing with an angle problem, it might be a tool to look into.  If you want lengths, then you still might want circles, but you would be applying power of a point, and circumcircles may not necessarily be the best idea (incircles, maybe).  As for auxiliary lines, I am nearly as clueless as you  :P

You can do a similar construction to prove stewart's theorem.

i am student teacher and i want learn more about strategies in teaching mathematics.... my lesson now is about oblique triangle... what strategies i need to do?

That's a completely different question, give it its own thread.

Of course, on a suitably forum.

[color=darkred][b]Lemma.[/b] Let $ ABC$ be a triangle. Then exists the identity $ \boxed{b\cdot\cos C+c\cdot\cos B=a}$.[/color]

[color=darkblue][b]Proof.[/b] Suppose w.l.o.g. $ 90^{\circ}\not\in\{B,C\}$. Denote the projection $ D$ of the vertex $ A$ on the sideline $ BC$. Appears two cases.

$ 1\blacktriangleright D\in (BC)$. In this case, $ BC=DB+DC$ and $ \{\begin{array}{c}DB=c\cdot\cos B\\\\ DC=b\cdot\cos C\end{array}\|$ $ \implies$ $ a=b\cdot\cos C+c\cdot\cos B$.

$ 2\blacktriangleright C\not\in [BC]$. In this case $ \{\begin{array}{c}C\in (BD)\implies\|\begin{array}{c}m(\widehat{ACD})=180^{\circ}-C\implies\cos\widehat{ACD}=-\cos C\\\\ BC=BD-CD\ ,\ BD=c\cdot\cos B\ ,\ CD=-b\cdot\cos C\end{array}\\\\ B\in (DC)\implies\|\begin{array}{c}m(\widehat{ABD})=180^{\circ}-B\implies\cos\widehat{ABD}=-\cos B\\\\ BC=CD-BD\ ,\ CD=b\cdot\cos C\ ,\ BD=-c\cdot\cos C\end{array}\end{array}\|$ $ \implies a=b\cdot\cos C+c\cdot\cos B$.

[b][u]Proof of the Cosinus' theorem.[/u][/b] $ \{\begin{array}{c}a=b\cdot\cos C+c\cdot\cos B\\\\ b=c\cdot\cos A+a\cdot\cos C\\\\ c=a\cdot\cos B+b\cdot\cos A\end{array}\|$ $ \begin{array}{cc}\bigodot &-a\\\\ \bigodot & b\\\\ \bigodot & c\end{array}\bigoplus\implies-a^{2}+b^{2}+c^{2}=2bc\cdot\cos A$ $ \implies\boxed{a^{2}=b^{2}+c^{2}-2bc\cdot\cos A}$.[/color]


how much material did you get to prepare with?

in other words, how much preparation material did you have access to?

(in general i'm talking to those who made NATS)

Well originally I had about 5-6 national tests, but after my team bought 2 and I looked and asked around for hours upon hours, I found about 16.
In state and chapter I only had 2001-2007.

But it's my first and only year at mathcounts.

i have nats and states 1990~2007 and chapters 2001~2007, and nats team 1984~1989
i also have some AMC 10, 12, and AIME, msotly years 2001~2007

Sprint, Target, and Team since 1992 or before (can't remember), and AMC 10, 12, and AIME since 2001.
I didn't use much of the AMC and AIME stuff though...

[quote="frost13"]
[b]17[/b] years of Nationals tests![/quote]

i have all the nats tests all the way back to 1990. i think thats 19 tests?

XD i did all of them in a week...

I have Sprint, Target, Team, Countdown, ans answers from 2004- 2008...

I MEAN 2007. :ninja:

and: 1991, 1992, 1998 (?) and 2000(?) :D


How many rounds of USAMTS have you participated in [i]this year[/i] (Year 15) so far?

Also, no discussing Round 3 problems until...uh...a few days.

I didn't find out about it until the third round, and I did a few, and then got lazy and procrastinated, and never sent my answers in, and this sentence is too long....

This is my first year taking USAMTS. Although the test is much easier than what I anticipated, I still did horribly on first two rounds (last minute test taker!). I think I'm going to get a 63.

3 rounds of course, since only 3 rounds are out this year.

Oh, I get what you mean. That score is what I believe I 'll get after three rounds.


how do you do the problem:

what is the sum of all the positive multiples of 6 that are less than 200?

please include explantion.

[hide="Click to reveal hidden text"]Then this is really an arithmetic sequence, starting with 6.

Just by looking at it, you'll realize the largest multiple of 6 under 200 is 198. Using $a_n=a_1+d(n-1)$, substitute it all in to get $198=6+6(n-1)$

Simplify to get $198=6n$ which finally gets $n=33$

So 198 is the 33rd term in this sequence (you probably could just realize that instantly though).

Use the formula $S_n=\frac{n} {2}(a_1+a_n)$:

$S_{33}=\frac{33} {2}(6+198)$
$S_{33}=\frac{33} {2}(204)$
$S_{33}=3366$[/hide]


probability
algorithm
search
irrational number

[quote="furious"][quote="Athena"]more ideas:
youve discovered the end of pi
[/quote]
lol.
i've discovered that the final digit of pi is 7.  :D[/quote]

Can you prove that it isn't 7?

Hey, no fair, what happened to my comment?

[quote="JBL"]Base $pi$, the last digit of $pi$ is 1.[/quote]

Actually, what that really reminds me of is jokes about 30 mph cars at MOP -- "A car travelling 30 mph is only travelling 1 mph mod 29," and so on.

I have narrowed the down the option of the last digits of pi.  It has been a hard time but it is one of these numbers, 1/2/3/4/5/6/7/8/9  :D lol.  It can't be zero cause the last number would be the number be4.  notice the hard work lol.  I don't think it can be done.

elvenchamp777

Well, you can cross 7 off of that list. So now it's just down to 1/2/3/4/5/6/8/9 Yay. 1/8 chance of guessing correctly.

Still waiting for Simon's proof. ;-) No pressure, Simon. (gj on this week's Individ btw. I was away when u IMed)

Here are some simple-sounding problems about the digits of $\pi$.  Show that there are an infinite number of 7's in the decimal expansion of $\pi$.  Prove or disprove: Beyond a certain point, the decimal expansion of $\pi$ just consists of 0's and 1's.

For that second one, can I use the binary expansion instead?  :):)

[quote="furious"][quote="JBL"]Base $pi$, the last digit of $pi$ is 1.[/quote]

wouldn't pi be 10 in base pi, making 0 be the last digit of pi?[/quote]
Ya lol.  I'll just say the last digit is x where x is a number between 1 and 9 inclusively.  Let you minds wonder.  (If I dumb it down for myself it makes me feel proud).  And why can you rule out 7?

[quote="yif man12"]Because Simon says so.[/quote]:D:D Ok then we can rule out 6 too :)

Um, does an irrational number have an end?

Um, nah.

I say pi ends in $i$. As many people have already determined smartly before.

[color=darkred]Admin edit: this kind of language is not permitted on the forums. Please behave![/color]

*slaps miraculouspostmaster in the face*

grr. :P

Hi all. Just registered.

Here's a useful link though:

http://crd.lbl.gov/~dhbailey/

...u can compute the Nth Pi digit on the fly, without computing any of the previous. (spigot algorithm for pi and new formula)

Best,

E.

If I recall correctly, their method can compute the nth bit of $\pi$ quickly, but not the nth (decimal) digit.

Yes they compute the Nth digit in bits and hex.

The string search in the first 4 billion digits is fun.  :)

http://pisearch.lbl.gov/

Best,

E.

Actually, it seems that a method has been developed that can compute the nth digit of pi in any base in $O(n^2)$

[url=http://fabrice.bellard.free.fr/pi/pi_n2/pi_n2.html]See link[/url]

Proof that the last digit of pi is both 4 and 5:
Assume that the last digit of Pi is not 4 nor 5.
But Pi is irrational, so it is a non-repeating, non-terminating decimal.
Therefore, there is no last digit.
Therefore, the last digit is not not 4 nor 5.
Therefore, the last digit is both 4 and 5.

[quote="diophantus"]Didn't this same problem happen in Star Trek?  A rebellious computer system had them trapped inside a ship but was still doing computations for the prisoners.  Ther prisoners asked it to find the last digit of pi.  After the computer had overheated and destroyed itself, the prisoners escaped.  If the operating system of a hyperspatial ship cannot find this digit, why would we be able to?[/quote]

Good point. I always base my assumptions on television series.

[quote="diophantus"]Didn't this same problem happen in Star Trek?  A rebellious computer system had them trapped inside a ship but was still doing computations for the prisoners.  Ther prisoners asked it to find the last digit of pi.  After the computer had overheated and destroyed itself, the prisoners escaped.  If the operating system of a hyperspatial ship cannot find this digit, why would we be able to?[/quote]

Indeed.  Star Trek: The Original Series. Spock: 'Computer: calculate, to the last digit, of the decimal value or pi.' if I remember correctly...

is base pi even a thing. like how can you have non natural bases.


algorithm
number theory proposed
number theory

A unit fraction is a rational number with numerator 1.  Is it true that every positive rational number can be written as the sum of distinct positive unit fractions?  If so, how do you do it; if not, why not?

I am sure that any fraction can be shown in the form of sum of unit fraction... below is my algorithm to do it.
let us suppose any fraction $ \frac{a}{b}$.
so we will write above as $ \frac{a}{b}\equal{}(1\plus{}\frac{1}{b})(\frac{a}{b\plus{}1})$
and then we will recursively apply above till $ \frac{a}{b\plus{}k}\equal{}\frac{1}{c}$
and then we multiply all and hence the sum of rational numbers we get is the our answer.

EXAMPLE
$ \frac{5}{16}\equal{}(1\plus{}\frac{1}{16})\frac{5}{17}$
and then we write $ \frac{5}{17}$ using above algo.
and in the end we will get the following
$ \frac{5}{16}\equal{}(1\plus{}\frac{1}{16})(1\plus{}\frac{1}{17})(1\plus{}\frac{1}{18})(1\plus{}\frac{1}{19})\frac{5}{20}$
and $ \frac{5}{20}$ can be written as $ \frac{1}{4}$.

Hence our answer is

$ \frac{5}{16}\equal{}(1\plus{}\frac{1}{16})(1\plus{}\frac{1}{17})(1\plus{}\frac{1}{18})(1\plus{}\frac{1}{19})\frac{1}{4}$
 :roll:


calculus
integration
number theory open
number theory

$a(n)$ sequence of elements in {0;1}.Assume that the real number
$\sum_{n\geq 0}\frac{a(n)}{3^n}$ is irrational, then it is transcendental.

This is a special case of Borel's problem : any algebraic irrationnal number should be normal in any integral bases. In particular all digits should appears with the same frequency.
Note that very recently Y. Bugeaud and F Lucas proved (announced in CRAS Paris) that algebraic irrationnal numbers can not have the sequence of their digits generated by a finite automaton.

Yes Pierre , I discovered this site through animath forum ;)


conics
ellipse
rotation
symmetry
geometry
geometric transformation
linear algebra

Let $A$ be the smallest positive angle measure, in degrees, through which the ellipse $5x^{2}+6xy+5y^{2}-8=0$ can be rotated counterclockwise to make the major axis vertical. Let $B$ be the square of the distance between the foci of the ellipse. Find $A+B$.

[hide="something that might help"]
use a calculator

[/hide]

[hide]
The equation is symmetric, thus it has symmetry over the line $y=x$ and its axes are at a $45^\circ$ angle to the x- and y-axes. Apply a rotation matrix for that angle, counterclockwise, to get the rotated ellipse ${\frac{(x^{'})^{2}}{4}+\frac{(y^{'})^{2}}{1}=1}$. (EDIT:) But this has a horizontal major axis, so we need to give the conic another $90^\circ$ spin, meaning that the total counter-clockwise rotation is $135^\circ$. Then, the square of the distance between foci equal to $\left(2\sqrt{4-1}\right)^{2}=12$, thus we have $135+12=\boxed{147}$.[/hide]

(EDIT: I forgot that the denominators are $a$ and $b$ [i]squared[/i], not to the first; my answer now agrees with [b]rnwang2[/b]'s answer key.)

[quote="Altheman"][quote="cincodemayo5590"][hide]
The equation is symmetric, thus it has symmetry over the line $y=x$ and its axes are at a $45^\circ$ angle to the x- and y-axes. Apply a rotation matrix for that angle, counterclockwise, to get the rotated ellipse ${\frac{(x^{'})^{2}}{1}+\frac{(y^{'})^{2}}{4}=1}$, which has the square of the distance between foci equal to $\left(2\sqrt{4^{2}-1^{2}}\right)^{2}=60$, thus we have $45+60=\boxed{105}$.[/hide][/quote]

woot go NT precalc

I did not recognize that the numbers worked out well, but I would still say that this problem is annoying without a calculator. There is a general rotation matrix that is a pain to do...[/quote]

It wasn't that bad, since its just $x=\frac{x^{'}+y^{'}}{\sqrt{2}}$ etc... but rofl, i remember that mr. kim was like... "oh the tangent is undefined so it MUST be a $45^\circ$ rotation!!11one," and he didn't say anything at all about how the equation is symmetric. Yay mr. kim!!

[quote="rnwang2"]The answer sheet says it is $\boxed{147}$.[/quote]

I've got a feeling that your answer sheet is wrong.  :)  (That is, unless you can find something wrong with my solution... I might have messed up in my arithmetic, but the nice rotated equation makes me think otherwise.)

EDIT: Oops, there is something wrong; a $45^\circ$ rotation gives an ellipse with horizontal major axis, so we need another $90^\circ$ turn to get the correct ellipse. The answer still isn't 147, though...

EDIT 2: Oops, again. Now it agrees with the answer key.

[quote="rnwang2"]The answer sheet says it is $\boxed{147}$.[/quote]

Do you have a source of this question

[quote="tjhance"][quote="cincodemayo5590"][hide]
The equation is symmetric, thus it has symmetry over the line $y=x$ and its axes are at a $45^\circ$ angle to the x- and y-axes. Apply a rotation matrix for that angle, counterclockwise, to get the rotated ellipse ${\frac{(x^{'})^{2}}{4}+\frac{(y^{'})^{2}}{1}=1}$. (EDIT:) But this has a horizontal major axis, so we need to give the conic another $90^\circ$ spin, meaning that the total counter-clockwise rotation is $135^\circ$. Then, the square of the distance between foci equal to $\left(2\sqrt{4^{2}-1^{2}}\right)^{2}=60$, thus we have $135+60=\boxed{195}$.[/hide][/quote]

The distance between the foci is $2\sqrt{4-1}$, not $2\sqrt{4^{2}-1^{2}}$

fixing this mistake gives us the correct answer.[/quote]

Ooh right, that's the mistake, duh. Thanks.  :oops:


An isosceles triangle with $ AB=AC=7$ and $ BC=4$ is inscribed in a circle. Find the radius of this circle.

Note: using the fact that $ [ABC]=\frac{abc}{4R}$ is cheating, because the proof of this requires the law of sines.

[hide="hint"]the altitude of $ ABC$ extended past $ BC$ is the diameter of the circle.[/hide]

[hide="hint 2"]power of a point[/hide]

[hide="alternative to hint 2"]similar right triangles[/hide][/hide]

By power of a point:

[hide] Let the point where the altitude from A meets BC be D. Then by power of a point, 2*2 = sqrt(49-4) *x.

By solving for x, you get x=4/(3sqrt5), then radius  = x + sqrt(49-4) divided by two = 49sqrt5/ 30 [/hide]

[hide]The altitude to the base is $ 3\sqrt{5}$ by Pythagorean Theorem. Using Power of a Point,
$ 3\sqrt{5}*(2r-3\sqrt{5})=4$
$ r=\frac{49\sqrt{5}}{30}$[/hide]

[quote="abacadaea"]i looked at the hints and i realized my solution will be wrong but ill give it a shot.

I used the pythagoreon theroem to get the altitude to the base was $ 3\sqrt{5}$. I let the position where the altitude intersects the base be point $ D$. I let the center of the circle be $ O$. i let $ \line seg{BO}$(the radius) be $ x$ so $ DO$ is $ 3\sqrt{5}-x$. Using the pythagorean thereom, it is clear that $ 4+45-3x\sqrt{5}+x^{2}=x^{2}$. solving for $ x$ gets us $ x=\frac{49\sqrt{5}}{15}$(i hope i did that right). Therefore that radius is $ \frac{49\sqrt{5}}{15}$.[/quote]

No, you did everything right except for one thing. When you square $ (3\sqrt{5}-x)$, you get $ 45-6x\sqrt{5}+x^{2}$, not $ 45-3x\sqrt{5}+x^{2}$. So, the answer should be $ \frac{49\sqrt{5}}{30}$.


function
calculus
derivative
integration
limit
real analysis
real analysis unsolved

a. Let $ F$ be absolutely continuous on $ [c,d]$ and $ g$ be absolutely continuous on $ [a,b]$ with $ c \le g \le d$. Then $ F \circ g$ is absolutely continuous on $ [a,b]$.

b. Let $ E \equal{} \{x: g'(x)\equal{}0\}$. Then $ m(g(E)) \equal{} 0$.

b. We can freely reuse my proof for Sard's lemma from http://www.mathlinks.ro/viewtopic.php?highlight=neglijable&t=20251 . Indeed, $ g'(x)$ is a function with bounded variation, so it has at most countable set of break points, exclude this set from consideration. Consider only critical points $ x$ s.t. $ g'(t)$ is continuous in p. $ t \equal{} x$. Then we can construct intervals $ I_n(x)$ from my proof, so we will succeed.

[Edit: wrong statement about bounded variation of $ g'$]

It's not true, in view of the fundamental theorem of calculus. Take any integrable function not of bounded variation and form its indefinite integral. In this way, we can easily get a $ C^1$ function whose derivative is not of bounded variation. So I don't know where the claim comes from. And in this case, the derivative is only defined a.e, so the set where it's not defined may for example be countable or larger, and then we can easily make it of not bounded variation. So the claim is both preposterous and makes little sense

[quote="Kalle"]I wrote a simple proof of b) here http://www.mathlinks.ro/Forum/viewtopic.php?t=196722

Look a few posts down[/quote]
Thank you and is it because $ T_a^x = \int_a^x{|g'|dx} = P_a^x + N_a^x$ that $ g'(x) = 0 \Longrightarrow P_a^x' = 0$ so that we only need to prove the monotonic case?

[quote="yangzhe1990"][quote="Myth"]Indeed, $ g'(x)$ is a function with bounded variation[/quote]
can you explan me why?[/quote]
Sorry, dunno how I came to it  :maybe:

Anyway we can use the following Lebesgue's theorem:
for all functions $ f\in L_{loc}(a,b)$ the identity $ \lim_{h\to 0}\frac{1}{h}\int_{0}^{h}f(x\plus{}s)ds\equal{}f(x)$ holds a.e. on $ [a,b]$.

Applying it for $ g'(x)$ we obtain that my proof reusable for case of absolutely continuous functions.


Let $(x_n)$ : $x_1=a$ and $x_{n+1}=\frac{2x_n^3}{3x_n^2-1}$
 Find $a$ such that : there exists $\lim_{n\to \infty}x_n$

$a = 1$ gives $\lim_{n \to \infty} x_n = 1$.  Or do we find all such $a$?

Edit:  $\lim_{n \to \infty} x_n = 0, \pm 1$ for all $a \neq \pm \sqrt{ \frac{1}{3} }$.

Proof: (Well, more of a demonstration)

$x_{n+1} = \frac{ 2x_n^3}{3x_n^2 - 1} = x_n \frac{2}{3 - \frac{1}{x_n^2}}$

For $a \ge 1$, the coefficient of $x_n$ is always less than or equal to (when $a = 1$) 1, so $x_n$ is a decreasing sequence and it is easily shown that $x_{\infty} = 1$ stabilizes the sequence.

For other $a$ the behavior is bounded by several constants, but generally either it becomes larger than one and reverts to the above case, becomes close to $-1$ and goes there, or becomes close to $0$ and goes there.  In any case, it is always bounded because whenever $|x_k| > 1$ it begins to decrease to $\pm 1$ by the above argument.

$x_{n+1} = \frac{2x_n^3}{3x_n^2 - 1} \approx x_n$
$2x_n^3 = 3x_n^3 - x_n$
$x_n^3 - x_n = 0$
$x_n (x_n^2 - 1) = 0$
$x_n = 0, \pm 1$

But, you didn't find all values of a that this sequence is identifiable.
Your solution is not complete.

Well, okay.  If any of the values of $a_n$ reach $\pm \sqrt{ \frac{1}{3} }$ then the denominator of the next term is zero and the limit does not exist.

The calculation of these values requires finding $a_{n} = f(a_{n+1})$ which involves solving a general cubic.  Supposing, though, that such a function exists, the sequence $b_1 = \pm \sqrt{ \frac{1}{3} }, b_n = f(b_{n-1})$ would generate all values of $a$ that cause the limit not to exist.

I mean that how  you identify the sequence (b_n) clearly.Changing variable b_n=.... for instance.

[quote="zhaobin"]t0rajir0u,sorry,I can't understand your solution.
can you explain?or can write what is your conclusion first?[/quote]

Conclusion:  Define $b_1 = \pm \sqrt{ \frac{1}{3} }, b_n = f(b_{n-1})$ such that $x_n = f(x_{n+1})$, then the values of $b_k$ are the values of $x_1 = a$ for which the limit does not exist.

Demonstration:  $x_{n+1} = \frac{2x_n^3}{3x_n^2 - 1} = x_n \frac{2}{3 - \frac{1}{x_n^2} }$

For $a > 1$ the fraction on the right is less than one and will continue decreasing until $n \gg 1, x_n \approx 1$.

For $a < -1$ a similar argument gives $\lim_{n \to \infty} x_n = -1$.

For values in between, it depends based on several bounds, one of which is $a = \pm \sqrt{ \frac{1}{3} }$.  Note that if $a$ has this value, the denominator of the fraction on the right is zero, and $x_2$ is not defined.  Similarly, if $a$ has such a value that $x_2 = \pm \sqrt{ \frac{1}{3} }$, then $x_3$ will not be defined, etc.

The behavior is complicated $-1 < a < 1$ - sometimes it will tend to $1$, sometimes $-1$, sometimes $0$, and working out the specific behavior is annoying and depends on the values of $b_n$, which I cannot calculate because finding $x_{n} = f(x_{n+1})$ requires solving a cubic.

I realize this isn't a very good solution  :(


Factor completely. If the polynomial is prime, say so: 27x^3 + 64

I belive I answered this before... :(

[hide]$ (3x\plus{}4)(9x^2 \minus{}12x \plus{}16)$[/hide]

[url=http://www.math.unt.edu/mathlab/emathlab/differenceoftwocubes.htm]Sum and Difference of Two Cubes[/url]


I think this problem is quite boring and so we have no solutions. But because this is in the <<Contests>> we should post one.

Well, what's the ideal solution here? Or is it just a boring casework? (writing $n$ in different forms considering that it has 16 divisors)

This problem is very easy.It's in Problem and solution around the world 1999-2000.

Mind writing your solution tkhtn please? :)

Sorry! This problem is in Balkan,Junio,2002.
And here is solution:
If$d_2>2$ then all divisorsof n are odd but $d_4+d_6$ is divisor of$n$ and it's even,contradiction],if$d_3=3$ then 6 is divisor of n$\to$ or $d_5=6$ or $d_6=6$or $d_4=6$.
Assume $d_4=6$ then $d_2+d_4=8$ is divisor of$n \to 4$is divisor of$n \to d_4=4$,contradiction.
Assume $d_5=6$ then $d_6=(d_2+d_4)d_6>d_6$,contradiction..........
And $N=2002$. sory because I have go to bed.
See you agains.

HM :D . continue. I think this problem must be complete today.
Assume $d_6=6$ then $d_4=4;d_5=5$.But $d_5=(d_2+d_4)d_6>d_6$,contradiction.
$\to d_3>3\to$3 isn't a divisor of n.
Because $(d_2+d_4)d_6$ is a divisor of n $\to (d_2+d_4)=d_4+2$ is divisor of n.if $d_4 +1$ is a divisor of n then we have three consecutive divisors $\to$ 3 is a divisor of n, contradiction $\to d_5=d_4+2$and $d_5=3k+1\to d_5={1,4,7,10,13 or 16}$.Obvious $d_5\ge 5$.
Assume $d_5=7\to d_3=4,d_4=5\to 10,14$is divisors of n $\to d_7\le 10$.
But $d_7=(d_2+d_4)d_6>(2+5)7=49$,contradiction.
Other case similar and last:$d_5=13$ is true$\to d_4=11\to d_3={4 or 7}$.
Now is very easy. :oops:  :P  :lol:  :D  :)

Could you explain why this equality holds

$d_5=(d_2+d_4)d_6$ because we only know that $d_{d_5}=(d_2+d_4)d_6$

BTW, do not answer with a word only.

Ok so if all the divisors are odd, then $d_2+d_4 = \text{Even}$ implying that a divisor is even which is an obvious contradiction. Thus $d_2 = 2$.

Now assume $d_3 = 4$. Note that this means $8$ must be a factor of $n$ (just check Binomial-Theorems solution to see all the possible combinations and note that we can't have a square of a prime in our prime factorization).

If $d_4=8$ this implies $10d_6 = d_{d_5}$ implying $5$ is a factor, which is impossible. If $d_5=8$ then if $d_4 = 5$ this implies $5+2=7$ is a factor, if $d_4=6$, this implies $6=2*3$, or $3$ is a factor, and if $d_4=7$, this implies $7+2=9=3*3$ that 3 is a factor.

If $d_6=8$ then we can see that the only possibility is $d_4=5$ and $d_5=7$ (as discussed above). A quick check shows that this does not work. And if we let $d_7=8$ this means we have to find $3$ primes between $4$ and $8$ even though there are only 2 => another contradiction.

We know that $d_3$ has to be a prime number now, as if it was composite, then there exists a prime factor greater than $2$ in it which is less than it but greater than 2.

Now, we simply mush through cases.

If $d_3=3$, this implies either $d_4=5$ or $d_4=6$. If $d_4=6$ this assumes $6+2 =8$ is a factor, which it's not since $4$ is not a factor. Thus this means $d_4=5, _5=6$ and since $2+5 = 7$ we have that $d_6=7$. Using B-T's chart we see the only way we can have 4 prime factors and 16 factors is $2*3*5*7$. A quick check shows that this does not work. Thus $d_3 \ne 3$

If $d_3=5$ we see that $d_4$ is either $7$ or $10$. If $d_4=7$ then $d_2+d_4=9$ is a factor implying 3 is a factor, which it isn't. If $d_4=10$ this again implies $2+10 = 12$ is a factor, again implying 3 is a factor which it is not. Thus $d_3 \ne 5$.

If $d_1=1, d_2 =2, d_3=7$ then $d_4=11,13,$ or $14$. If $d_4=13$ this implies $13+2=15$ is a factor, implying 3 and 5 are factors which they are not. If $d_4=14$ this implies $14+2 = 16$ is a factor, implying false factors of $4$ and $8$. Thus the only possibility is $d_4=11$ which implies $d_5= 13$. Again, the only way to have these 4 prime factors is $2*7*11*13$. A quick check shows this works. Thus one solution is $2002$.

If $d_1=1, d_2=2, d_3=11$, then $d_4= 13,17,19,22$. As before, we quickly show $d_4=13,19,22$ don't work, making the only possibility $d_4=17$. But we have the restriction that $d_5 \le 16$ (or else $d_{d_5}$ isn't a factor of $n$). Since $d_5 > d_4 =17$ this is a contradiction.

Finally, $d_1=1, d_2=2, d_3=13$. The next prime number of $13$ is $17 \ge 16$. Thus this means that $d_5 > 17$ which doesn't work. And subsequent primes will not work either.

Thus the only solution is 2002.

Sorry this was rushed I may come back and edit given that I have to go somewhere around right now

[quote=Valentin Vornicu]Find all positive integers which have exactly 16 positive divisors $1 = d_1 < d_2 < \ldots < d_{16} =n$ such that the divisor $d_k$, where $k = d_5$, equals $(d_2 + d_4) d_6$.[/quote]

Let $n=\prod_{i=1}^{a} (p_i)^{b_i}$ then
No. Of divisors of $n=\prod_{i=1}^{a}(1+b_i)=16$
So $1\leq a \leq 4$ Hence there are some cases..
Case 1-: when $a=1$ then $n=(p_1)^{15}$ then $d_5=(p_1)^4$ and as $d_{d_5}=d_6(d_4+d_2) \implies d_{(p_1)^4}=(p_1)^6(p_1^3+1)$ which is not possible.

Case 2-: when $a=2$ then $n=p_1^{b_1}*p_2^{b_2}$
Then $(1+b_1)(1+b_2)=16=2*8=4*4$
1-:  let $b_1=1, b_2=7$  then $n=p_1*p_2^7$ and let $p_1>p_2$ then $d_2=p_2, d_4=p_2^2$ and as
$d_2+d_4|n \implies p_2(p_2+1)|n$ which is not possible.
2-:  when $b_1=b_2=3$ then $n=p_1^3*p_2^3$ and let $p_1>p_2$ then again $d_2=p_2, d_4=p_2^2$ and as $d_2+d_4|n \implies p_2(p_2+1)|n$ which is not possible.
     So no solution in this case.
Case 3-: when $a=3$ then $n=p_1^{b_1}*p_2^{b_2}*p_3^{b_3}$ then $(1+b_1)(1+b_2)(1+b_3)=16=2*2*4$
1-: when $b_1=b_2=1, b_3=3$ then $n=p_1*p_2*p_3^3$ and atleast 1 prime divisor should be even then let $p_1=2$ and $p_2>p_3$
Then $d_2=2, d_4=p_2, d_5=2*p_3, d_6=2*p_2$
Then $6<2*p_3\leq 16 \implies p_3=5$ or $7$
And also $2*p_2(2+p_2)|2*p_2*p_3^3\implies 2+p_2=p_3^2$
if $p_3=5$ then $p_2=23$ but
Then $d_6=p_3^2=25$  hence $p_3\neq 5$
If $p_3=7$ then $p_2=47$ but then $d_5=p_2=46$ hence $p_3\neq 7$.
Hence no solution in this case.
Case 4-: when $a=4$ clearly then $b_1=b_2=b_3=b_4=1$ and $n=2*p_2*p_3*p_4$
 WLOG $p_1>p_2>p_3$
1-: then If $d_2=2, d_4=p_2, d_5=p_1, d_6=2*p_3$
Then $2*p_3(p_2+2)|2*p_2*p_3*p_1$ which is only possible if $p_2+2=p_1$ and also $6<p_1\leq 16$ possible values of $p_1\in \{7, 11,13\}$
If $p_1=7$ then $p_3=3, p_2=5$ but then $d_5\neq 7$ so $p_1\neq 7$
If $p_1=11$ then also $p_2=5, p_3=3$ is not possible. So checking $p_2=7, p_3=5$ gives $d_5=10$ not $11$ so $p_1\neq 11$.
If $p_1=13$ then $p_2=11, p_3=7$ works as $2*p_2>p_1$ so $d_5=p_1$. Hence $\boxed{n=2002}$ is a solution.
And now still some subcases of case 4 is left which is when $2*p_i<p_k$ for $i, k\leq 4$ which can easily be solved by above reasoning and gives No solution.
So $\boxed{n=2002}$ is the only solution.


Richard is hitchhiking from Decatur, AL, to Amherst, Va. The probability that he will se a car within the next 20 minutes is 609/625. What is the probability that he will se a car in the next 5 minutes? Assume the probability of seeing a car at any moment is uniform for the entire 20 minutes.

[hide]$\frac{609}{2500}$? There's probably some trick in the question that I didn't catch. [/hide]

Yeah it's a uniform distribution, therefore you just divide by four.

But does this belong on the Mandelbrot? :?

Errr...no you can't just divide by four.
[hide="hint"]Look at the complement, ie. the probability that he does not see a car in the next 20 minutes.[/hide]

[hide="Answer"]Probability that he does not see a car in 20 minutes is $\frac{16}{625}$, so probability that he does not see car in the first five minutes is $\frac{4}{625}$, and the chance that he does see a car in the first five minutes is $\frac{621}{625}$. If that's right, I'm guessing it's because he can see more than one car in that time...[/hide]

JesusFreak, that doesn't make sense.  How could he have a greater probability of seeing a car in the next 5 minutes than seeing one in the next 20 minutes?

[hide]chance of not seeing one in 20 mins is 16/625, so by randomness i think the answer's
$\frac{\displaystyle\sum_{k=1}^{16}}{\displaystyle\sum_{k=1}^{15}+\displaystyle\sum_{k=1}^{14}+\displaystyle\sum_{k=1}^{13}+\displaystyle\sum_{k=1}^{12}+\displaystyle\sum_{k=1}^{11}+\displaystyle\sum_{k=1}^{10}+\displaystyle\sum_{k=1}^{9}+\displaystyle\sum_{k=1}^{8}+\displaystyle\sum_{k=1}^{7}+\displaystyle\sum_{k=1}^{6}+\displaystyle\sum_{k=1}^{5}+\displaystyle\sum_{k=1}^{4}+\displaystyle\sum_{k=1}^{3}+\displaystyle\sum_{k=1}^{2}+\displaystyle\sum_{k=1}^{1}}$
which is $\frac{136}{680}$ which is $\frac{1}{5}$, so $\frac{1}{5}$ is the probability he won't see a car in the next five minutes; $\frac{4}{5}$ is probability he will see a car in the next 5 minutes
[/hide]