vector
probability
function
calculus
calculus open

I'm certain that someone's looked at this problem before, but I haven't found a reference yet.

What is the maximum possible value of min{P(X>Y), P(Y>Z), P(Z>X)}, where X, Y, and Z are independent random variables?

An upper bound of 2/3 is pretty easy to get, and I've constructed an example where 5/9 is attained, but I don't know where the actual bound lies (I suspect 5/9).

Here's an example where all three probabilities are $\frac{\sqrt{5}-1}2\approx .618$:
$X=0$ always
$P(Y=-1)=t$
$P(Y=2)=1-t$
$P(Z=1)=t$
$P(Z=-2)=1-t$
where $t=\frac{\sqrt{5}-1}2$.
Then clearly $P(X>Y)=P(Z>X)=t$ and $P(Y>Z)=1-t+t(1-t)=1-t^2=t$.

If I don't mistake, jmerry's example is optimal. The minimum of the three probabilities is always not greater than $t$. My proof  consists of 3 steps.

1) It suffices to consider random variables taking just 2 different values.

2) If $X$ takes values $x_1>x_2$ with probabilities $1-p$ and $p$ and similarly for $Y$ and $Z$ (using letters $q$ and $r$ respectively), then the worst possible ordering (up to cyclic rearrangements of $X,Y,Z$) of the values is $x_1>y_1>z_1>x_2>y_2>z_2$.

3) For this ordering, the three probabilities are given by $1-p(1-q)$, $1-q(1-r)$ and $p(1-r)$. It is impossible to have all three numbers strictly greater than $t$.

I'll post a full proof later if nobody finds a simpler approach...

[quote="fedja"] I'll post a full proof later if nobody finds a simpler approach...[/quote]OK, here goes.

Step 1:
First of all, we may assume that $X$, $Y$ and $Z$ take finitely many values. Indeed, choose a large integer $n$ and replace $X$ by $\widetilde X$ that equals $-n$ if $X<n$, $\frac kn$ if $\frac kn\leq X<\frac{k+1}n$, $k=-n^2,\dots, n^2$ and $n$ if $X\geq n$ and similarly for $Y$ and $Z$. It is not hard to check that $P\{\widetilde X>\widetilde Y\}$ tends to $P\{X>Y\}$ as $n\to\infty$. Now let $x_1>x_2>\dots>x_m$ be the set of values taken by the random variable $X$ and let $p_1,\dots,p_m>0$ be the corresponding probabilities. If $m>2$, we shall change the probabilities $p_j$ in such a way that one of them becomes $0$ and the probabilities $P\{X>Y\}$,  $P\{Y>Z\}$, $P\{Z>X\}$ do not decrease. Note that, due to independence of $X$ and $Y$, $P\{X>Y\}=\sum_j p_j P\{Y<x_j\}$ and similarly for $Z$. Thus our problem becomes to change $p_j$ such that the linear form $\sum_j p_j$ remains $1$ and two other fixed linear forms of $p_j$ do not decrease. If $m>3$, we can even move $p_j$ so that all three linear forms stay the same. (we have more variables than equations so the set of solutions forms a non-trivial linear subspace in $\mathbb R^m$). If $m=3$, we can no longer fix all three linear forms, but we still can choose the non-trivial increments $\Delta p_j$ so that $\sum_j \Delta p_j=0$ (this gives us a plane in $\mathbb R^3$) and two other forms are non-negative (this gives us an intersection of two half-planes on our plane and the intersection of any 2 half-planes contains a non-zero vector (provided that their boundaries contain the origin, which is the case in our situation)). So, we can reduce the number of different values of $X$ to $2$. Now the same can be done for $Y$ and, finally, for $Z$.

Step2:
Consider now some possible ordering of $x_1,\dots,z_2$. If in this ordering we have some $x$ immediately followed by some $z$, we can exchange them, increasing the probability that $Z>X$ and keeping two other probabilities unchanged. So, we may assume that $z$ never follows $x$. Similarly $x$ never follows $y$ and $y$ never follows $z$. Due to cyclic simmetry, we may assume that the largest value is $x_1$. It cannot be followed by $z_1$. Also, it cannot be followed by $x_2$ (otherwise $Z<X$ with probability $1$). Therefore, the next element in our ordering must be $y_1$. Reasoning this way again and again, we see that only $2$ orderings are possible: $x_1>y_1>z_1>x_2>y_2>z_2$ and $x_1>y_1>z_1>z_2>x_2>y_2$. But in the second case we may replace $z_1$ and $z_2$ by a common value $z$ attained with probability $1$ and add a dummy value $z'>x_1$ that is taken with probability $0$, thus reducing the second case to a partial case of (a cyclic rearrangement of) the first case.

Step 3:  Suppose that all three probabilities are greater than $t$. Then, since $p(1-r)>t$, we must have $p>t$ and $1-r>\frac tp$. Since  $1-q(1-r)>t$ and $1-t^2=t$, we conclude that $q<pt$. But then $p(1-q)>p(1-pt)\geq t^2$ because $p\mapsto p(1-pt)$ is a concave function of $p$ and it takes the value $t^2$ at $p=t$ and at $p=1$. Thus, the first probability does not exceed $1-t^2=t$ - contradiction.

The end.

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NERDS! NERDS! NERDS! NERDS! NERDS! NERDS! NERDS! NERDS! NERDS! NERDS! NERDS! NERDS! NERDS! NERDS! NERDS! NERDS! NERDS! NERDS! NERDS! NERDS! NERDS! NERDS! NERDS! NERDS! NERDS! NERDS! NERDS! NERDS! NERDS! NERDS! NERDS! NERDS! NERDS! NERDS! NERDS! NERDS! NERDS! NERDS! NERDS! NERDS! NERDS! NERDS! NERDS![/hide][/quote]
huge SPAM detected  :spam:
take action Valentin  :yoda:   :police:   :agent:   :starwars:   :pilot:


calculus
integration
logarithms
trigonometry
calculus computations

$ \int (\frac {1 \minus{} \sqrt {x}}{1 \plus{} \sqrt {x}})^{\frac {1}{2}}.\frac{1}{x} dx$

Ans-$ \minus{} 2\ln \frac {1 \minus{} \sqrt {1 \minus{} x}}{\sqrt {x}} \minus{} 2 \sin^{ \minus{} 1} \sqrt {x} \plus{} C$

put $ x\equal{}\cos^2\,2\theta$ to get

$ \int\;\sqrt{\;\dfrac{2\sin^2\,\theta}{2\cos^2\,\theta} \;}\left(\dfrac{\minus{}4\sin\,2\theta\;\cos\,2\theta\;\textbf d\theta}{\cos^2\,2\theta}\right)\equal{}4\int\dfrac{\minus{}\sin\,\theta}{\cos\,\theta}\left(\dfrac{\sin\,2\theta}{\cos\,2\theta}\right)\;\textbf d\theta$

$ \equal{}4\int\dfrac{\left(1\minus{}2\sin^2\,\theta\right)\minus{}1}{1\minus{}2\sin^2\,\theta}\;\textbf d\theta\equal{}4\int\;\left(1\minus{}\sec\,2\theta\right)\;\textbf d\theta$

$ \equal{}4\theta\minus{}2\ln\left(\sec\,2\theta\plus{}\tan\,2\theta\right)\plus{}K\equal{}\boxed{2\left[\cos^{\minus{}1}\,\sqrt x\minus{}\ln\left(\dfrac{1\plus{}\sqrt{\;1\minus{}x^2\;}}{\sqrt x}\right)\right]\plus{}K}$


Find all fuction \[ f: Q^+ \to Q^+\] such that:

1) \[ f(x) + f(1/x)=1, \forall x \in Q^+\] and

2) \[ f (1+2x) = \fr{1}{2} f(x), \forall x \in Q^+\]

\[ f(x) = \dfrac{1}{x+1} \] is an unique  solution?

I think 2)$ f(1\plus{}2x)\equal{}\frac{f(x)}{2}.$
For $ x\equal{}1$ we get $ f(x)\equal{}\frac 12 \equal{}\frac{1}{1\plus{}x}$. For $ x\equal{}\frac 12, f(x)\equal{}y$ we get $ f(2)\equal{}\frac y2 \equal{}1\minus{}y\to y\equal{}\frac 23$.
Let $ S\equal{}\{x|f(x)\equal{}\frac{1}{1\plus{}x}\}$, $ R\equal{}\{\frac{1}{1\plus{}x},x\in S\},y\in R\to y\in (0,1)$. $ 1,\frac 12 \in S\to \frac 12, \frac 23 \in R.$
(1) If $ x\in S$, then $ f(\frac 1x )\equal{}1\minus{}\frac 1x \equal{}\frac{1}{1\plus{}\frac 1x}\to \frac 1x \in S$ or $ y\in R\to 1\minus{}y\in R$.
(2) If $ x\in S$, then $ f(1\plus{}2x)\equal{}\frac{1}{2\plus{}2x}\to 1\plus{}2x\in S$ or $ y\in R\to \frac y2 \in R$.
If $ f$ continiosly from (1-2) we get $ R\equal{}(0,1)$ or $ f(x)\equal{}\frac{1}{1\plus{}x}$ for all $ x\in Q_\plus{}$.

[quote="Rust"]I think 2)$ f(1 \plus{} 2x) \equal{} \frac {f(x)}{2}.$
For $ x \equal{} 1$ we get $ f(x) \equal{} \frac 12 \equal{} \frac {1}{1 \plus{} x}$. For $ x \equal{} \frac 12, f(x) \equal{} y$ we get $ f(2) \equal{} \frac y2 \equal{} 1 \minus{} y\to y \equal{} \frac 23$.
Let $ S \equal{} \{x|f(x) \equal{} \frac {1}{1 \plus{} x}\}$, $ R \equal{} \{\frac {1}{1 \plus{} x},x\in S\},y\in R\to y\in (0,1)$. $ 1,\frac 12 \in S\to \frac 12, \frac 23 \in R.$
(1) If $ x\in S$, then $ f(\frac 1x ) \equal{} 1 \minus{} \frac 1x \equal{} \frac {1}{1 \plus{} \frac 1x}\to \frac 1x \in S$ or $ y\in R\to 1 \minus{} y\in R$.
(2) If $ x\in S$, then $ f(1 \plus{} 2x) \equal{} \frac {1}{2 \plus{} 2x}\to 1 \plus{} 2x\in S$ or $ y\in R\to \frac y2 \in R$.
If $ f$ continiosly from (1-2) we get $ R \equal{} (0,1)$ or $ f(x) \equal{} \frac {1}{1 \plus{} x}$ for all $ x\in Q_ \plus{}$.[/quote]

Can u please explain clearly the last line "If $ f$ continiosly from (1-2) we get $ R \equal{} (0,1)$ or $ f(x) \equal{} \frac {1}{1 \plus{} x}$ for all $ x\in Q_ \plus{}$."


calculus
integration
limit
trigonometry
inequalities
calculus computations

Evaluate the limit \[L = \lim_{y \rightarrow 0}\frac{1}{y}\int_{0}^{\pi}\tan{(y\sin x)}\,dx\]

From $t \leq \tan t \leq t^{3}+t$ we have $y \, \sin x \leq \tan (y \, \sin x) \leq y^{3}\sin^{3}x+y \sin x$

integrating the inequality we get \[\int_{0}^{\pi}y \sin x \, dx \leq \int_{0}^{\pi}\tan(y \, \sin x) \, dx \leq \int_{0}^{\pi}y \sin x( y^{2}\sin^{2}x+1) \, dx\] which is \[2y \leq \int_{0}^{\pi}\tan(y \sin x) \, dx \leq \frac{4}{3}y^{3}+2y\] so \[2 \leq \frac{1}{y }\int_{0}^{\pi}\tan(y \sin x) \, dx \leq \frac{4}{3}y^{2}+2\] and we have $\lim_{y \rightarrow 0}\frac{1}{y }\int_{0}^{\pi}\tan(y \sin x) \, dx =2$


There are many ways that function can be defined and one of them is "odd, and even."

The function is defined even if:

$f(x) = f(-x)$

The function is defined odd if:

$f(x) = -f(x)$

It's hard to explain them without the actual problem so I'll show some.

$x^2+1$

Is this EVEN or ODD or NEITHER?

Put the negative in the place of x.

$(-x)^2 + 1 = x^2 + 1$ Therefore, this is even.

$x^3-2$

Is this EVEN or ODD or NEITHER?

Again, same thing.

$(-x)^3-2 = -x^3-2$ This case, the function is neither. The reason is that if you look at the $-x^3-2$ and $x^3-2$, the -2 didn't change its sign. And it CAN't. So, it is neither.

Let's look at one more.

$x^{99999}$

$(-x)^{99999} = -x^{99999}$

So, this function is odd.

Problems:

1. $g(x) = x^2-x+2$, even, odd, or neither?
2. $f(x) = x^4-x^2-7$, even, odd, or neither?
3. $k(x) = x^7+7x^5-2$, even, odd, or neither?
4. $p(x) = x^{11} - 2x^{10}$, even, odd, or neither?

Answers

Don't look at them until you're done!

[hide="number_one"]

$(-x)^2-(-x)+2 = x^2+x+2$ Therefore, neither
[/hide]

[hide="number_two"]

$(-x)^4-(-x)^2-7 = x^4-x^2-7$ Congruent to the orginial equation, so it's EVEN.
[/hide]

[hide="number_three"]

$(-x)^7-7(-x)^5-2 = -x^7+7x^5-2$ Neither[/hide]

[hide="number_four"]

$(-x)^{11} - 2(x)^{10} = -x^{11} - 2x^{10}$ Neither[/hide]

The odd and even function can also be defined using the graph.

[hide]$g(x) = x^2 - x + 2
g(-x) = (-x)^2 + x + 2 = x^2 + x + 2 =/= -g(x) =/= g(x)$

neither.

$f(x) = x^4 - x^2 + 7
f(-x) = (-x)^4 - (-x)^2 + 7 = x^4 - x^2 + 7 = f(x)$

even.

$k(x) = x^7 + 7x^5 + 2
k(-x) = (-x)^7 + 7(-x)^5 + 2 = -x^7 - 7x^5 + 2 =/= -k(x) =/= k(x)$

neither. (so close though!)

$p(x) = x^{11} - 2x^{10}
p(-x) = (-x)^{11} - 2(-x)^{10} = -x^{11} - 2x^{10} =/= -p(x) =/= p(x)$

neither.
I hope I got these right..
As an extension, could I say as a rule, even functions are symetrrical about a y-axis and odd functions... rotated by pi around the origin?[/hide]

[quote="mikewd"]As an extension, could I say as a rule, even functions are symetrrical about a y-axis and odd functions... rotated by pi around the origin?[/quote]

That is correct.  Another trick: when they are polynomials, a function is even if it has only even exponents of x.  For example, $4x^6-12x^2+10x^0$.  It is odd if it has only odd exponents, for example $12x^{11}-8x^3+x^{-3}$  Neither if it has both even and odd.

M.

As the famous mathematician  observerd:

 [quote]It does seem [b]odd[/b] that some functions  have to be rotated by pi to get  symmetry  :)  [/quote]

By the way another  useful thing to know is that every function can be written as a some of  an even function and an odd function.

Eg:   y(x) =   x^2+3x+ 4

is neither odd nor even but it is sum of

Y1 =  x^2+4
and Y2 = 3x
(Y=Y1+Y2)
where Y1 is even and Y2 is  odd.

Challange Question:   Find the two function, one odd, one even such that their sum is:

 f(x)  =  (x+1)^2

[hide]We can do FOIL, and get $f(x) = x^2+2x+1$.  SInce $x^2+1 = (-x)^2+1$, $f(x) = x^2+1$ is an even function.  And since $2x = -2(-x)$, $f(x) = 2x$ is an odd function.  The functions $f(x) = x^2+1$ and $f(x) = 2x$ add up to $f(x) = {(x+1)}^2$, and are even and odd respectively.$

I was just wondering, could you also dfine even and odd functions as a function where the exponents are even, and a function where the exponents are odd respectively?[/hide]

No, because there are lots of functions that are even and odd but aren't polynomials.  For example, the function $2^x + \frac{1}{2^x}$ is not a polynomial, but it is an even function.  For polynomials, though, what you said is true -- even polynomials have only even exponents.

[hide]this is too hard and confusing..It sounds easy though[/hide]

Let me add my own challenge problem: for the function $f(x) = 2^x$, write it at $f(x) = e(x) + o(x)$ where $e(x)$ is even and $o(x)$ is odd.  (This is fairly tricky, I think.  You're going to have to figure out how to use the evenness and oddness properties to solve for the functions e and o.)

[quote="Silverfalcon"]The function is defined odd if:
$f(x) = -f(x)$
[/quote]
This should read:

$f(x)=-f(-x)$

Answer:[hide]First, $2^x=e(x)+o(x)$
Also, $2^{-x}=e(-x)+o(-x)=e(x)-o(x)$
Add both equations up:
$2^x+2^{-x}=2e(x)$
$e(x)=2^{x-1}+2^{-x-1}$
Then since $2^x=e(x)+o(x)$ you solve for o(x) to be
$o(x)=2^{x-1}-2^{-x-1}$
[/hide]

Additional problems:
1. Can a function be both even and odd?
2. If o(x) is an odd function and e(x) is an even function, then o(x)*e(x) is:
A) Even B) Odd C) Neither D) Both?
3. How about o(x)*o(x)*e(x)?


This year, I'm taking the Comprehensive. I tend to suck at the practice problems though. Can anyone give me tips on how to prepare?

To start with, I did lots of practice exams.  Afterward, I looked through the official solutions to all the problems I missed to make sure I got them all.  My dad also helped me to do the practice tests and solve the problems.  All of the exams are located on their website, so they are very easy to have access to.  I got 15th at the comprehensive contest doing this.

Also, learn lots of common trig ratios, and formulas that are similar.  Know why the work in addition to the formula so if you forget the formula, you can derive it.

Thanks  :D

I already had the practice exams since last year. Now I need to do them  :wink:

For you... practice AIMEs. They seem to be the most useful things to practice on in general, and will help you at almost all tests. Look at some comprehensive tests too, but do so sparingly. And of course read books. This is just general info that will prepare you well for any competition. If you're interested exclusively in the comprehensive test, b-flat's advice may be better.

Thanks. My AIME score needs to improve too (5 last year)...
 :blush:


function calculus integration trigonometry Euler geometry rectangle

Say you wanted to estimate $ \sum_{k \equal{} a}^{b} f(k)$ .

You might just take the definite integral of f(x) dx from a-1/2 to b+1/2.

But what if you wanted more accuracy? You could make values at integers "count" more.

Making integers count more is easy with the cosine. So I came up with this formula. It works best for a function that grows in magnitude instead of one that gets smaller.

[img]http://i17.photobucket.com/albums/b66/kenyai/intsum4.png[/img]

Is this formula valid and useful for most functions?
And most importantly, how do you choose the best c?

Offhand, the denominator could vanish if $ f(x)$ is "orthogonal enough" to $ \cos^{2n} (\pi x)$ (an obvious candidate is a sum of sines), so I'm not sure how useful this technique is.  The main technique I know for estimating a sum with an integral (or vice versa) is the [url=http://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula]Euler-Maclaurin formula[/url].

The reason for the f(c) and the denominator is that f(c)/denominator gives an estimate for what factor you would have to multiply by to make the "spike" at an integer x equal to a rectangle of width 1 and height f(x). The only assumption is that the spikes are all about the same size compared to such a rectangle.

To see how this works with ln(x): [url]http://i17.photobucket.com/albums/b66/kenyai/LogWithCosineSquared.png[/url]

Anyway, this method might be better if you don't want to find a power series for the function or don't want to differentiate for some reason and thus are unable to use Euler-Maclaurin.


Let $ b$ be a natural number. Find all $ 2002$-tuples $ (a_1,a_2,...,a_{2002})$, of natural numbers such that: $ \displaystyle\sum_{j\equal{}1}^{2002} a_j^{a_j}\equal{}2002b^b.$


Does anyone have the book "Problem solving strategies" by A.Engel?I'm interested in getting it!

don't worry it's being translated as we speak in Romanian. it will be available in about 3 months.

Heh, I have a photocopied version of the book, which was photocopied from a photocopy of the book to begin with... :P

That just says how popular the book is ;)

It's a bit toooooo late to write on this topic, but what about the Romanian version of the Problem Solving Strategies?
When will it be ready, because many people would buy it in Romanian :):D

I am not sure about that "many people will buy it in Romania" thing. latest sales in book at the NMO show a dramatic decrease in Math Olympiad book sales.

There are few that can really use that book, as it is not quite the begginner's book, and also it's not that good for an advanced guy ...

Well, in Germany the book is very popular, particularly cuz two of the four problems of the 2nd Round Bundeswettbewerb Mathematik (German National Mathematics Competition) 2003 were from this book! I have never bought it myself, but I have it borrowed from Engel himself  :D  :D  :D

  dg


The numbers $1,2,3,...,20$ are written. A person will erase two numbers $a$ and $b$ and write $a+b-1$ instead. After repeating this step $19$ times, there will be only one number left. What is that number?

[hide]Erase the two ends (1, 20; 2, 19; 3, 18; etc) 10 times to get 10 20's. Erasing a pair of 20's five times will leave you with 5 39's. Now erase two pairs of 39's to get 1 39 and 2 77's. Pair the 77's to get 153. Now combing the 153 and the 39, to get 191. [/hide]

Isn't this  a special approach...( a special case).. I believe this is a standard beginning IMO level problem involving use of invariants.... :lol:

Yeah, it's OK as long as the question doesn't ask you to show your work  :D

Yes, the question is such that it will always give this answer. So, I guess it is OK for the BAsic level MCQs


Prove that the equation
cos(cos(cos(cos(x))) = sin(sin(sin(sin(x)))
has no solution.


[hide]
$\frac{1}{abc} \le \frac{\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}}{3}$ by AM-GM.

Do the same for $a,b,d$; $a,c,d$; and $b,c,d$. Then sum them up.[/hide]

[quote="ddziabenko"]Prove that for all positive reals $ a,b,c,d$, we have $ \frac {a \plus{} b \plus{} c \plus{} d}{abcd}\leq \frac {1}{a^{3}} \plus{} \frac {1}{b^{3}} \plus{} \frac {1}{c^{3}} \plus{} \frac {1}{d^{3}}$[/quote]
Lol.. This is just rearrangement on $ \frac {1}{a}, \frac {1}{b}, \frac {1}{c}, \frac {1}{d}$.....
Hölder can kill it too:
$ \sum_{cyc} \frac{1}{a^3} \equal{} \left ( \sum_{cyc} \frac{1}{a^3} \right )^{1/3}\left ( \sum_{cyc} \frac{1}{b^3} \right )^{1/3}\left ( \sum_{cyc} \frac{1}{c^3} \right )^{1/3} \ge \sum_{cyc} \frac{1}{abc} \equal{} \frac{a\plus{}b\plus{}c\plus{}d}{abcd}$

[quote="ddziabenko"]Prove that for all positive reals $a,b,c,d$, we have $\frac{a+b+c+d}{abcd}\leq \frac{1}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}}+\frac{1}{d^{3}}$[/quote]
The generalization $\frac{a_1+a_2+a_3+\cdots +a_n}{a_1a_2a_3\cdots a_n}\le \frac{1}{a_1^{n-1}}+\frac{1}{a_2^{n-1}}+\frac{1}{a_3^{n-1}}+\cdots +\frac{1}{a_n^{n-1}}$ also holds, where $a_i>0\ \forall i\in \{1,2,3,...,n\}$.


From number 2 and number 5, we get:
\[ \cos{A}+\cos{B}+\cos{C} \leq \frac{3}{2} \]

A stab at number 2

[hide] Use AM-GM to show that the expression $\displaystyle\sin{\frac{A}{2}}\sin{\frac{B}{2}}\sin{\frac{C}{2}}$ is maximized when $A=B=C=60$, so $\displaystyle\sin{\frac{A}{2}}\sin{\frac{B}{2}}\sin{\frac{C}{2}}\le (\frac{1}{2})^3$ or $\displaystyle\frac{1}{8}$.[/hide]

number 4 can be proven using a similar method.

I think that it should be AM-GM + Jensen.

Major Edit: I posted this unfinished and I was on the run. I now have time to make it work, sorry for any confusion.

$\boxed{\cos{A} +\cos{B}+\cos{C}=1+4\sin{\frac{A}{2}}\sin{\frac{B}{2}}\sin{\frac{C}{2}}}$
I proved this [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=187198&highlight=#187198]here.[/url]

Using double-angle idenitity for cosine, we get:
$1 - 2\sin^2\frac{A}{2} +1 - 2\sin^2\frac{B}{2} + 1 - 2\sin\frac{C}{2} = 1 + 4\sin{\frac{A}{2}}\sin{\frac{B}{2}}\sin{\frac{C}{2}}$

$\Rightarrow -2\sin^2\frac{A}{2} - 2\sin^2\frac{B}{2} - 2\sin\frac{C}{2} =-2 + 4\sin{\frac{A}{2}}\sin{\frac{B}{2}}\sin{\frac{C}{2}}$

$\boxed{\Rightarrow (\sin{\frac{A}{2}})^2+(\sin{\frac{B}{2}})^2+(\sin{\frac{C}{2}})^2 +2\sin{\frac{A}{2}}\sin{\frac{B}{2}}\sin{\frac{C}{2}} = 1}$

Using the same method as I did in my last post, we get:
[hide="A METHOD!"]
$\displaystyle\sum_{cyc}\sin A = S$.
Now, $2S = \displaystyle\sum_{cyc}(\sin A + \sin B) = \displaystyle\sum_{cyc}\left(2\cos\left(\frac{A-B}{2}\right)\sin\left(\frac{A+B}{2}\right)\right)$

$= \displaystyle\sum_{cyc}2\sin\left(\frac{\pi - C}{2}\right)\left(\cos\frac{A}{2}\cos\frac{B}{2}+\sin\frac{A}{2}\sin\frac{B}{2}\right) = \displaystyle\sum_{cyc}2\left(\cos\frac{C}{2}\cos\frac{A}{2}\cos\frac{B}{2}+\cos\frac{C}{2}\sin\frac{A}{2}\sin\frac{B}{2}\right)$

$\Rightarrow S = \displaystyle\sum_{cyc}\left(\cos\frac{C}{2}\cos\frac{A}{2}\cos\frac{B}{2}+\cos\frac{C}{2}\sin\frac{A}{2}\sin\frac{B}{2}\right)$

$= 3\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2} + \displaystyle\sum_{cyc}\cos\frac{C}{2}\sin\frac{A}{2}\sin\frac{B}{2}$

Now, say $T = \displaystyle\sum_{cyc}\cos\frac{C}{2}\sin\frac{A}{2}\sin\frac{B}{2}$ and then we can say:

$2T = \displaystyle\sum_{cyc}\left(\cos\frac{C}{2}\sin\frac{A}{2}\sin\frac{B}{2}+\cos\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\right)$

$= \displaystyle\sum_{cyc}\sin\frac{B}{2}\left(\cos\frac{C}{2}\sin\frac{A}{2}+\cos\frac{A}{2}\sin\frac{C}{2}\right) = \displaystyle\sum_{cyc}\sin\frac{B}{2}\left(\sin\left(\frac{A+C}{2}\right)\right)$

$= \displaystyle\sum_{cyc}\sin\frac{B}{2}\sin\left(\frac{\pi - B}{2}\right)= \displaystyle\sum_{cyc} \sin\frac{B}{2}\cos\frac{B}{2}$

$\Rightarrow 4T = \displaystyle\sum_{cyc} 2\sin\frac{B}{2}\cos\frac{B}{2} = \displaystyle\sum_{cyc} \sin B = S$

Say, $P = \cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}$

$\Rightarrow T + 3P = S = 4T \Rightarrow 3P = 3T \Rightarrow S = 4T = 4P$

Which leaves us with:
[/hide]
$\sin{A} +\sin{B}+\sin{C}=4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}$
I don't know why I did this? I will use it though to expand later.

Since $\displaystyle\frac{A}{2}$,$\frac{B}{2}$, and $\frac{C}{2}$ all lie on the interval $\left(0, \frac{\pi}{2}\right)$ we know sine and cosine are concave on those three angles, so by Jensen's we get:
$\frac{1}{3}\sin\frac{A}{2} + \frac{1}{3}\sin\frac{B}{2} + \frac{1}{3}\sin\frac{C}{2} \leq \sin\left(\frac{1}{3}\frac{A}{2} + \frac{1}{3}\frac{B}{2} + \frac{1}{3}\frac{C}{2}\right) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$
and
$\frac{1}{3}\cos\frac{A}{2} + \frac{1}{3}\cos\frac{B}{2} + \frac{1}{3}\cos\frac{C}{2} \leq \cos\left(\frac{1}{3}\frac{A}{2} + \frac{1}{3}\frac{B}{2} + \frac{1}{3}\frac{C}{2}\right) = \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$
We also know each must be positive on this interval, so we can then apply AM-GM to the numbers to get:
$\sqrt[3]{\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}} \leq \frac{1}{3}\sin\frac{A}{2} + \frac{1}{3}\sin\frac{B}{2} + \frac{1}{3}\sin\frac{C}{2} \leq \frac{1}{2}$
$\boxed{\Rightarrow \sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2} \leq \left(\frac{1}{2}\right)^3 = \frac{1}{8}}$
and
$\sqrt[3]{\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}} \leq \frac{1}{3}\cos\frac{A}{2} + \frac{1}{3}\cos\frac{B}{2} + \frac{1}{3}\cos\frac{C}{2} \leq \frac{\sqrt{3}}{2}$
$\boxed{\Rightarrow \cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2} \leq \left(\frac{\sqrt{3}}{2}\right)^3 = \frac{3\sqrt{3}}{8}}$

Using $\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2} \leq \frac{1}{8}$ we get:
$-2\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2} \geq -\frac{1}{4}$
$\Rightarrow 1 - 2\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2} \geq 1 - \frac{1}{4} = \frac{3}{4}$
From our other identity, it reduces to:
$\boxed{(\sin{\frac{A}{2}})^2+(\sin{\frac{B}{2}})^2+(\sin{\frac{C}{2}})^2 \geq \frac{3}{4}}$

I guess I didn't need to provide the identity with the sum of sines, but I did, so I'll use it now:
$\sin A + \sin B + \sin C = 4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2} \leq 4 \cdot \frac{3\sqrt{3}}{8} = \frac{3\sqrt{3}}{2}$
Which we don't need the identity for because sine is concave on the entire interval $(0, \pi)$ so we can use Jensen's anyways.

isn't your final solution a contradiction for some values of $(a,b,c)$ ?


[quote="Ihatepie"]In Okmulgee, Oklahoma you cannot take a picture of a man with a wooden leg. Why not?[/quote]

[hide="i have a question"]does this have anything to do with the fact that there is the word "ok" in the places, and also oak is a kind of wood on a tree?? cuz that is what i think, except i cannot really think of a reason to what this may have to do with the answer... :maybe:[/hide]

[quote="Ihatepie"][quote="Nawahd Werdna"][hide] :idea:  :idea: You cannot take a picture of a man with a wooden leg, because you cannot take a picture with a woodn leg, you must use a [b]camera[/b]... right? :D [/hide]

- NAWAHD WERDNA[/quote]

ur right

srry klutzycherry but i dont know how u thought of the oks but ur wrong[/quote]

it happens... :P


geometry
3D geometry
combinatorics unsolved
combinatorics

Calculate $ \binom{3n\plus{}1}{1}\plus{}\binom{3n\plus{}1}{4}\plus{}\cdots \plus{}\binom{3n\plus{}1}{3n\plus{}1}$

If $ \omega$ is a complex cube root of unity, then
\[ \frac{2^{3n\plus{}1}\plus{}\omega^2(1\plus{}\omega)^{3n\plus{}1}\plus{}\omega(1\plus{}\omega^2)^{3n\plus{}1}}{3}\equal{}\sum_{i\equal{}0}^{3n\plus{}1}{3n\plus{}1\choose i}\left[\frac{1\plus{}\omega^{i\plus{}2}\plus{}\omega^{2i\plus{}1}}{3}\right].\]
Now, if $ i\equal{}3n\plus{}1$, the term in brackets will be 1, otherwise it will be $ \frac{1\plus{}\omega\plus{}\omega^2}{3}\equal{}0$. So we have
\[ \frac{2^{3n\plus{}1}\plus{}\omega^2(1\plus{}\omega)^{3n\plus{}1}\plus{}\omega(1\plus{}\omega^2)^{3n\plus{}1}}{3}\equal{}\sum_{i\equal{}0}^n{3n\plus{}1\choose3i\plus{}1}.\]


quadratics
algebra
polynomial
number theory theorems
number theory

i  just  finished  learning  quadratic  reciprocity  and primitive  roots.  now  i  think  i need  some  problems  to  apply  my  new  tools but  i caNT find  any .  could  someone please  post  their  favorite  olympiad  problems  here  whose  solution involves  these  tools??  please!!! thanks  :P

I've been out of the olympiad system for a while now ( I am old...) but here is one of the nicest :

show that there are infinitely many primes p such that $p\equiv 1$ mod 3

hint : suppose there are only finitely many
for this think of the polynomial $x^2+3$ and now put 'a well chosen number' in it for x and consider the primes dividing its value

This is pretty nice: prove that there are no positive integers $x$ and $y$ such that \[ \frac{y^2 + y - 1}{10x \pm 3} \] is a positive integer.

A "brute force" solution:

Checking crongruence of y mod 10:
$y \equiv 1 \Rightarrow  y^2 \equiv 1 \Rightarrow  y^2 +y -1 \equiv 1 (mod 10)$
$y \equiv 2 \Rightarrow  y^2 \equiv 4 \Rightarrow  y^2 +y -1 \equiv 5 (mod 10)$
$y \equiv 3 \Rightarrow  y^2 \equiv 9 \Rightarrow  y^2 +y -1 \equiv 1 (mod 10)$
$y \equiv 4 \Rightarrow  y^2 \equiv 6 \Rightarrow  y^2 +y -1 \equiv 9 (mod 10)$
$y \equiv 5 \Rightarrow  y^2 \equiv 5 \Rightarrow  y^2 +y -1 \equiv 9 (mod 10)$
$y \equiv 6 \Rightarrow  y^2 \equiv 6 \Rightarrow  y^2 +y -1 \equiv 1 (mod 10)$
$y \equiv 7 \Rightarrow  y^2 \equiv 9 \Rightarrow  y^2 +y -1 \equiv 5 (mod 10)$
$y \equiv 8 \Rightarrow  y^2 \equiv 4 \Rightarrow  y^2 +y -1 \equiv 1 (mod 10)$
$y \equiv 9 \Rightarrow  y^2 \equiv 1 \Rightarrow  y^2 +y -1 \equiv 9 (mod 10)$
$y \equiv 0 \Rightarrow  y^2 \equiv 0 \Rightarrow  y^2 +y -1 \equiv 9 (mod 10)$

So $y^2 + y -1$ is never $\equiv 3$ or $\equiv -3 \equiv 7$

but that doesn't mean it can't have a divisor which is congruent to 3 or 7 (mod 10)!

Another cool problem: If $a$ and $b$ are integers, show that $\frac{a^2-2}{2b^2+3}$ cannot be an integer.  I think it was posted in Pre-Olympiad somewhere though... but there wasn't a solution simpler than using some quadratic reciprocity stuff.

[quote="perfect_radio"][quote="fredbel6"]show that there are infinitely many primes p such that $p\equiv 1$ mod 3[/quote]

how does your proof go? it seems interesting...[/quote]

suppose there are only finitely many of them:$p_1\cdots p_m$

make $M= p_1\cdots p_m$

now $M^2+3$ is a number greater than one, not divisible by three
so a prime p $\neq 3$ divides it, but it can't be a $p_i$

but obviously $-3$ is square mod p
quadratic reciprocity implies that if $p\neq 3$ this means : $p\equiv 1\bmod{3}$

contradiction

The same goes for $p\equiv 1 \bmod{4}$

in general by using cyclotomic polynomials you can do the case $p\equiv 1 \bmod{n}$ of dirichlet

Here is one interesting problem:
Let $F_{n}=2^{2^{n}}+1$. Prove that $F_n$ is prime iff $3^{\frac{F_n-1}{2}}\equiv-1(modF_n)$ :?:

In this problem everything you have to do is to look on the order of $3 \mod{F_n}$.

This problem is not hard but it is some criterion for finding Fermat's prime numbers :P

It's quite easy, but anyway it's a nice problem :).


[quote="M4RI0"]Two numbers are randomly selected from interval $I = [0, 1]$. Given $\alpha \in I$, what is the probability that the smaller of the two numbers does not exceed $a$?

Is the answer $(100\alpha)%
$, it just seems too easy. :|[/quote]

$\alpha$ is not one of the numbers, it is a parameter that you express the answer in terms of

[quote="M4RI0"]Two numbers are randomly selected from interval $I = [0, 1]$. Given $\alpha \in I$, what is the probability that the smaller of the two numbers does not exceed $\alpha$?

Is the answer $(100 \alpha)$%, it just seems too easy. :|[/quote]

$(100 \alpha)\% = \alpha$.  But no, this is not the answer.

[hide="Solution"]We consider the complement: what is the probability that the smaller of the two does exceed $\alpha$?  Well, then both numbers exceed $\alpha$, and the probability of that is $(1-\alpha)^{2}$.  So our answer is just $1-(1-\alpha)^{2}= 2\alpha-\alpha^{2}$.[/hide]


are there restrictions on the values of n and m?

Even if you consider all  integers .

Hint 1 - Easy to prove that  no negative integeres will do.
Hint 2 -  Not that hard to porve that
[hide]only solutions are  (0,1)(3,0)(4,2)  [/hide]

Interesting that it was in 1992 ... have seen this problem looooong ago !

Also if  you like to cheat  and look at  similar problem: ... look [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=3%5Ex%2B4%5Ey&t=19256]Here .[/url]

could u please explain how u got your answer

If you are still looking for a solution , hope this helps. We will not even restrict  for positive integers but will consider all integers ..(Of course for positive integers, the solution is even more simple)
For $n<0$  it is easy to prove [hide] that fractional part is greater than zero, and less than  $(1/2+1/3)$, so no solution  exists [/hide]
For $n=1$  only answer is  $2^3+3^0$  because..
[hide]$2^m = x^2-1 = (x+1)(x-1)$   hence both the factors have to be a power  of 2, and  only $2,4$ are two powers of 2  whose difference is 2 .. (the difference becomes larger. as m increases) . we have m=3,n=0  .[/hide]
For $n>0$ take mod 3,[hide] since all squares are 1  or 0 mod 3, m has to be [b]even [/b]
$2^m + 0 \equiv (-1)^m \equiv 1\bmod 3$  (This has to be 1 mod 3  , 0 is not possible as 2^m can not be 0 mod 3 ), hence m has to be even, let $m=2k$  so you have
$3^n = (x+2^k)(x-2^k)$
Both the factors have to be power of $3$  so let $(x+2^k)=3^a$ and $(x-2^k) = 3^b$ with $a+b=n$  and $a>b$
subtracting  we get  $3^a - 3^b = 2 \cdot 2^k = 2^{k+1}$
since  RHS is even  $b$ has to be zero (and  $a=n$)  we have:
$3^n - 1 = 2^{k+1}$
again when  $k=0$ the only solution is when $n=1$
if $k>0$ take mod 4,  RHS is zero, so  $0 \equiv 3^n-1 \equiv (-1)^n -1$  so $n$ has to be even too, let $n=2j$
so now we have:
$3^{2j} - 1 = 2^{k+1} = (3^j+1)(3^j-1)$
Again both factors have to be a power of 2  (same as a case not too far above) and thus $j=1$ or $n=2$  [/hide]

Thus only solutions are:

[hide]n=0, m=3  or $2^3+3^0 = 9$
n=1  m=0 or  $2^0+3^1 = 4$
[b]m=2 m=4 [/b] or $2^4+3^2 = 25$   [b](Only one with both m and n positive integer[/b]) [/hide]

By the way, this is a particular case of the  famous Beal's conjucture, so if any one proves a general case here, there is a million dollart prize waiting for that! :)

Sorry for reviving an age-old thread, but, the question says that $m,n$ are positive integers. Can '0' be claimed so?

I have a different approach to this. Using Fermat's Infinite Descent and primitive pythagorean triplets, we can show that $m \leq 3$ Then, using congruence and caseworks and smaller inequalities will give the answer. Mine is too long a approach, but different.

[quote=INMO 1992 P8]Determine all pairs $(m,n)$ of positive integers for which $2^{m} + 3^{n}$ is a perfect square.[/quote]
[b][color=#000]Solution:[/color][/b] $$2^m+3^n=k^2$$
Take $\pmod{3}$ $\implies$ $k^2 \equiv 2^m \equiv 0,1 \pmod{3}$. Now, Since, $2^m  ~ \cancel{\equiv} ~ 0 \pmod{3}$ $\implies$ $k^2 \equiv 1 \pmod{3}$ and $m=2m'$ for $m' \in \mathbb{N}$. Now, take $\pmod{4}$ $\implies$ $3^n \equiv k^2 \equiv 1 \pmod{4}$ $\implies$ $n=2n'$ for $n' \in \mathbb{N}$
$$3^n=(k~+~2^{m'})(k~-~2^{m'}) \implies \begin{cases} k+2^{m'}=3^a \\ k-2^{m'}=3^b \end{cases} \implies 2k=3^a+3^b$$
Obviously, $a \geq b$ $\implies$ $a=b+d$ for some $d$ $\implies$ $2k=3^b(3^d+1)$ $\implies$ $3 \mid 2k$ which is not possible, hence, $b=0$
$$2k=3^n+1$$
Now, $n=2n'$ $\implies $ $$2^m+~3^n=k^2 \implies 2^{m+2}=2^{2m'+2}=(3^n-1)^2 \implies 2^{m'+1}=(3^{n'}+~1)(3^{n'}-~1) $$
Hence, From [url=https://artofproblemsolving.com/community/c776104h1909855_solutions_to_the_diophantine_1]here[/url] $\implies$ $\boxed{(m,n) \equiv (4,2)} \qquad \blacksquare$

[quote=Rushil]Determine all pairs $(m,n)$ of positive integers for which $2^{m} + 3^{n}$ is a perfect square.[/quote]

Clearly checking mod $3,4$ gives $m,n$ even.Let $m=2a,n=2b$Thus we get $$(2^a)^2+(3^b)^2=k^2$$
Since $\text{gcd}(2^a,3^b)=1$ thus this pythagoras triplet is primitive.Hence $\exists p,q$ such that $2pq=2^a$ and $p^2-q^2=3^b$.The first equation gives $p,q$ are powers of and so even.Plugging in second LHS becomes difference of two powers of $2$ so is even whereas RHS is odd.Hence this is only possible when these differences are $2^2-2^0=3^1$.Hence $p=2,q=1,b=1$.Hence $a=2 \implies m=4,n=2$.Thus the only solution is $\boxed{(4,2)}$.$\square$

Let $2^m+3^n=x^2$.

Taking $\pmod 3$, we find $2^m \equiv x^2 \pmod 3$. Hence, we find $2^m \equiv 1 \pmod 3$.

[u]Claim 1:[/u] $m$ must be even.

[i]Proof 1:[/i] We can break into cases based on the parity of $m$.

Case 1: $m$ is odd.
Set $m=2a+1$, so $2^{2a+1} \equiv 1 \pmod 3 \rightarrow 2^{2a}\cdot2\equiv 1 \pmod 3$. However, $2^{2a}\cdot2\equiv 2 \pmod 3$, which is a contradiction, so $m$ cannot be odd.

Case 2: $m$ is even.
Set $m=2a$. We get $4^a \equiv 1 \pmod 3$, which holds.

Now, we can rewrite our equation as $4^a+3^n = x^2$. Taking $\pmod 4$, we get $3^n \equiv 1 \pmod 4$.
Using the same technique as above (breaking into cases), we can find that $n$ must also be even (call it $2b$).

With this information, we rewrite it as $4^a+9^b=x^2 \rightarrow 4^a=(x+3^b)(x-3^b)$. With this, we know that each of $x+3^b$ and $x-3^b$ are a power of $2$. We have the following system of equations, for some $c$ and $d$: $$x+3^b=2^c$$ $$x-3^b=2^d$$ From this, we can find that $d=1$ and $c$ is even, so let $c-1=2e$ for some $e$.

We can find that $3^b=(2^e+1)(2^e-1)$. Once again, using the fact that $2^e+1$ and $2^e-1$ are both powers of $3$, we set up a system (for some $f$ and $g$): $$2^e+1=3^f$$ $$2^e-1=3^g$$ Again, we can figure out that $f=0,g=1$, so $e=1$. Additionally $c=3$, so $b=1$.

We can now write our equation as $4^a+9=x^2$. We have a final system, $$x+2^a=9$$ $$x-2^a=1$$ From this, we find $x=5$, so $a=2$, so the only solution is $(m,n)=(4,2)$.

Solved with [i]RP3.1415[/i]

By inspection, $m,n \geq 2$ and taking mod $3,4$ knowing this gives that we may express $m=2x$ and $n=2y.$ So, $$4^x+9^y=k^2 \qquad (\spadesuit)$$ $$4^x=(k+3^y)(k-3^y).$$ Since both factors on the RHS are powers of two, we must have by EA that $$k-3^y=\gcd (k+3^y,k-3^y)=\gcd(k-3^y,2 \cdot 3^y).$$ Since $2\cdot 3^y$ contains at maximum one power of $2,$ we must have that $k-3^y$ contains at most one power of $2.$ It is easy to see that there cannot be zero powers of two, so $k-3^y=2.$ Thus, $$k+3^y=2^p (\heartsuit) =2+2\cdot 3^y=2(1+3^y)$$ $$\implies 1+3^y=2^{p-1}.$$ Taking this mod $3$ yields that $p-1$ is even, so we can express $p-1=2q$ for some integer $q.$ So, $$1+3^y=2^{2q} \qquad (\clubsuit)$$ $$3^y=(2^q+1)(2^q-1).$$ So, both factors on the RHS are powers of $3$ and so $$2^q-1=\gcd(2^q+1,2^q-1)=\gcd(2^q-1,2)=1.$$ Since $2^q-1$ is odd, we must have the final step above, implying that $q=1,$ which goes with $p=3.$ Substituting back into $\clubsuit$ gives $y=1.$ Substituting these back into $\heartsuit$ gives $k=5.$ Substituting back into $\spadesuit$ gives $x=2,$ yielding the final solution $(m,n)=\boxed{(4,2)}.$

Similar to the above solutions
[hide=solution]Let $2^m+3^n=x^2 .$ By $$\pmod 3\implies m \text{ is even}.$$ So $$3^n=(x-2^k)(x+2^k)\implies x-2^k=1,x+2^k=3^n, \text{ since} \gcd( x-2^k, x+2^k)|2^k\implies \gcd=( x-2^k, x+2^k)=1.$$

[b]Case 1:[/b] When $k$ is odd.
Then we get $$x=1+2^k\implies 3|x. \text{ But} 3| x+2^k.$$ So no solution.

[b]Case 2:[/b] When $k$ is even.
Note that $x\equiv 1 \pmod 4.$ Taking $\pmod 4$ on $$x+2^k=3^n\implies n\text{ is even}.$$
Hence, let $n=2n'.$ We get $$1+2^k=3^n-2^k\implies 3^n-1=2^{k+1}\implies (3^{n'}-1)(3^{n'}+1)=2^{k+1}$$ $$\implies n'=1\implies \boxed{n=2, k=4}$$
[/hide]


Let's say we have the equation $K^2 + t + sK = 0$ with roots $r_1,r_2$ such that $r_1 - 2r_2 = 2$ and $2r_1 - 3r_2 = 5$. Find the value of $s$.

[hide]Solve the system of equations so $r_1 = 4$ and $r_2 = 1$.

$(K - 4)(K - 1) = K^2 + 4 - 5K$ so $s = -5$[/hide]

yeah.
also after solving for r1,and r2,
you can say that the "s" is 4+1
In general the sum of roots of the equation ax2+bx+c is b/a


[quote="Beat"]x-real.
$2x^{2}-3x=1+2x\sqrt{x^{2}-3x}$[/quote]

It is equivalent to:

$x^{2}-3x-2x\sqrt{x^{2}-3x}+x^{2}=1$
$(\sqrt{x^{2}-3x}-x)^{2}=1$

We have $\sqrt{x^{2}-3x}-x=1$ or $\sqrt{x^{2}-3x}-x=-1$.

Squaring first we get $x=-\frac{1}{5}$ and from second we have $x=-1$

After checking we have that $x=-\frac{1}{5}$ is the only solution.


function
analytic geometry
number theory unsolved
number theory

I had some questions I don't know how to do it, could you help me solve it ? Thanks.

1. If 35! = 10333147966386144929*66651337523200000000, what digit does * represent?

2. If $36^{logp}-24^{logp} = p$, what is the right most digit of $3^{p-1}$?

3. How many lattice points lie on the function: $f(x) = \frac{7-x^{3}}{x}$? ( A lattice point is a point with integer coordinates )

More Eastlake Math Competition  :rotfl:

1.  Recall that the sum of the digits of a number divisible by $9$ is itself divisible by $9$.  Then just add.  :)  The answer is $\boxed{6}$ but I got it wrong on the real thing :|  Still the top scorer :)

3.  Divide to get $f(x) = \frac{7}{x} - x^2$.  This is an integer iff $x | 7$ from which we have $x = \pm 1, \pm 7$, so there are $\boxed{4}$ lattice points.

Should be moved,  I think...... ;)

[i]moderator: done ;)[/i]


This is frm the mathcounts forum but anyways kinda an extension

1. Is there a formula that tells how many ways are there to write a positive number n as the sum of consecutive integers (not including n itself as a case)

2. what if n is negative will it change?

3. If there isnt, how many ways are there to write 100 as the sum of consecutive integers (not including n)

4. what if all the consecutive integers have to be positive(not including n) ?

[hide="1&2"]If it is possible, then there are $div(n)-1$ ways. In other words, it is the number of divisors of $n$ minus 1. Think about why the divisor approach "works". For each "product", we can form an addition. Try to figure out a way to do this. If I tell you, you won't have the experience of deriving it.[/hide]
[hide="3"]I believe there are $8$ ways.[/hide]
[hide="4"]What do you mean, not including $n$? If the addends are positive, so is the sum. It is half the number of divisors of $n$. If $n$ is square, it is $floor(div(n))$ [/hide]

EDIT: Thanks, i_like_pie. I think there was something wrong with me when I wrote that. :fool:
EDIT: Thanks, Art of Owna

[quote="vishalarul"][hide="1&2"]If it is possible, then there are $2$ ways. If there aren't there are $0$ ways.

Triangular numbers (& negative triangular numbers) are the numbers with this property.[/hide]
[hide="3"]I believe there are $0$ ways. $100$ is not triangular.[/hide]
[hide="4"]What do you mean, not including $n$? If the addends are positive, so is the sum. Anyway, if it is triangular, then $1$. If not, then $0$.[/hide][/quote]
The integers don't have to include 1 and the number.

For example, $60=10+11+12+13+14$.

I think 100 there are 8 ways because you cant include 100 as a way to write 100 as the sum of consecutive numbers so 3*3-1=8

[quote="vishalarul"][hide="1&2"]If it is possible, then there are $div(n)-1$ ways. In other words, it is the number of divisors of $n$ minus 1. Think about why the divisor approach "works". For each "product", we can form an addition. Try to figure out a way to do this. If I tell you, you won't have the experience of deriving it.[/hide][/quote]

There are several reasons this is incorrect.

[hide="1, 2 (edited)"] The sum of $2k+1$ consecutive integers with middle term $a$ is $a(2k+1)$.  Any unique [i]positive[/i] choice of $k$ produces a unique sum.

The sum of $2k$ consecutive integers with middle terms $a, a+1$ is $k(2a+1)$.  Any unique choice of $a$ (as long as $2a+1$ is the [i]same sign as $n$[/i], so that $k$ is positive) produces a unique sum.

Therefore for any integer $n$ let $n = 2^{e}o$ where $o$ is the greatest odd divisor of $n$.  Let $d(n)$ denote the divisor function.  Then the number of ways to write $n$ as a sum of more than one consecutive integer is given by

$(d(o)-1)+d(o ) = \boxed{ 2d(o)-1 }$

Why?  The first term gives all the ways to choose $k$ such that $n$ can be written as a sum of $2k+1$ consecutive integers, whereas the second term gives all the ways to choose $a$ such that $n$ can be written as a sum of $2k$ consecutive integers with middle terms $a, a+1$.

This method is independent of the sign of $n$. [/hide]

[quote="vishalarul"][hide="3"]I believe there are $8$ ways.[/hide][/quote]

(Edited)  There are $5$.

\begin{eqnarray*}100 &=&-8+...+16 \\ &=& 18+...+22 \\ &=& 9+...+16 \\ &=&-17+...+22 \\ &=&-99+...+100 \\ \end{eqnarray*}

[quote="vishalarul"][hide="4"]What do you mean, not including $n$? If the addends are positive, so is the sum. It is half the number of divisors of $n$. If $n$ is square, it is $floor(div(n))$ [/hide][/quote]

(Edited)  Since there is a bijection between the expressions involving negative numbers and the ones involving only positive numbers (triangular numbers are a special case), I believe there are $d(o)-1$ ways in this case.

[quote="t0rajir0u"][quote="vishalarul"][hide="1&2"]If it is possible, then there are $div(n)-1$ ways. In other words, it is the number of divisors of $n$ minus 1. Think about why the divisor approach "works". For each "product", we can form an addition. Try to figure out a way to do this. If I tell you, you won't have the experience of deriving it.[/hide][/quote]

[b]There are several reasons this is incorrect.  For one thing, this is obviously false if $n$ is a power of two, which cannot be written as a sum of more than one consecutive integer[/b] [i]at all.[/i]

[hide="1, 2"] The sum of $2k+1$ consecutive integers with middle term $a$ is $a(2k+1)$.  Any unique [i]positive[/i] choice of $k$ produces a unique sum.

The sum of $2k$ consecutive integers with middle terms $a, a+1$ is $k(2a+1)$.  Any unique choice of $a$ (positive [i]or negative[/i]) produces a unique sum.

Therefore for any integer $n$ let $n = 2^{e}o$ where $o$ is the greatest odd divisor of $n$.  Let $d(n)$ denote the divisor function.  Then the number of ways to write $n$ as a sum of more than one consecutive integer is given by

$(d(o)-1)+(2 d(o ) = \boxed{ 3d(o)-1 }$

Why?  The first term gives all the ways to choose $k$ such that $n$ can be written as a sum of $2k+1$ consecutive integers, whereas the second term gives all the ways to choose $a$ such that $n$ can be written as a sum of $2k$ consecutive integers with middle terms $a, a+1$.

This method is independent of the sign of $n$. [/hide]

[quote="vishalarul"][hide="3"]I believe there are $8$ ways.[/hide][/quote]

Coincidentally, this is the correct numerical answer, but your method fails for other numbers.

[quote="vishalarul"][hide="4"]What do you mean, not including $n$? If the addends are positive, so is the sum. It is half the number of divisors of $n$. If $n$ is square, it is $floor(div(n))$ [/hide][/quote]

Also incorrect.  This case is more complicated than you give it credit for, and as of yet I have not found a concise description of this function.[/quote]
but coundt something like 2 be written as -1+0+1+2?

Sorry, my solution was a bit off.  It turns out that, yes, setting $a = 0$ in the second case allows the positive powers of two to be written as a sum of consecutive integers in one way.   :oops:

My original point still stands, which was that powers of two have many more divisors than they have ways of being written as sums of consecutive integers.

[quote="t0rajir0u"][quote="vishalarul"][hide="1&2"]If it is possible, then there are $div(n)-1$ ways. In other words, it is the number of divisors of $n$ minus 1. Think about why the divisor approach "works". For each "product", we can form an addition. Try to figure out a way to do this. If I tell you, you won't have the experience of deriving it.[/hide][/quote]

There are several reasons this is incorrect.  For one thing, this is obviously false if $n$ is a power of two, which cannot be written as a sum of more than one consecutive integer [i]at all.[/i]

[hide="1, 2 (edited)"] The sum of $2k+1$ consecutive integers with middle term $a$ is $a(2k+1)$.  Any unique [i]positive[/i] choice of $k$ produces a unique sum.

The sum of $2k$ consecutive integers with middle terms $a, a+1$ is $k(2a+1)$.  Any unique choice of $a$ (as long as $2a+1$ is the [i]same sign as $n$[/i], so that $k$ is positive) produces a unique sum.

Therefore for any integer $n$ let $n = 2^{e}o$ where $o$ is the greatest odd divisor of $n$.  Let $d(n)$ denote the divisor function.  Then the number of ways to write $n$ as a sum of more than one consecutive integer is given by

$(d(o)-1)+d(o ) = \boxed{ 2d(o)-1 }$

Why?  The first term gives all the ways to choose $k$ such that $n$ can be written as a sum of $2k+1$ consecutive integers, whereas the second term gives all the ways to choose $a$ such that $n$ can be written as a sum of $2k$ consecutive integers with middle terms $a, a+1$.

This method is independent of the sign of $n$. [/hide]

[quote="vishalarul"][hide="3"]I believe there are $8$ ways.[/hide][/quote]

(Edited)  There are $5$.

\begin{eqnarray*}100 &=&-8+...+16 \\ &=& 18+...+22 \\ &=& 9+...+16 \\ &=&-17+...+22 \\ &=&-99+...+100 \\ \end{eqnarray*}

[quote="vishalarul"][hide="4"]What do you mean, not including $n$? If the addends are positive, so is the sum. It is half the number of divisors of $n$. If $n$ is square, it is $floor(div(n))$ [/hide][/quote]

(Edited)  Since there is a bijection between the expressions involving negative numbers and the ones involving only positive numbers (triangular numbers are a special case), I believe there are $d(o)-1$ ways in this case.[/quote]

Thank you very much