#1 We choose 2 numbers in between 1 and 100, inclusive, such that their difference is 7 and their product is a multiple of 5. How many ways can we do this?

#2 How many 7-digit numbers are multiples of 388 and end in 388?

#3 Bob chose four distinct digits from the set $ \{1,2,3,4,5,6,7,8,9\}$.With them he formed all the possible four digit number and added all the numbers formed. The result was $ 193,314$. Find the four digits Bob chose.

[hide="For #1:"]
This should work...

Let the two numbers be $ a$ and $ b$ such that $ a > b$.

We know that $ a \minus{} b \equal{} 7$ and $ ab\equiv 0\pmod{5}$, since only one of the two numbers can be a multiple of $ 5$ we can start with cases, one in which $ a$ is a multiple of $ 5$ and the other where $ b$ is a multiple of $ 5$.

Case 1: ($ a$ is a multiple of $ 5$)

 The lowest value for $ a$ that satisfies the conditions is $ 10 \equal{} 5(2)$ where $ b$ is $ 3$.

 The highest value for $ a$ that satisfies the conditions is $ 100 \equal{} 5(20)$ where $ b$ is $ 93$.

So there are a total of $ 20 \minus{} 2 \plus{} 1 \equal{} 19$ cases that work.

Case 2: ($ b$ is a multiple of $ 5$)

 The lowest value for $ b$ that satisfies the conditions is $ 5$ where $ a$ is $ 12$.

 The highest value for $ b$ that satisfies the conditions is $ 90 \equal{} 5(18)$ where $ a$ is $ 97$.

So there are a total of $ 18$ cases that work.

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

In total there are $ 19 \plus{} 18 \equal{} \boxed{37}$ cases that work.
[/hide]

Hey tony are you going to the Olimpiada de Mayo this saturday??

UNCOOL, U POSTED THAT JUST 1 MIN BEFORE I FINISHED TYPING  :furious:
For #2:
[hide]$ lcd(1000, 388) \equal{} 97000$
$ 10000/97 \equal{} 103.0.......$
thus there are 10 such numbers
EDIT: the first digit of the number can be 0. so it should be 9000/97=92.78.....so 92 such numbers[/hide]

what [i]what [/i] is?
anyways, i think tonypr's solution for #1 was wrong...
For #1:
[hide]one of the numbers must be a multiple of 5
first choose the number that must be a multiple of 5.  there are 20 such numbers
for 5, 95, and 100, there is 1 choice for the 2nd number.
for the 17 numbers remaining, there are 2 choices each.
thus there are $ 3 \plus{} (2*17) \equal{} 37$ ways total[/hide]

[hide="#3"]$ 6666(a\plus{}b\plus{}c\plus{}d)\equal{}193314\iff a\plus{}b\plus{}c\plus{}d\equal{}29$

Maximum possible sum of distinct digits is $ 6\plus{}7\plus{}8\plus{}9\equal{}30$, hence the only possibility to obtain $ 29$ is $ (5,7,8,9)$[/hide]

[quote="modularmarc101"]Could you explain what that is?[/quote]

He was looking for the lowest multiple of $ 388$ that ends in 3 zeros so that when added to a number that works like $ 3880388$ it will still be a multiple of $ 388$ and end in $ 388$

So any number in the form of $ 3880388 + 97000k$ ends in $ 388$ and is a multiple of $ 388$.

We want to find $ 1000000 < 3880388 + 97000k < 10000000$

Since we are only dealing with the thousands we don't need to worry about the $ 388$:

\begin{eqnarray*} 1000 & < & 3880 + 97k < 100000 \\
1000 & < & 97(40 + k) < 10000 \\
10 & < & 40 + k < 103 \\
- 30 & < & k < 63 \end{eqnarray*}

From this we see that there are $ \boxed{92}$ solutions.

The lowest of which is $ 3880388 - 29\cdot 97000 = 3880388 - 2813000 = 1067388$
The highest is $ 3880388 + 62\cdot 97 = 3880388 + 6014000 = 9894388$

To Modularmarc101, I'll try, it's that I have a competition that same day but it's in the univerisity in mayaguez so it shouldn't be that hard to be there.

To abcak, the solution may look different but our solutions are in a way the same, just that I counted the cases for the options separately while you did them together.

[quote="abcak"]nice one for #3, Farenhajt.

Is it kinda like adding it long if you know what i mean?[/quote]

Well, if Bob could do it, so can we ;)

(I'm just imagining various "Bobs" from problems like this, who set up on long addition of $ 24$ four-digit numbers just for the... fun of it :P)


Find all continuous functions $ f: \mathbb{R}\to\mathbb{R}$ such that $ f(x)\minus{}f(y)$ is rational for all real numbers such that $ x\minus{}y$ is rational.

[hide="Solution"]$ f(x) \equal{} qx \plus{} c$, where $ q\in \mathbb{Q}, \ c\in \mathbb{R}$[/hide]

[hide="Idea"]Intermediate Value Theorem and density of set $ \left\{\frac {k}{2^{m}}, \ k\in \mathbb{Z}, m \in \mathbb{N}\right\}$ in $ \mathbb{R}$.[/hide]

[hide="Proof"]For each $ n\in \mathbb{N}$ define $ g_{n}(x) \equal{} f\left(x \plus{} \tfrac{1}{2^{n}}\right) \minus{} f(x)$. Then $ g_{n}$ is continous and $ \forall x\in \mathbb{R} \quad g_{n}(x) \in \mathbb{Q}$, so from Intermediate Value Theorem, and from density of irrational numbers in $ \mathbb{R}$, we obtain that $ g_{n}$ must be constant function. Let $ g_{n}(x) \equal{} s_{n}$ and $ S \equal{} \left\{\frac {k}{2^{m}}, \ k \in \mathbb{Z}, m \in \mathbb{N}\right\}$. Since $ f\left(\tfrac{k}{2^{m}}\right) \equal{} ks_{m} \plus{} f(0)$ and  $ s_{n \plus{} m} \equal{} \frac {s_{m}}{2^{n}}$, we have $ \forall t\in S \quad f(t) \equal{} ts_{0} \plus{} f(0)$.  Now density of $ S$ in $ \mathbb{R}$ implies that equality holds also for all $ t \in \mathbb{R}$, because $ f$ is continous, so the only solutions are $ f(x) \equal{} qx \plus{} c$, where $ q\in \mathbb{Q}, \ c\in \mathbb{R}$. [/hide]


What is the smallest possible real value of $ x^2 \plus{} 8x$?

hello, the smallest possible real value is $ \minus{}16$, it will be attained for $ x\equal{}\minus{}4$,because
$ x^2\plus{}8x\equal{}(x\minus{}4)^2\minus{}16$.
Sonnhard.

[hide="Alternately"]The minimum of the expression $ ax^2 \plus{} bx \plus{} c$, where $ a$, $ b$, and $ c$ are real numbers and $ a\ne 0$, occurs when $ x \equal{} \minus{} \frac {b}{2a}$.  So the minimum of this expression is when $ x \equal{} \minus{} \frac {8}{2(1)} \equal{} \minus{} 4$.  Thus, the minimum value of $ x^2 \plus{} 8x$ is $ ( \minus{} 4)^2 \plus{} 8( \minus{} 4) \equal{} 16 \minus{} 32 \equal{} \boxed{ \minus{} 16}$.[/hide]


It really does seem like the result should be true, though, doesn't it?  But as the pentagon shows, we can have $ \omega(G) \cdot \alpha(G) < |G|$.  One fix is $ \chi(G) \cdot \alpha(G) \geq |G|$, and this fact is obvious.  The last time I thought about this, I think that I decided that graph capacities could be a lot easier to understand if $ \omega \cdot \alpha \geq |G|$.  Alas.

[quote="madadpoor"]The order of the simple graph G is P. K is the max number such that G has K nonadjacent nodes (pairwise). B is the min number such that  the nodes of G can be divded in B group(s) such that all nodes in each group is adjacent (pairwise). can we assert that K=B? prove your assertion. ( we can prove that K<= B )[/quote]

[hide="here is my solution"]The result is false and there is a simple counterexample with $ P\equal{}5$.  Let $ G$ have the $ 5$ vertices $ V_{1}$, $ V_{2}$, $ V_{3}$, $ V_{4}$, and $ V_{5}$.  Let $ G$ have exactly $ 5$ edges:  $ V_{1}V_{2}$, $ V_{2}V_{3}$, $ V_{3}V_{4}$, $ V_{4}V_{5}$, and $ V_{5}V_{1}$.  It is fairly easy to see that for this graph $ G$ we have $ K\equal{}2$ but $ B\equal{}3$.[/hide]


The polynomial $p(x) = x^3 + ax^2 + bx + c$ has three real positive roots, whereas at least two of them are distinct. Prove that we have:

\[a^2 + b^2 + 18c > 0.\]

$P(x)=(x-p)(x-q)(x-r)$ with $p\not=q$, $p,q,r \in \mathbb{R}^+_0$

c=-pqr
b=pq+qr+rp
a=-p-q-r

So we need to prove $(p+q+r)^2 + (pq+qr+rp)^2 - 18pqr > 0$, which follows directly from AM-GM. (strict inequality since at least 2 are different)


Like for example 5 to the power of 2 and 2 5ths.
And how do I do, for example, 5 to the power of .65. How do I do those?
Thanks.

In the case of the first problem, you have $ 5^\frac{12}{5}\equal{}\sqrt[5]{5^{12}}\equal{}25\sqrt[5]{25}$.
Expressions like $ a^\frac{b}{c}$ can be written as $ \sqrt[c]{a^b}$.

In the second problem you could convert $ .65$ to a fraction ($ .65\equal{}\frac{65}{100}\equal{}\frac{13}{20}$) so that you have $ 5^\frac{13}{20}\equal{}\sqrt[20]{5^{13}}$ which can't be simplified any further.


Three boys agree to divide a bag of marbles in the following manner.  The first boy takes one more than half the marbles.  The second takes a third of the number remaining.  The third boy finds that he is left with twice as many marbles as the second boy.  The original number of marbles:

$ \textbf{(A)}\ \text{is none of the following} \qquad
\textbf{(B)}\ \text{cannot be determined from the given data}\\
\textbf{(C)}\ \text{is 20 or 26} \qquad
\textbf{(D)}\ \text{is 14 or 32} \qquad
\textbf{(E)}\ \text{is 8 or 38}$

[quote="KMST"][quote="Nandubetax"]Its $ D$ just by applying the numbers to the Problem.[/quote]Why couldn't the original number of marbles be $ 8$?
The first boy would take $ 4 \plus{} 1 \equal{} 5$, leaving $ 3$ marbles, $ 1$ for the second boy and $ 2$ for the third one.

Why couldn't the original number of marbles be $ 20$?
The first boy would take $ 10 \plus{} 1 \equal{} 11$, leaving $ 9$ marbles, $ 3$ for the second boy and $ 6$ for the third one.

Could it be that there were $ 68$ marbles? Or $ 602$?

I'm sure [hide="that"] the number remaining after the first boy took his share was a multiple of $ 3$.[/hide]

[hide="and that "]the original number was even.[/hide][/quote]This raises a good question: can we actually determine the exact number of marbles?
[hide="Solution"]Let $ M$ be the original number of marbles. Then, the first boy gets $ \frac{M}2\plus{}1\equal{}\frac{M\plus{}2}2$ marbles and there are $ \frac{M}2\minus{}1\equal{}\frac{M\minus{}2}2$ marbles left for the two other boys. The second boy gets $ \frac13\left(\frac{M\minus{}2}2\right)\equal{}\frac{M\minus{}2}6$ and the third boy gets the remaining marbles, $ \frac23\left(\frac{M\minus{}2}2\right)$. So, the expression $ \frac{M\minus{}2}6$ must be an integer and thus $ M$ is of the form $ 6k\plus{}2$. There are no more additional information from the problem that would help you determine how many marbles there were in the bag. Thus, the answer is $ \bf (B)$.

Note that $ \bf (C), (D), (E)$ are all numbers of the form $ 6k\plus{}2$.[/hide]


Hi,
   I want to draw a consecutive polygon of similar side inside a given polygon (of any number of sides) at a specified distance (varies).
When the space inside the polygon is not enough to draw another polygon, I should stop drawing polygon.

Can anyone tell me how to detect whether the a polygon at specified distance can be drawn inside the polygon or not.

Refer attached figure for more idea.

[quote="pleurestique"]you didn't say that your polygon must be regular. but if your polygon isn't regular, drawing lines parallel to its sides all at the same distance doesn't necessarily produce a polygon similar to it. so we'll assume your polygon is regular. then:

find the length of the apothem of the regular polygon and divide by your given distance, then take the floor - that's how many polygons you can have before you have to stop.[/quote]

Sorry for not providing sufficient information

My Polygon may be either regular or irregular. Refer Fig Sample 1: In that polygon drawn in yellow should not be drawn. (it is different from the given shape,in that case i should stop draw polygon)

Refer Sample 2: I am drawing parallel linse for the given polygon and find the intersection point and draw polygon with all the intersection point as vertex of polygon.

Kindly do help to solve the problem

if you want help then you'd better make sure you've expressed your problem clearly.


calculus
integration
inequalities
calculus computations

Let $ a,\ b$ be constant numers such that $ a>b>0$. Find the volume of the set $ R$ in Eucledian space which satisfies the following system of inequalities:

$ x\geq 0,\ y\geq 0,\ z\geq 0,\ max(x,\ y,\ z)\leq a,\ x\plus{}y\plus{}z\minus{}min(x,\ y,\ z)\leq a\plus{}b$.


How can I get any previous Mathcounts competitions besides the 2003-2004 competitions?


The characteristic function to an interval $ [a,b] \subseteq [\minus{} \pi, \pi]$ is an even or an odd function?

If the subset (sub-interval) which you choose $ [a,b]$ from $ [\minus{}\pi ,\pi]$ is symmetric about $ 0$ then the the indicator function is an even function . :roll:


linear algebra
matrix
linear algebra unsolved

Let $ a,b,c$ are positive real numbers and  $ X,Y,Z \in M_n{\mathbb{(C)}}$ are Hermitian matrices and positive define such that :
$ X^nY^n \equal{} Y^nX^n$,
$ X \plus{} Y \plus{} Z \equal{} (a \plus{} b \plus{} c)I_n$,
$ a^2X \plus{} b^2Y \plus{} c^2Z \plus{} (abc)I_n \equal{} 4XYZ$
Prove that : $ X \equal{} \frac {b \plus{} c}{2}I_n$,$ Y \equal{} \frac {c \plus{} a}{2}I_n$,$ Z \equal{} \frac {a \plus{} b}{2}I_n$
 $ I_n$ is $ n$x$ n$ identity matrix . $ n \in \mathbb{N^{*}}$

[quote="QuyBac"]$ X^nY^n \equal{} Y^nX^n$,
[/quote]Since $ X,Y>0$, the condition implies that $ X,Y$ commute.
[quote]$ X \plus{} Y \plus{} Z \equal{} (a \plus{} b \plus{} c)I_n$,[/quote]... such that $ Z$ commutes with $ X,Y$. So assume that everything is diagonal and solve the exercise simply over the reals.


I don't understand the definitions of the sets :( . First I thought a_1,...,a_k are the elements of A but what are the elements of B then. Please explain it.

$A$ and $B$ are sets of sequences with those properties.

All right. Sorry for the idiot question but I'm not in best form today.

Misha

Nice and easy:
Take $k$ odd. By considering the substitutions $b_{i}=a_{i}+i$, since $k|1+2+...+k$ for odd $k$, it's clear there is a bijection between $A$ and
$A'=\left\{ 1\leq b_{1} < ... < b_{k} \leq 2k : k | b_{1}+...+b_{k} \right\}$ and a bijection between $B$ and $B'=\left\{ 1\leq b_{1} < ... < b_{k} = 2k : k | b_{1}+...+b_{k} \right\}$, so $|A|=|A'|$ and $|B|=|B'|$.
If $F$ is the family of all subsets of $[2k]$ with $k$ elements whose sum of elements is divisible with $k$, then $M \in F \Rightarrow [2k]-M \in F$ so the number of $M \in F$ s.t. $2k \in M$ is equal to the number of $M \in F$ s.t. $2k \notin M$. In other words, $|B'|=|A'-B'| \Leftrightarrow |A'|=2|B'|$ so $|A|=2|B|$. For even $k$ it's the same thing.


Let k be a positive integer such that

$((k!)!)!\leq{((2006k^{2})!)!}$

Find, with proof, the number of possible values of k.

Couldnt you just ignore the last two factorials, and then solve

$k!\leq{2006k^2}$   ?

$(k-1)!\leq{2006k}$

I'm not even sure if this last line does anything.

Yeah, your on the right track. Now just explain how you know that it is valid to ignore the outside factorials.

So basically we just have to show that $a! \leq b!$ if and only if $a \leq b$

so...

Since $a! = a(a-1)!$ and we're given that $a \geq 1$, we know that

$a! \geq (a-1)!$, so, (I don't know how to put this into a more rigorous proof), the result follows.

Does that cover everything or are there holes?

well your argument can constitute part of the "if" portion of the proof (where you assume $a\leq{b}$ and prove the other assertion). You should use the same idea, but start by showing that $(a+1)!\geq{a!}$. Then, say something about "similarly" and write $(a+2)!\geq{(a+1)!}$. Then use this pattern to inductively write a string of inequalities that relate b to a $b!\geq{(b-1)!}\geq{\ldots}\geq{(a+1)!}\geq{a!}$

Think about the "only if" portion of the proof (where we assume $a!\leq{b!}$ and try to prove the other assertion)


There is a reasonable way to do this... but I'm leaving it for an MC student.

What grade are you in? Not 6, 7, or 8th is all I know.

[quote="236factorial"]Express the following infinetly repeating hexadecimal (base 16) into a decimal fraction: 0.1999DECDECDEC......

I think that there is a special way to do this, but I don't know it.
Is this too hard or too easy for Mathcounts?[/quote]

Simplify it into base 10... and you get 0.2000453453453.... the rest is self explanatory, I think...

[hide] 0.2=1/5.... subtract 1/5 and you get 0.0000453453453...=$\frac{151}{3330000}$, add 1/5 to that and you get: $\frac{666151}{3330000}$...[/hide]

Edit: woops, I messed up...

I think simplifying repeating decimals into base 10 is too hard for Mathcounts problems.

[quote="236factorial"]Is this too hard or too easy for Mathcounts?[/quote]
Too hard.  This is also a messy, ugly problem that doesn't come out nicely.  [What is the source of this problem?]

However, the concepts are within the scope of MATHCOUNTS.

Let's split this repeating hexadecimal into two parts, then follow the techniques you National competitors (should) already know for simplifying repeating decimals:

$0.1999\hbox{DECDEC}\ldots_{16}=0.1999999\ldots_{16}+0.0000453453\ldots_{16}$

Now, $0.1\overline{9}_{16}=\frac{1_{16}}{10_{16}}+\frac{9_{16}}{\hbox{F0}_{16}}=\frac{24_{10}}{240_{10}}=\frac{1}{10}$

$0.0000\overline{\hbox{453}}_{16}=\frac{\hbox{453}_{16}}{\hbox{FFF0000}_{16}}$

Convert to decimal.  (I did the first one for you.)  Then add the fractions.

This problem comes from an ABACUS Math International Challenge for grades 5 and 6.
I don't think that many 5th and 6th graders can understand base changes very well.

5th and 6th graders? That seems unlikely, despite being international, unless this was the hardest problem.

I feel sorry for those poor 5th and 6th graders... that's cruel and unusual punishment! *sigh* forced to solve harder-than-mathcounts math problems at such a young age *sigh*  :D

Yeah, their problems are harder than many AIME problems, from what Valentin Vornicu has posted.

You just noticed?

As an additional note, you should see what the kids in Russia do!

There's a reason the Eastern European countries started the IMO! Their teams always make many of the top 5.

[quote="tarquin"]You should see what the kids in Romania do ...[/quote]
Would you believe a 6th grader from Romania is credited on the ABACUS Web Site as the author of this problem?

Pssh. In Korea, I'm only a little advanced. And I'm hopefully taking multivariable calc next year.

And how old are you? I know a guy who took Statistics in 8th grade, and now, in 11th grade, in our high school that has a highly gifted magnet which offers up through Discrete Math and Linear Algebra, he doesn't have any math to do.

I skipped a grade and I'm still 2 years ahead in math.  Next year I might skip pre-cal (which I kinda did over the summer last year) and go to calc.

I'm in eighth grade and taking Math Analysis (precal).


number theory
relatively prime
number theory proposed

Prove that if $N$ is relatively prime with 10 then $N^{1}01$ ends with three digits, which are also the last three digits of N.

davron l.


ratio
geometry
combinatorial geometry
geometry proposed

Consider five points in the plane, no three collinear. The convex hull of this points has area $S$. Prove that there exist three points of them that form a triangle with area at most $\frac{5-\sqrt 5}{10}S$

can anybody post a solution for this one

This one was proposed by Zhen Wang and Ji Chen, see: [url=http://menet.math.ecnu.edu.cn/teaching/]Mathematics Teaching (Shanghai)[/url], 1988, No.2, p.40, Problem 155. The solution see 1988, No.3, p.41.

In March, 1989, Paul Erdos visited Colorado Springs. He offered $\$10$ for the solution during his talk at the University of Colorado, and as a coauthor, Alexander Soifer immediately offered to match it. See:

Alexander Soifer, [url=http://www.mathpropress.com/books/problemBooks/Triangle/]How Does One Cut a Triangle?[/url], Center for Excellence in Mathematics, Colorado Springs: 1990, p. 113.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Hello,
Troubling over a series of rather simple problems.

(i)  Example of a group $G$ with infinite order such that every proper subgroup is of infinite index.

(ii)  Example of a group $G$ with infinite order such that every nontrvial subgroup is of finite index.

(iii)  Example of a group $G$ such that there exist nontrival subgroups of infinite index and proper subgroups of finite index.

My thoughts on the above are as follows:

(i)  $\mathbb{Q}$  (Unsure as if this is correct as since $|\mathbb Q| = |\mathbb Z|$)

(ii) $\mathbb{Z}$  (I believe this is true since every group $2\mathbb{Z}, 3\mathbb{Z}, \cdots$ is isomorphic to $\mathbb Z_2,\mathbb Z_3, \cdots,$ respectively.)

(iii) $\mathbb{Z}_k\times\mathbb{Z}, k\in\mathbb N\backslash\left\{0\right\}$  (Consider the subgroup $(0,\mathbb Z)$ is infinite and $(\mathbb Z_k, 0)$ is finite.)

Am I on the right track?  If not, could you please advise me on my poor logic!

Thanks,
JNG

nZ is not isomorphic to Zn, but Z/nZ is isomorphic to Zn.
(f: Z onto Zn defined by f(a) = â is a surjective homomorphism of groups with kernel nZ)

Sorry, that is quite true.  Poor notation on my part.

-JNG


[quote="Interval"]What is the basic difference between geometry and trigonometry at the high school level?

Thanks![/quote]
Well, geometry includes triangles, circles and stuff. But mostly triangles. Trigonometry is the study of angles. You can learn more about trig from this http://mathworld.wolfram.com/Trigonometry.html

[quote="Interval"]What is the basic difference between geometry and trigonometry at the high school level?

Thanks![/quote]

geometry is dealing with lines and circles and basic proofs.  Geometry also includes [size=75]a little[/size] bit of trig like 30-60-90 triangles and such.  you get into a lot of depth in trig in high school

I read that geometry means TO MEASURE THE EARTH.

Is this true?

Also, trig is basically the study of angles and triangle sides in general, right?


Prove that nineth power of a number is of the form 19m or 19m+-1 where m is a non negative integer

[hide]Clearly, any numbers of the form $ (19k)^9$ (which is $ 0 \pmod {19}$) is equivalent to $ 0 \pmod {19}$, so then we look at the factorings of $ x^9 \minus{} 1$ and $ x^9 \plus{} 1$:

$ x^9 \minus{} 1 \equal{} (x \minus{} 1)(x^2 \plus{} x \plus{} 1)(x^6 \plus{} x^3 \plus{} 1)$
$ x^9 \plus{} 1 \equal{} (x \plus{} 1)(x^2 \minus{} x \plus{} 1)(x^6 \minus{} x^3 \plus{} 1)$

So clearly, any number of the form $ 19k \pm 1$ (which is $ 1$ or $ 18 \pmod{19}$) follows this rule because then the $ (x \pm 1)$ will be divisible by 19.

Also, anything in the form $ 19k\pm \{7,8,11,12\}$ (which are $ 7,8,11,12 \pmod{19}$) will follow this rule because then the $ x^2\pm x \plus{} 1$ term will be divisible by 19.

And finally, those numbers for which $ x^3\equiv 7,8,11,12 \pmod{19}$ also follow the rule because the last factor is just equal to the second factor with $ x^3$ instead of $ x$. Those are $ \{2,3,4,5,6,9,10,13,14,15,16,17\}$, which complete all the remainders mod 19.

So in conclusion, the unavailability of Fermat's Little Theorem compels us to write a convoluted solution in place of what would've been a short one.[/hide]

Using Fermat's Little:

[hide]If the number is a multiple of 19, then its nineth power is a multiple of 19. Otherwise, by Fermat's Little, we have x^18-1 = (x^9-1)(x^9+1) = 0, which implies that x^9 is plus or minus 1.[/hide]

Or we could brutally calculate all values of 1,2,3...,19 all to the power of 9 and then find the residue...
It wouldn't take really long


$Question 1$
There is a $100 \times 10$ rectangular grid. All the $1000$ points are coloured either red, green or blue.

Prove that there exists a rectangle in the plane such that all four of its vert the same colour. Or, prove that it is not a necessary condition

$Question 2$
A frog performs a random walk on the integers. At each stage it jumps either one integer forwards or one integer back with equal probability. It starts its random walk from zero. Find the probability that it reaches $+10$ before it reaches $-20$.

N.2:
see here http://www.usamts.org/Forms/ExpectedValue.pdf

actually i'm not sure that helps

i'm not sure this is correct solution, here is my idea:
p(-20)=p(-10)^2 (easily seen)
So we find prob it reaches p(-10) resets, then reaches it again before p(10)
we can reduce problem to prob. it reaches -2 before 1
P=1\2 (1/2+1/2P) P=1/3


[quote="Ravi B"]On the MathLinks skin, I noticed that the Resources button brings you to the same location as clicking on Links.  (I was expecting the button to send me to the AoPS Resources page.)  Is that the way it is supposed to be?[/quote]I didn't have time to complete everything on the skin, but that's where it's supposed to lead. The MathLinks skin is desgined to be more helpfull to one advanced user looking for advanced problems, so the links are adaptive.


Let $ x,y,z$ be non-negative real numbers satisfy $ xy \plus{} yz \plus{} zx \equal{} 1$.Prove that
$ \frac {1}{{1 \plus{} yz \plus{} {x^2}}} \plus{} \frac {1}{{1 \plus{} zx \plus{} {y^2}}} \plus{} \frac {1}{{1 \plus{} xy \plus{} {z^2}}} \le \frac {9}{5}$

\[ \Longleftrightarrow \sum{\left(\frac {3}{5} \minus{} \frac {1}{1 \plus{} yz \plus{} x^2}\right)}\geq 0\]

\[ \Longleftrightarrow\sum{\frac {(x \minus{} y)(x \minus{} z) \plus{} x(x \minus{} y) \plus{} x(x \minus{} z)}{1 \plus{} yz \plus{} x^2}}\geq 0\]

\[ \Longleftrightarrow\sum{\frac {1}{1 \plus{} yz \plus{} x^2}(x \minus{} y)(x \minus{} z)} \plus{} \sum{(x \minus{} y)\left(\frac {x}{1 \plus{} yz \plus{} x^2} \minus{} \frac {y}{1 \plus{} zx \plus{} y^2}\right)}\geq 0\]

\[ \Longleftrightarrow\sum{\frac {1}{1 \plus{} yz \plus{} x^2}(x \minus{} y)(x \minus{} z)} \plus{} \sum{\frac {2z(x \plus{} y)}{(1 \plus{} yz \plus{} x^2)(1 \plus{} zx \plus{} y^2)}(x \minus{} y)^2}\geq 0\]
But
\[ \sum{\frac {1}{1 \plus{} yz \plus{} x^2}(x \minus{} y)(x \minus{} z)}\geq 0\]
it's true by $ V \minus{} V Schur$,and the second part is obvious positive.
I hope there isn't any mistakes.....

I am sorry I made some typo,now I correct them. :blush:

[quote="Tourish"]
\[ \Longleftrightarrow \sum{\left(\frac {3}{5} \minus{} \frac {1}{1 \plus{} yz \plus{} x^2}\right)}\geq 0\]

\[ \Longleftrightarrow\sum{\frac {(x \minus{} y)(x \minus{} z) \plus{} x(x \minus{} y) \plus{} x(x \minus{} z)}{1 \plus{} yz \plus{} x^2}}\geq 0\]

\[ \Longleftrightarrow\sum{\frac {1}{1 \plus{} yz \plus{} x^2}(x \minus{} y)(y \minus{} z)} \plus{} \sum{(x \minus{} y)\left(\frac {x}{1 \plus{} yz \plus{} x^2} \minus{} \frac {y}{1 \plus{} zx \plus{} y^2}\right)}\geq 0\]

\[ \Longleftrightarrow\sum{\frac {1}{1 \plus{} yz \plus{} x^2}(x \minus{} y)(y \minus{} z)} \plus{} \sum{\frac {2z(x \plus{} y)}{(1 \plus{} yz \plus{} x^2)(1 \plus{} zx \plus{} y^2)}(x \minus{} y)^2}\geq 0\]
But
\[ \sum{\frac {1}{1 \plus{} yz \plus{} x^2}(x \minus{} y)(y \minus{} z)}\geq 0\]
it's true by $ V \minus{} V Schur$,and the second part is obvious positive.
I hope there isn't any mistakes.....[/quote]
Your proof is very nice,Tourish. I think this is very natural way to prove it.
[hide="P/s"]I will wait for a proof by Cauchy-Schwarz inequality.[/hide]

[quote="Tourish"]
\[ \Longleftrightarrow\sum{\frac {(x \minus{} y)(x \minus{} z) \plus{} x(x \minus{} y) \plus{} x(x \minus{} z)}{1 \plus{} yz \plus{} x^2}}\geq 0\]
[/quote]

nice step!

where did you use the caushy shwarz ineq.?

Can someone give me another proof by Cauchy-Schwarz?
And the double inequality which I've posted in this link:

http://www.mathlinks.ro/viewtopic.php?t=317946


Hi please could someone help me into solving this equations , please I will be very happy to know the process of solving ,, :(  :(  :(

Thank you very much!!!!

6x + 7  - x -  11   = 0
     2              2

and the second one:

x + 3  = 1 + x + 1
 x -2            2(x-3)

For the first one you add all like terms. So 6x+x=7x. And 7-11=-4.So we have 7x-4=0.Add the four on the other side so we have 7x=4. Then divide both sides by seven to isolate the variable and x=4/7.

The second one we do the same thing. So x+3=x+2. Subtract the x from both sides and we get 3=2, which can't be correct so no solutions or empty set. I may have done this wrong though.
Whats about the x-2 2(x-3)?

Hmm, I believe the two questions are two system of equations. Perhaps you could fix any typos?

Thank you very much my friends,,, :blush:   Iam sorry but I cant find a way to copy paste from m.word please find the attached pdf file with the problematic equations on this link http://www.mathlinks.ro/Forum/viewtopic.php?t=121429

thank you very much!!!!

I'm gonna go out on a limb and assume that your last expression "x -2 2(x-3)" was supposed to be the equation $x-2=2(x-3)$ even if my assumption is incorrect, this will be a sample solution to a two-step algebraic equation from me to you. To solve for $x$ you would first distribute the $2$ on the right side of the equation which would give $x-2=2x-3*2$ then isolate your variable (x in this case) to one side of the equation and have your constant(s) on the other side. $x-2-x=2x-6-x \rightarrow-2=x-6 \rightarrow x-6+6=-2+6 \rightarrow x=4$


[url=http://mathworld.wolfram.com]MathWorld[/url] has recieved an email from [url=http://www.mersenne.org]GIMPS[/url] (Great Internet Mersenne Prime Search) reporting that a 42nd Mersenne Prime has been found!  Click [url=http://mathworld.wolfram.com/news/2005-02-18/mersenne/]here[/url] for the article and more info on Mersenne Primes.

(Note: All 3 links will take you outside of AoPS)

Cool! Thanks for sharing. By the way, it was great to meet you and Sly Si in person at the MATHCOUNTS chapter meet.

An update:

The GIMPS website has now officially reported that they have found a potential new mersenne prime.  They are in the process of verifying that it is indeed prime, and they will hopefully have announced the actual number and the name of the person that found it by the weekend.

A few notes about the candidate (assuming it is indeed prime): it will be the largest known prime number, but it is not long enough to win the US$100,000 prize for finding the first 10,000,000-digit prime number.  Also, this will be the 8th mersenne prime found by GIMPS, all of which were the largest known primes in the world upon their discovery.

Another update:

GIMPS has announced that its new mersenne prime has been independently verified to be prime; the number is [size=150]2^25,964,951-1[/size].

yeah... I was at the Rice Math contest and the guest lecutrer mentioned that 42nd mersenne prime was found on 18th. did they say how many digits?? How many more until they can give out the 100grand for the 100 billion or 100 million digit mersenne prime?

Seems I have to put my GIMPS program back in the background, just in case :D :D

and now there seems to be another one

on december 16 (or 15?), 2005, a new mersenne prime was found.

it's still being verified

see http://www.mersenne.org

Nowadays, which is the bigger prime found????


Hey guys, I plan on double-majoring in chemistry and economics. Which colleges would you suggest I apply to, not considering the Ivy leagues? From what I've heard, Duke, U of Chicago, Northwestern are good choices? Any others?

U Chicago is definitely a good choice.  I go to Hopkins and I've heard that the econ department is one of the best in the nation, as well as the chemistry department.

I am wondering why don't you want to apply to Ivy League? I have heard they are very good :)  :maybe:

The Ivy League is nothing more than an athletic conference.  Yes, the schools are quite good but there are plenty of non-Ivy schools that are just as good if not better.

yeah i know that... im wondering with which sport a person will be accepted easily in Ivy league universities? soccer, basketball, tennis,swimming,etc

In most cases, you have to be recruited by the coach if you want to actually get on a top team.  Some teams are walk-ons, others are not.  And if you are not being recruited, it doesn't really matter which sport you play-rather what you get out of it.  And no one gets accepted easily.


algebra
polynomial
limit
algebra unsolved

Let $1,\dots,999$ be the roots of $P(x)$,  $deg P(x)=999$. Does there exist two polynomials $S(x), \deg S(x)=37$, $Q(x), \deg Q(x)=27$ such that $P(x)=Q(S(x))$?

Let $1<\alpha_{1}<\cdots<\alpha_{r}<999$ be the real roots of $S'$ between $1$ and $999$ ($r\leq \deg S'=36$), such that $S$ is strictly increasing or decreasing on each of the intervals $[1,\alpha_{1}],(\alpha_{1},\alpha_{2}],\ldots,(\alpha_{r-1},\alpha_{r}],(\alpha_{r},999]$. This gives $r+1\leq 37$ strictly increasing or decreasing sequences (possibly empty):

$S(1),S(2),\ldots,S([\alpha_{1}])$
$S([\alpha_{1}]+1),S([\alpha_{1}]+2),\ldots,S([\alpha_{2}])$
$\vdots$
$S([\alpha_{r}]+1),S([\alpha_{r}]+2),\ldots,S(999)$

Now $Q(S(k))=P(k)=0$ for all $k=1,\ldots,999$, and there are at most $\deg Q=27$ different roots of $Q$. Therefore $S$ can take at most $27$ different values on $\{1,2,\ldots,999\}$. And so each of the $r+1\leq 37$ sequences above must have length $\leq 27$. But these sequences cover all $999=27\cdot 37$ numbers $1,2,\ldots,999$. So we must have $r+1=37$ sequences, each of length exactly $27$, and which all run through all $27$ different roots of $Q$. In particular $[\alpha_{1}]=27$ and $[\alpha_{2}]=54$. Now if $S$ were both increasing or decreasing on $[1,\alpha_{1}]$ and $(\alpha_{1},\alpha_{2}]$, then we would have $54$ different values $S(1),S(2),\ldots,S(54)$. Contradiction!
Therefore the sequences $S(1),S(2),\ldots,S(27)$ and $S(28),S(29),\ldots,S(54)$ run through the roots of $Q$ in converse order. And this gives $S(k)=S(55-k)$ for all $k=1,\ldots,27$.
But now the polynomial $S(X)-S(55-X)$ has degree $\leq 37$ and $54$ different roots $1,2,\ldots,54$. And then $S(X)=S(55-X)$ and $\lim_{X\rightarrow+\infty}S(X)=\lim_{X\rightarrow+\infty}S(55-X)=\lim_{X\rightarrow+\infty}S(-X)$, which contradicts $\deg S=37$ odd.


This might have been posted somewhere more general, since this is his site, but Mr. Rusczyk (the Las Vegas site coordinator) passed out from the heat in the testing room.  He was probably ok after getting water, but if not, get well soon!

They said it was also from food poisoning, wasn't it?

I thought he said something about food--heat and food poisoning, then.  That sounds more serious, so again, get well soon.

Does anyone know whether the bad food was the cafeteria food or something he'd eaten before?

the cafetira food was defenatly not the best; but considering how much i ate i think we can well assume it was probably not from there :D  :D

i was excited when i heard Mr. Rusczyk was the chair for the LV site.
anyways...i hope Mr. Rusczyk gets well soon.

Ouch, heat plus food poisoning (or so rumor has it).  Either way, best wishes for a quick and pleasant recovery.

Yeah, I hope he gets better. Heat exhaustion is awful.

I am glad that GA went to Penn state. Penguins like me do worse in heat.

Mr. Rusczyk got food poisoning at LV? I feel lucky for just throwing up my breakfast at Penn state. The food there was edible, just not digestable.

Hopefully he gets better.

I hope you feel better soon Mr. Rusczyk!  The food at PennState was OK, I didn't really like it.  But I stocked up on donuts!

according to what i read in Mr. Crawford's and Mr. Rusczyk's blog; he (Mr. Rusczyk) is fine :)

Yay!  He is better!

I didn't know that he passed out.

At Penn State, the food was O.K. (not good, but O.K.), and there was PLENTY of water (it rained all day Friday and part of Saturday).  I'm glad at Penn State, because nothing MAJOR happened there (apart from a couple minor elevator incedents...)

Excellent!  That's a relief.  :-)

[quote="10000th User"]Mr. Rusczyk is OK, he is back!!! He just made a post here

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=527965#p527965]http://www.mathlinks.ro/Forum/viewtopic.php?p=527965#p527965[/url][/quote]

i went to penn state with the GA arml team. we ate at some cracker-barrel style restaurant near penn state that gave me food poisoning. the food tasted great when i was eating it, but later i found myself so sick of the taste. even when i think of it now i get this urge to throw up. i never want to go to cracker-barrel again in my life.

Thanks for the sentiments.  Just to clarify - I didn't quite pass out, though I came close at times.  It was food poisoning, and I can't blame UNLV, as I never had time to hit the cafeteria there - I ate at various restaurants or random snacks we picked up for coaches' meetings or the grading room.  Fortunately, there were many experienced ARML people there and an army of volunteers, so I wasn't needed Saturday.


linear algebra
matrix
algebra
polynomial
induction
complex numbers
linear algebra unsolved

Let $ A \in Mn(\mathbb{R})$and for all $ X\in Mn(\mathbb{R}): \det(A\plus{}X)\equal{}\det(X)$ .Prove that $ A$ is zero.

let $ p(x) \equal{} \det (xI \minus{} A)$
we have for any real $ x$,
$ p(x) \equal{} \det (xI \minus{} A) \equal{} \det (xI \minus{} A \plus{} A) \equal{} x^n$
Thus $ p(x) \minus{} x^n$ has infinite roots.
Which means $ p(x) \equal{} x^n$ for $ x \in \mathbb{C}$
Thus the eigenvalue of $ A$ is only $ 0$
We can just consider $ A$ is the jordan block case.
And if $ A \ne 0$ to construct $ X$ to find a contradiction :) .

But I still have problem,if we change the condition that $ A$ is a complex matrix how to slove it(In this case,you can't assume that $ A$ is just the Jordan block):?: ,which is:
Let $ A \in Mn(\mathbb{C})$and for all $ X\in Mn(\mathbb{R}): \det(A \plus{} X) \equal{} \det(X)$ .Prove that $ A$ is zero.

Edit:I have an idea to slove this:
If we consider $ \det (A \plus{} X) \minus{} \det X$ is an polynomial of $ X$ which have $ n^2$ variables.
Then the polynomial is zero for all $ X \in Mn(\mathbb{R})$
Which means $ \det (A \plus{} X) \minus{} \det X \equal{} 0$ is an identity.
(Because:
We can use induction(on $ n$) to prove that an polynomial $ p(x)$ with $ n$ variables have all  $ p(x) \equal{} 0,x\in \mathbb{R}^n$ then $ p(x) \equal{} 0$)
Then we have, $ \det(A \plus{} X) \equal{} \det(X)$,for all $ X \in Mn(\mathbb{C})$
The rest is similar to above.

Is my approach feasible?

When I get back to the dormitory,I find the following nice theorem will still be true if we use something named Hilbert Nullstellenatz.(Which can be proved by induction)

Let $ A \in M_n(\mathbb{C})$ and for all $ X$ with entities $ x_{ij} \in \{a,b\}$($ a,b$ are any fixed distinct complex numbers),we have $ \det (A \plus{} X) \equal{} \det (X)$,then$ A \equal{} 0$.

Isn't it nicer :lol: ?


Putnam
integration
trigonometry
function
calculus
induction
college contests

Putnam 1985

A5) Let $I_{m}=\int_{0}^{2\pi}\cos (x)\cos (2x)\cdots\cos (mx) dx$. For which integers $m$, $1\leq m\leq 10$ is $I_{m}\neq 0$?

I made a plot with maple to help me figure it out, the plot is of the functions $y=\prod_{k=1}^{m}\cos (kx),$ for $1\leq m\leq 10,$ the blue curves are the ones whose integral over $\left[ 0,2\pi\right]$ is non-zero, and the red ones have zero as the value of its integral.

[url=http://123pichosting.com/viewer.php?id=2117Putnam_1985_A5.jpg][img]http://123pichosting.com/thumbs/2117Putnam_1985_A5.jpg[/img][/url]

[hide="my solution"]Set $f_{m}(x)=\prod_{k=1}^{m}\cos (kx),$ for $1\leq m\leq 10.$ Recall that $\cos (x)=\frac{1}{2}(e^{ix}+e^{-ix})$ so that we may write $f_{m}(x)=\prod_{k=1}^{m}\frac{1}{2}(e^{ikx}+e^{-ikx})$. Notice that \[f_{m}(x+\pi )=\prod_{k=1}^{m}\frac{1}{2}(e^{ik(x+\pi)}+e^{-ik(x+\pi)}) = \prod_{k=1}^{m}\frac{1}{2}(e^{ik\pi}e^{ikx}+e^{-ik\pi}e^{-ikx})\] here I note that $e^{ik\pi}= (e^{i\pi})^{k}=(-1)^{k}$ and that $e^{-ik\pi}= (e^{i\pi})^{-k}=(-1)^{-k}=(-1)^{k}$ so we have \[f_{m}(x+\pi )=\prod_{k=1}^{m}\frac{(-1)^{k}}{2}(e^{ikx}+e^{-ikx}) =(-1)^{\sum_{k=1}^{m}k}\prod_{k=1}^{m}\frac{1}{2}(e^{ikx}+e^{-ikx})\]  \[=(-1)^{\frac{m(m+1)}{2}}\prod_{k=1}^{m}\cos (kx)=(-1)^{\frac{m(m+1)}{2}}f_{m}(x)\] ok, so here is the punchline, \[I_{m}=\int_{0}^{2\pi}f_{m}(x)dx= \int_{0}^{\pi}f_{m}(x)dx+\int_{\pi}^{2\pi}f_{m}(x)dx\]  \[= \int_{0}^{\pi}f_{m}(x)dx+\int_{0}^{\pi}f_{m}(x+\pi )dx= \int_{0}^{\pi}\left[ f_{m}(x)+f_{m}(x+\pi )\right] dx\]  \[= \int_{0}^{\pi}\left[ f_{m}(x)+(-1)^{\frac{m(m+1)}{2}}f_{m}(x)\right] dx= \int_{0}^{\pi}f_{m}(x)\left[ 1+(-1)^{\frac{m(m+1)}{2}}\right] dx\] which is non-zero for $m=3,4,7,8$ since $(-1)^{\frac{m(m+1)}{2}}=-1,-1,1,1,-1,-1,1,1,\ldots$ for $m=1,2,3,\ldots$.[/hide]

Can you set up a recurrence from $I_{m+1}$ to $I_{m}$?
I will look for a way :wink:

The [hide="answer"]$I_{n}$ is nonzero if and only if $1+2+\cdots+m=\frac{m(m+1)}2$ is odd, which occurs whenever $n$ is equivalent to 1 or 2 mod 4.[/hide]

For the idea behind that answer, see this [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=103871]closely related problem[/url]

(All right, it's basically equivalent to your hidden solution)

question, how do you know that $2\int_{0}^{\pi}f_{m}(x)dx\neq 0$?

I have the following solution, which does not use $\cos x=\frac{e^{ix}+e^{-ix}}{2}$.
Let $c_k(x)=cos(kx)$ for each integer $k$. By standard trigonometric identities, $c_kc_l=\frac{c_{k+l}+c_{k-l}}{2}$. Multiplying by $c_m$ yields $c_kc_lc_m=\frac{c_{k+l}c_m+c_{k-l}c_m}{2}=\frac{c_{k+l+m}+c_{k+l-m}+c_{k-l+m}+c_{k-l-m}}{4}$.

Continuing this way, and using the fact that $c_k=c_{-k}$ we can prove by (a very neat) induction that

$2\prod_{i=1}^m{c_{k_i}}=\frac{\sum_b{c_{b}}}{2^{m-1}}$

Where the sum on the right hand side is taken over all possible $b=\pm k_1\pm k_2\pm \ldots \pm k_m$. In our problem, $k_i=i$.

Since $\int_0^{2\pi} c_k(x) \,dx= 2\pi$ iff $k=0$ and is $0$ otherwise, we have $I_m=\frac{\sum_b{\int_0^{2\pi} c_b(x)\,dx}}{2^m}\neq 0 $ iff for at least one $b$ we have $\int_0^{2\pi} c_b(x)\,dx \neq 0$, which happens iff we can choose signs in $\pm 1 \pm 2 \pm \ldots \pm m$ that yield $0$.

By parity, we may discard $m=1,2,5,6,9,10$ as in the other solutions. For the other numbers we use $1-2+3$, $1-2-3+4$, $-1+2-3-4+5-6+7$ and $1-2-3+4+5-6-7+8$.


1)  Solve for $x,y \in \mathbb{Z}$, $y \neq 0$:

$x^2 + y^3 = (x+y)^2$

2) Again, solve for $x,y \in \mathbb{Z}$, $y \neq 0$:

$x^3 + y^3= (x-y)^4$

[hide="Problem 1"]$x^{2}+y^{3}=\left(x+y\right)^{2}\implies y^{2}+y-2x=0$

$\implies y=-\frac{1}{2}\pm\frac{\sqrt{1+8x}}{2}$

Then $8x=\left(k-1\right)\left(k+1\right)$, so there are an infinite number of solutions.[/hide]


Some chessmen are placed on an $ n$ by $ n$ chessboard, with positive integral $ n$, such that:

1. Every square without a chess piece shares a side with one that has a chess piece.
2.Given any two squares that contain chessmen, there is a sequence of squares containing chessmen, starting and ending with the given squares, such that every two consecutive squares share a side.

Prove that at least $ \frac{n^2\minus{}2}{3}$ chessmen are on the board.

[hide="my solution"]
Let us set $ x$ as the number of chessmen. We know that there are $ n^2$ squares that must either be covered by a chess piece or adjacent to a square that is covered by a chess piece because of rule 1.

We can "tile" the board with chessmen by first laying down one, and then laying another adjacent to that one, and another adjacent to that one, et cetera, thus adhering to rule 2.

The first chess piece "tiles" the square, and the four adjacent squares around it. All of the chessmen that are added after that one "tile" a maximum of three squares after the first. So, the maximum number of squares that are covered can be represented in terms of $ x$; the initial five "tiled" by the original chess piece, plus the three "tiled" by every next chess piece:

$ 5 \plus{} 3(x \minus{} 1)$

Then, for sufficient $ x$,

$ 5 \plus{} 3(x\minus{}1) \geq n^2$

Solving for $ x$, we have

$ x \geq \frac{n^2\minus{}2}{3}$,

proving our original theorem.
[/hide]

Source: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=340035#340035]USAMO 1999, Problem 1[/url]


Is there a certain amount of questions you need to finish, before getting videos? I already did 62+ problems. I have done 100+ now, and I have only gotten 2 more videos. Is there a "Complete 110+ problems, get 2 xtra videos?" I like the videos, and want to get more. Just a question :)

You need to pass subjects in order to access more videos.  So, if you just answer "1" for 60 problems to get to the videos, you'll only have a few videos you can see before you actually start doing the problems, getting them right, passing some subjects, and earning access to more videos.

Unless, of course, the answer to all 60 problems IS 1.

I don't think that's a good idea. You should click on [list]Video lessons[/list] after every problem.

Since AoPS is nonprofit, they don't want random people who aren't even trying to learn taking advantage of their videos, so you have to answer a certain amount of correct answers to watch.

This thread is like 4 years old. Why was it bumped?

I don't know! But the videos are helpful. I think that this type of security is great, anyway.


conics
geometry
3D geometry
ellipse
parabola

can you guys help me the question is attached!   :huh:
thanks in advance!

In general, the type of conic section is determined by the angle of the plane, and whether or not it passes through the vertex.  Let's denote the angle the plane forms with the base of the cone (in degrees) $\theta$, and the angle that the base of the cone forms with its side (also in degrees) $\alpha$.

First, the conic sections when the plane does not pass through the vertex:  if $\theta = 0$, this conic section will be a circle; if $0<\theta<\alpha$, the conic will be an ellipse; If $\theta=\alpha$, then the conic will be a parabola; If $\alpha<\theta\le90$, then the conic will be a hyperbola.

Now I will cover the degenerate conic sections, which happen when the plane passes through the origin:  if $0\le\theta<\alpha$, then the conic section will be a single point; if $\theta=\alpha$, then the conic section will be a single line; if $\alpha<\theta\le90$, then the conic section will be a pair of intersecting lines.

If you have further questions, feel free to ask.


search
number theory unsolved
number theory

$ a,b$ are odd integers
\[ \begin{array}{l} a_1 \equal{} a \\
a_2 \equal{} b \\
\vdots \\
\vdots \\
a_n \equal{} largest ..odd..divisor..of(a_{n \minus{} 1} \plus{} a_{n \minus{} 2} ) \\
\end{array}
\]
Prove that there exists  a natural number $ N$ such that
\[ \forall k \ge N,a_k \equal{} \gcd (a,b)
\]

Posted before. See here:

[url]http://www.mathlinks.ro/viewtopic.php?search_id=1870291191&t=58207[/url]


My next forum game is coming up. Sign up for it if you know you will be on AoPS at least once every two days! You must have joined AoPS on or before November 19, 2009 to join this game, and you must have been on AoPS for at least a year to take spots 11 through 15. Provide your gender, and pick a number between 10 and 99 inclusive that no user who has previously signed up has picked.

1. monkeygirl13 [17 - 85,52,42,57] - female
2. westiepaw [37 - 77,55,71,82] - female
3. wrightboy10 [52 - 10,59,58,10] - male
4. Richard_Min [99 - 33,18,42,93] - male
5. Butler [50 - 79,83,38,62] - male
6. randomguy64 [42 - 46,86,82,66] - male
7. tennis123 [21 - 44,38,61,46] - female
8. jjfun1 [33 - 78,63,25,98] - female
9. andrewjjiang97 [13 - 70,11,65,95] - male
10. vahalla [59 - 52,11,62,28] - male
11. ksun48 [98 - 93,27,42,87] - male
12. AIME15 [62 - 21,83,38,18] - male
13. aleph0 [61 - 77,56,24,80] - male
14. Yoshi [97 - 77,46,20,88] - male
15. bluecarneal [15 - 46,43,79,48] - male

Wow, I am indeed surprised that within half an hour so many people replied, it wasn't like this for LTM, you know. Also, no need to keep track of who's in; I'll update the first post for you. As for the question on what ATTOS is about, the reason I have a separate signup page is for the uncertainty in the air...NEVR FEER TZE UNNOEN.

1.
2.
3.
4.
5.
6.
7.
8.
9. andrewjjiang97, male, (13)
10. vahalla, male? (59)
11. ksun48, ????? (98)
12. AIME15, male, (62)
13. aleph0, male, (61)
14.
15. bluecarneal, male (15)
butler, pick spot, male, (50)

@phire: sorry i had to do that

[quote="PhireKaLk6781"]By the way, because I know someone will ask, those four numbers in the first post after your number are randomly generated.[/quote]

what are they for?

1. monkeygirl13 (17) female
2.
3.
4.
5. piegragh, male
6.
7. tennis123 (3) female
8.
9. andrewjjiang97, male, (13)
10. vahalla, male? (59)
11. ksun48, ????? (98)
12. AIME15, male, (62)
13. aleph0, male, (61)
14.
15. bluecarneal, male (15)
butler, pick spot, male, (50)

Current replacements list:
piegraph
spaceguy524

At
The
T_ _ _ _ _ _ _
Of
S_ _ _ _ _ _

This makes sense, but what are the other two words?

I want to say "at the threshold of silence"

He told me that the second letter of the word starting with s was p, but "threshold" sounds pretty good.

spartan?
I can't think of anymore sp 7 letter words.

EDIT: forgot I had a crossword puzzle solver.

there are a lot more, but none would be relevant.

I used a crossword puzzle to read off possibilities. They were in alphabetical order. He said I was close when I typed "spheres" but farther when I typed "spirits".

At the Threshold of Spiders? Idk, that might work.  :P

[quote="aleph0"]He (PhireKalk) told me that the second letter of the word starting with s was p, but "threshold" sounds pretty good.[/quote]

So the last word probably isn't 'silence'.

At The Threshold Of Spleens? lol not

At The Threshold of Sp_ _ _ _ _

Threshold doesn't have the correct amount of letters, so it can't be it.

Wow, I could probably even make a game just for guessing the name of the next game at this rate.

At the Threshold of Silence, does sound like the name of a game, or a book.

I thought I was right....


Hi people... :)
I just found an inequality in a paper of the Peruvian training ....

I'd LIKE TO know if this is an original constraction!! :?
It has an introduction ...

Summarizing....

Let $x,y,z > 0$  real numbers,
we know that:

$\frac{a^3 + b^3 + c^3 }{abc} \geq 3$, (directly MA-MG)

but the left part could be re-writed like that:

\[ \frac{a^3 + b^3 + c^3 }{abc} \geq (\frac{a}{b} + \frac{b}{c} + \frac{c}{a}) + (\frac{a}{b} - \frac{b}{a}) + (\frac{b}{c} - \frac{c}{b}) + (\frac{c}{a} - \frac{a}{c}).  \]

Prove this...

Carlos Bravo :)
LIMA-PERU

sincerely i have not seen this inequality before, but i can't guarantee that this is original. it's not so difficult, indeed is very nice and simple, here's my solution

multiply by $abc$ to obtain that your inequality is equivalent to
$a^3+b^3+c^3+a^2b+ac^2+b^2c\geq2(ab^2+a^2c+bc^2)$,
but by am-gm we have that
$a^3+ac^2\geq2a^2c$,   $b^3+a^2b%Error. "geqab" is a bad command.
^2$, and $c^3+b^2c\geq2bc^2$,
so we're done!  :D


ΝΔΟ--->Αν $ 24/xy\plus{}1$ τοτε $ 24/x\plus{}y$ $ x,y$ ανηκουν στους ακεραιους.
(με ενδιαφερει πολυ να δω διαφορετικες λυσεις)

Σαν μια πρωτη λυση προφανως $ gcd(x,24) \equal{} 1$ και $ x(y \plus{} x) \equal{} xy \plus{} x^2 \equiv xy \plus{} 1 \equiv 0 \mod 24$ απο οπου $ 24|x \plus{} y$

Σαν μια δευτερη προσεγγιση, αλλα οχι με καμια μεγαλη διαφορα, μπορεις να θεωρησεις το τριωνυμο με ριζες τα $ x,y$ και να το χειριστεις ως εξης.
Eστω οτι ειναι το $ P(x) \equal{} x^2 \plus{} bx \plus{} c$ με $ 24|c \plus{} 1$ .Toτε Μodulo 24 ισχυει $ P(x) \equal{} bx \plus{} c \plus{} 1\equal{}bx$ απο οπου το ζητουμενο.

Εστω οτι $ x \plus{} y \equal{} a$ , $ x \minus{} y \equal{} b$ τοτε η ασκηση μας μετασχηματιζεται ως εξης:
NΔΟ-->ΑΝ $ 24/(a^2 \minus{} b^2 \plus{} 4)/4$ τοτε και $ 24/a$
αν ο $ 3$ δεν διαιρει τον $ a$ τοτε $ a^2 \equal{} 1mod4 \equal{} > b^2 \equal{} 2mod3$ ATOΠΟ Αρα $ a \equal{} 0mod3$
Μενει να ερευνησουμε αν διαιρειτε με το $ 32$
εχουμε οτι $ 32/a^2 \minus{} b^2 \plus{} 4$ 8a δειξω οτι και $ 32/a$
επειδη $ 32/a^2 \minus{} b^2 \plus{} 4 \equal{} > a \equal{} bmod2$
εστω οτι $ a \equal{} 8k \plus{} z$ , $ b \equal{} 8k \plus{} m$ τοτε $ a^2 \minus{} b^2 \plus{} 4 \equal{} z^2 \minus{} m^2 \plus{} 4mod32$ με $ z,m \equal{} {0,1,2,3,4,5,6,7,8}$
τελος επειδη $ z \equal{} mmod2$ ευκολα βλεπουμε οτι $ z \equal{} 0$ αρα $ 8/a$ .


[b] Let $M$ be a positive integer , $M\le 100$ .  Using only $M$ perfect one ohm   resistors, construct  a resistance of [/b]  [color=red] [b] E [/b] [/color]  [b] ohms ,
 were  [/b]  [color=red] [b] E [/b] [/color]  [b] approximate ,,  $e$ "  (Napier's constant) with at least five decimals (only series-parallel   circuits  are allowed).  [/b]
See also :
 [1] Problem E2459 proposed by A.A.Mullin in Amer.Math.Monthly (1974)
 [2] Problem 393 from Matematicki Vesnik 13(28)(1976),solution in      15(1978).

I can do six decimal places with M=18, but I don't use only paralell-series circuits. R is approximately 2.71828103... if you label vertices 0 to 13 and join the pairs

0 7
0 4
1 3
1 5
2 8
2 10
2 4
3 7
3 11
3 4
5 8
5 9
5 10
9 11
10 4
11 4
4 6
6 12

The requested resistance is measured between nodes 0 and 12.

Bus,
how did you come about doing that? I'm afraid I'm not very good with circuits, is there a method for figuring out those resistances?

(I'm kubus, not bus:)

To calculate resistance of a non-parallel-series circuit, you can either do some clever tricks (triangle-star exchange, cutting and pasting parts, ...) or simply write the potentials (voltages) in the nodes as N unknows and use the Ohm's and Kirchhoff's laws to write N linear equations. Then just solve. I wrote a program some time ago to do this for me, I've started a thread about finding a scheme with resistance equal to $\pi \approx 3.14159...$ a couple of months ago.

If you only want parallel-series circuits, you can make use of the continued fraction of e:
$e \approx 2+\frac{1}{1+\frac{1}{2+\frac{1}{1+\frac{1}{1+\frac{1}{4+\cdots}}}}}$
This way, with 21 resistors, you can get right 7 decimal digits.

Whoops, I just saw the flag, and he's been posting more frequently lately... sorry  :blush:

And, thanks for the reply =)

[quote="kubus"]This way, with 21 resistors, you can get right 7 decimal digits.[/quote]
Hi kubus, can you prove that  [color=red][b] 21 [/b] [/color]  is [b] the minimal number of resistors [/b]  in  order to obtain an approximation for $e$ with at least [color=red][b] 7 [/b] [/color] decimals ?

I certainly can not prove this, moreover, I doubt it is the minimal number of resistors.

A kind of "proof" would be to write a program to generate ALL the circuits with 1,2,3,4, ... resistors and see what is the minimum number. This could be really time-consuming, but I'll think about it.

Both of my examples were just first that came to my mind (or first spat out by the computer). I am sure you can do better in both cases.


Prove that if a, b, c are the lengths of the sides of a triangle then

3/2 <= a/(b + c) + b/(c + a) + c/(a + b) < 2.

This one is easy 8-)

Still working hard on those ineg

* 3/2 <= a/(b + c) + b/(c + a) + c/(a + b)

I think it works for every a,b,c No  :?:

Wolog a >= b >= c
then 1/a+b <= 1/a+c <= 1/b+c

By Chebyshev :

a/(b + c) + b/(c + a) + c/(a + b) >= 1/3 (a+b+c)(1/a+b + 1/a+c + 1/b+c))

By Cauchy :  (:Sigma: ai)(:Sigma: 1/ai) >= n:^2: (if n ai)

So (1/a+b + 1/a+c + 1/b+c))(a+b + a+c + b+c) >= 9

a/(b + c) + b/(c + a) + c/(a + b) >= 1/3 (a+b+c)*9/2/(a+b+c) = 3/2

* a/(b + c) + b/(c + a) + c/(a + b) < 2

s = (a+b+c)/2, then b+c > s, c+a > s, b+a > s for a triangle.
so a/(b + c) + b/(c + a) + c/(a + b) < (a+b+c)/s = 2

This one was not the most difficult I think  :)

Indeed. The first inequality is true for any a, b, c > 0.
It is called "Nesbitt inequality".
Remember this one since it is very often useful !

Numerous proofs are available :
Cauchy ;
Rearrangement ;
Chebyshef ...

The right inequality is also quite easy 8-)


calculus geometry trigonometry function conics vector logarithms

I'm just curious: How many of you took Pre-Calculus in 9th grade? Just wondering, 'cause I want to know how you can deal with taking such a hard subject!!!

I took precal in grade 9. I found it to be mostly a repeat of material I had learned in the past few years (algebra, algebra 2, and geometry), although that may be because I had a really good geometry teacher and I took algebra 2 twice because public schools are stupid, so I had a good foundation in those subjects.

Our classes were 50 min, 5 days/week. I probably played about 3 hours of calculator tetris every week in that class.

I am actually taking Precalculus right now in the 7th grade. It really is review if you have taken Algebra, Algebra II, and Geometry like Xevarion said.

It's possible that at his school, the precal curriculum does include a lot of new material.

Some important things that were in my precal curriculum were trig (memorize those trig identities!) and getting familiar with functions like exponentials and log and rational functions.

If you haven't seen much about those functions before in geometry and algebra, learning that material could be hard. Make sure you study that stuff and understand how to manipulate things abstractly and also the concrete meaning of what you're doing.

Another tip: always do your homework. Pay attention in class (unless you're totally sure you already understand all the material from the lesson).

I took it... never. I did do calculus in tenth grade, but I skipped precalculus before that. The key is that I taught myself a lot of algebra and trig while I was bored in geometry.

I took it in 9th grade after having skipped Algebra II. I learned a great deal in that class, in particular about trig identities, law of cosines, law of sines, conic sections, vectors, complex numbers, and e. I think I worked harder (and learned more) during that year than in any other year I was in high school. Doing a LOT of AHSMEs (now called AMCs) helped reinforce the concepts that I was learning in precal at the time.

The book that we used is still popular today - Brown and Robbins' [u]Advanced Mathematics[/u], although I think the book has gone through a lot of changes since then.

9th grade as well. I had already taken the EPGY Precalculus course which I found extremely difficult, so I decided a review would be worthwhile. Boy, was I wrong. Never trust public school math education. I did manage to almost never have HW from that class though, since I always did it during class.

I never actually took precalc - I took Honors Geometry in 9th grade (they laughed it off when I asked to skip; although the next year I managed to yell my way into calculus). Honestly, a lot of precalc is a direct extension of algebra II. The main things that may be new are dealing with exponentials, dealing with logarithms, and dealing with trig.

Exponentials and logarithms are really the same thing; it just takes some seeing to get used to. As for trig, I was having a bit of trouble at first. Then, my dad got out his old math book, which had a page full of trig identities and said, derive all of these out yourself, and if you make it to the end, you must have figured trig out. And it really helped.


Let $ f: \mathbb{N} \rightarrow \mathbb{N}$ be a function such that $ f(n\plus{}1) > f(n)$ and $ f(f(n)) \equal{} 3n$ for all $ n$. Evaluate $ f(2001)$.

Well how did you get to that function? :huh:

[quote="jimhu"][hide]
Consider $ f(n) \equal{} \lfloor \sqrt {3}n \rfloor$ The result follows
[/hide][/quote]
Sorry, how does it follow?  $ \left\lfloor\sqrt3\lfloor\sqrt3\cdot1\rfloor\right\rfloor\equal{}1<3\equal{}f(f(1))$.

[hide="WRONG Solution - DO NOT DISILLUSION YOURSELF!"]
Note that $ f(f(1)) \equal{} 3$
Now, if $ f(1) \equal{} x$ and $ f(x) \equal{} 3$, clearly $ x$ must be $ 2$. Then $ f(3) \equal{} 3(2) \equal{} 6$, and for every $ a_n$, $ a_n \equal{} 3a_{n \minus{} 2}$. Thus if $ a_1 \equal{} 1$, $ \{a_n\} \equal{} \{1, 2, 3, 6, 9, 12, \dots\}$. From this, it follows that  $ f(x) \equal{} \begin{cases} \frac 32 x & \text{if } 2\ |\ x \\
2x & \text{if } 2\not|\ x \end{cases}$

Then it becomes apparent that $ f(2001) \equal{} \boxed {4002}$.
[/hide]

Tisk Tisk yongie, it nshould be 9,12 not 9,18. This is the most natural approach to solving the problem. Namely, solving enough numbers to find a pattern.

f(1) cannot equal 1, since then f(f(1))=1
If $ f(1)\equal{}k\ge3$ then $ 3\equal{}f(f(1))\equal{}f(k)\ge f(3)>f(1)\equal{}k$ contradiction
So f(1)=2
And therefore f(2)=f(f(1))=3
f(3)=f(f(2))=6
f(6)=f(f(3))=9
Which means that f(4)=7 and f(5)=8 since they are squeezed in between f(3) and f(6).
f(7)=12
f(8)=15
f(9)=18
f(12)=21
So f(10)=19 f(11)=20
f(15)=24
So f(13)=22 f(14)=23
f(18)=27
So f(16)=25 f(17)=26
f(19)=30
f(20)=33

1 2
2 3
3 6
4 7
5 8
6 9
7 12
8 15
9 18
10 19
11 20
12 21
13 22
14 23
15 24
16 25
17 26
18 27
19 30
20 33

Prove by induction that the differences between consecutive numbers goes 1 3 1 1 1 3 3 3 1 1 1 1 1 1 1 1 1 3 3 3 3 3 3 3 3 3 Where there are 3 times as many as previous (use induction on the powers of 3 should be useful). Then bash out 2001. This should work.

Oh... :blush: Never bothered to look at $ f(4)$... *whacks himself for being stupid again* :oops:

(just kidding)

Continuing his list:

1 2
2 3
3 6
4 7
5 8
6 9
7 12
8 15
9 18
10 19
11 20
12 21
13 22
14 23
15 24
16 25
17 26
18 27
19 30
20 33
21 36
22 39
23 42
24 45
25 48
26 51
27 54
28 55
29 56
30 57
31 58
32 59

Using induction, we can see that if $ 2 \cdot 3^k < n < 3^{k \plus{} 1}$, then $ f(n) \equal{} 3^{k \plus{} 1} \plus{} 3(n \minus{} 2 \cdot 3^k)$, or $ 3n \minus{} 3^{k \plus{} 1}$. (Where $ k$ is a natural number.) We note that $ 2 \cdot 3^6 < 2001 < 3^7$, so it follows that $ f(2001) \equal{} 6003 \minus{} 2187 \equal{} \boxed {3816}$.

P.S. My name is Yongyi, not yongie... :|


17.In a college 240 fruits are distributed among the students. Find the number of students who take all the three fruits.  Followng are the observations:

i.Every student takes at least one fruit – mango, banana or apple and at most one fruit of each type.

ii.Ratio of number of students who take only a mango to those who take only an apple to those who take only a banana is 9:4:3
iii.Ratio of number of students taking banana and an apple but not a mango to those taking an apple and a mango but not a banana to those taking a mango and a banana but not an apple is 4:6:3

iv.Ratio of Number of students taking at least mango, at least banana, at least apple is 5:3:4

[hide="Good hint"]Use a Venn-Diagram. That really helps on problems like these. [/hide]

Can someone solve the problem? When I tried it with my method, it won't work.  :(

even better than a venn diagram is a generalized venn diagram- without having to draw circles!! the mathematical name for this is [b]The Principle of Inclusion-Exclusion[/b] of PIE (try searching the forums for it...)

EDIT: better yet, there is an entry in the AoPS wiki!!

I don't think that will work because we only have ratios now.

It works fine.  We just interpret, for example, $9: 4: 3$ to mean the quantities $9x, 4x, 3x$ and we continue assigning variables until we get a (rather large but solvable) system.

Yay to rather large but solvable systems!!
and computer lol...

can you solve it??? i think the question is wrong....not sure

$M: B: A: M\cap B: B\cap A: M\cap A: M\cap B\cap A=9: 3: 4: 3: 4: 6: 2$

So, 240 fruit is impossible.

[hide="Outline"] Draw the Venn Diagram. Fill in $9,3,4$. From iii, fill in $3x, 4x, 6x$. Call the center $y$.
From iv, we get the easily solved system:
$3+7x+y=3z$
$9+9x+y=5z$
$4+10x+y=4z$ [/hide]


There was a death and the victim was identified as A, who recently was married. The police went to the crime scene and they reported the death as a suicide.

Later that day, after the police left, detective B, hired by A's friend who thought it was a murder, searched the crime scene and found a note the police missed.

It read,

"4,3: 8,1:_: 9,1: 2,1: 7,4:_: 6,1:9,3:_: 9,1: 4,3: 3,3: 3,2: !"

Detective B took out his cell phone and started dialing the police to tell them about his findings. Once B opened the phone to dial, he immediately screamed out, "I Solved  It!"

Who was the murderer and how did B find out?

[hide="Hint"]
[code]
 1    2    3
     ABC  DEF

 4    5    6
GHI  JKL  MNO

 7    8    9
PQRS TUV WXYZ

      0
[/code][/hide]
[hide="Answer"]
IT_WAS_MY_WIFE!
[/hide]


What are bots and what role do they serve on the board?

[[AoPS_forums_tutorial#Bot]]

That explains a lot.

[quote="Temperal"]http://en.wikipedia.org/wiki/Internet_bot

Also, a user called Yahoo Slurp! was mistaken for a bot once.[/quote]
Was that because he made some posts?


calculus
integration
calculus computations

Compute the integral: $\displaystyle \int \frac{x^{3}+x+1}{x^{4}+x^{2}+1} dx.$

$\frac{x^3+x+1}{x^4+x^2+1} = \frac{1}{2(x^2-x+1)} + \frac{2x+1}{2(x^2+x+1)}$.

For the first integral on the right,

$\int \frac{dx}{x^2-x+1} = \int \frac{d(x-\frac{1}{2})}{(x-\frac{1}{2})^2+\frac{3}{4}} = \frac{1}{\sqrt{\frac34}}\arctan \frac{x-\frac{1}{2}}{\sqrt{\frac34}} + c$

and for the second one, notice that $(x^2+x+1)' = 2x + 1$ hence it is $\frac{1}{2}\ln(x^2+x+1) + c$.

Therefore the result is

$\frac{1}{2}\ln(x^2+x+1) + \frac{1}{\sqrt3}\arctan \frac{2x-1}{\sqrt3} + c$

$\text{I saw a different }(but\;ridiculously\;longer)\text{ method . . .}$

$\displaystyle{\text{We have: }\int \frac{x^3 + x + 1}{x^4 + x^2 + 1}\;dx  }$

$\text{Multiply top and bottom by }x^2 - 1:$

$\displaystyle{\frac{x^2-1}{x^2-1}\cdot\frac{x^3+x+1}{x^4+x^2+1} \;= \;\frac{x^5-2x^4+x^2-x-1}{x^6-1} \;= \;\frac{x^5 - 2x^3 + x^2 - x - 1}{(x-1)(x+1)(x^2-x+1)(x^2+x+1)}   }$

$\text{ . . . and apply Partial Fractions.}$

Well the first method applied partial fractions from the beginning....


logarithms
floor function
limit
calculus
integration
calculus computations

1. Prove that the sequence $ x_n\equal{} \left\{1\plus{} \frac{1}{2}\plus{} \frac{1}{3}\plus{} \dots \plus{} \frac{1}{n} \right\}\minus{} \left\{ \ln{n} \right\}$ is divergent where $ \{ x \}\equal{} x \minus{} \lfloor x \rfloor$ is the fractional part of $ x \in \mathbb{R}$.

First show(or assume that we know) that $ \lim \sum\frac{1}{n}\minus{}\ln n$ exists. Then your statement follows from the fact that $ [\sum \frac 1n] \minus{}[\ln n]$ takes values 1 and 0 infinitely often: indeed, if the limit of the difference of fractional parts exists, then the same would be true for integer parts.

We don't quite need convergence- just that $ \sum_{k\equal{}1}^n \frac1k\minus{}\ln n$ is bounded away from any integers for large $ n$. This is easy with standard integral estimates. Then the two parts increment at different times, and we get infinitely many jumps up and down.

[quote="Yustas"] $ [\sum \frac 1n] \minus{} [\ln n]$ takes values 1 and 0 infinitely often.[/quote]

I`ll apreciate very much if you can show a rigurous proof for this part. :D

Thanks


I found several problems with the AMC and AIME state statistics

For example, on the amc 12b, it seems like Wendy Hou, who got the 133.5, was classified as a male.

In addition, what's really weird are the AIME I statistics.

Apparently, it says that there was 2 sophomores who got 10s, no sophomores with 9's, and 1 sophomore with an 8. So that means at most 3 people made the USAMO who are sophomores.

Florida Results for sophomores in AIME I are:

10's: Rui Yang, Gaku, Liu, Wendy Hou
9's: Chenyu Lin,  Saiem Gilani
8's: Marcel Nunez

So the 8 is right, and Wendy was included as a female 10. However, only one male, either gaku or rui, was reported as a 10, and neither myself nor saiem were accounted for. Marcel was reported, however.

Since all 6 of us made the USAMO, there must be some problem with the statistics.

The statistics, or the bubbling skill of the participants? I forgot to bubble my gender on the AMC10B.

yea I remember I was specifically instructed at some point not to bubble in grade, gender, age as there wasn't any point

for the AIME, I'm positive I bubbled in right.

By looking at the USAMO qualifiers list and the AIME statistics, the two conflicts.

Seriously, I think people have taken it a bit to the extreme. The AMC is trying their best, I'm sure they will realize their mistakes and fix them as soon as possible. It's not easy running a competition of this size! I for one am thankful for all the work they put it to keep this set of competitions the best and most prestigious of all :)


algebra
polynomial
Vieta
number theory proposed
number theory

Show that there exist integers $ j,k,l,m,n$ greater than $ 100$ such that:

$ j^2\plus{}k^2\plus{}l^2\plus{}m^2\plus{}n^2\equal{}jklmn\minus{}12.$

It has a solution (2,2,2,2,2) and we can see the equation about j.

And we can use vieta jumping
Then, (2,2,2,2,2^4-2) also can be a solution

repeat it.

we can take j=k=l=m=n & find j = 2

So initial seed of solution is (2,2,2,2,2)

WLOG 2 <= j <= k <= l <= m <= n

Using vieta we can see (i,j) is solution for quadratic equation and i  = klmn - j >= 8n -j >= 7n so new root is bigger than "n"

So (klmn-j,k,l,m,n) is solution , because of symmetry and root flipping (k,l,m,n, klmn-j) is the solution.. Applying on this seed

(2,2,2,2,2) ==>  ( 2,2,2,2,14) ==> (2,2,2,14,110) == > (2,2,14,110, 6158) ...... we can easily see infinitely many solution beyond 100 can be constructed using this path


Does anyone know(if there is a place) where I can find a timeline for mathcounts? i.e. chapter, state, and national competition dates? Because there's stuff I'm thinking about signing up for a field trip and history day and I want to make sure it wouldn't clash with anything.

Well, it depends on your chapter and state, so you have to contact them.  And since the National Location is TBA, I think it's date is too.

[quote="maybach"]  And since the National Location is TBA, I think it's date is too.[/quote]

It's in Orlando.

oh.  I thought that was unofficial and could change........

But check with your state tennis123

Look [url=https://mathcounts.org/Page.aspx?pid=294]here[/url] for important dates (yes, nationals are in Orlando May 6-9). You can check [url=http://mathcounts.org/Document.Doc?id=411]here[/url] to see if your school has registered. When registration closes and it gets close to the State Competition, they will post its time, date, and place on the MC website.

EDIT: Sorry, fixed the link.

Yes. I think it might be for 2011 too. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=294840]this[/url] thread.


geometry
3D geometry
tetrahedron
parallelogram
vector
arithmetic sequence
Pythagorean Theorem

A trirectangular tetrahedron has the property that the areas of its four faces are in arithmetic progression and the smallest has area 4 + 3:rt2:. Compute the volume of the tetrahedron.

(A trirectangular tetrahedron is one in which the three edges emanating from one of the vertices are all perpendicular to each other.)

(Thanks for the catch, Fiery.  I copied this from Intermediate since no one there was doing it.)

Just a first observation: the area of the face which isn't a right triangle is the largest. I hope I'll be back for more :D

Really, this isn't a "nice" problem.  You have to know something about trirectangular tetrahedra in order to make it not a dreadful mess to do.

I think the 4+3 :rt2: side length scared people away. :)

By de Gua's Theorem, we know that the sum of the squares of the three faces adjacent to the trihedral angle equals the square of the other face, so that

(4 + 3*Sqrt[2])^2 + (4 + 3*Sqrt[2] + d)^2 + (4 + 3*Sqrt[2] + 2d)^2 = (4 + 3*Sqrt[2] + 3d)^2

Solving, we find d=3+2 :rt2:

The sides are thus 4+3 :rt2: , 7+5 :rt2: , 10+7 :rt2: , so

V = (1/6)(4+3 :rt2: )( 7+5 :rt2: )( 10+7 :rt2:)
V = 577/3 + 136*Sqrt(2)

By the way, what is the source of the problem?

How had you heard of de Gua's Theorem? I hadn't heard of it until you mentioned it.

The source of the problem is my own twisted mind.  I hadn't ever heard of de Gua's theorem by name, but it was given to me as a fun excercize at one point, and I think I may have posted a question about it on the board.  (Is there any deeper reason relating to Pythagorean Theorem that this should be true?  And consensus seemed to be not.)

I was wondering about a deeper reason relating to the Pythagorean Theorem as well. I imagine that there should be, but I don't see it immediately. Heron's Theorem just seems too messy.

Well, the proof of de Gua's is pretty straightforward (directly from Heron's, though some algebra required). I actually looked up the theorem first before proving it though, since Joel mentioned:

[quote]You have to know something about trirectangular tetrahedra in order to make it not a dreadful mess to do.[/quote]

A faster way to prove it is to use the fact that the area of a parallelogram spanned by vectors v1 and v2 is ||v1 x v2||. So the area of a triangle with those vectors as sides is 1/2*||v1 x v2||. The theorem is not hard to prove from there.

I do, however, believe that there's an intuitive arguement for why it is true. I seem to recall discovering it myself when I was in high school - before I knew about Heron's Formula or cross products. But I can't for the life of me remember how.

Nice problem by the way Joel!

By the way, why did you choose an area of 4+3 :rt2: ? Why not just like an area of 2 or something? :)

Because I checked it first with an area of 1, and you get a whole bunch of things like :sqrt:(4 + 3 :rt2:) or some such thing, and by multiplying all the edge lengths through I could make it go away, with the downside of multiplying the area by something not too pretty to get that.

I just proved it using Hero's and areas -- it's pretty straighforward if a bit messy.  I don't know how else one would prove it.

Try it via the way I suggested above JBL - using cross products. It turns out very nicely.

I think I remember the way that I found it in high school. I think I expressed the volume of the tetrahedron in different ways, using different triangles as bases, and set them equal or manipulated them in some way. The key is to find the altitude from that face to the origin.


As an expression for $ T_n$, I have worked out:
[hide]$ T_n \equal{} \frac {[(n \plus{} 1)!]^4(2^{4n \plus{} 2})}{[(2n \plus{} 2)!]^2(n \plus{} 1)}$

I don't know how to continue if I try to use this.

The algebra for getting this formula could be shown at a later time.  (EDIT)

[/hide]


Έχουμε την ορίζουσα
α β γ δ
β γ δ α
γ δ α β
δ α β γ

Αν προσθέσουμε όλες τις γραμμές στην πρώτη έχουμε:

α+β+γ+δ  α+β+γ+δ  α+β+γ+δ  α+β+γ+δ
β               γ               δ           α
γ               δ               α           β
δ               α               β           γ

που ισούται με
(α+β+γ+δ)*det(1 1 1 1
                      β γ δ α
                      γ δ α β
                      δ α β γ)

που ισούται με
(α+β+γ+δ)det(1 0 0 0
                    β γ-β δ-β α-β
                    γ δ-γ α-γ β-γ
                    δ α-δ β-δ γ-δ)

στο τέλος οι σημειώσεις μου γράφουν οτι προκύπτει ότι η ορίζουσα είναι ίση με

[(β-δ)^2+(α-γ)^2][(β+δ)^2-(α+γ)^2]

αυτό όμως πως προέκυψε? δεν μπορώ να καταλάβω τις πράξεις έκανε. Μπορέι κανείς να βοηθήσει? :maybe:

Σίγουρα λοίπουν αρκετές πράξεις. Ίσως να σας το άφησε σαν άσκηση για το σπίτι;

Υπάρχει πάντως ένα κόλπο που μας σώζει σε αρκετές περιπτώσεις. Η ορίζουσα είναι το γινόμενο των ιδιοτιμών, άρα αν μπορούμε να βρούμε τις ιδιοτιμές τότε βρήκαμε και την ορίζουσα. (Θέλει μόνο προσοχή στην πολλαπλότητα των ιδιοτιμών)

Εδώ, "εύκολα" βλέπουμε ότι το (1,1,1,1) είναι ιδιοδιάνυσμα με ιδιοτιμή α+β+γ+δ. Κάπως πιο δύσκολο είναι να παρατηρήσουμε ότι το (1,i,-1,-i) είναι και αυτό ιδιοδιάνυσμα με ιδιοτιμή α + βi - γ - δi. Θα αφήσω την εύρεση των άλλων δύο ιδιοτιμών σαν άσκηση.


Blank (for my own reasons) Please delete this topic.

Never mind... I did something wrong...

Is there a way to prove lateral area of cone = 1/2pl?

Lateral surface area right? Well the formula I remember it by is $\pi rs$ where r is the radius of the cone and s is the slant height. Your formula is basically the same thing because (I'm guessing that by perimeter you mean circumference) $ r = \frac {C}{2\pi} $ where C is the circumference so $\pi*\frac {C}{2\pi} * s = \frac 12 C s $ is the lateral surface area.

[quote="Farazy"]I know that there are 90 degrees...but I don't know how to prove it.  I am trying to prove that the lateral area of a cone is 1/2pl where p is the perimeter and l is the slant height.[/quote]

Really? Try making a cone with a height approaching 0 and the slant height approaching the radius.

Oh and to prove that $ \pi rs $ is the lateral surface area of a cone.
[hide]Stretch out the cone to be a certain part (piece of pie) of a circle. Let $s$ be the slant height of the circle, and $x$ be whatever fraction of the circle the cone is. Also know that $s$ is equal to the radius of the circle. The lateral surface area is the area of the piece of pie the cone became, which is also the area of the circle times whatever fraction of the circle the cone is. This can be said as $ s^2 * \pi * x $. The radius of the cone is equal to the circumference of the cone divided by two times pi. Know that the circumference of the cone is now a certain part, $x$, of the circumference of a circle. The radius, $r$, of the cone is equal to $ \frac {2s * \pi * x}{2 \pi} = sx$. This can be seen in the previous equation for the lateral surface area: $s * s * x * \pi$, substituting $r$ in, we get $\pi rs $.[/hide]


[quote="tpjmroc"]find the value of
$ \sum^{2n}_{p} p(C^{p}_{2n})^{2}$[/quote]

What is exactly the index of this sum ?
What are exactly bounds ?

Must we understand $ \sum^{2n}_{k\equal{}p} p(C^{p}_{2n})^{2}$ and so the answer is $ (2n\plus{}1\minus{}p)p(C^{p}_{2n})^{2}$

Must we understand $ \sum^{2n}_{p\equal{}0} p(C^{p}_{2n})^{2}$ but then why $ 2n$? and not $ n$?

Must we understand $ \sum^{2n}_{\text{even }p\equal{}0} p(C^{p}_{2n})^{2}$ but then why not $ \sum^{n}_{p\equal{}0} 2p(C^{2p}_{2n})^{2}$ ?

Thanks for any precision...

[quote="pco"][quote="tpjmroc"]find the value of
$ \sum^{2n}_{p} p(C^{p}_{2n})^{2}$[/quote]

Must we understand $ \sum^{2n}_{k \equal{} p} p(C^{p}_{2n})^{2}$ and so the answer is $ (2n \plus{} 1 \minus{} p)p(C^{p}_{2n})^{2}$

[/quote]

where do we have  `k` there?

If the sum is from $ k \equal{} 0$ to $ 2n$, then it's a repetition of http://www.mathlinks.ro/viewtopic.php?t=148948 . The key is $ p\binom{m}{p} \equal{} m\binom{m \minus{} 1}{p \minus{} 1} \equal{} m\binom{m \minus{} 1}{m \minus{} p}$.

Neither $ \left(\sum_{k\equal{}0}^n k\binom{2n}{2k}^2\right)_{n\in\mathbb{N}}$ nor $ \left(\sum_{k\equal{}0}^n 2k\binom{2n}{2k}^2\right)_{n\in\mathbb{N}}$ nor $ \left(\sum_{k\equal{}0}^n k\binom{n}{2k}^2\right)_{n\in\mathbb{N}}$ nor $ \left(\sum_{k\equal{}0}^n 2k\binom{n}{2k}^2\right)_{n\in\mathbb{N}}$ is in Sloane's encyclopedia.

  darij


I wonder if this site supports a rather powerful math syntax - LaTeX.

Please enlighten me on this.

Not at the moment, but the admin was working on it a while ago I think. I don't know whether the addition of a latex capability is still planned.

By the way, this question should be moved to the right forum.


inequalities
LaTeX
algorithm
inequalities proposed
111

How to prove ∑a/[a+(a^2+3bc)^1/2]<=1  where a b c are positive

maybe we can use   Vornicu schur to prove the less strong problem I had proposed  before.∑a*(a^2+3bc)^1/2>=2(ab+bc+ca)

any idea    I am waiting for its solution  Thanks

[quote="stevenwang1992"]any one helps me ?thanks :?:[/quote]
i think you should learn LATEX before you post problem ! :blush:

I guess so too.But I really donnot have time. :blush:

[quote="stevenwang1992"]How to prove $ \sum\frac {a}{a \plus{} \sqrt {a^2 \plus{} 3bc}}\leq1$  where $ a, b, c$ are positive[/quote]$ 1\geq\sum\frac {a}{\sqrt {a^2 \plus{} 3bc} \plus{} a} \equal{} \sum\frac {a(\sqrt {a^2 \plus{} 3bc} \minus{} a)}{3bc}\Longleftrightarrow$

$ \sum{a^2\sqrt {a^2 \plus{} 3bc}}\leq a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc.$

Using Cauchy inequlity, we have

$ \left(\sum{a^2\sqrt {a^2 \plus{} 3bc}}\right)^2\leq \sum{\frac {a^2}{b^2 \plus{} c^2 \minus{} bc \plus{} ca \plus{} ab}}\sum{a^2(a^2 \plus{} 3bc)(b^2 \plus{} c^2 \minus{} bc \plus{} ca \plus{} ab)}$

$ \equal{} (a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc)^2 \minus{} \frac {a^2b^2c^2(a \plus{} b \plus{} c)^2(a^2 \plus{} b^2 \plus{} c^2)(a^2 \plus{} b^2 \plus{} c^2 \minus{} bc \minus{} ca \minus{} ab)}{\prod{(b^2 \plus{} c^2 \minus{} bc \plus{} ca \plus{} ab)}}$

$ \leq (a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc)^2.$

It's my old problem.
[quote="Ji Chen"]
Using Cauchy inequlity, we have

$ \left(\sum{a^2\sqrt {a^2 \plus{} 3bc}}\right)^2\leq \sum{\frac {a^2}{b^2 \plus{} c^2 \minus{} bc \plus{} ca \plus{} ab}}\sum{a^2(a^2 \plus{} 3bc)(b^2 \plus{} c^2 \minus{} bc \plus{} ca \plus{} ab)}$

$ \equal{} (a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc)^2 \minus{} \frac {a^2b^2c^2(a \plus{} b \plus{} c)^2(a^2 \plus{} b^2 \plus{} c^2)(a^2 \plus{} b^2 \plus{} c^2 \minus{} bc \minus{} ca \minus{} ab)}{\prod{(b^2 \plus{} c^2 \minus{} bc \plus{} ca \plus{} ab)}}$
[/quote]
Veru nice identity, Jichen. Thank you very much for your nice proof  :lol:

Dear Ji Chen,

I would like to know, if possible, how you found this wonderful formula.

Did you use a special algorithm by computer?

Thank you very much. :)

[quote="Ji Chen"]$ \sum{\frac {a^2}{b^2 \plus{} c^2 \minus{} bc \plus{} ca \plus{} ab}}\sum{a^2(a^2 \plus{} 3bc)(b^2 \plus{} c^2 \minus{} bc \plus{} ca \plus{} ab)}$

$ \equal{} (a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc)^2 \minus{} \frac {a^2b^2c^2(a \plus{} b \plus{} c)^2(a^2 \plus{} b^2 \plus{} c^2)(a^2 \plus{} b^2 \plus{} c^2 \minus{} bc \minus{} ca \minus{} ab)}{\prod{(b^2 \plus{} c^2 \minus{} bc \plus{} ca \plus{} ab)}}$

[/quote]

Is there  a simple way to do it like sos?thanks :blush:

Dear arqady,

I checked your proof and it is correct but also NEAT, CLEAR and ELEGANT :wink:

Congratulations for this proof of a very hard inequality :)

[quote="arqady"][quote="stevenwang1992"]How to prove $ \sum_{cyc}\frac {a}{a \plus{} \sqrt {a^2 \plus{} 3bc}}\leq1$  where a b c are positive[/quote]
Here is my proof:
$ \sum_{cyc}\frac {a}{a \plus{} \sqrt {a^2 \plus{} 3bc}}\leq1\Leftrightarrow\sum_{cyc}a^2\sqrt {a^2 \plus{} 3bc}\leq a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc\Leftrightarrow$( after squaring )
$ \Leftrightarrow2\sum_{cyc}a^2b^2\sqrt {(a^2 \plus{} 3bc)(b^2 \plus{} 3ac)}\leq\sum_{cyc}(2a^3b^3 \plus{} 3a^4bc \plus{} 3a^2b^2c^2)\Leftrightarrow$( after squaring )
$ \Leftrightarrow\sum_{cyc}(9a^6 \plus{} 62a^4bc \plus{} 18a^3b^3 \plus{} 39a^2b^2c^2)\geq$
$ \geq8\sum_{cyc}a^2(a^2 \plus{} 3bc)\sqrt {(b^2 \plus{} 3ac)(c^2 \plus{} 3ab)}.$
But $ 8\sum_{cyc}a^2(a^2 \plus{} 3bc)\sqrt {(b^2 \plus{} 3ac)(c^2 \plus{} 3ab)}\leq4\sum_{cyc}a^2(a^2 \plus{} 3bc)(b^2 \plus{} c^2 \plus{} 3ab \plus{} 3ac).$
Hence, it remains to prove that
$ \sum_{cyc}(9a^6 \plus{} 62a^4bc \plus{} 18a^3b^3 \plus{} 39a^2b^2c^2)\geq$
$ \geq4\sum_{cyc}a^2(a^2 \plus{} 3bc)(b^2 \plus{} c^2 \plus{} 3ab \plus{} 3ac).$
But $ \sum_{cyc}(9a^6 \plus{} 62a^4bc \plus{} 18a^3b^3 \plus{} 39a^2b^2c^2)\geq$
$ \geq4\sum_{cyc}a^2(a^2 \plus{} 3bc)(b^2 \plus{} c^2 \plus{} 3ab \plus{} 3ac)\Leftrightarrow$
$ \Leftrightarrow\sum_{sym}(4.5a^6 \minus{} 12a^5b \minus{} 4a^4b^2 \plus{} 9a^3b^3 \plus{} 31a^4bc \minus{} 48a^3b^2c \plus{} 19.5a^2b^2c^2)\geq0\Leftrightarrow$
$ \Leftrightarrow6\left(\sum_{cyc}(a^3 \minus{} a^2b \minus{} a^2c \plus{} abc)\right)^2 \plus{}$
$ \plus{} \sum_{sym}(1.5a^6 \plus{} 2a^4b^2 \minus{} 3a^3b^3 \plus{} 7a^4bc \minus{} 12a^3b^2c \plus{} 4.5a^2b^2c^2)\geq0,$ which is obvious.[/quote]

[quote="Ji Chen"][quote="stevenwang1992"]How to prove $ \sum\frac {a}{a \plus{} \sqrt {a^2 \plus{} 3bc}}\leq1$  where $ a, b, c$ are positive[/quote][/quote]
The following is my proof:

Let $ x\equal{}\frac{a}{a\plus{}\sqrt{a^2\plus{}3bc}},y\equal{}\frac{b}{b\plus{}\sqrt{b^2\plus{}3ca}},z\equal{}\frac{c}{c\plus{}\sqrt{c^2\plus{}3ab}}.$ Then $ x,y,z <\frac{1}{2},$ and we find that $ \frac{bc}{a^2}\equal{}\frac{1\minus{}2x}{3x^2},\frac{ca}{b^2}\equal{}\frac{1\minus{}2y}{3y^2},\frac{ab}{c^2}\equal{}\frac{1\minus{}2z}{3z^2},$ from which it follows that
$ (1\minus{}2x)(1\minus{}2y)(1\minus{}2z) \equal{}27x^2y^2z^2.$
We are required to prove $ x\plus{}y\plus{}z \le 1.$ The method we choose to do this work is contradiction. Let us suppose that there exists some positive numbers $ x,y,z <\frac{1}{2}$ so that $ x\plus{}y\plus{}z>1.$ By this assumption, we have
$ (1\minus{}2x)(1\minus{}2y)(1\minus{}2z) <[(x\plus{}y\plus{}z)\minus{}2x][(x\plus{}y\plus{}z)\minus{}2y][(x\plus{}y\plus{}z)\minus{}2z]\equal{}(x\plus{}y\minus{}z)(y\plus{}z\minus{}x)(z\plus{}x\minus{}y)<(x\plus{}y\minus{}z)(y\plus{}z\minus{}x)(z\plus{}x\minus{}y)(x\plus{}y\plus{}z)^3 \equal{}[2(x^2y^2\plus{}y^2z^2\plus{}z^2x^2)\minus{}(x^4\plus{}y^4\plus{}z^4)](x\plus{}y\plus{}z)^2 \le 3[2(x^2y^2\plus{}y^2z^2\plus{}z^2x^2)\minus{}x^4\minus{}y^4\minus{}z^4](x^2\plus{}y^2\plus{}z^2).$
On the other hand, from the Schur's Inequality (the special case third degree), we have
$ [2(x^2y^2\plus{}y^2z^2\plus{}z^2x^2)\minus{}(x^4\plus{}y^4\plus{}z^4)](x^2\plus{}y^2\plus{}z^2) \le 9x^2y^2z^2,$
so from the above estimation, we get
$ (1\minus{}2x)(1\minus{}2y)(1\minus{}2z)<27x^2y^2z^2,$
which is contradiction, and hence our proof is completed.

[quote="manlio"]Dear arqady,

I checked your proof and it is correct but also NEAT, CLEAR and ELEGANT :wink:

Congratulations for this proof of a very hard inequality :)

[/quote]
Thank you!

[i]A very good solution , Can , Congratulation . :lol: [/i]


I am not sure, but I think I solved it :
We will not allow the first player to pass the second row.
We take the numbering of collumns
0 1 2 3 4 5

Now put the numbers
0 1 2 3 4 5
4 5 0 1 2 3
In the first 2 rows of another table.
Any time the first player puts a number in one of x, we put a number in the other x so that the relationship (bigger, smaller ) between the same numbers in both rows is same. Because nr are rational, we can do it.
Now, obviously, biggest number will not be neighbours :P

[quote="DusT"]
Now, obviously, biggest number will not be neighbours :P[/quote]

Obviously?

$\begin{array}{cccccc}
0&2&4&6&8&\mathbf{10}\\
9&11&1&3&5&\mathbf{15}
\end{array}&$

Here's a way:

Denote columns as c1, c2, c3, c4, c5, c6.

Show that Bob has the ability to do one of the following in any row:
(the basic strategy for Bob is to always place numbers in the row Alice put hers)
#1 Force the black square to be in one of c1, c2, or c3.
#2 Force the black square to be in one of c4, c5, or c6.
#3 Force the black square to be in one of c1, c2, c5, c6.

Now let Bob execute #1 on row 1, #3 on row 2, and #2 on row 3.

If, in the second row, black square is in one of c1, c2, then it cannot be connected to row 3. (case 1)

If, in the second row, black square is in one of c5, c6, then it cannot be connected to row 1. (case 2)

In either case, Bob wins.

_: Grid boxes
X: Possible locations of black square

Case 1:
X X X_ _ _
X X _ _ _ _
_ _ _ X X X

Case 2:
X X X_ _ _
_ _ _ _ X X
_ _ _ X X X

So in fact, Bob wins in a 6x3 grid.

Your solution is very similar to my one. Only my one is based on more explicit results:
1. Bob can force the maximum in a row to lie in a group of 3 squares he chooses.
2 Bob reaches the maximum in row ! to be on c1,c2,c3, on row 2 on c4,c5,c6, on row 3 on c1,c2,c6.
this is his winning strategy

Yeah, obiously!
Reply tu lugash
0 2 4 6 8 10
9 11 1 3 5 15
The second array is not sorted in the same way.
It should be
0 2 4 6 8 10
11 15 1 3 5 9

Hmm... DusT

How do you take care of cases in which Alice places her first number in the first row but the second number the second row, so that Bob cannot always place his number according to

012345
450123

?

For example, let's say the first two moves are as follows:

_ _ 2 _ _ _
_ _ _ _ 3 _
_ _ _ _ _ _

(Alice places 2, Bob places 3)

_ _ 2 _ _ _
1 _ _ _ 3 _
_ _ 4 _ _ _

(Alices places 1, Bob places 4)

_ _ 2 _ 5 _
1 _ _ _ 3 _
_ _ 4 _ _ _

Now Alice places 5, but Bob's counterspot is already taken.

Sorry if I explain wrong , but I mean whean Alice puts a number in square x, Bob puts the number in the other x, so that the arrangement of the numbers remains the same :P
Hope no it's clear

Yes, but my point was that what if the other "x" is already taken?

Isn't it obvious that there are only 2 letters of the same type, so when the 1st player takes one, we take the other, so either both are taken, or none is taken.
Now, it is solved I think :)

Isn't this sorta obvious? Bob always makes the last move. He knows which square will be black in every row except the one in which the last empty square is. If he sees that this square will complete the line, he can always choose a sufficiently small number. If it won't, then we can always choose a sufficiently large number.

No, that doesn't work -- given the black squares in the first five rows, there might be more than one possible position for a black square in the sixth row that Alice can use to draw her line.

I think that the confusion here is that a "line" doesn't mean a straight line. It means a path from the top to the bottom where adjacent squares share a vertex.

If the first square in rows 1 through 5 and the second square in row 6 is colored black, there exists a line (in the Euclidean sense) that connects the top edge to the bottom edge and passes only though black squares.

Is it necessary for each player to choose a rational number??????

[hide="Slightly different"]
We claim in each row, Bob can choose, prior to the game's beginning, three "desired" squares such that one of those will turn out black. To do this, if he plays in the same row as Alice every turn. If Alice plays in one of the desired black squares, he plays a smaller number in an undesired square. If she plays in an un-desired square, he plays in a desired one.

Then, using chessboard notation, have Bob desire a1, a2, a3, b1, b2, b6, c4, c5, c6. Clearly, Bob wins.[/hide]

I think I am interpreting this problem wrong or this is trivial. Kindly clarify:

The grid has 36 (even) number of squares so Bob will make the last move. Clearly he can fill the number such that if the greatest numbers of the other rows lie in that column, he can choose a sufficiently small number. If not, a sufficiently large number suffices.

In the 6 by 6 grid, shade in the first 3 squares in the first and fourth rows, the first, 5th, and 6th squares in the 2nd and 5th rows, and the 3rd, 4th, and 5th squares in the 3rd and 6th rows. If the black squares were located in any of these shaded squares, a path wouldn't exist. Hence, Bob can win if he can force the black squares to be in these shaded squares. To do this, if Alice writes a number in a shaded square, Bob writes a smaller number in an unshaded square in the same row. If Alice writes a number in an unshaded square, but writes a larger number in a shaded square in the same row.

[quote=billzhao]Alice and Bob play a game on a 6 by 6 grid.  On his or her turn, a player chooses a rational number not yet appearing in the grid and writes it in an empty square of the grid.  Alice goes first and then the players alternate.  When all squares have numbers written in them, in each row, the square with the greatest number in that row is colored black.  Alice wins if she can then draw a line from the top of the grid to the bottom of the grid that stays in black squares, and Bob wins if she can't.  (If two squares share a vertex, Alice can draw a line from one to the other that stays in those two squares.)  Find, with proof, a winning strategy for one of the players.[/quote]

Bob wins no matter what.
(Refer to the figure)Bob's strategy is to make sure that no two black squares share a vertice.He puts numbers in any row in an increasing order.

$\;\;\;\;\; \bullet $ If Alice decides to follow along,by trying to disrupt Bob's strategy,then Bob starts going for $3 \cdot 3$ squares,there are $14$ of these.He changes the Corner's rather than the middle if Alice has fixed the middle and if Alice hasn't fixed the middle he gors for the middle.
Alice can't disrupt it;there are an even number of squares i.e Bob finishes it.

$\;\;\;\;\; \bullet $ If Alice decides not to play along then Bob goes for squares of the same parity;this is a clever strategy i.e Alice can't give Bob a last minute surprise;If after some point she decides to play along then Bob is safe;All the odd parity/even parity squares are reeserved so in order to move to the bottom of the board Alice would need to use an square of the parity Bob choose but this is impossible since all of these are white squares.

not sure if this works but any line going through the top and to the bottom has to pass through only one square in at least one row

also bob can make sure that his turns are the turns that color squares black

so on each of these turns he has two choices for what the black square will be

and im pretty sure this implies bob will always win

[hide = Solution] This took me far longer than it should have.

I claim that Bob can always win.
Consider the $2 \times 6$ board below:
\[
\begin{bmatrix}
1 & 2 & 3 & 4 & 5 & 6 \\
4 & 5 & 6 & 1 & 2 & 3
\end{bmatrix}.
\]
The strategy is that if Alice colors $x$ Bob colors the other $x$. We then have that with this coloring Alice may never obtain a path where two black squares share a vertex. We then have for the remaining $4 \times 6$ portion of the board, Bob may play in the same way. $\blacksquare$ [/hide]

We claim that Bob always wins. Bob's winning strategy is to always place in the same row as Alice. Consider any row, and we claim that Bob can limit the black square's position to three of these squares. If Alice places a number in one of the squares Bob wants black, then Bob can place a smaller number in a square he wants white, and vice versa. Thus, our claim is true.

Bob can limit row 1's to the rightmost three columns, row 2's to the leftmost two columns and rightmost column, and row 3's to the leftmost three columns. Starting from the top row, there is no way to draw a line that gets past row $3,$ so we're done.

Divide the board into 12 "zones", each of which is a set of 3 (not necessarily connected) cells, which we call $A1,A2,B1,B2,C1,C2,D1,D2,E1,E2,F1,F2$, in the following manner:

$$\begin{matrix}
A1 & A1 & A1 & A2 & A2 & A2\\
B2 & B2 & B2 & B1 & B1 & B1\\
C1 & C1 & C2 & C2 & C2 & C1\\
D1 & D1 & D1 & D2 & D2 & D2\\
E2 & E2 & E2 & E1 & E1 & E1\\
F1 & F1 & F2 & F2 & F2 & F1\\

\end{matrix}$$

Bob can then use the following strategy:\\

If Alice writes a number in a "1" zone, Bob comes up with a number smaller than all numbers on the board (which is always possible) and writes it in a "2" zone of the same letter as Alice's zone.\\
If Alice writes a number in a "2" zone, Bob comes up with a number larger than all numbers on the board (which is again possible) and writes it in a "1" zone of the same letter as Alice's zone.\\

Note that the only time when the largest number in a row can be in a "2" zone is right after Alice writes a number in that zone, but using this strategy Bob will immediately overturn that by placing a larger number in the corresponding "1" zone. Thus, by the end of the process, all black cells are in "1" zones. This clearly prevents Alice from winning, so Bob wins are we are done.

We claim that $\boxed{\text{Bob}}$ is the winner.

The key in this problem is that Bob can always prevent Alice from taking any $3$ squares he chooses. Indeed say on one of Alice's turns she plays in a square he wishes to be white. Then he can simply play a higher number in a square that he wishes to be black and so on. Now the finish is obvious. Restrict Alice to the square $(1, 2, 3)$ in row $1$. On row $2$ restrict Alice to the squares $(1, 5, 6)$. Then on the third row restrict her to the square $(4, 5, 6)$. Clearly Alice cannot win.


Let a,b is the positive integer numbers, (a,b)=1. Prove that for all number greater than ab-a-b can be expressed in the form ax+by(x,y are positive  integer numbers) :D

Hi TRAN THAI HUNG. See here: http://en.wikipedia.org/wiki/Coin_problem

and here: http://www.math.ups.edu/~mspivey/FrobFinal.pdf

Hope this is helpful.

Thanks a lot for your problem, dear friend.

In fact, a friend of mine gave me another similar problem about a month ago:
$ a,b\in\mathbb N; a > b$. Define $ S \equal{} \{ax \plus{} by \ | \ x,y\in\mathbb N_0\}$
If $ S$ misses exactly $ 35$ integers [b]including[/b] $ 58$; what are the values of $ a$ and $ b$?


what is the remainder when 3^25586 is divided by 7?

What about solving it without using mod 7?

you could always do this:

(3^0)/7= remainder 1
(3^1)/7= remainder 3
(3^2)/7= remainder 2
(3^3)/7= remainder 6
(3^4)/7= remainder 4
(3^5)/7= remainder 5

and just keep going and find a pattern.
it's pretty much the same thing as using mods, but it doesn't look as scary.

Thanks. It almost the same, but is better for people that don't know mods... that's what I wanted.... thanks so much!

Using Fermat's Little Theorem, your life becomes a lot easier :lol: :
3^6 congruent to 1 (mod 7)
Raising both sides to the 4264th power,
3^25584 congruent to 1 mod 7
Multiplying both side by 3^2,
3^25586 congruent to 3^2 (mod 7) congruent to 2 (mod 7)
The answer is 2.

I would agree with tim1234133 to get 2 mod 7 so remainder is 2


group theory
superior algebra
superior algebra unsolved

59.  A cyclic group of order 15 has an element $ x$ such that the set $ \{x^3, x^5, x^9\}$ has exactly two elements.  How many elements are in the set $ \{x^{13n}: n\mbox{ is a positive integer} \}$?

60.  If $ S$ is a ring with the property that $ s\equal{}s^2$ for each $ s$ in $ S$, which of the following must be true?

I. $ s\plus{}s\equal{}0$ for each $ s$ in $ S$.

II.  $ (s\plus{}t)^2 \equal{} s^2 \plus{} t^2$ for each $ s,t$ in $ S$.

III.  $ S$ is commutative.

59. Obviously x is not e, so it cannot be $ x^3\equal{}x^5$ or $ x^5\equal{}x^9$. Then it must be $ x^3\equal{}x^9$, ie
$ x^6\equal{}e$. Thus x must be of order 3 and $ x^{13n}$ can be x, $ x^2$ or e, which are 3 diff. elements.

60.  I. $ s\plus{}s\equal{}(s\plus{}s)^2\equal{}s^2\plus{}s^2\plus{}s^2\plus{}s^2\equal{}s\plus{}s\plus{}s\plus{}s$, so it must be $ s\plus{}s\equal{}0$
II. $ (s\plus{}t)^2\equal{}s\plus{}t\equal{}s^2\plus{}t^2$
III. $ (s\plus{}t)^2\equal{}s^2\plus{}st\plus{}ts\plus{}t^2\equal{}s^2\plus{}t^2$. thus we have $ st\plus{}ts\equal{}0$, but $ st\plus{}st\equal{}0$, so
 it follows that $ st\equal{}ts$\,.
All claims are true.