For $k > 1$ we have

${\textstyle a_{k}+b_{k}+c_{k}+d_{k}= (a_{k-1}-b_{k-1})+(b_{k-1}-c_{k-1})+(c_{k-1}-d_{k-1})+(d_{k-1}-a_{k-1}) = 0.}$

Hence $a+b+c+d = a_{1997}+b_{1997}+c_{1997}+d_{1997}= 0$, i.e. $-a = b+c+d$. In other words

$(1) \;\; bc-ad = bc+d(b+c+d) = (b+d)(c+d),$

$(2) \;\; bd-ac = bd+c(b+c+d) = (b+c)(c+d),$

$(3) \;\; cd-ab = cd+b(b+c+d) = (b+c)(b+d).$

Let $x = \mid b+c\mid , \; y = \mid b+d\mid $ and $z = \mid c+d\mid .$ Then according to (1)-(3)

$(4) \;\; \mid bc-ad\mid \: = yz,$

$(5) \;\; \mid ac-bd\mid \: = xz,$

$(6) \;\; \mid ab-cd\mid \: = xy.$

Assume that all three integers $\mid bc-ad\mid , \; \mid ac-bd\mid $ and $\mid ab-cd\mid $ are prime (*). Then by the equations (4)-(6) $x=y=1, \; x=z=1$ or $y=z=1$, which implies that $xy=1, \; xz=1$ or $yz=1$ respectively. This contradiction shows that the assumption (*) is false.

[quote="Hawk Tiger"]What does SIMO stands for?[/quote]
Singapore Mathematics Olympiad.


Find (if exists) $ a_1,a_2,...,a_{19}\in\mathbb{N}$ such that every natural number $ \leq 999999$ can be expressed as the sum of $ a_{i_1},...,a_{i_k}$ with $ k\leq 19$ and $ i_1,...,i_k\leq 19$

There are $ 2^{19}\minus{}1$ non empty subsets $ A\equal{}(a_{i_1},...,a_{i_k})$, therefore there are no more different $ 2^{19}\minus{}1$ sums $ S(A)\equal{}\sum_{a_i\in F} a_i$. But $ 2^{19}\minus{}1<999999$.

[quote="Rust"]There are $ 2^{19} \minus{} 1$ non empty subsets $ A \equal{} (a_{i_1},...,a_{i_k})$, therefore there are no more different $ 2^{19} \minus{} 1$ sums $ S(A) \equal{} \sum_{a_i\in F} a_i$. But $ 2^{19} \minus{} 1 < 999999$.[/quote]

I think you're wrong : it is possible to have for example $ a_1\plus{}a_1\plus{}a_1\plus{}a_1\plus{}a_1\plus{}a_1\plus{}a_1$ $ \plus{}a_1\plus{}a_1\plus{}a_1\plus{}a_1\plus{}a_1\plus{}a_1\plus{}a_1\plus{}a_1\plus{}a_1\plus{}a_1\plus{}a_1\plus{}a_1$ and so we have more than $ 2^{19}$ possibilities.

[quote="pco"][quote="Rust"]There are $ 2^{19} \minus{} 1$ non empty subsets $ A \equal{} (a_{i_1},...,a_{i_k})$, therefore there are no more different $ 2^{19} \minus{} 1$ sums $ S(A) \equal{} \sum_{a_i\in F} a_i$. But $ 2^{19} \minus{} 1 < 999999$.[/quote]

I think you're wrong : it is possible to have for example $ a_1 \plus{} a_1 \plus{} a_1 \plus{} a_1 \plus{} a_1 \plus{} a_1 \plus{} a_1$ $ \plus{} a_1 \plus{} a_1 \plus{} a_1 \plus{} a_1 \plus{} a_1 \plus{} a_1 \plus{} a_1 \plus{} a_1 \plus{} a_1 \plus{} a_1 \plus{} a_1 \plus{} a_1$ and so we have more than $ 2^{19}$ possibilities.[/quote]

May be thats not what he is looking because you need 1 to be present. So, just only 1 is enough to represent all numbers.

Only 19 numbers are allowed be added together. Note the condition $ k \leq 19$.

[quote="specialformath"]Note that $ k\leq 19$ and $ i_1\neq i_2 \neq ... \neq i_k$ :)[/quote]

I suppose that $ i_1\neq i_2 \neq ... \neq i_k$ means $ i_m\neq i_n$ $ \forall m\neq n\in[1,19]$.

Then (with this new information), Rust's demo is OK and no solution exists.


geometry
3D geometry
dodecahedron
power of a point
\/closed

One of the best parts of the revamped site are the Realplayer clips on the left side of the screen. They really help to explain some things, such as the area of a dodecahedron formula.

Suggestion: Would it be possible to add some Realplayer clips that explain some of the geometry theorems like the Power of a Point and other theorems that we might have missed in geometry class but might come in handy? Thanks.

Real player? I thought they were in flash...


How does one derive the general form of the quadratic equation?

[quote="sagar_astakala"]DEar tjhance

The question is to deriving the quadratic equation. But you have given the derivation to find the solutions of the equation. :(

The quadratic equation with roots A and B is

x^2 - x(A+B) + AB = 0

thank you[/quote]

yea....but it was pretty clear that interval meant quadratic formula

...

also nice with the multiplying by 4a...i was wondering how to get around the |a| stuff...

[quote="Altheman"]
yea....but it was pretty clear that interval meant quadratic formula
[/quote]
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=130929


At sunrise, two drivers started driving (in a straight line) towards each other, each at their own constant speed. One started at pt. A and went towards B while the other did the opposite. They met at noon but did not stop; each continued driving maintaining their speed and direction. The first driver arrived at pt. B at 4:00 p.m., and the other driver arrived at Pt. A at 9:00 p.m. At what time did the sun rise that day

Post your actual solution or show your analysis. I don't have the answer on me at the moment.

[hide]
Call the fraction of the total distance covered by the first and second person $A$ and $B$, respectively. Let $x$ be the number of hours between midnight and sunrise.

$(a+b)(12-x)=1$
$(a)(12+4-x)=1$
$(b)(12+9-x)=1$

Just solve this set of equations I think, unless I messed up somewhere.

[/hide]


Find the minimum distance from points of:

1) y equals the negative square of x (y= - (x squared)) to the line

2) 4x + 3y - 8 = 0

It is well known that the distance from a point $ (m,n)$ to a line $ Ax\plus{}By\plus{}C\equal{}0$ is $ \frac{|Am\plus{}Bn\plus{}C|}{\sqrt{A^2\plus{}B^2}}$. All points on $ y\equal{}\minus{}x^2$ can be expressed as $ \left(x,\minus{}x^2\right)$. The distance from this point to the line is $ \frac{|4x\minus{}3x^2\minus{}8|}5\equal{}\frac{|3\left(x\minus{}\frac23\right)^2\plus{}\frac{20}3|}5$. The quadratic in the numerator doesn't have any real roots, which means it never equals $ 0$. Thus, the minimum for the quadratic is $ \frac{20}3$, and our answer is $ \frac43$.


ratio
geometry
similar triangles
geometry solved

Let ABC be an acute triangle with circumcentre O. Let P and Q be points on AC and BC respectively such that AP/PQ = BC/AB and BQ/PQ = AC/AB.

Prove that O, P, Q and C are concyclic.

I couldn't do this one, am I missing something?

See http://www.mathlinks.ro/Forum/viewtopic.php?t=30195 . By the way, it's not from the Austrian-Polish 2002.

  darij

dear arne i finally solved it.but as u know imnot so well in english.
and i need also the picture of it to show u what i have don.
to solve it i used inversion . and petalomy (batlamios).the inversion was C(O,R).
but if u really want the complete solution,i will try to write it for u as soon as i can ;) .

Just try to write it here, preferably without too many typos, to make life easier :)
I am sure we will understand (even without a diagram).

of course i will write it as soon as i can.
do u u self try to solve it ?
i want to know what did u do during u thinking.

I have been playing with ratios, trying to construct all sorts of similar triangles etc, trying to use spiral similarities; I also sort of managed to locate the centre of the circle passing through O, P, Q and C (assuming that O, P, Q, C are concyclic :() and that there is just one such circle but I didn't make a lot of progress.

im ashame to say it arneh:
do u know what do we call theline wich is tha bigest chord of acircle :(

thanks alot arneh. now i meet u to solve my new problem ;)


probability
Ross Mathematics Program
calculus
graph theory
combinatorics theorems
combinatorics

Hello. I am starting in combinatorics and I am looking for some books suitable to an olympic training.
I found:

"Principles and Techniques in Combinatorics" by Chen Chuan-Chong and Koh Khee-Meng
"A Path to Combinatorics for Undergraduates" by Titu Andreescu and Zuming Feng

About Graph Teory:

"Combinatorics and Graph Theory" (Undergraduate) by John M. Harris, Jeffry L. Hirst, and Michael J. Mossinghoff

Do you think that this book could be helpful? Is it too difficult for a high-school student?
Do I need any other book?

Thanks a lot. :)

Well i dont know whether this topic should be here at all :maybe:

I can assure you that path to combinatorics by Titu Andreescu and Zuming Feng can be read on your own :) but its a nice book for starters (Esp when he starts talking abt his favourite ESPN Football commentator Lee Croso :D)

Douglas West's book on graph theory is really good as is Reinhard Diestel's. You could look these books up sometime. I believe Diestel is available online.

[quote="SML"]Could you tell me any book about probability? :?:[/quote]

Sheldon Ross, A first course in probability
It requires calculus


calculus
derivative
absolute value
calculus computations

a particle moves along the y-axis so it's velocity is v(t) = 1-arctan(e^x).
Is the speed increasing or decreasing at t=2?

Can you provide me the explanations too?

$ a(t)\equal{}v'(t)\equal{}\minus{}\frac{e^{x}}{1\plus{}e^{2x}}$.  Then $ v'(2)\equal{}a(2)\equal{}\minus{}\frac{e^{2}}{1\plus{}e^{4}}$.  Since the derivative is negative, it is decreasing.

Isn't the velocity is the speed's derivative?

That sentence doesn't really make sense, but no.  Velocity is the derivative of position, acceleration is the derivative of velocity, and speed is the magnitude of the velocity.

[quote="JRav"]$ a(t) \equal{} v'(t) \equal{} \minus{} \frac {e^{x}}{1 \plus{} e^{2x}}$.  Then $ v'(2) \equal{} a(2) \equal{} \minus{} \frac {e^{2}}{1 \plus{} e^{4}}$.  Since the derivative is negative, it is decreasing.[/quote]
Since the derivative is negative, the velocity is decreasing. But that doesn't mean anything about the speed.

What matters is that the velocity and acceleration share the same sign. If the acceleration was negative, the particle could still be speeding up if the velocity was negative.

So, to find out if the particle is increasing or decreasing in speed, you have to compare the signs of the velocity and $ t \equal{} 2$ and the acceleration (derivative of velocity) at $ t \equal{} 2$.

$ v(2) \equal{} \minus{} .4362783$
$ v'(2) \equal{} \minus{} .1329011$

Since the velocity and acceleration are both negative, the particle is actually [b]speeding up[/b]. I think that's right at least. My explanation is correct, I'm not sure if my actual calculations are, though. :P

If you're calculations are correct, then you should be right and speed is increasing.  I misread the question.  However, since the velocity is decreasing, that indicates that it is turning around but increasing the rate at which it does so.

[quote="cdymdcool"]if a'(t) is negative then it's concave down, and v(t) is negative than it should be decreasing!
How is it increasing?[/quote]
You're correct, the [b]velocity[/b] is decreasing. But speed is the [b]absolute value[/b] of velocity. So, the velocity is getting more negative, but the speed is growing more positive.

[quote="cdymdcool"]SO you are saying that speed will never be negative, since it's the absolute value of the velocity?[/quote]
Does a car's speedometer show negative speeds?

(The answer is you are correct- speed cannot be negative.)

Let $ x(t)$ be position, $ v(t)$ be velocity, $ s(t)$ be speed, and $ a(t)$ be acceleration.

Then:

$ v(t) \equal{} x'(t)$

$ a(t) \equal{} v'(t) \equal{} x''(t)$

And

$ s(t) \equal{} |v(t)|$, which means that:

$ s'(t) \equal{} \frac{v(t) \cdot v'(t)}{|v(t)|}$, therefore, to determine the sign of the rate of change in the speed, determine the product of the signs of $ v(t)$ and $ v'(t)$.


$2n-1$ points are marked on a line, with each point coloured black or white. For each point we add the number of white points to the right of that point and the number of black points to the left of the that point and doing this for each point, we obtain $2n-1$ such numbers. Suppose in this set of numbers, exactly one number occurs an odd number of times. Find all possibilities for this number.

any ideas on this problem people?
i am terrible at combinatorics and have no idea.
a solution would be very much appreciated.

Suppose you have a white dot followed immediately by a black one (or vice versa). What happens if you remove both?

EDIT:I tried to post the proof but I found a mistake, sorry.

Suppose you have two consecutives points of different colors, then if you switch them, the parity of the number of times each number appears doesnt change. So you can suppose there are $k$ black points on the right and $n-k$ white points on the left.

Uh, ok. Slightly different than mine...

My idea was: if you remove both, then you delete the same value twice (and parity doesn't change) and decrease every surviving value by 1. Every situation can be reduced to k dots of the same colour. This situation fulfills the hypotheses iff k=1 etc...


There are $2n$ different numbers in a row. By one move we can interchange any two numbers or interchange any $3$ numbers cyclically (choose $a,b,c$ and place $a$ instead of $b$, $b$ instead of $c$, $c$ instead of $a$). What is the minimal number of moves that is always sufficient to arrange the numbers in increasing order ?

1) One can easy prove that it is possible to properly place at least two numbers. So we need at most $n$ interchanges. :D
2) Consider permutation $(2n,2n-1,n+1,n,...,2,1)$. Divide row into two parts: first $n$ numbers and second $n$ numbers. Then there are $2n$ numbers, which are not in parts they supposed to be. Moreover, each turn we can place at most two numbers into appropriate part. So we need at least $n$ turns.  :D


Using each of the eight digits 1, 3, 4, 5, 6, 7, 8 and 9 exactly once, a three-digit number A, two two-digit numbers B and C, B < C, and a one digit number D are formed. The numbers are such that A + D = B + C = 143. In how many ways can this be done?

1. D must be 13x. 6 choices of x. (4,9), (5,8), (6,7). The other in the pair will be A.
2. 4 choices left for the choice of the ones digit one of B/C, this forces the ones digit of the other number to be the other in the pair.
3. 2 choices left for the tens digit to complete one number. The other tens digit is forced to be the other in the pair. All numbers are seen to be used 1. B/C will be forced as the smaller number will be B.

Thus 48 ways.


MATHCOUNTS
induction
trigonometry
LaTeX
AMC 10
AMC 8

I'm thinking about making a mock AMC 10 over winter break. Maybe easy enough as practice for beginners. I won't include problems which obviously exceed AMC 10 level. Anyone interested?

I'll definitely do this.

 :D Hope I can get around 130's. :)

I'm glad that some people are interested. If enough people are, it will happen after winter break, probably the weekend after it. Since, I am going on a short vacation, as many probably are. Besides that, the test will be standard, with standard rules and problem formats. If you aren't familiar with rules, go to the AMC forum and check them out.

Just curious, are you going to make the questions or are you going to take them out from the math contests?

I did one (MSC) and it wasn't easy to create a problem. Of course, that's why math is so fun though. :P

I will attempt to the best of my ability to make unique questions, but it depends on my time constraint. I may have to look at old problems and derive my problems from those. Of course, they won't be exactly alike, but the idea behind them might be the same.

[quote="white_horse_king88"]I'm thinking about making a mock AMC 10 over winter break. Maybe easy enough as practice for beginners.[/quote]

This is, of course, an excellent idea. More power to you for working up a practice test.

But there seems to be an inherent contradiction in the plan. A mock AMC 10 would a test above the level of AMC 8 and MATHCOUNTS, that is above the level of this Getting Started Forum. Or a test that is easy enough to be practice for beginners would be a test at the AMC 8 level.

I suppose you could draft the practice test, and then PM a few people to figure out what level it is at. An AMC 8 level practice test would be excellent for this forum. An AMC 10 level practice test would be excellent too--but for the Intermediate Forum.

Ok, I'll try to make it more suitable for AMC 8 then. Thanks for the tips! I'll send it to some people when I'm done for proof-reading and some revisions if necessary.

I'm not fully agreed with tokenadult.

In your words, the AMC-10 is in Intermediate. If so, then means I, who is taking AMC-10 and usually get around 100-112 or around that score, should be in Intermediate. But in the Intermediate Forum, the usual materials discussed are from books like Art and Craft of the Problem Solving or Problem Strategies (I think I saw this book on there, not quite sure) which is certainly above in the level of AMC-10.

Now, I'm not saying you're wrong but I'm just wondering, then this AMC-10 should be in Intermediate Forum and only AMC-8 is eligible to be here? That is a matter that's hard to determine because I remember seeing the posts concerning the Mathematical Induction in this forum, and if AMC-10 is in Intermediate, then I should say that proof should be intermediate and many of the users should be more tended to Intermediate then what they are in Getting Started.

Again, this matter is always hard to discuss due to the high intelligence of the users.

[quote="Silverfalcon"]But in the Intermediate Forum, the usual materials discussed are from books like Art and Craft of the Problem Solving or Problem Strategies (I think I saw this book on there, not quite sure) which is certainly above in the level of AMC-10.[/quote]

Yes, the problem is that most of the problem-solving forums here get too many problems posted to them that are too hard.  :(  That is why we have the odd situation here in Getting Started of challenge problem threads in which many of the people answering problems are undergraduate math majors or college graduates. I need to do my part to introduce more EASY problems to this forum--which will be inviting to the many registered users of AoPS who never post, because they don't find problems at their beginning level even in this Getting Started Forum.

This issue of having challenge problems that are too hard on most forums arises for a happy reason: everyone who participates here regularly rapidly gets better at solving problems. So after a while, people forget how difficult problems are to true beginners. They hang out on the forum they are used to hanging out on, but they post harder and harder problems the longer they stay. I'm happy to have anyone post problems to this Getting Started Forum; all I ask is that they set the problems they pose to about the level of AMC 8 or MATHCOUNTS problems, which presuppose no proofs, and presuppose very little curriculum content beyond what is found in a prealgebra course in United States schools. (AMC 8 and MATHCOUNTS are both about at a sixth-grade level for problem-solving contests in China, where problem-solving begins at a young age.)

I've conferred with Mr. Rusczyk about this, and the general idea is to make sure there are plenty of problems at ALL levels, but especially at the very easiest levels, to welcome people to this site. There is certainly an honored place for problems that involve geometric proofs, or for problems that involve trigonometry, or for problems that get deeply into number theory. But that place would be at the Intermediate Forum level, or at the Pre-Olympiad level. AMC 10 is certainly Intermediate (the word means "in between" after all) in level between the AMC 8 and AMC 12, and thus only the AMC 8 is well characterized as Getting Started level.

I appreciate the participation of all the young people on the forums who are so eager to advance their mathematical abilities to posing problems to one another. I would simply point out that the population of lurking members of AoPs is much larger than posting members of AoPS. (The typical member has a post count of 0.) My advantage as a busy father who didn't study math as an undergraduate is that I still remember what it is like to not be able to solve many mathematical problems--I've never had much practice. I am about at the AMC 8 or MATHCOUNTS level personally--I'd probably score about the same on the AMC 10 as you, maybe lower. I will try to scrounge up time to post more problems at an EASY Getting Started level, and I will continue to "promote" problems from here to the Intermediate Forum, which also needs more easy problems. I just have to find the time to open up some of the books I have (in English and in Chinese) and transcribe and adapt problems from them.

Good luck on your AMC 10.

Thank you, tokenadult.

Yes, as a moderator myself, I'll try to pick out the easy problems. I remember that when I came here, I felt "outside of the whole."

I'll make a thread with the links to the good problems within the Art of Problem Solving.

To whitehorseking88, sorry for the offtopic discussion!

Oh I wouldn't be concerned by being good or not. This won't be AMC 10 level, for the most part, so I wouldn't worry at all. Just take it for experience, and make sure to ask questions about problems you don't understand.

Oh yeah, can someone please tell me how to make a latex file on Adobe Reader? I don't know how to create a file which people can access by a link. Would it be ok if I sent the actual problems to someone who knows how to do this? And they create the file?

[quote="tokenadult"]About the LaTeX file, I think this site's [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php]LaTeX section[/url] has a lot of information about free file-conversion utitilies. I have to learn all that stuff this year. Let's see together what works to make a readable file for everyone.[/quote]

Personally I don't like GO2PDF, which is what's there, because it creates a line of text at the bottom of every page. I recommend CutePDF, which is also a virtual printer.

And you can't make files with Adobe Reader, you need to actually buy Adobe Acrobat.

Ok, this is done. Anyone who wants to take it can. It'll be next week because AMC I is this week.


AMC
AIME
geometry
ratio
trigonometry
absolute value
number theory

hi, here are some dropped AIME problems that never made the actual test..enjoy

note..when i say a number $abc$, i mean $a,b,c$ are the digits.

$(1.)$ Determine the number of five digit integers $37abc$ (in base $10$), such that $37abc, 37bca, 37cab$ is divisible by $37$.

$(2.)$ Determine the value of $10(x+y)$, given that $[x]+[y]+y=43.8$, and $x+y-[x]=18.4$, where $[$ $]$ means greatest integer less than or equal to $x$ or $y$.

$(3.)$ In the $3 \times 3$ array shown at the bottom, each letter stands for a distinct non zero integer. Each of the three digit numbers $abc$, $def$, $ghi$, $adg$, $beh$, $cfi$ and $aei$ is divisible by $11$. what is the maximum value of the three digit number $ceg$?

a b c
d e f
g h i

$(4.)$ In $\triangle{ABC}$, let $D$ and $E$ be points on sides $AC$ and $BC$, respectively, so that $DE$ is parallel to $AB$. Let $P$ be the intersection of $AE$ and $BD$, and assume that the area of $\triangle{ABP}=36$ and the area of $\triangle{EDP}=25$. Find the area of $\triangle{ABC}$

$(5.)$ Let $x$ and $y$ be real numbers which differ in absolute value and satisfy the equations:

$x^3=13x+3y$
$y^3=3x+13y$

Find $(x^2-y^2)^2$

$(6.)$ For each arrangement of the ten numbers $1,4,7,10,13,16,19,22,25,28$ around the circle, let $N$ denote the largest of the ten sums obtained by adding each number to the sum of its neighbors. What is the smallest value of $N$ that can be obtained?

$(7.)$ Find $x^2-yz$ if $x,y,z$ are distinct positive integers such that $y^2-zx=-103$ and $z^2-xy=22$

$(8.)$ In $\triangle{ABC}$, the diameter $AB=30$ and let $AC$ and $BC$ be two secants of the triangle intersecting the circle at points $D$ and $E$ respectively. If $AD=AC/3$ and $BE=BC/4$, what is the area of $\triangle{ABC}$?

$(9.)$ Define a positive integer $N$ to have the reversed sum property if there is a three digit number $abc$ with neither $a$ nor $c$ equal to zero and such that $abc+cba=N$. How many positive integers have the reversed sum property.

$(10.)$ Point $P$ is located inside $\triangle{ABC}$ so that $\angle{PAB}=\angle{PBC}=\angle{PCA}$. The sides of the triangle are $AB=13, BC=14, CA=15$. The tangent of $\angle{PAB}$ may be expressed in the form $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Compute $m+n$.

$(11.)$ Let us call a positive integer $n$ "fortunate" if the sum of its digits is a multiple of $7$, and "superfortunate" if it is fortunate and if none of the numbers $n+1,n+2,...,n+12$ is fortunate. Find the smallest superfortunate number.

$(12.)$ Let $S$ be a subset of $[1,2,3,....,1000]$ so that if $a$ and $b$ are any (not necessarily distinct) elements of $S$, then their product $a \cdot b$, is not an element of $S$. What is the maximum number of elements of $S$.

$(13.)$ Some of the squares of a $3 \times 3$ board are colored black in such a way that in each row and in each column there is at least one black square. Regarding symmetries of the same coloring as distinct, find the total number of ways in which one can color the board.

$(14.)$ In a round robin tournament, occasionally a cyclic $3-$set occurs, that is, a set of $[a,b,c]$ of teams where $a$ beats $b$, $b$ beats $c$, and $c$ beats $a$. If $23$ teams play a round robin (without any ties), what is the greatest number of cyclic $3-$sets that can occur?

$(15.)$ Define a normal series to be an expression $a_{1} \pm a_{2} \pm...\pm a_{n}$, where each $\pm$ is either a $+$ or a $-$, and the operations are performed in the normal order from left to right. Define a reverse series to be an expression $R: a_{1} \pm 1_{2} \pm a_{3} \pm...\pm a_{n}$, where again each $\pm$ is either a $+$ or a $-$, but now the operations are grouped from left to right. For instance,

$R: 3-4-6+5=3-(4-(6+5))=3-4+6+5=10$

As just illustrated, each reverse series is equivalent to a normal series with the same terms in the same order, but with some of the signs changed. Conversely, each normal series has an equivalent reversed series. If the normal series $a_{1} \pm a_{2} \pm a_{3} \pm...\pm a_{1000}$ has $600$ minuses among its $999$ signs, at most how many minuses does the equivalent reverse series have?

[quote="mysmartmouth"]Is the answer t number 1 [hide]12?[/hide]

I'm not sure at all, though.  If it's correct, I'll post a solution.[/quote]

No, that's not correct.

Here's the answers I got so far:

[hide]1) 28
2) 274
4) 396
5) 133
7) 172
8) 540
9) 170
10) 463

All the other ones I'm really shaky about :( [/hide]

If anyone would like me to post the solutions to these I will

[quote="Iversonfan2005"][quote="mysmartmouth"]Can you post a solution to number 1?[/quote]

Let $p=abc$, $q=bca$, and $r=cab$. Then $10p-q=999a, 10q-r=999b, 10r-p=999c$, and $999 \equiv 0 \mod{37}$. With this, we see that if any of $p, q, r$ are divisible by $37$, the other two variables are as well. So if the first two digits of a five digit number are $37$, then we can find out how many numbers between $000$ and $999$ are multiples of $37$ (including $000$ itself), and there are $\boxed{28}$.[/quote]
$bca$ wasnt included in the problem, it says $acb$

sorry, i wrote it wrong by accident on the problem :blush:

[quote="mysmartmouth"]Hmmm, it seemed fishy to me.  And is number 3 written correctly?[/quote]

yes

Here are the solutions to the ones I know how to do:

[hide="1"]Let $p=abc$, $q=bca$, and $r=cab$. Then $10p-q=999a, 10q-r=999b, 10r-p=999c$, and $999 \equiv 0 \mod{37}$. With this, we see that if any of $p, q, r$ are divisible by $37$, the other two variables are as well. So if the first two digits of a five digit number are $37$, then we can find out how many numbers between $000$ and $999$ are multiples of $37$ (including $000$ itself), and there are $\boxed{28}$.[/hide]

[hide="2"]If we let $x=[x]+a$ and $y=[y]+b$, where $a$ and $b$ are the decimal parts of $x$ and $y$ respectively, then we substitute into the equations:

$[x]+[y]+[y]+b=43.8$
$[x]+a+[y]+b-[x]=18.4$

Then $b$ must be $.8$ and $a$ must be $.6$.

Then the equations become:

$[x]+2[y]=43$
$[y]=17$

So $[y]=17$ and $[x]=9$, so $x=9.6$ and $y=17.8$, and $10(x+y)=\boxed{274}$[/hide]
[hide="4"]
Let $PX$ be the altitude of $\triangle{PBA}$, and let $YP$ be the altitude of $\triangle{DEP}$. Because $\triangle{DEP} \sim \triangle{PBA}$, the ratio of their sides or altitudes is the square root of the ratio of their areas.

So $\frac{DE}{AB}=\frac{YP}{PX}=\sqrt{\frac{25}{36}}=\frac{5}{6}$

Let $AB=6x$ and $DE=5x$, and $PX=6y$, and $YP=5y$. Because $\triangle{ACB} \sim \triangle{DCE}$, we know that the ratio of their sides are also $\frac{5}{6}$, and the ratio of their areas is $\frac{25}{36}$. If $CZ$ is the altitude of $\triangle{DCE}$, then $CX$ is the altitude of $\triangle{ACB}$. Because $\frac{CZ}{CX}+\frac{5}{6}$, and by letting $CZ=a$, and knowing that $CX=CZ+YX$, and $YX=5y+6y=11y$, we have that:

$\frac{CZ}{CX}=\frac{a}{11y+a}=\frac{5}{6} \Longrightarrow a=55y=CZ$.

Then, we have that $[\triangle{DEP}]=\frac{5x \cdot 5y}{y}=25 \Longrightarrow xy=2$. Then $[\triangle{ACB}]=\frac{6x \cdot 66y}{2}=\boxed{396}$[/hide]

[hide="5"]$(1) x^3=13x+3y$
$(2) y^3=3x+13y$

$(1)-(2)$ gives $x^2-y^3=10x-10y=10(x-y)=(x-y)(x^2+xy+y^2) \Longrightarrow (x+y)^2-xy=10$

$(1)+(2)$ gives $x^3+y^3=16x+16y=16(x+y)=(x+y)(x^2-xy+y^2) \Longrightarrow (x+y)^2-3xy=16$

$(3) (x+y)^2-xy=10$
$(4) (x+y)^2-3xy=16$

$(3)-(4)$ gives $2xy=-6 \Longrightarrow xy=-3$. Plugging into the simplified form of $(1)-(2)$, which was $x^2+xy+y^2=10$, we get $x^2+y^2=13$.

Because $(x^2-y^2)^2=(x^2+y^2)^2-4(xy)^2=\boxed{133}$[/hide]
[hide="7"]
$(1) x^2-yz=?$
$(2) y^2-zx=-103$
$(3) z^2-xy=22$

$(3)-(2)$ gives us (after factoring) $(x+y+z)(z-y)=5^3$. If we let $x+y+z=25$ and $z-y=5$, then through trial and error, $y=3, z=8, x=14$, and $x^2-yz=\boxed{172}$[/hide]
[hide="8"]
After drawing in $AE$ and $BD$, $\triangle{DAB}$ and $\triangle{AEB}$ are right, since their hypotenuses are the diameter of the circle. Then, if we let $EB=y$, then $CE=3y$, and if we let $AD=x$, then $DC=2x$. Since $AE$ and $DB$ are the altitudes of $\triangle{ACB}$, and through Pythagorean theorem:

$DB^2=30^2-x^2=(4y)^2-(2x)^2$
$AE^2=30^2-y^2=(3x)^2-(3y)^2$

Then, using the first equation, $x^2=\frac{16y^2-30^2}{3}$ and using the second, $x^2=\frac{30^2+8y^2}{9}$. Solving, we have $y=\sqrt{90}$ and $x=\sqrt{180}$, and $DB=\sqrt{720}$.

$[\triangle{ACB}]=\frac{3x \cdot DB}{2}=\boxed{540}$[/hide]

[hide="9"]The number $(abc)$ is in the form $100a+10b+c$, the number $(cba)$ is in the form $100c+10b+a$, thus $N=101(a+c)+20b$. Thus $1\leq a+c\leq 18$ and $0\leq b\leq 9$, there are $17$ values for $a+c$ and $10$ values for $b$, thus there are $\boxed{170}$ distinct numbers.

Remark: This problem is very similar to the one gauss202 solved on this forum.[/hide]

[hide="10"]Let $\angle{PAB}, \angle{PBC}$, and $\angle{PCA}=\theta$ and let $PA=x, PB=y$, and $PC=z$. Using law of cosines, we have:

$y^2=x^2+13^2-2(13)(x)(\cos{\theta})$
$x^2=z^2+15^2-2(15)(z)(\cos{\theta})$
$z^2=y^2+14^2-2(14)(y)(\cos{\theta})$

Adding all three equations and factoring and rearranging, we have:

$2\cos{\theta}(13x+14y+15z)=13^2+14^2+15^2=590$

Also, $[\triangle{ABC}]=[\triangle{APB}]+[\triangle{PCB}]+[\triangle{PCA}]=\frac{1}{2} \sin{\theta}(13x+14y+15z)$

To find $[\triangle{ABC}]$, we use Heron's formula and get $[\triangle{ABC}]=84$. Now, we have:

$(1) 2\cos{\theta}(13x+14y+15z)=13^2+14^2+15^2=590$
$(2) \frac{1}{2} \sin{\theta}(13x+14y+15z)$

$(2)/(1)$ yields $\frac{\tan{\theta}}{4}=\frac{84}{590}$ and $\tan{\theta}=\frac{168}{295}=\frac{m}{n}$, thus $m+n=\boxed{463}$
[/hide]

If anyone has solutions to the other problems, please post them. Unfortunately, the book I got these problems from have only the answers, but not the solutions.

BTW, the answers are here just in case you want to check your answers before posting solutions:

[hide]3.734
6.48
11.993
12.969
13.265
14.506
15.799[/hide]

Out of curiosity, where did you get these problems?

[quote="chess64"]Probably one of the contest books published by MAA.[/quote]

yes, in fact, contest problem book V

anyone... :maybe:, I really would like to know how to do the ones that are unanswered :wink:

all 15 problems have been posted in separate threads in HSM, I just created the related post collection

[url=https://artofproblemsolving.com/community/c3734496_198388_15_dropped_aime_problems_]here[/url]

enjoy  / start solving

more USA mocks [url=https://artofproblemsolving.com/community/c2439870_usa_mocks]here [/url] (also mentioned [url=https://artofproblemsolving.com/community/c5h3200761p29225908]here[/url])
more AIME mocks [url=https://artofproblemsolving.com/community/c2439872_aime_mocks]here[/url],
more User-created contests  [url=https://artofproblemsolving.com/community/c2435719_user_created_contests]here[/url] (contains all above)


We call a number [i]perfect[/i] if the sum of its positive integer divisors(including $1$ and $n$) equals $2n$. Determine all [i]perfect[/i] numbers $n$ for which $n-1$ and $n+1$ are prime numbers.

An odd perfect number does not fit since $2| n + 1$ and $n = 1$ is not perfect.
For $p > 2$ we have
\begin{eqnarray*} 2^{p - 1} \cdot \left( 2^p - 1 \right) - 1 & \equiv & ( - 1 )^{p - 1} \cdot \left( ( - 1 )^p - 1 \right) - 1\\ & \equiv & 1 \cdot \left( - 2 \right) - 1\\ 2^{p - 1} \cdot ( 2^p - 1 ) - 1 & \equiv & 0 ( \mbox { mod } 3 ) \end{eqnarray*}
But $2^{p - 1} \cdot ( 2^p - 1 ) - 1 \geq 27$ is supposed prime. Impossible.
The case $p=2$ gives the desired solution, which is $n=6$ [corrected a slight mistake]

Quite same aprroach is the following :
The number n will have the form $2^{k-1}(2^k-1)$  where $2^k-1$ is a prime number .from the problem $n-1,n+1$ are primes so  $3|n$ .So $3|(2^{k-1}(2^k-1))$  so
$2^k-1=3$ so k=2 and n=6 is the unique solution

If n=2k+1 then one of n-1 or n+1 isnt a prime number.So 2 divides n.If 3 doesnt divide n then one of n-1 or n+1 isnt a prime number.So 3 divides n.Hence 6 divides n.If n=6 we get a solution.Let n=6r(r>=2).Then 2n=12r=3r+2r+r+2+3+6+1+n.Hence2+3+6+1=0.Contradiction.If some of the divisors match(for example 3r=6)we can check that there is no solution.

[quote="Hawk Tiger"]It has already been known that if n be a even perfect number then n must have the form :$2^{p-1}(2^p-1)$ where $p,2^p-1$both are primes.[/quote]

It's the Euler's theory  ;)

[quote="Hawk Tiger"]It has already been known that if n be a even perfect number then n must have the form :$2^{p-1}(2^p-1)$ where $p,2^p-1$both are primes.[/quote]
Since this solves 90% the problem, you have to prove that.

Yes ,you are right,freemind.I just think the others should have the chance to solve this one by themselves,So I didn't post the solution.

Theorem:
If $a$  is a perfect even number then there exist $n$  such that
$a=2^{n-1}(2^n-1)$

Proof :
Because $a$ is even it will have the form $a=2^{n-1}b$  where $b$ is an odd number .We have $s(b)=\frac{2^nb}{s(2^{n-1}}$  where $s(k)$ is the sum of divisors of $k$ .
So $2^n-1|b$  and $s(b)=b+\frac{b}{2^n-1}$  . From this
$\frac{b}{2^n-1}=1$  and the conclusion follows

Obviously n=6k, k is a positive integer.If k is bigger than 1 then the number 1+k+2k+3k is bigger than 6k=n and 1,k,2k,3k are divisors of n so it contradicts to n is perfect so k=1 and n=6.It is clear that 6 is perfect :D

Which is the correct and small solution? I dont understand

6 is the only solution.

If the number is bigger than 6, it's of form 6k, because primes are only 6k+1 and 6k-1. But 6k can't be perfect because 1,k,2k,3k,6k are its distinct divisors of 6k (k>1) and their sum is bigger than 12k, so k=1

Is this what you wanted?

many thank bugi this is very simle solution

The first perfect number is $n = 6,$ which satisfies the conditions, so assume that $n > 6.$ We claim that no perfect number $n > 6$ has both $n-1$ and $n+1$ prime.

Suppose otherwise; then, since $n-1$ and $n+1$ cannot be multiples of $3,$ $n$ must be a multiple of $3.$ Similarly, $n$ must be even since $n-1$ and $n+1$ must both be odd. Then, $n$ is a multiple of $6$. Since $n > 6,$ the numbers $n, \dfrac{n}{2}, \dfrac{n}{3}, \dfrac{n}{6}$, and $1$ are pairwise [i]distinct[/i] and divisors of $n,$ so the sum of the divisors of $n$ is at least \[n + \dfrac{n}{2} + \dfrac{n}{3} + \dfrac{n}{6} + 1 = 2n+1.\] Thus, the sum of the divisors of $n$ is strictly greater than $2n,$ so $n$ cannot be perfect, a contradiction. We conclude that $n = \boxed{6}$ is the only such integer. $\square$

(Similar to Bugi's.)


a,b numere complexe a.i. $|a+b| \ge 2$ si $|a+b| \ge 1+|ab|$.Aratati ca :
$|a^{n+1}+b^{n+1}| \ge |a^{n}+b^{n}|$


[b]Problem[/b] .Let $ a,b,c$ be non-negative real numbers such that $ a\plus{}b\plus{}c\equal{}3$
Prove that

$ \frac{a}{\sqrt{b^2c^3\plus{}1}}\plus{}\frac{b}{\sqrt{c^2a^3\plus{}1}}\plus{}\frac{c}{\sqrt{a^2b^3\plus{}1}} \ge \frac{3\sqrt{2}}{2}$

 :)

[quote="vuthanhtu_hd"][b]Problem[/b] .Let $ a,b,c$ be non-negative real numbers such that $ a \plus{} b \plus{} c \equal{} 3$
Prove that

$ \frac {a}{\sqrt {b^2c^3 \plus{} 1}} \plus{} \frac {b}{\sqrt {c^2a^3 \plus{} 1}} \plus{} \frac {c}{\sqrt {a^2b^3 \plus{} 1}} \ge \frac {3\sqrt {2}}{2}$

 :)[/quote]
By Holder $ \left(\sum_{cyc}\frac {a}{\sqrt {b^2c^3 \plus{} 1}}\right)^2\sum_{cyc}a(b^2c^3 \plus{} 1)\geq(a \plus{} b \plus{} c)^3.$
Hence, it remains to prove that $ (a \plus{} b \plus{} c)^6\geq243abc(a^2c \plus{} b^2a \plus{} c^2b),$ which  follows from
$ a^2c \plus{} b^2a \plus{} c^2b\leq\frac {4}{27}(a \plus{} b \plus{} c)^3 \minus{} abc.$
The following inequality is also true.
Let $ k,$  $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a \plus{} b \plus{} c \equal{} 3.$ Prove that
\[ \frac {a}{\sqrt {b^2c^3 \plus{} k}} \plus{} \frac {b}{\sqrt {c^2a^3 \plus{} k}} \plus{} \frac {c}{\sqrt {a^2b^3 \plus{} k}} \ge \frac {3}{\sqrt {k \plus{} 1}}\]
What about the following?
Let   $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a \plus{} b \plus{} c \equal{} 3.$ Prove that
\[ \frac {a}{\sqrt {bc^4 \plus{} 1}} \plus{} \frac {b}{\sqrt {ca^4 \plus{} 1}} \plus{} \frac {c}{\sqrt {ab^4 \plus{} 1}} \ge \frac {3}{\sqrt2}\]

[quote="arqady"][quote="vuthanhtu_hd"][b]Problem[/b] .Let $ a,b,c$ be non-negative real numbers such that $ a \plus{} b \plus{} c \equal{} 3$
Prove that

$ \frac {a}{\sqrt {b^2c^3 \plus{} 1}} \plus{} \frac {b}{\sqrt {c^2a^3 \plus{} 1}} \plus{} \frac {c}{\sqrt {a^2b^3 \plus{} 1}} \ge \frac {3\sqrt {2}}{2}$

 :)[/quote]
By Holder $ \left(\sum_{cyc}\frac {a}{\sqrt {b^2c^3 \plus{} 1}}\right)^2\sum_{cyc}a(b^2c^3 \plus{} 1)\geq(a \plus{} b \plus{} c)^3.$
Hence, it remains to prove that $ (a \plus{} b \plus{} c)^6\geq243abc(a^2c \plus{} b^2a \plus{} c^2b),$ which  follows from
$ a^2c \plus{} b^2a \plus{} c^2b\leq\frac {4}{27}(a \plus{} b \plus{} c)^3 \minus{} abc.$
The following inequality is also true.
Let $ k,$  $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a \plus{} b \plus{} c \equal{} 3.$ Prove that
\[ \frac {a}{\sqrt {b^2c^3 \plus{} k}} \plus{} \frac {b}{\sqrt {c^2a^3 \plus{} k}} \plus{} \frac {c}{\sqrt {a^2b^3 \plus{} k}} \ge \frac {3}{\sqrt {k \plus{} 1}}\]
[/quote]
Dear arqady.Can you prove my new problem? It's stronger than old problem
[b]New problem[/b].Let $ a,b,c$ be non-negative real numbers such that $ a \plus{} b \plus{} c \equal{} 3$
Prove that
$ \sum \frac {a}{\sqrt {c(b^2 \plus{} c^2)^2 \plus{} 4}} \ge \frac {3\sqrt {2}}{4}$  :)

[quote="arqady"]
What about the following?
Let   $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a \plus{} b \plus{} c \equal{} 3.$ Prove that
\[ \frac {a}{\sqrt {bc^4 \plus{} 1}} \plus{} \frac {b}{\sqrt {ca^4 \plus{} 1}} \plus{} \frac {c}{\sqrt {ab^4 \plus{} 1}} \ge \frac {3}{\sqrt2}\]
[/quote]
Thanks for your problem  :wink:
I think it's true.I will try to prove it.

[quote="arqady"]
What about the following?
Let   $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a \plus{} b \plus{} c \equal{} 3.$ Prove that
\[ \frac {a}{\sqrt {bc^4 \plus{} 1}} \plus{} \frac {b}{\sqrt {ca^4 \plus{} 1}} \plus{} \frac {c}{\sqrt {ab^4 \plus{} 1}} \ge \frac {3}{\sqrt2}\]
[/quote]
Dear arqady, my cyclic lenma works here  :)
We only need to prove: $ abc\sum ab(a^2 \plus{} b^2) \plus{} \sum a^2 \plus{} \sum ab \le 12$
With $ a\plus{}b\plus{}c\equal{}3$
pqr or your uvw method can finish the proof  :)


In a 6x4 shaded grid of squares, 12 of them are to be shaded so that there are two shaded squares in each row and three shaded squares in each column. Let N be the number of shadings with this property and find the remainder when N is divided by 1000.

man this was on the aime '97 except 4 by 4...

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=393645#p393645[/url]

See my solution at http://www.artofproblemsolving.com/Wiki/index.php/2007_AIME_I_Problems/Problem_10


[quote="Euclid_great"][quote="brian_gold"]($ n\in N$)Prove $ \frac {1}{n \plus{} 1}(\frac {1}{1} \plus{} \frac {1}{3} \plus{} ... \plus{} \frac {1}{2n \plus{} 1}) > \frac {1}{n}(\frac {1}{2} \plus{} \frac {1}{4} \plus{} ... \plus{} \frac {1}{2n})$[/quote]

I think that should be $ \frac {1}{n \plus{} 1}(\frac {1}{1} \plus{} \frac {1}{3} \plus{} ... \plus{} \frac {1}{\color{red}2n \minus{} 1}) > \frac {1}{n}(\frac {1}{2} \plus{} \frac {1}{4} \plus{} ... \plus{} \frac {1}{2n})$ unless and until that is a typo.[/quote]
Try it for $ n \equal{} 1$..

edited:

Sorry the correct question should be as follows:

$ \frac {1}{n \plus{} 1}(\frac {1}{1} \plus{} \frac {1}{3} \plus{} ... \plus{} \frac {1}{\color{red}2n \minus{} 1}) {\color{red} \ge} \frac {1}{n}(\frac {1}{2} \plus{} \frac {1}{4} \plus{} ... \plus{} \frac {1}{2n})$


I took this problem in an old book. I have a question, but let me post problem first.

Determine all function $f: \mathbb{R}\to\mathbb{R}$ such that:
\[ f(x^5-y^5)=x^2f(x^3)-y^2f(y^3)  \]

[quote="Nameless"]This problem is in a very popular book in VietNam. But its solution is very long. Do you have a short solution Lovasz ?
If you have please post here[/quote]

Nameless, after some substutions, we come up this:
\[ f(x^5-y^5) = f(x^5)-f(y^5)  \]
And now, this is my "question":
[i]Why don't the author apply Cauchy function to get a simple proof, while I see many books with long solution as you said above ?[/i]

Could you tell me the name of popular book in Vietnam ,Nameless?
Thank you.

I love math

[quote="mps"]Could you tell me the name of popular book in Vietnam ,Nameless?
Thank you.[/quote]

May it be [i]Functional problems in Mathematical Olympiads[/i]?

Lovasz ,what's the name of the author ? And where can I found that book?


trivial doubts:)

1)How does thermal stability of carbonates vary across a group? across a period?..
I think it should decrease across the group from the following logic..as you go down the group attraction betwwen the ions and hence lattice energy decreases.
therefore the carbonate should decompose easily and be less stable....
however in books and on the net it is given the other way round...thermal stability increases down the group

2) Hydration energy of lithium is greatest amongst the alkali metals....since smaller size leads to greater electrostatic attraction....however at the same time it will cause crowding of water molecules near its surface..hence very few molecules of water will be able to hydrate it.....why doesn't this reduce the hydration energy?

I would be grateful if any1 could answer these questions..
Thanks a lot..!

yes, the stability of carbonates increases down the group even though you would expect the lattice energy to decrease down the group. dont forget that the lattice energy of the products of decomposition of carbonates are metal oxides and CO2. and the stability(lattice energy) of oxides down the group decreases more than that for the corresponding carbonates. also note that the metal ions of higher atomic number are similar in size to the carbonate ion and fit better in the crystal structure. the metal ions with smaller atomic mass are more polarizing than the metal ions with large atomic mass (fajan's rule) which means that they can attract the O- of carbonate ion better than the larger ions and hence are less stable as the stability mainly depends on the metal oxygen bond and not on the metal ion- carbonate ion attraction.
across a period the stability decreases, again for the same reason.


We take $ x,y,z>0$ such that $ \sqrt{xy}\plus{}\sqrt{yz}\plus{}\sqrt{zx}\equal{}1$.Prove that  $ 8xyz\geq (1\minus{}x)(1\minus{}y)(1\minus{}z)$.

[quote="skylover"]We take $ x,y,z > 0$ such that $ \sqrt {xy} \plus{} \sqrt {yz} \plus{} \sqrt {zx} \equal{} 1$.Prove that  $ 8xyz\geq (1 \minus{} x)(1 \minus{} y)(1 \minus{} z)$.[/quote]
Let $ x\equal{}a^2,$ $ y\equal{}b^2$ and $ z\equal{}c^2,$ where $ a,$ $ b$ and $ c$ are non-negative. Then your inequality is equivalent to $ \sum_{sym}(a^4b^2\minus{}a^3b^3\plus{}a^4bc\minus{}2a^3b^2c\plus{}a^2b^2c^2)\geq0,$ which true by Schur.


[quote="math0boy"]$ a,b,c>0$  prove that

$ (a\plus{}b\minus{}c)^2(b\plus{}c\minus{}a)^2(c\plus{}a\minus{}b)^2\equal{}>(a^2\plus{}b^2\minus{}c^2)(b^2\plus{}c^2\minus{}a^2)(c^2\plus{}a^2\minus{}b^2)$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=18908


Prove that in the following  sequence there is no any prime number:

$10001,\,100010001,\,1000100010001,\cdots$

Generalize this problem.

All terms give by : $x_{n}=\frac{10^{5n}-1}{10^{5}-1},n=2,3,...$
If $p|n$ then $x_{p}|x_{n}$. If n is even $x_{2}|x_{n}$, but $x_{2}$ is not prime.
If p is odd prime, then $x_{p}=\frac{(10^{p}-1)(10^{4p}+10^{3p}+10^{2p}+10^{p}+1)}{10^{5}-1}$. If p=5 we check, that $x_{5}$ is not prime. Because $10^{p}-1\not =10^{5}-1 (p\not =5)$ and second divisor greater, than $10^{5}-1$ we have $x_{p}$ is composite. Therefore for all n $x_{n}$ is composite.


Is it possible to use a $\LaTeX$ packege in this forum?  If so, how?

I don't think so. Only administartors can do this. But I think that all math's packages are loaded.

One more question:  is it possible to use eqnarray*?

The amsmath packages are loaded so you can use anything that uses it, such as matrices.

Eqnarray does work here but is now obsolete and replaced by the amsmath align environment (if you're interested then [url=http://tug.org/pracjourn/2006-4/madsen]Avoid eqnarray![/url] explains why).

Align is quite similar but needs less & to use. For example:
[code]\par
\begin{align}
  f(x) &= \int\sin x\: \text{d}x\\
  g(x) &= \int \frac{1}{x}\: \text{d}x
\end{align}[/code]
\begin{align}f(x) &= \int\sin x\: \text{d}x\\ g(x) &= \int \frac{1}{x}\: \text{d}x \end{align}
or without equation numbers use align* so [code]\par
\begin{align*}
  f(x) &= \int\sin x\: \text{d}x\\
  g(x) &= \int \frac{1}{x}\: \text{d}x
\end{align*}[/code] gives
\begin{align*}f(x) &= \int\sin x\: \text{d}x\\ g(x) &= \int \frac{1}{x}\: \text{d}x \end{align*}

(There's a small bug in the forum which means you have to add \par before \begin{align} until it's fixed.)

Thanks for the tip.  I'll keep that in mind.  Also, you may want to sticky a note about it, because I was probably not the only person going around using eqnarray.

It's in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=122157]Metapost & eqnarray[/url] but that's not very obvious. In fact I am wondering whether this with other common questions should be in a FAQ. I'll start a thread asking for questions.


function
inequalities
integration
geometry
geometric transformation
real analysis
real analysis theorems

For any measurable function f: [0,1] -> C prove that

lim_p->oo ||f||_p = ||f||_oo

Thank you

Ok, I found out. I proved it with 2 inequalities.Thanks anyway still don't all the details in my other post hough...

[quote]Ok, I found out. I proved it with 2 inequalities.[/quote]

I guess one of the inequalities is
||f||_p $\leq$ ||f||_oo .

What is the other inequality?

Sorry was in a hurry before.Here is my solution for anyone interested...

Let c>0 and set A_c = {x : |f(x)| > ||f||_\infty - c}. Then 0 < |A_c| <= 1.

When then have

\int_{A_c} (||f||_\infty - c)^p dx <= \int_{A_c} |f|^p dx <= \int_{[0,1]} |f|^p dx.

So that

[\int_{A_c} (||f||_\infty - c)^p dx]^(1/p) <= [\int_{[0,1]} |f|^p dx]^(1/p)

i.e.

(||f||_\infty - c) |A_c|^(1/p) <= ||f||_p

Letting p tend to infinity we get (noting that |A_c| > 0),

||f||_\infty - c <= lim_{p->\infty} ||f||_p <= ||f||_\infty.

Now let c tend to 0.

Translation service:

[quote="nfrer"]Sorry was in a hurry before.Here is my solution for anyone interested...

Let $c>0$ and set $A_c = \{x : |f(x)| > \|f\|_\infty - c\}$. Then $0 < |A_c| \le 1$.

When then have

$\int_{A_c} (\|f\|_\infty - c)^p\, dx \le \int_{A_c} |f|^p\, dx \le \int_{[0,1]} |f|^p\, dx$.

So that

$[\int_{A_c} (\|f\|_\infty - c)^p dx]^{\frac1p} \le [\int_{[0,1]} |f|^p\, dx]^{\frac1p}$

i.e.

$(\|f\|_\infty - c) |A_c|^{\frac1p} \le \|f\|_p$

Letting p tend to infinity we get (noting that $|A_c| > 0$),

$\|f\|_\infty - c \le \lim_{p\to\infty} \|f\|_p \le \|f\|_\infty$.

Now let $c$ tend to $0$.[/quote]
It's actually pretty easy- adding dollar signs is almost enough already.


floor function
complementary counting
combinatorics proposed
combinatorics

Let $ m,n$ be positive integers with $ m \le n$, and let $ F$ be a family of m-element subsets of $ \{ 1,2,...,n \}$ satisfying $ A \cap B \not\equal{} \emptyset$ for all $ A,B \in F$. Determine the maximum possible number of elements in $ F$.

If $ m > \left\lfloor \frac {n}{2}\right\rfloor$, then there must be a common element in any two sets, so the answer is just $ \binom{n}{m}$

If $ m\le \left\lfloor \frac {n}{2}\right\rfloor$, then we choose a subset of $ n$ of size $ 2m \minus{} 1$ and choose every subset of this to get $ \binom{2m \minus{} 1}{m}$ regardless of $ n$. We claim this is the best we can do. Suppose we had a set with $ k$ elements outside of our subset. Then it could have only $ m \minus{} k$ elements within the set, so there would be $ m \plus{} k \minus{} 1$ elements inside the set which would not be contained in it. Hence $ \binom{m \plus{} k \minus{} 1}{m}\ge 1$ sets within our subset would have to be removed, so this is the maximum, as claimed.

Your argument for $ \binom{2m \minus{} 1}{m}$ as a maximum is completely unconvincing.  It's also certainly wrong, since we can get $ \binom{n \minus{} 1}{m \minus{} 1} > \binom{2m \minus{} 1}{m \minus{} 1} \equal{} \binom{2m\minus{} 1}{m}$ when $ n > 2m$ by taking all $ m$-sets containing the element $ n$.

I feel slightly ridiculous now, both because my argument was such blatant nonsense and because I just realized we can simply count these as follows: Consider one set $ S\in F$; then every other set in $ F$ must contain an element of $ S$ and conversely every subset of $ \mathbb{N}_{n \plus{} 1}$ containing an element of $ S$ must satisfy this; thus we simply count all those with exactly one element in $ S$ (which is $ m\binom{n \minus{} m}{m \minus{} 1}$) then exactly two ($ \binom{m}{2}\binom{n \minus{} m}{m \minus{} 2}$) etc. so we get $ \sum_{i \equal{} 1}^{m}\binom{m}{i}\binom{n \minus{} m}{m \minus{} i}$ which by Vandermonde's identity is just $ \binom{n}{m} \minus{} \binom{n \minus{} m}{m}$; note the latter term is zero when $ m > \left\lfloor \frac {n}{2}\right\rfloor$, so in that case we have that it is $ \binom{n}{m}$, as before.

EDIT: Did I say slightly? I meant absolutely. As before, count them, but this time use complementary counting to get $ \binom{n}{m} \minus{} \binom{n \minus{} m}{m}$ directly. At least now I'll remember how to prove Vandermonde's.

Well, it gets worse: those $ \binom{n \minus{} m}{m}$ subsets all intersect the set $ S$ you chose, but how do you know they intersect each other?

Argh; I suppose I was supposing that any two shared a common element of $ S$ or something of that sort. Well, any two sets with at least $ \left\lfloor \frac{m}{2}\right\rfloor\plus{}1$ elements in $ S$ or at least $ \left\lfloor\frac{n\minus{}m}{2}\right\rfloor \plus{}1$ elements not in $ S$ will have a common element, and sets with smaller numbers in either of the two groups would then necessarily not share elements with one of these, but the fact that the sets can share elements both in and out of $ S$ seems to complicate matters, so I don't think considering a particular set helps.

In fact I really have no idea how to approach this problem now. Considering individual elements as members of intersections doesn't seem to lead anywhere, and I don't really know what the answer would be. Would you be so kind as to provide a hint, JBL?

I always find it fascinating when a problem that appears in every good combinatorics book makes it into a contest. This one is the Erdös-Ko-Rado theorem:
http://www.mathlinks.ro/viewtopic.php?t=202508 (no proof, but the Wikipedia page has one)
http://www.mathlinks.ro/viewtopic.php?t=129197 (cckek completely misunderstood the question, but Severius posted the right solution)

  darij


Let S be a collection of subsets of {1, 2, ... , n} of size 3. Any two distinct elements of S have at most one common element. Prove that
n(n-4)/6 \leq |S| < n(n-1)/6

Hint : use Problem An&Bn

The minoration is absurd. Or are you asking for a maximal collection?

Pierre.

[quote="A1lqdSchool"]Any two distinct elements of S have at most one common element.[/quote]

Looks like a contradictio...

+Yes, pbornsztein's right!
+Peter VDD , you note that S is the set!

oh yes, the elements ARE sets here :D

lol sorry was confused :P


The inequality is (a,b,c>0, sums cyclic) $\sum\frac{b+c}{a^{2}}\geq \sum\frac{1}{a}$.

I cannot find an equality case. I got
$\sum\frac{1}{a}>\sum\frac{1}{a+b}$ (since b>0), so $\sum\frac{b+c}{a^{2}}> (\frac{\sum\frac{1}{b+c}}{\sum\frac{1}{a}})(\sum\frac{b+c}{a^{2}}) \geq \sum\frac{1}{a}$ by CS.

Have I gone wrong, or was the "or equal" a misprint? Thanks.

I think you're right. In fact, some inequalities can have no equality condition even if the inequality contains equality expression. Because $2 \geq 1$ means that 2 is bigger than 1 or equal to 1, not that "there should be some case in which 2 is equal to 1."

$\sum_{cyc}\frac{b+c}{a^{2}}= \sum_{cyc}(\frac{b}{a^{2}}+\frac{a}{b^{2}}) \geq \sum_{cyc}(\frac{1}{a}+\frac{1}{b}) = 2 \cdot \sum_{cyc}\frac{1}{a}$

So I think the "or equal" is right, there is just a "2" missing.

Ah that would explain it. I thought it was too trivial. Thanks. :D


Anyone interested in an AIM mathcounts math jam type thing?
I'd be willing to host one in an AIM chatroom, for people who want to test their speed.
This evening I'd be able to, or probably some time Saturday/Sunday, but right now, please respond if you're interested and if you'd be able to do it around 8 EST this evening.

*By the way, the problems I'd use would be probably from saab or from older mathcounts tests*

People who are willing who's Screen name I know:

Biffanddoc
Noneoftheabove
Easyaspi
Pianoforte
Lucy4eva
*someone else who I know off AOPS*

That'd be really cool.  I would definitely do that except for the fact I don't' have aim :blush: .  I'll get aim pretty soon and am also willing to host/participate in these events.

Also, please either post, or PM me your AIM sn.
Even if its visible to me, please post or PM it.

And also post if you could do it tonight at 8 EST (in 1.5 hours)

Edit:
I appreciate that people are willing, to participate, but please, if I don't know your sn, at least PM me it.
Also, please tell me if you would be able to participate TONIGHT.

o yeah im gonna try (about 10 min right?)
but i have office hours, so im probably gonna be slow

OK, math jam is over.
Pretty fun, I'll probably be doing one this sunday.
I'll start a new topic for that, just for clarity.

Top 3 Ranking:

noneoftheabove
biffanddoc
pianoforte

There were about 8 participants in all. I am hoping for about 10 on Sunday.

If anyone wants the whole conversation, and yes, there's a lot of spam and stuff, PM me.
biffanddoc and pianoforte tied for 2nd, so i did a tiebreaker, and while piano answered first, he forgot a key aspect.

Good job guys!

O YEAH  :winner_second:  :winner_second:

I'd be up for one on Sunday. Possibly Saturday.

CONGRATS noneoftheabove and biffanddoc!

YAY :winner_third:  :winner_third:


Prove that for $a,b,c$ positive reals

$a^2+b^2+c^2+ab+bc+ca \geq \sqrt{2(ab(a+b)+bc(b+c)+ca(c+a))(a+b+c)}$

Would you believe me if I said that the inequality is equivalent to $a^4+b^4+c^4 \geq a^2b^2+b^2c^2+c^2a^2$ upon squaring and expanding?

How did you manage to transform this simple inequality into such hidden form?

I solved it in a very simple way by substitution

$x=b+c$ and similar.

It reduces after easy steps at the obvious $(\sum xy)^2 \geq 3xyz(x+y+z)$


induction
algebra
function
domain
algorithm
complex numbers
number theory

I was playing around with trying to factorize complex numbers into primes - don't ask - but after giving up there I came back to simple positive integers. How can you [b]prove[/b] that a positive integer has a unique prime factorization?

I think I can get it down to proving p|ab implies p|a or p|b for a prime p and positive integers a,b. I can't even prove this! Is this some sort of axiom or is it provable?(without assuming unique prime factorizations and the such.) Its obviously true.. but I just can't think of what to do..

Now I can see why people in the past could have huge problems proving things that are so obvious.

With Bachet-Bezout for instance.

if p does not divide a then (a,p) = 1

By Bachet-Bezout, there exist u,v with au + pv = 1 (*)

We have ab = kp, multiplying (*) by b gives b = p (v + k) so p | b.

I think you left out a u in there somewhere i/3, but the gist of it is right.

TripleM, if you got it down to that point then I think you're on the right track. As for proving unique factorization in complex numbers, I assume you mean complex numbers with integer real and imaginary parts. Here's a hint to get you started.

For any complex number, a = u + iv, define |a| = u:^2: + v:^2:, and call any number with |a| = 1 a unit. Then if a = bc, where a, b, and c are non-units in the complex integers, show that |b| < |a| and |c| < |a|. Now try and show that any complex integer, a, is either a unit or can be written as the product of a finite number of primes by induction on the size of |a|. Once you get that, then you just have to show that this product is unique (up to unit factors). To do this you'll need something similar to what i/3 used, but for complex integers.

The fun part is then to figure out what exactly the prime elements *are* in this ring.

H. Davenport also makes an interesting point about unique factorization by looking at the natural numbers that are 1 mod 4. These are closed under multiplication, but do they have unique factorization within themselves? If not, why?

They don't possess prime factorization. I think 441 is the smallest number that causes these problems. 441=9*49=21*21.

But can you tell why? What step in the proof of unique factorization in the natural numbers fails here?

Just giving this a bump because I think the question about the 1 mod 4 numbers is interesting.

The Gaussian integers don't have unique prime factorizations either.

ok here is the path that works for (pretty much) any unique (prime) factored domain:

division algorithm (or some variation of it) => ax + by = 1 (bachet-bezout someone called it... not too sure about that name but w/e)
=> a|bc, (a,b)=1 -> a|c. => unique prime factorization.

two primes are considered equivalent in unique prime factorization if they are associates. (i.e. a is an associate of b (denoted a ~ b) iff a|b and b|a. This statement is equivalent to a = bu for some unit u. A unit is an element of the system which has an inverse k s.t. uk=1)

The set of gaussian integers (denoted Z[i]) has a form of the division algorithm.

Investigate "norms". In Z[sqrt(d)], the norm of a + b*sqrt(d) = a^2 - d*b^2.
(specifically if x is an element of Z[sqrt(d)] the norm of x is
x * conj(x). Norms are always in Z, allowing you to convert difficult relationships in other systems to easier ones in the integers.]

Prove if x,y are in Z[sqrt(d)], N(x)N(y)=N(xy) where N(x) denotes the norm of x.

Try proving unique factorization in Z (the integers) first, by following the path from division algorithm to UPF.
Try only to use theorems you can prove directly from the axioms of the integers. Its a good exercise.

There is a type of division algorithm in Z[i]. Can you find it (hint: consider norms).

BTW gaussian integers DO have unique prime factorization.

I would also like to add that
you cannot use modular arithmetic to prove
the ax+by=1 statement.

This is because you would have to do division in a modulus to pull it off, as far as I see. And that requirs you have the
a|bc, (a,b)=1 => a|c theorem. Which follows from the ax+by=1 theorem. Hello circular logic.

Once you figure out that the complex integers have unique factorization, the fun part is then figuring out what the primes are.

oh yeah that 3-part condition set...

[hide]

I. 1+i, 1-i, i-1, i+1 (because two is annoying)

II. x where N(x) = p, and p is a rational prime congruent to 1 (mod 4)

III. p where p is a rational prime congruent to 3 (mod 4)

(rational prime means a prime in Z)

[/hide]


inequalities
function
limit
topology
Functional Analysis
real analysis
real analysis solved

$f:[a,b]->[a,b]$ and $|f(x)-f(y)| < |x-y|$ for every x and y in [a,b].Let $x_0$ in [a,b] and $f(x_n_+_1)=f(x_n)$ .Show that  $x_n$  is convergent.

[quote="adyo"] $f(x_n_+_1)=f(x_n)$ [/quote]
You mean $x_{n+1}=f(x_n)$ isn't it ?

We talked about this one before, you also have a generalisation

$E$ metric space compact $f:E\rightarrow E$ such that for any
$x \neq y$ $\Rightarrow$ $d(f(x),f(y))< d(x,y)$
Then $f$ has a unique fixed point in $E$

Unicity if there is exist $a\neq b$ two fixed points $f(a)=a, f(b)=b$
with the given inequality you get $a=b$

Existence suppose for any $a\in E, d(a,f(a))> 0$

let $g(a)=d(a,f(a))$ the map from $E \rightarrow ]0;\+\infty[$ continous on the compact $E$, the infinimum is reached, call it $b$
$0 < d(b,f(b)) \leq d(a,f(a))$ since the distance between $b$ and $f(b)$
is $>0$ you have $b\neq f(b)$, apply the hypothesis $d(f(b),f(f(b))) < d(b,f(b))$
you get a contradiction since $b$ was the infinimum of $g$

Clearly, the sequence $a_n=|x_{n+1}-x_n|$ is strictly decreasing, and since it's bounded, it must converge to some limit $\ge 0$.

As Moubinool said, the function has a unique fixed point. It has a fixed point because of a well-known theorem (any continuous function from a closed ball to itself has a fixed point), and it must be unique because otherwise we would have $|f(x_1)-f(x_2)|=|x_1-x_2|$ for two distinct fixed points, and this is false.

Let the fixed point be called $x_0$. If $x_n\in [a,x_0]$ then $x_{n+1}\in [x_0,b]$ and vice-versa. This means that one of the subsequences $x_{2n+1},x_{2n}$ is strictly increasing to a limit $\ell_1\le x_0$, and the other one is strictly decreasing to a limit $\ell_2\ge x_0$.

We then get, by taking limits, that $f(\ell_1)=\ell_2,\ f(\ell_2)=\ell_1$, and this gives $\ell_1=\ell_2$.

there is one similar problem:
if E is a metric compact space and f a function from E to E such that
d(f(x),f(y))>=d(x,y)
then f is an isometry ..

I was solving last problem about isometry three weeks ago with my students, but we weren't able to do it for 20 minutes. Then I read solution in my functional analysis problem book and discovered that solution is wrong. :?
So what about proof, alekk?

Just a humor.
I gave test to my students one week ago:

Check compactness in C[0,1] for following sets:
1) $M_1=\{x\in C^1[0,1]\,|\,x(0)=0,\ |x(t)x'(t)|\leq 1\ \forall\,t\in[0,1]\}$
2) $M_2=\{x\in C^2[0,1]\,|\,x(0)=0,\ |x''(t)|\leq |x(t)|\ \forall\,t\in[0,1]\}$

Quotation from student's script:
[quote]
...
1) Boundedness of $M_1$ is evident.
...
2) Boundedness of $M_2$ is far from evidence, [i]but it can be proved in the same way as for[/i] $M_1$
...
[/quote]
 :lol:  :lol:  :lol:

Pretty good, Myth, especially since $M_2$ is unbounded. (Just let $x_n(t)=nt$.) If I'm grading that paper, do I now believe your student when he claims that the boundedness of $M_1$ is "evident"? Even if it is true that $M_1$ is bounded, since $|x^2(t)|\le 2t\le 2$? Of course, $M_1$ is not closed, hence not compact.

Example: let $x_n(t) = \sqrt{n}t$ for $0\le t\le\frac1n$ and $x_n(t)=\sqrt{2t-\frac1n}$ for $\frac1n\le t\le 1$. This is a sequence of functions in $M_1$ that converges uniformly to $\sqrt{2t}$, which is in $C[0,1]$ but not in $M_1$.

You passed my test  :)  :)  :)
Did you solve alekk's problem about isometry?

here is a solution :

we want to prove that f is an isometry. Consider two arbitrary points a and b in E and the sequences
a_n=f(a_(n-1)) ; a_0=a
b_n=f(b_(n-1)) ; b_0=b
B is a metric compact space so there is a strictly increasing function g so that
a_(g(n))->A and b_(g(n))->B. Of course since f is an isometry d(a,b)=d(A,B), but we don't know that yet. We are going to prove this fact:
let e>0 a real. There exist N0 large enough such that for n>N0 :
d(a, a_(g(n+1)-g(n)) )<=d(a_(g(n)), a_(g(n+1)))<=e
d(b, b_(g(n+1)-g(n)) )<=d(b_(g(n)), b_(g(n+1)))<=e
hence, with k=g(n+1)-g(n),
d(a,b)<=d(f(a),f(b))<=...<=d(a_k, b_k)
         <=d(a_k,a) + d(a,b) + d(b,b_k) <= d(a,b) + 2e
but this is true for every e then we can conclude that:
d(a,b)<=d(f(a),f(b))<=d(a, b)
and
d(a,b)=d(f(a),f(b))

do you think it's correct ?

I think your solution is correct. Thank you very much!

By mean value theorem, there  exist $c$ such that $\left|\frac{f(x_n)-f(\alpha)}{x_n-\alpha}\right|=f'(c)$ where $a<x_n<c<\alpha<b$ or $a<\alpha <c<x_n<b$.

Let $r=f'(c)$,we have $|x_{n+1}-\alpha|=|f(x_n)-f(\alpha)|=\left|\frac{f(x_n)-f(\alpha)}{x_n-\alpha}(x_n-\alpha)\right|=|r||x_n-\alpha|$.From given condition,
$|r|<1$ so we obtain $|x_{n+1}-\alpha|<r|x_n-\alpha|$.Using this repeatingly,yielding $0<|x_n-\alpha|<r^n |x_1-\alpha|$.Since ${\lim_{n\to\infty} r^n}=0$ so by Archimedes' axiom $\lim_{n\to\infty} |x_n-\alpha|=0\Longleftrightarrow \lim_{n\to\infty} x_n=\alpha$.

Just as expected,this solution is not proof,do you think so,Moderator Myth.
Because in the beginning,the existence of $\lim_{n\to\infty} x_n$ is assumed.
 :blush:  :P

kunny


find the smallest integer $x$ for which
\[\frac1{2005}<\frac{y}{x}<\frac1{2004}\]
and $y$ is some integer

[hide]Well, since $x$ must be an integer, we can start multiplying top and bottom of all three expressions by successive integers. The first one we try is $2$, so we have $\frac{2}{4010}<\frac{y}{x}<\frac{2}{4008}$. So, $x=4009$.[/hide]

$1/2005<y/x<1/2004$, so you know that x has to be an integer in between $2004$ and $2005$, however since there are no integers between those two numbers then you can multiply both these numbers by 2, getting $4010$ and $4008$ thus getting $4009$ and $1/4009$ fits the inequality so $x=4009$ is the answer.

[quote="Palytoxin"]that's why i posted this problem in Getting Started because i remember the first time i saw this problem, i didn't know where to start to tackle this problem. afterwards, i figured out a general way of solving this and applied it and it worked! ill let others post their solutions before i tell answer, i guess you guys already figured it out...  :D[/quote]

Did you solve it the same way Palytoxin or did you have a different way?

this problem can be solved easily without using my method but i did it in different way, instead i did it the hard way, i solved a more general problem:

prove that there exists a unique smallest positive integer $n$ such that, for positive integers $a,b,c,d,m$ and $(a,b)=1,(c,d)=1$,
\[\frac{a}{b}<\frac{m}{n}<\frac{c}{d}\]

[size=200][color=red]PS:[/color][/size] please [b]DO NOT[/b] solve this problem here, this is [b]NOT[/b] Getting Started if you read forum's guidelines, if you know solution, even if you think it's easy, dont solve it here. i'm just following forum guidelines so [b]DO NOT POST YOUR SOLUTION HERE[/b], do it in Intermediate Topics forum or something. thank you


AMC AIME ARML USA(J)MO USAMO MATHCOUNTS calculus

4 (if the posted answers are right)

I thought it was hard this year; I did better last year and did a bunch of work in between.

I should have had seven, but with two careless mistakes, I got a five. :(

i'm looking at a six, which is exactly what i got last year. lame.

If I still want to make ARML (of course I do), I will have to really make up for a 164 USAMO index.

8. Whee. I budgeted 3 mistakes for myself so I'd improve my score and sure enough missed #1 and #4 stupidly and missed another completely, so I got precisely my goal.

I got an 8 :(   I don't think the test was necessarily harder, but I hope it was (because I want to make USAMO by a comfortable margin), and I hope there was more room for oversight.

And by the way, the answers are posted [url=http://www.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/2007-aa/07-AIMEanswers.shtml]here[/url] on the AMC website.

Edit :  Whoa, actually, after looking at the AMC forum, I think that this was probably harder than last year's.

(15 questions on test)-(5 that I didn't know how to do)-(4 careless errors on seperate problems)-(2 problems that I missed a case on) = 4 :wallbash_red:  :noo:  :surrender:

Index = 133.5+10*4=173.5

Chance of making USAMO = 0

Chance of making ARML = 0.6 ish

Funny, I did better on last year's AIME with about 5 hours of sleep and no idea of what to expect than on this year's AIME with 8 hours of sleep and plenty of practice over the last year.

Just 0.6??? How do you figure that? BTW, what are you doing after school next Monday? I've got another practice for the Dickerson/Dodgen kids at Walton and would love for either you or Miles to come by!

Fulton
(I've got to get enough posts to stop being a new member! :-)

And I do assume that 0.6 is 60% NOT 0.6% :-) (Wheee, 2 posts in a day!)

I am attending our math team's final practice of the year on Monday from 3:30 to 4:15. I may be able to attend depending on when the practice starts.

Now that I have seen other people's scores on the AIME and have considered the fact that I have placed highly at several tournaments, I guess that my ARML chances are more like 0.75 to 0.80 or maybe more. And yes that is a decimal and not a percentage.

BTW: Congrats on Walton winning the spring tournament. I guess not having Miles or Rohan significantly lowered Lassiter's team score.

After seeing other scores, I haven't completely lost hope for making ARML. I need to remember that ARML is the top 30 in the state and that I was easily top 30 in Kennesaw, which my math teacher says most people in the state took. I guess I should just start worrying about GCTM state.

[quote="gtf"](Wheee, 2 posts in a day!)[/quote]

Eh, Mr. Fulton, this is a friendly moderator of a different forum. It is usually ill-advised to post after oneself without anyone else in between. Otherwise we shall have to bring you up to the excellent Mr. Tong for him to moderate your conversations. :P :rotfl:

santosh, i'll punch you in the face...

not make ARML, gees. you make me sick.

What Amy said!  :) As a coach, I can't say that anyone will automatically make the team, but let's see... a sophomore that made the A team as a freshman.... I think you've got a decent chance!

The MathCounts practice at Walton will start at 4:30 and end about 6:30. If you could make it and talk to them, that would be outstanding!

Thanks for the congratulations. I was really hoping that we could catch Wheeler for 2nd, but the first tournament killed us. We were 5th with a bunch of our good folks out for a full Orchestra concert. (Orchestra and part of the band.) Congratulations on 2nd for the year (and closing on Phillip!)

Billy, you should probably report me. This is now my 10th post total and I'm probably abusing it. (Report me to Howard? Isn't that ironic!  :rotfl: )

(Arrrgggh! 10 posts and STILL a new member!)

i'm missing something. (I think 25 and you stop being a new member)

i think i got roughly a 7 or an 8 (don't remember what i put), but yeah dont think i made usamo again =( i'm kind of embarrassed at how badly my math has gone downhill since like freshman year......whatever we'll see.

Naw, you're probably not missing anything. It was late (after several late nights) and I was typing a Calculus test as well as answering a bunch of email questions from my students so my brain activity was pretty low (and started pretty low to start with!)

(Only 14 more posts! :-)
Fulton

[quote="Benjamin Hu"]15 questions. (8 I knew how to do)-(1 where I made a stupid computation mistake) -(1 where I made another stupid mistake)=6 problems done correctly.
index: 99(darn)+10*6=159...
No chance of making USAMO...
Some chance of making ARML...
Wow...This AIME was much harder than the previous ones.[/quote]
This year, a 6 may well be enough for USAMO.  Nobody really knows.  You're in ninth grade, right?

[quote="Benjamin Hu"]Yes, I am in ninth grade.[/quote]

You'll make it.

Well, I thought your AMC score was more of like a 120 or so... but I just realized that every student who makes a 6+ this time who is a freshie or sophomore doesn't get in. Sorry  :(

But you'll still make ARML (with 99.99% confidence) and pwn it up there.

Twenty-five Georgia people beat me on the AIME and eleven tied me. Eleven people who beat me and five who tied me will be graduating next year. Out of those whose grade is less than or equal to mine, six people beat me and two people tied me. Now I know that if I am in ARML range, I am at the very bottom, and if I make it, I will likely be somewhere around the thirty-third pick. It may help that I've been recognized at three JV and four varsity tournaments this year, but I placed pretty low among those who were recognized at many of them. It all comes down to next Saturday I guess. :prays:

I take making ARML way too seriously. :lol: I know that no one cares about what I had to say above. I guess I'm pretty much writing it for myself. That's pretty stupid, I have a brain for thinking to myself.

Eh, the good thing is that Mr. Fulton knows you, and you've probably gained recognition among the coaches by now if you've scored well at like 4 or 5 tourneys. I remember last year that I did pretty okay at all the tournaments (top 20 most of the time except UGA) and I didn't even place until I got to JV State (4th). And you've done like 10 times better than I did last year, and I'm sure you'll beat my score last year for Varsity State. So you have a huge chance at ARML. Don't worry  :)


integration
calculus
function
probability and stats

This should be a standard question...which i cannot solve. :(

Let $ X$ be a non-negative random variable, show that for any $ 0<p<\infty$, \[ E(X^p)\equal{}p\int_{0}^{\infty} \lambda^{p\minus{}1}P(X>\lambda)d\lambda.\]

Thanks in advance.

Try using the identity
\[ X^p \equal{} \int_0^X p \lambda^{p\minus{}1} \, d \lambda.
\]

:blush: sorry... my calculus not v good..

maybe that it is easier to see it with a more general statement. For example, suppose that $ f$ is a derivable function such that $ f(0)\equal{}0$. Then
$ E(f(X)) \equal{} \int_{\Omega} f(X(\omega)) d \omega \equal{}  \int_{\Omega} \int_{\lambda \equal{} 0}^{X(\omega)} f'(\lambda) d \lambda d \omega$
Hence
$ E(f(X)) \equal{} \int_{\Omega} \int_{\lambda \equal{} 0}^{\infty} f'(\lambda) 1_{ \{ \lambda \leq X(\omega) \} } d \lambda d \omega$
and I think you guessed what it remains to do to conclude  :)

ooops. sorry. forgot to say i got it.

thanks for the help!


[quote="Captain Sugar"]a bank gives out PIN numbers to its customers.  they are each 12 numbers long.  the last two numbers of the PIN number must be the last two numbers of the year it was issued. (e.g. a PIN issued in 2001 would end in 01.) the first two numbers must be the number of the month which the customer was born in.  how many PIN numbers are possible?

and the slightly harder part:
somehow the bank still exists in the year 10000.  they decide to continue issuing PIN numbers to new customers, but these new PIN numbers are 24 numbers long, with the last 6 numbers being the year (e.g. the year 10000 is represented by 010000). the first two numbers are the month which the customer was born in, just like the old ones. they keep the older PIN numbers, though, and use them with the new ones. now how many PIN numbers are possible?[/quote]

[hide="1"]the first two numbers have 12 choices and the last ten can be any digit. so it is $\boxed{120000000000}$[/hide]

for number 2, do you count the old ones [i]and[/i] the new ones or just the new ones?

FYI- PIN means personal identification number, so PIN number is redundant. :D
If you mean that how many combinations of old and new PINs they can have, it's [hide] the number of old PINs is $12\times10^{10}$. The number of new PINs is $12\times10^{22}$. So, the number of combinations is $144\times10^{32}=\boxed{14400000000000000000000000000000000}$. :D [/hide]

[hide="1"]
There are 10*10=100 ways for the last two numbers.
There are 12 ways for the first 2 numbers.
That's 4 numbers. Then the 8 left are
$10^8$ and with the last 2, it is $10^{10}$
So, the answer is:
$12\cdot10^{10}=120000000000$[/hide]

[hide="#1"]
12 ways for the first two numbers and 10 ways for the rest. $\boxed{12 * 10^{10}}$.
[/hide]
[hide="#2"]
12 ways for first two numbers and 10 ways for the rest. $12*10^{22}$ ways. With the old ones, it is $\boxed{120000000000120000000000}$.[/hide]

What year did the bank start issuing the old numbers?

[quote="Captain Sugar"][quote="Ignite168"][quote="Captain Sugar"]just about everyone keeps forgetting that the last two numbers in the first PIN number problem have to be the last two numbers of the year, and the last six numbers have to be for problem two.

you count the old ones and the new ones.[/quote]

Well you would need to know what time period the bank issued old numbers and what time period they issued new numbers.[/quote]

they started issuing new numbers in the year 10000, like it says in the problem ;)[/quote]

Until 99999999999999999999999999999999999... ? Did they issue old numbers for every year before 10000 without any hiatus?

it doesn't matter the year, it only matters how many possible there are. the last year for the second problem would be 999999, after then the bank would issue new numbers.

[color=darkblue][hide="answer to part 1"]417654129152000[/hide][/color]


inequalities
LaTeX
calculus
calculus computations

Let a,b,c>0 .Show that [(a/b)+(b/c)+(c/a)] squared is greater than (a+b+c)(1/a+1/b+1/c).
Sorry for Latex absence but im writing on someones else pc

[quote="etaratoris"]Let $ a,\ b\ ,c>0$ .Show that $ \left(\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\right)^2\geq \left(a\plus{}b\plus{}c\right)\left(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\right)$.[/quote]

[quote="kunny"][quote="etaratoris"]Let $ a,\ b\ ,c > 0$ .Show that $ \left(\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\right)^2\geq \left(a \plus{} b \plus{} c\right)\left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right)$.[/quote][/quote]

Anybody has idea? :blush:

Let $ \frac {a}{b} \equal{} x$, $ \frac {b}{c} \equal{} y$, $ \frac {c}{a} \equal{} z$, expanding, we have
$ \sum x^2 \minus{} x \plus{} \frac {1}{x} \ge 3$
$ x^2 \minus{} x \plus{} \frac {1}{x} \ge 1$
$ (x\plus{}1)(x \minus{} 1)^2 \ge 0$
Which is true, equality when a=b=c.

This is an inequality from [color=blue]IRAN 2005 NMO[/color]. It is also posted here: [url]http://www.mathlinks.ro/viewtopic.php?p=316311#316311[/url]


Does anyone know of any books that have combinatorial/discrete geometry problems like proving Helly's theorem, or like show that any convex geometric shape can be inscribed in a rectangle with no more than twice it's area or #4 on 2008 USAMO?

If you want a proof of Helly's theorem, see [url=http://en.wikipedia.org/wiki/Helly%27s_theorem]here[/url].  It does require that you know some analysis, though.

No actually I know the proof I just want some books that have these sorts of problems.


For any positive integer k, let f(k) be the number of elements in the set {k+1,k+2,k+3,....,2k} whose base 2 representation has precisely three 1s.
(a) Prove that, for each positive integer m, there exists atleast one positive integer k such that f(k)=m
(b) Determine all positive integers m for which there exists exactly oe k with f(k)=m


A new automative engine lubricant claims its use will save $ 10\%$ of the fuel cost of the vehicle in which it is used. An automobile tire company makes the same claim about a new radial tire design. If both are used with the same car, what is the theoretical percent of fuel savings?

But we are asked for FUEL SAVINGS.

Answer: 1-81%=[b]19%[/b]


A square has an area of 100 cm^2. An equilateral triangle has the same side length. Compute the perimeter of the triangle.

[hide="Solution"]
The square has $ \plus{}\sqrt{100}\equal{}10\ \text{cm}$ side length.
If the triangle has the same length, it's perimeter would be $ 3(10)\equal{}\boxed{30\ \text{cm}}$
[/hide]

Since the square has an area of 100 cm^2, the length of one side would equal the square root of 100, which is 10. The equilateral triangle has the same side length, the sides would be 10 cm each, so the perimeter would be 3 * 10, 30 cm.


Let $x_i, 1 \leq i \leq n$ be real numbers with $n \geq 3$. Let $p$ and $q$ be their symmetric sum of degree 1 and 2 respectively. Prove that:

1) $p^2 \cdot \frac{n-1}{n}-2q \geq 0$

2) $\left|x_i - \frac{p}{n}\right| \leq \sqrt{p^2 - \frac{2nq}{n-1}} \cdot \frac{n-1}{n}$ for every meaningful $i$.

"symmetric sum of degree 1 and 2 respectively"- what is that?


superior algebra
superior algebra unsolved

Let  $ p$ be a prime number. Can we construct a minimal dense subset $ \mathcal{U}$ of the $ p$-adic integers $ \bf{Z}_p$ (i.e. for every other dense subset $ \mathcal{V}$ we have that $ \mathcal{U} \subsetneq \mathcal{V} \subset \bf{Z}$).

No. If $ U$ is dense, so is $ U\plus{}v$ for any $ v \in \mathbb Z_p$.

Ok, thanx.

Can we describe such dense subsets? Like:

$ \mathcal{U} \equal{} \{n \in \bf{Z} : \textrm{ n satisfies a finite number of properties } \}$

The original exercise I solved was: "find the $ p$ such that the set of all natural numbers ending on $ 2007$ is dense in $ \bf{Z}_p$". So I wondered if we could describe all dense subsets (or some at least) with similar simple properties on its elements.

Stated otherwise: "Can we find a necessary and sufficient condition on a subset $ \mathcal{U}$ of $ \bf{Z}$ such that it is dense in $ \bf{Z}_p$"

The obvious condition is: $ A \subset \mathbb Z$ is dense in $ \mathbb Z_p$ iff for any $ m,n \in \mathbb N$ there is an $ a \in A$ such that $ a \equiv m \mod p^n$.


group theory
superior algebra
superior algebra unsolved

[quote="mikejones"]for this problem I have the folowing answer:
Let G1 and G2 be groups.
a) Define pie : G1XG2 -->G1 by pie 1 (a1,a2)=a1, for all (a1,a2) in G1XG2 and define Pie 2: G1XG2--> G1 by pie 2(a1,a2)=a2, for all (a1,a2) in G1XG2 and. Show that both pie 1 and 2 are Homomorphisms.

b) Let G be any group, and let Ru: G-->G1XG2 be a function. Show that Ru is a group homomorphism if and only if pie1 Ru and pie2 Ru are both group homomorphism.

A)I would like to know if I first verify that pie(ab)=pie1(a)pie1(b) for any a,b in G1XG2 if a=(a1, a2) and b=(b1, b2) then, as it is defined ab=(a1b1, a2b2). SO is this correct?
B) Would I do it the same way or how would I do it?[/quote]

A)You are correct. By definition it suffices to prove that $pie_1(ab)=pie_1(a)pie_1(b)$ for any a,b in ${G_1}\times{G_2}$, which is obvious by the definition of "product" in ${G_1}\times{G_2}$.

B)The "IF" part is obvious, since the composition of two homomorphisms is also a homomorphism.
The "ONLY IF" part: Notice that $Ru(g)$ can be written $(pie_1\circ Ru(g),pie_2\circ Ru(g))$ in ${G_1}\times{G_2}$, then it is easy to verify that $Ru(g_1g_2)=Ru(g_1)Ru(g_2)$ because $pie_1\circ Ru$ and $pie_2\circ Ru$ are both homomorphisms.


The arithmetic mean of 12 numbers is what percent of the sum of the same 12 numbers?  Express your answer as a mixed numeral.

[hide]
Well, the mean of n numbers is the sum over 10, and n=12 in this case, so:

\[\frac {1}{12}=8\frac{1}{3}\%\]

Seems pretty easy for a state target, but whatever.[/hide]

Looks good.  I think you wrote 10 when you meant n, but otherwise fine.

Yup. I really should proofread these first . . .  :blush:


[quote="Virgil Nicula"]$\{x,y\}\subset (0.1)\ ,\ x+y=1\Longrightarrow 2\left(x^2+y^2\right)^3\ge \left(\frac{xy}{2xy-1}\right)^2\ .$[/quote]

First: $(2xy-1)^2=(x^2+y^2)^2$

So, we can rewrite the inequality as: $2(x^2+y^2)^5 \geq (xy)^2$

From the power mean inequality we have: $\sqrt{\frac{x^2+y^2}{2}}\geq \frac{x+y}{2} \geq \sqrt{xy}$

Then we get: $x^2+y^2 \geq \frac{1}{2}$ ; $xy \leq \frac{1}{4}$

So we can prove: $2(x^2+y^2)^5 \geq \frac{1}{16} \geq (xy)^2$


i need to make some paper on tied vectors.
they're also called bound vectors.
but i cant find on google anything good.
i need some help
i mean ...some courses or good links where i may read about them

Maybe you already have that, but, anyway... :

http://www.ucl.ac.uk/Mathematics/geomath/vecsnb/MHvecslnk52.html
http://www.netcomuk.co.uk/~jenolive/vect1.html
http://wiki.beyondunreal.com/wiki/Vector


[b]Q.Find all values of 'n' for which the equation[/b]
       [b]    n {x} = [ x ] has exactly ten(10) solutions............where {x} denotes fractional part of x i.e = x - [x]    and [x] represents the smallest integer less than or equal to x[/u][/b]

From $0\leqslant\{x\}<1$ we get $0\leqslant [x]<n\iff 0\leqslant [x]\leqslant n-1$

From $x=[x]+\{x\}$ and $\{x\}=\frac{[x]}{n}$ it follows $x=\frac{n+1}{n}[x].$ Therefore, every value of $\lbrack x\rbrack$ corresponds to exactly one value of $x$. Since there are $n$ possible values for $\lbrack x\rbrack$, we conclude that $n=10.$

[b]Q.Find all values of 'n'(real number) for which the equation[/b]
       [b]    n {x} = [ x ] has exactly ten(10) solutions............where {x} denotes fractional part of x i.e = x - [x]    and [x] represents the smallest integer less than or equal to x[/u][/b]  [quote="rohitsingh0812"] well farenhajt.... your solution is not wrong...but i would like to quote that n belongs to the set of real numbers and not only to that of natural numbers...so you had to find all the values of n for which the question holds....[/quote]

Then it should've been said so in the first place (that $n$ is real).

OK, from $0\leqslant [x]<n$ and $x=\frac{n+1}{n}[x]$ (which hold whether or not $n$ is integer) we conclude that, in order for equation to have exactly $10$ solutions, maximal possible value for $\lbrack x\rbrack$ is $9$, hence the desired set of values for $n$ is $9<n\leqslant 10$


[quote="Kamyar1991"]very nice!!![/quote]
Let $ a\equal{}b\equal{}1,c\equal{}0$, the inequality reversed.

the variable are in R+!!!!!!!!!!  :wink:

I think you have learned "lime"  ^^. :D.$ a\equal{}b\equal{}1;c \rightarrow 0$

i agree with u!!  but the lemma when is correct that we have ab>=1 then in main qestion  we must have abc>=1

I think it's still wrong,take $ a \equal{} 10,b \equal{} 1,c \equal{} 0.1$

You're wrong again ^^ .If $ abc\equal{}1$ it's wrong ;) .You should check your ineq carefully before sent to ML ^^

sorry. in iran 5 days later we had second round olympiad and for this i didnt sleep for 48 hours so i a little confused. we must had ab>=1 & ac >=1 and bc>=1 :blush:  :blush:  :blush:


real analysis
calculus
integration
function
real analysis theorems

Let's start with its construction:

[quote]$\displaystyle \bullet \, \, \,$ If $\displaystyle f$ is a simple function, i.e. if $\displaystyle f$ can be written as $\displaystyle f = \sum_{k=1}^{n} c_k \chi_{A_k}$, where $\displaystyle c_k \in \mathbb R$, $\displaystyle A_k \in \mathfrak B$, $\displaystyle \forall k \in \overline{1,n}$, and $\displaystyle n \in \mathbb N$, then define \[ \displaystyle \int_X f \, d\mu : = \sum_{k=1}^n c_k \int_{X} \chi_{A_k} \, d \mu = \sum_{k=1}^{n} c_k \mu \left( A_k \right) .  \][/quote]

How do you prove that the integral is uniquely defined? (A simple function can be written in more than one way as a linear combination of characteristic functions of measurable sets; how do you prove that the value stays the same?)

I have other questions, but this is the most "basic".

Supose $f=\sum_j\alpha_j\chi_{A_j}=\sum_k\beta_k\chi_{B_k}$

We can, WLOG, insist that the sets $A_j$ be pairwise disjoint, and similarly that $B_k$ be pairwise disjoint.

Both sums are finite sums.

Consider the sets $E_{jk}=A_j\cap B_k.$ If $x\in E_{jk},$ then $f(x)=\alpha_j=\beta_k.$ That is, if $E_{jk}\ne\emptyset,$ then $\alpha_j=\beta_k.$

Then $\sum_j\alpha_j\mu(A_j)=\sum_j\alpha_j\mu\left(\bigcup_k E_{jk}\right)$

Let's use the addititivity - finite additivity - of $\mu$:

$=\sum_j\sum_k\alpha_j\mu(E_{jk})=\sum_k\sum_j\alpha_j\mu(E_{jk}).$

The only pairs $j,k$ which contribute anything nonzero to the sum are those for which $E_jk$ has positive measure, which is contained in those $E_{jk}$ which are nonempty. Whenever that happens, $\alpha_j=\beta_k.$ Thus our sum is

$=\sum_k\sum_j\beta_k\mu(E_{jk})= \sum_k\beta_k\sum_j\mu\left(\bigcup_j E_{jk} \right)=\sum_k\beta_k\mu(B_k).$


function
percent
analytic geometry
graphing lines
slope

x:     0   1   2
f(x): a   24  b
(respectively)

The table above shows some values for the function f. If f is a linear function, what is the value of a + b?

Age 6: 45 inches; Age 12: 60 inches. (I summed the question since it had a graph)
The person's height at the age of 12 was what percent greater than her height at the age of 6?

The least integer of a set of consecutive integers is -25. If the sum of these integers is 26, how many integers are in this set?

[hide="#1"]Since $ f(x)$ is linear, it must have the same slope between any two points. You are given points $ P_1(0,a), P_2(1,24), P_3(2,b)$. Since slope$ (P_1P_2) \equal{}$ slope$ (P_2P_3)$,  you have $ \frac {24 \minus{} a}{1 \minus{} 0} \equal{} \frac {b \minus{} 24}{2 \minus{} 1}$. Hence,  $ 24 \minus{} a \equal{} b \minus{} 24$, and $ a \plus{} b \equal{} \boxed{48}$.[/hide]

[hide="#3"]Since the sum is positive and the least consecutive number is $ \minus{}25$, the set must have positive numbers. It is obvious that $ \minus{}25\minus{}24\minus{}\cdots\minus{}1\plus{}0\plus{}1\plus{}2\plus{}\cdots\plus{}25\plus{}26\equal{}26$ since the numbers less than $ 26$ cancel out. So, this set is $ \{\minus{}25,\minus{}24,\minus{}23,...,\minus{}1,0,1,...,25,26\}$, which has $ 25\times 2\plus{}2\equal{}\boxed{52}$ numbers.[/hide]

this might not belong in HSB, but ill answer them anyway

#2:
60 is __% greater than 45
is the question
suppose 60 is p% greater than 45
then 60=(100+p)% * 45 (this is basically the definition of percent more)
[hide="see if you can sol;ve that b4 looking here"]
$ \frac43=(100+p)%
$
$ \frac{400}{3}=(100+p)$
$ \frac{100}{3}=p$
[/hide]

number 3: suppose the last number is x
the sum of the numbers from -25 to 0 is simply -(1+2+3 ... +25)= $ -\left(\frac{(25)(26)}{2}\right)=-325$
and the sum of the numbers from 1 to x is $ \frac{(x)(x+1)}{2}$
so $ -325+\frac{(x)(x+1)}{2}=26 \implies \frac{(x)(x+1)}{2}=351 \implies (x)(x+1)=702$

[hide="try to solve it yourself b4 looking here"]there are a few methods you can use from here
$ x^2+x=702 \implies x^2+x+\frac14=\frac{2809}{4} \implies \left(x+\frac12\right)^2 = \left(\frac{53}{2}\right)^2 \implies x+\frac12= \pm \frac{53}{2} \implies x=26, -27$
but -27 is extraneous so $ \boxed{x=26}$[/hide]

I'm not too sure about either one. Considering they're SAT questions, answer choices would have been nice  :)

[hide="#2"]
$ \frac{x}{100}45=60$

$ 45x=6000$

$ x=\frac{6000}{45}\approx{133%}
$

So, 60 is approximately 33% greater than 45.
[/hide]

[hide="#3"]
$ \sum_{k=-25}^{n}k=\frac{n(n-25)}{2}=26$

$ n(n-25)=52$

$ n^2-25n-52=0$

$ \frac{25\pm\sqrt{25^2-4(-52)}}{2}\approx{27}$

$ 27-(-25)=\fbox{52}$

52 numbers in the series.
[/hide]

oh yeah... srry since the series for #3 goes from -25 to 26, there are 26-(-25)+1 terms or 52 terms
and yes, $ \frac{100}{3} \approx 33$

[quote="archimedes1"][quote="grn_trtle"][hide="#3"]
$ \sum_{k \equal{} \minus{} 25}^{n}k \equal{} \frac {n(n \minus{} 25)}{2} \equal{} 26$

$ n(n \minus{} 25) \equal{} 52$

$ n^2 \minus{} 25n \minus{} 52 \equal{} 0$

$ \frac {25\pm\sqrt {25^2 \minus{} 4( \minus{} 52)}}{2}\approx{27}$

$ 27 \minus{} ( \minus{} 25) \equal{} \fbox{52}$

52 numbers in the series.
[/hide][/quote]Hmm, instead of chugging with a formula, we could think about what the sum from negative 25 up to positive 25 is, and then add 26.[/quote]

Yes, that's the "traditional" way to do it... My solution was a little bit flawed I think (the end result for x was something like 26.9), but I ended up getting 52 anyway.


[quote="mathgeniuse^ln(x)"]I think this is the right place to post this

A set contains five integers.  When distinct elements of this set are added together, two at a time, a complete list of different sums that result is: 637, 669, 794, 915, 919, 951, 1040, 1072, 1197.

What is the largest of the five integers in the set?[/quote]
There are only 9 sums, but $\binom{5}{2}=10$.  :maybe:

If they gave you the tenth sum, it would be too easy.

You should read the whole question, and yes if I gave 10 sums, it would be easy.

Are you sure you got the numbers right?

I'm getting strange results... :huh:

the numbers are correct, the problem is stated correctly...
[hide="hint"]let a,b,c,d,e and test out the cases[/hide]

Yes, that is what you should do.

Notice what the problem stated.

If you still don't get it...

[hide][quote="mathgeniuse^ln(x)"]
A set contains five integers.  When distinct elements of this set are added together, two at a time, a complete list of [b]different [/b]sums that result is: \[/quote][/hide]

If you don't understand the question, try to understand what Klebian is trying to imply.

[quote="AdorableAddict"][quote="Ignite168"][quote="AdorableAddict"][hide="Answer"]The missing sum is $794$ and the the largest number is $659$.[/hide]
I will post solution if it is correct.[/quote]
[hide="Hint"]
There is NO missing sum, it's a COMPLETE list.[/hide][/quote]
What I meant by missing sum is the sum that was repeated. As in, two pairs result in that exact same sum.[/quote]

Woops, I read that as the largest number was the first number you posted and the sum was the 2nd number, although it's easy to see that wouldn't work, I wasn't thinking.  :blush:  For people that still don't get it though, my hint is sorta a major one.

The only thing I don't get though is that if one of the elements was repeated, there should only be 7 sums.

say the double sum is n, then 4 of the numbers are a,n-a,b,n-b

[quote="AdorableAddict"][hide="Answer"]The missing sum is $794$ and the the largest number is $659$.[/hide]
I will post solution if it is correct.[/quote]

I believe that is correct.  Post your solution if you want to.

OK here's the [hide="Solution"]
Denote the five numbers as $a, b, c, d,$ and $e$ such that $a \le b \le c \le d \le e$.
Then $a+b$ would produce the smallest sum, which is $637$, and likewise, $d+e = 1197$.

Now suppose the ''missing'' sum is $x$, then adding the nine known sums and $x$ would result in the sum of all sums of pairs, or namely \[x+637+669+794+915+919+951+1040+1072+1197 = (a+b)+(a+c)+(a+d)+(a+e)+(b+c)+(b+d)+(b+e)+(c+d)+(c+e)+(d+e)\]  \[x+8194 = 4(a+b+c+d+e)\] And since $(a+b)+(d+e) = 637+1197 = 1834$, we can replace that in the equation above: \[x+8194 = 4(1834+c)\]  \[x+858 = 4c\] Now we can clearly see that $x \equiv 2 \text{ mod }4$, and since $x$ is among the given sums, $x = 794$, the only sum that gives a remainder of $2$ when divided by $4$.

Plugging x back into $x+8194 = 4(1834+c)$, we can get $c = 413$.
To find the largest number, note that the second largest sum, $1072$, can only be produced by $c+e$. Thus, $c+e = 1072 \Rightarrow e = 1072-c = 1072-413 = \boxed{659}$
[/hide]

Adorable Addict are you saying that by using two distinct pairs of numbers, that both pairs would add up to $794$?
If it does, how do you know that it's $794$ that appears twice and not any of the rest of the numbers?

We know that x has to be 2 (mod 4) because x+858=4c. (x+2=0(mod 4)). x is 794 because it's the only one that is congruent to 2 (mod 4), in other words the only one that [b]can[/b] be x.

I understand laughinghead505. I am not very well versed in modular math, but after studying a web site explaining the basics of modular math it does make sense now.


A couple plans to have three children. The birth of a boy or girl is
equally likely. What is the probability that all three children will be
the same sex? Express your answer as a common fraction.

After one child is born, the other two must have the same gender, so the probability is $ \frac{1}{4}$.

There are 8 possibilities, and the only ways for all the children to have the same gender is either bbb or ggg. Therefore, the chance that all are the smae gender is 1/4

This is very similar to flipping a coin.  Unless, of course, there is a slight possibility...


$ sin(x)\plus{}sin(2x)\plus{}sin(3x)\plus{}...\plus{}sin(nx) \equal{} ?$

Multiply $ 2\sin \frac{x}{2}$ or $ 2\cos \frac{x}{2}$ by left the side and you will see Telescope.

or You can use Complex number.


algebra
polynomial
quadratics
linear algebra
matrix
conics
ellipse

Let's define a polynomial P(x) to be P(x) = x^2 + 2*p*xy + y^2, where 0 <= p <= 1. Is there any interesting property of this polynomial?

Please notice that when p = 1, this polynomial becomes P(x) = (x+y)^2. Can we say that as p goes from 0 to 1, this polynomial becomes more and more factorizable?

I ask this question because polynomials like this show up at a lot of places in my research.

Thanks

[quote="Kent Merryfield"]We should say $ P(x,y)$ rather than $ P(x).$ This has a name: $ P(x,y)$ is a [i]quadratic form[/i]. (In fact the natural home for this discussion is really the linear algebra forum, but I'll leave it here for the time being.)

Here's a matrix representation: $ P(x,y) \equal{} x^2 \plus{} 2pxy \plus{} y^2 \equal{} \begin{bmatrix}x & y\end{bmatrix} \begin{bmatrix}1 & p \\
p & 1\end{bmatrix}\begin{bmatrix}x \\
y\end{bmatrix}.$

For $ \minus{} 1 < p < 1,$ this particular quadratic form is [i]positive definite[/i]. That is, $ P(x,y) > 0$ for all $ (x,y)\ne(0,0).$

Positive definite quadratic forms are associated with ellipses. Fix $ c > 0.$ The set $ P(x,y) \equal{} c$ is an ellipse. For $ p \equal{} 0,$ it is a circle; as $ p\to 1$ (or as $ p\to \minus{} 1$,) the ellipse gets longer and more eccentric; in the limit as $ p \equal{} 1$ it becomes degenerate.

There's an association between quadratic forms (positive definite or at least positive semidefinite) and jointly distributed random variables. In this particular case, think of two random variables each with standard deviation $ 1$ and with correlation coefficient $ p.$[/quote]

Thank you for your reply. Can you be more specific about the association between quadratic forms and jointly distributed random variables? I can only think of bivariate Gaussian distribution whose joint density function has a term there which is in this form ...