[quote="Slizzel"]A friend send me this IQ test. It takes 40 minutes ... I didn't manage to finish all the questions (I did 31 of 39) and my score was  [hide="..."]113[/hide]. Is that good? Hope that's true because I did an online IQ test and giving random answers I obtain 130. That was false anyway :rotfl:

[url=http://www.math-linfo.as.ro/test.exe]IQ test Download[/url]

p.s.: [i]I uploaded the test on one of my ftp accounts. You cand download it from there. I did this because I couldn't upload an *.exe file as an atachament.[/i][/quote]There is no small amount of controversy over what IQ actually measures. In many circumstances IQ tests can extremely biased for cultural or other reasons. Good performance on a standard IQ test only seems to correlate well with one's socio-economic status. IQ does not indicate future success, and to my mind if it does not do that then it isn't of any value.

That said, in 1905 French psychologists Binet and Simon proposed an intelligence test that measured a person's "mental age." In 1912 the German William Stern divided mental age by chronological age to obtain an intelligence quotient. In 1916 Lewis Terman of Stanford University used a scale where IQ was multiplied by 100. The test that Terman developed became known as the Stanford-Binet test. Finally in 1939 David Wechsler, a Romanian born American psychologist, defined IQ as a normally distributed random variable having a mean = 100 and standard deviation = 15.  Using Wechsler's scale an IQ in the range (85, 115) is average.

However, one ought to be suspicious of online IQ tests. There are no standards for item selection or scoring or anything and thus the scores you obtain from one test to the next vary widely.

i got a score of [hide]120, pretty bad considering half the questions were really easy.[/hide]

I got something like 105. I did like two questions later on and got <65 :roll:

This goes in the round table where most of the other IQ discussions are.

Sorry.  :blush: Lately I'm posting only in High-School and Games & Fun factory :P

Watch as I score a 65 since most of you guys are smarter than me.

EDIT: 105

I generally dislike the idea of a person's numerical IQ, being born out of the eugenics and social darwinism movements. I like to take them though, and this one was pretty interesting. I like the way the problems are all in exactly the same format. That one's ability on this type of question measures his or her intelligence is dubious, but it certainly does give a good indication of some kind of intelligence.

Anyway, I got a 132. The time they give you on this test is odd, though. I spent 15 minutes on all but two questions and had 25 minutes to ponder the remaining two. It took about 7 or 8 minutes to get answers for them. On a well-designed IQ test one should not be able to afford 3 to 4 minutes on a single question! Unless of course there are few, difficult questions, but this wasn't the case.

I'm glad you liked the test. :) Somebody sent me it by e-mail. I lost some minutes because I was talking with that person on Yahoo! Messenger while I was solving it :).

[quote="white_horse_king88"]I think that was a challenging IQ test. Nicely original. I liked it.

I got a 127, way below expectations... but it says 140 is mensa... so.[/quote]A little bit of perspective is needed to interpret that result, white_horse.  First, presuming you are a member of a random population -- which might be a stretch since the only people who really have sufficient interest in their IQ's to worry about their particular number are those who by any reasonable qualitative definition are highly intelligent to begin with -- and second, assuming that the test (I didn't bother to take it) is scored according to a Wechsler scale, an IQ = 127 puts you in the 96th percentile. (An IQ = 140 is the 99th percentile.)

The main problem with these online IQ tests, already mentioned in the thread, is their complete lack of validation.

For much thought-provoking reading about IQ testing, see the online bibliography

[url=http://learninfreedom.org/iqbooks.html]Books on IQ and Human Intelligence[/url]

where books that mention most of the problems of IQ tests are recommended for your reading pleasure.

Online IQ tests are usually just ego-building.

Although this one looks a bit like ego-hurting, I wouldn't trust anything like this online.

If you want to get tested, I'm sure your school has some sort of gifted program and usually they'll do an IQ-like test to see if you qualify (at least that's what I did I think). Or you could probably schedule a test with mensa if you were really gung-ho about it.

But, all in all, it's just a number. It's not going to help you to get into a better school, get a better job, or have a better life when you find it out. It could actually cause you to get worse if you get a lower-than-expected score because you may mentally block yourselves because you deem yourself not 'smart enough'.

I got above end 140 (resp 137, 160 for the other two standard deviations). Whatever that means... Fun problems!

[quote="Peter VDD"]this one is from http://www.iqtest.dk .

I got 149 on it.[/quote]The thing you need to know now is out of people who have taken that test, how many have scored likewise?

On the website there's a nice applet telling you where you are in the population. The website also considers age.

Anyone able to tell me if this is right? (from the applet file)

You can see the order of my questions in the attachement.

1 G
2 D
3 D
4 F
5 F
6 H
7 E
8 D
9 G
10 H
11 H
12 D
13 D
14 C
15 B
16 H
17 E
18 A
19 F
20 H
21 G
22 E
23 E
24 H
25 A
26 B
27 B
28 B
29 F
30 E
31 B
32 G
33 F
34 G
35 H
36 G
37 A - but also doubting to B...?
38 G
39 A

[quote="Derek"]I took the first 18 questions then randomly guessed on the rest because I had to go to the bathroom and I got a 104.[/quote]

I did that to second time :)

I got 114,lower than I expect.But I still think I am smart. :D

Here's an interesting collection:

[url=http://www.eskimo.com/~miyaguch/]www.eskimo.com/~miyaguch/[/url]

look at some of the "Hoeflin" tests. They are supposedly able to seperate out people in the 99.9999th percentile. I think this is B.S., considering how much harder math contests are, and how many people do well on those.

I find that many of the problems are very easy (like the verbal analogies, many of which are dumb), and some of the math problems can easily be solved with Polya's theory of counting. Some are more interesting, however, like some of the sequences in the "power test", tile laying problems and the solid intersection ones.

What do you think?

[quote="pianoforte"]I got 132 (standard deviation 16, what ever that means), with 17 min remaining. :D[/quote]pianoforte,

A standard deviation is measure of how widely the population varies around its mean. On a bell curve 68% of the population falls within the interval of average $\pm$ 1 standard deviation. If the mean (average) is 100, as it is with most IQ tests, then you scored precisely 2 standard deviations above the mean. That implies you are in the 98 percentile according to the test you took.

I got:

124 on the first
138 on the http://www.iqtest.dk

Whatever those numbers may mean, I do not think they measure
[i][b]'intelligence'[/b][/i] (whatever [i][b]that[/b][/i] might mean!)

A very intelligent person on another forum where I'm a member  puts the following question at his signature line: [b]"Are human beings more [i]'intelligent'[/i] than yeast?" [/b]

When you think seriously and honestly on this question, it would appear that, as a species, we are a lot [i]less[/i] intelligent than yeast - to judge by the godawful mess we have made of our world thus far!

[To find out more about the mess we are making in our world, check out http://www.dieoff.org, and think of participating in the discussions at  'AlasBabylon' (a group at YahooGroups)].

GSC

[quote="gsc"]When you think seriously and honestly on this question, it would appear that, as a species, we are a lot [i]less[/i] intelligent than yeast - to judge by the godawful mess we have made of our world thus far!

[To find out more about the mess we are making in our world, check out http://www.dieoff.org, and think of participating in the discussions at  'AlasBabylon' (a group a YahooGroups)].[/quote]

I don't see the relevance? :? This test measures individual intelligence compared to the population average... what has this to do with the politics messing up our world? [as the politics are completely independent of our personal IQ]

"Measure" is not the correct word to use in regard to IQ tests. "Estimate" is a bit more accurate, but even then one must prove that there is some underlying construct that has real-world validity before taking IQ scores very seriously. IQ scores are based on "samples of learned behavior," in the well-formed phrase of Alan S. Kaufman, and sometimes those samples are not very representative of the test-taker's behavior in circumstances outside the testing room.

Those tests are not at all accurate.  I know my IQ is at least 150, but they tell me its less than 130 (go figure)

[quote="Peter VDD"][

I don't see the relevance? :? This test measures individual intelligence compared to the population average... what has this to do with the politics messing up our world? [as the politics are completely independent of our personal IQ][/quote]]

[b]Relevance[/b]:
The underlying issue is that the word [i][b]intelligence[/b][/i] bears with it all kinds of connotations.  Let me discuss a few of them here:

1.  [i][b]Intelligence[/b][/i] generally connotes [i][b]wisdom[/b][/i] as well (or at least the ability to be wise).

2. We - members of the human race - pride ourselves on being the *Masters of the Universe*, the most *intelligent* of all species (and so on and so forth, practically ad infinitum, until verily reels the mind).  I.e. we regularly define standards of *intelligence* between species and we like to place ourselves at the very top of the heap.  I claim we need to think again about that.

3. *Intelligence* means, to me, some quality that would lead to what I'd like to define as *intelligent behaviour*.  Intelligent behaviour should imply, I'd believe, behaving so as not to create a mess.

4.  As a species, we have been behaving rather *unintelligently* (in my view) - as what we have been doing on earth is precisely to create a godawful mess on it (whether this be caused by politics or by genetics).

5.  There seems to be somewhat a disconnect between 2 and 4 above (to my mind, at least).

6.  There may well be a disconnect, I believe, between what we measure as *intelligence* (via IQ and other tests) and whatever we [i][b]should[/b][/i] term as *true intelligence*.

7. Politics is part of our behaviour.  Our political behaviour in groups should be included in any measure of *intelligence* (when we measure 'speci'-al intelligence - i.e. intelligence as a species).

8.  In view of the above, I believe we need to ask ourselves, very seriously, the question: [i][b]"Are humans more intelligent than yeast?"[/b][/i] (Taking into consideration all the connotations to most of us of the word *intelligence*).

9.  Quite apart from 'speci'-al intelligence, we also may need to rethink this business of *intelligence* per se, this quality that is claimed to represent "individual intelligence compared to the population average".

I could go for that if we were to remove the word *intelligence* [u][i][b](along with all that it connotes to our minds)[/b][/i][/u] and say that we are just measuring, say, "individual bahoozabachahooeyfump compared with the population average bahoozabachahooeyfump".

(I think it may be the disconnect between *intelligent behaviour* and *what we have been doing on earth* that is actually causing the problems in my mind).

GSC[/i][/b]

While it is statistically interesting, i.e. entertaining if not psychologically relevant, the whole business of IQ is elitist and unnecessary. In developed nations elementary education is compulsory. Participation in which brings one's native intelligence to light in most cases. Therefore, IQ testing in the developed world is redundant; and in developing nations the results of such tests are typically skewed implying that they have little if any validity when applied to such societies.

Since IQ is redundant in some cases and invalid in all others, all a high score can impart is bragging rights to an individual who is probably predisposed toward arrogance in the first place.

One seldom sees anyone with an IQ of 90 reporting what their IQ is, assuming that they know it, even though such a score is perfectly normal.

[quote="Dr. No"]One seldom sees anyone with an IQ of 90 reporting what their IQ is, assuming that they know it, even though such a score is perfectly normal.[/quote]Someone who's bad at football will also not report their scores in a football test, that's perfectly normal. Why does that make football irrelevant? [even though everyone gets the football basics in school and therefor football talents will not stay unremarked there]

Earlier, I had reported a score of 138 on the test at http://www.iqtest.dk

I tried out the test twice more (trying as far as possible to forget the responses I had made previously).

Score on second try - 124.
Third try -  114.

Does this mean that in the space of just one day or so my 'IQ' has somehow degraded from near-Mensa level to something just above the mean of 100?  The next time around will my IQ be only 94 or lower??

[i][b]ALAS! Woe is me!  (Or, should that be Wee is Moe, as in the Three Stooges?)[/b][/i]  (But somehow, I really do not feel any less *intelligent* than I did yesterday).

Or, does it just mean that I was mentally tired the second and third times around, and my scores are actually indicative only of my 'mental alertness' at the time of taking the test?

Or, is it that I 'did it deliberately'?  (I do not believe this was the case).

Or what?

GSC

[quote="samath"]I'm wondering how much that test actually bases the "score" on how many questinos you got.  It seems to be pretty random to me.[/quote]It's not just a right or wrong. Sometimes you spotted one pattern, but didn't spot the better pattern, and you still get more than for a random answer. I guess.

On the 'IQ test' that he had posted (which started this thread), I observe that "slizzel" got a score of 113 when he actually did the test, and a score of 130 when he filled in random answers!!  [b][i]So what is his real IQ??[/i][/b]  (Is there such a thing at all?)

I did that test once, and I got a score of 124, if I remember right.
On the test at http://www.iqtest.dk, I got the following scores:

Trial No. 1 - 138
Trial No. 2 - 124
Trial No. 3 - 114
Trial No. 4 - 128
Trial No. 5 - 104
Trial No. 6 - 136 (A short time ago).

Each time, I had done the test as *honestly* as I could, trying not to remember my responses on the previous trials.  I've not done the test filling in responses randomly - maybe I should try that out next?  Would [i][b]that[/b][/i] give me a better measure of my 'IQ' than when I do it 'honestly' ??!!

I believe these IQ tests are generally based on bogus concepts in any case, because we really do not know what actually constitutes [i][b]'human intelligence'[/b][/i] in the first place!

In any case, the ability to recognise such patterns as are provided in this kind of 'IQ test' surely can give us only a measure of one of the many abilities and potentials that constitute what we should consider to be 'real intelligence'.  Any measure of real intelligence must include, in some way, some measure of how well the person tested would measure up to the challenges he/she faces in life (and that is very difficult to do - if possible at all).   I understand that Einstein was considered by his teachers to be of inferior intellectual powers - one does not know how he would have measured up on these IQ tests as I think such did not exist during his youth.

Professor Howard Gardner of Harvard University had a concept he calls 'multiple intelligences', which I believe may be indicative of how such an issue of 'measuring human beings' may be approached - but even that is, I believe, of extremely limited value.  For one thing, while it's true enough that "a leopard does not change its spots" (i.e., it does not become an elephant or a camel), human beings often do grow and change considerably during their lives, in many of their characteristics and abilities.

Very broadly (if we [i][b]MUST[/b][/i] 'measure' people, which I am generally loath to do), my feeling is that  a proper definition of "intelligence" (as one possibly desirable human characteristic) should not be limited to:

-- skills in recognising such patterns alone;
-- nor to skills in math alone;
-- nor to skills in using words alone (prose/poetry/oral rhetoric each being different ways of using words; and then there are different languages - and what about the skills or ability to pick up languages that are not one's mother tongue?);
-- nor to skills in making others laugh (or cry);
-- nor to skills in manipulating others by way of persuasion,
-- .....

I really do not believe that any test that can be given in a limited period of a day or two (or maybe even a week or two) can measure all such skills and abilities that should constitute an acceptable idea of [i][b]'human intelligence'.[/b][/i]

Specifically in the case of the test at http://www.iqtest.dk, the choices provided for question No. 19 [i][b]do NOT include[/b][/i] what appears to me to be the correct pattern -  [i][b]I find this to be rather amazing[/b][/i].   I've not tested out the other questions for 'correctness of choices provided'- but a few of them seem to me to be rather suspect (as do some of the questions themselves).

GSC

[quote="gsc"]I did that test once, and I got a score of 124, if I remember right.
On the test at http://www.iqtest.dk, I got the following scores:

Trial No. 1 - 138
Trial No. 2 - 124
Trial No. 3 - 114
Trial No. 4 - 128
Trial No. 5 - 104
Trial No. 6 - 136 (A short time ago).[/quote]It measures your capability to spot patterns etc. at a certain instant. Maybe you were not quite awake at try 5 and were in top form for try 1 and 6?

[quote="gsc"]Each time, I had done the test as *honestly* as I could, trying not to remember my responses on the previous trials.  I've not done the test filling in responses randomly - maybe I should try that out next?  Would [i][b]that[/b][/i] give me a better measure of my 'IQ' than when I do it 'honestly' ??!![/quote]I did for you. When guessing [i]really[/i] randomly you get a "below 73". ;)


trigonometry
conics
ellipse
geometry
geometry unsolved

[quote="ZHANGWENZHONGKK"]In $ \vartriangle ABC$ , $ B_1$ is on $ AB$ and $ C_1$ is on $ AC$ . If $ BB_1 \plus{} B_1 C \equal{} BC_1 \plus{} C_1 C$ ,prove that $ \vartriangle ABC_1$ and $ \vartriangle AB_1 C$ have the same circumcircle.[/quote]
I haven't read this problem carefully, but it maybe wrong  :?:
$ 2R(ABC_1).\sin A\equal{}BC_1, 2R(ACB_1).\sin A\equal{}CB_1$.
If $ R(ABC_1)\equal{}R(ACB_1)$, then $ BC_1\equal{}CB_1\Rightarrow BB_1\equal{}CC_1$.
W can construct an elipse with foci B,C cuts AB,AC at D,E, so BD+DC=BE+EA, but BD is not equal to CE when $ AB\ne AC$

i think he want to say "have the same [i]incircle[/i]"


There are two types of tiles -- one that is 2 units wide and 1 unit long, and the other that is shaped like an L and covers three square units. You are allowed to rotate the tiles when you lay them.Calculate the number of different ways of tiling rooms with the following dimensions --

1) 2 X 8
2) 2 X 10
3) 2 X 12

Are the tiles unique or the same? E.g. do permutations of vertical 2x1 tiles forming an 8x2 room count? Are the tiles considered symmetric or is it a different combination if you flip one 2x1 tile around? I'm guessing the answer to both of these is no, but if you painted each 1x1 area of each tile a different color there would be far more combinations.

Well, I'm sorry to say that your guess was wrong......the tiles need not be unique, and well, rotating a 2 X 1 tile by $180^\circ$ gives you the same combination....but if two L shaped tiles are together and form a 3 X 3  square, and this whole square is rotated by $90\circ$,it is a different combination. To make it even clearer, a 2 X 3 room can be tiled in 5 ways.


Can someone explain how to do the input and output for the USACO in C/C++? I just want to be able to read a specific line of a file and store it to a string.

ifstream fin("(filename)");
ofstream fout("(filename)");

use fin and fout like you would use cin and cout

OK thank you. So you put those two lines at the beginning of the code, and then you can just use fin and fout and they will read and write on that file? Does the filename actually go in parentheses?

right, it will tell you in the assingment what the filename is. so if the in file is rect1.in it would be

ifstream fin("rect1.in");

say you want to get an integer input

int a;
fin>>a;

and you have the value for a


In Mrs. Craig's statistics class there are twice as many $ 11^\text{th}$-grade students as there are $ 9^\text{th}$ and $ 10^\text{th}$-grade students combined. There are twice as many $ 12^\text{th}$-grade students as $ 9^\text{th}$-grade students, and the number of $ 11^\text{th}$-grade students is ten times the number of $ 12^\text{th}$-grade students. If there are 32 students altogether in Mrs. Craig's statistics class, how many $ 12^\text{th}$-grade students are there?

[hide="SOLUTION"]Let the number of $ 9^\text{th}$ graders in the class be $ x$. Thus $ 2x$ twelfth-graders, and $ 20x$ eleventh-graders. Because there are only $ 32$ students total, $ x$ must be $ 1$. There are thus a total of $ 2x \equal{} \boxed{2}$ students. [/hide]

Hi;

The problem becomes a 4 x 4 set of simul equations:

a11=2 (a9+a10)
a12=2 a9
a11=10 a12
a11+a12+a9+a10 = 32

Which are easy to solve:

9th graders = 1
10th graders = 9
11th graders = 20
12th graders =2

Um, KingSmasher's solution is way much easier, if you want easy, and leads to a correct answer.

Hi PhireKaLk6781;

My answer is correct. Let's look at the problem.

There are twice as many 11 th graders as 9th and 10 th combined.(means add 9th and 10th)

20 = 2 *(9+1) correct.

There are twice as many 12 th graders as 9th graders.

2 = 2*1 correct.

The number of 11th grade students is 10 times the number of 12 th grade students.

20 = 10 * 2 correct.

32 students all together.

1+9+20+2 = 32 correct.

So where am I wrong? There are 2 12th graders.

You are not wrong. Your answer agrees with that of KingSmasher's, it's just that his is faster.

[quote="izzy"]
1. Do not post a second solution to a problem unless

-- You arrived at a different answer.

-- You have a different way of getting to the answer.

-- You noticed a mistake someone made.

-- You wish to clarify something that was unclear in someone else's post. [/quote]

i'm pretty sure that bobbym's solution fits the 2nd -- though, so i don't see anything wrong.
also, just guess+check for cd problems. # of 11th graders<32
#11th graders is 10 times that # 12th graders
#12th graders is twice of #9th
so #12th graders is even, but less than or equal to 3, so it has to be 2. 0 would make problem messed up, but you can check it to make sure.

if this was a countdown, bobbym would have lost
you're algebra is good but the other guy has logic

Yes, sir! mathyoda is right!!!

Oh and my signiture!!! Never forget that!!!

R@D!O@CT!VÉ

P.S. ╥ tastes good.


Greetings everyone from Latvia!

I hope i posted my problem at the right section. If no, I am sorry. I understand everything from a-z in algebra. But geometry is my weak side. So i am asking for help here. Probably better would say - begging. So my exercise.

Cylinder height is [b]6cm[/b] and it is crossed with plane which is parallel it's axis. Plane is situated in [b]3cm [/b]distance from axis and breaks off from basis [b]8cm[/b] tall hord. I need to calculate cylinder volume.

V=πr2 X height ( H )

My english isn't that good when i am writing some math exercises. Sorry about mistakes.[/code]

I attached sketch. I think this cylinder should look like this. A[size=75]1[/size]D is 8 cm, ZX is height ( h ) 6cm, and there is gap between plane and and axis - 3cm.

Site where you can download my sketch: http://yy.lv/upload/index.php?ACT=4&f=13908&ext=BMP

No worries it is a virus free site.

Yes, I don't know how to draw in paint :)

You basis of the cylinder gives you the answer: you'll find a right triangle with sides 3 and 4, so the ather one is...5 ! It's your radius!
I'll make a figure for you !

You just saved my life. Thank you so much!

We are welcome ! And greetings from France and Romania  !


function
combinatorics proposed
combinatorics

Place a positive real $a_{i,j}$ at each lattice $(i,j)$ of the plane such that $a_{i,j}=\frac{a_{i+1,j}+a_{i,j+1}+a_{i-1,j}+a_{i,j-1}}{4}$ for all $i,j$ .prove that all $a_{i,j}$ are equal.

i heared really fantastic proof based on convexity :? is it the proof that you know ?

Actually, I wasn't solving it, but my supervisor often say me that it is something like maximum principle.

Can you tell your fantastic proof? Please!

[quote="Myth"]Louville'e theorem. It is really well known, but can anyone give a reference?[/quote]

 I do know the Liouville's theorem for the harmonic function:
[i] Every positive harmonic function on the plane must be constant![/i]
 Now we have to prove a discrete version...
 I wonder if we can find an analogous proof!!
 But anyway, how wonderful mathematics is!!  :lol:

I have called to my supervisor today. And he told me his proof.

You want to say something?? Like main ideas.... :D

Actually, it was proof for http://www.mathlinks.ro/Forum/viewtopic.php?t=5327 , where we demand boundedness.

[quote="Myth"]Someone should write solution at last.
This theorem appeared on forum at least 4 times. No one solution was proposed![/quote]

 So... :D


The sum of all possible values (n from 0 to infinity) of the number one less than the     f(n)th root of n converges to an easily expressable number - create an f(n)

Challenge (I don't know if this is possible): create an f(n) that outputs a rational number or even an integer.

[hide="Solution"]
We seek $f(n)$ such that

$\sum_{k=1}^{\infty} \left( k^{ \frac{1}{ f(k) } } - 1 \right)$ converges to a rational or an integer.

Let $g(k) = k^{ \frac{1}{ f(k) } }$.  We can simply pick $g(k) = 1 + \frac{1}{2^k}$ and the sum converges to $1$.  Then

$\frac{1}{f(k)} = \frac{ \ln (1 + \frac{1}{2^k} ) }{ \ln k}$
$f(k) = \boxed{ \frac{\ln k}{\ln (1 + \frac{1}{2^k} ) } }$. [/hide]

What about an f(n) with an outcome of e?

Let's generalize to a limit of $r$.  Extending our previous solution, let $g(k) = 1 + \frac{r}{2^k}$.  Then

$f(k) = \frac{\ln k}{\ln (1 + \frac{r}{2^k} )}$

Note that this is, of course, far from the only such function.


[quote="madadpoor"]It seems be incorrect.
This qoutient is equal to $ 2610.77$.[/quote]
Oh I'm sorry there was a mistake in the third exponents its should $ (7836)^2$ not $ (7836)^3$

Thanks for your nice solution Kouichi Nakagawa  :)


$ x^4 \plus{} x^3 \plus{} x^2 \plus{} x \plus{} 1 \equal{} 0\to \frac{x^5\minus{}1}{x\minus{}1}\equal{}0\to x\equal{}exp(\frac{2\pi ik}{5},k\equal{}1,2,3,4.$


For a positive integer n, let P(n) be the product of the nonzero base 10 digits of n . Call n prodigitious if P(n) divides n . Show that one can not have a sequence of fourteen consecutive positive integers that are all prodigitious

Let's assume that $x_1,...,x_{14}$ is such a sequence.
Then, among the 10 consecutive numbers $x_2,...,x_{11}$ at least one, say $k$, divisible by 10, and therefore has unit digit 0.
Note that $k-1,k+1,k+2,k+3$ are all terms of the given sequence.
We have $P(k)$ divides $k$.

But, for $i=1,2,3$, the number $k+i$ has the same digits as $k$ excpet the unit digit which is $i$.
Thus $P(k+i)=i \times P(k)$. (1)

In particular, $P(k+1)=P(k)$ divides $k+1$. Thus $P(k)$ divides $(k+1)-k=1$, from which we deduce that $P(k)=1$ so that $k=10^{a_n}+...+10^{a_1}$ where $a_n > ... > a_1 \geq 1$.
Then, from (1), $P(k+3)=3$ divides $k+3$, and then divides $k$.

But $k-1 = 10^{a_n}+...+10^{a_2} + 99...9$, where there are $a_1$ digits 9.
It follows that $P(k-1)=9^{a_1} = 0$mod[3]. Thus, 3 divides $P(k-1)$ and then divides $k-1$, which means that both $k$ and $k-1$ are divisible by 3. A contradiction.

Pierre.


floor function
inequalities
calculus
integration

Find the number of positive integers $x$ such that \[ \left[ \frac{x}{99} \right] = \left[ \frac{x}{101} \right] .  \]

[quote="Peter VDD"]What exactly do the brackets mean here?[/quote]
Title says 'gint' and means 'g' 'int', and so he meant greatest integer :D

We have that $\frac{x}{99} - 1 < \left\lfloor \frac{x}{99}\right\rfloor = \left\lfloor \frac{x}{101}\right\rfloor$.

We want to find the highest multiple of $99$ for which the inequality holds.

So we have $k - 1 < \left\lfloor \frac{99}{101}k\right\rfloor = \left\lfloor k - \frac{2k}{101}\right\rfloor$.

Hence $k = 50$ and the maximum $x$ is $4949$. There are $4949$ solutions.

[quote="paladin8"]If I'm understanding correctly, neither of your solutions takes into account that numbers like $x = 100$, for example, fail?[/quote] :wallbash_red:  aaarghhh!, yes, that's the problem with my way of trial and error, i was aiming for big numbers and the maximum and missed trivial things like that :( Well, then essentially we have to subtract the number of integers between the succesive multiples of $99$ and $101$ and the multiples of $99$ themselves from $4949$.

ACTUALLY I GOT AN ANSWER OF (49)^2!!! :maybe:

rahul1990gupta is right there are 2499 solutions

[quote=kk108][quote=Maharjun][b]Please patiently read[/b]

[hide]
$ [k]$ implies greatest integer not greater than x i.e
$ [2.4] = 2 \\
\& [ - 2.3] = - 3$

If $ [k] = n \implies n\leq k < (n + 1)$

now, let us take $ [{x \over 99}] = [{x \over 101}] = n$ Where $ n \in N + \{0\}$ because x is positive

$ \implies n \leq {x \over 99} < (n + 1)$ [b]AND[/b] $ \implies n \leq {x \over 101} < (n + 1)$

$ \implies x \in [99n,99(n + 1)) \cap [101n,101(n + 1))$

$ \implies x \in [101n,99(n + 1))$

$ \implies 101n \leq x < 99(n + 1)$

there can be a value of x satisfying this condition only until $ 101n < 99(n + 1)$

${ 101n < 99(n + 1) \\
 \\
\& \implies 101n - 99n < 99 \\
 \\
\& \implies n < {99 \over 2} = 49.5 \\
 \\
\& \implies n \in \{0,1,2,3,4,5............49}\}$

for $ n = k$ for some $ k\leq 49,k \geq 0$ we have -

$ 101k \leq x < 99(k + 1)$

Since $ 99(k + 1)$ is excluded, number of values for $ x$ are -

$ 99(k + 1) - 101k = 99 - 2k$

therefore, total values for x are

$ (\sum\limits_{k = 0}^{49} 99 - 2k) - 1$--------------------[since the value of x = 0 is excluded because x is positve, not non zero]

$ = 50 \cdot 99 - 2({49(49 + 1) \over 2}) - 1 \\ \& = 50 \cdot 50 - 1 \\ \& = 2500 - 1 = 2499$

thus there are [u][i][b]2499 positive integral values of x[/b][/i][/u][/hide][/quote]

I didn't quite understand the last parts of this solution ; can someone kindly explain me ?[/quote]

Well, the sum of 50 99s is 50 * 99, and the sum of 2*0 + 2*1 + 2*2...2*49 = 2*(0 + 1 ... 48 + 49) = 2 * (49 * 50 / 2) = 49 * 50 So 50 * 99 - 50 * 49 -1 = 50 * 50 - 1.

[hide=Solution]Let $x=99q_1+r_1=101q_2+r_2$ where $r_1$ and $r_2$ are non-negative integers less than $99$ and $101$, respectively. Also, $q_1, q_2$ are non-negative.
$[\frac{x}{99}]=q_1=[\frac{x}{101}]=q_2=q$ (suppose)
So $x=99q+r_1=101q+r_2$.
$2q=r_1-r_2$
This difference can have a maximum value of $98$, which will give $q=49$.
Therefore $q\le49$.
For a particular value of $q$, we have $2q\le r_1\le98$, which gives $99-2q$ possible non-negative pairs $(r_1, r_2)$.
Hence, the total number of cases is $\sum_{q=0}^{49}\left(99-2q\right)=\left(99\bullet49\right)-\left(49\bullet50\right)=2500$.
Here, we have also considered the case where $q=0$ and $r_1=r_2=0$, which yields $x=0$. Since, $x$ is positive, we reject this case and our final answer is $2500-1=2499$. [/hide]
Hope it helps :)


inequalities
triangle inequality
IMC
college contests

It's obvious for $n=1$, and $n=1$ solves any other case as follows:

$|z_i|\le 1\Rightarrow |\frac {\sum z_i}n|\le \frac {\sum |z_i|}n\le 1\Rightarrow \frac {\sum z_i}n\in D$, so we can find $z\in D$ s.t. $|z-\frac {\sum z_i}n|\ge 1$.

We now have $\sum |z-z_i|\ge n|z-\frac {\sum z_i}n|\ge n$ and we're done.

[quote="Valentin Vornicu"]Let $D$ be the closed unit disk in the plane, and let $z_1,z_2,\ldots,z_n$ be fixed points in $D$. Prove that there exists a point $z$ in $D$ such that the sum of the distances from $z$ to each of the $n$ points is greater or equal than $n$.[/quote]

I have found a nice solution
Let P_1 be a point on the circumference and P_2 its diametrally opposed. Let A and B be the sum of the distances from P_1 and P_2 to the n fixed points respectively. By triangle inequality we have P_1z_i + P_2z_i >= 2  for i=1,2,...,n.
Summing over all is we have A + B >= 2n so A or B must be greater than or equal than n, this means that at least one between P_1 and P_2 is the z we were looking for.


Okay, i've resolved to learn how to program this summer, but I'm not sure what to do.  I tried reading "C programming language" but that didn't work out.  From the posts of some of you (Snowstorm, JSteinhardt,...) I'm thinking of signing up for a course from ICT([url=http://ict.org/]http://ict.org/[/url]), CTY([url=http://www.jhu.edu/cty/]www.jhu.edu/cty/[/url], for distance education), or EPGY([url=http://epgy.stanford.edu/]epgy.stanford.edu/[/url]).  I'm leaning towards epgy right now because it uses the exact same curriculum, book, and compiler as it's cty counterpart, plus it's a bit cheaper.  Just wanted to know if anyone had any final suggestions.
btw, i'm in 11th grade

[quote="quantum leap"]Okay, i've resolved to learn how to program this summer, but I'm not sure what to do.  I tried reading "C programming language" but that didn't work out.  From the posts of some of you (Snowstorm, JSteinhardt,...) I'm thinking of signing up for a course from ICT, CTY, or EPGY.  I'm leaning towards epgy right now because it uses the exact same curriculum, book, and compiler as it's cty counterpart, plus it's a bit cheaper.  Just wanted to know if anyone had any final suggestions.[/quote]

I liked the curriculum and faculty at CTY a lot, never heard of the other two. The people at CTY seemed kinda old to me (although I did this as a rising 7th/8th grader, so if you are a rising Freshman or Sophomore or something maybe it will be more tolerable).


inequalities
function
inequalities proposed

Given a positive integer.Let $x_{i}$ be a real sequence satisfied:
(n-1)$\sum x_{i}^2$=$(\sum x_{i})^2$ i=1 to n.
Show that all $x_{i}$ have the same sign.

for $x_i=0$, we have the inequality $\displaystyle (n-1)(\sum x_i^2)\geq (\sum x_i)^2$. furthermore, the arithmetic mean of both zeroes of the equality is positive, so both zeroes are positive(else one would be positive and one negative, so the function had to be negative at 0).

Peter


For how many positive integer values of $n$ is $1!+2!+3!+...+n!$ a perfect square?

A) none
B) 1
C) 2
D) 3
E) infinitely many

[hide](C) After 3!, this equals 3 mod 5. But 3 is not a quadratic residue mod 5, hence our only solutions are 1 and 3.[/hide]


geometry
perpendicular bisector
geometry unsolved

[color=darkred]In a scalene triangle $ABC,$ points $A',\,B',\,C'$ are the intersection points of the internal bisectors of angles $A,\,B,\,C$ with the opposite sides, respectively. Let $BC$ meet the perpendicular bisector of $AA'$ at $A'',\, CA$ meet the perpendicular bisector of $BB'$ at $B',$ and $AB$ meet the perpendicular bisector of $CC'$ at $C''.$ Prove that $A'',\, B''$ and $C''$ are collinear.[/color]

Also posted here:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=17064[/url]


The sequence of real numbers $ u_1,u_2,u_3,\ldots$ is such that $ u_1 \equal{} 1$ and $ u_{n \plus{} 1} \equal{} \frac {5u_n \plus{} 4}{u_n \plus{} 2}$ for all $ n \geq 1$. Prove that $ u_nn < 4$ for all $ n \geq 1$. [hide="Solution"] The proof is by induction. Let $ P(n)$ be the statement that $ u_nn < 4$, $ n \in \mathbb{N},n\geq 1$. For $ n \equal{} 1$, the LHS is equal to $ u_1(1) \equal{} 1$ by definition; since $ 1 < 4$, $ P(1)$ is true. Now, assume that $ P(k)$ is true for all $ k \in \mathbb{N},1 \leq k \leq m$ for some $ m \in \mathbb{N}$ ($ m \geq 1$). Then $ u_{k \plus{} 1}(k \plus{} 1) \equal{} \frac {5u_k \plus{} 4}{u_k \plus{} 2}(k \plus{} 1) \equal{} \frac{5u_kk \plus{} 4k \plus{} 5u_k\plus{}4}{u_k \plus{} 2} < \frac{20\plus{}4k\plus{}20/k\plus{}4}{4/k \plus{} 2} < \frac{2k^2\plus{}12k\plus{}10}{k \plus{} 2}$. Now at this point I want to make use of the result that $ 2k^2 \plus{} 12k \plus{} 10 < 4(k \plus{} 2)$ has no real solutions in $ k$. I will not prove this here, for it may be proven simply by considering a discriminant. Now we have $ u_{k \plus{} 1}(k \plus{} 1) < 4$, and $ P(k \plus{} 1)$ is true. $ P(1)$ is true, and the truth of all $ P(k)$, $ 1 \leq k \leq m$ for some $ m \in \mathbb{N}$ and $ k \in \mathbb{N}$, implies the truth of $ P(k \plus{} 1)$, therefore, by mathematical induction, $ P(n)$ is true for all $ n\in \mathbb{N},n \geq 1$. [/hide]

We have the following blatant counter-example: $ u_1 \equal{} 1 \implies u_2 \equal{} 3 \implies 2 u_2 \equal{} 6$

Your error comes right from the first inequality, where you put an upper bound on the numerator and the denominator. This doesn't show anything. For example, $ \frac{4}{3} < \frac{111}{7}$ and $ \frac{4}{3} > \frac{7}{111}$, but in both cases, the numerator and denominator on the right side are higher than the respective ones on the left side!

The more likely problem is to show that $ u_n < 4$, which is very straightforward

[hide="Hint"]$ \frac {5x \plus{} 4}{x \plus{} 2} \equal{} 5 \minus{} \frac {6}{x \plus{} 2}$[/hide]

[hide="INDUCTION!!!"]
[hide="Base case"]
To prove that $ u_1 < 4$, we require the use of Fermat's Last Theorem. Since $ a^4 \plus{} b^4 \equal{} c^4$ has no solutions, we have that $ u_1<4$.
[/hide]
So blah assume that $ u_k < 4$, then $ u_{k \plus{} 1} \equal{} 5 \minus{} \frac 6{u_k \plus{} 2}$, and hmm $ u_k < 4$ so $ u_k \plus{} 2 < 6$, so $ \frac 6{u_k \plus{} 2} > 1$, so $ 5 \minus{} \frac 6{u_k \plus{} 2} < 4$ and oh darn we're done.
[/hide]

And also, I think:
 $ 2k^{2} \plus{} 12k \plus{} 10 < 4(k \plus{} 2)$ has no solutions in $ k$
contradicts your argument?


Baron Minchausen talked to a mathematician. Baron said that in his country from any town one can reach any other town by a road. Also, if one makes a circular trip from any town, one passes through an odd number of other towns. By this, as an answer to the mathematician's question, baron said that each town is counted as many times as it is passed through. Baron also added that the same number of roads start at each town in his country, except for the town where he was born, at which a smaller number of roads. Then the mathematician said that baron lied. How did he conclude that ?

[quote]Bebakhshid ke saal ro engilisi ferestadam zira engilisish ,engilisiye ravoniye[/quote]

 :P


Hi ! Let $ x,y,z > 0$ and $ xyz\equal{}8$ prove that :
$ \frac{1}{\sqrt{1\plus{}x^3}} \plus{} \frac{1}{\sqrt{1\plus{}y^3}} \plus{} \frac{1}{\sqrt{1\plus{}z^3}} \geq 1$

[quote="mahanmath"]Hi ! Let $ x,y,z > 0$ and $ xyz \equal{} 8$ prove that :
$ \frac {1}{\sqrt {1 \plus{} x^3}} \plus{} \frac {1}{\sqrt {1 \plus{} y^3}} \plus{} \frac {1}{\sqrt {1 \plus{} z^3}} \geq 1$[/quote]
After using $ \sqrt {x^3 \plus{} 1}\leq\frac {x^2 \plus{} 2}{2}$ and  full expanding use AM-GM. :wink:

[quote="alex2008"][quote="Obel1x"][quote="arqady"]
After using $ \sqrt {x^3 \plus{} 1}\leq\frac {x^2 \plus{} 2}{2}$ and  full expanding use AM-GM. :wink:[/quote]

Actually it is $ \sqrt {x^3 \plus{} 1}\leq\frac {x^3 \plus{} 2}{2}$[/quote]

This doesn't solve the problem , but arqady's solves it ;)[/quote]

Sorry then.
[hide="PS:"] He didn't show the solution of it,that's why I thought he used Am-Gm and made that typo  :)[/hide]

[quote="Obel1x"][/quote][quote="alex2008"][quote="Obel1x"][quote="arqady"]
After using $ \sqrt {x^3 + 1}\leq\frac {x^2 + 2}{2}$ and  full expanding use AM-GM. :wink:[/quote]

Actually it is $ \sqrt {x^3 + 1}\leq\frac {x^3 + 2}{2}$[/quote]

This doesn't solve the problem , but arqady's solves it ;)[/quote][quote="Obel1x"]

Sorry then.
[hide="PS:"] He didn't show the solution of it,that's why I thought he used Am-Gm and made that typo  :)[/hide][/quote]

He used AM-GM to prove it as :

\[ \sqrt{x^3+1}=\sqrt{(x+1)(x^2-x+1)}\le \frac{x+1+x^2-x+1}{2}=\frac{x^2+2}{2}\];)


Five sailors are stranded on a desert island, and decide to go out and get coconuts to eat. When they get back it's pretty late, so they agree to split up the coconuts in the morning.

Well, the first one doesn't trust the others, and gets up in the middle of the night to get his share early. But he's pretty honest, and divides the coconuts into five equal piles, and throws one to a monkey to make the division even. He takes the coconuts back to his shelter and goes back to sleep.

The second one doesn't trust the others either, and gets up to get his share early. He divides the coconuts remaining (he doesn't know the first one has taken his share) and throws one to the same monkey to make the division even. He takes the coconuts back to his shelter and goes back to sleep.

The same thing happens with the next three sailors.

In the morning the five sailors find out that a whole army of monkeys have taken the coconuts in the shelters (because they saw the sailors go back) and were just in time to chase the monkeys away from the original pile. So they divide the coconuts into five piles, one for each, and once again throw a coconut to the monkeys (who at this time are pretty happy).

The question is:

What is the smallest number of coconuts that could have been in the original pile?

I remember this from an abacus problem. I think there had to be a lot, but I havent' found the answer yet.

Hint: What does the final pile have to be divisible by? So the fifth pile is congruent to 1 mod what? Work backward.

There are infinitely many possibilities. Just find the smallest possible (nonnegative) one.


combinatorics
IMO Shortlist
counting
IMO Longlist

Given $m+n$ numbers: $a_i,$ $i = 1,2, \ldots, m,$ $b_j$, $j = 1,2, \ldots, n,$ determine the number of pairs $(a_i,b_j)$ for which $|i-j| \geq k,$ where $k$ is a non-negative integer.

Provided $k\ge 1$ answer is $\sum_{i=1}^{m}A_i+\sum_{i=1}^{n}B_i$ where,
$A_i=\begin{cases} m-k-i,  \text{if} \ \ m-k-i>0 \\ 0,\text{otherwise}\end{cases}$, and $B_i=\begin{cases}n-k-i,\text{if} \ \ n-k-i>0 \\ 0,\text{otherwise}\end{cases}$.

If $k=0$ then the answer is simply $mn$.


inequalities
function
logarithms
inequalities unsolved

Let a,b,c,d be positive numbers and st:[tex]a+b+c+d=1[/tex]
Prove that:
[tex]\frac{abcd}{(a+b+c)(a+b+d)(a+c+d)(b+c+d)}\leq\ \frac{16}{81}(abc+bcd+cda+dab)[/tex]

Nice inequality on direct usage of AM_GM.

$(a+b+c)\geq 3(abc)^{1/3}$ (others the same)

$abc+bcd+cda+abd\geq 4(abcd)^{3/4}$

$abcd \leq (1/4) ^4$

The last sign was reversed.I think you made a mistake!

If you look carefully at the inequalities given till now in Mathlinks contest, you will find a solution for an even stronger inequality than yours. ;)

God help,I really want to know the proof of the ineq above.

Anyone who can prove it?I am waiting for the proof. :?

[quote="prowler"]Nice inequality on direct usage of AM_GM.

$(a+b+c)\geq 3(abc)^{1/3}$ (others the same)

$abc+bcd+cda+abd\geq 4(abcd)^{3/4}$

$abcd \leq (1/4)^{4}$[/quote]
There's a mistake.
$(a+b+c)(a+b+d)(a+c+d)(b+c+d)\geq 3^{4}abcd$
Therefore,
$1/(a+b+c)(a+b+d)(a+c+d)(b+c+d)=1/81$
Also we have $16(abc+bcd+acd+abd)/81\geq 64(abcd)^{3/4}/81$
But how do you proof that the last inequality is true?? :?:

This inequality is not easy but I think maybe Ji Chen's method does no good for those inequality learners :wink:

[quote="suweijie"]harazi,please show me the link,the ineq is too hard for me,but I really want to know the proof of it. :(  :?  :(  :?[/quote]
Actually, see
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=19097[/url]
And set $f(x)=\ln x$, we'll get
$(a+b+c)(b+c+d)(c+d+a)(d+a+b)\geqslant\frac{3^{4}}{4^{8/3}}(abcd)^{1/3}$
So the problem comes down to
$\frac{3^{4}}{4^{8/3}}(abcd)^{1/3}\geqslant\frac{81abcd}{16(abc+bcd+cda+dab)}\\ \Longleftrightarrow(abc+bcd+cda+dab)^{3}\geqslant16a^{2}b^{2}c^{2}d^{2}(a+b+c+d)\\ \Longleftrightarrow\frac{x+y+z+w}{4}\geqslant\left(\frac{xyz+yzw+zwx+wxy}{4}\right)^{1/3}$
Where $x=\frac{1}{a},y=\frac{1}{b},z=\frac{1}{c},w=\frac{1}{d}$


[quote="GameBot"]A model of a square pyramid is constructed using tennis balls. There are 16 balls in the bottom layer, and 1 ball in the top layer. In all other layers, each ball rests on 4 other balls. How many balls will be needed to form this pyramid?[/quote]

30. :)

16 in the bottom layer, 9 in the next, then 4, then 1. Each time there is one less on each side, resulting in 4^2, 3^2, 2^2, 1^2.

Perfect squares!  :lol:


When I'm reading the posts here , some of the things that are latexed dissapear .(do not show up) ... is there something wrong with my computer or is this a more global problem?

Sometimes I get the same thing but it usually comes back.  It's probably with your computer.  On sites that use $ \text{\TeX}$, it usually takes some time for everything on the page to load, it's not exclusive to AoPS.


Okay, so I got lost several times last year.  So, naturally, I go to look for a map.  But there are 9 different maps.

http://maps.ucsc.edu/

Which maps will be useful?  Which won't?

For UTDers, Titu sent us an email with information about travel information and courses. Try emailing him to see if he did the same for UCSCers.

Okay, I will, but did he send you a map of exactly where your dorms are?

p.s.  Lesson learned: Don't wander campus alone.  It's easier to get lost and harder to get found.

i second that ps...except for whenever i got found, it was by zuming's parents(aka my teachers)...so it was kinda awkward...


calculus
derivative
function
calculus computations

The following question can be solved using inequality. However, we are interested in a calculus solution.

Given: x , y >0; x+y <=1 . Find  Min,  Max of:
\[ S \equal{} 1/(x^2 \plus{} y^2) \plus{} 1/xy \plus{} xy \minus{} 2008
\]

Since this is a constrained, min/max problems, you'll want to use Lagrange multipliers.

Let $ f(x,y)\equal{}S$ as you defined it, and $ g(x,y)\equal{}x\plus{}y$.  Then solve
$ \nabla f(\textbf{x})\equal{}\lambda\nabla g(\textbf{x})$ to get your critical points.

[quote="toanviet"]The following question can be solved using inequality. However, we are interested in a calculus solution.

Given: x , y >0; x+y <=1 . Find  Min,  Max of:
\[ S = 1/(x^2 + y^2) + 1/xy + xy - 2008
\]
[/quote]

First we need to compute any relative extrema that might be within the region

So seenig that

$ \frac{\partial{S}}{\partial{x}}=\frac{-2x}{(x^2+y^2)^2}-\frac{1}{x^2y}+y$

and

$ \frac{\partial{S}}{\partial{y}}=\frac{-2y}{(x^2+y^2)^}-\frac{1}{y^2x}+x$

Now seeing that

$ \not\exists(x,y)\backepsilon\frac{\partial{S}}{\partial{x}}(x,y)=\frac{\partial{S}}{\partial{y}}(x,y)=0\wedge{x+y}\leq{1}$

But seeing that $ (0,1)$ and $ (1,0)$ make both partial derivatives undefined and sum to be less than or equal to zero, we can conclude that those are two critical points.

So now we must consider the ede of the region, we do this by lagrange multipliers, now letting $ g(x,y)=x+y-1$

We must solve the following system

$ \left\{\begin{array}{c}f(x,y)_x=g(x,y)_x\lambda\\f(x,y)_y=g(x,y)_y\lambda\\g(x,y)=0\end{array}\right\}$

Test all points that satsify that and the two points mentioned above in the original function, the biggest is the max, and the smallest is the min.


[hide]

Umm... the six-digit numbers that don't have a single even digit must be composed of the digits 1, 3, 5, 7, or 9...

five choices for the first digit, five choices for the second digit, five choices for the third digit, five choices for the fourth digit, five choices for the fifth digit, and five choice for the sixth digit.

5^6 = 15625

There are 900000 six-digit numbers, so 900000 - 15625 = 884375.

[/hide]

[quote="gameworld7"]there's only 899999 6-digit numbers. i think but that's what i think. i might be wrong[/quote]

Actually there are 900000 6 digit numbers, and the problem was asking for how many 6 digit numbers have at least one even digits, not how 6 digit numbers there are total.

[quote="A+MATH"]What do you mean by six digit numbers, do you mean with a even number or just six digit ones?[/quote]

I mean numbers with 6 digits.

Hello, rrusczyk!

[quote]How many 6 digit numbers have at least one even digit?[/quote]

As h_s_potter2002 pointed out: there are 900,000 six-digit numbers.

The opposite of "at least one even" is "no even" or "[b][i]all odd[/b][/i]".

How many six-digits number have all odd digits?
There are five choices for the first digit: {1,3,5,7,9}
and five choices for the second digit . . . etc.
Hence, there are $5^6 = 15,625$ six-digit numbers with all odd digits.

Therefore, there are $900,000 - 15,625 = 884,375$ six-digit numbers with some even digits.

*tears out hair*
*grumbles about people not using spoilers*
*realizes that you can't LaTeX in spoilers*
*calms down*

[hide]ok so there are 999999 - 99999 = 900000 possible 6 digit numbers.  We know or at least hope we know that there are 5 primes, 1, 3, 5, 7, 9.  Therefore, 5 exp 6 is the total number of numbers which don't have any even digits.  So 900000 - (5 exp 6) = 900000 - 15625 = 884375[/hide]

1, 3, 5, 7 and 9 are odd not prime...3, 5 and 7 are prime but 1 and 9 are not.

Instead of using 5 exp 6 you could do 5^6 which is used more often.  Or if you want it to be really nice use $\LaTeX$ which looks like $5^6$...to do this you type 5^6 and enclose it with dollar signs:

[code]
$5^6$[/code]


limit
integration
calculus
calculus computations

Evaluate the following limit:

$ \lim_{n\rightarrow \infty} \left(\frac{1}{\sqrt{n}\sqrt{n\plus{}1}}\plus{}\frac{1}{\sqrt{n}\sqrt{n\plus{}2}}\plus{}\cdots\plus{}\frac{1}{\sqrt{n}{\sqrt{n\plus{}n}}}\right)$.

$ \sqrt {n}\sqrt {n \plus{} k} \equal{} n\left(1 \plus{} \sqrt{\frac {k}{n}}\right)\rightarrow \text{Limit of Riemman Sum}$.

Not exactly that, but that's the idea. $ \frac1{\sqrt{n}\cdot\sqrt{n\plus{}k}}\equal{}\frac1n\cdot\frac1{\sqrt{1\plus{}\frac{k}{n}}}$. Therefore, using a uniform partition,
$ \lim_{n\to\infty}\sum_{k\equal{}1}^n \frac1n\cdot\frac1{\sqrt{1\plus{}\frac{k}{n}}} \equal{} \int_1^2 \frac1{\sqrt{x}}\,dx \equal{} \left[ 2x^{\frac12}\right]_1^2 \equal{} 2\sqrt{2}\minus{}2$.

That's correct. I just give a hint.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=118716&search_id=56017944]Practice[/url]


Seven judges graded Joshua's science project, and the mean of those seven scores was 8.0. Scores could be any integer value 1-10. When the highest and lowest scores were removed, the mean of the five remaining scores was 7.8. What was the least possible score that could have been removed?

This means that the sum of the judges' scores was 56. When the highest and lowest scores were removed, the sum became 39. So the highest+lowest is 56-39=17. The lowest possible score that could be taken was 7.

EDIT: Typo, thanks Mew.

BY izzys logic H+L=17. L is clearly minimized when H is maximized. H is maximized at 10, making L=7

Um, I put 7 and it says the correct answer is 6. Can you mods please take a look at that and change it? Thanks

how is the answer 6?
I got seven and i checked a bunch of times.
:o :o :o :o


integration
function
Functional Analysis
real analysis
real analysis unsolved

I'm having problems proving Picard-Lindelöf theorem. I have the sequence $ y_{k\plus{}1} (t) \equal{} y_0 \plus{} \int_{t_0}^t F(s, y_k(s))ds$ and  want to prove that this converges. I define $ y_{k\plus{}1}: \equal{}T(y_k)$. Now I would like to use Banachs fix point theorem, but this requires that T is a strict contraction. From here I can't come any further.

Two things:

The first is that you're going to have to adjust which Banach space you're talking about. You want $ C[t_0\minus{}\delta,t_0\plus{}\delta]$ - actually a closed subset even of that, with the values of the function $ y$ confined to some bounded interval - for some $ \delta>0,$ but you're going to have to wait to choose $ \delta$ until you can find the conditions you need for the map to be a contraction. It probably won't be a contraction for all $ \delta,$ just for sufficiently small $ \delta.$

The second is that you do have (and must have) conditions on $ F.$ The most significant is that $ F$ is a Lipschitz function of its second argument. That is, $ |F(x,u)\minus{}F(x,v)|\le M|u\minus{}v|$ for a specified $ M.$

Then, using $ \|\cdot\|$ to mean the sup norm on the interval $ [t_0\minus{}\delta,t_0\plus{}\delta],$ we get the following:

$ T(u)(t)\minus{}T(v)(t)\equal{}\left[y_0\plus{}\int_{t_0}^tF(s,u(s))\,ds\right]\minus{}
\left[y_0\plus{}\int_{t_0}^tF(s,v(s))\,ds\right]$

$ T(u)(t)\minus{}T(v)(t)\equal{}\int_{t_0}^t(F(s,u(s))\minus{}F(s,v(s)))\,ds$

$ |T(u)(t)\minus{}T(v)(t)|\le\left|\int_{t_0}^tM|u(s)\minus{}v(s)|\,ds\right|
\le|t\minus{}t_0|M\|u\minus{}v\|.$

We still have to take the sup over $ t$ to get $ \|T(u)\minus{}T(v)\|,$ and there are some other considerations, like making sure that $ (t,T(u)(t))$ lands within the region for which the Lipschitz condition on $ F$ is valid. Your strict contraction is going to come from choosing $ \delta$ so that $ \delta M<1.$


modular arithmetic
geometry
number theory
algebra unsolved
algebra

I think it will be very usefull for everybody...

1. If you mean that the $ a_i$ are all the positive integers less than $ 10^{2007}$ that are prime to $ 10^{2007}$ then we may assume that $ a_1<a_2<\dots<a_n$ and we have $ a_i\plus{}a_{n\minus{}i\plus{}1}\equal{}10^{2007}$ for all $ i$. Thus $ \sum_{i\equal{}1}^na_i\equal{}\frac{10^{2007}}{2}$. But if you mean that $ (a_i)$ is any sequence of positive integers satisfying the two conditions then the intended sum is not definite.

4. If we have $ a_m\equal{}a_1^k$ for some $ m$ then $ a_1\plus{}(m\minus{}1)d\equal{}a_1^k\implies (m\minus{}1)d\equal{}a_1^k\minus{}a_1$. Now since $ (a_1,d)\equal{}1$, there exists $ q$ such that $ a_1^q\equiv 1\pmod d$ which implies $ d\mid a_1^{q\plus{}1}\minus{}a_1$. So we can take $ k\equal{}q\plus{}1$ and $ m\minus{}1\equal{}\frac{a_1^{q\plus{}1}\minus{}a_1}{d}$.

hi, this is my first message  :oops:

1. given $ a_i < m$ and $ (a_i,m)=1 \forall i$. so, if $ m=\prod_{i=1}^{k}{p_i^{\alpha_i}}$ with $ p_i$ prime and $ \alpha_i>0 \forall i$ so (after nayel observation) $ \displaystyle \sum_{i=1}^{n}{a_i}=\frac{m{\phi{(m)}}}{2}$ where $ \phi{(m)}=m (\prod_{1}^{k}{1-\frac{1}{p_i}})$ is the number of $ a_i$

3. forall ABC not collinear exist D,E,F such that ABCD, ABCE, ABCF are parallelogramm. for example, ABCD. P is a point inside(!!) ABCD. from P we draw parallel line to AB and BC through P. these lines intersect AB,BC,CD,DA in G,H,I,L. we get L,M,N,O on those lines such that LG=GP,MH=HP,NI=IP,OL=LP.
for each choise of P we get LMNO  convex quadrilateral and A,B,C are three midpont  of those sides

paolo

1. Don't post problems in an image like this.  Post using text (or LaTeX).

2. Generally, you should only have one problem in a post, especially when you are dealing with problems in different subjects (i.e. algebra, geometry, number theory, etc.).


4 questions:

How many of these answers are equal to 0?
How many of these answers are equal to 1?
How many of these answers are equal to 2?
How many of these answers are equal to 3?

Fill in the numbers so that each answer is correct. There are two solutions.

---

And here's a variation. As far as I know there's only one solution here.

100 questions:

How many of these answers are equal to 0?
How many of these answers are equal to 1?
How many of these answers are equal to 2?
etc.
etc.
etc.
How many of these answers are equal to 98?
How many of these answers are equal to 99?

This problem is quite confusing. At first, I couldn't understand it.

[hide="I got"]$2,0,2,0$
$1,2,1,0$[/hide]
[hide="And"]$96,2,1,0,0,...,0,0,1,0,0,0$[/hide]

Isnt this from a Mathcounts Masters Round? Except they changed the wording around a bit?

I got the idea from a puzzle in Games magazine.


delta ABC,a=BC, b=CA, c=AB,
prove that
$ \frac {(bC \minus{} cB)^{2} \plus{} (cA \minus{} aC)^{2} \plus{} (aB \minus{} bA)^{2}}{(a \plus{} b \plus{} c)^{2}} \le \frac {(A\plus{}B\plus{}C)^{2}}{4}$

What about A,B,C? Are they any positive reals?

[quote="lasha"]What about $ A,B,C$? Are they any positive reals?[/quote]

$ A,B,C$ may refer to the measure of the angles of $ \Delta ABC$, in radians... but the wording is quite confusing, so I'm not sure. :ddr:

[quote="earldbest"]
$ A,B,C$ may refer to the measure of the angles of $ \Delta ABC$, in radians... but the wording is quite confusing, so I'm not sure. :ddr:[/quote]
sorry for my confusing word.
you are right.$ A,B,C$ refer to the measure of the angles of $ \Delta ABC$,
and $ a,b,c$ refer to length of side.


geometry
3D geometry
pyramid
Pythagorean Theorem

What is the area, in square units, of the four triangular faces of a right, square-based pyramid that has base edges measuring 6 units and lateral edges measuring 5 units?

[quote=james4l]The lateral edges measure 5 and the base measures 6, so the height, by Pythagorean Theorem, is 4. Therefore, each triangular face is area of 12, so 4*12 = 48 is our answer.[/quote]

You need to slow down a more.

you can make a right triangle with the lateral edge, half the base, and the height. Since this triangle has hypotenuse 5 and leg 6/2 = 3, the height is 4. Thus, the volume is $\frac{1}{3}Bh = \frac{1}{3}*6^2*4 = 48$


I'm pretty sure this is not Calculus, this is middle school math.

1 Kg = 1 L

$ 10^6$ mg = $ 10^3$ ml

Now you can make a porportion,

$ \frac{250}{10^6} \equal{} \frac{?}{10^3}$

$ ? \equal{} \frac{25}{100} \equal{} \boxed{\frac{1}{4}}$

1 liter has 1000 milliliters in it, and 1 kg has 1000 milligrams in it. so if 1=1, then 250/1000 = 250/1000. 250mg = 250mL.

Where did my zeroes go? I thought of grams. add 3 0s to the mgs.

[quote="pascal12"][quote="arbala"]1kg=1L
1000000mg=1000ml

1mg=1000/1000000=1/1000

250mg=250*1/1000= 1/4
250mg=1/4ml[/quote]
[quote="GoldenRatioOfPhi"]:ninja: 25/100=1/4 :ddr:[/quote]

Hey guys, I just noticed you're new here. For future reminder, please don't post solutions that have already been posted unless you have something new to say or a different method, and especially, don't revive topics. This thread was already over 2 weeks old and solved.[/quote]

Thanks for you information i newly join and your post help me.


How do you type the arc symbol in Latex?

\arc{} doesn't work and it's the only logical one... i think...

Several variations are discussed in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=18587]this thread[/url] (starting at post #6).
There's not really a nice way to do it.


Find all regular $n$-gons with the following properties:
(a) a diagonal is equal to the sum of two other diagonals
(b) a diagonal is equal to the sum of a side and another diagonal

A question: it is needed both conditions?
Better: it is equivalent to this reformulation:

Find all regular n-gon satisfying both conditions below:

1 - There are three diagonals, one of them is equal to sum of other two;
2 - There are two diagonals whose difference is a side.


geometry
trapezoid
inequalities unsolved
inequalities

Let $ABCD$ be a quadrilateral inscribed in a circle of radius $R$ and circumscribed in a circle of radius $r$.  If $s$ is the semiperimeter we have

$ s^2 \geq 8r( \sqrt{4R^2 +r^2} -r) $

with equality if and only if the quadrilateral is an isosceles trapezium

Did you get the solution of this problem?


function
Euler
number theory
prime factorization
number theory proposed

For a positive integer $ n$, $ f(n)$ denotes the sum of all the exponents in the prime factorization of $ n$.

Determine the values of $ \sum_{n \equal{} 1}^{\infty} \frac {( \minus{} 1)^{f(n)}}{n}$ and $ \sum_{n \equal{} 1}^{\infty} \frac {( \minus{} 1)^{f(n)}}{n^2}$

$ a(n) \equal{} (\minus{}1)^{f(n)}$ is an arithmetic function with value $ \minus{}1$ at every prime, hence $ \sum \frac{a(n)}{n^s}$ has Euler product

$ \prod_{p} \left( \frac{1}{1 \plus{} \frac{1}{p^s}} \right)$

which diverges as $ s \to 1$.  I'm not sure what its value when $ s \equal{} 2$ is.

$ \prod_{p} \left( \frac {1}{1 \plus{} \frac {1}{p^s}} \right) \cdot \prod_{p} \left( \frac {1}{1 \minus{} \frac {1}{p^s}} \right) \equal{} \prod_{p} \left( \frac {1}{1 \minus{} \frac {1}{p^{2s}}} \right)$, thus $ \prod_{p} \left( \frac {1}{1 \plus{} \frac {1}{p^s}} \right) \equal{} \frac {\zeta(2s)}{\zeta(s)}$.
So the answer for $ s \equal{} 2$ is $ \frac {\pi^2}{15}$.

Right.

@t0rajir0u: Your wording makes it sound like the first sum diverges.
It is true that $ \prod_{p} \left( 1 \plus{} \frac {1}{p}\right)$ diverges. But $ \prod_{p} \left( \frac {1}{1 \plus{} \frac {1}{p}} \right)$ does not. It is equal to $ 0$.
The first sum equals $ 0$.

Sorry; the convention is that an infinite product diverges if it approaches $ 0$ but none of its terms are equal to $ 0$, but that doesn't translate into the sum diverging to infinity, of course.  My apologies.

No problem. I was not aware of the convention. That's all. I should be sorry.


Αυτό το ερώτημα είναι μέρος μιας άσκησης από το αγγλικό φόρουμ :

[color=red][b]ΑΣΚΗΣΗ[/b][/color]

 [b][color=red]Δίνεται τρίγωνο ΑΒΓ, η διάμεσος ΑZ και τα ύψη ΒΔ , ΓΕ. Η κάθετη στην ΑΖ στο Α τέμνει τις ευθείες ΒΔ , ΓΕ στα Μ,Ν. Να αποδειχθεί ότι ΑΜ=ΑΝ.[/color][/b]

[b]Σχόλιο[/b]

 Η λύση που έχω δει βασίζεται σε αρμονικές τετράδες , αρμονικές δέσμες και πολικές. Αυτό όμως το ζητούμενο με τα τόσο απλά δεδομένα φαίνεται περίεργο να μην βγαίνει και πιο απλά. Αν βρείτε κάτι, το βάζετε εδώ να το δούμε όλοι.

 Μπάμπης

[b][color=red]edit.[/color][/b]Έβαλα ΑΖ τη διάμεσο

Δεν ξέρω αν είναι το στυλ της λύσης που ζητάμε… αλλά αν δε μου ξέφυγε κάτι βγαίνει εύκολα με αναλυτική γεωμετρία.

Βολεύει να πάρουμε κατακόρυφο άξονα την ευθεία ΑΜ, οριζόντιο την κάθετο της ΑΜ και αρχή αξόνων το Α.
Οι συντεταγμένες των σημείων είναι:
Α (0,0)
Μ (0,μ)
Β (-α, μ-λ)
Γ (α, μ+λ)
Οι 2 τελευταίες σχέσεις έχουν επιλεγεί έτσι ώστε το Μ να είναι μέσο της ΒΓ.

Βρίσκεις τις εξισώσεις των ευθειών (παίζοντας με συντελεστές διεύθυνσης της καθέτου και διερχόμενο σημείο) και τέλος βρίσκεις τα σημεία τομής των ΒΔ και ΓΕ με τον οριζόντιο άξονα. Οι τετμημένες τους είναι αντίθετες και άρα ισαπέχουν από το Α.

Μπάμπη, ονόμασες δύο διαφορετικά σημεία ως Μ. Στην παρακάτω λύση θα γράφω ΑΖ για την διάμεσο.

Από νόμο ημιτόνων στο ΑΒΖ παίρνω ημ(ΒΑΖ)/(ΒΖ) = ημ(ΒΖΑ)/(ΑΒ). Στο τρίγωνο ΑΕΝ έχω συν(ΕΑΝ) = ημ(ΒΑΖ) = (ΑΕ)/(ΑΝ). Άρα (ΑΝ) = (ΑΕ)/ημ(ΒΑΖ) =  (ΑΕ)(ΑΒ)/[(ΒΖ)ημ(ΒΖΑ)]. Ομοίως, (ΑΜ) = (ΑΔ)(ΑΓ)/[(ΓΖ)ημ(ΓΖΑ)]. Επειδή ΒΖ=ΓΖ και  ημ(ΒΖΑ) = ημ(ΓΖΑ), αρκεί να δείξω ότι (ΑΕ)(ΑΒ) = (ΑΔ)(ΑΓ). Αλλά συν(ΒΑΓ) = (ΑΔ)/(ΑΒ) = (ΑΕ)/(ΑΓ)

Συνθετικά η άσκηση πράγματι φαίνεται αρκετά δύσκολη να βγει απλά και φαίνεται να χρειάζεται αρμονικά και πολικές (ή τουλάχιστον έμμεσα). Η άσκηση είναι άμεσα συνδεδεμένη με θεώρημα πεταλούδας για τον κύκλο με διάμετρο τη $ BC$ και ειδικά στην περίπτωση που η Α είναι αμβλεία η λύση είναι άμεση με τη χρήση του θεωρήματος αυτού. (Στην περίπτωση της οξείας πιστεύω ότι θα βγει αν χρησιμοποιήσουμε το παραπάνω αποτέλεσμα για το ορθόκεντρο του τριγώνου... αλλά δεν είμαι και σίγουρος αν βγαίνει).

Πάντως θέλω να βάλω και 1 ακόμα ερωτηματάκι για το σχήμα, χωρίς όμως να είμαι 100% σίγουρος ότι ισχύει :blush:
Να αποδείξετε ότι αν οι [b]ΔΕ, ΒΓ[/b] τέμνουν την [b]ΜΝ[/b] (δηλαδή την κάθετη στην διάμεσο) στα [b]Κ,Λ[/b] αντίστοιχα, τότε [b]ΑΚ=ΑΛ[/b]

Δια του $ B$ φέρνουμε ευθεία παράλληλη προς την ευθεία $ CE\perp AB,$ η οποία τέμνει την $ AH,$ όπου $ H$ είναι το ορθόκεντρο του $ \bigtriangleup ABC,$ στο σημείο $ X$ και έστω $ Q$ το σημείο επί της ευθείας $ CE,$ ώστε $ XQ\parallel BD\perp AC.$

Από το παραλληλόγραμμο $ BHQX$ έχουμε $ BP \equal{} PQ,$ όπου $ P\equiv BQ\cap HX$ και αρκεί να αποδείξουμε ότι $ BQ\perp AZ,$ όπου $ Z$ είναι το μέσον της πλευράς $ BC.$

Είναι εύκολο να δούμε ότι τα τρίγωνα $ \bigtriangleup QHX,\ \bigtriangleup ABC$ είναι όμοια γιατί έχουν $ \angle H \equal{} \angle B$ και $ \angle X \equal{} \angle C.$

Επί πλέον, τα τρίγωνα αυτά είναι συσχετισμένα έτσι ώστε οι ομόλογες πλευρές τους να είναι μεταξύ τους κάθετες $ ($ $ QH\perp AB,$ $ QX\perp AC,$ $ HX\perp BC$ $ )$ και επομένως έχουμε $ QP\perp AZ$  $ ,(1)$ ως οι διάμεσες δια των ομόλογων κορυφών τους $ Q,\ A,$ αντιστοίχως.

Από $ (1)$ και $ MN\perp AZ$ προκύπτει ότι $ BQ\parallel MN,$ από όπου με το [b][size=100]θεώρημα Θαλή[/size][/b] ( λόγω $ BP \equal{} PQ$ ), συμπεραίνουμε ότι $ AM \equal{} AN$ και η πρόταση έχει αποδειχθεί.

$ \bullet$ Η απόδειξη αυτή αφιερώνεται στον [b][size=100]Δημήτρη Παπαδημητρίου[/size][/b].

 Κώστας Βήττας.

[quote="bilot4"]... Να αποδείξετε ότι αν οι [b]ΔΕ, ΒΓ[/b] τέμνουν την [b]ΜΝ[/b] (δηλαδή την κάθετη στην διάμεσο) στα [b]Κ,Λ[/b] αντίστοιχα, τότε [b]ΑΚ=ΑΛ[/b][/quote]
[hide]Νομίζω ότι αυτό αληθεύει και μπορεί να αποδειχθεί με βάση το [b][size=100]θεώρημα Μενελάου[/size][/b], στο τρίγωνο $ \bigtriangleup HMN$ όπου έχει αποδειχθεί ότι $ AM \equal{} AN,$ με διατέμνουσες τις ευθείες $ DE,\ BC,\ AB$ και $ AC$ $ ($ σχήμα [b][size=100]t=240616[/size][/b] $ ).$[/hide]
Κώστας Βήττας.

[quote="vittasko"]
$ \bullet$ Η απόδειξη αυτή αφιερώνεται στον [b][size=100]Δημήτρη Παπαδημητρίου[/size][/b].
 Κώστας Βήττας.[/quote]

Σας ευχαριστώ πάρα πολύ για την αφιέρωση κ. Βήττα. Η λύση είναι υπέροχη και βασισμένη σε πολύ απλά θεωρήματα! :clap2:

Αναφορικά με το ερώτημα που έβαλα νομίζω πως αποτελεί γενίκευση του θεωρήματος της πεταλούδας για τα σημεία που είναι εκτός κύκλου και όπως πολύ σωστά τονίζετε αποδεικνύεται και με θεώρημα Μενελάου. Μπορείτε να βρείτε αυτή τη γενίκευση εδώ:
[i]hrcak.srce.hr/file/1635[/i]


linear algebra
matrix
analytic geometry
calculus
derivative
advanced fields
advanced fields unsolved

does anyone know how to show that the levi-civita tensor (not the symbol) is really a tensor - disregarding one sign? i don't even know where to start with this proof. any help would be GREATLY appreciated.

Show that it is a multilinear map from $\phi: V\times ... \times V\times V^{*}\times ... \times V^{*}\rightarrow\mathbb{R}$ where $V^{*}$ is the dual space to $V$. I'm pretty sure that in the Levi-civita case $V=\mathbb{R}^{3}$.

As for multilinearity, show $\phi(v_{1}, ... ,\alpha v_{k}+\alpha' v_{k}', ...)=\alpha\phi (v_{1}, ..., v_{k}, ...)+\alpha'\phi(v_{1}, ... , v_{k}', ...)$ for all $\alpha, \alpha'\in\mathbb{R}$ and $v_{k}, v_{k}'\in V$.

here are the levi-civita tensor components defined:

e_(mu_1mu_2...mu_n) = (sqrt(det(g)))(e*_(mu_1mu_2...mu_n)

where g is the metric, and e*_(mu_1mu_2...mu_n) is the levi-civita symbol.

to prove that the lc tensor is a tensor i believe i only have to show that it transforms as a tensor, in other words that it obeys the tensor transformation law, given below for a (k,l)-tensor:

T^(mu'_1...mu'_k)_(nu'_1...nu'_l) = (dx^mu'_1/dx^mu_1)...(dx^mu'_k/dx^mu_k)(dx^nu_1/dx^nu'_1)...(dx^nu_l/dx^nu'_l)T^(mu_1...mu_k)_(nu_1...nu_l)

det(g) transforms as follows:

det(g(x^mu')) = (J^-2)det(g(x^mu))

where J is the Jacobian matrix (det(dx^mu'/dx^mu))

this follows easily from taking the determinant of both sides of the following transformation law for the metric g:

g_mu'nu' = (dx^mu/dx^mu')(dx^nu/dx^nu')g_munu

the levi-civita symbol (e*_(mu_1mu_2...mu_n)) is defined as follows:

                                       { +1     if (mu_1mu_2...mu_n) is an even permutation of 123...(n-1)n
e*_(mu_1mu_2...mu_n) = { -1      if (mu_1mu_2...mu_n) is an odd permutation of 123...(n-1)n
                                       { 0       otherwise

it is defined this way in any coordinate system. therefore it should obey the transformation e*_(mu'_1mu'_2...mu'_n) = e*_(mu_1mu_2...mu_n). however in my gen. rel. book they state that the symbol transforms in the following way:

e*_(mu'_1mu'_2...mu'_n) = Je*_(mu_1mu_2...mu_n)(dx^mu_1/dx^mu'_1)(dx^mu_2/dx^mu'_2)...(dx^mu_n/dx^mu'_n)

or in other words that it transforms exactly as a tensor but only with the Jacobian out front. they derive this from the fact that any determinant of an n x n matrix M^mu_mu' obeys the following property

(e*_(mu'_1mu'_2...mu'_n))det(M) = (e*_(mu_1mu_2...mu_n))(M^(mu_1)_mu'_1)(M^(mu_2)_mu'_2)...(M^(mu_n)_mu'_n)

where i guess M^(mu_i)_(mu'_i) is the iith component of the M matrix. they then set M^mu_mu' to equal (dx^mu/dx^mu') (which is the inverse of the Jacobian matrix) and multiply through by the inverse of the determinant to get the Jacobian on the other side, as above.

 I don't understand how they can say that the lc symbol is "defined not to change under coordinate transformations" and then define its transformation as above, (tensor with jacobian), less than half a page later, unless the Jacobian happens to be the inverse of that string of products of derivatives at the end. which it doesn't, i don't think. if someone could help me understand this that would be great.

i also really need to figure out how to put equations in here.


We have a body $A$ on an inclined plane of angle $\alpha$ and $\mu = \frac{\sqrt{3}}{3}$.

a) Find $\alpha$ s.t. the descent of the body is uniform.

b) Let $\alpha = 60^{\circ}$. At the top of the plane we have an ideal pulley. Another body $B$ is tied to $A$ via the pulley. It has mass $m_B = \frac14 m_A$. Find the speed of the system, when $B$ has risen $h = 0.5m$. (the system is big enough to ensure that until then the two bodies mentain their direction of movement)

[color=blue]
a)Because the descent of the body is uniform so
$\tan\alpha=\mu=\frac{\sqrt{3}}{3}\Longrightarrow \alpha=30^0$
b)The vertical displacement of mass A is $h\sin\alpha$
The law of conservation of energy
$\large 0=\frac{1}{2}(m_A+m_B)v^2+m_Bgh-m_Agh\sin\alpha$

$\large \Longrightarrow v=\sqrt{\frac{2}{5}gh(4\sin\alpha-1)}$[/color]

#1: Somehow I feel it's a typicall Romanian problem. Am I right?

It's a stupid "BAC" problem, which I couldn't solve given that my knowledge in physics is very, very LOW. (romanian "bata")

I suggest that you open a physica book and study... in case you still need physics :)


Find all solutions to the set of equations $x^2+y^2=xy+13$, $x+y=2xy-17$.

Pfff  :wallbash: Apparently I can't solve quadratics.

Edited.

Yep, (3,4) and (4,3) definitely work.  Discriminant should be 625 up there.  Solutions end up being xy = 12 and xy = 23/4.


algebra
polynomial
function
algebra proposed

$ f(x) \equal{}\frac{x^{1998}}{1\plus{}x^{2}}$ is not polinomial. But for $ P(x)\equiv 0$ we have $ |f(x)\minus{}P(x)| < x^{1998}$ (c=1).

It is true problem:
Vietnamese TST 1998 : Assume that the function $ f(x)$ defined  on $ R$ has the following property : for every positive number $ c$ there exists a polynomial  $ P_{c}(x)$ with real coefficients such that
                                $ | f(x)\minus{}P_{c}(x) |\leq cx^{1}998$ for all $ x\in R$
Prove that  $ f(x)$ is also a polynomial with real function

[quote="Lee Sang Hoon"]Vietnam TST 1998
prove that function f:R->R that satisfies following condition is polynomial.
[condition] there exists c>0 and polynomial P(x) such that |f(x)-P(x)|<=cx^1998[/quote]
Please post your solution here http://www.mathlinks.ro/Forum/viewtopic.php?p=268334#268334  :wink:  .


This is a problem I've discovered ...

The date for the MOP this year was put out as June 11th (I think) but my school doesn't get out until the 25th, and I don't think I can miss that much school, even though I would really like to go if I got invited.

Might anybody else have this problem?

And does anyone have any advice about what I should do?

I have that problem.  The construction of an extention to one of our schools pushed schools back.  I wouldn't know what to do and I probably don't have to worry about it.  (I posted because I have the same problem).  I probably would just beg the school to let me go.

Well, first worry about making MOP, because those dates are tentative anyway (they're the same as last year's, without any adjustment for day of week).

[quote="paladin8"]Bah, who cares about school if you can go to MOP.[/quote]
My thinking exactly. Take the finals early and get the heck out of there.

If I made MOP, I'd do the same thing... but if the principal said no, you can't go, I'd probably #1: light begging #2: begging #3: crying #4: I'd smack him. If all fails, then you could ask Mr. Feng about what to do, I guess...

Just say that you got gonorria or something and then skip the last month of school :D .

or just go to a school that releases before that day...

[quote="Some people"]certain things[/quote]
Please try to be less... infantile.

Your school will probably understand.  But the dates will probably change anyway, and until the USAMO is over, anyhow, your time would probably be better spent doing problems so as to make MOP more likely for you than worrying about what will happen if you do get in.

before worrying about this, at least make MOP first.  :wink:

Two weeks shouldn't be much trouble, since you can inform your school and teachers in advance. I missed a similar amount of school the two times I went to MOP- all I had to do was arrange to take some finals early.
The second time, my schedule was mostly AP classes- those are really easy to get out of early, since they're basically over in May.

Yeah, I had to miss over a week of school last year, but it wasn't too much of a problem. Anyways, MOP is so much more important than school, as long as you maintain your grade . . .

My school district has a history of not letting me do anything I want ....

They like things done their way, and as far as they're concerned, if I want to go, I can.  I'll just fail all my classes, and they won't care if I make stuff up or not.

I had to miss a week of school last year for MOP.  The teachers were ok about it....as long as you took extra exams beforehand :(.

[quote="tokenadult"]It sounds like you should get into Running Start or the U Dub early entrance program, if your school district is so inflexible.

Anyway, don't miss MOP. Colleges know MOP is a bigger deal than high school.[/quote]

Actually, you should make sure that you don't fail your classes because you skipped your finals (it won't get that bad, right?), or if you've really really pissed off your teachers/administrations. Generally speaking, being a MOPper is much much better than merely doing well in high school. But if you end up having teachers who aren't willing to write brilliant recommendation letters for you (I doubt this would be a problem in this case), or failing an entire semester of classes, you'd be better off going to school and making the grades than making the IMO (as far as colleges are concerned).

There are many people who made MOP (I've used this example before, a black MOPper) but didn't get into any top colleges because of his grades. But hey, if your guidance counselor is willing to personally clear up the matter for you with the admissions people, it could count as extenuating circumstances, but this is hopeful thinking of course.

Anyway, I like how you already assumed you're making MOP. That kind of confidence will keep you focusd on the USAMO  :)

Well, I'm in 9th grade and I've already taken it once, so I should do pretty well.  I'm hoping.

Actually, the counselors we have are the worst of the district:mad: they seem unable to get anything right.

I'm concerened that if I did go, my grades would suffer so much that in the end it wouldn't be worth it.

I know I won't make it, but I have finals that week, and I am afraid of missing any classes for they will be having inportant info.

I also hope that School ends right before Mop Begins but I don't know.

You really need to find yourself another school if your school doesn't appreciate the importance of MOP. Please tell your parents I said so--I'm a parent who homeschools his children rather than let school limit their opportunities.

[quote="tokenadult"]'m a parent who homeschools his children rather than let school limit their opportunities.[/quote]

Yes, finding a school that fosters interest in anything (math included) is a necessity.  So if yours does not, then find one that does, or look into homeschooling options.  I am fortunate enough to go to a private school that places an emphasis on academics, but only one public high school in my area has a strong math team (that would be Bellevue HS).  You need support from your teachers and your school in general.  If you are not getting that, there is a problem.

Homeschool=pure joy

My school is quite ghetto.

VERY UNSPECTACULAR

I had to miss a couple of weeks of school for MOP two years ago. My school was very good about it; I was able to take most of my exams [i]late[/i] (and in whatever order I wanted to even :P)


We see that the area of the square is $ 9$, and the area of the triangle created by drawing the line is $ \frac{2\cdot3}{2}\equal{}3$, so $ \boxed{\frac{2}{3}}$ of the area of the square is above the line.


function
conics
parabola
inequalities proposed
inequalities

Find the minimum value of  $(1+u)(1+v)(1+w)$ where $0<u,v,w\le{\frac{7}{16}}$ and $u+v+w=1.$

[quote="hungkhtn"]The Minimum take if $a=b=7/16,c=1/8$
It's because $f(x)=ln(1+x)$  is non-convex function.[/quote]
This function is actually concave but then you will find the maximum ... :?

To RaduSorici: Yes, hungkhtn meant exactly that the function is concave (I think he has just forgot the word) and of course it attains it 's minimum at the endpoints, what do you mean?
Thank you for your solution, hungkhtn, actually I did this in a little bit another way. If we fix $u$ , then we get a parabola, which attains it's minimum at the endpoints .

:blush: I'm forgot the word,sorry.My E is very bad!
Armo,You are think same me.For the concave function,the minimun take in end point!


superior algebra
superior algebra theorems

I know the positive integers and the rationals are isomorphic to the integers. And I guess the Gaussian integers would be as well. What are some other sets that are isomorphic to the integers?

Isomorphic in what sense? By the usual algebraic standards, no two of those structures are isomorphic.

In the sense that a bijection exists between one set and the other? Maybe I am confusing this definition.

The word "isomorphism" always has to be interpreted relative to some kind of structure you want the isomorphism to preserve, and it is usually not used to just mean "bijection."  (Technically, "set isomorphism" can be interpreted to mean "bijection," but this is not the usual terminology by historical precedent.)

A set that is in bijection with the integers is called [url=http://en.wikipedia.org/wiki/Countable_set]countable[/url].  Two sets that are in bijection with each other are called [url=http://en.wikipedia.org/wiki/Equinumerosity]equinumerous, equipollent, or equipotent[/url] or are said to have the same [url=http://en.wikipedia.org/wiki/Cardinality]cardinality[/url].  There are many standard results known about what types of sets are countable and about how to make more countable sets out of a given collection of countable sets; for example, the finite Cartesian product or countable union of countable sets is countable.  My favorite countability result is [url=http://qchu.wordpress.com/2009/09/21/some-countability-proofs/]here[/url].

jmerry didn't understand your meaning because all of the sets you listed are also [url=http://en.wikipedia.org/wiki/Ring_%28mathematics%29]rings[/url], and an isomorphism of rings (with identity) needs to preserve the identity, addition, and multiplication.


[quote="bilarev"]Find the least real number $m$ such that with all $5$ equilaterial triangles with sum of areas $m$ we can cover an equilaterial triangle with side 1.

[i]O. Mushkarov, N. Nikolov[/i][/quote]

Can you precise "cover"? Can triangles overlapp?

[quote="A. S. K."]Can you precise "cover"? Can triangles overlapp?[/quote]
I think that the triangles can overlapp!

I think the answer is $ 2$ but I cannot prove it.
Also I'm not sure about this.

Nobody solved it? I want a solution for this;;;;

The answer is $1.25$ so your configuration is wrong.
This with $5$ triangles with area $0.25$, the four ones do the job already and the fifth is overlap.
But it is easy to chose the vertices and endpoints to see it can't with less area.

The answer is indeed $2$. For the lower bound let us take triangles of areas $1-\epsilon,1-\epsilon,0,0,0$.
For the upper bound: Let $A \geq B \geq C \geq D \geq E$ be the sequence of areas.
We put triangles $A,B,C$ in vertices of our triangle.
If they cover our triangle then we are done.  So we assume they don't. Let us observe that each pair of triangles among $A,B,C$ overlaps: i.e. $\sqrt{B}+\sqrt{C} \geq 1$. (Otherwise $2=A+B+C+D+E < 1 + (\sqrt{B}+ \sqrt{C})^2<2$).

Now let $X,Y,Z$ be areas of intersections of pairs of $A,B,C$. Of course we have $X + Y \leq A, Y+Z \leq B, Z+X\leq C$ therefore
$A+B+C \geq 2(X+Y+Z)$ so at least  $(A+B+C)/2$ of our initial triangle is cover.
The uncover area is a triangle of area at most $1-(A+B+C)/2 = (D+E)/2 \leq D$. So we can cover it using $D$ triangle, and we are done.


Let $ f(n)\in\mathbb{N}\;,\;\forall x\in\mathbb{N}$ and

1, $ f(1) \equal{} 1\;f(2) \equal{} 2$

2,$ f(n \plus{} 2) \equal{} f(n \plus{} 2 \minus{} f(n \plus{} 1)) \plus{} f(n \plus{} 1 \minus{} f(n))\;,\;\forall x\in\mathbb{N}$

a, Prove that $ 0\le f(n \plus{} 1 \minus{} f(n)\le 1$ and if $ n$ is odd then $ f(n \plus{} 1) \equal{} f(n) \plus{} 1$

b,Find all $ n$ so that $ f(n) \equal{} 2^{10} \plus{} 1$

I guess sth is not quite right with your problem statement (brackets ?), e.g., $ f(7)\equal{}f(6)\equal{}f(5)\equal{}4$, contradicting second half of statement a)

It took me a while to figure it out.

Claim: $ \forall k\equal{}2,3,\ldots,$ and $ i\equal{}1,\ldots,k$ and $ j\equal{}1,\ldots,2^{k\minus{}1}\minus{}k$ we have $ f(2^k\minus{}i)\equal{}2^{k\minus{}1}$ and $ f(2^k\minus{}k\minus{}j)\equal{}2^{k\minus{}1}\minus{}f(j)$

Proof: Induction on $ n$, (which can be uniquely represented by the above description). The induction start $ k\equal{}2,3$ resp. $ n\leq 7$ is checked by hand. So assume $ n>7,k>3$ from now on, then

$ f(2^k\minus{}i)\equal{}f(2^k\minus{}i\minus{}f(2^k\minus{}i\minus{}1))\plus{}f(2^k\minus{}i\minus{}1\minus{}f(2^k\minus{}i\minus{}2))$

Case 1: $ i\leq k\minus{}2$ gives from the induction hypothesis $ f(2^k\minus{}i)\equal{}f(2^k\minus{}i\minus{}2^{k\minus{}1})\plus{}f(2^k\minus{}i\minus{}1\minus{}2^{k\minus{}1})\equal{}f(2^{k\minus{}1}\minus{}i)\plus{}f(2^{k\minus{}1}\minus{}i\minus{}1)\equal{}2^{k\minus{}2}\plus{}2^{k\minus{}2}\equal{}2^{k\minus{}1}$

Case 2: $ i\equal{}k\minus{}1$, this yields $ f(2^k\minus{}i)\equal{}f(2^k\minus{}k\plus{}1\minus{}2^{k\minus{}1}\plus{}f(2^k\minus{}k\minus{}2^{k\minus{}1}\plus{}1)\equal{}2^{k\minus{}1}$

Case 3: $ i\equal{}k$ determines $ f(2^k\minus{}i)\equal{}f(2^k\minus{}k\minus{}2^{k\minus{}1}\plus{}1))\plus{}f(2^k\minus{}k\minus{}1\minus{}2^{k\minus{}1}\plus{}2)\equal{}2^{k\minus{}1}$

Similarly, one verifies the case $ f(2^k\minus{}i\minus{}j)$.

The claim immediately confirms the first half of statement a) about the monotonicity (the second part was wrongly stated, see earlier reply).

Moreover, we have for $ k\geq 2$

$ f(2^k)\equal{}f(2^{k\plus{}1}\minus{}(k\plus{}1)\minus{}(2^k\minus{}(k\plus{}1)))\equal{}2^k\minus{}f(2^k\minus{}k\minus{}1)\\
\equal{}2^k\minus{}(2^{k\minus{}1}\minus{}f(1))\equal{}2^{k\minus{}1}\plus{}1$.

Since $ f(2^k\minus{}1)\equal{}2^{k\minus{}1}$ all solutions to $ f(n)\equal{}2^{10}\plus{}1$ must fulfill $ n\geq 2^{11}$ by a).

and for $ k\geq 3$ we calculate

$ f(2^k\plus{}1)\equal{}f(2^{k\plus{}1}\minus{}(k\plus{}1)\minus{}(2^k\minus{}(k\plus{}2)))\equal{}2^k\minus{}f(2^k\minus{}(k\plus{}2)\equal{}\\
2^k\minus{}(2^{k\minus{}1}\minus{}f(2))\equal{}2^{k\minus{}1}\plus{}2$

ie. $ n\equal{}2^{11}$ is the only solution

Where is a problem' solution ,  :) . I have posted it on Unsolved box but it isn't quite hard ( i also have poted another problem on different box so i must post it on here )  :lol:

pls can u elaborate your last post [b]quangpbc[/b], your statement was not fully clear. P.S. the proof is complete


Let $ x,y,z,t>0$ and $ x^9\plus{}y^9\plus{}z^9\plus{}t^9\equal{}4$.Prove:
$ \sum\frac{xyz}{xyzt\plus{}t^2} \le \frac{2}{x^2y^2z^2t^2}$

[quote="CSS-MU"]Let $ x,y,z,t > 0$ and $ x^9 \plus{} y^9 \plus{} z^9 \plus{} t^9 \equal{} 4$.Prove:
$ \sum \frac {xyz}{xyzt \plus{} t^2} \le \frac {2}{x^2y^2z^2t^2}$[/quote]

$ LHS \equal{} \sum \frac { (xyz \plus{} 1)xyz}{(xyz \plus{} t^2)(xyz \plus{} 1)} \le \sum \frac {(xyz \plus{} 1)xyz}{(xyz \plus{} t)^2}$
$ \le \frac {(xyz \plus{} 1)xyz}{4xyz}$
next prove that
$ \frac {\sum(xyz \plus{} 1)xyz}{4xyzt} \le \frac {2}{x^2y^2z^2t^2}$
$ \Leftrightarrow \sum (xyz)^2 \plus{} \sum xyz \leq 8$

Let$ x^9 \plus{} y^9 \plus{} z^9 \plus{} t^9 \equal{} 4$ so
$ 4 \ge \sum x^6 \ge \sum (xyz)^2 and 4 \ge \sum x^3 \ge \sum xyz$
---------------------------------------------Because
$ x^6 \plus{} y^6 \plus{} z^6 \ge 3(xyz)^2$
$ x^3 \plus{} y^3 \plus{} z^3 \ge 3xyz$

[quote="bigbang195"][quote="CSS-MU"]Let $ x,y,z,t > 0$ and $ x^9 \plus{} y^9 \plus{} z^9 \plus{} t^9 \equal{} 4$.Prove:
$ \sum\frac {xyz}{xyzt \plus{} t^2} \le \frac {2}{x^2y^2z^2t^2}$[/quote]

$ LHS \equal{} \sum \frac {(xyz \plus{} 1)xyz}{(xyz \plus{} t^2)(xyz \plus{} 1)} \le \sum \frac {(xyz \plus{} 1)xyz}{(xyz \plus{} t)^2}$
$ \le \frac {\sum (xyz \plus{} 1)xyz}{4xyzt}$
next prove that
$ \frac {(xyz \plus{} 1)xyz}{4xyzt} \le \frac {2}{x^2y^2z^2t^2}$
$ \Leftrightarrow \sum (xyz)^2 \plus{} \sum xyz \le 8$

Let$ x^9 \plus{} y^9 \plus{} z^9 \plus{} t^9 \equal{} 4$ so
$ 4 \ge \sum x^6 \ge \sum (xyz)^2 and 4 \ge \sum x^3 \ge \sum xyz$
---------------------------------------------Because
$ x^6 \plus{} y^6 \plus{} z^6 \ge 3(xyz)^2$
$ x^3 \plus{} y^3 \plus{} z^3 \ge 3xyz$[/quote]


Sometimes I have difficulty accessing the site.  One second it's working fine, then another, I cant go anywhere on the site cause the pages won't load.  Is anyone else having this problem?  I dont think that its just because of my internet, cause when I have this problem, I open other browser windows and have no problem opening other sites.

This usually happens late at night / early morning on eastern time.   I dont recall having this problem during the day.  And i dont recall there being an unusual amount of traffic; oftentimes when this happens it's just me or 2 or 3 people on the site.

[quote="rrusczyk"]Is it just the last couple days?  Our web hosting provider was having a problem over the last 2-3 days.  They tell me that it will be all fixed by the end of this weekend.[/quote]

Yes, just in the last couple of days I have noticed that speed is much worse. Glad to hear that they are working on fixing it.

This hasn't started now, it's been happening for almost a month. Currently it's much better than the beginning of September. The server went down almost daily and it didn't just pertain to nighttime.
Also this doesn't have anything to do with the browser you're using or the OS since the server itself goes down everytime.

hmm...its happened to me too. But it started in August. I rarely get the problem though...about once a week


Please submit feedback about the quality of the problems in this competition.

Tips on how to improve them would be appreciated as well.  :)

[quote="Altheman"]i think they are good problems, but at the same time, i have seen (and solved) at least 5 of them before, so originality is an issue[/quote]Yeah, that issue will be fixed next round.  ;)

Good problems, but a lot of them are pretty easy. I guess it's because it's the first round... :) But #10 doesn't seem like the hardest to me. ;)

Regarding difficulty level, I'm going to get an idea of everybody's abilities after the first round is submitted. The difficult of problems will be increased (likely) or decreased (maybe not...) so that it's challenging, but fun for everybody.  :)

Final found will be harder than the normal rounds.  ;)

The problems are good, just a few mistakes on the original question sheet, nothing that couldn't be sorted out though :) !

They're at a good difficulty level at the moment, nothing too hard.


[quote="xxxxtt"]Solve the following equation

(x+1)^9 - x^10 =511[/quote]

==========
The equation has trivial sln x_1=1 and another real solution x_2 = 5.063416747. But I dont know how to "solve" it. By me, we can't solve this equation algebraically. Anyone have idea?

Namdung

Oh sorry, i misread the text of the problem.It says solve in Z and its trivial


$ (1)$ Solve the equation in positive integers: $ x^{2}\plus{}y^{2}\plus{}z^{2}\plus{}u^{2}\equal{} 2xyzu$.
$ (2)$     $ x^{2}\plus{}y^{2}\plus{}z^{2}\equal{}2xyz$.

[hide="2"] Suppose $ 2^{k}$ is the greatest power of two dividing $ x, y$ and $ z$, where $ k\ge 0$.  Then let $ x \equal{} 2^{k}a$, $ y \equal{} 2^{k}b$, $ z \equal{} 2^{k}c$.  We have

$ a^{2}\plus{}b^{2}\plus{}c^{2}\equal{} 2^{k\plus{}1}abc$
Clearly, $ a, b, c$ are not all odd, and neither are they all even.  So let $ a$ be even and let $ c$ be odd.  Take the above equation modulo $ 2^{k\plus{}2}$

$ a^{2}\plus{}b^{2}\plus{}c^{2}\equiv 0\mod{2^{k\plus{}2}}$
Then, because $ 4|2^{k\plus{}2}$,
$ \implies b^{2}\plus{}c^{2}\equiv 0\mod{4}$
$ \implies b^{2}\equiv\minus{}1\mod{4}$

Imposshibibble!  Hence, there are no solutions.[/hide]

[hide="1"]
Again, let $ x\equal{}2^{k}a$, $ y\equal{}2^{k}b$, $ z\equal{}2^{k}c$, $ w\equal{}2^{k}d$.  Then, if any of $ a, b, c, d$ is even, the greatest power of two dividing the left hand side is $ 2^{k\plus{}1}$ (taking the squares modulo $ 4$ again, also consider at least one of $ a, b, c, d$ is odd), while the greatest power of two dividing the right hand side is at least $ 2^{k\plus{}2}$.  Therefore, $ a, b, c, d$ are all odd.  Thus, $ a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}d^{2}\equiv 4\mod{8}$, so the greatest power of two dividing the left hand side is $ 2^{k\plus{}2}$.  But the greatest power of two dividing the right hand side is $ 2^{k\plus{}1}$.  Imposshibibble!  Hence there are no solutions.[/hide]

I am sayin' that the equations all have empty solution sets. Because wherever you look, you see there will be a paradox. It all depend on that it is impossible if a perfect square is divisible by 2, but not by 4. Thanks. Good wishes.


Let be $ a,b,c>0$ such that $ abc\equal{}1$ . Show that :

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

[quote="alex2008"]Let be $ a,b,c > 0$ such that $ abc \equal{} 1$ . Show that :

$ \frac {1}{a^2 \plus{} 2b^2 \plus{} 3} \plus{} \frac {1}{b^2 \plus{} 2c^2 \plus{} 3} \plus{} \frac {1}{c^2 \plus{} 2a^2 \plus{} 3}\le \frac {1}{2}$[/quote]
It's obvious after expanding.

:)
[hide]
By AMGM
$ \sum{\dfrac{1}{a^2\plus{}2b^2\plus{}3}}\le\sum{\dfrac{1}{2ab\plus{}2b\plus{}2}}\equal{}$

$ \dfrac{1}{2}\left(\dfrac{1}{ab\plus{}b\plus{}1}\plus{}\dfrac{ab}{ab\plus{}b\plus{}1}\plus{}\dfrac{b}{ab\plus{}b\plus{}1}\right)\equal{}\dfrac{1}{2}$

With equality only if $ a\equal{}b\equal{}c\equal{}1$
[/hide]


A tank holds 500 gallons of brine.  Brine containing 2 pounds per gallon of salt flows into the tank at the rate of 5 gallons per minute, and the mixture, kept uniform, flows out at the rate of 10 gallons per minute.  If the maximum amount of salt is found in the tank at the end of 20 minutes, what was the initial salt content of the tank?

My doubt is that... I am asked for the initial content of salt of the tank, does it mean that I should add that answer to the amount of brine (500)?

Here's my solution so far... Please correct me if I am wrong [:D]

Let Q be the amount of salt present in the tank at any time t.

dQ/dt = 10 - 10Q/ (500 - 5t)

Is my starting equation correct?

Differential equations (at least at this level) IS calculus, so I'm moving this problem to the calculus computations and tutorials section.

Your differential equation is correct.

You don't know the initial condition. "Brine" is just water with some salt in it. You don't know how much salt - that's the initial condition. You're being asked to reconstruct what the initial condition must have been from the condition that the maximum amount happens at $t=20.$

Don't add anything extra.


function
geometry
integration
Functional Analysis
complex analysis

Here is, I think a very quick , problem:

Let $f(z)=\prod_{n=1}^{\infty}\frac{1+z^{n}}{1+z^{2n}}$.

The Hardy-Hilbert space is the Hilbert space of analytic functions over the unit disk, which are square integrable with respect to the area measure, i.e., $f\in H$ if and only if

$\int_{D}|f(z)|^{2}dA(z)<\infty$.

Is $f$ an element of the Hardy space?

You mean [url=http://mathworld.wolfram.com/BergmanSpace.html]Bergman space[/url], not [url=http://mathworld.wolfram.com/HardySpace.html]Hardy space[/url].

you are right, as it is writen( I was thinking at one space and I was writing the definition of another), that is the definition of the Bergman space,

but, let`s study the  membership in both spaces.

Let $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$. By Parseval's identity, $f$ belongs to the Hardy space iff $\sum_{n=0}^{\infty}|a_{n}|^{2}<\infty$, and it belongs to the Bergman space iff $\sum_{n=0}^{\infty}\frac{|a_{n}|^{2}}{n+1}<\infty$. It appears that this $f$ has infinitely many unimodular coefficients, which would exclude it from the Hardy space. Don't know about Bergman..

This is the proof I was thinking of. I think it does not belong to Bergman since all of the coefficients are greater than or equal to 1 from which it follows that the sum is $\infty$. It also follows if I remember corectly that the function is not a Bloch element.

But if the coefficients are very sparse, the Bergman norm might still be finite...  :maybe:

Some coefficients are zero (the first is the $x^{3}$ coefficient) but the nonzero coefficients are very common.

One term $\frac{1+x^{n}}{1+x^{2n}}$ of the product is $1+x^{n}-x^{2n}-x^{3n}+x^{4n}+x^{5n}-x^{6n}-x^{7n}+\cdots$. The coefficient $a_{k}$ of $x^{k}$ in the product is $\sum_{d|k}f(d)$, where $f(d)=1$ if $d\equiv 0\text{ or }1\mod 4$ and $f(d)=-1$ if $d\equiv 2\text{ or }3\mod 4$. Since $f$ is multiplicative for odd $d$, $a_{k}$ is multiplicative for odd $k$. At prime powers $k=p^{r}$, $a_{k}=r+1$ if $p\equiv 1\mod 4$ and $a_{k}=\begin{cases}0& r\text{ is odd}\\ 1&r\text{ is even}\end{cases}$ if $p\equiv 3\mod 4$. This is nonzero if and only if $k$ can be written as a sum of squares.

For even $k$, it's easier. Whatever the odd factors contribute, it's overwhelmed by the even factors. For odd $m$, $a_{2m}$ is negative unless all prime factors of $m$ are 1 mod 4, and $a_{4m}$ is positive unless $a_{m}$ is zero. Then $a_{2^{r}m}>0$ for all $r\ge3$.

The simplest condition here: $a_{k}$ is nonzero if $k$ is $0$ or $6$ mod 8. That's definitely not sparse enough.

\[f(z)=(1+z)(1+z^{3})(1+z^{5})\ldots \]
And it's clear that all coefficients $\geqslant1$, but that by $z^{2}$.

Ah. I see what happened to me; my argument actually uses the sum $\sum_{n=1}^{\infty}\frac{1+x^{n}}{1+x^{2n}}-1$. That has some zero and negative coefficients.

In the original product, both the numerator and the denominator converge unconditionally as formal power series and as functions on the disk; that allows us to rearrange and cancel terms.


Duke college Princeton HMMT Stanford geometry

?? - Duke
Nov 18 - UGA
Dec 9 - Vestavia Hills
Dec 16 - Princeton
Feb 24 - HMMT
?? - Furman

What else?

Yeah, we have a big list now in Tipton's room.

I think most schools don't go all over the place like we do though, so no one else seems interested.  :(

blah freshmens don't go to most of the tournaments
our first one being vestavia itself =(

Columbus High School in Columbus, GA (somewhat close to Vestavia) is hosting a large tournament.

Here is the link
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=627754#627754
We are trying to get a lot of strong Atlanta teams.

[quote="LuCy4EvA"]yeah i think we're (vestavia) already going to that.[/quote]
Ooh, Vestavia is going to Columbus High Invitational?  <jumps for joy and makes mental note to attend Columbus High Invitational at any cost>  I love going to tournaments where Vestavia will be.  I am always thirsty for revenge... :D

haha, who are you and what school are you from?
what did vestavia ever do to you?

[quote="drunner2007"]he's thinking about the GACS tournament from last year. if i remember correctly, of the top 10 placers at the tournament, vestavia had 9! boysoprano was the only GA contestant to break into the top 10[/quote]
Yes, this is so.  (It was only Alg. II division, though.  Grr, I was fourth, but they changed me to second after the awards had been given out when another error was discovered in the test...  And the only other Georgian in the top 25 or so of Alg. II also went to my school :o )  I am Miles Edwards.

oh haha.
well you won't be seeing us at GACs this year...

[quote="LuCy4EvA"]well you won't be seeing us at GACs this year...[/quote]

yayayayyyyyyy

[quote="drunner2007"]what did you not like about GACS?
i've never been to it[/quote]

tests are terrible, ciphering is terrible, anything involving asking me a math question is bad. the only redeeming quality of the tournament is the size of the trophies

Stanford and Rice have tournaments somewhat along the lines of Duke and HMMT, but I don't think there's a date yet.  When I know, I'll make it known.

Although I think the people going to Duke, HMMT, and places like that are probably not going to be in the target audience for Stanford's tournament.

sorry to bring back old post, but how hard is Vestavia?

Like, what are usually the range of scores in each section that get top 25 or something.

Algebra II first place is usually in the 60's, and it drops pretty quick.

Geometry can vary; usually first is somewhere in 60's or 70's, but it can be as low as 50's.

Comprehensive usually goes up to about 90 = first place, but it drops really quick.

so basically, all the scores can be very high in 1st place, but drop by a lot quickly

So if 1st place in geometry was like 70, 10th place might be like 50 or something

If i got a 30 or 40, I would likely not get into the top 25 or 30 or so rite