In New Zealand, we're usually told that we do very well at the IMO "for a small country".  I decided to test this proud tale to see if it was true.  After all, I'm sure the rest of the mini-countries at the IMO have heard very similar claims.

Basically, I used a model which assumed that gaining a point at the IMO was a "rare event", so that the results that all students from a particular country would obtain if they sat the IMO would follow a Poisson distribution.  I then wrote a program that took data on 2004 IMO results and country populations from the internet, and calculated the proportion of a country's students that that country's IMO team accounted for (in most cases 6/[country's school-age population], adjusted when there were fewer than 6 on the team, and estimating the country's school-age population at 1/4 of the total population.)  It's then pretty easy for a computer to calculate what the average IMO score across all the country's students "would" have been, given that the top 1/10,000 or 1/100,000 or whatever of them scored [whatever their average was].  The resultant "average" makes for some interesting rankings, even if it seems unlikely that the average student in any of these countries would really score this high.  (I don't think Poisson is, in the end, a very good model for something as random as the IMO.)

Anyway, here are [hide="The Results"]
 Av..  Country
 7.06         Bulgaria
 6.01         Mongolia
 6.01         Hungary
 5.55         Moldova
 5.54         Singapore
 5.37         Israel
 4.72         Belarus
 4.53         Georgia
 4.15         Slovakia
 4.11         Taiwan                 (10th)
 3.91         Hong Kong
 3.81         Macau
 3.8          Romania
 3.77         Serbia and Montenegro
 3.6          Armenia
 3.58         Greece
 3.28         Kazakhstan
 3.28         Estonia
 2.94         Czech Rep
 2.73         Croatia                 (20th)
 2.69         Australia
 2.56         Luxembourg
 2.48         Ukraine
 2.31         Korea
 2.24         Poland
 2.24         Canada
 2.18         FYROM
 2.07         Belgium
 2.06         Slovenia
 2            Iran                        (30th)
 1.99         Vietnam
 1.77         Colombia
 1.72         Sweden
 1.68         Azerbaijan
 1.66         Latvia
 1.64         Lithuania
 1.6          UK
 1.54         South Africa
 1.46         Kyrgystan
 1.41         Russia                  (40th)
 1.39         Japan
 1.29         Morocco
 1.27         Turkey
 1.27         Germany
 1.26         Argentina
1.25         Albania
 1.24         Uzbekistan
 1.2          New Zealand
 1.16         Cyprus
 1.14         Norway                (50th)
 1.13         Peru
 1.09         Thailand
 1.09         Switzerland
 1.09         Cuba
 1.06         France
 1.01         Austria
.99          Turkmenistan
 .98          Puerto Rico
 .91          USA
 .87          Uruguay               (60th)
 .87          Ireland
 .85          Finland
 .75          Mexico
 .75          Denmark
 .75          Brazil
 .74          Netherlands
 .68          Italy
 .65          Iceland
 .62          Spain
 .58          Bosnia and Herzegovina  (70th)
 .49          Trinidad and Tobago
 .28          China
 .26          Venezuela
 .26          Malaysia
 .24          Sri Lanka
 .24          Indonesia
 .24          India
 .21          Tunisia
 .15          Portugal
 .14          Paraguay            (80th)
 .09          Mozambique
 .03          Philippines
 .01          Ecuador
 0            Saudi Arabia
 0            Kuwait
[/hide]And as for New Zealand?  Well, 48th is a slight improvement on 58th, but I don't think it's quite the encouraging figure we imagined.

Well, it's nothing surprise to see bulgaria to be atop in the rankings. Even in real imo they did that two years ago.
For moldova, they are having a very very strong team this year.... so that I predict they will be top 10 or even top 5 this year, so again.. it's not surprising.
Perhaps the most surprising thing to me is that Hong Kong ranks the 11th.  :P

[quote="Stijn"]Well your system fails a bit on the big countries with high scores. China gets almost perfect results and has only .28? I'm afraid they won't get higher than 1.00 in your system even if they score 6x42...[/quote]

she's comparing the results based on population... that's why the ranking of china is so low because china has too many people..

This is a very interesting study. How about including scores from other years as well? :)

edit: I misunderstood siuhochung, which makes this a useless post...  :blush:

It is a very nice idea to try and make a ranking which takes country sizes into account.
However I think taking a Poisson distribution does not accurately represent the truth, since Poisson distributions have small tales (i.e. the probability of getting a really high score decrease exponentially with the score). However it frequently happens on the IMO that one person from a team scores more than twice as much as the next best contestant of that country. in 2004 this happened for Argentina (34/17), Azerbeidzjan (24/12), Cyprus (21/10), Denmark (20/9), Ecuador (13/1), Finland (22/9), Ireland (23/8), Saudi-Arabia (2/1), Kuwait (5/0), Luxemburg (25/9), Paraguay (8/3), Philipines (7/3), Puerto Rico (28/9). However the probabilities of this happening with small taled probability-distributions is very slim.

Tonight I might try and find a better fitting distribution and maybe redo the math.

I agree, Tasar, Poisson doesn't fit perfectly.  (Another reason:  at the IMO people are just as likely to get, say, 18 as 19; the odd point here and there is just chance.  On Poisson, each point higher is an order of magnitude less likely.)  I can't think of a better model myself, though (admittedly I'm not exactly an expert in statistics :lol:) except for possibly the current one improved by some "smoothing".  Possibly the method is unfair on large countries, but some big ones like Russia still manage to do pretty well.

Anyway, improvements are welcome; do you want the data I used?

Valentin, I'd love to see previous years' results too, but, given that the IMO is next month, I don't exactly have the time at the moment.  (I did all this number-crunching several months ago but didn't get around to posting it.)  If you want to do it yourself, you're most welcome.

Biggest surprise, in my opinion:  Mongolia!  Who knew it was so brilliant?  :D I didn't, anyway.

[quote="Tasar"]However it frequently happens on the IMO that one person from a team scores more than twice as much as the next best contestant of that country.[/quote]

This reminds me of china mo this year. The team of a province has a perfect score (126/126), then a gold medalist (93/126), then a ?? (forgot, should be bronze - silver), and a weaker member (9/126)..

[quote="Valentin Vornicu"](offtopic) siuhochung, do you know if wpolly qualified for the IMO in the end?  :?[/quote]

he didn't. The China team was posted in the Chinese forum here around a month ago. wpolly said he was the 8th.

I thought about a good method, and talked to my former statistics professor. He came up with the very crazy idea of trying to estimate the percentage of students in a country who would score 0. The rest of them would score points according to some fixed distribution. This method is actually used for some purposes but estimating the amount of people scoring zero by looking at those that don't seems a bit silly. However I tried it and it gave a reasonable result. The distribution for those that don't score zero is some adjusted exponential distribution, where the higher the scores the harder it is to score an extra point (I estimated this by looking at the scores of countries of those people scoring more than the lowest score of the team). Subsequently I calculated how much point each country would get if there were 10.000.000 inhabitants (a rough rounding of the geometric mean of the population sizes of the participating countries).
One nice feature is that China and Trinidad & Tobago can score almost identical, whereras there populations differ a factor 1000 (but this also happened on Agrippina's list).
The fraction of people scoring more than zero in a country would according to my calculations differ from about 1/10.000 to 1/10.000.000.

The results is generally not much different to Agrippina's, and New Zealand rises to 43rd (from Agrippina's 48st). This is still not very good, but I think New Zealand's leaders might be kidding themselves for being good for a small country, since half of the smaller countries (in population) score more points (without population correction) on the IMO.

The scores:
[hide]1. Bulgaria 200
2. Hungary 187
3. Taiwan 171
4. Macao 168
5. Mongolia 167
6. Moldova 160
Singapore 160
8. Israel 159
9. Rumania 156
10. Belorus 153
11. Vietnam 145
12. Georgia 143
13. Russia 142
14. Estonia 140
15. Luxemburg 138
16. Iceland 137
17. Ukraine 135
18. Slovakia 135
19. United States 132
20. Armenia 131
21. Iran 130
Serbia & Montenegro 130

Hong Kong 130
24. South-Korea 126
25. Greece 124
Cyprus 124
27. Kazakhstan 121
28. Japan 118
29. Macedonia 117
30. Slovenia 115
31. Croatia 112
32. Czech Republic 108
33. Latvia 106
      Australia
      Poland
36. China 102
37. Canada 100
38. Trinidad & Tobago 98
39. Lithuania 95
40. Albania 88
41. Belgium 85
42. UK 84
      New Zealand 84
44. Kyrgystan 83
45. Colombia 81
46. Uruguay 80
47. Azerbeidzjan 79
      Norway 79
49. Sweden 78
50. Ireland 76
51. Puerto Rico 75
52. Turkmenistan 74
53. Germany 70
54. Finland 69
      Bosnia-Herzegovina 69
56. South Africa 67
57. Switzerland 66
58. Denmark 65
59. Turkey 62
60. Austria 61
61. Morocco 52
62. Argentina 50
63. Uzbekistan 48
64. Brazil 47
65. Peru 46
66. Cuba 44
67. Thailand 42
68. France 38
69. The Netherlands 37
70. Tunesia 31
71. Portugal 25
72. Mexico 24
73. Kuwait 21
74. Paraguay 20
75. India 18
76. Italy 17
     Spain 17
78. Sri Lanka 16
79. Maleysia 14
80. Ecuador 10
81. Mozambique 7
82. Venezuela 6
83. Indonesia 3
84. Philippines 2
      Saudi-Arabia 2
[/hide]

That's very neat.  Particularly impressive that your method doesn't seem to unduly penalise the very big countries (Russia, the US and China still have good scores.)  So what's the name of this distribution?  And how do you work out the proportion of students that score 0?

Anyway, I'm content now that we've brought New Zealand up to 43rd.  Any ranking that puts us within cooee of Australia is fine with me . . . .  the remaining 15% we can blame on average temperature, or on number of flies per capita.

Actually, your post made me think Agrippina, and I made a mistake and didn't do what I wanted to do  :blush: . So I guess I will have to redo the math. What I wanted to do is create a one parameter family of distributions, which describes the distribution $X$ of how one trained(!) student from some country would score, given by
\[P(X=0)=p, \qquad P(X=k)= (1-p)P(Y=k), (k>0), \]
where $Y$ is another fixed random variable (taking values in 1,2,..,43). The one variable in this system is of course $p$.
Subsequently I estimated $p$ (actually by taking the value of $p$ for which the expectation of the sum of the scores oof the six best (or more precisely: the number of contestants) contestants would be the score they obtained). Having estimated $p$ it is again easy to calculate the expected score of that countries best 6 students  if the the country had 10.000.000 inhabitants.
The question remains which distribution to take for $Y$. I looked at the results of the IMO and estimated the probabilty of getting exactly $x$ points given you get at least $x$ points by looking at the fraction of students scoring x, when there is at least one member of their team scoring at most $x$. Smoothing this function I got that this probability is about $.85-x/400$. Here I made a mistake, for I did not pick $Y$ to satisfy that $P(Y=x~|~Y\geq x)=.85-x/400$, but I took $P(Y=x) = N \prod_{k=1}^x (.85-k/400)$. This in analogy with the exponential distribution, where due to some mathematical miracle it does work if $P(Y=k~|~Y\geq k)$ is constant. Indeed the method I wanted to use should give a higher probability for scoring 42, then 41, which I guess is true, however the one I took, doesn't. For a large region of the support (I guess up to 35 points or so) the difference shouldn't be too big though.
  If I have the time I might actually redo the calculations, though I think that will not be until August (and at that time analyzing the results of Mexico is probably more interesting).

  There is a very simple quick calculation of what the score for 10.000.000 inhabitants should be, given we take $Y$ and exponential distribution with exponent $.8$ (i.e. $P(Y=k) = 1/2 * .8^k$). Then the adjusted score should be about
score obtained + 6*(log(population)-log(10.000.000))/log(4/5).
This formula definitely is not perfect, but very easy to use and gives a reasonable ranking (if you except some countries scoring negative, that is ;) ).

BTW I don't think my distribution has a name, but to me that is never a problem.

Edited to correct the rough estimate formula, since formulas with logarithms should always be independent of the base.

I've got a better idea of what you mean now, but I still don't understand the last bit where in the method you "wanted to use" (:P) you create a distribution for the ones who don't score 0.  For example, how could you find that $P(Y=x~|~Y\geq x)=.85-\frac{x}{400}$, when this would make $P(Y=x~|~Y\geq x)$ decrease as $x$ increases?  Shouldn't it be increasing?  And shouldn't $P(Y=42~|~Y\geq 42)$ be $1$, rather than your model's $.85 - \frac{42}{400}$?

Anyway, if it has no name then it can go into the literature as Tasar's Distribution.

You are right that $P(Y=x~|~Y\geq x)$ should increase, I "wanted" $P(Y>x~|~Y\geq x) = .85-x/400$ (for $x<42$, note the difference to the initial formula) and for $x=42$ I would like of course $P(Y>42~|~Y\geq 42)=0$.  Indeed it is clear that $P(Y=x~|~Y\geq x)$ is not approximately 80%, for then one would almost always score exactly as much, or nearly that, as your teammates, whereas in reality the best contestant of a country usually scores one or even two dozen more points than the worst contestant of that country. It amazed me though that the increase in how hard it is to get one extra point is not very high (it ranges from 85% manages to score one extra point from 0, to 75% manages to get 41 instead of 40 (but, then again, most people have already failed to get 40)). My estimation had to use data from good, well-trained countries to estimate this probability for high scores and from low-scoring (presumably less hard training) countries for the low scores, since they are the only ones that obtain those scores, so maybe this introduced a methodical error. This error is however intrinsic in assuming that $Y$ has the same distribution for all countries and that $p$ is the only variable.

I could do some intitial calculations with the "distribution I wanted", and found that the difference in the normalized score for 10.000.000 inhabitants between using this new distribution and using the old one is negligible, in fact for most countries I checked it was only a few hundredths (0.01) of a point.

Nice job guys (and girl :))

Hello Tasar, are you Fokko, or Johan, or someone else... ?

Good Luck in Mexico, Heather!

Hi Arne, thanks for the nice words. I'm not Johan (did you know he hates statistics?), but Fokko.

Due to progressive automation of the calculation process, I was able to make a new list in about half an  hour, and the difference to my previous one is huge. Without rounding Romania's adjusted score changes from 156,48 to 156,51, and hence gains a point after rounding and all other countries' scores remain the same after rounding.

The next step would be looking at other years, so I can show that the Netherlands didn't just score more points than Belgium (if we did that) because we were bigger, but because we were brighter (there must be some years in which this was true, perhaps 1983?).

I like the statistics  :P  Well....I guess I am a little biased (looks at flag).

Haha Fokko, there certainly were a few years where you beat us recently... :)

Belgium - Netherlands in the last 17 years
2004: 86 - 53
2003: 70 - 30
2002: 58 - 55 (Tight!)
2001: 25 - 42
2000: 51 - 60
1999: 51 - 74
1998: 71 - 62
1997: 88 - 94
1996: 75 - 26
1995: 83 - 85 (Tight!)
1994: 105 - 99
1993: 55 - 58 (Tight again!)
1992: 100 - 71
1991: 121 - 73
1990: (Belgium didn't compete)
1989: 104 - 92
1988: 76 - 85
In the last 17 years, Belgium "won" 9 times and the Netherlands won 8 times. It has always been tight and it will probably always be like that, but recently Belgium has been doing a little bit better. Of course we are the smaller country ;)

These look just like Basketball scores  :roll:  :lol:

This is fun and interesting, though I really don't feel like diving into statistics myself :)

[quote="Tasar"]The next step would be looking at other years, so I can show that the Netherlands didn't just score more points than Belgium (if we did that) because we were bigger, but because we were brighter (there must be some years in which this was true, perhaps 1983?).[/quote]
I read this as a challenge, Tasar :). I accept it with a smile, and I am sure my teammates will do the same :roll:

Arne, why did you stop your list in 1988? Is it because apart from 1986 (in which the Netherlands did not compete) Belgium was beaten by the Netherlands ever since Belgium began competing again in 1977? Let me finish the list:
Belgium vs. The Netherlands at the IMO:
1987: 74-146
1986: The Netherlands did not compete
1985: 60-72
1984: 56-93
1983: 31-143
1982: 50-92
1981: 139-219
1980: 102-172 (replacement international mathematical competition in Luxemburg)
1979: 66-130
1978: Belgium did not compete
1977: 33-185
1970-1976: Belgium did not compete
1969: 57-51

Therefore, in years that both our countries participated it is still 16-10 in the dutch advantage. Now, I don't want to say anything about the future, but I wouldn't call all old results tight (33-185 tight?). However the last 10 years it has generally been quite close, and so it should be interesting to see what happens this year.

I just didn't have the results you have, otherwise I would have posted them Fokko. You can't really say that it was tight in Japan or Athens but this year it might be much tighter. Wish your team good luck from me (especially Sietske & Johan).

2005: Belgium 74 (1SM 1BM 2HM) - Netherlands 62 (2BM 2HM)
Belgium wins again! But congratz to the Dutch team, and especially concratz because 2005/2 was proposed by the Netherlands.

Fokko (Tasar), Johan (Thabit) or anybody else - could you please post the names of the contestants with their scores and contestant codes so that I can find out who got which result? The contestant codes alone don't really help me.

NETHERLANDS
Score	P1	P2	P3	P4	P5	P6	= Final	Medal -- Name
NLD1	0 	7 	0 	0 	0 	2 	= 9 	Honor -- Sjoerd Boersma
NLD2	7 	2 	0 	2 	5 	1 	= 17 	Bronze -- Johan Konter
NLD3	2 	7 	0 	0 	7 	1 	= 17 	Bronze  -- Jinbi Jin
NLD4	0 	4 	0 	1 	0 	0 	= 5 	 -- Johannes Steenstra
NLD5	2 	1 	0 	7 	1 	0 	= 11 	Honor -- Sietske Tacoma
NLD6	0 	1 	0 	0 	0 	2 	= 3 	 -- Anne de haan
TOTAL	62

I've processed this year's results using Tasar's method; the final ranking is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=46759]here[/url].  I changed the process for calculating the distribution of non-zero scores slightly, calculating $P(X>n/X\geq n)$ using only those countries whose minimum score was less than (not less than or equal to) $n$.

Thanks for the kind wishes, Arne; I won a silver, and was delighted. :lol:

Rankings of various countries of interest:  New Zealand 20 (hurray!), Belgium 42, Netherlands 55.

[quote="fleeting_guest"][quote="Agrippina"] Biggest surprise, in my opinion:  Mongolia!  Who knew it was so brilliant?  :D I didn't, anyway.[/quote]

Mongolia and Iran have extremely young populations (more than half under 30).  If one uses instead of total population, the student population of those countries, or population aged 15-19, the statistics for young countries would drop and the dinosaurs of Western Europe would do better.[/quote]
but the population of mongolia is at present 2.6 million people.   :lol:


function
inequalities
inequalities solved

for real numbers $x,y,z >0$ and $x+y+z=1$.
prove that :
$\frac{x}{\sqrt{1-x}} +\frac{y}{\sqrt{1-y}}+ \frac{z}{\sqrt{1-z}} \geq \sqrt{3/2}$.

Let the function
$f(x)=\frac{x}{\sqrt{1-x}}$ . We  have $f''(x)>0$ .

Use Jensen and get the derised ineq.


kind of reminds me of chaos theory...depending on the functions and initial condition it could converge, settle into a pattern, or have chaotic behavior.

One thing you can try is to find a fixed point of $f$ (i.e. a point $x$ so that $f(x)=x$). If it tends to a limit, it must tend to a fixed point. You can then use some bounding argument to show that it actually approaches that fixed point. I think you need something like $|f'(x)|<1$ in some neighborhood of a fixed point for convergence to it.

mathkid wrote:I was wondering if it is possible to calculate the limit to infinity of a recursion, assuming that it converges.

for example.. if a function is iterated to infinty like f(f(f(... x))), and it converges, could we calculate the limit?

I thought of this when learning about newton's method for approximation.

If the number of iterations when applying the method approached infinity, we could solve polynomials. I'm sure if this was possible, it would have been discovered.

I don't know if i'm using the right terminology when talking about recursion, but it sounds like an interesting idea.

Any thoughts, or am I just crazy?

Assuming that the sequence converges, evaluation of the limit boils down to solving the equation

f(L) = L where L = the desired limit.

The above eqn. can be arrived from the folowing theorem.

If sequence u_n converges to limit L, the sequence u_(n-1) also converges to L.

Now in order to solve f(L) = L , u need some additional conditions on the monotonicity etc. of the function.

Assuming that f is bounded above by say M,

we have |f(L)| <= M,from which we can proceed towards proving that f(L)=L = M

Ex. if f(x) = sin(x)

Then the limit L is 0 for sin(L) = L has the unique solution L=0[/hide]


function
abstract algebra
superior algebra
superior algebra solved

Need some help of finding the order of p bar in
(Z/(p^n-1)Z)* where p is a prime and n is positive integer, and how to deduce that n l phi(p^n-1)... phi is Euler's function
Thx

The order of $p$, who belongs to this group, is clearly n; whence by Lagrange's theorem $n | \phi(p^{n}-1)$ (because $x$ belongs to the group iff $gcd(x,p^{n}-1)=1$, meaning the group has $\phi (p^{n}-1)$ elements).


prove that the first 2006 digits after the decimal point in the following expression are all the same...

$(6+\sqrt{35})^{2006}$

[hide="remark"]
infact one can show the first 2006 digits would be all 9's[/hide]

[quote="ZetaX"] giving all those nines.[/quote]

im sorry but can you explain better?? :?


geometry
rectangle
combinatorics proposed
combinatorics

A tromino is a figure that is obtained from a square $2\times 2$ by removing one of it's corners.
Now, we $2005$ non-overlaping trominos are placed in the plane. We know that for any tromino, the missing corner (the one we removed from the $2\times 2$ square) is covered by another tromino in the tiling. Prove that we can remove $k$  so that this property is preserved. Also, $1\leq k <2005$
A very nice problem...
[Edit, since it raised many unclarities]

Can they overlap? Or maybe that's not important?

It is enough to prove our result for "conected" figures:

We do it by contradiction:

Suppose $T$ is a conected figure, pick its downest tromino, it is easy to see, its missing square is on top, or otherwise there would be a $2*3$ "good" figure. Now since there are no $2*3$ "good" figures, it is easy to see all trominoes belonging to $T$, have the missing square on top. But then the upper tromino of $T$ also have its missing square on top, but the square that covers it must then form a $2*3$ "good figure with it. Contradiction!

So we conclude, in every finite, conected part, there is a $2*3$ rectangle.

Hi, Pascual!
I think there is a little mistake...
$0110$
$0312$
$3322$
I tried to show an example when not all the figures have the missing square in the top...That is, I think, not true....
Figure 1 has nothing missing in the top, while 2 & 3, beign lowest, don't have the square in the top row...
I also tend to believe that all the "good" figures have a $2\times 3$ square, but I'm not sure...

Ok, thanks for editing the initial message :).

Consider the digraph having our pieces as vertices (label them $v_1,\ldots,v_{2005}$), and having an arc from $v_i$ to $v_j$ iff $v_i$ covers the empty square of $v_j$. Each vertex has in-degree exactly $1$, and out-degree belonging to $\{0,1,2,3\}$. This means that there are precisely $2005$ arcs. We have to show that there is a vertex with out-degree zero.

Assume all vertices have degree $\ge 1$. Since there are exactly as many arcs as there are vertices, it means that each vertex has both out- and in-degree exactly $1$. In turn, this means that our graph is broken up in disjoint circuits. Since $2005$ is odd, there must be a circuit with an odd number of vertices. We have to show that this is impossible, i.e. that there cannot be any odd circuits.

Now let's think about a circuit as being a sequence of squares in the plane, s.t. each square has exactly two corners covered by the other squares. The distance between the centers of two consecutive squares in this circuit is $\sqrt 2$ (I'm assuming, just like Pascual, that two consecutive pieces do not form a $2\times 3$ rectangle; in such a situation, the distance between the centers of the two squares would be $1$). If we assume the center of the first one is (even,even), then the next one is (odd,odd), then the next one is (even,even), and so on. This means that we have to have an even number to get back where we started.

Hope it's not wrong. I had quite a few attempts which all turned out to be mistakes :).

Sorry for my stupid mistake! Grobber, nice proof!

Actually, the idea of the official solution is the same!
They assume there are no $2\times 3$ rectangles and prove there are no odd cycles...
I wonder if it would be possible to have a tiling without $2\times 3$ rectangles...I tried, but not too hard though, to build an example and failed...
Maybe someone else would succeed here :)

Of course, there are lots of such tilings :). Put four trominoes on the plane s.t. they form something like a swastika :).


function
inequalities unsolved
inequalities

find all function continuous  $f(x)$ :$R\to R$ and follow property   :
$f(x-y)f(y-z)f(z-x) + 8 = 0$

we have f(x)=0  isn't a  satisfying  function .
 we have f(0)=-2 .let g(x) =f(x)/-2 .
 hence g(x-y)g(y-z)g(z-x)=1 -> g(x)g(y)=g(x+y)
   because g(x) continuous then g(x)=a^x (a>0).
 hence f(x)=-2.a^x(a>0)


Two fair 20-sided dice are rolled. The faces on each die are numbered from 1 to 20. If neither of the dice shows a number less than or equal to n, where n is an integer, the probability that at least one of the dice shows a 20 is 11/36. What is n?

The phrase "at least" is generally a clue to use complementary probability, which is what we'll try to use here. The probability of not seeing any twenties is equivalent to 36/36-11/36=25/36. We can immediately see that 25/36=(5/6)^2. Is this only a coincidence, or could this be the foundation of a correct hunch? Let's see. Since we have two dice, we can calculate the probability of not rolling a twenty on once dice and multiply it by the probability of not rolling a twenty on the other. In this case, the probabilities are the same, so P(Not 20)=5/6, as 5/6*5/6=25/36. (P(Not 20) is the probability that we don't roll a 20.) We now have that P(Not 20) = 5/6. We must now express P(Not 20) in terms of N, so we can solve for N. Well, the total number of numbers we can roll is 20-N. Excluding 20, this becomes 20-N-1. We can see that the probability of P(Not 20) is equivalent to (20-N-1)/(20-N), which in turn equals 5/6. So, we have the equation (19-N)/(20-N)=5/6. Solving for N, we find that N=14. Let us check the logic of this. If N=14, then our set of valid rolls is {15, 16, 17, 18, 19, 20}. We can immediately see that the probability of not rolling 20 equals 5/6. (5/6)^2 equals 25/36, and 1-25/36 = 11/36. So we see that n=14 is correct.


combinatorics
graph theory
Extremal Graph Theory
Extremal combinatorics
IMO Shortlist

155 birds $ P_1, \ldots, P_{155}$ are sitting down on the boundary of a circle $ C.$ Two birds $ P_i, P_j$ are mutually visible if the angle at centre $ m(\cdot)$ of their positions $ m(P_iP_j) \leq 10^{\circ}.$  Find the smallest number of mutually visible pairs of birds, i.e. minimal set of pairs $ \{x,y\}$ of mutually visible pairs of birds with $ x,y \in \{P_1, \ldots, P_{155}\}.$ One assumes that a position (point) on $ C$ can be occupied simultaneously by several birds, e.g. all possible birds.

Make graph $G'$ with 155 vertices(birds) like this:

Two vertices (birds) are connected $\Leftrightarrow $ when they do [u]not[/u] see each other.

Therefore we have no clique $K_{36}'$ in graph. By Turan, we have

\[\varepsilon' \leq {k-2\over 2(k-1)}\cdot n^2 -{r(k-1-r)\over 2(k-1)} = 11665\]
where $k= 36$ and $r$ remainder if $n$ divide with  $k-1$, so $r=15$.

As we know upon Turan configuration with $\varepsilon' = 11665$
exist, smallest number of bird pairs seeing each other is 270.

     But if we use the result https://en.wikipedia.org/wiki/Tur%C3%A1n%27s_theorem.
     We have $|E|\le \lfloor \frac {k-1}{2k}n^2 \rfloor=11669$,which is not $11665$,could someone explain why this happens?Confused.THANKS

[quote=NoYoo-hooForMe][quote="Number1"]As we know upon Turan configuration with $\varepsilon' = 11665$
exist, smallest number of bird pairs seeing each other is 270.[/quote]

Isn't it necessary to actually exhibit an arrangement of birds achieving the minimum of $270$?  Just because Turan guarantees a graph satisfying a particular property exists, it doesn't necessarily follow that there is an arrangement of birds corresponding to that graph.

Fortunately it is not hard to give a specific arrangement of birds.  Divide the birds into $35$ groups and place the groups equidistant on the circle--so each group is separated by just over $10$ degrees of arc and the birds can see only other birds in the same group.  There will be $20$ groups of $4$ birds and $15$ groups of $5$ birds ($20*4+15*5=155$).  The total number of bird pairs that can see each other is thus:  $20\dbinom{4}{2}+15\dbinom{5}{2}=270$.[/quote]

Turan s graph has maximum no of edges no containing K36 in this case.
notice each edge denotes a bird pair not seeing each other .
there are 155*154/2 pairs of birds out of which at most maximum 11669 pairs are not mutually visible.
hence the min pairs that are always visible are 155*154/2 - 11669 =266 pairs.


vector
abstract algebra
superior algebra
superior algebra unsolved

Let $ V,\ W$ be finitely dimensional vector space over complex field and $ f: V\rightarrow V,\ g: W\rightarrow W$ be linear transformation respectively.
Suppose that $ f$ and $ g$ don't have same eigen values. If the linear mapping $ \phi : V\rightarrow W$ satisfies that $ \phi f\equal{}g\phi$, prove that $ \phi \equal{}0$.

if $ f(x) \equal{}\lambda x$ then $ g\phi(x) \equal{}\phi f(x) \equal{}\lambda\phi(x)$.  By hypothesis ($ f,g$ have disjoint eigenvalues), $ \phi(x) \equal{} 0$.  So $ \phi$ is zero on the eigenvectors of $ f$.  So if $ f$ is semisimple then you're done.  Otherwise note that $ \phi (f\minus{}\lambda\text{ id}) \equal{} (g\minus{}\lambda\text{ id})\phi$, so $ \phi (f\minus{}\lambda\text{ id})^{k}\equal{} (g\minus{}\lambda\text{ id})^{k}\phi$ for every positive integer $ k$.

if $ (f\minus{}\lambda\text{ id})^{k}x \equal{} 0$ then $ (g\minus{}\lambda\text{ id})^{k}\phi(x) \equal{} 0$.  But by hypothesis that $ f,g$ have disjoint eigenvalues, again, $ (g\minus{}\lambda\text{ id})^{k}$ has no 0 as an eigenvalue.  So again $ \phi(x) \equal{} 0$.  That is, $ \phi$ is zero on the kernels of $ (f\minus{}\lambda\text{ id})^{k}$ as $ \lambda$ goes thru the eigenvalues of $ f$ and $ k$ runs thru the positive integers.  But these kernels span $ V$.  so $ \phi \equal{} 0$


geometry
combinatorics proposed
combinatorics

Suppose that $\{ D_n \}_{n \geq 1}$ is a finite sequence of disks in the plane whose total area is less than $1$. Prove that it is possible to rearrange the disks s.t. they are disjoint from each other and are all contained inside a disk of area $4$.

i will give a prize of $50$ eurocents for the one who will solve this problem...  :roll:

There is a square of area $\frac 8\pi$ inside the large disk of area $4$, and each small disk can be fitted inside a square s.t. the areas of all the squares sum up to $\le \frac 4\pi$. Now look for a proof posted by pbornsztein on the forum of the fact that several squares with total area $\le\frac 12$ can always be fitted without overlapping inside a square of area $1$, and you're done :).

I'm not sure if I understand what grobber means exactly, but I think that what he says is affected by the fact that if we have two circles, and their circumscribed squares (with parallel sides, of course) are included in one another, then it doesn't follow that the circles are included in one another. For this, take a big circle and its circumscribed square, and then take a small circle externally tangent to the big circle and internally tangent to the big circle's circumscribed square. I remember trying to solve this in the same way and failing. I would very much like to see a complete solution!

How about reading the post above one more time? :) You take the square [b]inscribed[/b] in the large disk, and the squares [b]circumscribed[/b] around the small disks.

i also had grobber's idea, but gave up on it quite quickly, because it seemed like a dead end...

the solution is here: http://www.mathlinks.ro/Forum/viewtopic.php?p=21783#21783

Yes, thank you very much grobber, you are right! My mistake! :lol: Don't take it personal :-)


probability
algebra
polynomial
combinatorics unsolved
combinatorics

We are given $ n$ dice. For every $ i$ and $ j$ with $ 1\leq j\leq 6$ and $ 1\leq i \leq n$, let $ p_{i,j}$ be the probability that throwing the $ i$-th dice results in $ j$.
We throw all $ n$ dice (each of them once). Let $ S$ be the sum of the appearing numbers. (Of course, $ n\leq S \leq 6n$.) Prove that there exist $ m$ and $ m'$ such that
$ \text{probability}\left(S \equal{} m\right) \neq \text{probability}\left(S \equal{} m'\right)$, $ n\leq m\leq 6n$, $ n\leq m' \leq 6n$.

i know its stange solution but I'm searching for another sol :maybe:

i guess n>1...

xp(i,1) + x^2p(i,2) + ... + x^6 p(i,6) is defined as ther characteristic polynomial of the dice i. All the possible outcomes of throwing n dices are considered by the product of the n characteristic polynomials, namely x^k in the product has coefficient c, then c is the result of sum of the products of probabilities over all possible sets of dices that sum up k. No p(i,6) can't be zero for otherwise all the p(i,j) must be zero since 6n can't be generated.

so if the probabilities are all equal then the resulting product polynomial say c(x^n+x^(n+1)+... + x^(6n) ), then it's divisible by 1+x+..+x^(5n), so it has all the 5nth roots of unity as roots with exception of 1. So for every such root of unity it is also a root of some of the carachteristic polynomial. We can muliply out a x out of all charachteristic polynomials, then it's clear that for all i p(i,1) + xp(i,2) + ... + x^5 p(i,6) has complex unities as roots by counting the number of roots in both sides. Then p(i,1) is 1 away from the complex 0, and its a positive real number, so it's 1, and all the other coefficients must be zero since the sum of the coefficients in one polynomial is 1. This yields an absurd recalling the above observation.

I wonder now whether this is the strange solution  :maybe:

psykyk

sorry no n>1 stuff a stupid mistake of mine, the proof works for n>=1

oh! the solution that I saw is the same. any way thanx ,but no one knows other sol?

[quote="psykyk"]i guess n>1...

 it's clear that for all i p(i,1) + xp(i,2) + ... + x^5 p(i,6) has complex unities as roots by counting the number of roots in both sides. Then p(i,1) is 1 away from the complex 0, and its a positive real number, so it's 1, and all the other coefficients must be zero since the sum of the coefficients in one polynomial is 1.

psykyk[/quote]

Can Psykyk or anyone explain how we get p(i,1) is 1 away from the complex 0?
Thanks.


induction
function
algebra proposed
algebra

Find all f:N->N s.t.

i) f is strictly increasing on N;
ii) f(2n)=f(n)+n for all n\in N;
iii) f(n) prime => n prime.

[quote="grobber"]Find all $f: \mathbb{N} \to \mathbb{N}$ s.t.

i) $f$ is strictly increasing on $\mathbb{N}$;
ii) $f(2n)=f(n)+n$ for all $n \in \mathbb{N}$;
iii) $f(n)$ prime  $\Rightarrow$ $n$ prime.[/quote]

Here I'd like to use [url=http://mathworld.wolfram.com/DirichletsTheorem.html]Dirichlet's theorem[/url].

Let $f(1)=a$.
Then, thanks to [b]ii)[/b] : $f(2)=a+1$ and $f(2^2)=a+1+2=a+3$, and, via induction : $f(2^n)=a+2^n-1$.
Thanks to [b]i)[/b], we have that : $f(i)=a+i-1$.
Thanks to [b]iii)[/b], we know that : if $a+i-1$ is prime then $i$ is prime.
Let's suppose that $a \ge 3$ :
Let $q$ be a prime number greater than $a-1$, then $gcd(q,a-1) = 1$.
Hence the series $qm+(a-1)$ (with $m \in \mathbb{N}$) an infinite number of primes. Hence $qm$ is a prime number, contradiction.
We thus deduce that : $a = 0$ or $a = 1$ or $a = 2$ :
$\bullet$ For $a = 0$ :  $f(5) = 4$, but since $5$ is prime wheras $4$ is not, it is not possible.
$\bullet$ For $a = 1$ : $f(n) = n$, which is possible.
$\bullet$ For $a = 2$ : $f(5) = 6$, but since $5$ is prime wheras $6$ is not, it is not possible.

So, we get that the only function satisfying these relations is :
$\boxed{f(n) = n}$.

 :D

Oh, nice!  :D

Thanks a lot [b]Enescu[/b]. :)

[quote=enescu]You don't need Dirichlet's theorem. After obtaining $f(n)=a+n-1$ for all $n$, suppose $a>1$ and consider $a$ consecutive composite numbers, e.g. $(a+1)!+2,(a+1)!+3,\ldots, (a+1)!+a+1.$ Let $p$ be the least prime greater than $(a+1)!+a+1$. Then $p-a+1$ must be a prime, and that is a contradiction :)[/quote]

what is contradiction here ???? can you pls tell ......

If $p\le (a+1)!+2a$, then $p-a+1$ is composite.
Else, $p-a+1>(a+1)!+a+1$. In this case, we have a prime larger than $(a+1)!+a+1$ that is less than $p$, contradicting its minimality.

[quote=TLP.39]
Consider the set $A=\{(a+1)!+i:2\le i\le a+1\} $.
If $p\le (a+1)!+2a$, then $p-a+1\in A $ but we know that $i\mid (a+1)!+i $ so it's composite.
Else, $p-a+1>(a+1)!+a+1$. In this case, we have a prime larger than $(a+1)!+a+1$ that is less than $p$, contradicting its minimality.[/quote]
@below I just try to give you a compressive explanation.
-----
We have the set $A=\{(a+1)!+2,(a+1)!+3,\cdots,(a+1)!+a+1\} $ so if $p\le (a+1)!+2a $ we must have $p-a+1\le ((a+1)!+2a)-a+1=(a+1)!+a+1$ so this combined with $(a+1)!+a+1 <p $ implies $p-a+1\in A $. But notice that each element in $A $ is of the form $(a+1)!+i $ and $2\le i\le a+1$ so $i\mid (a+1)!+i $ and this number can't be prime.

[quote=Al3jandro0000][quote=TLP.39]
Consider the set $A=\{(a+1)!+i:2\le i\le a+1\} $.
If $p\le (a+1)!+2a$, then $p-a+1\in A $ but we know that $i\mid (a+1)!+i $ so it's composite.
Else, $p-a+1>(a+1)!+a+1$. In this case, we have a prime larger than $(a+1)!+a+1$ that is less than $p$, contradicting its minimality.[/quote][/quote]

What does this mean ?


function
number theory proposed
number theory

Find all $f : N \longmapsto N$ that there exist $k \in N$ and a prime $p$ that:
$\forall n \geq k \ f(n+p)=f(n)$ and also if $m \mid n$ then $f(m+1) \mid f(n)+1$

We can show that $f(n)\in\{1,2\}$ for all $n\geq 2$. Right? Futher, apply some reasonings to obtain that $f(pk+i)=1$ for $i=2,3,...,p$, $f(pk+1)=1$ or $f(pk+1)=2$ for all $k$, $f(1)$ is an arbitrary number. Or something like that. I am not sure. Excuse me all nonsenses I have written here.

Suppose $n \geq k$ and $p$ doesn't divide $n-1$. You see there is $k$ that $n-1\ |\ n+kp$. So $f(n)\ |\ f(n+kp)+1$ but we know that $f(n)=f(n+kp)$, Therefore $f(n)\ |\ 1$ and $f(n)=1$. Consider an arbitary $n \neq 1$, We know that $n-1\ |\ (n-1)kp$, So $f(n)\ |\ f\big((n-1)kp\big)+1=2$. So for every $n\neq 1,\ f(n) \in \{1,2\}$. Now we have 2 cases:
a ) $f(n)=2, \forall n\geq k$ and $p\ |\ n-1$, Consider $n$ that $p$ doesn't divide $n-1$. There is $m$ that $n-1|m$ and $p|m-1$. So $f(n)\ |\ f(m)+1=3$ and $f(n)=1$. So $f(n)=1$ for all  $n \geq k$ or $m$ doesn't divide $n$; And is arbitarily defined for every $n<k$ and $p\ |\ n-1$
b ) $f(n)=1, \forall n\geq k$ and $p\ |\ n-1$, In this case $f(n)=1, \forall n \geq k$, and if we suppose: $S=\{a\ |\ f(a)=2 \}$ then it doesn't exist $m,n \in S $ that $m-1\ |\ n$. So all of functions with the problem properties are defined like this: Suppose $S$ is  a finite subset of natural numbers that there doesn't exist $m,n\in S$ that $m-1\ |\ n$ and for $n>1,\ f(n)=2$ iff $n\in S$. $f(1)$ is arbitarily defined with this condition that $f(2)\ |\ f(1)+1$

Note $f(px)=f(pk)$ for all $x\geq k$ now taking $m=pk-1,n=(pk)(pk-1)\implies f(pk)=1$ now take $m=px-1$ where $x<k$ and $n=pt(px-1)$ so again we get $f(px)=f(pk)=1$ for all $x\in\mathbb N$. Now taking $m=px,n=pkx$ we've $f(px+1)|2$ now so for $p|n-1$ we've $f(n)=1$ or $2$ now taking $(n-1,p)=1$ and by PHP there exist a large $k$ such that $n-1|kp-1$ now so taking $m=n-1$ we've $f(n)|f(n+kp)+1\implies f(n)=1$. So we've $f(n)=1$ for all $(n-1,p)\neq 1$ otherwise our $f$ is one or two. Now obviously we can't take $m,n$ such that $m\equiv 0 (mod p)$ and $n\equiv 1(mod p)$ and thus $f(px+1)$ can take any of one or two. Also $f(1)$ can be taken as arbitrary.

First let's deal only with large $m$:
Case 1: $m \not \equiv 0 \pmod{p}$. Say $m \mid tp+1 \implies m \mid m+tp+1$ (it's obvious there exists such $t$).
$P(m,m+tp+1): f(m+1) \mid f(m+tp+1)+1 \implies f(m+1) \mid f(m+1)+1 \implies f(m+1)=1$ (using $f(m+p)=f(m)$ for large $m$).
Thus $f(m)=1$ for $m \not \equiv 1 \pmod{p}$.
Taking $m=tp$ now (aka $m \equiv 0 \pmod{p}$), $P(tp,ltp): f(tp+1) \mid f(tlp)+1 \implies f(tp+1) \mid 2 \implies f(tp+1)=1$ or $2$.
Thus $f(m)=1$ for $ m \not \equiv 1 \pmod{p}$, $f(m)=1, m \equiv 1 \pmod{p}$, and $f(m)=1$ are only 2 solutions for large $m$.

Time to tackle small $m$:
Case 1: Continuation of function  $f(m)=1$ for $ m \not \equiv 1 \pmod{p}$, $f(m)=1, m \equiv 1 \pmod{p}$, $m$ large.
Taking small $m \not \equiv 0 \pmod{p}$:
$P(m,tm)(\text{gcd}(t,m)=1,t$ large):$f(m+1) \mid 2$. $P(m,tpm): f(m+1) \mid 3$.
Thus $f(m)=1$ for small $m$, $m \not \equiv 1 \pmod{p}$.
Taking $m=np \equiv 0 \pmod{p}$, $P(m,tm): f(np+1) \mid f(npt)+1 \implies f(m+1) \mid 2$.
Thus $f(m)=1$ or $2$ (pointwise) for small $m \equiv 1 \pmod{p}$.

Case 2: Continuation of function $f(m)=1$ for large $m$.
Note $f(m+1) \mid f(tm)+1 \implies f(m+1) \mid 2 \implies f(m)=1$ or $2$ (pointwise).

Thus our 2 solution sets, speaking broadly, are:
1. $f(n)=1$ for all $n \geq k$, $k$ chosen arbitrarily, $f(n)=1$ or $2$ for $n<k$, chosen arbitrarily for each $n$ (Clearly this works).
2. $f(n)=1$ for all $n \geq k$ where $n \not \equiv 1 \pmod{p}$, $f(n)=2$ for all $n \geq k$ where $n \equiv 1 \pmod{p}$, $f(n)=1$ or $2$ for all $n<k$, chosen arbitrarily. Again, this should work.

Possible source of error: I took this [hide=condition]  $f(m+1) \mid f(n)+1$ [/hide]to hold only for large $n$, as it was included in the same statement. It could be a misinterpretation which if fixed could make some of the pathological cases disappear, but I think it's intended as I interpreted it though...


what is the minimum no of digits to be placed in a 9/9 sudoku for a unique solution?

is it ever possible without repeatation of any given digit?

Ha, ha I don't think that this problem can be solved with "bare hands". A computer could be useful...

[url=http://en.wikipedia.org/wiki/Mathematics_of_Sudoku]See here[/url]. (This applies to your other topic also)


Noticed this post in another forum under lightbulb jokes, and thought it was hilarious and was sooo typical of forums. So here it is:

How many forum members does it takes to change a light bulb?

1 to change the light bulb and to post that the light bulb has been changed

14 to share similar experiences of changing light bulbs and how the light bulb could have been changed differently

7 to caution about the dangers of changing light bulbs

1 to move it to the Lighting section

2 to argue that it should be moved to the Electricals section

7 to point out spelling/grammar errors in posts about changing light bulbs

5 to flame the spell checkers

3 to correct spelling/grammar flames

6 to argue over whether it's "lightbulb" or "light bulb" ... another 6 to condemn those 6 as stupid

2 industry professionals to inform the group that the proper term is "lamp"

15 know-it-alls who claim they were in the industry, and that "light bulb" is perfectly correct

19 to post that this forum is not about light bulbs and to please take this discussion to a lightbulb forum

11 to defend the posting to this forum saying that we all use light bulbs and therefore the posts are relevant to this forum

36 to debate which method of changing light bulbs is superior, where to buy the best light bulbs, what brand of light bulbs work best for this technique and what brands are faulty

7 to post URL's where one can see examples of different light bulbs

4 to post that the URL's were posted incorrectly and then post the corrected URL's

3 to post about links they found from the URL's that are relevant to this group which makes light bulbs relevant to this group

13 to link all posts to date, quote them in their entirety including all headers and signatures, and add "Me too"

5 to post to the group that they will no longer post because they cannot handle the light bulb controversy

4 to say "didn't we go through this already a short time ago?"

13 to say "do a Google search on light bulbs before posting questions about light bulbs"

1 forum lurker to respond to the original post 6 months from now and start it all over again.

ouch. i think i see myself under about... 8 of those categories.... at least... *grins sheepishly*

I know what you mean.  Thats one of the better ones I have heard.  Though my favorite lightbulb joke is:

How many catholics does it take to change a lightbulb?

1 to hold the ladder.
1 to turn the ladder.
3 to pray against the power of darkness.

How many Delta "Service" representatives does it take to return lost luggage?

1 to send it

6 to recover it from the wrong continent

5 to write hold music to last several hours while the disgruntled owner waits to speak to a human being

3 to deliver it

and all 15 to forget to apologize profusely for depriving the owner of clean clothing for a week

Sorry, but I'm very bitter about a recent experience with Delta Airlines.  My sincerest apologies to anyone offended.

I'm a little college-obsessed right now-incoming senior and all that, but i saw this college-lightbulb joke site earlier this year and thought it was pretty funny.  Especially the part about Harvard-a guy i know is going to Harvard and fits the description exactly:

[url]http://www.theprices.net/humor/classic/collegelightbulb.html[/url]


geometry perimeter circumcircle incenter

Let $ P$ be any point inside an acute triangle. Let $ D$ and $ d$ be, respectively, the maximum and minimum distances from $ P$ to any point on the perimeter of the triangle. Prove that $ D\geq 2d$ and determine when equality holds.

I believe $ D \ge R$, $ r ge d$, for an acute triangle the argument here is straightforward: for the first one, draw the circumcenter, then do some angle comparisons, similarly for the second use the incenter and use the fact that $ ad_{a}+bd_{b}+cd_{c}$ is constant.


There are plenty of problems I've encountered in standard math tests that have dealt with things related to calender/days of the year, and I've been wondering if there's a list of basic principles that can be applied to any calender-based problem?

For example, I believe that the max amount of x day of the week on yth day of the month (ex. Friday the 13ths, Wednesday the 27ths)in a year is 3, while the minimum is 1, given that y < 31. Principles like this all related to the rules of calenders would be really helpful if anyone had a list of them.

Calendar axioms are more along the line of 30 days hath September, April, June, and November.

I hate calendar problems.  There isn't any nice way to do them -- you're almost always best off just counting, or keeping one of those people who can figure out the day of the week of every day since the Gregorian calendar was instituted around, just in case.

[quote="ComplexZeta"]$365\equiv 1\pmod 7$.[/quote]

Ah, but $366\equiv 2\pmod 7$

$31\equiv 3\pmod 7$

$30\equiv 2\pmod 7$

$29\equiv 1\pmod 7$

$28\equiv 0\pmod 7$

Here's an amusing problem from the 1950 Putnam.  Prove that the probability that Christmas falls on a Wednesday is not 1/7.

The usual rule for leap years applies.  A leap year is a year that is divisible by 4 but not 100, or divisible by 400.

Ravi - . Of course, the cycle repeates in 400 year, but 400 is not divisible by 7.

Right, but first you have to calculate that 400 years contains an integer number of weeks.

It turns out that the probability that Christmas is on Wednesday is 57/400.  In a similar fashion, it turns out that the 13th of a month is (slightly) more likely to be on Friday than any other day of the year.

[quote="Ravi B"]Here's an amusing problem from the 1950 Putnam. Prove that the probability that Christmas falls on a Wednesday is not 1/7. [/quote]

This problem was also on the 2002 MMPC part II, which I solved :).


Given four points, $A,B,C,D$ on the circumference of a circle. $AC$ intersects with $BD$ (inside the circle), and the intersection is called $P$ $AP=6$, $CP=2$, $BP=3$, $DP=4$. Find the diameter of a circle.

It is not possible to find the circumcircle diameter. The quadrilateral diagonals AC, BD intersecting at the point P can form any angle. Since $PA \cdot PC = PB \cdot PD$ regardless of this angle, the resulting quadrilateral ABCD is always cyclic. The circumcircle diameter can be anywhere from AC = 8 (the longer diagonal) to infinity.

[quote="xxreddevilzxx"]A funny thing was that the problem came from Mu Alpha Theta, so i thought it was solvable.[/quote]
The problem says to $\text{find}$ the diameter. So it should include the $\text{range}$ of $d$.

[quote="Hexaditidom"]yetti, what do you use to make those diagrams?[/quote]
It is called the geometer's sketchpad.
I think Yetti is an expert sketcher, i don't know how could he draw such good images...

[quote="Hexaditidom"]yetti, what do you use to make those diagrams?[/quote]
I make the drawings with Geometer's Sketchpad, then copy and paste them into an older version of Adobe Photoshop and export the file as a transparent GIF. See my last reply in the thread [url=http://http://www.mathlinks.ro/Forum/viewtopic.php?t=34221]Geometrical problem[/url] and also a long discussion about [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=11055]Geometer's Sketchpad[/url]. As for converting the files into the JPG (cannot be transparent) or GIF formats, various graphics converters are available for download (although they usually work for a limited time, such as 2 weeks or a month, unless you buy a licence, usually about $\$20). An older version of Adobe Photoshop is also very cheap (I use the ancient version 5).

Yetti


number theory
least common multiple
superior algebra
superior algebra solved

Hi once again.

How could one prove that in general the order of a permutation is the lcm of the lengths of its cycles?  I'm sure there is some straight forward (and simple) thing I'm missing here.

Thank you very much.

Bailey

The cycles that make up a permutation commute with each other. You fix every element exactly when every cycle is back to its starting point.

Thanks, jmerry.  That makes sense to me and helps a lot.

Bailey


superior algebra
superior algebra theorems

Does someone knows where can i find some good information about Elliptic Curves? I need it for a presentation...ç

ok, and do you know where can i find something about the relation between Fermat Last Theorem and these curves??

Thanks

Try http://www.jmilne.org/math/CourseNotes/math679.dvi and http://math.stanford.edu/~lekheng/flt/wiles.pdf . I like elliptic curves too!

[quote="jmerry"]A very nice book (probably worth reading at some point): Rational Points on Elliptic Curves, by Silverman and Tate.[/quote]How would you estimate its level compared to "The Arithmetic of EC" and "Advanced Topics in the Arithmetic of EC"? Is there something in "Rational Points on EC" not covered in those other two books?

I don't know the other books. The attraction of the one I know is that it is accessible on a relatively elementary level.

There are various online resources listed at [url=http://dmoz.org/Science/Math/Number_Theory/Elliptic_Curves_and_Modular_Forms/]dmoz.org/Science/Math/Number_Theory/Elliptic_Curves_and_Modular_Forms/[/url].


algebra
polynomial
Ring Theory
superior algebra
superior algebra unsolved

Prove that any prime ideal in the polynomial ring Z[X] can be
generated by two generators. What can you say about the number of
generators of a non-prime ideal in Z[X]?

if $ R$ is a PID, $ Spec R[X]$ can be determined with the fibers of $ Spec R[X] \to Spec R$, because $ Spec R/p[X]$ ($ p$ prime) and $ Spec Q(R)[X]$ are known. the prime ideals of $ R[X]$ have thus the following form: $ (0)$, $ (p)$ for some prime $ p$, $ (p,f)$ for some prime $ p$ and monic $ f \in R[X]$ such that $ f$ is irreducible mod $ p$, and $ (f)$ for some irreducible $ f \in R[X]$ (i.e. $ f$ is primitive and irreducible in $ Q(R)[X]$). in particular, they are generated by two elements. I guess that $ (X^2\plus{}1, X^3\plus{}1, 2*X)$ cannot be generated by two elements.


function
limit
algebra
polynomial
algebra proposed

is this function injective?is it surjective?

f: R to R
f(x)= 2x^3-3x+2

 :blush:

Supposing $ f$ is injective it means that $ f(a)\equal{}f(b)$ $ \Rightarrow$ $ a\equal{}b$ (only) $ \forall a,b\in\mathbb R$
$ f(a)\equal{}f(b)$ $ \Leftrightarrow$ $ (a\minus{}b)(2a^{2}\plus{}2ab\plus{}2b^{2}\minus{}3)\equal{}0$ .But this does not imply only $ a\equal{}b$ for all $ a,b\in\mathbb R$ ,because
$ \Delta^{'}_{a}\equal{}\minus{}3(b^{2}\minus{}2)<0$ is not true $ \forall b\in\mathbb R$,Therefore $ f$ is non-injective.

 $ f$ is surjective $ \Leftrightarrow$ $ \forall y\in\mathbb R$ $ \exists$ $ x\in\mathbb R$ such that $ f(x)\equal{}y$
$ \Leftrightarrow$
$ \forall y\in\mathbb R$ $ \exists$ $ x\in\mathbb R$ such that $ x^{3}\equal{}\frac{3x\minus{}2\plus{}y}{2}$
Consider the functions $ g_{1,2}:\mathbb R\to\mathbb R$ , $ g_{1}(x)\equal{}x^{3}$ and $ g_{2}(x)\equal{}\frac{3x\minus{}2\plus{}y}{2}$
If we plot these functions $ g_{1}$ and $ g_{2}$ it's clear that they intersect $ \forall y\in\mathbb R$,because $ g_{1}$ "covers" all the values from $ \mathbb R$ and $ g_{2}$ it's a line.So the equiation $ g_{1}(x)\equal{}g_{2}(x)$ has a solutions $ \forall y\in\mathbb R$ ,meaning that $ f$ is surjective.

Another proof for the surjectivity would be the following:
$ f$ is clearly continous $ (1)$
$ \lim_{x\rightarrow\infty}f(x)\equal{}\infty$ ,$ (2)$
$ \lim_{x\rightarrow\minus{}\infty}f(x)\equal{}\minus{}\infty$ ,$ (3)$
From $ (1),(2),(3)$ $ \Rightarrow$ $ f$ is surjective.
 :lol:

[quote="greatestmaths"][color=darkred]The function $ f:\mathbb R\rightarrow\mathbb R$, where $ f(x) = 2x^{3}-3x+2$ is injective? Is it surjective?[/color][/quote]
[color=darkblue][b]Proof.[/b] Write $ f(x) = x(2x^{2}-3)+2$ and observe that $ f(x) = 2\Longleftrightarrow x\in\{\ -\sqrt{\frac{3}{2}}\ ,\ 0\ ,\ \sqrt{\frac{3}{2}}\ \}$, i.e. $ f\ :\ \{\begin{array}{cc}0 &\searrow\\\\ -\sqrt{\frac{3}{2}}&\rightarrow\\\\ \sqrt{\frac{3}{2}}&\nearrow\end{array}2$.

Therefore the function $ f$ [b]isn't injective[/b] because exists at least two different real numbers $ x_{1}= 0$, $ x_{2}=\sqrt{\frac{3}{2}}$ such that $ f(x_{1}) = f(x_{2}) = 2$.

For any $ \lambda\in\mathbb R$ the polynomial equation with real coefficients $ P(x)\equiv f(x)-\lambda = 0$ has at least a real root because the its degree is odd,

i.e. for any $ \lambda\in\mathbb R$ exists at least $ x_{\lambda}\in\mathbb R$ such that $ f(x_{\lambda}) =\lambda$ what means that the function $ f$ [b]is surjective[/b].

Indeed, the number of the its complex roots is even because for any $ z\in\mathbb C$, $ P(z) = 0\Longleftrightarrow$ $ P(\overline z) = 0$ ("odd" - "even" = "odd" and $ 0$ is even).

In conclusion, the function $ f$ [b]isn't injective and is surjective[/b]. The nice [b]Svejk[/b]'s proof is generally and very usually in such cases.[/color]


geometry
geometric transformation
reflection
floor function
ceiling function

[hide]for 5 digits, the digits can be
$AAAAA$,
$AABAA$,
$ABABA$,
$ABBBA$,
$ABCBA$.
For $AAAAA$ there are 9 choices for $A$ (you can't use $0$) $=9$ total.
For $AABAA$, $ABABA$, and $ABBBA$ there are $9$ choices for both $A$ and $B$ because you can use $0$ as $B$ but can't use $A$ for $B$. $9*9*3=243$.
For $ABCBA$, there can be $9$ choices for $A$, $9$ for $B$, and $8$, for $C$. $9*9*8=648$. $648 + 243 + 9 = \boxed {900}$
For 6 digits they can be in the form
$AAAAAA$,
$AABBAA$,
$ABBBBA$,
$ABAABA$, or
$ABCCBA$.
For $AAAAAA$ there are 9 choices for $A$ (you can't use $0$) $=9$ total.
For $AABBAA$, $ABBBBA$, and $ABAABA$ there are $9$ choices for both $A$ and $B$ because you can use $0$ as $B$ but can't use $A$ for $B$. $9*9*3=243$.
For $ABCCBA$, there can be $9$ choices for $A$, $9$ for $B$, and $8$, for $C$. $9*9*8=648$. $648 + 243 + 9 = \boxed{900}$. WOW its the exact same thing as for 5 digits.
But, it is different for 7.
The choices are:
$AAAAAAA$,
$AAABAAA$,
$AABBBAA$,
$AABABAA$,
$ABBBBBA$,
$ABBABBA$,
$ABAAABA$,
$ABABABA$,
$AABCBAA$,
$ABCACBA$,
$ABCCCBA$,
$ABCACBA$,
$ABBCBBA$,
$ABACABA$, and
$ABCDCBA$.
WOW thats a lot.
For the ones that just involve $A$, there are $9$ possible choices. (can't use $0$) There is only one combination with just $A$ so $9$.
For the ones with just $A$ and $B$, there are $9$ possible choices for $A$ and $9$ for $B$ (because $B$ can use $0$ but can't use $A$). There are $7$ that use just $A$ and $B$, and $9*9*7=567$.
For the ones that involve only $A$, $B$, and $C$, there are $9$ combinations for $A$, $9$ for $B$ and $8$ for $C$ (It cant use $A$ or $B$). There are $6$ combinations that use just $A$, $B$, and $C$, and $9*9*8*6=3888$.
For the ones involving $A$, $B$, $C$, and $D$, there are $9$ combinations for $A$, $9$ for $B$, $8$ for $C$, and $7$ for $D$. There is only one that uses $A$, $B$, $C$, and $D$ and $9*9*8*7=4536$. Add up the totals and get $9 + 567 + 3888 + 4536 = \boxed{9000}$. (Can you see a pattern between $5$ digits, $6$, and $7$??)
[/hide]
EDIT I FINISHED THE SOLUTIONS!!!

[hide]For 5-digit palindromes, we look at the first three digits. We only care about these 3 digits because the last two are dependent of these first 3 digits.
There are 900 three digit numbers. Just like 6-digit palindromes (last three digits are reflection of first three digit).

for 7-digit palindromes, we look at the first four digits, to get 9000 numbers. [/hide]

How about this: how many different palindromes are there with n digits?  :D

[quote="solafidefarms"]236!, that's a very pretty solution. Biffandoc, that may work, but 236!'s method is faster.[/quote]NOOOOOOOOOOO i thought I gave a great solution and now its the slow way.
[hide="answer to the other problem"]
it is $10^{[[n/2]]-1} *9$
where $[[n]]$ is the greatest integer $\leq n$
[/hide]

I got the same thing, but slightly simplified:

[hide]$9\cdot 10^{\lfloor \frac{n}{2} \rfloor}$[/hide]

JesusFreak is ${\lfloor \frac{n}{2} \rfloor}$ the same as $[[\frac n2]]-1$?

Whoops, my mistake. Your notation and mine are the same, but we both got it wrong. It should be:

$9\cdot 10^{\lceil \frac{n}{2} \rceil -1}$

Where $\lceil x \rceil$ denotes the least integer greater than or equal to $x$.

[quote="cardwacko9999"]sry this post is not finished... [hide]friend of mine[/hide] pressed tab enter...

will finish later

with n digits, there are
[hide][/hide] palindromes.[/quote]
what??????


[quote="chuchen"]1. How many base three numbers with anywhere from 1 to 5 digits use only 1s and 0s?
2. 314203 base5 divided by 23 base 5.
3.  In class we did two "experiments".  The first was pick a number, square it, and change it into the base 1 less than your original number.  The second was pick a number, cube it, and change it into the base 1 less than your original number.  If I said pick a number and just change that to the base 1 before it, you would find that you'd get the same thing all the time just like you did in the two "experiments".  What if you raised the original number to the zero power and then put it into the previous base.  That would give you 4 things to look at, raising to the 0 power, 1st power, 2nd power, and 3rd power.  First, can you predict what would happen if you raised 9 to 4th power and converted that to base 8?  Fill in the chart below and predict what the second to last digit would be if you raised 9 to the 9th power and then convert it to base 8?
0th power_____, 1st power_____, 2nd power_____
3rd power _____, 4th power_________

4.  Suppose we flip four coins simultaneously: a penny, a nickel, a dime, and a quarter.  What is the probability that at least 15 cent worth of coins come up heads?[/quote]

[hide="1"]There are $2^{5}=32$ ways to place the digits.  However, 00000 does not work so it is $\boxed{31}$.[/hide]

[hide="2"]$\frac{314203_{5}}{23_{5}}=\frac{3\cdot 5^{5}+1\cdot 5^{4}+4\cdot 5^{3}+2\cdot 5^{2}+0\cdot 5^{1}+3\cdot 5^{0}}{2\cdot 5^{1}+3\cdot 5^{0}}=$something[/hide]

[hide="4"]If the quarter is heads, the other coins can be anything. The probability of that is $\frac{1}{2}$. If the quarter is tails, the dime and nickel are heads, then the penny can be either head or tails. The probability of that is $\frac{1}{8}$. You can't make 15 with a penny and a nickel so that's all the ways. $\frac{1}{2}+\frac{1}{8}=\frac{5}{8}$[/hide]

[hide="3"] Take a number $n+1$.  Let's square this and convert it into base n.
\[(n+1)^{2}_{10}=(n^{2}+2n+1)_{10}=(n^{2}+2(n^{1})+1(n^{0}))_{10}=121_{n}\]
Let's look at the general case now:
\[(n+1)^{x}_{10}=(n^{x}+\binom{x}{1}n^{x-1}+\binom{x}{2}n^{x-2}...+\binom{x}{x-1}n^{1}+1)_{10}=(10^{n}+\binom{x}{1}10^{x-1}+\binom{x}{2}10^{x-2}...+\binom{x}{x-1}10^{1}+1)_{n}=11^{x}_{n}\]
This method is [b]quite[/b] complicated.  A simpler way to do it is to just see a pattern when bringing 9 to the third power in base 8, then the fourth power in base 8.  Eventually, it will become evident that $9^{x}_{10}=11^{x}_{8}$.  Plugging x in for 0, 1, 2, 3, and 4, we find that the answers are $1, 11, 121, 1331,$ and $14641$ respectively.  [/hide]


For how many ordered pairs of positive integers $(x,y),$ with $y<x\le 100,$ are both $\frac xy$ and $\frac{x+1}{y+1}$ integers?

[hide="Forced but pretty"]85. By going through some brute forcing of the problem, by letting y= a set number and finding all x's, it eventually becomes apparent that to find the x, the first x is y(y+1)+y with the common difference between x's being y(y+1). I haven't really tried shaping this up into showing why this is, but it shouldn't really be too hard. Starting with y=1, y(y+1)+y = 3. And then 3+y(y+1) = 5, etc. giving 49 pairs for y=1. For y=2, common difference of 6 with the first term being 6+2=8. This helps speed up the process, but it's still somewhat brute force.[/hide]

Yep that's right.  That's basically what I did and the official solution is also just like that.  Even though it may seem like brute force, it's a good lesson to learn to not be afraid to pound out problems on the AIME (or on any contest for that matter)...just as long as you don't get carried away...often times when you brute force on the AIME, you'll get caught on little arithmetic errors.  It takes a delicated balance between knowing when to look for a more elegant approach and just forcing it out ;)  But actually, on this one, there aren't many other things that you could really try which is why you would resort to 'brute forcing'.

[hide="Solution"]
We have $y | x$, so we write $x = ky$ for some integer $k$.  Then we have $\frac{ky+1}{y+1}$ is an integer.  Subtracting 1, we have $\frac{(k-1)y}{y+1}$ must also be an integer, and because $y$ and $y+1$ have no common factors, $k-1 = m(y+1)$ for some integer $m$.  This gives us $x = ( m(y+1) + 1)y = my(y+1) + y$ will produce solutions.

We seek $y(y+1) + y < 100$, which implies $y \le 9$.  The various equations that give us $x$-solutions are:

$x = 2m + 1$ (49 solutions)
$x = 6m + 2$ (16 solutions)
$x = 12m + 3$ (8 solutions)
$x = 20m + 4$ (4 solutions)
$x = 30m + 5$ (3 solutions)
$x = 42m + 6$ (2 solutions)
$x = 56m + 7$ (1 solution)
$x = 72m + 8$ (1 solution)
$x = 90m + 9$ (1 solution)

In total, we have $85$ solutions... eww, brute force  :(
[/hide]

[hide=progress]We know that $\frac{x}{y}$ is an integer, so denote $\frac{x}{y} = p$ for some integer $p$, which gives $x = py$. Then we must have $\frac{py+1}{y+1}$, or equivalently $p - \frac{p-1}{y+1}$ is an integer. This is equivalent to saying that $\frac{p-1}{y+1}$ must be an integer[/hide]


How would you convert a base 10 decimal number to base 4? Take 0.625 for instance...

[url=http://cs.furman.edu/digitaldomain/more/ch6/dec_frac_to_bin.htm]This site[/url] had a decimal --> binary "explanation" (more like step by step -- no logic behind it explanation), but could someone please explain how it works? Are there other ways to think about it? Could you somehow split up the 0.625 into 625 (a whole number) multiplied by 10 to the negative 3?

The aops vol 1 book says for converting base 10 whole numbers to a base x, you just see how many of each power of x "fit into" the base 10 number. That's a really great way to think about it, but I one of my friends showed me another way--where you take the base 10 number and divide it continuously by x until you eventually get a "division tower", so to say. The base 10 number in base x is the last quotient and all the remainders written consecutively right after it. I sort of see how the two relate, but can't explain it or see it clearly.

Let $x<1$ be the decimal number we want to convert from base 10 to base 4.

Its base-10 representation would be $x={d_1 \over 10}+{d_2 \over 10^2}+{d_3 \over 10^3}+\dots+{d_n \over 10^n}$, where $d_i$ are its base-10 digits (this is true because, by the definition, the first decimal represents tenths, the second hundredths, the third thousandths and so on).

In the same fashion, its base-4 representation would be of the form $x={a_1 \over 4}+{a_2 \over 4^2}+{a_3 \over 4^3}+\dots+{a_n \over 4^n}$, where $a_i$ are the new digits.

We should always keep in mind that the largest base-$M$ digit is $M-1$, therefore $0\leq a_i \leq 3$.

Now, let's see what happens if we multiply $x$ by 4. We get $4x={a_1}+{a_2 \over 4}+{a_3 \over 4^2}+\dots+{a_n \over 4^{n-1}}$. Integer part of this number is [u]exactly our first digit[/u], because the sum of the other terms can't exceed 1. (Take that for granted - the proof is easy and involves taking the sum of a geometric progression, but we don't really need it for what we're dealing with right now.)
Great, we've got the first decimal. Now what? Let's subtract it - what do we get? $4x-{a_1}={a_2 \over 4}+{a_3 \over 4^2}+\dots+{a_n \over 4^{n-1}}$ Obviously, if we multiply both sides by 4 again, we'll get $a_2$ as the integer part. Now it should be clear what's the general procedure:

1. Multiply $x$ by 4;
2. Take the integer part - that will be the first base-4 decimal
3. Subtract it to obtain the fractional part
4. Multiply the fractional part by 4 again, in order to get the second base-4 decimal as the integer part of the product
5. Subtract the integer part again
6. Keep repeating the procedure until the fractional part becomes zero (later we'll have a remark about this)
7. Write all the integer parts in the direct order (i.e. first computed first written) behind the decimal point, thus forming the base-4 representation of the number

In the case of $x=0.625$ it goes like this:
1. 0.625 x 4 = 2.5 - the first digit is 2, and we subtract it, getting 0.5
2. 0.5 x 4 = 2 - the second digit is 2, and we subtract it, getting 0. Since the fractional part is zero, the procedure is over.
Now we write the digits in the order of computing, i.e. $(0.625)_{10}=(0.22)_4$

Another example: Convert $x=0.84375$ to base 4.
Solution:
1. 0.84375 x 4 = 3.375 - first digit is 3; proceed with the fractional part
2. 0.375 x 4 = 1.5 - second digit is 1; proceed with the fractional part
3. 0.5 x 4 = 2 - third digit is 2; no fractional part

Therefore, $(0.84375)_{10}=(0.312)_4$

Important: The fractional part need not ever become zero, but in that case you will observe that the decimals repeat periodically - just as they can in base 10. For example, take $x=0.6$. You'll have 0.6 x 4 = [b][u]2[/u][/b].4, 0.4 x 4 = [b][u]1[/u][/b].6, 0.6 x 4 = [b][u]2[/u][/b].4, 0.4 x 4 = [b][u]1[/u][/b].6 and so on - we've got an infinite periodic decimal number $(0.21212121\dots)_4=(0.\overline{21})_4$. Just try to imagine any other procedure that would give you an infinite decimal number this easily.

Of course, all of the above holds for any other given base, not only for base 4. For example, let's convert 0.8272 to base 5. We have 0.8272 x 5 = [b][u]4[/u][/b].136, 0.136 x 5 = [u][b]0[/b][/u].68, 0.68 x 5 = [u][b]3[/b][/u].4, 0.4 x 5 = [u][b]2[/b][/u], giving $(0.8272)_{10} = (0.4032)_5$.

Hope all of this helped :)

PS. Yes, your friend is right - but only if you want to convert an [i]integer[/i] number. That can be done by obtaining the remainders in the chain division. The procedure would be completely reverse to the one given above, meaning that you (a) keep dividing instead of multiplying, (b) take the remainders instead of the integer parts, (c) write the digits in the reverse order (i.e. last computed first written) and (d) can never end up with infinitely many digits. This is so because in both cases we must compute the digit closest to the decimal point first, but in an integer that digit is the least significant, and in a decimal number it's the most significant, which produces completely opposite routes to the task. If you're interested in the logic behind the routine for integer numbers, take the hint from the above discussion and see what happens if $4^n a_n+{4^{n-1}}a_{n-1}+\dots+4^2 a_2+4a_1+a_0$ is divided by 4, keeping in mind that all the digits must be less than 4.


function
real analysis
calculus
integration
real analysis unsolved

Find a necessary and sufficient condition on the continuous function $g$ defined on $R$ with values in $R$ such that for any Lebesgue integrable function $f: [0,1]\to R$ we have that $g(f(x))$ is also Lebesgue integrable.

I think it's $g(x)=O(x)$. In other words, there has to be a constant $M>0$ such that $|g(x)|\le M|x|$ for all large $|x|$.

The sufficiency is clear:

If $|g(x)|\le M|x|,\ \forall |x|\ge t$ for some $t>0$, say, then $\int|g\circ f|$ breaks up as the sum between the integral over $f^{-1}((-t,t))$, which is finite because $g$ is bounded on $(-t,t)$, and that over the rest of $[0,1]$, which is not larger than $M\int|f|<\infty$.

Now the necessity (it's only going to be a sketch):

Assume WLOG that $g\ge 0$. Also, from now on, all real numbers which appear in series and such will be non-negative, in order to avoid using absolute values all the time. $\mu$ denotes the Lebesgue measure, and $\chi_{S},\ S\subset[0,1]$ will be the characteristic function of the measurable set $S$.

Pick $(a_{n})_{n},(b_{n})_{n}$ subject to the conditions $a_{n},b_{n}\ge 0,\ \sum a_{n}=1,\ \sum a_{n}b_{n}<\infty\ (*)$. We can certainly find disjoint measurable subsets $A_{n}$ of $[0,1]$ such that $\forall n,\ \mu(A_{n})=a_{n}$. Then consider $f=\sum b_{n}\chi_{A_{n}}$, which is in $L^{1}$. The integral of $g\circ f$ is $\sum a_{n}g(b_{n})$, which we know is finite. Hence for all sequences satisfying $(*)$, we have $\sum a_{n}g(b_{n})<\infty$. It should be easy to conclude now.

It's a bit easier by contradiction: suppose there is a sequence $(t_{n})_{n\ge 1}$ such that $|t_{n}|\ge 1$ and $|g(t_{n})|\ge 2^{n}|t_{n}|$. Define $f(x)=t_{n}$ for ${2^{-n}<x<2^{-n}+2^{-n}/t_{n}}$ and $f(x)=0$ otherwise. Then $\int_{0}^{1}|f(x)|dx=\sum_{n=1}^{\infty}2^{-n}=1$ but $\int_{0}^{1}|g(f(x))|dx\ge \sum_{n=1}^{\infty}1=\infty$.


geometry trapezoid rectangle inequalities

ABCD is a convex quadrilateral. X lies on the segment AB of the quadr and Y lies on the segment CD. If point W is the intersection of XC and BY, and point Z is the intersection of AY and DX, and AB is parallel to CD

A) prove that area XWYZ <=  1/4 area ABCD.
B) Is this true if AB is not parallel to CD

help!!!!

Its due monday so if someone could help me out before then that would be really nice of you!!

p.s. Please tell me if its too easy and I'll post them in more basic forums.

this is my hommeowrk for my fourth grade class at elementary school. if it is not okay, I'll stop posting here

this seems really hard for 4th grade... i doubt my geometry teacher could do this,... then again, she didn't prove anything in class...

I don't have much time, sorry, and i'm not that good at geometry, but I think you should try using simlar triangles. (keep in mind that ABCD is a trapezoid because of the parallel lines.)
Using the parallel lines and and vertical angles (ex: angle BWX is congruent to angle CWY since they're vertical angles, and angle YCW is congruent to angle BXW because of the parallel lines...you should first mark all congruent angles you see), its clear that triangles BXW and YCW are similar, and triangles AXZ and YDZ are similar. I hope this helps.

I'm not sure of this, but i think if you drop altititudes from point W to line AB and from point W to line CD, and then do the same for point Z (drop altitudes to AB and CD, btw these altitudes should be one strainght line, so i guess i couldve just said draw two perpendiculars through both AB and CD that go through W and Z respectively). If you name the 4 foots of the 4 altitudes you just drew M, N, O, P (it doesnt matter which order), then MNOP is a rectangle. My intuuition tells me that this rectangle is 1/2 the area of trapezoid ABCD, and my intuition also tells me that quadrialteral WXYZ is half the area of this rectangle, making its area 1/4 of ABCD's area. I havent proved this (cuz im in a rush right now), but you can try to prove this using simimlar tirangles again.

Sorry if my intuition is wrong.

I dont have time now to do part b

are u taking geometry already????? in 4th grade?... I knew like no math at all in 4th grade.

hope this helps

hehe. I've heard thing about your geometry teacher.... :ninja:

I think this would be okay here, since it involves a proof, and I think is above "Classroom Math level". The only problem is that if this is homework, we really should not be posting solutions for other people to cheat off of. However, if you post your own solution, we can help you along and correct you.  :lol:

fishy makes a good point, and that point was part of the reason i didnt work harder ont he prob.
btw, i love ur name, fishy

[quote="hurdler"]dynamo, your sentence doesnt make sense....[/quote]

Wait.... nevermind. For some reason I got u mixed up with someone else on AoPS...

Divide the quadrilateral into two trapezoids AXYD and XBCY by connecting points X and Y.

Now concentrating on one of the trapezoids, say AXYD.

In trapezoid AXYD, the following two facts are important in solving the problem:
(1) The area of the triangle AZD = The area of the triangle XYZ
Refer to the example: [url]http://www.idealmath.com/algebra/hottopics/Area/areatheorems2.htm[/url]
(2) Area of AZD / Area of DZY = Area of AXZ / area of ZXY
Refer to [url]http://www.idealmath.com/algebra/hottopics/Area/areatheorems.htm[/url]

Denote the area of AZD and XYZ as [b]A[/b], the area of AXZ as [b]A[/b]1 and the area of DZY as [b]A[/b]2

According to fact 2, we have
[b]A[/b] / [b]A[/b]2 = [b]A[/b]1 / [b]A[/b]
So [b]A[/b]^2 = [b]A[/b]1*[b]A[/b]2
[b]A[/b] = sqrt ([b]A[/b]1[b]A[/b]2)

Applying the AM-GM Inequality theorem, we get
[b]A[/b] = sqrt ([b]A[/b]1[b]A[/b]2) <= ([b]A[/b]1 + [b]A[/b]2)/2
2[b]A[/b] <= [b]A[/b]1 + [b]A[/b]2

Now we will show that [b]A[/b] <= (1/4) the area of trapezoid AXYD:
the area of trapezoid AXYD = [b]A[/b] + [b]A[/b] + [b]A[/b]1 + [b]A[/b]2 <= [b]A[/b] + [b]A[/b] + 2[b]A[/b] = 4[b]A[/b]
So we have [b]A[/b] <= (1/4) the area of trapezoid AXYD

i.e. Area of XYZ <= (1/4) the area of trapezoid AXYD

For the same reason, we have
Area of XYW <= (1/4) the area of trapezoid XBCY

Adding them together, we obtain:
Area of XYZ + Area of XYW <= (1/4)(the area of trapezoid AXYD + the area of trapezoid XBCY)
Area of quadrilateral XYZW <= (1/4) Area of quadrilateral ABCD


Find distinct digits $ A$ and $ B$ such that $ A47B$ is as large as possible and divisible by 36. Name the number.

For a number to be divisible by 36, it must be divisible by both 9 and 4. For it to be divisible by 9, the digits must add up to a multiple of 9. For it to be divisible by 4, the last two digits must be divisible by 4.

Let's look at $ B$ first. The possibilities are 2 and 6, since 72 and 76 are both divisible by 4. Now look at $ A$. If $ B \equal{} 2$, we can have 5. If $ B \equal{} 6$, we can have 1. Since we wish to maximize the number, our answer is $ \boxed{5472}$.


Prove some following inequality:
1. Prove that
$(x+y+xy)( x^{y-1}+y^{x-1})\geq  (\frac{1+x}{1-x})^{1-x}+(\frac{1+y}{1-y})^{1-y}$
where $0<x, y< 1$.
2. Prove that
$\frac{(1+x)\sqrt{x}}{x+2y-2xy}+\frac{(1+y)\sqrt{y}}{y+2x-2xy}> 1+x+y+\frac{2xy}{4+x+y}$
where $0< x, y< 1$.

Oh come on guys. I think they are not really hard.
I'm waiting for some solution.

Hix. No one interested?
So i'll post a easier. OK?
Prove that
$x^{x}(1-x)^{1-x}\geq \frac{(1+x)^{1-x}}{2+x}$
where $0<x <1$

oh yeah!!
I think it easy!!
ZEIKII
I will post my soln later


Eleven rectangular 9-inch by 12-inch pictures hang on a wall. How many square inches of the wall's surface area are covered by the pictures?


If the sum of two numbers is 1 and their product is 1, then the sum of their cubes is:
\[ \text{(A)}\ 2\ \ \ \ \text{(B)}\ -2 - \frac{3\sqrt{3}i}{4}\ \ \ \ \text{(C)}\ 0\ \ \ \ \text{(D)}\ -\frac{3\sqrt{3}i}{4}\ \ \ \ \text{(E)}\ -2 \]


function
geometry
algebra
polynomial
Ring Theory
superior algebra
superior algebra solved

Qualifying Examination
problem 2/Day 2/February 2004

Prove that each of the following rings  contains infinitely many prime ideals

(i) $\mathbb Z$

(ii) $\mathbb Q[X]$

(iii) The ring of regular functions on affine algebraic surface over $\mathbb C$
(assume standard results on algebraic geometry)

(iv) The countable direct product of copies of $R$ for any nonzero
commutative ring $R$ with unity

i) The prime ideals of $\mathbb Z$ are $p\mathbb Z$ for prime $p$, so that's that.

ii) There are infinitely many irreducible polynomials $f$ in $\mathbb Q[x]$ and $f\mathbb Q[x]$ work (they're even maximal ideals, so they're also prime ideals).

I don't have the tools to tackle iii) :)

iv) We can find infinitely many maximal ideals, which are also prime. Consider ideals of the form $I_n=\prod_{i\in\mathbb N^*} X_i$, where $X_i=R,\ \forall i\ne n,\ X_n$ is a maximal ideal of $R$ (it's well-known that there are maximal ideals if $R\ne\{0\}$).

Is this correct?

the maximal ideals contained in the ring of regular functions correspond to the points of the affine variety; if that one was finite, it had to be a point by irreducibility and i guess these ones were excluded by the problem or something. so we get infinitely many maximal ideals, so infinitely many maximal ideals.


Suppose that $ f(x\plus{}3)\equal{}3x^2\plus{}7x\plus{}4$ and $ f(x)\equal{}ax^2\plus{}bx\plus{}c$. What is $ a\plus{}b\plus{}c$?

$ \textbf{(A)}\minus{}\!1 \qquad
\textbf{(B)}\ 0 \qquad
\textbf{(C)}\ 1 \qquad
\textbf{(D)}\ 2 \qquad
\textbf{(E)}\ 3$


[hide] As state , $b>a$ . Here , $a+b=1$ , and if $a>\frac 12$ , $a+b>1$ contradict so $a<\frac 12$ .

Then , similarly if $b<\frac 12$ , $a+b<1$ which also contradict .So $b>\frac 12$

Next , we can see that $b-(a^2+b^2)=b(1-b)-a^2=ab-a^2=a(b-a)>0$ so we can say $b>a^2+b^2$ .

But $a^2+b^2=(a+b)^2-2ab=1-2ab>1-2\times\frac 14=\frac 12$ .

Combine all we get $a<\frac 12<a^2+b^2<b$ :P [/hide]

[quote="236factorial"]Given that $0<a<b<1$, and $a+b=1$, order $a, b, a^2+b^2, \frac{1}{2}$ from least to greatest.[/quote]

[hide]For questions like these, I just make up an easy value for $a$ and $b$ and evaluate.  :lol:

Let $a=\frac{1}{4}$
Let $b=\frac{3}{4}$

That satisfies all the conditions.

$a=\frac{1}{4}$
$b=\frac{3}{4}$
$a^2+b^2=c^2=\left(\frac{1}{4}\right)^2+\left(\frac{3}{4}\right)^2=\frac{1}{16}+\frac{9}{16}=\frac{10}{16}=\frac{5}{8}$
$\frac{1}{2}=\frac{1}{2}$

$\frac{1}{4}<\frac{1}{2}<\frac{5}{8}<\frac{3}{4}$

$a<\frac{1}{2}<a^2+b^2<b$[/hide]

[quote="236factorial"]Given that $0<a<b<1$, and $a+b=1$, order $a, b, a^2+b^2, \frac{1}{2}$ from least to greatest.[/quote][hide]Since $b>a$ and $a+b=1$, then $b>\frac{1}{2}>a$.
From AM-GM, $\frac{a+b}{2}\ge\sqrt{ab}$ or $ab\le\frac{1}{4}$.
Since $a^2+b^2=(a+b)^2-2ab$, we have $a^2+b^2>\frac{1}{2}$.
On the other hand, $a(b-a)>0\Rightarrow ab-a^2>0\Rightarrow b-a^2-b^2>0$, from which we get $b>a^2+b^2$.
So the order is $b>b^2+a^2>\frac{1}{2}>a$.[/hide]


Hi, i would be so gratefull if somebody could help me with this...

I have to find the Taylor expansion Ti(X) of a function F(x) in function of it's derivates... is it:

Ti(x) = f(mid(x)) + f'(mid(x)) (x-mid(X)) + ... + (1/(n!)) fn(x) (x-mid(x))^n

(note x is not an integer but an interval)                   ????????????????

thank you[/list]

[quote="popy"]Hi, i would be so gratefull if somebody could help me with this...

I have to find the Taylor expansion Ti(X) of a function F(x) in function of it's derivates... is it:

Ti(x) = f(mid(x)) + f'(mid(x)) (x-mid(X)) + ... + (1/(n!)) fn(x) (x-mid(x))^n

(note x is not an integer but an interval)                   ????????????????

thank you[/quote]

What exactly do you want help with. There is an obvius problem. $x$ is not an interval, it is a real number, that can move in the interval given by the ratius of convergence of the Taylor Series. But besides that, what else do you need, I think it is not clear, at least not for me!

BTW: This seems more Calculus than Advanced Fields.


Proove using mathematical induction:

1. F_n^2 - F_n-1^2 = F_n-2 x F_n+1
ie.  8^2 - 5^2 = 3 x 13 from Fibonacci Sequence

2. Find the general form of 8^2 = 5 x 13 - 1 and prove it using mathematical induction. These numbers are also from the Fibonacci Sequence.

Sorry, I don't know how to make the letters look professional...

Latex'd $F_n^2 - F_{n-1}^2 = F_{n-2} \times F_{n+1}$
It can be done without induction a lot faster, $F_n^2-F_{n-1}^2 = (F_n-F_{n-1})(F_n+F_{n-1})$ which, by definition of Fibonacci numbers is equivalent to $(F_{n-2})(F_{n+1})$.
As for the second one, the general form is $F_n^2 = F_{n-1} \times F_{n+1} - 1$.
I'll leave the induction up to you ;)


Given n 2D-points on plane each with x and y coordinates. The goal is to find the smallest distance between two points in the set.

It is easy to use brute force to compare each pair. However, that takes O(n^2) time.

I heard it is possible to reduce the time complexity to O(nlogn). Can anyone give a nice/simple method?


Prove that in a triangle: \[ 18Rr\le ab\plus{}ac\plus{}bc\]
where $ R$ is the radius of the circumcircle, $ r$ is the radius of the incircle and $ a$, $ b$, $ c$ are the lengths of the sides of the triangle.

$ 18Rr\le ab \plus{} ac \plus{} bc$ $ \Longleftrightarrow$ $ 18Rr\le p^2 \plus{} r^2 \plus{} 4Rr$ $ \Longleftrightarrow$ $ p^2\ge 14Rr \minus{} r^2$.

But $ \boxed {p^2\ge 16Rr \minus{} 5r^2}\ge 14Rr \minus{} r^2$ and we"ll obtain the required inequality.

[quote="dogcat"]Prove that in a triangle:
\[ 18Rr\le ab \plus{} ac \plus{} bc
\]
where $ R$ is the radius of the circumcircle, $ r$ is the radius of the incircle and $ a$, $ b$, $ c$ are the lengths of the sides of the triangle.[/quote]
[hide="Solution"]
By Cauchy, we have that $ (a\plus{}b\plus{}c)(bc\plus{}ac\plus{}ab)\ge (\sqrt{abc}\plus{}\sqrt{abc}\plus{}\sqrt{abc})^2\equal{}9abc$.  Thus, $ ab\plus{}bc\plus{}ac\ge \frac{9abc}{a\plus{}b\plus{}c}\equal{}\frac{9abc}{2s}$.  Since $ R\equal{}\frac{abc}{4[ABC]}$ and $ r\equal{}\frac{[ABC]}{s}$, we have that \[ 18Rr\equal{}18\cdot \frac{abc}{4[ABC]}\cdot \frac{[ABC]}{s}\equal{}\frac{9abc}{2s}\le ab\plus{}bc\plus{}ac\] as desired. [/hide]

[color=darkred] [size=117][b][u]Nice[/u], The QuattoMaster 6000[/b] ![/size] [/color] [color=darkblue]The your proof is more clearly such :

$ \left\|\begin{array}{c}
(a \plus{} b \plus{} c)\left(\frac 1a \plus{} \frac 1b \plus{} \frac 1c\right)\ge 9\ ,\ abc \equal{} 4RS \equal{} 4Rpr\Longrightarrow\\\\

ab \plus{} bc \plus{} ca\ge \frac {9abc}{a \plus{} b \plus{} c} \equal{} \frac {9\cdot 4Rpr}{2p}\Longrightarrow ab \plus{} bc \plus{} ca\ge 18Rr\end{array}\right\|$ .

In the my previous reply I wanted to show that the proposed inequality is

more weakly than $ \boxed {ab\plus{}bc\plus{}ca\ge 4r(5R\minus{}r)}\ge 18Rr$ because

$ R\ge 2r$ and $ ab\plus{}bc\plus{}ca\ge 4r(5R\minus{}r)$ $ \Longleftrightarrow$ $ p^2\plus{}5r^2\ge 16Rr$ .[/color]


Given positive numbers $ a_i$ such that product of any two (with different indecies) is at least 1. Prove that

$ \prod_{cyc} (a_i \plus{} a_{i \plus{} 1})^3 < \prod_{cyc} (a_i^2 \plus{} a_{i \plus{} 1}^2 \plus{} a_{i \plus{} 2}^2)^2$.

Thank you, zaya_yc! It is better then official solution, though without numerous typos it would be even more better.


Based on Table from Website
[hide="By Grade"]
6th-3
7th-7
8th-25
9th-69
10th-104
11th-221
12th-128
13th-1
[/hide]
[hide="By State"]
CA	89
NY	34
MA	31
IN	25
NJ	25
MD 	18
NH	18
TX	17
NC	15
CT	14
MI	14
FL	13
WA	13
GA	11
OH 	9
VA 	9
MN	7
PA	7
CO	6
IA	6
MO	5
OR	5
SC	5
NM	4
LA	3
TN	3
VT	3
AR	2
AZ	2
DE	2
HI	2
NV	2
OK	2
WI	2
AK	1
AL	1
DC	1
ID	1
KS	1
KY	1
ME	1
MS	1
MT	1
NE	1
RI	1
SD	1
UT	1
WV	1
WY	1
[/hide]

Note that 17/18 NH qualifiers are from Exeter  :wink:

[hide="Top Schools"]
Exeter	17
Montgomery Blair	10
Stuyvescant	10
AAST	9
DCDS	7
Lynbrook	7
Palo Alto	7
Andover	7
Saratoga	7
Windsor Plainsboro N+S	7
TJHSST	6
Fairview	5
HMGSHS	5
ILMSA	5
Lakeside	5
Lexington	5
Mission San Jose	5
Naperville 	5
West SHS	5

[/hide]

Thanks for compiling this!

But I have to say theres something wrong with IL and IN because ILs not on that list and I know IN has 14 qualifiers only

It's so depressing that 6th graders can get these problems while I rip my hair out trying to understand them. And I assume that the 4th-grade qualifier last year outside of the US who is now a 5th-grader made it in as well?  :(

For "Windsor Plainsboro 7" in schools, which west windsor plainsboro school is it?

edit-i looked at the list and it's both schools combined. 5 from South and 2 from North.

[quote="33286"]What is "Table from Website"?
I'm pretty sure that one of the 6th graders should be a 5th grader. You also missed a 13th grader.[/quote]

US Stats only, remember. The 5th grader's Canadian.

[quote="33286"]What is "Table from Website"?
I'm pretty sure that one of the 6th graders should be a 5th grader. You also missed a 13th grader.[/quote]

I didn't include the Canada kids. Its definitely 3 6th graders on the website.

[quote="Anaxerzia"][quote="ProtestanT"]For "Windsor Plainsboro 7" in schools, which west windsor plainsboro school is it?[/quote]

W Windsor Plainsboro HS-N[/quote]

I actually looked at the site now. it's 5 from South and 2 from North.

Woops! I guess I overlooked that... Ill just make it south...

Lmao @ CT being 10th in terms of USAMO qualifiers.

This is hilarious considering this is the first time EVER we're not near the very bottom of the list.  :rotfl:

First time CT as a whole has beaten TJ. (We beat all of VA this year :o)

GO WWP! haha u can combine us into 7.. we're same school district with 2 high schools lol
public school doing so well FTW

Why are there so many more juniors that qualified than seniors?

[quote="serialk11r"]What? There's no such thing as a "1/3 chance". A lot of people who make it to USAMO have no idea how to write a proper solution, so they get weeded out ;)[/quote]
so basically it's 1/2 chance of 70% and 1/2 chance of 0%, assuming half the people know how to write a proper solution, half don't, and you don't know which half you are :D

I found out there's 5 more "Wu"'s than there are "Michael"'s.
and there's one Michael Wu.
"Wu": 11 times
"Michael": 6 times

"John": 13 times (including in school names)
"Jonathan": 8 times
"Philip": 3 times
"Phillip": 25 times (including in school names)
"Ko": 18 times (including in school names)
"Kevin": 15 times

Almost all the Phillips come from the names of Andover and Exeter schools, so yeah thats distorted  :P

What the hell, there are that many "Ko"s around? Maybe we're just that good at math.

There are 17 "Qu"'s :D

This pwns the number of "Wu"'s

[quote="ProtestanT"]
"Phillip": 25 times ([b]including in school names[/b])[/quote]

I wonder where most of those come from.  HMMMMMMM.....

hey! that michael wu is from my school... hes a mad smart frosh

Wow you NC'ers have a ton of young talent... You guys will beast in the next 5 years o.o

Sheesh, look at CA: more than double the next highest number of qualifiers state. And like 60 of them are from San Francisco Bay Area. :D

@dnkywin: Harker MS had 4 qualifiers, I believe...

[b]Three International Students, YAY!![/b]

Last year, there was only one Taiwanese Student....

Wow, two more South Korean students!

Way to go~ :)

[quote="PI-Dimension"][quote="33286"]Sorry, I didn't see the part about "US stats only".

NC has 16 qualifiers now. :)[/quote]

Yay. :)

[hide="quick note"]
4 are middle schoolers o_O(including me! YAY). What's surprising is, this was highly based on AMC 10/12 rather than AIME... alot of people made alot of stupid mistakes, and missed the cutoff by 2-3 points which are MUCH smarter than me, who made it. =(. Oh well.
[/hide][/quote]

including someone who made blue MOP last year...


Indicate whether the statement is always true or sometimes false.  Justify your answer with a logical argument or counterexample.

(a) The expression $ tr(AA^T)$ and $ tr(A^TA)$ are always defined, regardless of the size of $ A$.
(b) $ tr(AA^T) \equal{} tr(A^TA)$ for every matrix $ A$.
(c) If $ A$ is a square matrix with two identical rows, then $ AA$ has two identical rows.
---
Should we take into account that the (first two) problems do not specify $ A$ to be a square matrix?
($ tr$ denotes trace)

It is very important that the first two questions don't specify that $ A$ is a square matrix.  The first question asks you whether $ AA^T$ and $ A^T A$ are always square, which is trivially true if $ A$ is square.

Questions of this sort tend to be constructed to trick you into thinking that something that is "obvious" is true when it is actually false, so a good general strategy is to try to construct counterexamples before assuming a statement is true.

Of course this was intended to apply to non-square matrices. For (a) and (b), we can say more. If $ A$ is $ n\times m$ and $ B$ is $ m\times n,$ then $ \text{tr}(AB)$ and $ \text{tr}(BA)$ are both defined and $ \text{tr}(AB)\equal{}\text{tr}(BA).$ This can be proved simply by writing out the appropriate definitions in summation notation and changing the order of summation, as appropriate.

In a similar way, we can say more on (c). If $ A$ (square or not) has two identical rows and $ AB$ is defined, then $ AB$ has two identical rows.


LaTeX
calculus
integration
number theory proposed
number theory

i'm sorry if this problem posted before
x,y:\in N and x^2-y^3=17.prove that:y^2+2x+2-composite number.
sorry for my english :blush:

With LaTeX:
Let $ x,y\in N$,such that $ x^{2}-y^{3}=17$.Prove that $ y^{2}+2x+2$ is composite number.

[hide="Solution"] Your [url=http://tnt.math.metro-u.ac.jp/simath/MORDELL/]Mordell's equation[/url] for $ k=17$, $ x^{2}= y^{3}+17$ has the positive integral solutions $ (x, y) \in \{ (5,2); (9,4); (23,8); (282,43); (375,52); (378661, 5234) \}$ (on the site the variables are reversed). Then $ y^{2}+2x+2 \in \{ 16, 36, 112, 2415, 3456, 28152080 \}$, obviously composite numbers (divisible by either $ 2$ or $ 5$).
However, I don't think that's the "official" solution (it seems quite random). [/hide]

I don't understand your solution :maybe: .How did you get that: Your Mordell's equation for k=17,x^2=y^3+17 has the positive real solutions $ (x,y)={(5,2);(9,4);(23,8);(282,43);(375,52);(378661,5234)}$Here is my solution:Suppose that $ y^{2}+2x+2$ is prime.$ x^{2}-y^{3}=17\Rightarrow x^{2}-16=y^{3}+1 \Rightarrow y^{2}-y+1=\frac{x^{2}-16}{y+1}$,so we have that $ y^{2}+2x+2=\frac{x^{2}-16}{y+1}+y+2x+1$,since $ y^{2}+2x+2=\frac{x^{2}-16}{y+1}+y+2x+1=\frac{x^{2}-16+y^{2}+y+2xy+2x+y+1}{y+1}=\frac{(y+x-3)(y+x+5)}{y+1}$
for $ x=1,2,3,4$ , $ y^{2}+2x+2$ is composite,so let $ x>4$,$ \Rightarrow y+x-3>y+1$ and $ y+x+5>y+1$.Let $ (y+x-3,y+1)=l$ since $ \frac{y+x-3}{l}|y^{2}+2x+2$ then there are two cases $ y+x-3=l$ or $ y+x-3=(y^{2}+2x+2)(l)$ in first case we have a contradiction to $ y+x-3>y+1\geq l$ in second we have that $ \frac{(y+x+5)l}{y+1}=1$,contradiction.

[url=http://tnt.math.metro-u.ac.jp/simath/MORDELL/]Click here[/url] for a database with solutions of Mordell's equation (which is the same link as in my post).

To PhilAndrew:i don't think that this can be useful,but anyway thanks. :lol:
To Birzhan: could you post your solution?


geometry
3D geometry
prism
analytic geometry
similar triangles

A cube of side 1 has P(0,0,1), Q(1,0,1), R(1,1,1), S(0,1,1), T(0,0,0), U(1,0,0), V(1,1,0), and W(0,1,0). The plane through P and the centers of faces TUVW and UQRV intersects UV at X. What is UX?

I did this analytic geometry style but I would like to find out the geometric solution...

In case anyone wonders, the Westpac is a bank which sponsors the Australian Mathematics Competitions for the Asia-Pacific region, and that S stands for Senior (grades 11-12); there are two more, the Intermediate (9-10) and the Junior (7-8). The questions are about AMC level.

Okay: let the center of face $ TWVU=A$ and the center of face $ QRVU=B$. Extend $ PB$ to intersect $ z=0$ at $ B'$. Just like copy the rectangular prism with PQB on it...it looks better if you draw it...but if you think of like half unit squares,

[code]A
----
|  |
----V
|  |
----
|  |
----B'[/code]

Then we have that $ X$ is the intersection of $ AB'$ and $ UV$...Just use like similar triangles to get like $ \frac{1}{2}*\frac{1}{3}$, then add $ \frac{1}{2}$ for an answer of $ \frac{2}{3}$.


Fiind date suprafata triunghiulara [ABC] si centrul ei de greutate G, sa se partitioneze aceasta suprafata in 7 suprafete de arii egale, folosind numai rigla negradata.

Constructia se poate deduce din figura atasata.