quadratics
greatest common divisor
geometry
geometric transformation
group theory

Let p be a prime number. Find how many sets X included in
{1,2,..,p-1} satisfies m,n from X with the property that  m*n (mod p) are from X.

thank you !

It doesn't say anything about this.
but I guess m<>n...

bye

Presumably one of them is the one with all elements that are quadratic residues (mod p), but I don't know where to go from there. I don't see any reason why m and n should be distinct.

Then we also have {1,2,4,8,...} and {1,3,9,..} and all the sets like these.. there seems to be quite a few..

For all you know, those two sets could be the same, and they could be the entire set too, if 2 and 3 are both generators of Z_p.

[quote="TripleM"]Usually in this type of problem m and n don't have to be distinct. It would make it a lot more 'nice'..[/quote]

That's not true in this case. The solution is "nice" if m and n are not assumed to be distinct. For those who know something about groups, that's a good way to approach this one.

I think this question is worth another look too. Even without knowing much about groups you might be able to notice something here.

This is basically [hide]wilson's theorem[/hide] right?

I'm pretty sure its a corollary off of that.

hmm. They both have to do with things mod p, but other than that I don't see the connection. But maybe there is. Explain more of what you're thinking. And even if it were true, how would it get you to the answer to the question?

Let p be a prime number. Find how many sets X included in
{1,2,..,p-1} satisfies m,n from X with the property that m*n (mod p) are from X.

Can someone please translate this into a bit clearer terms? The way I am reading this is that we are trying to find the number of sets x from the set 1 to (p-1) that when when all the elements of a set x is multiplied in mod p, we get the gcf (did he forget the parentheses?) of those elements?

A translation of this problem is: For each prime p, how many subsets of {1,.., p-1} are there which are closed under multiplication mod p. So if you take any two (not necessarily distinct) elements a, b of the subset, their product is again in the subset.

For example, for the prime 5 the subsets {1, 2, 3, 4}, {1, 4}, and {1} are all closed under multiplication, and these are the only ones. So in this case the answer is 3.

For the prime 7, the subsets {1, 2, 3, 4, 5, 6}, {1, 2, 4}, {1, 6}, and {1} are closed under multiplication, and these are the only ones. In this case the answer is 4.

For the prime 11, the subsets {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, {1, 3, 4, 5, 9}, {1, 9}, and {1} are the only ones closed under multiplication. So the answer for p=11 is 4.

I think you really have to notice something specific about multiplication modulo p to figure this one out. I'll give a hint and say, [hide]it's something that I mentioned recently in another thread in this forum under another problem.[/hide]

Nice forgotten problem.

I think I got it using group theory (especially [hide]cyclic (sub-)groups ![/hide]). Isn't it a bit heavy artillery

The number of such subsets is known but the explicit construction of every subsets ([hide]which is finding an explicit generator to (Z/pZ*,.)[/hide]) is an open problem, I think (I'm not sure). Do you confirm gauss202 ?


find all function $ f: N\toN - {1}$ that $ \forall n \in N$ :
\[ f(n)+f(n+1)=f(n+2)f(n+3)-168\]

Subtract both sides of the original equation from both sides of $ f(n\plus{}1)\plus{}f(n\plus{}2) \equal{} f(n\plus{}3)f(n\plus{}4) \minus{} 168$, one gets
$ f(n\plus{}2) \minus{} f(n) \equal{} f(n\plus{}3) [f(n\plus{}4) \minus{} f(n\plus{}2)]$.

Then, notice that $ f(n\plus{}4) \minus{} f(n\plus{}2) \leq f(n\plus{}2) \minus{} f(n)$. The sequence of differences decreases/increase and bounded below/above by 0 (depends on the values of $ f(3)\minus{}f(1)$ and $ f(4)\minus{}f(2)$). Then it must have limit which means this sequence will be eventually constant. Call it $ L$. But if $ L>0$, the values of $ f(n\plus{}3) \equal{} 1$ for all $ n>N$. And then $ f(n\plus{}2)\minus{}f(n) \equal{} 1 \minus{} 1 \equal{} 0 \forall n>N\plus{}3$.

Otherwise $ L\equal{}0$ and $ f(n\plus{}2)\equal{}f(n) \equal{} C$ forall $ n>N$. Plug in the original equation, $ C \equal{} 14$.

One can construct the original sequence from its final behavior.


[quote="box of rocks"]True or false for the following scenario?

You consider the set of true/false tests which have only an odd number
of questions.  In any of these true/false tests that you are given, you are going to randomly mark one answer for each question.  The chances that you
score a majority of the questions correctly in any of the tests** is 50%.

** That is, you have randomly answered questions from ONE test, BUT it could have been chosen from the set of ANY of the odd-numbered tests.[/quote]
[hide="solution"]
True. Although any individual test tested has a less than 50% chance of scoring the majority, the graph of $ y \equal{} f(x)$ where $ f(x)$ is the average of all the chances of scoring a majority of each odd number test with $ 1$ questions up to $ x$ questions approaches $ 0.5$ as $ x$ approaches infinity. Since there are tests with up to an infinite number of questions in the set of true/false tests which only have an odd number of questions, $ x$ would be infinity, and $ f(x)$ would arrive at the limit, $ 0.5$
[/hide]

Huh? The probability of getting more than half of the questions right on any particular test is always exactly $ 50\%$.

The only reason not to answer with an unqualified yes is that there's no canonical probability distribution on the (countably infinite) set of all such tests. For the question to be well-posed, we must specify some distribution.
If we do, the answer is yes; the average value of a constant with respect to any probability measure is simply that constant.

[quote="jmerry"]Huh? The probability of getting more than half of the questions right on any particular test is always exactly $ 50\%$.
[/quote]
Hmm that's true...What was I thinking?
Either way, it's a $ \text{TRUE}$


Haley has enlarged a 3'' by 5'' picture so that both the length and width are tripled. By what number must the original area of the picture be multiplied to get the enlarged area?

The original picture is 15 inches squared. (3*5)
The enlarged is 9*15=135.
135/15=9.
Also, if the ratio of 1 side is 3, the ratio of the area is 3^2=9.
Or, if the sides of a rectangle are x and y, the enlarged area is 3x*3y=9xy/xy=9.


Let O be the circumcenter of a traingle ABC. Choose a point P, different from B and C. The circumcenter of ABC meets AP at a point N and circle whose diameter is AP at a point E (N,E are both different from A). BC and AE intersect at M. Prove that MN always passes through a fixed point.
( Thank you for your help!)

See at http://www.mathlinks.ro/Forum/viewtopic.php?t=184005

Kostas Vittas.

next time you want us to help you in cheating, at least state the problem correctly  :rotfl:

  dg


calculus
integration
geometry
trigonometry
calculus computations

Well, i think i am messing up some stuffs in this integral. I have 2 totally different answeres and can´t see any mistake to take just one of them, or just abandon them. So here we go:
Find the area of the object generated by the intersection of 2 cilinders: ~$x^{2}+y^{2}=1$ and $x^{2}+z^{2}=1$.
Maybe is too elementar, but i need to clarify this subject in my mind as soon as possible!

The intersection of that solid with the plane $x = x_{0}$ is a square of sidelength $2\sqrt{1-x_{0}^{2}}$, so the total surface area is $\int_{-1}^{1}8 \sqrt{1-x^{2}}dx$.

JBL, $8\sqrt{1-x^{2}}\,dx$ is the surface area of a box with square base, differential height $dx,$ and [b]vertical[/b] sides. The object we want has sides that tilt. Care to reconsider your answer?

Whoops.  It's a good thing I'm safely past all my calculus classes.  Does making it into $8\int_{0}^\pi \sqrt{1-\cos^{2}t}dt$ fix it?


The startup is next Thursday; since I have never participated in such a quick paced event, I have no experience for such a test. Can someone post or PM me a past test (I know there are 10 questions on the site, but 10 questions is not the best indicator of the experience of a 100 questions test)? Any past test(s) would be greatly appreciated.

I think only teachers have them, and I think this is the wrong forum :wink: This should be moved to Nat'l Assessment and Testing.

Oh, I just saw the word "national" and clicked. :blush:


A sequence is defined by $ x_1\equal{}1,x_2\equal{}4,x_{n\plus{}2}\equal{}4x_{n\plus{}1}\minus{}x_n$ for $ n\geq 1$. Find all natural numbers $ m$ such that $ 3x_n^2\plus{}m$ is a perfect square for all natural numbers $ n$.

$ (x_{n \plus{} 2} \minus{} 2x_{n \plus{} 1})^2 \equal{} (2x_{n \plus{} 1} \minus{} x_n)^2$
$ \Leftrightarrow x_{n \plus{} 2}^2 \minus{} 4x_{n \plus{} 2}x_{n \plus{} 1} \plus{} x_{n \plus{} 1}^2 \equal{} x_{n \plus{} 1}^2 \minus{} 4x_{n \plus{} 1}x_n \plus{} x_n^2$
Then $ x_{n \plus{} 1}^2 \minus{} 4x_{n \plus{} 1}x_n \plus{} x_n^2 \equal{} x_2^2 \minus{} 4x_2x_1 \plus{} x_1^2 \equal{} 1$
Then $ 3x_n^2 \plus{} 1 \equal{} x_{n \plus{} 1}^2 \minus{} 4x_{n \plus{} 1}x_n \plus{} 4x_n^2 \equal{} (x_{n \plus{} 1} \minus{} 2x_n)^2 \equal{} y_n^2$
If exist $ m\neq 1$ satisfy $ 3x_n^2 \plus{} m \equal{} z_n^2$ then $ m \minus{} 1 \equal{} (y_n \plus{} z_n)(z_n \minus{} y_n)$ for all $ n$
Then $ m \minus{} 1$ have infinity many of divisors $ y_n \plus{} z_n$. Contracdition! (Clearly that $ y_n\plus{}z_n$ is increasing)

PS: We also can prove this problem by finding the general formul of $ x_n$


1) A constant force of 8.0 N is exerted for 4.0 s on a 16-kg object initially at rest. The change in speed of this object will be: A) 0.5 m/s B) 2 m/s C) 4 m/s D) 8 m/s E) 32 m/s

2) When a certain force is applied to the 1 kg standard mass its acceleration is 5.0 m/s2. When the same force is applied to another object its acceleration is one-fifth as much. The mass of the object is: A) 0.2 kg B) 0.5 kg C) 1.0 kg D) 5.0 kg E) 10.0 kg

 :blush:  :blush:  I have no idea about these 2 problems...? please leave your solution... I am sorry that i cannot post a problem that can entertain you :blush:  :blush: [/img]

[hide]1st one is 4m/s.use conservation of momentum [/hide][hide]2nd one is 5kg.[/hide]

[hide="More detailed solutions"]
1)Using $ F\equal{}M \cdot A$, the acceleration on the object is $ \frac{8.0 \, N}{16\, kg}\equal{}0.5\frac{m}{s^2}$.  So the change in velocity is the acceleration times the time, which is $ 0.5 \cdot 4\equal{}2 \frac{m}{s}$

2)Again using $ F\equal{}M \cdot A$, the force applied to the object is 5 N.  On the second object the acceleration is $ 1\frac{m}{s^2}$, so its mass is 5 kg. (An easier way to say this is that if the force is constant and the acceleration is one-fifth as much, the mass must be 5 times as much to compensate).
[/hide]

aadil
Shouldn't the first one by 2 m / s?

F = m a
F t = m v
Unless I'm doing something completely messed up I think
8 * 4 / 16 = 2 ?

aadil

There is no conservation of momentum. I think you mean to use the "Impulse Theorem" , in that case, $ F \Delta t \equal{}  m \Delta v\;.$


Hello

How do you prove this?

Prove Helly's theorem:
Given a list of convex sets in the plane K_1 , K_2 , ... , K_n with the property that every 3 sets in the list have a common point, show without using Radon's theorem that all the sets have a common point.

It can be easily proved by induction with basis $n=4$.

Right, I tried that approach but got stuck. Care to help?

Let's prove inductive basis for $n=4$.

Every three sets have a common point so we can define a point $A_i$ by stating that $A_i\in K_i\cap K_{i+1} \cap K_{i+2}$. Here and further all the indices are taken $\mod 4$. Note that $A_i,A_{i+1},A_{i+2}\in K_{i+2}$. Then by the definition of the convex set $\Delta A_i A_{i+1} A_{i+2}$ lies within the set $K_{i+2}$. Note that our arguments are not based on an indice $i$, so the recent statement is true $\forall i=0,1,2,3$. There are two cases now. In the first one the $4$ points form a convex quadrilateral in the plane, so the point of the intersection of its diagonals $O$ is on a side of every $\Delta A_i A_{i+1} A_{i+2}$, so we have $O\in K_{i}, \forall i=0,1,2,3$ as desired. The second case is that one point, say, $A_0$ lies within the triangle formed by the other $3$ points, namely within $\Delta A_1A_2A_3$. But in this case we have that $A_0\in K_{i}, \forall i=0,1,2,3$, as desired. So we have proved the inductive basis $n=4$.

Let us proceed with our proof having $n\ge 5$ points. Let's define $K'=K_n\cap K_{n-1}$. By inductive basis the statement that every $3$ sets from $K_1,K_2,...,K_{n-2},K'$ have a common point is true as $K'$ is convex being an intersection of $2$ convex sets. But now we have $n-1$ convex sets, every three of which have a common point, so by inductive hypothesis they have a common point. That completes our proof.

This message should be in "Combinatorics",I think.

there is a general n-mensional version..

Here's a quick proof for the general version (it uses Radon's Theorem though, which Clo didn't want to use :)):

It suffices to prove it in the case when we have $n+2$ sets. After that, we use induction, just like Xixas did. Let $A_1,\ldots,A_{n+2}$ be our sets, and let $\tilde A_i$ be the intersection of all of them except for $A_i$ (for all $i\in\overline{1,n+2}$). For each $i$, take a point $X_i\in \tilde A_i$. According to Radon's Theorem, $\{X_i\}_i$ can be partitioned in two non-empty subsets with intersecting convex hulls, and it's easy to see that a point common to these two convex hulls must lie in all the $A_i$'s.


calculus
integration
algebra
system of equations

My friends, I have a system of equations which I can not solve (I know the desired solution)
The system has 4 equations following and accept only integral solution

14(x+y)+xy=25
6z+36w+zw=5
14+x=6+w
14+y=36+z

the solution should be
x=3
y=-1
w=11
z=-23

can any of you show a way to get that solution T_T.

Your equations have infinitely many solutions.  My personal favorite is $(w, x, y, z) = (215, 207,-13,-35)$.

I know it has infinite solution T_T. However, the [b]desired [/b]solution is the one I posted. Can you show me how you get that solution? I know how to get the solutions by guessing (getting factors of 221), I need to have an algebra way to solve T_T.


induction
combinatorics open
combinatorics

$ \bullet$ Is there a closed formula for computing the sum $ \sum_{k \equal{} 0}^{n}k^2\cdot\binom{n}{k}$ ?

[quote="VSelkoR"]$ \bullet$ Is there a close formula for computing the sum $ \sum_{k \equal{} 0}^{n}k^2\cdot\binom{n}{k}$ ?[/quote]
look here i alredy posted a solution:

http://www.mathlinks.ro/Forum/viewtopic.php?t=302957

hello, you can prove by induction
$ \sum_{k\equal{}0}^{n}k^2\binom{n}{k}\equal{}2^{n\minus{}2}n(n\plus{}1)$.
Sonnhard.


function
real analysis
real analysis solved

Is there any continuous function $f:\mathbb{R}\to\mathbb{R}$ such that $f(f(x))=2^x$ ?;)

It was recently solved by fedja, in a topic started by harazi, except that instead of $2^x$ there was $e^x$.


The base and circumcircle of a triangle are given.Find the locus of the centroid.

Obviously your locus is the circle homothetic of the circumcircle in (P,1/3) where P is the midpoint of the base.
Kind regards.


It is true that every sequence of real numbers xn wich hasnt limit includes two subsequences yn,zn who have  distinct limits y,z?
If not what conditions must satisfy xn such that the proposition is true?

Well....it may depends on the definition you have of the limit of a sequence...If (x_n) tends to +00, does it implies that it has a limit or not? (hmm...the answer seems to be in the question...).
In the affirmative, then the answer at your question is yes.

in fact if xn is bounded then it contains at least a convergent subsequence. Obviously the terms remaining form another infinite sequence (otherwise the initial sequence would be convergent) and as this second sub-sequence is also bounded, it also contains a sub-sub-sequence convergent to another point.

if xn is not bounded, then it obviously contains a sequence which tends to \pm \infty . the remaining sequence is either bounded, in which case it contains two sub-sequences (as shown above) or it is also not bounded, which means that it also contains a sub-subsequence which tends to \pm \infty

[quote="Valentin Vornicu"]if xn is not bounded, then it obviously contains a sequence which tends to \pm \infty . the remaining sequence is either bounded, in which case it contains two sub-sequences (as shown above) or it is also not bounded, which means that it also contains a sub-subsequence which tends to \pm \infty[/quote]

the last  \pm \infty obviously points to the opposite of the first  \pm \infty


function
topology
real analysis
real analysis theorems

Hello,

I am looking for a certain theorem of Weierstrass concerning uniform convergence of continuous functions.  While I am familiar with Weierstrass and uniform convergence, I had never seen that form and the theorem in question in my syllabus had no references nor proof.

It was along these lines :

If a sequence of functions $f_n$ in $C_{0}(\Omega)$ , thus continuous on the open set $\Omega\subseteq\mathbb{R}^n$ , is such that on  every compactum $K\subseteq \Omega$ : $sup_{x\in K} |f_{j}(x)-f_{m}(x)|$ can be arbitrary small if j and m are big enough (where the demanded size of j and m only depends upon the compactum and how small you want it to be, not x),
then there is a unique continuus function on $\Omega$ : f, such that on every compactum, the restriction converge uniform to that function

It could also be that it worked with an ascending chain of compacta filling $\Omega$ up

Does this sound familiar?  Does it have some name?  Is there a link where I can read more, perhaps a precise proof?  The only hard part, if this IS the correct statement of the theorem, would be that the function you obtain by pointwise convergence is unique?

This sounds like that theorem of Egorov. [url=http://mathworld.wolfram.com/EgorovsTheorem.html]Mathworld[/url] has a brief summary.

Thanks but I doubt it is that theorem?  My professor seemed certain it was weierstrass


function
calculus
integration
real analysis
real analysis unsolved

$F:=\left\{f \text{ continuous }, f:[0,1]\to[0,2]| \int_0^1f(x)dx=1\right\}$.
a)What are the values of $\int_0^1xf(x)dx$ when $f\in F$.
b)Find $f\in F$ if $f^2\in F$ also.

The answer for the first one is $(0,1)$, I think. Let $F(x)=\int^x_0 f(t)dt$. Then $\int^1_0 xf(x)dx=1-\int^1_0 F(x)dx$, where $F:[0,1]\to [0,1]$ is increasing and derivable. I think it should be easy enough to provide examples which show that $\int^1_0 F(x)dx$ can come arbitrarily close to $0$ and $1$.

As for the second one, cute trick :). We get $\int^1_0 (f(x)-1)^2dx=0$, so guess what $f$ is :).

for the first one is wrong...  the second.. of course  :D

[quote="Kuba"]for the first one is wrong[/quote]

Of course it is :). I forgot the condition that $f$ only takes values in $[0,2]$, unfortunately :). How about the range $(\frac 14,\frac 34)$?


Prove that for each prime p there exists a prime q such that (n^p)-p isn't multiple of q for all n positive integer.


Among the positive integers that can be expressed as the sum of 2005 consecutive integers, which occupies the 2005th position when arranged in order?

[i]Roland Hablutzel, Venezuela[/i]
[hide="Remark"]The original question was: Among the positive integers that can be expressed as the sum of 2004 consecutive integers, and also as the sum of 2005 consecutive integers, which occupies the 2005th position when arranged in order?[/hide]

don't tell me it's really 2005+...+4009
If yes, this topic should be placed in intermediate forum.

[quote="Jutaro"]Well don't blame me, blame Centro!!  :? So i think that, as an Olympiad problem, this problem deserves a place here, but if it's more convenient to place it in the intermediate forum, it doesn't matter :)[/quote]
Oh no! I was tricked by the question!
We really need some checking to see how many "a"s are there to make -a-(a-1)-....+0+... is positive.

[quote=panamath][quote="Pascual2005"]$(k-1002)+(k-1001)...+k+...(k+1001)+(k+1002)=2005k$ so every number of the form $2005k$ can be written in that form, and only such numbers do. hence the answer to the first problem is $2005^2$

As for the secound question we have: then sum of 2004 consecutive integers is the sum of $2005$ minus the last term. so these sums are equal to $2005k-k-1002=1002(2k-1)$ So numbers that can be written in both ways are those numbers of the form $(2005)(1002)(2t-1)$ so the rest is easy.[/quote]

I think the answer to the problem is $2005 \cdot 2004$ not $2005^2$, I assume, you missed the case $k=0$[/quote]

$k=0$ gives 0, which is not a positive integer.


AMC
USA(J)MO
USAMO
inequalities
inequalities proposed

Given that $a,b,c,d,e$ are real numbers such that \[a+b+c+d+e=8\] \[a^2+b^2+c^2+d^2+e^2=16\] Determin the minimum value of $e$.

It would a nice if you'd PMed the solution as well as posting it.

I don't know if this solution is correct

From the first equation and using Cauchy-Scwarz inequality we obtain
\begin{eqnarray*}
8-e &=& a+b+c+d\\
& \leq & 2\sqrt{a^2+b^2+c^2+d^2}\\
& = & 2\sqrt{16-e^2}
\end{eqnarray*}
From the second equation $e\in[-4,4]$ and it follows that $8-e>0$ and we can square the above inequality to obtain
\[
64-16e+e^2\leq 64-4e^2\Leftrightarrow 5e(e-16/5)\leq 0
\]
So $e\in[0,16/5]$.


hei, Are there any chinese mathlinkers will go to NTU?

and my curiousity is "must Chinese take the ENTRANCE test?" : :rotfl: D

Do you mean the university in Singapore?

I will go to NUS this August.
Many of my friends will go to NTU. :D


If the sum of three prime nos. is 100 and the difference of 2 of them is 36,find the prime nos.

$a + b + c = 100$
$a + 2c = 64$
$2c = 64 - a$ since $a$ must be even, but the only prime even is $2$ so
$a = 2$, $c = 31$, $b = 67$

[hide="solution"]
$a+b+c=100$
$a-c=36$
Equating,
$b+2c=64$
$b=64-2c$
$b=2{32-c}$
Now $b$ has to have only $1$ more factor which is $1$ So,
$c=31$
$b=2$
$a=67$
[/hide]

[hide="Easy solution"]
Parity says that one number has to be 2.

2a + 36 = 98

a = 31, so the numbers are 2, 31, and 67.
[/hide]

[quote="riddler"]What is parity?[/quote]

Something about even and odd numbers, as everyone used here.

[quote="riddler"]Everyone???? I dont think I used parity.[/quote]

Sorry, I didn't see your solution  :blush:

[quote="Ars"]Riddler I personally do not have a clue how you got answer after seeing the solution and you used parity.[/quote]

I think riddler's solutin is totally unique from the others(although i still cant really understand the part where the factor of b comes)

Whats parity?I just used simple logic that one of them has to be even(accordin to an equation already given above),so one is 2.Rest u can find out easily,as everyone's done.

[quote]I just used simple logic that one of them has to be even(accordin to an equation already given above),so one is 2.Rest u can find out easily,as everyone's done.[/quote]
That basically is parity


This is from the number theory chapter, vol.2.

Note that 1990 can be 'turned into a square' by adding a digit on its right, and some digits on its left; i.e., $419904=648^{2}$. Prove that 1991 cannot be turned into a square by the same procedure; i.e., there are no digits d, x, y such that yx1991d is a pefect square.

[hide]The last digit of a perfect square can only be:
0,1,4,5,6,9

When expressed in modular 100 (because the tens digit must be 1):

0: is dicounted because if a square ends with a 0, it ends with two 0's (other than 0^2)

1: (10x+1)^2=100x^2+20x+1, also discounted because 1 is odd and 20x will give an even tens (even is x is negative)

4: (10x+2)^2=100x^2+40x+4, also dicounted for the same reason as 1

5: (10x+5)^2=100x^2+100x+25,  also dicounted for the same reason as 1

6: 4^2, possible

9: (10x+3)^2=100x^2+60x+9,  also dicounted for the same reason as 1

Now let's express it in modular 1000

16: (100x+4)^2=10000x^2+800x+16 which is discounted because the hundreds digit is even[/hide]

Let $T(n)$ be the sum of the digits in the natural number $n$. Assume there exists a integer square $x^{2}$ which 5 last digits are $1991d \; (*)$. Hence

$(1) \;\; x^{2}\; \equiv \; 1991d \pmod{10^{5}}$

We have

$n^{2}\; \equiv \; 0, 1, 4, 5, 6, 9 \pmod{10}$

for any integer $n$. Therefore

$(2) \;\; d \in \{0, 1, 4, 5, 6, 9\}= S_{1}.$

by (1). Futhermore

$x^{2}\; \equiv \; T(x^{2}) \; \equiv \; T(1991d) \; \equiv \; d+2 \pmod{9},$

giving

$d \; \equiv \; x^{2}-2 \equiv \; 2, 5, 7, 8 \pmod{9}.$

Consequently

$(3) \;\; d \in \{2, 5, 7, 8\}= S_{2}.$

Combining (2) and (3) we obtain

$d \in S_{1}\cap S_{2}= \{5\},$

i.e. $d = 5$. According to (1)

$x^{2}\; \equiv \; 15 \; \equiv \; 3 \pmod{4},$

which is impossible since $n^{2}\equiv 0,1 \pmod{4}$ for every integer $n$. This contradiction shows that the assumption $(*)$ is false.


See all the problems from 5-th Kyiv math festival [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=506789#p506789]here[/url]

Squirrels $A$ and $B$ have $360$ nuts. $A$ divides these nuts into five non-empty heaps and $B$ chooses three heaps. If the total number of nuts in these heaps is divisible by the total number of nuts in other two heaps then $A$ wins. Otherwise $B$ wins. Which of the squirrels has a winning strategy?

$A$ wins. The squirrel splits the nuts into 5 heaps with $60$,$60$,$60$,$60$,$120$ nuts in them. Then $B$ can choose the three heaps and because of symmetry, have only 2 cases:
1. $B$ chooses 3 heaps with $60$ nuts in each of them. then:
$120+60$ divides $60+60+60$. :)
2. $B$ chooses 2 heaps with $60$ nuts and one with $120$ nuts. then: $60+60$ divides $120+60+60$.
So $A$ has a winning strategy and we are done. :lol:  :)

what will happen if we replace $360$ with $n$?


dar mosalase ABC roye azlae AB va AC be tartib noghate M va N ra tory entekhab mikonim ke BM=CN.mahal barkhord BN ba
CM ra P minamim.az P khati movazie nimsaze daroony ras A ,rasm mikonim ta AC ra dar Q ghat konad.sabet konid CQ=AB

fek konam soorate soalet naghese,ehtemalan manzooret ine:

$ BM\equal{}CN\equal{}MN$

doroste? :maybe:

na soorat soal doroste(tooye halate khas chekesh kardam)   :!:  :!:

age farz konim khate movazie gozara az P zele BC ra dar P ghat konad kafie sabet konim ke CP=DB (in natijegiri badihei).

valy che :D tori

are hagh ba shomas niazi be oon sharte nis,vali khob ba oon sharte ham khodesh masale ghashangie... :D

baba dige chishod,kesei nabood in soalo hale kone :?:

in soal ba neveshtane menelaoos hal mishe.age bekhayd rahe daghightaresho mitoonam begam

man khodamam ye karaee kardam va fekr mikonam dorost bashe vally age mitoonid shomam eidehay haletoono bigid shayad ba rahe man fargh dashte bashe

dar zemn age kesy rahe hale hedesy ro dare onvan kone

bale midoonestam hal nashod :)

in soal ye kami be nazar sade miad valy vaghtee miay paye neveshtan mibiny ke ..........

ba in hal man halesho peida kardam  valy mikham bebinam rahehal shomaha chetorie :wink:

see:
http://www.mathlinks.ro/Forum/viewtopic.php?p=415147
http://www.mathlinks.ro/Forum/topic-5873.html
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1290563072&t=72806
http://www.mathlinks.ro/Forum/topic-19182.html
http://www.mathlinks.ro/Forum/topic-66410.html
 :agent:  :agent:  :agent:

rahe halle khodetoon chie? vali be nazare man  ke rahe halli sadetar az menelaoos nadare :)

are man ham az menelaus estefade kardam ama be shekli digar

age farz konim ke khat movazi zele BC ra dar T ghat konad kafist sabet konim ke CT=DB

pas hala kafist sabet konim ke gharineye D nesbat be vasat BC ba noghate P va Q ham khat ast ke ba 2 bar menelaus hal mishe :wink:

[quote="m bahador"]aval az hame tashakor mikonam be khater in hame rahe hal :roll:

but rahe hale man ham kheyli sadetare va ham ghashangtar

 :wink:[/quote]

sade tarin rahe hal neshane zibatarin nist :wink:

agar mikhahi yek hendese dane vagheE bashi,bayad dideto avaz koni,

hameye rahe hal ha ziba hastand,moghayese nakon azizam :wink:

ishalla ke movafagh bashi

kheili mamnoon Mr ASHEGH az in goftey zibatoon :yup:

vaghean motshakeram ke didamo be hendese avaz kardid :wink:


its given in brochure of DCE that students studying in Delhi will get 85% seats and outsiders will get 15%
so being an outsider can i give that exam?
plz tell me.

not all outsiders r allowed to give the exam <u must hav an Indian link>
Moreover,
[hide]the rest 15% seats r for AIEEE  :P
but wait i m not cent pc sure[/hide]

by outsider i mean not of Delhi.so plz tell me whether i can give that exam being a student of region other than delhi.

i think u will hav to give AIEEE  for admission :)

so it means that DCE is meant only for students of Delhi.but it can't be true as other states have thier CET too and they allow outsiders also.plz............. let me know if u r also an aspirant for DCE or DPMT

i [b]m[/b] an aspirant of DPMT :P
and it is only for those who pass 12th from Delhi :P

why not DCE?
i thought there is choice between Computer science and Biology.i m in bio.but i can give both engineering and medical.
well,thanks a lot, for sharing information. :)

ok.I got the information.
CEE is conducted for the students studying in the national capital.
Just like DPMT
others need to give AIEEE


Hi, here it is:
There is a math competition whcih participiants have to solve six tasks. The available marks are X, Y, Z or W. It transpired that for each pair of participants we can indicate two tasks , that in each of them participant A got different mark from participant B. Delimit the highest number of participants for which this situation is possible.

Do you know how to do it?

[b]This is from current Polish MO![/b]

You should NOT have posted this problem before 10th of October.

locked until october 10. Also klex, your flag says armenia but your email address ends in .pl?
[I hope you were not aware that this should not be posted...]


If there are two numbers, M and N, where $M\equiv 3\pmod{9}$ and $N\equiv 9\pmod{27}$, what would be the residue of MN in modulo 81?

[hide="Solution"]\begin{eqnarray*}M=9k+3\land N=27l+9 &\iff& MN=243kl+81k+81l+27\\ &\iff& MN=81(3kl+k+l)+27\\ &\iff& MN\equiv 27\pmod{81}\end{eqnarray*}[/hide]

[hide] For this problem, you could say $M=3$, $N=9$. $MN=\boxed{27}$.  :lol: [/hide]


Two circles with centres $ O_1,O_2$ are of radii 3 and 2 respectively.

$ \overline{AB}$ is thieir common tangent.

Given that the difference between  the length of $ \overline{O_1O_2}$ and the length $ \overline{AB}$   is $ 1$.

Find the length of $ \overline{AB}$.

Joining $ O_{1}$ to $ A$ and $ O_{2}$ to $ B$, we have two similar right-angled triangles.

Let $ AB \equal{} x$, then by Pythagoras,

$ \sqrt{4 \plus{} \frac{4x^{2}}{25}} \plus{} \sqrt{9\plus{}\frac{9x^{2}}{25}} \equal{} x\plus{}1$.

Getting a common denominator of $ 5$ on the LHS, bringing one radical to the right, squaring and so on, we get $ x\equal{}12$.

A solution with a little less math:

[hide]Create a rectangle $ ABO_2C$
$ BO_2\equal{}AC$, creating right $ \triangle O_2O_1C$ and using pythag (and plugging in what we know) we find (with $ AB\equal{}x$):
$ (x)^2\plus{}5^2\equal{}(x\plus{}1)^2$
$ x^2\plus{}25\equal{}x^2\plus{}2x\plus{}1$
$ 24\equal{}2x$
$ x\equal{}12$[/hide]


Let a,b,c,d be four nonnegative real numbers satisfying the conditions:
2(ab+bc+ca+ad+bd+cd)+abc+bcd+cda+abd=16
Prove that:
a+b+c+d \geq 2/3  *(ab+bc+cd+da+ac+bd)

You can find a solution in Kiran Kedlaya's wonderful inequality course.
See the links section.


Given that a,b,c are positive integers such that a^2 + b^2 = c^2.
show that
 ab/(a+b+c) is also an integer.

[hide="solution"]
We can write $ (a,b,c)\equal{}(k(m^2\minus{}n^2),2kmn,k(m^2\plus{}n^2))$ with $ k,m,n\in \mathbb{N}$.

Hence,$ \frac{ab}{a\plus{}b\plus{}c}\equal{}\frac{2k^2mn(m^2\minus{}n^2)}{2km^2\plus{}2kmn}\equal{}\frac{kn(m\minus{}n)(m\plus{}n)}{m\plus{}n}\equal{}kn(m\minus{}n)$.[/hide]

$ a^2\plus{}b^2\equal{}c^2\Leftrightarrow a^2\plus{}2ab\plus{}b^2\equal{}c^2\plus{}2ab\Leftrightarrow (a\plus{}b)^2\minus{}c^2\equal{}2ab\Leftrightarrow$

$ \frac{(a\plus{}b\plus{}c)(a\plus{}b\minus{}c)}{2}\equal{}ab$

$ \Rightarrow \frac{ab}{a\plus{}b\plus{}c}\equal{}\frac{a\plus{}b\minus{}c}{2}$

I)$ a\equal{}2k,b\equal{}2q\Rightarrow c\equal{}2p , k,q,p\in \mathbb{Z}\Rightarrow \frac{a\plus{}b\minus{}c}{2}\equal{}k\plus{}q\minus{}p\in \mathbb{Z}$

II)$ a\equal{}2k\plus{}1,b\equal{}2q$ or $ a\equal{}2k,b\equal{}2q\plus{}1\Rightarrow c\equal{}2p\plus{}1\Rightarrow a\equal{}k\plus{}q\minus{}p\in \mathbb{Z}$

III)$ a\equal{}2k\plus{}1,b\equal{}2q\plus{}1\Rightarrow c\equal{}2p\Rightarrow k\plus{}q\plus{}1\minus{}p\in \mathbb{Z}$

@Anavel_gato
M and N are amicable right? so they would always be integers?

@ Alex2008
Thats nice and simple.

The question actually was to prove that the inradius of any right angled triangle with integral sides has the inradius with an integeral length aswell.
Now can we prove this for the excircles?


algebra
polynomial
Galois Theory
superior algebra
superior algebra unsolved

Does there exist a polynomial $f \in \mathbb{Z}[X]$ such that $f$ is irreducible over $\mathbb{Q}$, and such that $f$ is reducible over $\mathbb{Z}/p\mathbb{Z}$  for all primes $p$? :D

Yes, check this one for example $f(x)=x^4+1$
According to the proof: $x^4+1=(x^2+1)^2-2x^2=(x^2-1)^2+2x^2=x^4-(-1)$ and it would be sufficient to prove that $f(x)$ is irreducible for $p>2$ prime if we manage to prove that either $2$, or $-1$, or $-2$ is a square in $\mathbb{Z}_p$. But the application $x \to x^2$ sends to each $x \in Z_p$ , exactly one of the $\frac{p-1}{2}$'th roots of unity, which form a group. Element of $Z_p$ which are not squares are exactly the roots of $x^{\frac{p-1}{2}}=-1$, so if both $2$ and $-2$ are not squares then their product $4$ must be a square in $Z_p$, thus $2$ is also square in $Z_p$.
It is easy to deal with the case $p=2$ separately.

My algebra professor mentioned the polynomials $X^4 + 6X^2 + 1$ and $X^4 - 2X^2 + 9$ :)

Could be stupid, but: There is clearly none of degree $2$, but does there exist such a polynomial with degree $3$¿

My professor mentioned that the smallest possible degree of such a polynomial is 4.

(I don't have a clue how to prove that.)

ZetaX, why is it not possible for polynomials of degree two?

Let the polynomial be $ax^2+bx+c$.
Now reducibilty means nothing different than having a root.
Then after substitution we get a polynomial of the type $y^2-a$ with $a$ being rational, and the claim is: if this polynomial has a root modulo every prime not dividing $2abc$, then $a$ is a square of a rational number. But after extending the Legendre symbol to rationals, the prove from http://www.mathlinks.ro/Forum/viewtopic.php?t=64322 should work.

[quote="ZetaX"]Now reducibilty means nothing different than having a root.[/quote]

This holds for polynomials of degree 3 as well, but I don't see how we could use that.

With Chebotarev density theorem...However I will look for an easier way, but i think there isn't.

Because if there where a cubic polinomyal, then its Galois group over the rational would be $S_3$ because it is a transitive soubgroup of $S_3$. This implies that there are many primes such that the polinomial is Irreducible in $Z_p$.


topology
geometry
rectangle
abstract algebra
group theory
irrational number
advanced fields

show that $\pi_1(\mathbb{R}^2-\mathbb{Q}^2)$ is uncountable.

One can form a series of loops $|x-1/n|+|y|=a/2^n$, where $a$ is irrational.  The total length of these loops is finite.  For any subset $S$ of the positive integers consider a path moving to the left from $(e,0)$ to $(0,0)$ that encircles the $n$th loop if $n \in S$, and otherwise only passes through the top half of the loop on its way to $0$.  The path can be arranged to avoid rational points, using the same idea of moving on lines with $x \pm y$ irrational. The homotopy classes of these uncountably many paths, or of the loops in $\pi_1$ formed by returning from $0$ to $(1,0)$, are all distinct.

You can't base the fundamental group at the origin, because that isn't even in the set.

If we base the group at $(r,0)$, the rectangles $x=r$ or $x=s$ or $|y|=r$, for $s$ another irrational number, should all be different elements of the fundamental group; the winding number around $(q,0)$ is constant under homotopy when $q$ is rational, and rational numbers are dense. There are already uncountably many of these loops.

could you describe this group more ,is it abelian?
or about its covering space :roll:

[quote="sam-n"]could you describe this group more ,is it abelian?[/quote]
Of course not- it's easy to see that it contains large free subgroups. In fact, the subgroup generated by my rectangles is free on that uncountable generating set. The crossings of the real axis can't move or be reordered.

[quote="sam-n"]or about its covering space[/quote]
What makes you think there is one?

oops,I want to apply usual approach to decide about the fundamental group ,but if we say that it is semilocally simply connected at least we know there is one .


The units digit of a six-digit number is removed, leaving a five-digit number. The removed units digit is then placed at the far left of the five-digit number, making a new six-digit number. If the new number is $ \frac{1}{3}$ of the original number, what is the sum of the digits of the original number?

Can anyone show this can be done using algebra?

Original number:
10a+b
New number:
a+100000b

(a+100000b)*3=10a+b
3a+300000b=10a+b
299999b=7a
42857b=a

Turns out for 0<b<7 it is just a rearrangement of the digits {1,4,2,8,5,7}

Sum = [b]27.[/b]

Where did you get 1,4,2,8,5,7 , because 42857*2 =85714 which doesn't include the 2?


Real numbers x, y, a satisfy the equations
x + y = x^3 + y^3 = x^5 + y^5 = a.
Find all possible values of a.

[quote="enescu"]Not exactly. See [url]http://www.imosuisse.ch/skripte/smo/smo2003/smo2003L.pdf[/url][/quote]
It is not in Enlish. :(  :(


The incircle of a triangle  $ABC$  touches side $AB$  at point $D$.Let $E$  a point on side $AC$ . Prove that the incircles of the triangles $ADE, BCE , BDE$  have a common tangent.

[u]Babis[/u]

see http://www.mathlinks.ro/Forum/viewtopic.php?t=81808 .

  darij


Prove that every 4n Fibonacci number is divisible by 3.

Consider the sequence modulo 3:

1, 1, 2, 0, 2, 2, 1, 0, 1, 1, 2, 0, 2, 2, 1, 0...

The period is 8 so the result follows.

How should I prove it using induction and the closed form, though? I have to write a computer program for it, and it is a little difficult using modulus.

The induction proof isn't even necessary, the period thing is trivial. You see, after 8 terms, the whole sequence starts over again, and keeps doing that, because of the recursive definition. (I thought I'd moved this because it's too easy for pre-olympiad?
Or was I dreaming that?)


Yeah, either positive, negative, or possibly zero.  Just some constant.

Hint -
[hide]Exponentials work, that was the inspiration for this problem, but I wanted to try and derive that they were the only things that work (if its true), besides degenerate cases such as constant fucntion when b=1.  Unfortunately, there are also cases such as the sine curve for example if b=1 and a=2pi, but since sine curves can be written like exponentials with imaginary coefficients I imagine it might take an exponential form anyway.  I don't know the answer, and was wondeirng if anyone else could figure it out.[/hide]

[hide="Answer"]$f(x)=c^x$, $c$ a constant.  Specifically, $c=b^{\frac{1}{a}}$.[/hide]

That doesn't give [hide="All possible functions"]
The actual solution is given by a function of the form $f(x) = b^{ \lfloor \frac{x}{a} \rfloor} g(x)$ where $g(x) = g(x + a)$.  This covers all possible solutions, including $f(x) = 0$.  $g(x)$ can be any function with period $a$.

Edit:  Note that $b = 1, g(x) = \sin x$ gives the $\sin x$ solution mentioned earlier, and also that $g(x) = b^{ \frac{x}{a} - \lfloor \frac{x}{a} \rfloor }$ gives the expected exponential solution.
[/hide]


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Please N.T.Tuan help me, I am very interesed in register. I only want that you explain the things that I don't understand.[/quote]
That is all! :)


probability
geometry
3D geometry
sphere
search
AMC 10
number theory

I recently did the 1999 National sprint round and got a 13  :blush: . It's harder than it looks and all of the questions take forever. I have 9 problems that I either didn't know how to do or wanted a faster solution, so I'll split this into two topics. Thanks in advance.

1. A positive three-digit integer is formed by choosing 3 different digits from the set {0,2,4,6,8}. The first digit cannot be 0. What is the probability that the number is divisible by 3? Express your answer as a common fraction.

SOLVED

[hide] I got 5/24 but the answer key says 5/12.[/hide]

3. There are four red balls and two white balls in a jar. One ball is randomly removed and replaced with a ball of the opposite color. The jar is then shaken, and one ball is randomly selected. What is the probability that this ball is red? Express your answer as a common fraction.

SOLVED

[hide]I have no clue how to do this, the answer key says 11/18.[/hide]

9. A sphere is inscribed in a cube with edge length 9 inches. Then, a smaller cube is inscribed in the sphere. How many cubic inches are in the volume of the inscribed cube? Express your answer in simplest radical form.

SOLVED

[hide] I have no clue how to do this, their answer was (81)radical3[/hide]

12. What is the smallest value of x such that |5x-1|=|3x+2|?
Express your answer as a common fraction.

SOLVED

[hide] I have no clue how to do this the answer key said -1/8[/hide]

13. if f(x) = (4x+1)/3, what is the value of (f^-1(1))^-1?
SOLVED

[hide] I got 5/3, the answer key says 2[/hide]

16. What is the sum of all positive two-digit integers with exactly 12 positive factors?

[hide] I can do this, it just takes too long. The answer is 402.[/hide]

Look here for the other problems: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1031228#1031228[list]

On number 12, you can square both sides to eliminate the absolute value bars.

#9 was like #21 on the AMC 10 last year.
Basically, the length of the internal diagonal of the smaller square is 9. Now you can use Pythagoras.
I bet on #13 you though f^-1(1) is the reciprocal of f(1). [b]It's not[/b]. f^-1(1) asks for which value of x is f(x) 1.

EDIT: Casework for 16,15,3, and 1. (There are probably faster solutions, but casework works)

I can't do a couple of them, but I can do some.
12[hide]
We have either $ 5x\minus{}1\equal{}3x\plus{}2$ or $ 5x\minus{}1\equal{}\minus{}(3x\plus{}2)$
The one on the right simplifies to $ 2x\equal{}3$ or $ x\equal{}\frac{3}{2}$
The one on the left simplifies to $ 5x\minus{}1\equal{}\minus{}3x\minus{}2$, then $ 8x\equal{}\minus{}1$, and finally $ x\equal{}\minus{}\frac{1}{8}$.
Since we are looking for the smallest one, we see $ x\equal{}\minus{}\frac{1}{8}$ is the answer.[/hide]

13[hide]
So you are talking about finding $ \frac{1}{g(1)}$ where g(x) is the inverse of f(x). We can solve for g(x) algebraically, by switching the x and the y. We would get $ x\equal{}\frac{4y\plus{}1}{3}$, which simplifies to $ 3x\equal{}4y\plus{}1$, $ 4y\equal{}3x\minus{}1$, and $ y\equal{}\frac{3x\minus{}1}{4}$. We now have g(x). Plug 1 into the equation and get $ \frac{3\minus{}1}{4}$, which is just $ \frac{1}{2}$ hiding. Since we are taking the inverse of this number we get $ \frac{1}{\frac{1}{2}}$ or 2.
Q.E.D.[/hide]

On number 3, first you should find out the probabilaty of getting a red ball then red ball, and then you should add the probability of white ball and red ball:

$ \frac{2}{3}\bullet\frac{1}{2}\plus{}\frac13\bullet\frac56\equal{}\frac13\plus{}\frac{5}{18}\equal{}\frac{11}{18}$

number 16 means that after u apply the number of factor theorem u either get 3*4 or 2*6 or 3*2*2(im not sure how 2 explain the number of factors sum1 else can do it for me)
that means if p , q , and r are primes, u got p^3*q^2 or p^5*q or p^2*q*r
for p^3*q^2 the only 2 digit # that fits this format is [b]72[/b]
for p^5*q the only 2 digit # that fits this format is [b]96[/b]
for p^2*q*r the only 2 digit #'s that fit this format are [b]60 , 84 , and 90[/b]
60+72+84+90+96 is . . .
[b]402[/b]

someone please verify this solution please

for number 3 i will split it into 2 cases:
u pick a red , or u pick a white

there is a 4/6 chance of picking a red and then there are 3 red and 3 white. this means there will be a 1/2 chance of picking a red then.

this means 4/6*1/2=[b]1/3[/b]

but were not done yet!
there is a 2/6 chance or picking a white and then there are 5 red and 1 white. this means there will be a 5/6 chance or picking a red then.

this means 2/6*5/6=[b]5/18[/b]

add the 2 cases together:  5/18+1/3=5/18+6/18=[b]11/18[/b]

i would also like this solution verified, thank you

for number 1 (i know, im going backwards) the sum of the numbers in the set has to be a multiple of 3

the possibilities are 0,2,4 0,4,8 2,4,6 and 4,6,8

the 0,2,4 and 0,4,8 cases appear 4 times and the 2,4,6 and 4,6,8 cases appear 6 times for a total of [b]20[/b] spossibilitie

there are a total of 4 possibilities for the first digit (not 0) 4 for the next and 3 for the last this gives a total of [b]36[/b] total possibilities

good/bad=20/36=[b]5/12[/b]

i would also like this solution to be verified

I'll verify Bobby's (lol) solution, except using latex...

[hide="16"]The number of factors of a positive number = ... go look it up in the search bar... It's been asked alot.

But... one more time.
For a number that has a prime factorization of $ p_1^k*p_2^m*p_3^n*\hdots$
The number of total factors is $ (k\plus{}1)(m\plus{}1)(n\plus{}1)\hdots$.  This is because that to find all combinations, we multiply all the exponents together.  However, since $ p_n^0$ is also a factor, we need to include all of those.

Thus, for a number to have 12 positive factors, we require that $ (k\plus{}1)(m\plus{}1)(n\plus{}1)\hdots\equal{}12$

The factors of 12 are $ 1,2,3,4,6,$ and $ 12$

We could have one power of $ 11$, but the smallest such number is $ 2^11\equal{}2048$, which obviously doesn't work.

For an integer in the form $ p_1*p_2^5$, we obviously wish to minimize $ p_2$.  So let $ p_2\equal{}2,p_1\equal{}3$
The integer is then $ 96$

For $ p_1^3*p_2^2$, $ p_1\equal{}2,p_2\equal{}3$ yields one answer, $ 72$.

However, we can also have that two other factors multiply to $ 4$.
So $ p_1^2*p_2*p_3$.  First, we try $ p_1\equal{}2$ and $ p_2\equal{}3,p_3\equal{}5$, this gives us $ 60$.
We can also have that $ p_2\equal{}3,p_3\equal{}7$, which gives us $ 84$.

The last one is $ p_1\equal{}3,p_2\equal{}2,p_3\equal{}5$.  This gives us $ 90$.

The sum is... $ 402$[/hide]

Just as bobby said... two cases

[hide="3"]
Case 1: The ball replaced was red

Now there are $ 3$ of each color.

The probability that the first ball was red: $ \frac{2}{3}$
The probability that the second ball is red: $ \frac{1}{2}$

Total probability: $ \frac{1}{3}$

Case 2: The ball replaced was white

Probability of first ball being white: $ \frac{1}{3}$
Probability that second ball is red: $ \frac{5}{6}$

Total: $ \frac{5}{18}$

Sum of the two probabilities: $ \frac{11}{18}$[/hide]

[quote="abacadaea"][hide="for number 1..."]
4*4*3 ways to chose a number. with a 0, there were 12 numbers, without a zero, there were 8 numbers. 20/48=5/12[/hide][/quote]

 :blush: I kept on doing 4*4*3*2 because I mis-did it ;)

btw, I finished target, and ended with a grand total of 21 with sprint and target   :wink:

I had a 19 on like the 1992 or 1993 last year :lol:

they make those nats tests really hard, don't they  :)

[quote="xpmath"]I think the first 15 or so aren't too bad compared to the last 15 at state, but the last few are usually pretty hard.

37 on the 1998 or something today :)[/quote]

on the nats test?!?!?! The top score is usually around 36.

anyways, The funny thing is that I got 8 out of the 10 last ones right.

I got 25/12. Doing AMC 12 practice does help :wink:
Uh, the top score isn't usually around 36...is it?Plus, I tend to do atleast 3 to 4 points better on practices.

Actually, this is it.

 In the 2005 Nationals, the highest score was a 39, and a score of 30 sufficed for placement in the top twelve, qualifying for the Countdown Round.


Does anyone know how much colleges care about the math contests that many of us are involved in?  For example, how good would a USAMO qualification look on my resume?  A MOP?  An IMO?  Not to say that I've done any of these, but as they are possibilities (however unlikely) of the future, I'm curious as to whether colleges would care (of course, this is if I were to go into engineering or economics or something math-related).

The [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9596]USAMO and College Applications thread[/url] has scrolled down the topic list a bit, but this is still a live issue, so let's add some new replies to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9596]the original thread[/url].


Prove that for every integer$ a_i$,

$ gcd(a_1, a_2, \cdots, a_n) \equal{} gcd \left( gcd(a_1, a_2, \cdots, a_{n \minus{} 1}), a_n \right)$.

Thanks

Let $ k\equal{}gcd(a_1, a_2, \cdots, a_n)$, then $ k|a_1,a_2,...,a_n$ $ \Rightarrow k|R.H.S.$

Now let $ R.H.S.\equal{}nk$, then $ nk|gcd(a_1,a_2,...,a_{n\minus{}1}),a_n \Rightarrow nk|a_1,a_2,...,a_n$
$ \Rightarrow nk|gcd(a_1,a_2,...,a_n)\equal{}k \Rightarrow n\equal{}1$, so $ R.H.S.\equal{}k$ also, done.

I think this should be moved to High School Basics


modular arithmetic
number theory
prime numbers

Find all prime numbers of the form 100 · · · 001, where the total number of zeros between the
first and last digits is even.

$ 1\plus{}10^{2n\plus{}1}$ mod 11 is $ 1\plus{}(\minus{}1)^{2n\plus{}1}\equal{}1\minus{}1\equal{}0$

Therefore it is divisible by 11.

Just continuing what rofler said,

the only prime that is $ \equiv 0 \pmod {11}$ is 11.

So the solution is $ \fbox{11}$.


Subj. :wink:

http://www.hms.gr/seemous08/xindex.php?id=2

We were organizing a USA team but then the math departments at MIT and Harvard got mad at us because we were getting money ourselves rather than from our department and now it seems we won't go after all. :mad: (But maybe this isn't exactly what happened; I wasn't directly involved.)

National team of Ukraine
Leader - Dmytro Mitin - Kyiv Taras Shevchenko National University
Deputy Leader - Nataliia Zhuravlova - Ministry of Education and Science of Ukraine
Contestant - Ivan Feshchenko - Kyiv Taras Shevchenko National University
Contestant - Ivan Iurchenko - Kyiv Taras Shevchenko National University
Contestant - Danylo Radchenko - Kyiv Taras Shevchenko National University
Contestant - Volodymyr Medvid - Kyiv Taras Shevchenko National University
Contestant - Artem Starchenko - Donetsk National University
Contestant - Stanislav Nikolaienko - Donetsk National University

Team of Kyiv Taras Shevchenko National University
Leader - Andrii Bondarenko
Contestant - Ivan Feshchenko
Contestant - Ivan Iurchenko
Contestant - Danylo Radchenko
Contestant - Volodymyr Medvid
Contestant - Andriy Arman
Contestant - Maksym Tantsiura

I know that a lot of people was angry with the jury of some problems. in especial a participant from Ukraine.

[quote="AdrianDD"]Does anyone has the results? Thank you.[/quote]

See the files attached.

Professor Giannopoulos published the problems and solutions as PDF-file at his homepage:
http://users.uoa.gr/~apgiannop/seemous08_exam.pdf


group theory
abstract algebra
algebra
polynomial
inequalities
logarithms
superior algebra

Let $ G$ be a finite group with $ n$ elements. What type of bounds can be put on $ |S|$ where $ S \equal{}\{H| H\leq G\}$. For example $ S\leq 2^{n}$, but this is trivial. How about $ S\leq 2n$.

Absolutely not. Consider $ (\mathbb{Z}/2\mathbb{Z})^{n}$; its subgroups are its subspaces. The number of these subgroups:
$ n=0$: 1
$ n=1$: 2
$ n=2$: 5
$ n=3$: 16
$ n=4$: 67
$ n=5$: 374
$ n=6$: 2825
$ n=7$: 29212
$ n=8$: 417199
$ n=9$: 8283458
$ n=10$: 229755605

That's a quarter of a billion subgroups for a group with a thousand elements. There is no polynomial bound; this sequence grows faster than any polynomial in $ 2^{n}$.

[quote="jmerry"]Absolutely not.[/quote]
I understand what you wrote I dont understand what you mean by that? Are you saying my inequality is wrong? It cant be because the number of subsets is $ 2^{n}$ (okay, okay, I should have written $ 2^{n}\minus{}1$ because we dont consider empty sets).

EDIT: Oops, I know now what you mean. Next time I try to take time to understand your post.

But can you improve the $ 2^{n}\minus{}1$ bound?

A simple estimate: Let $ G(n)$ be the maximum number. Then $ \log G(n)<(\log n)^{2}$, where the logarithms are base 2. (as long as $ n$ isn't tiny)
This comes from building up from the bottom: there are at most $ n\minus{}1$ minimal subgroups, at most $ n\minus{}2$ subgroups directly above each of those, and so on. At the last step, there's only one choice.
In practice, there's a ton of double-counting. The example I gave has $ \log G(n)$ asymptotically similar to $ \frac{1}2(\log n)^{2}$, and I'm pretty sure it's the maximum.

That asymptotic estimate for my example wasn't very good; I was overestimating the big terms in the middle. A revised version:

$ \log G(n) \equal{}\frac{1}{4}(\log n)^{2}\plus{}c(n)\plus{}o(1)$, where $ c(n)\approx\begin{cases}2.8820499&n\text{is even}\\2.8820462&n\text{is odd}\end{cases}$.

All logarithms are base 2.

[Errors fixed]

@jmerry:

How goes that together with the previous statement, that G(n) grows exponentially and not polynomiell? With your approximation one would have $ G(n)\equal{}2^{c\plus{}o(1)}n^{1/4}$ but this isn't exponential.


A box contains 100 balls of which r are red.  Suppose they are drawn one at a time without replacement. Determine the foll:
a) prob. the first ball drawn is red--->  r/100
b) prob. the 50th ball drawn is red--> (100-r)!*r / 100!
c) prob. the last ball drawn will be red...

I'm not sure if b) is correct which is why i posted it. About c), my answer didn't make any sense to me so I would appreciate some input.

Why don't you start with this: what's the probability that the second ball is red?

By the way, you might want to make sure that you check your answer to each part by trying out $ r \equal{} 0$ and $ r \equal{} 100$.  This won't guarantee that an answer is correct, but it will help you avoid many wrong answers.

I thought about my answer to part two and got something different.

I understand what you're saying about testing r=100 and r=0.  I suppose in part b, I would have to ensure that at r=51, there is no way to arrange the balls so that the 50th ball is the first red one.

I figure that the total number of ways of arranging the balls is 100! and that should be the denominator.

In the numerator, I'm looking for the total number of ways to arrange the balls such that the 50th ball is the first red one.

If r>50, I conclude that there is no way to do this, thus I assume that r is less than or equal to 50.

for the first slot, i have (100-r) choices, for the second, (100-r-1), and the third I have (100-r-2) etc.
When I get to the 50th slot, I have r choices.

Thus (100-r)(100-r-1)...(100-r-48)(r)(100-r-50)(100-r-51)....

This gives me    [(100-r)! / (100-r-49) ]* r in the numerator.

Then I get   [ (100-r)!/ 100!*(100-r-49)]*r

plugging in r=51 will not give 0 which i know is what it's supposed to be. so i'm assuming this formula only works for r<51.

[quote="Keno20"]I understand what you're saying about testing r=100 and r=0.  I suppose in part b, I would have to ensure that at r=51, there is no way to arrange the balls so that the 50th ball is the first red one.[/quote]  This doesn't appear to have anything to do with the question you asked in the first post :maybe: You asked for the probability that the fiftieth ball is [i]a[/i] red ball, not that the fiftieth ball is [i]the first[/i] red ball.

The argument you've given (for the wrong problem) is almost right -- but how many choices do you have for the 51st ball drawn?  (You appear to be claiming it's $ 50 \minus{} r$.)


The entries of a $ 4\times 4$ unit square grid are to be filled with elements from the set $ S \equal{} \{1,2,3,4\}$ such that if $ a,b,c,d$ belong to the same row or the same column then we have $ \{a,b,c,d\} \equal{} S$. At the beginning some of the entries are given. What is the minimum possible number of entries that must be given in order to guarantee a unique solution of the board?

[hide="For example"]
The following board has a unique solution:

[code]
1 2 3 _             1 2 3 4
_ 3 4 _             2 3 4 1
_ _ 1 2             3 4 1 2
4 _ _ _             4 1 2 3
[/code]

But the following has multiple solutions:

[code]
1 2 3 4             1 2 3 4                1 2 3 4
2 3 4 1             2 3 4 1                2 3 4 1
_ _ _ _             3 4 1 2                4 1 2 3
_ _ _ _             4 1 2 3                3 4 1 2
[/code]

But in this case the answer [i]could be[/i] $ 8$ (not that this is the actual answer), since we do get a unique solution of the board if $ 8$ [i]suitable[/i] entries are given.[/hide]

[hide="Solution"]$ 4$ is the minimum.

[code]1.2.
.3..
..1.
....[/code]
gives a unique solution.
[code]1423
2341
3214
4132[/code]

We'll show three clues is not possible, which is sufficient to show any lower number is also not possible. If any two rows or two columns have no clues, they can be interchanged to give different solutions. So there is at most one column and at most one row with no clues. This means that if any two of the three clues are in the same row and column, we have a problem.

Thus, by interchanging rows and columns, we can assume WLOG that the clues are in these positions
[code]x...
.y..
..z.
....[/code]

Suppose two of these clues are the same number. Then there are two numbers for which there are no clues. In any solution these numbers can be switched, so there can't be a unique solution. So all the clues are different numbers. WLOG they are $ 1,2,3$, and we can interchange rows and columns in the board above so that the clues are arranged as follows.
[code]1...
.2..
..3.
....[/code]

And we don't get a unique solution from this board.

Now, what if we replace $ 4$ with $ n$?[/hide]