If $ a,b,c$ are positive real numbers, then

$ \frac a{b}\plus{}\frac b{c}\plus{}\frac c{a} \ge 3\plus{}\frac {(a\minus{}c)^2}{ab\plus{}bc\plus{}ca}$.

It is my inequality in the book "Old and New Inequalities 2", dear Vasc.

Nice, Can. Indeed, I found now this inequality in "Old and New Inequalities 2".
This inequality is stronger that Belarus 1998, which can be written as

$ \frac a{b}\plus{}\frac b{c}\plus{}\frac c{a} \ge 3\plus{}\frac {(a\minus{}c)^2}{b^2\plus{}ab\plus{}bc\plus{}ca}$.

I have a solution to your inequality without Cauchy-Schwarz Inequality, only with AM-GM Inequality. Can somebody find it?

[quote="rachid"][quote="Vasc"]If $ a,b,c$ are positive real numbers, then

$ \frac a{b} + \frac b{c} + \frac c{a} \ge 3 + \frac {(a - c)^2}{ab + bc + ca}$.[/quote]
[quote="Vasc"]I have a solution to your inequality without Cauchy-Schwarz Inequality, only with AM-GM Inequality. Can somebody find it?[/quote]
Here is my AM-GM's proof:

Clearly,the inequality is equivalent to :

$ b^2 + \frac {ab^2}{c} + \frac {a^2c}{b} + \frac {bc^2}{a} \geq 2ab + 2bc$

We know that :

${ \frac {bc^2}{a} \geq 2bc - ab}$,

Hence,It suffices to prove that :

$ b^2 + \frac {ab^2}{c} + \frac {a^2c}{b} \geq 3ab$ which is only AM-GM

Done ![/quote]
Very nice your proof, rachid. The following proof is much longer than yours, but we dont need to expand the term $ (ab+bc+ca)\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right).$

I think that the following inequality holds, too:
$ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a} \ge 3\plus{}\frac{6(a\minus{}c)^2}{(a\plus{}b\plus{}c)^2}.$
Can someone find a nice proof?

[quote="can_hang2007"][quote="rachid"][quote="Vasc"]If $ a,b,c$ are positive real numbers, then

$ \frac a{b} + \frac b{c} + \frac c{a} \ge 3 + \frac {(a - c)^2}{ab + bc + ca}$.[/quote]
[quote="Vasc"]I have a solution to your inequality without Cauchy-Schwarz Inequality, only with AM-GM Inequality. Can somebody find it?[/quote]
Here is my AM-GM's proof:

Clearly,the inequality is equivalent to :

$ b^2 + \frac {ab^2}{c} + \frac {a^2c}{b} + \frac {bc^2}{a} \geq 2ab + 2bc$

We know that :

${ \frac {bc^2}{a} \geq 2bc - ab}$,

Hence,It suffices to prove that :

$ b^2 + \frac {ab^2}{c} + \frac {a^2c}{b} \geq 3ab$ which is only AM-GM

Done ![/quote]
Very nice your proof, rachid. The following proof is much longer than yours, but we dont need to expand the term $ (ab + bc + ca)\left(\frac {a}{b} + \frac {b}{c} + \frac {c}{a}\right).$[/quote]
very nice proof can_hang! :omighty:

[quote="can_hang2007"]I think that the following inequality holds, too:
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \ge 3 \plus{} \frac {6(a \minus{} c)^2}{(a \plus{} b \plus{} c)^2}.$
Can someone find a nice proof?[/quote]
I think the following is easier, but also too hard.

$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \ge 3 \plus{} \frac {2(a \minus{} c)^2}{a^2\plus{}b^2\plus{}c^2}.$

[quote="Vasc"][quote="can_hang2007"]I think that the following inequality holds, too:
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \ge 3 \plus{} \frac {6(a \minus{} c)^2}{(a \plus{} b \plus{} c)^2}.$
Can someone find a nice proof?[/quote]
I think the following is easier, but also too hard.

$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \ge 3 \plus{} \frac {2(a \minus{} c)^2}{a^2 \plus{} b^2 \plus{} c^2}.$[/quote]
Yeah, it is easier and here is my ugly proof:

Since $ (a \minus{} c)^{2} \equal{} [(a \minus{} b) \plus{} (b \minus{} c)]^{2}\leq 2(a \minus{} b)^{2} \plus{} 2(b \minus{} c)^{2},$ we have
\[ 3(a \minus{} c)^{2}\leq 2[(a \minus{} b)^{2} \plus{} (b \minus{} c)^{2} \plus{} (c \minus{} a)^{2}] \equal{} 4(a^{2} \plus{} b^{2} \plus{} c^{2} \minus{} ab \minus{} bc \minus{} ca).\]
Therefore
\[ \frac {2(a \minus{} c)^{2}}{a^{2} \plus{} b^{2} \plus{} c^{2}}\leq \frac {8}{3}\cdot \frac {a^{2} \plus{} b^{2} \plus{} c^{2} \minus{} ab \minus{} bc \minus{} ca}{a^{2} \plus{} b^{2} \plus{} c^{2}}.\]
It suffices to prove that
\[ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\geq 3 \plus{} \frac {8}{3}\cdot \frac {a^{2} \plus{} b^{2} \plus{} c^{2} \minus{} ab \minus{} bc \minus{} ca}{a^{2} \plus{} b^{2} \plus{} c^{2}}.\]
This is a cyclic inequality with $ a,b,c,$ so we may assume without loss of generality that $ b$ is the number between $ a$ and $ c.$

[i]Case 1.[/i] $ a\geq b\geq c.$ Since
\[ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \minus{} 3 \equal{} \frac {(a \minus{} b)^{2}}{ab} \plus{} \frac {(a \minus{} c)(b \minus{} c)}{ac},\]
and
\[ a^{2} \plus{} b^{2} \plus{} c^{2} \minus{} ab \minus{} bc \minus{} ca \equal{} (a \minus{} b)^{2} \plus{} (a \minus{} c)(b \minus{} c),\]
one can write our inequality in the form $ M(a \minus{} b)^{2} \plus{} N(a \minus{} c)(b \minus{} c)\geq 0,$ where
\[ M \equal{} \frac {1}{ab} \minus{} \frac {8}{3(a^{2} \plus{} b^{2} \plus{} c^{2})},\quad\text{and}\quad N \equal{} \frac {1}{ac} \minus{} \frac {8}{3(a^{2} \plus{} b^{2} \plus{} c^{2})}.\]
Notice now that $ N\geq 0$ for $ a\geq b\geq c.$

[i]Case 1.1.[/i] $ b \minus{} c\leq \dfrac{a \minus{} b}{4}.$ In this case, we have $ c\geq \dfrac{5b \minus{} a}{4},$ and we will prove that $ M\geq 0,$ i.e.
\[ 3(a^{2} \plus{} b^{2} \plus{} c^{2}) \minus{} 8ab\geq 0.\]
If $ a\geq 5b,$ then this is clearly true. Otherwise, we have
\[ 3(a^{2} \plus{} b^{2} \plus{} c^{2}) \minus{} 8ab\geq 3\left[ a^{2} \plus{} b^{2} \plus{} \frac {(5b \minus{} a)^{2}}{16}\right] \minus{} 8ab \equal{} \frac {51a^{2} \plus{} 123b^{2} \minus{} 158ab}{16}\geq 0.\]
Now since $ M,N\geq 0,(a \minus{} b)^{2}\geq 0$ and $ (a \minus{} c)(b \minus{} c)\geq 0,$ we see that the inequality
\[ M(a \minus{} b)^{2} \plus{} N(a \minus{} c)(b \minus{} c)\geq 0\]
is obviously true.

[i]Case 1.2.[/i] $ b \minus{} c\geq \dfrac{a \minus{} b}{4}.$ In this case, we have $ a \minus{} c\geq \dfrac{5(a \minus{} b)}{4}\geq 0$ yields
\[ (a \minus{} c)(b \minus{} c)\geq \frac {5}{16}(a \minus{} b)^{2}.\]
It follows that
\[ M(a \minus{} b)^{2} \plus{} N(a \minus{} c)(b \minus{} c)\geq \left( M \plus{} \frac {5}{16}N\right) (a \minus{} b)^{2},\]
and so, it suffices to prove that
\[ M \plus{} \frac {5}{16}N\geq 0.\]
This inequality is equivalent to
\[ \frac {16}{ab} \plus{} \frac {5}{ac}\geq \frac {56}{a^{2} \plus{} b^{2} \plus{} c^{2}}.\]
Applying the AM-GM Inequality, we have
\[ \frac {{16}}{{ab}} \plus{} \frac {5}{{ac}} \equal{} \frac {8}{{ab}} \plus{} \frac {8}{{ab}} \plus{} \frac {5}{{ac}}\geq \frac {{12\sqrt [3]{5}}}{\sqrt [3]{{{a^{3}}{b^{2}}c}}},\]
and also,
\[ {a^{2}} \plus{} {b^{2}} \plus{} {c^{2}} \equal{} 3\cdot \frac {{{a^{2}}}}{3} \plus{} 2\cdot \frac {{{b^{2}}}}{2} \plus{} {c^{2}}\geq \frac {6}{\sqrt [6]{{{3^{3}}{2^{2}}}}}\sqrt [6]{{{a^{6}}{b^{4}}{c^{2}}}} \equal{} \frac {6}{\sqrt [6]{{{3^{3}}{2^{2}}}}}\sqrt [3]{{{a^{3}}{b^{2}}c}} > \frac {{14}}{{3\sqrt [3]{5}}}\sqrt [3]{{{a^{3}}{b^{2}}c}}.\]
Therefore
\[ \frac {{16}}{{ab}} \plus{} \frac {5}{{ac}}\geq \frac {{12\sqrt [3]{5}}}{\sqrt [3]{{{a^{3}}{b^{2}}c}}} > \frac {{12\sqrt [3]{5}}}{\dfrac{3\sqrt [3]{5}}{14}(a^{2} \plus{} b^{2} \plus{} c^{2})} \equal{} \frac {56}{a^{2} \plus{} b^{2} \plus{} c^{2}}.\]
[i]Case 2.[/i] $ c\geq b\geq a.$ Since
\[ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \minus{} \frac {c}{b} \minus{} \frac {b}{a} \minus{} \frac {a}{c} \equal{} \frac {(c \minus{} b)(c \minus{} a)(b \minus{} a)}{abc}\geq 0,\]
it suffices to prove that
\[ \frac {c}{b} \plus{} \frac {b}{a} \plus{} \frac {a}{c}\geq 3 \plus{} \frac {8}{3}\cdot \frac {a^{2} \plus{} b^{2} \plus{} c^{2} \minus{} ab \minus{} bc \minus{} ca}{a^{2} \plus{} b^{2} \plus{} c^{2}}.\]
Applying the above case, for any $ x,y,z > 0$ with $ x\geq y\geq z,$ we have
\[ \frac {x}{y} \plus{} \frac {y}{z} \plus{} \frac {z}{x}\geq 3 \plus{} \frac {8}{3}\cdot \frac {x^{2} \plus{} y^{2} \plus{} z^{2} \minus{} xy \minus{} yz \minus{} zx}{x^{2} \plus{} y^{2} \plus{} z^{2}}.\]
Substituting $ x \equal{} c,y \equal{} b,z \equal{} a$ in this inequality, we get
\[ \frac {c}{b} \plus{} \frac {b}{a} \plus{} \frac {a}{c}\geq 3 \plus{} \frac {8}{3}\cdot \frac {a^{2} \plus{} b^{2} \plus{} c^{2} \minus{} ab \minus{} bc \minus{} ca}{a^{2} \plus{} b^{2} \plus{} c^{2}}.\]
The proof is completed. It is easy to see that the equality holds if and only if $ a \equal{} b \equal{} c.$

Very interesting solution, Can-hang.

Is the following inequality known or easy?

If $ a,b,c$ are positive real numbers such that $ ab\plus{}bc\plus{}ca\equal{}3$, then

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

[quote="Vasc"]Very interesting solution, Can-hang.

Is the following inequality known or easy?

If $ a,b,c$ are positive real numbers such that $ ab \plus{} bc \plus{} ca \equal{} 3$, then

$ \frac {a^2}{a^2 \plus{} b \plus{} c} \plus{} \frac {b^2}{a^2 \plus{} b \plus{} c} \plus{} \frac {c^2}{a^2 \plus{} b \plus{} c} \ge 1$.[/quote]
It is not known and not easy. It is only nice.  :lol:
Here is my solution for it:

Using the Cauchy Schwarz Inequality, we have
$ \sum \frac {a^2}{a^2 \plus{} b \plus{} c} \ge \frac {\left(\sum a^{3/2}\right)^2}{\sum a(a^2 \plus{} b \plus{} c)} \equal{} \frac {a^3 \plus{} b^3 \plus{} c^3 \plus{} 2\sum (ab)^{3/2}}{a^3 \plus{} b^3 \plus{} c^3 \plus{} 6},$
and thus, it suffices to prove that
$ (ab)^{3/2} \plus{} (bc)^{3/2} \plus{} (ca)^{3/2} \ge 3,$
which is true according to the Power Mean Inequality.

Nice, Can_Hang. This is also my solution. :P

the can_hang inequaliy can be turned into 2 variables inequality ,

$ \frac{x\minus{}1}{2} \plus{} \frac{2}{y\minus{}1} \plus{} \frac{y\minus{}1}{x\minus{}1} \geq   3\plus{} \frac{ (x\minus{}y)^2}{(x\plus{}y)^2}$  , with $ x,y>1$  ,I feel that there is a nice proof for it  :P

[quote=sqing]The following inequality is also true.
Let  $ a,b,c$ are positive real numbers.  Prove that$$ \frac a{b}\plus{}\frac b{c}\plus{}\frac c{a} \ge 2\plus{}\frac {a^2+b^2+c^2}{ab\plus{}bc\plus{}ca}.$$[/quote]

Prove this one:

$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq \frac{a^2+b^2+c^2}{ab+bc+ac}+\frac{6(a^2+b^2+c^2)}{(a+b+c)^2}$

The following inequalities are also true.
Let  $ a,b,c$ are positive real numbers such that $abc=1$ .  Prove that$$ \frac a{b}\plus{}\frac b{c}\plus{}\frac c{a} \ge 2\plus{}\frac {a^2+b^2+c^2}{a\plus{}b\plus{}c},$$ $$ \frac a{b}\plus{}\frac b{c}\plus{}\frac c{a} \ge \sqrt{3(a^2+b^2+c^2)}.$$

$$ \frac a{b}+\frac b{c}+\frac c{a} \ge \sqrt{3(a^2+b^2+c^2)}.$$
$$\iff \sum_{cyc} (a^4c^2+2a^2b)\geq 3\sum_{cyc} a^2$$,by taking both square and $abc=1$.
Indeed,
$$a^4c^2+a^2b+a^2b=3\sqrt[3]{a^6(a^2b^2c^2)}= 3a^2$$,

Similarly, it is true for $b$ and $c$.
QED.

@sqing
Are you sure?

$$ \frac a{b}+\frac b{c}+\frac c{a} \ge 2+\frac {a^2+b^2+c^2}{a+b+c},$$
Why not in homogeneous form?

http://www.artofproblemsolving.com/community/c6h1149455p5432855：
Let $a,\,b,\,c$ are positive real number. Prove that
\[\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geqslant 3 + \frac{31}{5}\left(\frac{c-a}{a+b+c}\right)^2.\]

[quote=Vasc]I think the following is easier, but also too hard.
$$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \ge 3 \plus{} \frac {2(a \minus{} c)^2}{a^2\plus{}b^2\plus{}c^2}.$$[/quote]
The inequality stronger is true
\[\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geqslant 3 + \frac{21(c-a)^2}{10(a^2+b^2+c^2)}.\]


[quote="Phelpedo"]How many subsets of the set $\{ 1,2,3,4,...,30\}$ have the property that the sum of the elements of the subset is greater than 232?[/quote]

[hide]Well, the total sum of all the elements is 465.

Notice that if we chose any subset and its complement, one of the two will have to sum to a number greater than or equal to 233, the other will have to sum to a number less than or equal to 232, since the closest they can be is 232 and 233, so one of the two sets will have to have a sum greater than 232.

So we know that there are 2^30 subsets, but only half of the sets work, so the answer is

$2^{29}=536870912$[/hide]


function
combinatorics unsolved
combinatorics

Use ordinary generating functions to determine the number of non-negative integer solutions to:

a²+b²+c²=190 (a, b, c are non-negative integers).

I would know how to do this if it was a + b + c = n, I just can't do it with the powers. Please help!

Austin

[hide="Maybe"]We know 1, 1, 1, 1... has a GF of $ 1\plus{}x\plus{}x^2\plus{}x^3... \equal{} \frac{1}{1\minus{}x}$.
Consider what happens when you differentiate.  You get $ 1\plus{}2x\plus{}3x^2\plus{}4x^3... \equal{} \frac{1}{(1\minus{}x)^2}$.
If we multiply by x, we can shift the whole series over one:  $ 0\plus{}1x\plus{}2x^2\plus{}3x^3... \equal{} \frac{x}{(1\minus{}x)^2}$.
Now we can differentiate again: $ 1\plus{}4x\plus{}9x^2\plus{}16x^3... \equal{} \frac{1\plus{}x}{(1\minus{}x)^3}$.
And multiply by x again: $ 0\plus{}1x\plus{}4x^2\plus{}9x^3... \equal{} \frac{(1\plus{}x)(x)}{(1\minus{}x)^3}$.

So now we have a GF for the perfect squares.  Does that help?
[/hide]

[quote="mysmartmouth"]So now we have a GF for the perfect squares.  Does that help?[/quote]

No ;).  You need to consider $ \sum_{n \geq 0} x^{n^2}$, which can be written in terms of an elliptic function ... basically, the direction here ("use ordinary generating functions") is unhelpful.  The relevant sequence is [url=http://www.research.att.com/~njas/sequences/A002102]A002102[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].


inequalities
geometry
3D geometry
algebra proposed
algebra

fact:0<a*b*c<=1
prove a/b+b/c+c/a=>a+b+c hint:use cauchy

I have a solution, but it's not really very nice.

Let the left hand side expression be $X$.

By Cauchy,
$X (ab + bc + ca) \geq (a+b+c)^2$
$X (a+b+c) \geq X abc(a+b+c) = X (abc^2 + ab^2c + a^2bc) \geq (ab+bc + ca)^2$

Take the first inequality, squared, times the second inequality, to get
$X^3 (ab+bc+ca)^2 (a+b+c) \geq (ab+bc+ca)^2 (a+b+c)^4$
Canceling and take cube roots, we are done.

Use \[2\frac{a}{b} + \frac{b}{c} \geq 3\sqrt[3]{\frac{a^2}{bc}} \geq 3a\] and the similar inequalities.


algebra
polynomial
Ring Theory
superior algebra
superior algebra unsolved

I'm looking for examples of finit fields with 8 elements, and 4 elements.
Thx

I don't know if you are used to quotients of polynomial rings to get field extensions, thus a more direct try (in fact the same):

Let $\zeta$ be a solution of $x^{2}+x+1=0$, so something of type a third root of unity.
Then $0, 1 , \zeta , 1+\zeta$ build, together with the condition above a field $\mod 2$. Just check.

Similar, we can get a field from solutions of $x^{3}+x+1$.

Thanks for the quick answer :)
I read the article [url=http://en.wikipedia.org/wiki/Finite_field]Finite Field[/url] And I didn't understand how to construct this field. I checked the operations gived for the $\mathbb{F}_{4}$, but for the $\mathbb{F}_{8}$ I can't do the same.

Please don't check field axioms or write down cayley tables, this is just stupid and gives no insight into the structure.

If $K$ is a field and $f$ is an irreducible polynomial over $K$, then $L : =K[x]/(f)$ is also a field (because $f$ is prime, so $(f)$ is a prime ideal and hence $(f)$ a maximal ideal since $K[x]$ is a PID). If $n$ is the degree of $f$, the monomials $x^{0}, ... , x^{n-1}$ are obviously a $K$-basis of $L$. In particular, if $K$ has $q$ elements, $L$ has $q^{n}$ elements.

Apply this to $K = F_{2}, f = x^{2}+x+1$ to get $F_{4}$; and $K = F_{2}, f = x^{3}+x+1$ yields $F_{8}$.


w is a circle without center and we want to find center . we can use compasses only.
how can we find the center ?

the centers of the arcs 1,2,...,6 are A,B,C,D,E and F respectively.
and the radius of them are AB,BA,CA(=k),DA,EA and FA respectively.
and its easy to prove that the point O is the center! :D


inequalities
algebra
system of equations
algebra proposed

hello, i have found the following solutions:
$ x\equal{}2,y\equal{}1$ or $ x\equal{}\frac{2}{3},y\equal{}\minus{}\frac{1}{3}$ or $ x\equal{}\frac{95}{24}\plus{}\frac{1}{24}\sqrt{95}i,y\equal{}\minus{}\frac{73}{24}\plus{}\frac{1}{24}\sqrt{95}i$ or $ \equal{}\frac{95}{24}\minus{}\frac{1}{24}\sqrt{95}i,y\equal{}\minus{}\frac{73}{24}\minus{}\frac{1}{24}\sqrt{95}i$
Sonnhard.

Let $a=x+y+\frac{1}{x+y}$ and $b=x-y$. Then we have an equivalent system of equations $\begin{cases} 3a^2+b^2=\frac{103}{3}\\ a+b=\frac{13}{3}\end{cases}$. This implies that $3a^2+\left(\frac{13}{3}-a\right)^2=\frac{103}{3}\implies 18a^2-39a-70=(3a-10)(6a+7)=0$, so $a=\frac{10}{3},-\frac{7}{6}$. Now, by AM-GM Inequality, when $x+y>0$, we have $a\geq 2$; when $x+y<0, a\leq -2$. Thus, the only value of $a$ that satisfy one of these inequalities is $a=\frac{10}{3}$. Consequently, we have $b=x-y=1, x+y=3,\frac{1}{3}$. Hence, we get $(x,y)=(2,1),\left(\frac{2}{3}-\frac{1}{3}\right)$.


parameterization
calculus
integration
probability
function
probability and stats

Hi, If anyone can give any help on this question at all it would be
greatly appreciated: Suppose that I want to generate a random sample
from some distribution with p.d.f. f and c.d.f. F, censored at the
right by a sample from distribution with p.d.f. u and c.d.f. U. As
usual F and U have certain parameters that determine scale, location,
e.t.c. By generating n observations from F, n observations from U and
taking the elementwise minimum from both samples should result in a
sample from F censored at the right from U.

I'm trying to find a rule to connect the parameter of U (the
parameters of F can be assumed known) and the precentage k of
censored observations. This is equivalent to P(X>Y) = k/100, where X~F
and Y~U. Given that the samples are i.i.d and drawn independently from
each other, then the joint p.d.f. from X and Y is simply the product
of f and u. Suppose that the range of X and Y is from 0 to infinity.

Then, P(X>Y) is a double integral with range 0<x<infinity and 0<y<x
and the integrand being the product of the two p.d.f.'s. Equating with
k/100 should give the required rule, which can then be translated to
an equation from which the value of the parameter can be derived.

I noticed that this works for, say X~exp(1) and Y~exp(m), where we
want to find m so that 30% of the observations are censored, but not
for any other combination that I tried, e.g. X~weibull(0.5, 0.8) and
Y~uniform(0,m), e.t.c. (calculations have been checked over and over
again many times).

Probably I'm doing something wrong, so can anybody help in identifying
the mistake ? Thanks in advance,

Dimitris

I do not see any mistake in your logic. The probability that $ X$ is censored by $ U$ is, indeed, $ \iint_{\{x>y>0\}}f(x)u(y)\,dx\,dy$. Perhaps, you should show your computations for some simple example and explain why you think that the result you obtained "doesn't work".

Thanks Fedja for having a look at my problem. A simple application of the rule is as follows: Suppose that we want to generate a random sample from the $ W(1, 2)$ distribution. In general the $ W(a,b)$ distribution has p.d.f.:
$ f(x) \equal{} ab^ax^{a \minus{} 1}e^{ \minus{} (bx)^a}$, $ x>0$, $ a > 0$ shape and $ b > 0$ scale parameter.
The censoring distribution is assumed to be the uniform $ U(0, \mu)$ distribution (p.d.f. $ u(x) \equal{} 1/\mu, 0 < x < \mu$). Now
$ P(X > Y) \equal{} \int_0^{ \plus{} \infty} 2e^{ \minus{} 2x} \left (\int_0^x \frac {1}{\mu} dy \right )\,dx \equal{} \frac {2}{\mu}\int_0^{ \plus{} \infty} xe^{ \minus{} 2x} \,dx$
 $ \equal{} \minus{} \frac {2}{\mu}\frac {1}{2}\int_0^{ \plus{} \infty} x \left (e^{ \minus{} 2x}\right )'dx \equal{} \minus{} \frac {1}{\mu} \left \{ x e^{ \minus{} 2x}\Big |_0^{ \plus{} \infty} \minus{} \int_0^{ \plus{} \infty} e^{ \minus{} 2x} \,dx \right \}$
From Hospital's rule $ x e^{ \minus{} 2x} \rightarrow 0$ as $ x \rightarrow \plus{} \infty$, thus
$ P(X > Y) \equal{} \frac {1}{\mu} \int_0^{ \plus{} \infty} e^{ \minus{} 2x} \,dx \equal{} \frac {1}{2\mu}$. Hence, if we want $ \lambda \%$ of the observations to be censored
$ \frac {1}{2\mu} \equal{} \frac {\lambda}{100} \Leftrightarrow 100 \equal{} 2\mu \lambda \Leftrightarrow \mu \equal{} \frac {50}{\lambda}$.
For example if we take $ \lambda \equal{} 10$, this gives $ \mu \equal{} 5$, i.e. we should generate an equal sized sample from $ U(0,5)$. Unfortunately this doesn't seem to work, as it results to an approximately 35% censoring. (In generating the samples I used the default S-plus version 6.1 functions for random number generation.). I'm really curious to find out what's going wrong so any ideas would be very much appreciated,

Dimitris

Ah, but your computation is obviously wrong: $ \int_0^x \frac 1\mu dx$ is [b]not[/b] $ \int_0^x u(x)\,dx$ when $ x>\mu$!

[This post was mostly composed before fedja posted- it's the same idea, but more elaborated]

[quote="DimitrisNo11"]$ P(X > Y) \equal{} \int_0^{ \plus{} \infty} 2e^{ \minus{} 2x} \left (\int_0^x \frac {1}{\mu} dy \right )\,dx \equal{} \frac {2}{\mu}\int_0^{ \plus{} \infty} xe^{ \minus{} 2x} \,dx$[/quote]
This is incorrect, because $ U$ is only uniform on $ [0,\mu]$ and not all of $ [0,\infty]$. A corrected version:
$ \int_0^{\mu}2e^{\minus{}2x}\left (\int_0^x \frac {1}{\mu} dy \right )\,dx\plus{}\int_{\mu}^{\infty}2e^{\minus{}2x}\left (\int_0^{\mu} \frac {1}{\mu} dy\right)\,dx$
$ \equal{}\frac1{\mu}\int_0^{\mu}2xe^{\minus{}2x}\,dx\plus{}\int_{\mu}^{\infty}2e^{\minus{}2x}\,dx \equal{}\frac1{\mu}\left[\minus{}xe^{\minus{}2x}\minus{}\frac12e^{\minus{}2x}\right]_0^{\mu}\plus{} \left[\vphantom{\frac12}\minus{}e^{\minus{}2x}\right]_{\mu}^{\infty}$
$ \equal{}\frac1{2\mu}\left(1\minus{}e^{\minus{}2\mu}\right)$

35% sounds like you tried $ \mu\equal{}1$. $ \mu\equal{}5$ should give you slightly less than 10%, but the difference isn't something you'd spot in a reasonable sample.

Yes, now, when jmerry mentioned it, I see that your computational mistake cannot alone be responsible for such a big difference. Since I'm 100% sure that the "theoretical part" is correct, you have to check your numerical simulations. Check 3 things: 1) that the X sample is, indeed, distributed correctly; 2) that U is distributed correctly, and 3) that X and U are, indeed, independent. I suspect that you have wrong  X or U in your simulation but it may happen that somehow the independence fails and then, of course, you may get completely crazy results.

Thanks for spotting this mistake in my computation jmerry and fedja. Oddly enough, in this an other examples I tried, the actual value of $ \mu$ is close enough to the solution obtained by breaking the integral in two parts as in jmerry's post above. The biggest problem though was possibly a bug in the built-in random generator for the weibull distribution of the package I use. I just tried the exponential distribution equivalent, i.e. I used a random number generator for exp(2) instead of weibull(1,2) and everything turned out OK.
Thanks a lot for your help


I please all members of ML who know something about the problems and solution for this competition to write here a link or give some information. A greek student asked me about it, but unfortunately I can not find in net a source.
  Thank you in advance - Babis(Greece)

Saludos
Aqui te envio dos  links
Competencia mediterranea del 2008
http://www.oei.es/oim/revistaoim/numero32/CMM08PrbsAlt.pdf

Competencia mediterranea del 2009
http://www.oei.es/oim/revistaoim/numero35/MEDITERRNEA2009.pdf

Espero que te sirva

______________________________________________________________________________
Centro Juvenil de Entrenamiento Olimpico
http://www.cjeo-peru.com


function
induction
real analysis
real analysis unsolved

Suppose $ f$ and $ g$ are continuous functions on $ [0,1]$, with $ 0\leq f(x)\leq 1$ and $ 0\leq g(x)\leq 1$ for all $ x\in [0,1]$. Suppose further that $ f$ is increasing and that $ f\circ g \equal{} g\circ f$. Show that there exists a number $ l$, such that $ f(l)\equal{}g(l)\equal{}l$.

Right, I'm not sure about all of this... Working with this sequence I've worked with is a bit confusing. Graphing this helped me a lot, so I hope all possible errors will be fixable. Please let me know if you have any doubts about this solution, because vou're probably right. :lol:

Note that $ f^n$ means $ f\circ f\circ ... \circ f$, $ n$ times, obviously.

[hide="Solution (I hope; v2)"]Since $ g$ is continuous and $ g(1) \minus{} 1\leq 0\leq g(0) \minus{} 0$, $ g$ obviously has a fixed point ($ t\in\left[0,1\right]$ such that $ g(t) \equal{} t$). Now, we have $ f(t) \equal{} f(g(t)) \equal{} g(f(t))$, which means that $ f(t)$ is a fixed point of $ g$. Using obvious induction we get further relations: $ g(f^n(t)) \equal{} f(g(f^{n \minus{} 1}(t))) \equal{} f^n(t)$, meaning all $ f^n(t)$ are fixed points of $ g$.

We can WLOG assume that $ f(t)\geq t$. Then, using the fact that $ f$ increases, we get $ f^n(t)\geq f^{n \minus{} 1}(t)$.  If $ f^n(t) \equal{} f^{n \minus{} 1}(t)$ for some $ n$, we can take $ l \equal{} f^{n \minus{} 1}(t)$ and that $ l$ is a fixed point of both functions. Otherwise, we have $ f^n(t) > f^{n \minus{} 1}(t)$ and sequence $ t$, $ f(t)$, $ f^2(t)$ is certainly infinite.

Since sequence $ t$, $ f(t)$, $ f^2(t)$, ..., is bounded from above, it must have both a supremum and a limit in $ l\leq 1$. If $ f(l) > l$, since $ f$ is continuous, there also must exist some $ n\in\mathbb{N}_0$ such that $ f(f^n(t)) > l$. However, that is an obvious contradiction. Since $ f(l)\geq l$ ($ f(x) \minus{} x\geq 0$ for all members of our sequence), we have that $ f(l) \equal{} l$.

Also $ g(l) \equal{} l$ because $ g(x) \minus{} x$ is continuous and we can let $ x$'s to be members of the sequence $ t$, $ f(t)$, $ f^2(t)$, ..., where for each of them $ g(x) \minus{} x \equal{} 0$ holds. Thus, we are done.[/hide]


(a) Find a set $A$ of ten positive integers such that no six distinct elements of $A$ have a sum which is divisible by $6$.

(b) Is it possible to find such a set if "ten" is replaced by "eleven"?

As far as I know, "set" implies no repeating elements. Moreover, you don't have "six distinct elements" there.

[hide]
a. Well the idea is that we have $5$ elements that are $1 \pmod{6}$ and $5$ that are $0 \pmod{6}$. The set $\{1, 6, 7, 12, 13, 18, 19, 24, 25, 30\}$ sufficies.

b. Don't know a solution without casework at the moment...[/hide]

[hide="b"]if there are two numbers for each partity mod 3, that sum is equal to 0 mod 6, otherwise
1) the partity that occurs the least occurs no times, then by pp, there exists 6 of a partity, but that is divisible by 6

2) the partity that occurs the least occurs 1 time, the other partities cannot occur 6 times, so they both must occur 5 times
now it is just a bit of casework...[/hide]


algebra
polynomial
Rational Root Theorem

I'm having so much trouble with factor theorem extended, especially with factor equations fully. For example: 2x^3 + x^2 -x-1. Can someone help me step by step?

In which set do you want to factorize?  :?

[quote="king_23"]I'm having so much trouble with factor theorem extended, especially with factor equations fully. For example: $2x^3 + x^2 -x-1$. Can someone help me step by step?[/quote]

Can this equation even be factored? To find factor this equation completely, you need to find the possible rational roots from the rational root theorem, and then use synthetic division to test each possible root to see which is a factor of the equation.

The possible rational roots of this equation are $\pm1, \pm\frac{1}{2}$. Not one of them gives a remainder of 0 after synthetic division.


At Central Middle School the 108 students who take the AMC 8 meet in the evening to talk about problems and eat an average of two cookies apiece. Walter and Gretel are baking Bonnie's Best Bar Cookies this year. Their recipe, which makes a pan of 15 cookies, list these items:\\
$ 1\frac{1}{2}$ cups of flour, 2 eggs, 3 tablespoons butter,
$ \frac{3}{4}$ cups sugar, and 1 package of chocolate drops. They
will make only full recipes, no partial recipes.
\\ \\
The drummer gets sick. The concert is cancelled. Walter and Gretel
must make enough pans of cookies to supply 216 cookies. There are 8
tablespoons in a stick of butter. How many sticks of butter will be
needed? (Some butter may be left over, of course.)

Since 15 cookies require 3 tablespoons,
216 cookies require 43.2 tablespoons.
Since a stick of butter contains 8 tbs,
the answer is 6.


That basically means it is one less than a multiple of all those numbers. The smallest number where that works is $ 2520$. $ 2520\minus{}1\equal{}2519$

[quote="FullMetalNaruto"]What is the smallest whole number such that, when 10.
divided by each of 10,9,8, 7, ... ,2, gives a remainder
of 9,8, 7, 6, ... ,1, respectively?

 :stretcher:  gosh, my head hurts. :)[/quote]
[hide]We want a number that is equivalent to 1 in each modulo (1-9). This is equivalent to finding the LCM of the numbers 1 through 9 and then subtracting 1. Thus the LCM of 1-9 is 2520 and subtracting 1 gives 2519.
[/hide]

If I understand you correctly, I think you want the solution of this set of congruences. You use the chinese remainder theorme to solve them or you can take them to a webpage which will do it for you.

$ \\ x \equiv 9 \mod 10 \\ x \equiv 8 \mod 9 \\ x \equiv 7 \mod 8 \\ x \equiv 6 \mod 7 \\ x \equiv 4 \mod 5 \\  x \equiv 3 \mod 4 \\ x \equiv 2 \mod 3 \\ x \equiv 1 \mod 2 $

x = 2519

[quote="bobbym"]You use the chinese remainder theorme to solve them or you can take them to a webpage which will do it for you.[/quote]

Yeah, or notice that each of those congruences is equivalent to $ x\equiv \minus{}1$, so the smallest $ x$ will obviously be $ \text{lcm}(10,\ldots,2)\minus{}1$ (because the LCM is the smallest number congruent to 0 for all of those).


I think it's still wrong maybe you do need additional conditions try $a=b=\frac{1}{\sqrt 2}$

[quote="MoHaMeD"]$a=b=\frac{1}{\sqrt 2}$[/quote]

With those values, does the inequality turn into an equality???????

Haven't checked these values, but still, [b]shobber[/b]'s counter-example works with the added condition...

[hide="Here is the first method."]$a^2+b^2=1\ ,\ a>0\ ,\ b>0\Longrightarrow \left(\exists\right)t\in (0,1)\ ,\ a=\frac{2t}{1+t^2}\ ,\ b=\frac{1-t^2}{1+t^2}\ .$
${\frac{1-a}{b}+\frac{1-b}{a}\ge 2(\sqrt 2 -1)\Longleftrightarrow\frac{1-\frac{2t}{1+t^2}}{\frac{1-t^2}{1+t^2}}+\frac{1-\frac{1-t^2}{1+t^2}}{\frac{2t}{1+t^2}}\ge 2(\sqrt 2-1})\Longleftrightarrow$
$\frac{1-t}{1+t}+t\ge 2(\sqrt 2-1)\Longleftrightarrow t^2-2\left(\sqrt 2-1\right)t+\left(3-2\sqrt 2\right)\ge 0\Longleftrightarrow$
$\left[t-\left(\sqrt 2-1\right)\right]^2\ge 0$ what is true. We have equality if and only if $t=\sqrt 2-1$,
i.e. $a=b=\frac{\sqrt 2}{2}\ .$[/hide]

My solution is
$\frac{1-a}{b}+\frac{1-b}{a}+2=\frac{1-(a-b)}{b}+\frac{1-(b-a)}{a}\geq 2\sqrt{\frac{1-(a-b)^2}{ab}=2\sqrt 2}$
this way it's done .

My solution is
$\frac{1-a}{b}+\frac{1-b}{a}+2=\frac{1-(a-b)}{b}+\frac{1-(b-a)}{a}\geq 2\sqrt{\frac{1-(a-b)^2}{ab}}=2\sqrt 2$
this way it's done .

Thanks ,I really casual with your answer


1. A metal $ M$ forms the compound $ MCl_{3}$ which is a soluble salt. A stable current of $ 1.0 A$ is passed through solution of this compound for $ \text{1 hour 20 minutes 25 seconds}$. $ 0.87g$ os solid $ M$ is liberated.

Where is this solid liberated (anode or cathode) and what is the atomic mass?

2. A soluble salt $ MSO_{4}$ undergoes electrolysis using $ Pt$ electrodes. $ 0.0327g$ of $ M$ is liberated and it requires $ 0.02M$ of $ KOH$ solution, $ 50mL$ to neutralise the mixture.

What if the atomic mass of $ M$?

3. A soluble salt undergoes electrolysis and $ 0.32g$ of oxygen gas is obtained at the anode and $ 0.12g$ of a metal is obtained at the cathode. If the atomic mass of this metal is below $ 100$ what is its atomic mass?

[u]Just in case you didn't see[/u]: Carcul, don't be too eager to solve these problems. :lol:

[hide="hint"]
1.direct application of farady's first law knowing that the equivalent mass of the compound is one-third it's molar mass
2.1 mole of the salt requires two moles of $ KOH$ to neutralise so .find the no. of moles of $ KOH$ using $ n=\frac{MV}{1000}$
3.just a case by case analysis .there can be at  most 6 oxygen....
[/hide]

[hide="Answer for 1"]In the compound $ MCl_{3}$, because M is a metal then chlorine will be present in solution as chloride anions $ Cl^{-}$ and the metal M as (hydrated) tripositive cations $ M^{3+}$. So, the solid metal M must be liberated at the cathode, which is always the place of reduction:

$ M^{3+}(aq)+3e^{-}\longrightarrow M(s)$.

The total time in which the current was passed through the solution was $ \Delta t = 4825\, s$, and so the total charge supplied was $ Q = I \cdot \Delta t = 1.0\, C/s \times 4825\, s = 4825\, C$. Now let's use Faraday constant ($ F = 9.6485 \times 10^{4}\, C/mol$) to calculate the corresponding quantity of electrons:

$ n(e^{-}) = \frac{Q}{F}= \frac{4825\, C}{9.6485 \times 10^{4}\, C/mol}= 0.050\, mol$.

From the stoichiometry of the cathodic half-reaction it follows that 3 moles of electrons "produce" one mole of metal M. So the quantity of M produced was

$ n(M) = \frac{0.050\, mol}{3}= 0.017\, mol$,

and so the molar mass of M is $ M = \frac{0.87\, g}{0.017\, mol}= 52.2\, g/mol$ (the metal is probably Cromium).
[/hide]


Princeton college trigonometry calculus limit ARML function

[quote="jb05"]All from last year's tests:

1. Find the smallest positive integer that is a multiple of 18 and whose digits can only be 4 or 7.
2. Find all integers n such that $ 7n^{3}-3n^{2}-3n-1$ is a perfect cube.
3. There is a circle c centered about the origin of radius 1. There are circles $ c_{1}$, ..., $ c_{6}$, each of radius r, such that each circle is completely inside c and tangent to it and its two neighbors. There is a circle d which is tangent to c, $ c_{1}$, and $ c_{2}$, but does not intersect any of these circles. What is the radius of circle d?
4. Points P and Q are located inside square ABCD so that DP is parallel to QB and DP=PQ=QB. Determine the minimum possible value of the angle ADP.
5. Express $ \sqrt{7+r\sqrt{3}}+\sqrt{7-4\sqrt{3}}$ in simplest possible form.
6. Find the greatest integer less than the number $ \sum_{i=1}^{1000000}\frac{1}{\sqrt{i}}$.
7. The curve $ y=x^{4}+2x^{3}-11x^{2}-13x+35$ has a bitangent (a line tangent to the curve at two points). What is the equation of the bitangent?

More later when I have time.[/quote]

Number 7: Set $ y = ax+b,$ and note that when everything is moved to one side, the side that is non-zero is a square of a monic quadratic.

Hmm... these are some pretty nice problems. Do the Princeton kids make up the problems?

#1's really easy:

The number has to be divisible by $ 2$ and $ 9,$ so we know that the last digit of the desired number is $ 4.$ Since the digital sum of the number must be divisible by $ 9,$ we find that $ 7+7+4 = 18$ is the smallest such possible value, so our number is $ 774.$

Lastly, as a note which really has nothing to do with math, I feel sorry for your guys's forum... it gets assaulted a lot  :( Maybe Arnav can clear things up again  :P

Yes. Finally some math and not spam. I will post my solutions when I have time. Thanks jb05! :D

2. $ 7n^{3}-3n^{2}-3n-1 = (2n)^{3}-(n+1)^{3}$ and by Fermat's last theorem there are no integer solutions. :rotfl:

@not_trig: You are wrong.  Fermat's Last Theorem clearly states that if $ x^{n}+y^{n}= z^{n}$ for $ n > 2$ and $ x,y,z,n$ integers, then it follows that $ xyz = 0$.  Let $ m$ be such that $ m^{3}+\left(n+1\right)^{3}= \left(2n\right)^{3}$.

Now we must have $ m\left(n+1\right)\left(2n\right) = 0$.  If $ n+1 = 0$ we have $ n =-1$.  If $ 2n = 0$ we have $ n = 0$.  If $ m = 0$ we have $ 2n = n+1 \Leftrightarrow n = 1$.  Thus $ n = \left\{-1,0,1\right\}$.

[quote="Karth"]Hmm... these are some pretty nice problems. Do the Princeton kids make up the problems?[/quote]

I guess. But they're nice because I was pretty selective. Those were from individual tests. But I was kind of in a hurry, so here are calculus (mostly series) and combo (actually called advanced topics, and it contained a couple non-combo problems last year) problems and some team problems:

8. Find the smallest positive integer n such that 2n+1 and 3n+1 are both squares.
9. How many unordered pairs of positive integers (a,b) can be the legs of a right triangle with hypotenuse 340.
10. S is a subset of {1,2,3,...,100}. What is the maximum number of elements of S such that the product of no two elements is a perfect square?
11. Evaluate $ \sum_{n+1}^{\infty}\frac{1}{n^{2}(n+1)}$.
12. What is the greatest possible number of edges in a planar graph with 12 vertices?
13. Aaron has a coin that is slightly unbalanced. The odds of getting heads are 60%. What are the odds that if he flips it endlessly, at some point during his flipping he has three more tails than heads?
14. A stick of length ten is marked with nine evenly spaced marks (so each is one unit apart). An ant is placed at every mark and at the endpoints, randomly facing either left or right. Suddenly, all the ants start walking simultaneously at a rate of one unit per second. If two ants collide head-on, they immediately reverse direction (which takes no time). Ants fall of the stick as soon as they walk past the endpoints (so the two on the end don't fall off immediately unless they are facing outwards). On average, how long will it take, in seconds, until all the ants fall off the stick?
15. What is the maximum number of vertices in a graph with the property that each vertex has order three and the distance between any two vertices is at most two?
16. Evaluate the limit $ \lim_{n\to \infty}\sum_{k=n}^{2n}\frac{(-1)^{k}}k$.
17. Evaluate $ \sum_{n=1}^{\infty}\frac{n^{2}}{(n-1)!}$.
18. Evaluate $ \sum_{n=0}^{\infty}\frac{1}{(n+1)2^{n}}$.

I'll let people relax and do these and then maybe I'll post some power round stuff. Or you could always just look for the entire test, but it's nice to have the good problems filtered out.

I would also note that #7 looks like a classic ARML problem with different numbers and #2 looks like an old MOP problem...

In #5, should the first term be $ \sqrt{7+4\sqrt{3}}$?  Because then the answer is just 4.

In #3, I need some clarification.  So we have that $ d$ is tangent to $ c, c_{1}, c_{2}$.  Now does $ d$ not intersect $ c,c_{1},c_{2}$ or does $ d$ not intersect $ c, c_{1}, c_{2}, \cdots, c_{6}$?  "these circles" is a little ambiguous.  But once this is cleared up, it should be a simple application of Descartes' Four-Circle Theorem.

[hide="solution to 11"]
$ \frac{1}{n^{3}+n^{2}}= \frac{A}{n}+\frac{B}{n^{2}}+\frac{C}{n+1}$
$ 1 = A\left(n^{2}+n\right)+B\left(n+1\right)+Cn^{2}$

$ n =-1$
$ 1 = 0+0+C$

$ n = 0$
$ 1 = 0+B+0$

$ n = 1$
$ 1 = 2A+2B+C \Leftrightarrow A =-1$.

So we have $ \sum_{n+1}^{\infty}\frac{1}{n^{2}\left(n+1\right)}= \sum_{n+1}^{\infty}\left(\frac{-1}{n}+\frac{1}{n^{2}}+\frac{1}{n+1}\right) = \frac{-1}{n+1}+\sum_{n+1}^{\infty}\frac{1}{n^{2}}$

now I don't know a nice closed form for the latter expression, but w/e.  One might want to look to see if the terms are actually the Taylor series for some bizarre function, but I dunno.  The Taylor series approach might also work for 16, 17, and 18.

Extension: Find $ \sum_{i=1}^{\infty}\sum_{n=i+1}^{\infty}\frac{1}{n^{2}\left(n+1\right)}$.
[/hide]

[quote="Anthony Cai"]I dont get number 4.  My answer is $ tan^{-1}(1/3)$ but I dont know if thats correct since its pretty odd to express it that way.[/quote]

I don't think that's correct. I got [hide]$ 15^\circ$[/hide] as my answer.

Whoa oops yea thanks
P and Q are INSIDE the square

I had P on AB and Q on DC and thats how i got that wrong answer.

[quote="Teki-Teki"]In #5, should the first term be $ \sqrt{7+4\sqrt{3}}$?  Because then the answer is just 4.[/quote]

Oops.

[quote="Teki-Teki"]In #3, I need some clarification.  So we have that $ d$ is tangent to $ c, c_{1}, c_{2}$.  Now does $ d$ not intersect $ c,c_{1},c_{2}$ or does $ d$ not intersect $ c, c_{1}, c_{2}, \cdots, c_{6}$?  "these circles" is a little ambiguous.  But once this is cleared up, it should be a simple application of Descartes' Four-Circle Theorem.[/quote]

I copied the problem word for word. It means it doesn't intersect any of the circles except $ c, c_{1}, c_{2}$ each at a single point. It is between those three. And yes, though before last year's USAMO, nobody actually knew Descartes' Four-Circle Theorem ;).

[quote="Teki-Teki"]
$ \frac{1}{n^{3}+n^{2}}= \frac{A}{n}+\frac{B}{n^{2}}+\frac{C}{n+1}$
$ 1 = A\left(n^{2}+n\right)+B\left(n+1\right)+Cn^{2}$

$ n =-1$
$ 1 = 0+0+C$

$ n = 0$
$ 1 = 0+B+0$

$ n = 1$
$ 1 = 2A+2B+C \Leftrightarrow A =-1$.

So we have $ \sum_{n+1}^{\infty}\frac{1}{n^{2}\left(n+1\right)}= \sum_{n+1}^{\infty}\left(\frac{-1}{n}+\frac{1}{n^{2}}+\frac{1}{n+1}\right) = \frac{-1}{n+1}+\sum_{n+1}^{\infty}\frac{1}{n^{2}}$

now I don't know a nice closed form for the latter expression, but w/e.  One might want to look to see if the terms are actually the Taylor series for some bizarre function, but I dunno.  The Taylor series approach might also work for 16, 17, and 18.

Extension: Find $ \sum_{i=1}^{\infty}\sum_{n=i+1}^{\infty}\frac{1}{n^{2}\left(n+1\right)}$.[/quote]

Sorry, another typo. The subscript should say $ n=1$, not $ n+1$ (I'm not sure what the latter would mean anyway... $ \sum_{n=n+1}^{\infty}blah$ is apparently 0 because no terms are in the sum?). So yes it is $ \frac{\pi^{2}}6-1$.

EDIT: Ignore all the junk I posted for the solution to the extension (I mean, it was right, it was just silly). It seems I have forgotten how to do simple things.

I interpreted it as, "Find a closed form in terms of $ n$ for ...", and ignored the discrepancy.

Correction: The subscript on the second summation in the extension should read $ n=i$.  It makes the problem slightly prettier.

There are still two possible values for the radius of $ d$, however.

[hide="Solution to 18"]
Consider the function $ f(x) =-2\ln{\left(1-\frac{x}{2}\right)}$.  The Taylor series expansion of $ g(x) = \ln{\left(1+x\right)}$ is well-known, and by simple substitution one can see that we can say:

$ f(x) = \sum_{n=0}^{\infty}\frac{x^{n+1}}{2^{n}\left(n+1\right)}$.

But then consider $ f(1)$.  We have $ f(1) = \sum_{n=0}^{\infty}\frac{1}{2^{n}\left(n+1\right)}=-2\ln{\frac{1}{2}}= 2\ln{2}$.
[/hide]

Idea for 17: Use Taylor series as in 18, but consider instead a function of the form $ f(x) = P(x)e^{x}$ where $ P(x)$ is a polynomial of $ x$ and take the derivative.  Then take the second derivative.

These are some nice problems; I'll have to look at them.  Sorry that I have not been able to do my moderating duties here because I was at Awesome Math.  I will try to fix what needs to be fixed (please pm me if you have anything specific and/or I miss something).

[quote="Teki-Teki"]There are still two possible values for the radius of $ d$, however.[/quote]

No. The arrangement of the $ c_{i}$ is unique, and a triple of circles has only two tangent circles, one of which intersects the others... So it is definitely unique.

Hey who all is on your teams this year... and how many teams are you all bringing?

At the questions: I dunno, it seemed like they recycled a bunch of AIME questions, and then adjusted a bunch of USAMO/IMO problems to fit a short answer format. It was mildly disconcerting, though it made the NT test pretty simple. I would not be terribly upset if they kept the same format on some of the questions. Pretty nice problems though. I'm not sure I like the new format... with no calculus. Sucky.

Well, the number theory test was particularly standard. I liked the so-called "advanced topics" test. I didn't realize they changed formats, though.

EDIT: Apparently, they [u]radically[/u] changed the format. So I'm not sure of much now.

Whoops: jb05's clarification actually does uniquely determine $ d$; I was wrong.

Further hint for 17: Consider $ f(x) = xe^{x}$.  Add $ f'(x)$ and $ f''(x)$, and use Taylor series to show that this is indeed the desired sum.

Here is a way to make Teki-teki's power series approach into a general method.

Observation: 99% of series problems on HMMT/Princeton calculus contests come from simple generating functions.

Suppose we wish to compute $ \sum a_{n}$.  To do so, we compute the function $ f(z)=\sum a_{n}z^{n}$.  Later we just find f(1).

Now just notice that $ \sum (n+1) \cdot a_{n}z^{n}=\sum \frac{d}{dz}(a_{n}z^{n+1})=\frac{d}{dz}(\sum a_{n}\cdot z^{n+1})$

We can also run this in reverse as $ \sum \frac{a_{n}}{n}z^{n}=\sum \int a_{n}z^{n-1}= \int \sum a_{n}z^{n-1}$

Using a combination of these tricks we can pull any rational function out from under a sum.  What we're left with will almost always be the power series for $ e^{z}$ or $ ln(z)$ or something else recognizable.

When using the integral method one has to be careful, because constants of integration pop up.  In this case it is best to leave everything as definite integrals.  Derivatives are much more easily handled.

Here is an example of the integral method in action on number 18:

$ \sum \frac{1}{(n+1)2^{n}}=\sum \int_{0}^{1}\frac{z^{n}}{2^{n}}= \int_{0}^{1}\sum \frac{z^{n}}{2^{n}}= \int_{0}^{1}\frac{1}{1-\frac{z}{2}}=2ln(2)$

Here is the derivative method applied to number 17:

We wish to calculate $ \sum_{1}^\infty \frac{n^{2}}{(n-1)!}z^{n-1}$

This is $ \frac{d}{dz}\sum_{1}^\infty \frac{n}{(n-1)!}z^{n}= \frac{d}{dz}( z \sum_{1}^\infty \frac{n}{(n-1)!}z^{n-1})$

Now we do it again to get $ \frac{d}{dz}(z \cdot \frac{d}{dz}( \sum_{1}^\infty \frac{z^{n}}{(n-1)!}))=\frac{d}{dz}(z \cdot \frac{d}{dz}(z \cdot e^{z}))$

The rest is easy.

Now if you can already see that the result will be a polynomial times $ e^{z}$, then teki-teki's way is slightly faster.  It is usually harder to predict what series will show up at the end if you must use the integral method.

To finish off #16, just note that the nth term in the sequence is strictly between $ \frac{-1}{n}$ and $ \frac{1}{n}$.  Thus the limit is $ 0$.

Fastest way for #17: $ \frac{(n+1)^{2}}{n!}=\frac{n}{(n-1)!}+\frac{2}{(n-1)!}+\frac{1}{n!}=\frac{1}{(n-2)!}+\frac{3}{(n-1)!}+\frac{1}{n!}$. Noting that the terms with a negative factorial on the bottom had a zero in the numerator at some point, this sums to $ 5e$.

And #16 is kind of silly, but the official solution is basically $ \sum_{i=1}^{\infty}\frac{(-1)^{i}}{i}$ converges, so any subsequence starting at an arbitrarily distant term converges to 0.

My point is... the integral/derivative method usually works, but is usually not the one-line solution.

Here is an interesting question from a former hmmt,

Every second, Andrea writes down a random digit uniformly chosen from the set {1,2,3,4}. She stops when the last two numbers she has written sum to a prime number. What is the probability that the last number she writes down is 1?

OK, I've solved a couple...

Is there a "slick" way to do #8?  The answer is trivial, but the only solution I have found is mod bashing to limit the searching  :( Actually, I have several mod bashing solutions, but whatever.

[hide="mod bashing on #8"]Let $ 2n+1=p^{2}$ and $ 3n+1=q^{2}$.  Thus, $ n=q^{2}-p^{2}\Rightarrow p^{2}=2q^{2}-2p^{2}+1$.

$ 3p^{2}=2q^{2}+1$

taking this mod 2, we see that $ p\equiv1\pmod2$, so let $ p=2k+1$.  Then, the equation becomes $ 12k^{2}+12k+3=2q^{2}+1$
$ 6k^{2}+6k+1=q^{2}$

Taking this mod 8, we can see that $ k\equiv0,4,7\pmod{8}$.

Checking these, we have:

If $ k=0$, then $ p=1$ and $ n=0$, which is a contradiction.

If $ k=4$, then $ p=9$, $ n=40$, and $ q=11$, so we are done.[/hide]

Note that you can limit the search a LOT more with mod bashing, but I left those parts out because they don't save you any time in finding the answer anyway.

I also have a solution to $ \#3$, I think, which gives
[hide]$ d=\frac{2\sqrt3-3}9$[/hide]
and I might post it later (got to go to bed now).

Just a comment - if you are specifically practicing for PUMaC this year, the format has changed.  There will, for example, be no calculus questions.  For a complete description of this year's PUMaC, see http://cgi.math.princeton.edu/mathclub/index.php/Princeton_Math_Competition

However, if you just want to do some cool math, by all means, try the calculus questions from last year. :)

[hide="solution to #3"][asy]unitsize(200); pair A,B,C,D; A=origin; B=(-1,-1.73250808); C=(1,-1.73250808); D=(0,-2.645561911); dot (B^^A^^C^^D); draw(A--B--C--A--D--B); draw(D--C); draw(circle(A,3)); draw(circle(B,1)); draw(circle(C,1)); draw(circle(D,0.3544380888)); label("\small { $O$} ",A,(0,1)); label("\small { $B$} ",B,(-.5,-.5)); label("\small { $C$} ",C,(.5,-.5)); label("\tiny { $D$} ",D,(0,-.7));[/asy]

Let the intersection of $ \overline{OD}$ and $ \overline{BC}$ be $ E$ (there is no room in the diagram).

Note that $ \overline{OD}\perp\overline{BC}$, and that $ m\angle BOC \equal{}\frac{360}{6}^{\circ}\equal{} 60^{\circ}$.  Thus, $ \frac{r}{OB}\equal{}\frac{1}{2}$, and $ r\plus{}OB \equal{} 1$, so $ r \equal{}\frac{1}{3}$.  From this, we can see that $ OE \equal{}\frac{\sqrt3}{3}$.

Note that $ 1 \equal{} OE\plus{}ED\plus{}d \equal{}\frac{\sqrt3}{3}\plus{}d\plus{}ED$ (*).

$ ED^{2}\plus{}EB^{2}\equal{} BD^{2}\Rightarrow\left(\frac{1}{3}\right)^{2}\plus{}ED^{2}\equal{} (\frac{1}{3}\plus{}d)^{2}$

$ ED \equal{}\sqrt{d^{2}\plus{}\frac{2d}{3}}$

Substituting into (*), we have:

$ 1\minus{}\frac{\sqrt3}{3}\equal{} d\plus{}\sqrt{d^{2}\plus{}\frac{2d}{3}}$

$ (1\minus{}\frac{\sqrt3}{3}\minus{}d)^{2}\equal{} d^{2}\plus{}\frac{2d}{3}$

$ (1\minus{}\frac{\sqrt3}{3})^{2}\minus{}2d(1\minus{}\frac{\sqrt3}{3}) \equal{}\frac{2d}{3}$

$ 2d\left(\frac{4}{3}\minus{}\frac{\sqrt3}{3}\right) \equal{}\frac{4}{3}\minus{}\frac{2\sqrt3}{3}$

$ d \equal{}\frac{\frac{4}{3}\minus{}\frac{2\sqrt3}{3}}{2\left(\frac{4}{3}\minus{}\frac{\sqrt3}{3}\right)}$

$ d \equal{}\frac{5\minus{}2\sqrt3}{13}$[/hide]

Hmm this is exactly what I did. But the official answer is $ \frac{14\minus{}6\sqrt{3}}{37}$. I know there was some error in the geometry solutions... maybe this was it. Also, major double typo here (but its fixed in the next statement):

[quote="b-flat"]$ ED^{2}\plus{}EB^{2}\equal{} BD^{2}\Rightarrow d^{2}\plus{}EB^{2}\equal{} (\frac{1}{3}\plus{}d)^{2}$[/quote]

[hide="as for #8"]take it mod 5 and 8 and you will find it must divide 40. And 40 works.[/hide]

And yes, we know the format has changed this year, but not until I had posted these problems.

@typo:  now fixed  :blush:

@8:  I was just wondering if anyone had a solution that would give all solutions; so if the problem asked for the 3rd solution, you would be fine.  I got that $ 40|n$ as well, however, you can't produce any more solutions easily given that...

Oh, I get what you're looking for.

Define $ a_{0}\equal{}1$, $ a_{1}\equal{}9$, and $ a_{k\plus{}2}\equal{}10a_{k\plus{}1}\minus{}a_{k}$. Define $ b$ with the same recursion and initial values 1, 11 instead of 1, 9. I think the integers $ n$ are exactly those satisfying $ 2n\plus{}1\equal{}a_{k}^{2}$ and $ 3n\plus{}1\equal{}b_{k}^{2}$ for some $ k$ (note that the conditions involving $ a$ and $ b$ are really the same condition, but are both included for completeness). It's easy to prove that for every $ k$, there is an integral $ n$ satisfying those conditions... [hide="as so:"]I claim that the set of all $ n$ satisfying the conditions of the problem contains at least an integer $ n_{k}$, for each $ k$, satisfying the two conditions. Since the characteristic equations of $ a$ and $ b$ are the same (say they have roots $ r_{1}$, $ r_{2}$), we may write $ a_{k}\equal{}cr_{1}^{k}\plus{}dr_{2}^{k}$ and $ b_{k}\equal{}er_{1}^{k}\plus{}fr_{1}^{k}$. Note that $ \frac{a_{k}^{2}\minus{}1}{2}$ is an integer ($ a_{k}$ is odd by a trivial induction), so if $ n_{k}$ exists, it is integral, and it exists if and only if $ \frac{a_{k}^{2}\minus{}1}{2}\equal{}\frac{b_{k}^{2}\minus{}1}{3}\Rightarrow 3a_{k}^{2}\minus{}1\equal{}2b_{k}^{2}$. Using our closed form definitions along with $ r_{1}r_{2}\equal{}1$, we want $ 3c^{2}r_{1}^{2k}\plus{}3d^{2}r_{2}^{2k}\plus{}6cd\minus{}1\equal{}2e^{2}r_{1}^{2k}\plus{}2f^{2}r_{2}^{2k}\plus{}4ef$. By computing $ c$, $ d$, $ e$, and $ f$, we verify this, as indeed $ 3c^{2}\equal{}2e^{2}$, $ 3d^{2}\equal{}2f^{2}$, and $ 6cd\minus{}1\equal{}4ef$.[/hide]
I cannot figure out how to prove that this yields every $ n$. It may not even be true, but I'm fairly sure it is. It is definitely true that the first few possible values of $ n$ are 0, 40, 3960, 388080, 38027920, 3726348120. I don't know if there are additional solutions beyond those (besides those generated by the formula).

Yeah, we write the questions. The calculus test was written by people who just finished probability theory; hence all the sums. This year they got rid of it.

So is NC/SC sending a team this year? I told Princeton to contact Noland at Enloe and Teague at NCSSM about organizing a trip, so you could contact them saying you're planning to go and wondering if they could get some sort of mass transportation (you can also contact john morrison who loves to laugh at people and call them funny names).

teague@ncssm.edu
morrison@ncssm.edu

sorry, dunno noland's email.

Umm everything is already planned. In fact, we should have already registered.

Noland and Teague (I don't know who Teague is) probably do not know this. Hopefully they aren't too confused.

Noland has no idea; Teague is some guy at NCSSM.  Seeing as I'm the only one from Enloe, and if Ray is going he's the only one from NCSSM, I think that Teague probably doesn't know.

Hey Pakman,

Note that this is the North Carolina Forum.  Why would I drive down to Florida just to meet up with a B team and then drive all the way back up through North Carolina along I-95 to New Jersey?

No, if I didn't have a team I would try to hook up with someone in Virginia or Maryland.  But at any rate, all (I think) of the people working on the problems here are on the NC A team.

you could alternatively not go to florida first

just sayin'

[quote="Pakman2012"]if uh, anyone doesn't have a team, you guys can come on the florida B team. let me know.[/quote]

We all have teams. I suggest you look in Florida to fill up the Florida B team. Or else change the name to "assorted people from everywhere" or some such.

[quote="not_trig"]So guys, should we get together in our respective subjects and do some "studying"?[/quote]

I will repeat my question, rephrased slightly differently: geometry people, who has a good problem?

(and NT people :wink: )

(and everyone else)

[quote="akech"]
$ \equal{} \frac {\pi^{2}}{6} \plus{} \lim_{m\to \infty}[{\frac {1}{m \plus{} 1} \minus{} 1}] \equal{} \minus{} \frac {\pi^{2}}{6}$[/quote]

$ \equal{}\frac{\pi^2}{6}\minus{}1\plus{}\lim_{n\rightarrow\infty} \frac{1}{m\plus{}1} \equal{} \frac{\pi^2}{6}\minus{}1\plus{}0 \equal{} \frac{\pi^2}{6}\minus{}1$ as was found about twenty-five posts ago.

Thank you for pointing that out. I do not know what I was thinking. Certainly it was a good mistake, don't you think so? Also I did not read other's people posts in full so I did not see the answer to this question.

For number 12: One can use the following fact: If G is a connected planar graph with the number of edges more than one, then the number of edges is less than or equal 3 times the number of vertices minus 6. So the answer to number twelve will be 30.


What about that?

The number's small enough to see that factors include primes: 43, 107, 179.


superior algebra
superior algebra unsolved

Let $ G$ be a finite semigroup.Prove that there exists $ a\in G$ such that $ a^2\equal{}a$

* It suffices to show this for some cyclic subsemigroup $ x, x^2, x^3,\ldots$.
* For some minimal $ i<j$, $ x^i \equal{} x^j$.
* Otherwise let $ k\equal{}l(j\minus{}i)$ be the largest multiple of $ j\minus{}i$ less than $ j$, so $ 2k \minus{} l(j\minus{}i) \equal{} k < j$ but $ 2k \minus{} (l\minus{}1)(j\minus{}i) \equal{} k \plus{} j\minus{}i \geq j$.
* Let $ a\equal{}x^k$.  Then $ a^2 \equal{} x^{2k} \equal{} x^{2k\minus{}l(j\minus{}i)} \equal{} x^k\equal{}a$.

Maybe it is a bit clearer way:
Let $ x^k \equal{} x^{k\plus{}d}$.
Choose $ i \in {0,\dots,d\minus{}1}$ such that $ d|k\plus{}i$ (can be always found).
$ (x^{k\plus{}i})^2\equal{}x^{k\plus{}i\plus{}k\plus{}i} \equal{} x^{k\plus{}i\plus{}nd} \equal{} x^{k\plus{}i}$.


Find the values of real value $a$ and  positive integer $n$ such that

$\left( \begin{array}{ccc}1& 2& a \\ 0& 1& 4 \\ 0& 0& 1 \end{array}\right)^{n}=\left( \begin{array}{ccc}1& 18& 2007 \\ 0& 1& 36 \\ 0& 0& 1 \end{array}\right).$

sorry, i have edited it.
is that right? :maybe:


what is the different betwin viola and violin...thanks ;)

The viola is thicker than the violin, and the sizes are measured differently (in a violin, they are numbered like full size, 1/2 size, 1/4 size, etc., and violas are measured by some other way).

also, the viola has the strings C, G, D, and A, while the violin has its strings lettered G, D, A, and E.

the viola plays lower notes than the violin. They're between the violin and the cello.

Violas are measured by the length of the main body- mine is 16 inches, a standard adult size. On this scale, a full-size violin would be about 12 or 13 inches.

The skills used to play either violin or viola are easily transferred, aside form the reading issue, and nearly all violists are former violinists. There are subtle differences- a viola responds more slowly than a violin, making technical passages significantly more difficult. On the other hand, a viola can create a richer tone than a violin, mostly because of its size and lower pitch.

A viola is a fifth lower than a violin, and an octave higher than a cello, so we get music transposed from both instruments. I prefer the cello music to the violin music, partly because it's easier to play on a viola.

jmerry referred to the reading issue. What's that all about? Violin music is written strictly in treble clef. (Extremely high passages may rarely get the "8va" notation.) Piano music is written in treble and bass clefs. But the home clef for a violist is something called alto clef. In this clef, middle C is the middle line. You could imagine it in a piano score pasted right in the middle so it overlaps both clefs. The lowest possible note on a viola, the C below middle C, is just slighlty below the bottom of that clef. High passages are frequently written with a clef change to treble clef.

It takes some getting used to; it took me several years.

However, if you as a music editor put a chunk of the viola part in treble clef  that doesn't have to be there, just because alto clef makes you uncomfortable, then the entire viola section would curse your name if they knew it.

One little thing violins and violas have in common is a bit of even-odd parity. If you're in the more common odd-numbered positions (1st, 3rd, or 5th position), then notes writted on lines are played with odd fingers and notes written on spaces are played with even fingers, including the "0" notation for open strings. 2nd, 4th, and 6th positions are harder to read in becuase they reverse parity. (This talk of odd-even parity makes no sense to a cellist.)

[quote="Lil_mad_dog"]and violas sound better when violinists are playing them[/quote]

Fixed   ;)


What is the least possible value of the expression $\frac{9a^2}{b} + \frac{243}{ab} + \frac{9b^2}{a}$ if $a$ and $b$ are positive integers?

[hide="hint"]

AM-GM works pretty well here :)
[/hide]

Wait, how do you arrive at 81 from that inequality. Sorry, I understand less about the AM-GM than either of you. Shyong may just get the gold medal for the NSW Math Competition this year.. haha.

[quote="robinhe"][hide]$\frac{\frac{9a^2}{b} + \frac{243}{ab} + \frac{9b^2}{a}}{3}\ge \sqrt[3]{\frac{9a^2}{b} \times \frac{243}{ab} \times \frac{9b^2}{a}}$ from the AM-GM inequality.

So $\frac{\frac{9a^2}{b} + \frac{243}{ab} + \frac{9b^2}{a}}{3}\ge \sqrt[3]{3^9}$

$\frac{\frac{9a^2}{b} + \frac{243}{ab} + \frac{9b^2}{a}}{3}\ge 27$

${\frac{9a^2}{b} + \frac{243}{ab} + \frac{9b^2}{a}}\ge 81$

Thus the minimum is $81$[/hide][/quote]

Nice

That's what I did too

[quote="surge"][quote="JavaMan"][hide]
This would be my first time using AM-GM, so please forgive mistakes :P

$27\frac{\frac{a^3}{ab} + \frac{27}{ab} + \frac{b^3}{ab}}{3} \ge 27(\frac{27a^3b^3}{a^3b^3})^{\frac{1}{3}}$

So it's minimum is $81$

[/hide][/quote]

How/Why was that 27 on the LHS randomly factored out?[/quote]

Hm, well you can factor a 9 out, so i factored that out, and multiplied by 3/3.

[quote="10000th User"]A word of caution for people using AM-GM 'mindlessly': you have to check if each of the summands can actually be equal to each other so that you can achieve minimum. In this problem, yes everything worked out nicely.[/quote]

An example of where it doesn't work would be good :)

[hide]
well for this I just noticed a and b were going to have to be 2 or 3 right away because if they were any higher the squared on the top would be to much and if they were 1 the the 243 would be too much

so I just checked both in a couple seconds and got 81[/hide]

Nice solution, robinhe. Btw, I noticed that the expression was symmetrical in a and b, so a=b=x for some x would give a minimum. So you can plug in values from a=b=1 onwards as Rowan did with 2 and 3.

are the questions from this years competition? if so, you are cheating...

what is the NSW maths competition anyway? is the UNSW?

Yes it is the UNSW. It's now known as the Mathematics Assessment from Educational Assessment Australia. So of the questions are interesting. That's from the past year questions... there's no way I could know this years'.

Just for sidenotes, there are several things to consider in this type of problem.

1. Make sure variables are positive reals. If they're not, AM-GM does not work (try it for yourself).
2. Some people might wonder, "How come I didn't think of that?" Well, in this case, it was pretty obvious because you want to simplify the variables and if they're multiplied, the job's accomplished, However, there are problems where additional investigations are necessary before going to use AM-GM.
3. For more practices, check out HMMT Algebra section. I remember seeing some AM-GM on earlier part of the exams.

What's the HMMT? Could you provide a link?


Hello everyone, I am new to this forum. The problem I am having is with a story problem. I actually already solved it, so I think. I just want to be sure that my answer is correct. I am pretty good at Algebra but story/word problems cause me difficulty.

The Question:
 A financial advisor suggests to a client that she invest part of her portfolio in municipal bonds earning 6.5% annual simple interest, and the remainder in corporate bonds earning 8.5% interest. How much should she be putting in the corporate bonds if she has 2,190 in interest each year?

The problem is incomplete. How much is the total amount she is investing?

When you find that piece of information (call it $ T$) you can set up the equation like this:

$ 0.065x+0.085(T-x)=2190$


A block of cheese in the shape of a rectangular solid measures $ 10$ cm by $ 13$ cm by $ 14$ cm. Ten slices are cut from the cheese. Each slice has a width of $ 1$ cm and is cut parallel to one face of the cheese. The individual slices are not necessarily parallel to each other. What is the maximum possible volume in cubic cm of the remaining block of cheese after ten slices have been cut off?

[hide="Not sure?"] To keep the max possible volume, shave off the side with the minimum area: so we go 10x13x14 to 10x13x13 to 10x12x13... eventually after 7 cuts to 10x10x10, and then after 10 to 9x9x9.
Thus our answer is [b]729[/b].

[/hide]

[hide="More rigorous solution"]
Define the lengths of the three sides $ 10$, $ 13$, and $ 14$ after we cut them as $ a$, $ b$, and $ c$ respectively. Since we make $ 10$ cuts, $ a\plus{}b\plus{}c\equal{}10\plus{}13\plus{}14\minus{}10\equal{}27$. By AM-GM, $ \frac{a\plus{}b\plus{}c}{3}\ge\sqrt[3]{abc}$ or $ abc\le729$. Equality is reached when $ a\equal{}b\equal{}c\equal{}9$, so $ \boxed{729}$ is the maximum.
[/hide]

[hide]For every slice, we decrease the length of some dimension, and only one dimension, by 1. Thus, the volume is $ (10 \minus{} a)(13 \minus{} b)(14 \minus{} c)$ where $ a \plus{} b \plus{} c \equal{} 10$. $ a \equal{} 1,b \equal{} 4,c \equal{} 5$ approximates a cube most closely (is a cube) so $ \boxed{729}$.[/hide]

[hide="same old same old"]well you know that the maximum volume of a figure whose dimensional sum is given is created when the figure is closest to a cube. each cut will subtract $ 1$ from the dimensional sum so its just : $ 13\plus{}14\plus{}10\minus{}10\equal{}27$ for the sum, now to make that a cube just make it $ 9\times9\times9 \equal{}\fbox{729}$[/hide]

[hide="Answer"]
We need to maximize the area, so we have $ (10\minus{}a)(13\minus{}b)(14\minus{}c)$, where we want $ 10\minus{}a$ to be close to $ 13\minus{}b$ and $ 14\minus{}c$. We can see that by subtracting from each of the sides, we have something like:

10x13x14
10x13x13
10x12x13
10x12x12
10x11x12
10x11x11
10x10x11
10x10x10
9x10x10
9x9x10
9x9x9

Answer is then $ \boxed{729}$.[/hide]

[hide=Solution]
We see that the volume of the cheese becomes $(10-x)(13-y)(14-z)$ for some $x$, $y$, and $z$ such that $x+y+z=10$.

We use AM-GM to get that $10-x+13-y+14-z\ge 3\sqrt[3]{(10-x)(13-y)(14-z)}$ so $37-(x+y+z)\ge 3\sqrt[3]{(10-x)(13-y)(14-z)}$.

We know that $x+y+z=10$ so we can substitute to get that
\begin{align*}
27 &\ge 3\sqrt[3]{(10-x)(13-y)(14-z)} \\
9 &\ge \sqrt[3]{(10-x)(13-y)(14-z)} \\
\boxed{729} &\ge (10-x)(13-y)(14-z). \\
\end{align*}
[/hide]

amctrivial grind

The total is $(10-x)(13-y)(14-z)$ with $x+y+z=10$.

By AM-GM, this is less than or equal to \[\left(\frac{10-x+13-y+14-z}{3}\right)^3=\left(\frac{27}{3}\right)^3=\boxed{729},\] which is achievable which $x=1,y=4,z=5$.


prove that(elementary prove)
$\lim_{n\rightarrow\infty}\frac{(2\sqrt[n]{n}-1)^{n}}{n^{2}}=1$


Thanks JBL.  I didn't know that is what I needed in order to get those symbols in there with out them changing after I post them.

So, are you saying my work is correct?

I wasn't saying that, but if I had been, I would have been telling the truth ;)

You can read about LaTeX in the LaTeX guide (there is a link at left in the AoPS skin) to get the basics (skip the parts about downloading things and formatting documents -- you don't need those on the forums; all you really need to know is some of the basic codes, like \cdot for $\cdot$ and \frac{a}{b} for $\frac{a}{b}$).  Then you just put your code between dollar signs and it comes out looking very snazzy.  You can practice in the LaTeX forum.


Yale college Harvard Stanford MIT Columbia percent

Hello guys,
This is my short resume:
Average GPA: 5.67 (the best is 6.00)
Participated in summer program
worked for an IT company for 3 months
EC:
Teaching Mathematics Club - Founder and President
Webdesign Club - Co-President
Member of Model United Nations
Member of the basketball and soccer team
Member of the Math team of my school (about 4-5 awards)
Publications of pics and works in the literature magazine of my school
Member of the Business Club
Member of the Snowboard Team

SAT 2150/2400
SAT II Physics 800, Math 800, Chemistry 800
AP English Literature - 5

Universities I consider:
Yale
Harvard
Stanford
MIT
Cornell
Brown
Cambridge
Columbia

Honestly, I would put them all at reaches.  You've provided a laundry list of EC's but unfortunately, they appear disjointed and with the exception of the 4-5 math awards, you haven't displayed any evidence of standing out.  In makes it look as though you are a serial joiner hoping to pad your resume.  Also, although your SAT scores are fine (not amazing, but certainly respectable), you have only one AP course.  Is that because your school only offers that one or did you choose not to take any others?

I'm not being mean, just honest.  Look for safety schools.

With admissions rates under 10 percent (well, some of them), there is no good reason to [i]expect[/i] to get into any one of the schools you listed, unless you are from an encouraged demographic group. This holds for even the best students.

Having said that, you are obviously not going to send that laundry-list resume to any college. I know it will look much more attractive when you do it for real. However, until you do it for real, we cannot give you any meaningful feedback. There are many resumes that, when reduced to a laundry list, will look like that. Here are some things you can do to elevate your resume and not be boring.

Focus on what you did, interesting experiences you've gained, results you have achieved (not necessarily in terms of awards), and skills that you have. For example:
- give some detail about the IT job. What were your responsibilities? Did you make any major contributions? Make sure to place this in a separate "Employment" category!
- include a few bullets regarding your responsibilities and innovations for math club and things that you were very actively involved in
- rather than saying that you were in a web design club (which sounds highly unimpressive), you can cite proficiency in HTML, CSS, JavaScript (maybe?) as skills, and link various webpages that you've created for yourself and for others.
- what is business club? Did anything come as a result of it? Did you make money off stocks? If you made anything substantial, write it here!
- snowboarding club sounds useless. Instead, list snowboarding as one of your major hobbies (yes, you can create a hobbies section on your resume, separate from "serious" (academic, athletic, etc) activities
- some schools accept attachments. If you have any interesting pics or literary works that you think are representative of your work, send them!

Catch the drift?

Oh, and retake the SAT's if you can, though your score is good as it stands. Aim to get at least a 2200, with emphasis on the Critical Reading section, which is the most important section. Someone who gets 800ma, 800wr, 700cr = 2300 will probably be a bit less impressive than someone who gets 750-all = 2250. While SAT's won't make or break your profile in themselves, they can provide that helpful boost if the person evaluating your application is on the edge.

Yes, I am pretty good in swimming and really talented in wrestling, but I am from Europe. I think this makes the recruiting pretty much impossible :?

[quote="tokenadult"]Are you good enough in any of your sports to be a recruited athlete?[/quote]

If a person is really good in a particular sport, but is from Eastern Europe, what is needed to be recruited?

I would say it's considerably harder.  You'd probably need international, or at least national, recognition.


[quote="eminem"]hey guys , get this within $1$ minute , ok ??? thats the maximum time the question deserves , its sooooooooooooooooooooooo stupid

$x,y,z$ are real and non-negative

$x+y+z=1$

find minimum value of (1+$1/x$)(1+$1/y$)(1+$1/z$)[/quote]

just holder....

$(1+\dfrac{1}{x})(1+\dfrac{1}{y})(1+\dfrac{1}{z})\geq(1+\dfrac{1}{\sqrt[3]{xyz}})^3\geq(1+\dfrac{3}{x+y+z})^3=64$

or.....

you expand $(1+\dfrac{1}{x})(1+\dfrac{1}{y})(1+\dfrac{1}{z})$ and use cauchy....

 :)  :D

[hide]I got 64.  Not sure if its right though.  I thought that maybe since it never specified that x,y, and z were are different numbers, I just put them equal to 1/3.
So 1/3+1/3+1/3=1
And
(1+1/1/3)(1+1/1/3)(1+1/1/3)
=(1+3)(1+3)(1+3)
=4^3
=64[/hide]

Is that the wrong way to solve this problem? :blush:   Is it even right?

[quote="riddler"]but they are integers? that is what real means isnt it? :?[/quote]

$x\in\mathbb{R}$ means that x is any number that is on the real line...it could be a decimal number.  In this case, $x\in\mathbb{R}$ and x is a nonnegative number.

Masoud Zargar

$(1+\dfrac{1}{x})(1+\dfrac{1}{y})(1+\dfrac{1}{z})$
$\dfrac{(x+1)(y+1)(z+1)}{xyz}$

application of AM-GM inequality to the numerator and the denominator yields the solution $64$

Or if we know intuitively that the product is maximized when the terms are equal, the problem is trivial.

An increase in one term results in a decrease in another term, which decreases the total product from the equality case.


It's not very difficult.You can use TREBUSEP and use ma^2+mb^2+mb^2=3/4(a^2+b^2+c^2) and ma+mb+mc=<9/2R

See also

http://www.mathlinks.ro/Forum/viewtopic.php?t=13032
http://www.mathlinks.ro/Forum/viewtopic.php?t=16023

  darij


[color=darkblue][u][b]Problem 1[/b][/u]

Let $ a\neq 0$ and $ 3a \plus{} 4b \plus{} 6c \equal{} 0$. Prove equation: $ ax^2 \plus{} bx \plus{} c \equal{} 0$ exist root $ x_0 \in (0;1)$

[u][b]Problem 2[/b][/u]

Let $ a\neq 0$ and $ 5a \plus{} 4b \plus{} 6c \equal{} 0$. Prove equation: $ ax^2 \plus{} bx \plus{} c \equal{} 0$ exist root $ x_0 \in (0;2)$[/color]

Oh! No body say about this problem? Is it hard?

[quote="thanhnam2902"][color=darkblue][u][b]Problem 1[/b][/u]

Let $ a\neq 0$ and $ 3a + 4b + 6c = 0$. Prove equation: $ ax^2 + bx + c = 0$ exist root $ x_0 \in (0;1)$

[/color][/quote]

Solution:

Let $ f(x)=ax^2+bx+c$

$ \bullet$ If $ c=0$, equation become $ ax^2+bx=0 \Longleftrightarrow  [ \begin {array}{c} x=0 \\ x = -\frac{b}{a}=\frac{3}{4} \in (0;1) \end {array}$

$ \bullet$ If $ c\neq 0$, we have $ f(0). f ( \frac{3}{4} )= c. (\frac{9a}{16}+\frac{3b}{4}+c )$
$ = \frac{c}{16}. ( {9a+12b+16c} ) =\frac{c}{16}. ( {-18c+16c} ) =-\frac{c^2}{8}<0$

$ \Longrightarrow f(x) = 0$ have root $ x_0 \in (0; \frac{3}{4}) \Longrightarrow x_0 \in (0; 1)$

Integration kills them, but we can't use it in this forum.

How to use integration?

(By the way, what's the source of the problems?)

Sorry, my idea couldn't apply to the problem. :blush:


algebra
polynomial
number theory
number theory unsolved

I am trying to solve a difficult - I think- problem in Number Theory, which is as follows:

Let a polynomial p(x). The number "roots mod m" ( =p(m) ) is the subset of Zm which satisfy p(x)=O(mod m). Prove that                  p(m*n)=p(m)*p(n) if gcd(m,n)=1.

If you have seen this problem in a book or a site please tell me what book (or site) it is.

Thank you!  :lol:

If i understood it correctly, it's only chinese remainder theorem.
(Since $p(x) \equiv 0 \mod m,n \Leftrightarrow p(x) \equiv 0 \mod mn$ )


Find all of the function's features, and an explanation of all unusual properties, and everything concerns limits and continuity.
The function is: f(x)=int(sin(x))

And also analyze the function property and end behavor, anyone can give me a clue?


Let x,y,z be positive real numbers satisfy
$ 2\sqrt{xy} \plus{} \sqrt{xz} \equal{}1$
find the minimum value of
$ S\equal{}\frac{3yz}{x}\plus{}\frac{4zx}{y}\plus{}\frac{5xy}{z}$

Maybe you can help me then.

By AM-GM for $ S$, $ S \ge 3 \sqrt [3]{60xyz}$.  Rewriting $ S \ge 3 (\frac {xy}{z} \plus{} \frac {yz}{x} \plus{} \frac {xz}{y}) \plus{} \frac {xy}{z} \plus{} \frac {2xy}{z}$ allows AM-GM on these two to get $ S \ge \sqrt [3]{xyz} \plus{} 2x\sqrt2$.

AM-GM for the original equation given yields $ \frac {1}{2} \ge \sqrt {2x\sqrt {yz}}$.  Simplifying yields $ \dfrac{1}{64x} \ge xyz$.

If we substitute so all variables are $ x$, we can solve and find the minimum, but I don't have a good method for solving for $ x$.

I found the another way to solve the problem, we have
 $ S\equal{}\frac{3yz}{x}\plus{}\frac{4zx}{y}\plus{}\frac{5xy}{z}\equal{}(\frac{4xy}{z}\plus{}\frac{2zx}{y}\plus{}\frac{2yz}{x})\plus{}(\frac{xy}{z}\plus{}\frac{2zx}{y}\plus{}\frac{yz}{x}) \geq (8\sqrt[8]{(\frac{xy}{z})^4(\frac{zx}{y})^2(\frac{yz}{x})^2} \plus{} 4\sqrt[4]{(\frac{xy}{z})(\frac{zx}{y})^2(\frac{yz}{x})})\equal{}4(2\sqrt{xy}\plus{}\sqrt{zx})\equal{}4$
The equality holds if and only if $ x\equal{}y\equal{}z\equal{}\frac{1}{3}$

I guess that is why you thought this had a good solution :)

[quote="dulldrake"]I found the another way to solve the problem, we have
 $ S \equal{} \frac {3yz}{x} \plus{} \frac {4zx}{y} \plus{} \frac {5xy}{z} \equal{} (\frac {4xy}{z} \plus{} \frac {2zx}{y} \plus{} \frac {2yz}{x}) \plus{} (\frac {xy}{z} \plus{} \frac {2zx}{y} \plus{} \frac {yz}{x}) \geq (8\sqrt [8]{(\frac {xy}{z})^4(\frac {zx}{y})^2(\frac {yz}{x})^2} \plus{} 4\sqrt [4]{(\frac {xy}{z})(\frac {zx}{y})^2(\frac {yz}{x})}) \equal{} 4(2\sqrt {xy} \plus{} \sqrt {zx}) \equal{} 4$
The equality holds if and only if $ x \equal{} y \equal{} z \equal{} \frac {1}{3}$[/quote]

Can you help me with your solution?  I'm pretty bad at inequalities.

All, he did was use Weighted AM-GM.

If you type in Google, you get it.


Find all continous functions $ f:\mathbb{R}\to\mathbb{R}$ so that :
$ f(x\plus{}2f(y))\equal{}f(x)\plus{}f(y)\plus{}y$

[quote="Cyberjunkie"]$ f(x\plus{}2f(y)) \equal{} f(x)\plus{}f(y)\plus{}y$  (1)
$ x\leftarrow y\minus{}2f(y)$ we obtain $ f(y\minus{}2f(y)) \equal{}\minus{}y$  (2)
in (1):$ y\leftarrow y\minus{}2f(y)$ we obtain $ f(x\minus{}2y) \equal{} f(x)\minus{}2f(y)$ (3)
$ x \equal{} y \equal{} 0$ in (3)$ \Rightarrow f(0) \equal{} 0. x \equal{} 2y$ we have $ f(2y) \equal{} 2f(y)$ and so (3) can be rewritten as $ f(x\minus{}2y) \equal{} f(x)\minus{}f(2y)$.we put here $ x\leftarrow x\plus{}y$ and $ y\leftarrow\frac{y}{2}$,and obtain $ f(x\plus{}y) \equal{} f(x)\plus{}f(y)$ This is $ f(x) \equal{} ax$ Considering this in (1) we have $ a(x\plus{}2ay) \equal{} ax\plus{}ay\plus{}y$ or $ 2a^{2}\equal{} a\plus{}1.a \equal{} 1$ or $ a \equal{}\minus{}\frac{1}{2}$[/quote]

Very nice solution :first: .Can you prove that $ f(x)$ is a surjective function?  :)


Find all functions $ f: (0,\infty)\longrightarrow (0,\infty)$ such that:
i) $ f(x) \in (1, \infty)$ for each $ x\in (0,1)$, and
ii) For all $x,y \in (0,\infty)$,
\[ f(xf(y)) \equal{} yf(x).\]

Let $ y \equal{} \frac {1}{f(x)}$, then exist $ z \equal{} xf(\frac {1}{f(x)})$, suth that $ f(z) \equal{} 1$.
$ y \equal{} z$ give $ f(x) \equal{} zf(x)\to z \equal{} 1$. Therefore $ f(\frac {1}{f(x)})\equiv \frac 1x.$
$ x \equal{} 1$ give $ f(f(y))\equiv y$. Therefore $ f$ is bijective and ($ y \equal{} \frac 1x$) $ f(\frac 1x) \equal{} \frac {1}{f(x)}$.
It give $ f(x) > 1,x < 1,$ and $ f(x) < 1,x > 1$, therefore $ f(x) \equal{} x$ only for $ x \equal{} 1$.
$ y \equal{} x$ give $ f(xf(x))\equiv xf(x)$. therefore $ xf(x)\equiv 1$ or $ f(x) \equal{} \frac 1x$.

[quote="Rust"] ...
It give $ f(0,1)\to (1,\infty), f(1,\infty)\to (0,1)$ and $ f(x) \equal{} x\to x \equal{} 1$.
...[/quote]

I liked this.
:)

[quote="Rust"]I edit my post.[/quote]

There was no irony at all in my post and I think your previous writing was very easy to understand.

Sorry if you thought I was ironic.  :oops:


We mean a traingle in $\mathbb Q^{n}$, 3 points that are not collinear in $\mathbb Q^{n}$

a) Suppose that $ABC$ is triangle in $\mathbb Q^{n}$. Prove that there is a triangle $A'B'C'$ in $\mathbb Q^{5}$ that $\angle B'A'C'=\angle BAC$.
b) Find a natural $m$ that for each traingle that can be embedded in $\mathbb Q^{n}$ it can be embedded in $\mathbb Q^{m}$.
c) Find a triangle that can be embedded in $\mathbb Q^{n}$ and no triangle similar to it can be embedded in $\mathbb Q^{3}$.
d) Find a natural $m'$ that for each traingle that can be embedded in $\mathbb Q^{n}$ then there is a triangle similar to it, that can be embedded in $\mathbb Q^{m}$.
You must prove the problem for $m=9$ and $m'=6$ to get complete mark. (Better results leads to additional mark.)

I think this one is inspired from a not-so-old article in the AMM where it was proved that :
let's say that a triangle is $\mathbb{Q}^{n}$-inscriptible if we may find a triangle similar to it and whose vertices belong to $\mathbb{Q}^{n}$.
Then :
1) A triangle is $\mathbb{Q}^{n}$-inscriptible for some $n$ if and only if the squares of the tangents of its non-right angles are rational numbers.
2) In that case, $n=5$ is always sufficient.
3) The triangles which are $\mathbb{Q}^{3}$-inscriptibles are exactly those which are $\mathbb{Q}^{4}$-inscriptible.

Pierre.


Prove that :

$ x^5\ge 27(2x\minus{}3)(x^2\minus{}3x\plus{}3)\ ,\ (\forall)x\ge 0$ .

Expanding we have $ x^5\minus{}54x^3\plus{}729x^2\minus{}405x\plus{}729 \geq 0$

For all $ x \geq 3\sqrt{6}$ , $ x^5 \geq 54x^3$ and $ 729x^2 > 405x$ so it is true.

When $ 0 \leq x \leq 1$, $ x^5\minus{}54x^3\plus{}729x^2\minus{}405x\plus{}729 > 729 \minus{} 54x^3 \minus{} 405x > 0$ so it is true.

When $ 1 \leq x \leq 3\sqrt{6}$ , we note that $ 729x^2\minus{}405x>0$, and that the minimum value of $ x^5\minus{}54x^3$ occurs when $ 5x^4\minus{}162x^2\equal{}x^2(5x^2\minus{}162)\equal{}0 \implies x^2\equal{}32.4$ at which point it is simple to show that $ x^5\minus{}54x^3<729$, so our desired quantity is again positive.

Thus, it is true for all nonnegative x.


modular arithmetic
Divisibility
number theory
IMO
IMO 1990
Laurentiu Panaitopol

Approach by [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=8090]maxal[/url]:

Let $ N \equal{} \{ n\in\mathbb{N} : 2^n\equiv \minus{} 1\pmod{n^2} \}$ be the set of all solutions and $ P \equal{} \{ p\text{ is prime} : \exists n\in N, p|n \}$ be the set of all prime factors of the solutions.

It is clear that the smallest element of $ P$ is 3.
Assume that $ P\ne\{3\}$ and let's try to determine the second smallest element $ q \equal{} \min (P\setminus\{3\})$.

Let $ n\in N$ be a multiple of $ q$. It is important to notice that $ 9\not|n$ (otherwise it is easy to get that any power of $ 3$ divides $ n$, a non-sense). Therefore, $ n \equal{} 3^t n'$ where $ t \equal{} 0$ or $ 1$ and $ n'$ does not have prime divisors smaller than $ q$.
Since $ 2^{2n}\equiv 1\pmod{q}$, the multiplicative order $ r \equal{} ord_q(2)$ of 2 modulo $ q$ divides $ 2n$. Moreover, $ r$ must be even, since otherwise we would have $ 2^n \equiv 1\pmod{q}$, a contradiction to the required $ 2^n \equiv \minus{} 1\pmod{q}$.
Since $ r < q$ we must have $ r \equal{} 2$ or $ 2\cdot 3 \equal{} 6$. But the numbers $ 2^2 \minus{} 1 \equal{} 3$ and $ 2^6 \minus{} 1 \equal{} 63$ deliver only one new prime factor $ 7$, implying that $ q \equal{} 7$. However, in this case $ r \equal{} ord_7(2) \equal{} 3$, a  contradiction. This contradiction proves that $ P \equal{} \{3\}$ and thus $ N \equal{} \{1,3\}$.

Let us assume n > 1. Obviously n is odd. Let p ≥ 3 be the smallest
prime divisor of n. In this case (p − 1, n) = 1. Since 2n + 1 | 22n − 1, we
have that p | 22n − 1. Thus it follows from Fermat’s little theorem and
elementary number theory that p | (22n − 1, 2p−1 − 1) = 2(2n,p−1) − 1.
Since (2n, p − 1) ≤ 2, it follows that p | 3 and hence p = 3.
Let us assume now that n is of the form n = 3kd, where 2, 3  d. We first
prove that k = 1.
Lemma. If 2m − 1 is divisible by 3r, then m is divisible by 3r−1.
Proof. This is the lemma from (SL97-14) with p = 3, a = 22, k = m,
α = 1, and β = r.
Since 32k divides n2 | 22n − 1, we can apply the lemma to m = 2n and
r = 2k to conclude that 32k−1 | n = 3kd. Hence k = 1.
Finally, let us assume d > 1 and let q be the smallest prime factor of d.
Obviously q is odd, q ≥ 5, and (n, q−1) ∈ {1, 3}. We then have q | 22n−1
and q | 2q−1 − 1. Consequently, q | 2(2n,q−1) − 1 = 22(n,q−1) − 1, which
divides 26 −1 = 63 = 32 · 7, so we must have q = 7. However, in that case
we obtain 7 | n | 2n + 1, which is a contradiction, since powers of two can
only be congruent to 1,2 and 4 modulo 7. It thus follows that d = 1 and
n = 3. Hence n > 1 ⇒ n = 3.
It is easily verified that n = 1 and n = 3 are indeed solutions. Hence these
are the only solutions.

[quote="ramakrishna"]$ \cdots$ Since 2n + 1 | 22n − 1, $ \cdots$[/quote]
I don't understand this :maybe:
perhaps you could use latex?

That's the official solution (see the attachment). ramakrishna copied it and so the LaTeX formulas got lost.

Another approach:

Let $ p$ be the smallest prime divisor of $ n$. It is easy to check that, $ n\equal{}3$ is obviously solution. Let $ p^{a} \parallel{} n$ . $ p | 2^{2n}\minus{}1$ and $ p | 2^{p\minus{}1} \minus{} 1$ (By the fermat's theorem), we obtain that $ p\equal{}3$. It is also obvious that n is an odd number.
Lemma: For all $ n$ positive integers, $ 2$ is a primitive root modulo $ 3^{n}$.
$ 3^{2a} | 2^{2n} \minus{} 1$. Using the lemma, we get that $ \phi(3^{2a}) | 2n$. Using the power of three, we obtain that $ 3^{2a\minus{}1} | 3^{a}$. This is only possible when $ a \geq 2a\minus{}1$. So $ a\equal{}1$. Now $ q$ be the second smallest prime divisor of $ n$. $ (2n,q\minus{}1) \equal{} 2 , 6$. If this equals to 2, we get $ q\equal{}3$ which is a contradiction.If  $ (2n,q\minus{}1) \equal{} 6$ then $ q|63$. We know that $ q$ is different from $ 3$. Hence $ q$ must be $ 7$. But this is impossible since
$ 2^{n} \plus{}1\equiv 2\mod 7$ when $ n$ is divisible by $ 3$. Hence the answer is $ n \equal{} 3$.

FelixD, where did you find the official solution ?

The "official" solution is that in the shortlist, http://www.artofproblemsolving.com/Forum/download/file.php?id=1042, pp. 45-46, which is somewhat different from those posted here.

[quote="orl"]Determine all integers $ n > 1$ such that
\[ \frac {2^n \plus{} 1}{n^2}
\]
is an integer.[/quote]

[hide]i 've just found it, i'm not sure if it's right. i wish u checked it for me[/hide]

because $n^2|2^n+1$ so $n$ is odd. then $n=\prod {{p_i}^{{\alpha _i}}} $ with $p_i $ odd

$n^2|2^n+1$ lead to $v_p(n^2)<v_p(2^n+1)$

but,use the lifting exponent we have $v_p(2^n+1)=v_p(2^n-(-1)^n)=v_p(3)+v_p(n)$

if $p_i \ge 5$ then $v_p(3)=0$.

so  that we have $v_p(n^2) \le v_p(n)$ wich $n=1$

if p=3 we have $v_3(n^2) \le 1+ v_3 (n) $ then $v_3(n)=0;1$ then $n=1,3$

summary,we have n=1,n=3  :!:

Generalisation: if $p$ is an odd prime, $n>1$ and $(p-1)^n+1$ is divisible by $n^{p-1}$, then $p=n=3$.

nnosipov,you missed the case $n=p=2$

It was a problem from Imo with the additional condition $n<2p$.But we can even eliminate the case $n<2p$ and much more generally prove the fact.

Some approaches of the following generalisation?
Let $a,b>1$ positive integers such that $\dfrac{a^b+1}{b^a}$ is an integer. Find all $a,b$.

[quote="Victory.US"]...but,use the lifting exponent we have $v_p(2^n+1)=v_p(2^n-(-1)^n)=v_p(3)+v_p(n)$...[/quote]

No, this is not true. If you want to use LTE you should make sure that $p|2+1=3,$ but this is not true for any prime $p,$ and so you can't use LTE here.

Well let me post my solution
[b]Lemma:[/b]
If $n|2^n+1,n=3^k$
[b]Proof:[/b]
Obviously $n$ odd,let $p$ be the smallest prime factor of $n,$then
$2^{2n}\equiv 1\mod p,2^{p-1}\equiv 1\mod p\implies 2^{gcd(2n,p-1)}\equiv 1\mod p$
Since $p$ is the smallest prime factor,no prime of $p-1$ will be shared with $n,$so $gcd(2n,p-1)=2$
Therefore $2^2\equiv 1\mod p\implies p=3$ and so let $n=3^kl$ where $3\not| l,$but if $l>1$ similarly the smallest prime factor of $l$ is $3,$contradiction.
Thus $n=3^k$
Now get back to original problem and we shall use Lte,so
$v_3(n^2)=2k$ and
$v_3(2^n+1)=k+1$
If $n>1,k\ge 1$ and we get $2k\le k+1\implies k\le 1$ implying $k=1$
Then the only solutions $n=\boxed {1,3}$

$\boxed{n=1}$ satisfy the divisibility. Let $n\ge 2 $
Obvious that $n$ is odd number.
Let $p_1$ is the smallest prime divisor of $n$.
From order of prime we have $p_1 -1 | 2$ so $p_1$ must be $3$
From $n$ is odd $LTE$ says \[\nu_3(n^2)\leq \nu_3(2^n+1)= \nu_3(n)+1\]
so $\nu_3(n)=1$. $\boxed{n=3}$ satisfy divisibility. So n have one more prime divisor.
 Let $p_2$ is the smallest prime divisor of $n$ after $3$.
Easy to see from order of prime $p_2=7$ and from $2^n+1\equiv 2\pmod{7}$ contradaction. All of the solutions are $\boxed{n=1,3}$

n is obviously odd, and 1,3 work; we prove they're the only ones, so henceforth assume $n>3$. Let $p$ is the smallest prime divisor of $n$. We have $$2^{2n}\equiv1\pmod p\implies\text{ord}_{p}2\mid\gcd(2n,p-1)=2\implies 2^2\equiv1\pmod p\implies p=3$$$$2\nu_3(n)=\nu_3(n^2)\le\nu_3(2^n+1)= \nu_3(n)+1\implies\nu_3(n)=1.$$ Suppose FTSOC n has another divisor, let the smallest one be q ($\ne3$). We have $$\gcd(2n,q-1)=\{1,2,3\}\stackrel{2^{\gcd(2n,q-1)}\equiv1\pmod q}{\implies}q=7\implies2^n+1\equiv2\pmod 7,$$ contradiction, so our only solutions are the claimed. $\blacksquare$

e take a prime $p$ such that $p | n$. We can see that \[2^n \equiv -1 \pmod{p} \Rightarrow 2^{2n} \equiv 1 \pmod{p}\] This means that the $ord_p(2) | \gcd(p-1, 2n)$. (both of them give $2^x \equiv 1 \pmod{p}$)

We look at the smallest prime $p$ (no such one could exist and the answer could be $\boxed{1}$) of $n$ we can obviously see that we must have $p = 3$ because every other smallest prime would lead to the order being $2$ (which only is only true of $3$). We now claim that $v_3(n) = 1$.

Let $n = 3x$. We conclude that $v_3(n) = 1$ because if say $v_3(n) = y > 1$ then $v_3(8^x + 1) = 2 + v_p(x) = y+1$. This means that we need $2y \leq y+1$ or $y = 1$.

Staying with $n = 3x$ we need $9x^2 | 8^x + 1$ however, for odd $x$ we already have the $9$ taken care of. Thus we need $x^2 |8^x + 1$. This means by similar logic as given in the beginning that for a prime $p$ such that $p|x$, $ord_p(8) | \gcd(p-1,2x)$

Now note that every smallest prime will give $\gcd(p-1, 2x) = 2$. Thus we need $64 \equiv 1 \pmod{p}$ the only prime $p$ besides $3$ that satisfies this is $7$. However, this can never happen because $7^x \equiv 1, 2, 4 \pmod{p}$ This means that our only other solution is $\boxed{3}$. $\blacksquare$

Clearly $n$ is odd. Consider the smallest prime factor $p | n$. Then $2^n \equiv -1 \pmod p$ so $$2 | ord_p(2) | \gcd(p-1, 2n).$$ Since $p$ is minimal, $p-1\nmid n$ so $$\gcd(p-1, 2n) |2 \implies ord_p(2) = 2 \implies 2^2 \equiv 1 \pmod 3 \implies p = 3\implies 3 | n$$ Note that $$2\nu_3(n) \leq \nu_3(2^n + 1) = 1 = \nu_3(n) \implies \nu_3(n) = 1.$$
In a similar fasion, consider the second smallest prime factor $q | n$, and we have $$2 | ord_q(2) | \gcd(p-1, 2n) |6$$. This means $ord_q(2) = 2, 6$, but the former is impossible since it would force $$2^2 \equiv 1 \pmod q \implies q = 3.$$ Thus $$2^6 \equiv 1 \pmod q \implies q = 7.$$ However, $2^3 \equiv 1 \pmod 7$ so the order is not $6$, a contradiction. Thus $n$ can only have at most 1 prime factor, which is $3$, so $n = 1, 3$.

https://acrobat.adobe.com/id/urn:aaid:sc:AP:5e232744-31e7-45fb-b85b-ee24dbfe5d4a

https://acrobat.adobe.com/id/urn:aaid:sc:AP:c6760a55-4f6b-446b-acf4-75750df43223

I was new to using it and could not express myself in writing. Please forgive me.
This is a direct method without mod .

First we clearly have $n$ odd as otherwise it fails $\pmod 4.$ Next notice that if $p \mid 2^n+1$ then letting $o$ be the order of $p$ we get $o \mid 2n,p-1$ and $o \nmid n.$ Let $p$ be the smallest odd prime factor of $n$ and we get $o \mid 2n,p-1$ but $o<p$ so from $o \mid 2n$ we get $o=2$ and $p=3.$ Now we use LTE to get $\nu_3(2^n+1)=\nu_3(n)+1 \ge 2\nu_3(n)$ so $\nu_3=1.$ Then letting $p$ be the second smallest odd prime factor of $n$ we have $o \mid 2n,p-1$ but this implies $o<p$ and since $o \mid 2n$ and $o \nmid n$ we get $o=2,6.$ We see $o=2$ gives $p=3$ which is impossible and $o=6$ gives no solutions so the only solution is $n=3$ (and trivial $n=1$).

First note that all even $n$ fail, and $n = 1$ can be checked to work. Now we wish to determine all $n$ such that $2^n \equiv -1 \pmod{n^2}$. Take the smallest prime $p \mid n$. Then we require,
\begin{align*}
2^{2n} \equiv 1 \pmod{p}
\end{align*}
Hence $\text{ord}_p(2) \mid \gcd(2n, p - 1)$. Note that $p-1$ and $n$ are coprime and hence $\gcd(2n, p - 1) = 2$, as $p > 2$. Now take cases.

First if $\text{ord}_p(2) = 1$ we require,
\begin{align*}
2 \equiv 1 \pmod{p}
\end{align*}
which obviously fails. Then we must have $\text{ord}_p(2) = 2$ and hence,
\begin{align*}
4 \equiv 1 \pmod{p} \implies p = 3
\end{align*}
Thus we have $n = 3k$ for some $k$. Then our desired condition becomes,
\begin{align*}
9k^2 \mid 8^k + 1
\end{align*}
Now we can use LTE to find $\nu_3(8^k + 1) = 2 + \nu_3(k)$. However we also have $\nu_3(9k^2) = 2 + 2\nu_3(k)$. Clearly then we must have,
\begin{align*}
2 + \nu_3(k) \geq 2 + 2\nu_3(k) \iff 0 \geq \nu_3(k)
\end{align*}
whence equality must occur. Thus $k$ is relatively prime to $3$. Now take the smallest prime $p \neq 3$ dividing $k$. Then we must have,
\begin{align*}
8^{2k} \equiv 1 \pmod{p}
\end{align*}
Then $\text{ord}_p(8) \mid \gcd(2k, p-1)$. Clearly $\gcd(k, p-1) = 1$ and hence $\text{ord}_p(8) \mid 2$. Taking cases once more we can easily see that both imply that $7 \mid k$. However if $7 \mid k$, then we can easily see that $8^k + 1 \equiv 2 \pmod{7}$, a contradiction. Thus $k$ has no prime divisors, and thus $k = 1$. Therefore $n = 3k = 3$.

Then our final solution set is $\boxed{n = 1 \text{ and } n = 3}$.

$n = 1$ obviously works, so assume $n \geq 2$ and let $p$ the smallest prime dividing $n$. Then $p^2 \mid 2^n + 1$, so $2^n \equiv -1 \pmod p$, and $2^{2n} \equiv 1 \pmod p$, so the order of $2 \pmod p$ divides $ \gcd(2n, p-1)$, so the order of $2 \pmod p$ is $2$. Then $2^2 \equiv 1 \pmod p$, so $p = 3$, and $n$ is odd.

$n = 3$ works. If $n > 3$, let $q$ be the second smallest prime dividing $n$. Then the order of $2 \pmod q$ divides $\gcd(2n, q-1)$. From LTE, $\nu _3(2^n +1) = \nu_3(n) + 1$ and $\nu_3(n^2) = 2 \nu_3(n)$, so $\nu_3(n) + 1 \geq 2 \nu_3(n)$ and $\nu_3(n) = 1$. Then $\gcd(2n, q-1)$ must equal $6$ or $3$, so $8 \equiv 1 \pmod q$ or $63 \equiv 1 \pmod q$, therefore $q = 7$. However, then $2^n + 1 \equiv 8^{n/3} + 1 \equiv 2 \pmod 7$, contradiction, so there are no solutions for $n > 3$.

Therefore, the only solutions are $n = 1$ and $n = 3$.

 :coolspeak:
$\textcolor{blue}{\text{Claim} \hspace{0.15cm}\nu_3(n)=1}$
Clearly $n$ odd; let $p$ be the smallest prime divisor of $n$ then $2^{2n}\equiv 1\pmod p \hspace{0.1cm}\text{and}\hspace{0.1cm} 2^{p-1}\equiv 1\pmod p$ so $\operatorname{Ord}_p(2)\mid 2\gcd\left(n, \frac{p-1}{2} \right)=2\Rightarrow \operatorname{Ord}_p(2)=2\Rightarrow p=3$, but by Lifting the Exponent $$2\nu_3(n)\leq\nu_3(2^n+1)=\nu_3(2+1)+\nu_3(n)=1+\nu_3(n)\Rightarrow \nu_3(n)=1$$
$\textcolor{blue}{\text{Claim} \hspace{0.15cm}\text{No other prime divides} \hspace{0.15cm}n}$
Let $q$ be the second smallest prime dividing $n$ then  $$\operatorname{Ord}_q(2)\mid 2\gcd\left(n, \frac{q-1}{2} \right)=\begin{cases} 2\cdot 3 \Rightarrow \operatorname{Ord}_q(2)=\begin{cases} 6 \Rightarrow q\mid 2^6-1=63=7\cdot 3^2 \\ 3 \Rightarrow q\mid 2^3-1=7 \end{cases} \\2\cdot 1\hspace{0.5cm} \text{Impossible}&\end{cases}$$
Therefore $q=7$ but $2^n+1\equiv 8^{\frac{n}{3}}+1\equiv 2\pmod 7$ a contradiction, so $\boxed{n=1, 3}$

I claim that the only solution is when $\boxed{n = 3}$.

Let $p$ be the smallest prime dividing $n$. Then \[2^{2n} \equiv 1 \pmod p \text{ and } 2^{p - 1} \equiv 1 \pmod p \implies o(2)_p \mid \gcd(2n, p - 1) \in \{1, 2\}. \] Now if $o(2)_p = 1$ we obtain no solution. If, however, $o(2)_p = 2$, then $\boxed{p = 3}$.

[color=#ff9a00][b]Lemma:[/b][/color] $2$ is always a primitive root modulo $3^N$ for any $N \in \mathbb{Z}_{>0}$.

If now $3^{\alpha} \parallel n$ for some $\alpha$, then $3^{2 \alpha} \parallel n^2$ and we have that \[2^{\phi(3^{2 \alpha})} \equiv \pmod{3^{2 \alpha}} \text{ and } 2^{2n} \equiv 1 \pmod{3^{2 \alpha}} \implies \phi(3^{2 \alpha}) \mid 2n \implies v_3(2n) \geq v_3(\phi(3^{2 \alpha})) \implies \alpha \geq 2 \alpha - 1 \implies \alpha = 1. \] Therefore $3 \parallel n$. If now $q$ is the second smallest prime that divides $n$, we must have again that $o(2)_q \mid \gcd(2n, q - 1) \in \{1, 2, 6\}$. Here $o(2)_q = 1, 2$ provides us with the information that $q = 1$ and $q = 3$, both of which are contradictions. Hence in our final option $o(2)_q = 6 \implies \boxed{q = 7}$. However notice that $2^n + 1 \equiv 8^N + 1 \equiv 2 \pmod 7$, which is impossible.

Therefore no other prime may divide $n$ but $3$, providing us with $\boxed{n = 3}$ as the only valid solution. $\blacksquare$

No $n>1$ condition given to me

-----

The answer is $\boxed{n=1,3}$. Begin by noting that $n=1$ works. Henceforth assume $n \neq 1$.

Clearly, $n$ is odd by parity arguments. Denote the smallest prime factor of $n$ as $p$. We have

\[p \mid 2^n+1 \implies \operatorname{ord}_p(2) \mid \gcd(p-1,2n)\]

If $p-1 \mid n$, this contradicts minimal condition, so we must have $\operatorname{ord}_p(2)=2$. This means that

\[2^2 \equiv 1  \pmod{p} \implies p=3.\]

Then,

\[\nu_3(n^2) = 2\nu_3(n) \le \nu_3(2^n+1) = \nu_3(2+1) + \nu_(n)\]
\[\iff \nu_3(n) \le 1.\]

We know $3$ divides $n$, so we have that exactly one power of $3$ divides $n$.

Denote the second-smallest prime dividing $n$ as $q$. We have

\[q \mid 2^n+1 \implies \operatorname{ord}_q(2) \mid \gcd(q-1,2n).\]

Thus, $\operatorname{ord}_q(2) = 2, 6$. Caseworking shows that $q=3,7$, but neither works. Hence, $n$ contains only prime factors of $3$. This implies the only solution is $n = 3$.

Suppose FTSOC $n$ has at least 2 distinct prime divisors.
[list]
[*] Let the smallest be $p_1 > 2$. Then
\[\operatorname{ord}_{p_1} 2 \mid \gcd(2n,p_1-1) = 2 \implies p_1 = 3.\]
In particular, LTE gives us
\[v_3(n^2) \leq v_3(2^n+1) \implies 2 v_3(n) \leq 1 + v_3(n) \implies v_3(n) = 1.\]
[*] Suppose the second smallest is $p_2 > 3$. Then
\[\operatorname{ord}_{p_2} 2 \mid \gcd(2n,p_2-1) = 2,6 \implies p_2 = 7.\]
However, 7 never divides $2^n+1$ for any $n$, contradiction.
[/list]

Hence we extract our only answers $n = \boxed{1,3}$. $\blacksquare$

[quote=ddot1]I think that's missing a little detail: How do you know that $3$ divides $n$ only once? Otherwise, you could have something like $$\gcd(2n,p_2-1)=18.$$[/quote]

Follows from the LTE statement - $0 < v_3(n) \leq 1$.

Oh, duh - sorry, I didn't read it carefully enough.


number theory
greatest common divisor
superior algebra
superior algebra unsolved

Relations R_1 and R_2 on a nonempty set X are different if there are x,y $ \in$ X such that exactly one of xR_1y and xR_2y is true.

R_1 (set of positive integers) where xR_1y iff gcd (x,y) >1

R_2 (set of real numbers) where xR_2y iff x-y $ \in$ Q.

If X contains n elements, how many different binary relations are there on X?

Any help would be welcome! thanx.

Just a thought. If we are talking about binary relations on X, it's base 2. So, could it be 2^n?

But I am not sure.

For each of the $ n^2$ elements in $ X \times X$ there is a choice to either include it in the relation or not, so ...


Consider an array of numbers of size $8 \times 8$. Each of the numbers in the array equals 1 or -1. "Doing a move" means that you pick any number in the array and you change the sign of all numbers which are in the same row or column as the number you picked. (This includes changing the sign of the "chosen" number itself.) Prove that one can transform any given array into an array containing numbers +1 only by performing this kind of moves repeatedly.

For each square X let the X-cross be the set of squares in the same row or column as X (including X), so that a move changes the sign of the squares of the X-cross for the chosen square X.

We say that a square X is odd if the number of minus signs on its cross is odd, and even otherwise. Now consider the following set of moves: for each square X, apply one move to X if it is odd, and none if it is even. This solves the problem.

To prove this gives the result, do the following: for each square X with a $-$, place a coin with an X written on it, on each square of the X-cross (so each square will, in the end, have as many coins as the number of $-$s on its cross). When you're done, for each square apply to it as many moves as the number of coins on it (and note that the above set of moves is the same as this taken mod 2, so it's equivalent). The point is that the moves of the X-coins add up to changing just X, since each square not on the X-cross has just 2 X-coins on its cross, each square on the X-cross but different from X has 8 X-coins on its cross, and X has 15 X-coins on its cross. Then it is obvious that this gives the required result.