[quote="agolsme"][quote="Joey"]You are throwing a dinner party.  There are 2 round tables. One seats 4 people, the other seats 6.  There is space to invite 9 people plus yourself. However there are 15 people that wish to go. Sony and Cherry refuse to attend together, but everyone else gets along fine.

1) How many guest lists are possible?

2)Once a guest list is chosen, how many seating plans are possible?[/quote]

[hide="1"]
We can use case work. Either Sony goes, or Cherry goes, or neither go.

Case 1(Sony goes):
$\binom{14}{10}\binom{10}{4}$
Case 2(Cherry goes):
same as case 1
Case 3(Neither go):
$\binom{13}{10}\binom{10}{4}$

So there are a possible $2\binom{14}{10}{10}{4} + \binom{13}{10}\binom{10}{4}$ choices

[/hide]

[hide="2"]
Well after you get the 10 people, you need to choose 4 of them for one table, while the rest go to another table and order them accordingly:
so $\binom{10}{4}3!5!$
[/hide]
[/hide][/quote]

Sorry but I was wondering, What exactly does the $\binom{10}{4}3!5!$ mean? Is that like 10 C 4 if you're using nCr?

For the first point

[hide]I suppose that "15 people" includes Sony, Cherry and me.

- Sony first table, Cherry second table,  3! 5! 5! / 13!

- Cherry first table, Sony second table, 3! 5! 5! / 13!

- Sony first table, Cherry absent, 3! 6! 4! / 13!

- Sony second table, Cherry absent, 4! 5! 4! / 13!

- Sony absent, Cherry first table, 3! 6! 4! / 13!

- Sony absent, Cherry second table, 4! 5! 4! / 13!

- Both absent, 4! 6! 3! / 13!

I think that can be simplified to

9 4! 5! 4! / 13![/hide]

Angel "Java" Lopez

[quote="Joey"]You are throwing a dinner party.  There are 2 round tables. One seats 4 people, the other seats 6.  There is space to invite 9 people plus yourself. However there are 15 people that wish to go. Sony and Cherry refuse to attend together, but everyone else gets along fine.

1) How many guest lists are possible?

2)Once a guest list is chosen, how many seating plans are possible?[/quote]

[hide="1?"] of course you will be attending, so ur really inviting 9 people out of 15 who wish to go.  when either sony or cherry are invited, then the other can't.  so it becomes 3 cases
9 choose 13 (neither sony nor cherry goes).
then, one for when sony goes, and one for when cherry goes.  since the conditions are the same, its 2 times (8 choose 13).  (one spot taken by u, one by sony or cherry, so only looking for 8).
add them together
(9 C 13) + 2(8 C 13)=(715)+2(2574)=715+5148=5863 guest lists possible.  [/hide]

[hide="2?"] 4C10 times permutations possible 3! (n-1)! since its circular.  other table is 6C10 times permutations possible 5!  yes? or no[/hide]

[hide]

My first solution is if either Sony attends or Cherry attends.

First of all, you have $14$ people, but can invite only $9$.

That means a group of $14$ and your inviting $9$ at a time.

If Sony doesn't come, the number of combinations possible is $_{14}C_9$

$_{14}C_9 = \frac{_{14}P_9}{9!} = 2002$

If Cherry  doesn't come, the number of combinations would also be $2002$

That would make a total number of $4004$ possible guest lists.

My second solution would be if neither Sony or Cherry attends.

That means a group of $13$ and your inviting $9$ at a time.

So $_{13}C_{9} = \frac{_{13}P_{9}}{9!} = 715$

The second part of this problem asks how many possible seating arrangements would there be for $10$

the number of possible seating arrangements for the table of four would be

$_{10}C_{4} = \frac{_{10}C_{4}}{4!} = 210$

the number of possible seating arrangements for the table of six would be

$_{10}C_{6} = \frac{_{10}C_{6}}{6!} = 210$

the total number of seating arrangements would be $210^2 = 44100$

[/hide]


probability
expected value
combinatorics proposed
combinatorics

For a positive integer $n$, let $p_k(n)$ be the number of permutations of the set $\{1,2,...,n\}$ having exactly $k$ fixed points. Determine the value of \[\sum_{k=0}^n (k - 1)^2 \cdot p_k(n).\]

Here's my solution using the probabilistic method.
[hide]This problem is equivalent to the following.  Let $\pi$ be a random permutation on $1..n$.  Let $k$ be the number of fixed points of $\pi$.  Compute the expected value of $(k-1)^2$, multiplied by $n!$.

We can express $k$ as the sum
\[k = \sum_{i = 1}^n [\pi(i) = i],\]
where [S] is the indicator of S: 1 if S is true and 0 otherwise.   By squaring we get
\[k^2 - k = \sum_{i \ne j} [\pi(i) = i \land \pi(j) = j].\]

If we take expected value of both sides of the first displayed equation, we get
\[{E(k) = \sum_{i = 1}^n \Pr[\pi(i) = i}] = \sum_{i = 1}^n \frac{1}{n} = 1.\]
(That's what the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=41&t=60752]IMO 1987 problem[/url] asked for in equivalent form: The expected number of fixed points of a random permutation is 1.)

If we take expected value of both sides of the second displayed equation, we get (for $n \ge 2$)
\[E(k^2 - k) = \sum_{i \ne j} \Pr[\pi(i) = i \land \pi(j) = j]
 = \sum_{i \ne j} \frac{1}{n(n-1)} = 1.\]

Thus (for $n \ge 2$)
\[E((k-1)^2) = E(k^2 - k) - E(k) + E(1) = 1 - 1 + 1 = 1.\]
So the answer is $n!$ (for $n \ge 2$). [/hide]


logarithms
calculus
integration
function
complex numbers
complex analysis
complex analysis solved

Prove that $\mathop{Res}_{u=0}\frac{(1+u)^n}{u^{n+2}}\ln\left(1+\frac{u^2}{1+u}\right)=0$ for all even $n\neq 6k+2$. :?

Wow! It is really nice (at least, my solution I have just found seems nice to me).

I will propose it for my university students. He-he-he.

[quote="Myth"]
I will propose it for my university students. He-he-he.[/quote]
Poor students!  :P  (actually I am just envious: my undergrads would, probably, never solve it :(). Anyway, here is a small [hide="hint"] Write the residue as a contour integral and make change of variable $v=\frac u{1+u}$[/hide]

Very nice problem, Myth! I have never seen anything like this before. Also, if you just  expand the logarithm into its Taylor series and compute the residue termwise, you'll get a nice identity $\sum_{k\geq 1}\frac{(-1)^k}{k}{n-k\choose k-1}=0 ($n is congruent to $0$ or $4$ modulo $6$). I wonder whether it has some simple combinatorial proof.

I forgot to insert a link for the intial problem http://www.mathlinks.ro/Forum/viewtopic.php?t=34892

Actually, I meant "purely" school solution, where you are only to use the fact that $Res$ is a coefficient of $u^{-1}$ and some small complex numbers tricks. In this way I got another unexpected combinatorial identity.

what year of undergraduates is this for? :)

To solve this problem you should only know that $-\ln (1-x)=x+\frac{x^2}2+\frac{x^3}{3}+...$ (besides what I said above). It is a first year of Univ education.

Since a week passed and nobody posted a solution, I'll post mine. It uses a change of variable trick and it is the first time I see a problem about finding a residue where a natural change of variable immediately leads to a solution (that's why I like this problem so much). On the other hand, it seems that Myth has some more straightforward solution...

So let $v=\frac u{1+u}$, or, which is the same, $u=\frac v{1-v}$. Note that when $u$ travels around zero once in a small loop, so does $v$ and vice versa. So, the residue is the same as the residue of the function we obtain after changing variable in the differential form $\frac{(1+u)^n}{u^{n+2}}\ln\left(1+\frac{u^2}{1+u}\right) du$. Note that $v+uv=u$ so $dv+udv+vdu=du$ or $\frac {du}{1+u}=\frac{dv}{1-v}$. Thus
$\frac{(1+u)^n}{u^{n+2}}\ln\left(1+\frac{u^2}{1+u}\right) du = \frac{(1+u)^{n+1}}{u^{n+2}}\ln\left(1+\frac{u^2}{1+u}\right) \frac{du}{1+u} \\ = \frac{1-v}{v^{n+2}}\ln\left(1+\frac{v^2}{1-v}\right) \frac{dv}{1-v}= \frac1{v^{n+2}}\ln\left(1+\frac{v^2}{1-v}\right) dv$.
This means that we need to find the coefficient at $v^{n+1}$ of the Taylor series of $\ln\left(1+\frac{v^2}{1-v}\right)$ at $0$. The last trick is to observe that $1+\frac{v^2}{1-v}=\frac{1-v+v^2}{1-v}=\frac{1+v^3}{1-v^2}$, so $\ln\left(1+\frac{v^2}{1-v}\right)=\ln(1+v^3)-\ln(1-v^2)$, from where the coefficients can easily be found.

Note by the way, that the last trick can be made for the original expression and results in a fancy combinatorial identity different from the identity the problem started with.


modular arithmetic
number theory
number theory proposed

Prove that :

$ 1111^{2222} \plus{} 2222^{3333} \plus{} 3333^{4444} \equiv 4 \pmod 7$

[hide="hint"] Use [b]Fermat's little theorem[/b] [/hide]


How many odd five-digit counting numbers can be formed by
choosing digits from the set $ \{1, 2, 3, 4, 5, 6, 7\}$ if digits can be repeated?

Yea, pretty much. Or you could use complementary counting, or whatever it's called, but that wouldn't do anything  :P

[quote=007math]If we do complementary counting, it is $ 7^5 \minus{} (7\times7\times7\times\times7\times3) \equal{} \boxed{9604}$[/quote]

Right.

[quote=ishankhare][hide=Answer]$\boxed{9604}$[/hide][/quote]
Can you post a different solution instead of a repeated answer? Thanks.

I don't get it. Can anyone explain why there are 4 sevens? Like, why is it $7*7*7*7*4$? There are 3 even numbers. Why isn't it $7*7*7*4$?

For the first place, we have $4$ options and for the remaining places, we have $7^4$ options.
Hence, the required no. is $4\cdot 7^4$.

yeah . There are 4 7's since there are 4 more digits to fill in after you fill in the ones place. :)

[quote=GeniusKid1221]ok. It still doesn't really make sense though. Is it 7 because there are 7 numbers in the list?[/quote]

For an number to be odd/even, it only depends on the last digit. So we could put any number into the first 4 digits, because that won't affect the evenness and oddness of a number. Then, for the last digit, there are only 4 digits we can put to make it odd, so that is why it is *4 there.

[quote=007math]If we do complementary counting, it is $ 7^5 \minus{} (7\times7\times7\times\times7\times3) \equal{} \boxed{9604}$[/quote]

[quote=AIME15]There are $ 7$ choices for the first digit, $ 7$ for the second, $ 7$ for the third, $ 7$ for the fourth, and $ 4$ for the fifth (it must be odd).  Thus, there are $ 7 \times 7 \times 7 \times 7 \times 4\equal{}\boxed{9604}$ total numbers.[/quote]

HIDE ANSWERS PLS.

complementary counting is subtracting the number of things you don't want from the total things.

[quote=kc5170][hide=Solution]We have 7 choices for the first, 7 for the second, and so on. This gives us $7^5=\boxed{9604}$[/hide][/quote]

Sorry for the bump but $7^5 \neq 9604$

[hide]
There are 7 options, 1, 2, 3, 4, 5, 6, and 7, for the first four digits, but there are only four options, 1, 3, 5, and 7, for the last digit. Thus, we get:
7 * 7 * 7 * 7 * 4 = 9604
[/hide]


Given that $a,b,c>0\/$.
Find the minimum value of $y=\sqrt{a^2+x^2} + \sqrt{b^2 + (c-x)^2}\/$.

[quote="kunny"]The answer is $\sqrt{(a+b)^2+c^2}$[/quote]
You are right...
Can you share your method??? I guess you are not using the differentiation... am i right :?  :?  :?

Yes. I didn't use differentiation.On $x-y$ plane, set $A(x,0),B(c,a),P(x,a)$ and $Q(c,a+b)\ (a,b,c>0)$, we have $\sqrt{a^2+x^2}+\sqrt{b^2+(c-x)^2}=OP+PQ\geq OQ$. the equality holds when $O,P,Q$ are collinear.

similar to mine one, but yours is nicer!!!


Solve equation

$ 2.3^{2(\log_2x\minus{}1)}\equal{}x^{1\plus{}log_23}\minus{}x^2$

hello, i think your equation can only be solved by a numerical method.
Sonnhard.

hello, ohh, i see you meant $ 2\cdot3^{2(\log_2x \minus{} 1)} \equal{} x^{1 \plus{} log_23} \minus{} x^2$?
Sonnhard.

hello, i have computed $ x_1\equal{}2$ or $ x_2\equal{}2^{\frac{\ln(2)}{\ln(3)\minus{}\ln(2)}}$.
Sonnhard.


Pick an integer a > 2.
Let u be the bigger root of the quadratic equation x^2 - ax + 1 = 0.
Compute numerically (preferably on a calculator or computer) some (positive integer) powers of u.
What do you notice about these values?  Why does this happen?

Ted

I didn't compute anything, but they should get arbitrarily close to integers as the powers get larger. We can prove this by Newton Sums (or induction for a less interesting proof, in my opinion). That's because the other root is less than 1, and so that part of it will get close to 0 as the powers get big. Specifically, if r and s are the roots, then r^n+s^n is an integer for each n. If s^n is getting close to 0, then r^n must be getting close to integers.


Here's a problem that might help with your conversions-

A mad scientist needs to conduct a strange experiment that only he really understands. He must fill a rectangular tank that measures 3 fathoms by 2 rods by 1/40 of a furlong. He fills it with a clear, odorless liquid with a density of about 3,678,675 grains per US hogshead.

How many kips of the substance does he require to fill the tank? Please round up to the nearest kip.

Anyone know?

hmm that's from neopets! you can find all conversions on the net.

xD

I remember seeing this :P . I know I got it right, but I'm not telling :P .


trigonometry
algebra
polynomial
number theory
relatively prime
prime factorization

For 0<x<90, find the sum of all rational x (in degrees) for which:

$\sin{x}=\sin{(x^{2})}$

(remind me not to listen to advice without carefully reading it.)

Actually, your original answers are correct.

Treething, why does $\sin{x^{2}}=\sin{(360C+180-x)}$?

Because sin(x) = sin(180-x), and sin(x) = sin(360+x).

Thus $\sin{x}=\sin{(360C+180-x)}$. I want to know why $\sin{x^{2}}=\sin{(360C+180-x)}$.

I appreciate the explanations, but could you be a bit more explicit?

[quote="bthings_2000"]Thus $\sin{x}=\sin{(360C+180-x)}$. I want to know why $\sin{x^{2}}=\sin{(360C+180-x)}$.[/quote]

Given in the problem condition that $\sin x = \sin x^{2}$.

[quote="Zellex"]I mis-read the problem in the beginning myself...and I didn't even listen to my advice. Anyway, it works then you set $x^{2}= 360k+180-x$, but $k$ has to be relatively big to get integers solutions...is there a faster way besides adding and attempting to factor?[/quote]
I've tried taking it mod 360, but that doesn't help :(

ok, then how do you get from    sin(x) = sin(360C + 180 - x)

to  x = {0, 1, 35, 44, 81}

without a calculator?

[hide="Solution"] We will make the assumption that $x$ is integer.  This assumption is justified because the fractional part of $x^{2}$ is too different from the fractional part of $x$.

First we should take care of the case $x = x^{2}-360k$.  This gives

$x(x-1) = 360k$

The trivial $k = 0$ gives $x = \boxed{0, 1}$.  For all other solutions, $k \neq 0$.

The RHS needs to be divided into the product of two integers that are relatively prime to each other.  $k = 360, 361$ produces irrelevant values of $x$, so we will concern ourselves with smaller values.  There are only a few ways to factor $360$ properly.  Let $k = ab$ where $gcd(a, b) = 1, a < b$.  Then

$360k = 8 \cdot 5 \cdot 9 \cdot k = 40a \cdot 9b = 72a \cdot 5b = 45a \cdot 8b$

$a = 2, b = 9$ produces two consecutive integers, giving us a solution $x = \boxed{81}$.  The others are too large.

The other case, which is quite tricky to discover, is $x = 360k+180-x^{2}$.  In this case,

$x(x+1) = 180(2k+1)$

We can perform a similar factoring.  Let $2k+1 = ab$ where $a, b$ are odd, $gcd(a, b) = 1, a < b$.

$180(2k+1) = 4 \cdot 5 \cdot 9 \cdot (2k+1) = 20a \cdot 9b = 36a \cdot 5b = 45a \cdot 4b$

This produces $a = 1, b = 7, 11$ which correspond to $x = \boxed{35, 44}$.  Again, the other solutions are too large.
[/hide]

Edit:  By the way, [b]PenguinIntegral[/b]... [hide="There is another way"] Of course, what we're really trying to do is solve a polynomial $\bmod m$.  There is a general method for this!

The idea is to solve the polynomial $\bmod m_{1}, m_{2}, ... m_{k}$ (which is the "relatively prime factorization" of $m$ such that $m_{1}\cdot m_{2}\cdot ... \cdot m_{k}= m$), to get bounds on $x \bmod m_{1}, m_{2}, ... m_{k}$, and then to combine all that information via the Chinese Remainder Theorem.

Of course, this method is a bit cumbersome for $m = 360$, which is why I did not use it (although it is a very straightforward way).  It's very useful for, say, $m = 35$.  [/hide]

excellent solution... but could you more explicitly justify the initial assumption that x is an integer?

[hide="Well..."] Simply consider the nature of our expressions for $k$, which must be an integer.  In the first case,

$k = \frac{x(x-1)}{360}$

And if $x$ is non-integer (and rational, as specified!) the numerator is clearly non-integer.  (As I said, $x$ and $x^{2}$ have different fractional parts.  More specifically, $x^{2}$ will have a larger denominator, so their difference cannot have a zero fractional part.)  In the second,

$k = \frac{x(x+1)}{360}-1$

Again, considering the fractional part of the numerator, we conclude that $x$ is integer. [/hide]


one marble is randomly drawn, and then NOT REPLACED. from a jar containing two white marbles and one black marble. A second marble is drawn. what is the probability of drawing a white one and then a black one ?

i need explain this .. but i dont know how :S ! pls helpme >.<

Probability of drawing a white one on the first draw=$ \frac{2}{3}$
Black one on second draw=$ \frac{1}{2}$

So it's simply $ \frac{2}{3}\times\frac{1}{2}\equal{}\boxed{\frac{1}{3}}$.

[quote="gauss1181"]Probability of drawing a white one on the first draw=$ \frac {2}{3}$
Black one on second draw=$ \frac {1}{2}$

So it's simply $ \frac {2}{3}\times\frac {1}{2} \equal{} \boxed{\frac {1}{3}}$.[/quote]

Tenkiu

a pizza shop has 9 differeent toppings for its pizzas. how many different pizzas can be made using 1, 2, or 3 toppings if no toppings is used twice on one pizza? explainme this

It would just be, using the combinations formula, $ C(9,1)\plus{}C(9,2)\plus{}C(9,3)\equal{}9\plus{}36\plus{}84\equal{}\boxed{129}$ pizzas.

[quote="gauss1181"]It would just be, using the combinations formula, $ C(9,1) \plus{} C(9,2) \plus{} C(9,3) \equal{} 9 \plus{} 36 \plus{} 84 \equal{} \boxed{129}$ pizzas.[/quote]yes but i need explain this ...

C(8,2) ? 8C2 ?

how many pairs of colors can be put together if there are 8 colors to choose from?

Assumably order doesn't matter, so $ \binom{8}{2}\equal{}\boxed{28}$.

Seeing that your name is "Espanol" it most likely means you speak spanish as a main language soo,

Sean $ c_1, c_2, ..., c_8$ los 8 colores, donde $ c_i$ es un color.

Solución 1:
Los pares de colores son estos:
$ (c_1,c_2),(c_1,c_3),\cdots,(c_1,c_8) \equal{} 7$ combinaciones
$ (c_2,c_3),(c_2,c_4),\cdots,(c_2,c_8) \equal{} 6$ combinaciones
$ \hdots$
$ (c_7,c_8) \equal{} 1$ combinaciones

y en total son $ 1\plus{}2\plus{}\cdots\plus{}7 \equal{} \dfrac{7\cdot 8}{2} \equal{} 24$

Solución 2:
Escogiendo dos colores sin repetir una combinación, se puede obtener que los combinaciones son "8 escoge 2", que es igual a $ \binom{8}{2} \equal{} \dfrac{8!}{(2!)(8\minus{}2)!} \equal{} \dfrac{8\cdot 7}{2} \equal{} \boxed{28}$

Donde $ \binom{n}{k} \equal{} \dfrac{n!}{(k!)(n\minus{}k)!}$

Yo quiero explicar una solucion en espanol. Dame mas problemas.


calculus
inequalities
rearrangement inequality
inequalities unsolved

hello, setting $b=ax,c=ay$ then we get
$y=\sqrt{\frac{3}{a^2}-x^2-1}$
and we obtain
$a^5(x^2+x^3y^2+y^3)\le 3$
without loss of generality we set $a=1$
and with $y=\sqrt{2-x^2}$ we find
$x^2+2x^3-x^5+(2-x^2)\sqrt{2-x^2}\le 3$
let
$f(x)=x^2+2x^3-x^5+(2-x^2)\sqrt{2-x^2}$
with calculus we find that $f(x)$ has a global maximima for $x=1$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]
without loss of generality we set $a=1$
[/quote]
No, we can not. :wink:

You proved that $a^3b^2+b^3c^2+c^3a^2\leq3$ for $a=1$ and $b^2+c^2=2$. What happens if $a\neq1$? :wink:
There is a nice proof of my inequality by Rearrangement.

[quote="Dr Sonnhard Graubner"]hello, i think it's not difficult to calculate this case.
Sonnhard.[/quote]
I think it is not difficult to say it. :wink: You are welcome to show us your proof.

[quote="sqing"]The following inequality is true .
$a>0,b>0,c>0,a^2+b^2+c^2=3.$ Prove that\[a^{\frac{5}{2}}b^2+b^{\frac{5}{2}}c^2+c^{\frac{5}{2}}a^2\leq3.\][/quote]
It is weaker than arqady's inequality  :wink:
Because \[ a^{\frac{5}{2}} \leq \frac{1}{2}(a^2+a^3)\] and
\[ a^2b^2+b^2c^2+c^2a^2 \leq 3.\]

[quote="arqady"]$a>0,b>0,c>0,a^2+b^2+c^2=3.$ Prove that

 $a^3b^2+b^3c^2+c^3a^2\leq3.$[/quote]
It is true by rearrangement inequality  :lol:

Thank dm_edogawasonichi.
The following inequality is true .
$a>0,b>0,c>0,a^2+b^2+c^2=3.$ Prove that\[a^{\frac{8}{3}}b^2+b^{\frac{8}{3}}c^2+c^{\frac{8}{3}}a^2\leq3.\]

It is also weaker. I see some posts of you is weaker than the first posts in many topics  :wink:  :oops:!


induction
number theory
Euler s Theorem
IMO 1971
phi function
IMO

Show that there exists an infinite set $ S$ of positive integers such that for any two distinct elements $ m$ and $ n$ of $ S$, the integers $ 2^m \minus{} 3$ and $ 2^n \minus{} 3$ are coprime.

[i]Alternative formulation.[/i] Prove that the set $ \left\{2^k\minus{}3\mid k\in\mathbb{N}\right\}$ contains an infinite subset such that any two distinct elements of this subset are coprime.

Since $2$ is coprime with all the prime divisors of $2^{a_i}-3,\ \forall i\in\overline{1,n}$, we can choose $a_{n+1}$ s.t. $2^{a_{n+1}}-1$ is divisible with all those primes, and $2^{a_{n+1}}-3$ will be coprime with $2^{a_i}-3,\ \forall i\in\overline{1,n}$, so we can construct the sequence recursively.

Why we know always exist $a_{n+1}$ such that $2^{a_{n+1}}-1$ is divisible with all the prime divisors of $2^{a_{1}}-3$ ?

thx..

[quote="Yosh..."]Why we know always exist $a_{n+1}$ such that $2^{a_{n+1}}-1$ is divisible with all the prime divisors of $2^{a_{1}}-3$ ?[/quote]
Probably you mean $2^{a_{n}}-3$, not just $2^{a_{1}}-3$. Wlog, assume $a_{n}\ge 3$. Then say $p_{1}, p_{2}, \ldots, p_{k}\in \mathbb{N}$ are all the (pairwise distinct) primes dividing $2^{a_{n}}-3$ and let $a_{n+1}= a_{n}\cdot \prod_{i=1}^{k}(p_{i}-1)$. Obviously $p_{i}$ is odd, for any $i \in \overline{1,k}$. So $\prod_{i=1}^{k}p_{i}$ divides $2^{a_{n+1}}-1$, by Fermat's little theorem, and $\gcd(2^{a_{n+1}}-3, 2^{a_{n}}-3) = 1$. Now, by induction, it follows $\gcd(2^{a_{m}}-3, 2^{a_{n}}-3) = 1$, for any distinct $m, n \in \mathbb{N}$.

Here is the same problem:
[b]1[/b]Consider set $ \{a^n\minus{}(a\plus{}1)\}$ where $ a\in N$
Prove that we can find an subset such that each term is relative prime to the other.


Let BM be the median relative to side AC of triangle ABC. Let P be a point on MC. If we consider the parallel to BM passing trough P and we denote with N and Q the point where it touches the sides BC and  AB prove that:

$ PN\plus{}PQ\equal{}2BM$

easy
we have $ \frac{CP}{CM}\equal{}\frac{NP}{BM}$ and $ \frac{AP}{AM}\equal{}\frac{PQ}{MB}$
we add the two equalities then the result follow.... :blush:

Let $ ABC$ be a triangle and let $ M$ be a given point which belong to the sideline $ AC$ .

Then exists the implication $ \boxed {\ \frac {\overline {AM}}{\overline {MC}}\equal{}\lambda\ \implies\ \overline {PN}\plus{}\lambda\cdot\overline {PQ}\equal{}(\lambda \plus{}1)\cdot\overline {MB}\ }$ .


inequalities
search
inequalities proposed

Prove that if $ a,b,c \ge 0$ then $ \frac{a}{\sqrt{a\plus{}b}}\plus{}\frac{b}{\sqrt{b\plus{}c}}\plus{}\frac{c}{\sqrt{c\plus{}a}} \le \frac{5}{4}\sqrt{a\plus{}b\plus{}c}$.

[quote="arqady"]I am sure that you have a proof by Cauchy - Schwartz inequality. :lol:[/quote]
Yeah, you're right, Aqardy. :) What's your solution for this nice inequality?

I remember your SUPER HARD and SUPER NICE solution posted at http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1041154634&t=152848   :wink:

[quote="polskimisiek"]Wow, would you please explain how did you think of sth like that, because I simply do not understand how a normal human can work out a solution like that :wink:
Also one more question, when you say 5 different ways, do you mean only elementary mathematics?[/quote]
Yeah, it's unbelievable but we do it indeed. :) Just use some note on using Cauchy Schwarz inequality only. :)
Unfortunatly, not all of my solution are elementary mathematics. But there is a solution with the $ p,q,r$ technique and I like it the most. :)

Can you prove similar inequalities with 4 variables and 5 variables?


parameterization
function
algebra unsolved
algebra

Find all values the parameter $ a$ and for which there exists a function $f: \mathbb R \to \mathbb R$ for which
\[f(x^2\plus{}y\plus{}f(y))\equal{}(f(x))^2\plus{}ay\]
for all $x,y \in \mathbb R$.

[quote="aktyw19"]Find all values the parameter $ a$ and for which there exists a function $ f: R \rightarrow R$
$ f(x^2 \plus{} y \plus{} f(y)) \equal{} (f(x))^2 \plus{} ay$.[/quote]

I'm sorry for my long rather ugly proof and I hope somebody will provide a nicer one  :blush:

Let $ P(x,y)$ be the assertion $ f(x^2 \plus{} y \plus{} f(y)) \equal{} f(x)^2 \plus{} ay$
Let $ c \equal{} f(0)$

If $ a \equal{} 0$, we have solutions, for example $ f(x) \equal{} 0$ $ \forall x$.
So we'll consider from now that $ a\ne 0$.

1) $ f(x)$ is a surjection
===============
Just choose $ y$ such that RHS take any value you want.

2) $ f(x)$ is an injection, so a bijection and $ f(0) \equal{} 0$ and $ f( \minus{} x) \equal{} \minus{} f(x)$ $ \forall x$
=================================================
2.1) $ f(u) \equal{} f(v)\le 0$ $ \implies$ $ u \equal{} v$
--------------------------------------------
Let $ x\le 0$. Since $ f(x)$ is a surjection, let $ y$ such that $ f(y) \equal{} x$.

Then $ P(\sqrt { \minus{} x},y)$ $ \implies$ $ x \equal{} f(\sqrt { \minus{} x})^2 \plus{} ay$ and so $ y \equal{} \frac {x \minus{} f(\sqrt { \minus{} x})^2}a$ and so $ f(u) \equal{} f(v) \equal{} x\le 0$ $ \implies$ $ u \equal{} v \equal{} \frac {x \minus{} f(\sqrt { \minus{} x})^2}a$

2.2) $ f(u) \equal{} f(v)$ $ \implies$ $ u^2 \equal{} v^2$
---------------------------------------------
Let $ f(u) \equal{} f(v)$. We can choose $ y$ such that $ f(u)^2 \plus{} ay \le 0$ and so $ f(u^2 \plus{} y \plus{} f(y)) \equal{} f(v^2 \plus{} y \plus{} f(y)) \equal{} f(u)^2 \plus{} ay\le 0$ and so $ u^2 \equal{} v^2$
Q.E.D.

2.3) $ f( \minus{} x) \equal{} \pm f(x)$
-----------------------
Just compare $ P(x,y)$ and $ P( \minus{} x,y)$

2.4) $ x \equal{} 0$ $ \iff$ $ f(x) \equal{} 0$
----------------------------
Let $ x < 0$ and $ \ne f(0)$. Since $ f(x)$ is a surjection, let $ u$ such that $ f(u) \equal{} x$. According to 2.3, either $ f( \minus{} u) \equal{} f(u)$, either $ f( \minus{} u) \equal{} \minus{} f(u)$
If $ f( \minus{} u) \equal{} f(u) \equal{} x < 0$, according to 2.1, we get $ \minus{} u \equal{} u$ and so $ u \equal{} 0$, which is wrong since $ f(u) \equal{} x\ne f(0)$
So $ f( \minus{} u) \equal{} \minus{} f(u)$. Then :

$ P(0,u)$ $ \implies$ $ f(u \plus{} f(u)) \equal{} c^2 \plus{} au$
$ P(0, \minus{} u)$ $ \implies$ $ f( \minus{} u \minus{} f(u)) \equal{} c^2 \minus{} au$ and so either $ c^2 \plus{} au \equal{} c^2 \minus{} au$ and so $ u \equal{} 0$, wrong, either $ c^2 \plus{} au \equal{} \minus{} c^2 \plus{} au$ and so $ c \equal{} 0$
So $ x \equal{} 0$ $ \implies$ $ f(x) \equal{} 0$

Now, if $ f(x) \equal{} 0$, we get $ f(x) \equal{} f(0)$ and, according to 2.2, $ x^2 \equal{} 0^2$ and so $ x \equal{} 0$.
Q.E.D

2.5) $ f( \minus{} x) \equal{} \minus{} f(x)$ $ \forall x$
-------------------------------
$ f( \minus{} 0) \equal{} \minus{} f(0) \equal{} 0$
Suppose now $ \exists u\ne 0$ such that $ f( \minus{} u) \equal{} f(u)\ne 0$. Then, since $ f(x)$ is a surjection, $ \exists w$ such that $ f(w) \equal{} \minus{} f(u)$
Then , if $ f( \minus{} w) \equal{} \minus{} f(w)$, we get that $ f( \minus{} w) \equal{} f(u)$ and so $ w^2 \equal{} u^2$ and $ w \equal{} \pm u$ which is impossible since $ f(u) \equal{} f( \minus{} u)$ while $ f(w) \equal{} \minus{} f(u)\ne f(u)$.
So $ f( \minus{} w) \equal{} f(w) \equal{} \minus{} f(u) \equal{} \minus{} f( \minus{} u)$ but this is also impossible since one of the two values $ f(u)$ or $ f(w)$ is negative :
If $ f(u) \equal{} f( \minus{} u) < 0$, according to 2.1, $ u \equal{} \minus{} u$ and so $ u \equal{} 0$, wrong
If $ f(w) \equal{} f( \minus{} w) < 0$, according to 2.1, $ w \equal{} \minus{} w$ and so $ w \equal{} 0$ and $ f(u) \equal{} \minus{} f(w) \equal{} 0$ and so, according to 2.4, $ u \equal{} 0$, wrong

So $ f( \minus{} u) \equal{} \minus{} f(u)$
Q.E.D.

2.6) $ f(x)$ is injective
------------------------
From 2.2, we get that $ f(u) \equal{} f(v)$ $ \implies$ $ u^2 \equal{} v^2$.
If $ u \equal{} \minus{} v$, we have then $ f(v) \equal{} \minus{} f(u) \equal{} f(u)$ and so $ u \equal{} v \equal{} 0$
So $ f(u) \equal{} f(v)$ $ \implies$ $ u \equal{} v$
Q.E.D.

3) $ f(x) \equal{} x$ $ \forall x$
=============
$ P(x,0)$ $ \implies$ $ f(x^2) \equal{} f(x)^2$
$ P(0,x)$ $ \implies$ $ f(x \plus{} f(x)) \equal{} ax$

So (transforming $ P(x,y)$) : $ f(x^2 \plus{} y \plus{} f(y)) \equal{} f(x^2) \plus{} f(y \plus{} f(y))$

$ P(0,\frac {f(x)}a)$ $ \implies$ $ f(\frac {f(x)}a \plus{} f(\frac {f(x)}a)) \equal{} f(x)$ and, since $ f(x)$ is injective : $ \frac {f(x)}a \plus{} f(\frac {f(x)}a) \equal{} x$

So $ x \plus{} f(x)$ is surjective and "$ f(x^2 \plus{} y \plus{} f(y)) \equal{} f(x^2) \plus{} f(y \plus{} f(y))$" may be written :

$ f(x \plus{} y) \equal{} f(x) \plus{} f(y)$ $ \forall x\ge 0$, $ \forall y$

And, since $ f( \minus{} x) \equal{} \minus{} f(x)$ : $ f(x \plus{} y) \equal{} f(x) \plus{} f(y)$ $ \forall x,y$

So we have a solution of Cauchy's equation which is $ \ge 0$ $ \forall x\ge 0$ (since $ f(x^2) \equal{} f(x)^2$)
So $ f(x) \equal{} bx$

Plugging back in the equation, we get $ b \equal{} 1$ (since $ a\ne 0$) and so $ a \equal{} 2$

4) Solution
=======
The required values are $ a\in\{0,2\}$


[quote="zaizai-hoang"][url]http://www.mathlinks.ro/Forum/viewtopic.php?p=851519#851519[/url]
You can see a little in here  :wink: But it is not the most best  :P
Now, let prove the following inequality. Maybe you can see what I was spoken  :wink:
Problem: [zaizai-hoang] [Problem of day - 30/5/2007]
[url]http://www.bdt.ch06k.com/[/url]
Let $a,b,c,d$ are real number such that $a^{2}+b^{2}+c^{2}+d^{2}=4$. Prove that:
\[\frac{ab}{5-ab}+\frac{bc}{5-bc}+\frac{cd}{5-cd}+\frac{da}{5-da}\le 1 \]
With Thomas Mildorf's idea, we can find the following inequality:
\[\frac{x}{5-x}\le \frac{1}{4}+k(x-1) \]
In here, $x=ab$.... The $k=\frac{5}{16}$ then we have:
\[\frac{x}{5-x}\le \frac{1}{4}+\frac{5(x-1)}{16}\]

\[\iff \frac{5(a-1)^{2}}{16(5-a)}\le 0 \]
It is not true  :P ......[/quote]

Hoang, you seem to be hiding around the bush. Here is my solution.

For ease of computation, notice that
\[\frac x{5-x}=\frac5{5-x}-1. \]
The desired inequality takes the form
\[\frac1{5-ab}+\frac1{5-bc}+\frac1{5-cd}+\frac1{5-da}\leq 1. \]
By AM-GM inequality,
\[4=a^{2}+b^{2}+c^{2}+d^{2}\geq a^{2}+b^{2}\geq2ab. \]
Thus, $\max(ab,bc,cd,da)\leq 2$. We need to find $k$ and $l$ such that for $0\leq x\leq 2$
\[\frac1{5-x}\leq k(x^{2}-1)+l(x-1)+\frac14, \]
which reads as follows after some manipulations
\[(x-1)\left[\frac1{4(5-x)}-k(x+1)-l\right]\leq0. \]
We wish to establish a relation between $k$ and $l$. This is done by taking into account the fact that equality occurs at $a=b=c=d=1$, accordingly $x=1$. $k$ and $l$ are determined such that the expression in the square bracket has $x-1$ as a factor. We have $\tfrac1{16}-2k=l$. Plugging this $l$ back into the expression we have
\[(x-1)\left[\frac1{4(5-x)}-kx+k-\frac1{16}\right]\leq0. \]
This is equivalent to
\[(x-1)^{2}\left[\frac1{4(5-x)}-4k\right]\leq0. \]
Again, we have to choose $k$ such that $16k \geq 1/(5-x)$ for all $0\leq x\leq2$. For this, $k=\tfrac1{48}$ is a valid value. With this $k$, we have $l=\tfrac1{48}$ too.

Thus, we have a new bound
\[\frac1{5-x}\leq \frac1{48}(x^{2}-1)+\frac1{48}(x-1)+\frac14, \]
or equivalently
\[\frac1{5-x}\leq \frac1{48}x^{2}+\frac1{48}x+\frac5{24}. \]
Using this estimation repeatedly for $ab, bc,cd,da$, it is now sufficient to prove that
\[(ab)^{2}+(bc)^{2}+(cd)^{2}+(da)^{2}+ab+bc+cd+da+40\leq 48, \]
which follows immediately from AM-GM inequality and the fact that $ab+bc+cd+da=(a+c)(b+d)$.

Q.E.D :lol:

[hide]To Hoang: You did not want to share the idea with me but I finally managed to work it out. This approach can solve many other inequalities, one of which is mine: http://www.mathlinks.ro/Forum/viewtopic.php?search_id=2134875935&t=138988 [/hide]

what do you think if I say that it is not necessary in this case?   :D
And here:
http://www.mathlinks.ro/Forum/viewtopic.php?t=138988
In your topic,the approach is not necessary  :P
 Congratulare you on new idea  :roll:

I do not quite understand what you mean, Hoang. Can you clarify?

Establishing new bounds in inequalities involes a number of techniques and ideas. This idea makes use of equality case, factoring,... Tangent line method, and this parabolic estimation works well for other class of inequalities with fractions. Anyway, we have not been able to handle with inequalities which have more than one equality cases.


algebra
polynomial
induction
function
limit
algebra unsolved

Prove that exists a polynomial $P(x) \in \mathbb{R}[x]$, such that:

$P(x^2+1)=P(x^2)+1$ and $P(x)=0$

and also show it is unique.

$x \in \mathbb{R}$?

Then $P(n) = n$ by induction for integers $n$, so the polynomial $P(x) - x$ has infinite zeroes, implying it is zero everywhere.  Thus, $P(x) = x$.  QED.

[quote="mathmanman"]You may want to have a look at : http://www.mathlinks.ro/Forum/viewtopic.php?t=39570 . ;)[/quote]

But how do you show there's only one solution if P(x)=0 :?:

[quote="M4RI0"]Is it the same for

$P(x^2+1)=P(x)^2+1$ and $P(x)=0$   :?:[/quote]

Basically.

Consider the sequence $a_0 = 0, a_{n+1} = a_0^2 + 1$.  It is trivial to show by induction that $P(a_k) = a_k$ for non-negative integers $k$, meaning that, again, the polynomial $P(x) - x$ has infinite zeros, so it must be zero everywhere.

[hide="Some supporting lemmas"][b]Lemma:[/b]  If a polynomial $P(x)$ of degree $n$ is zero for $n+1$ distinct values of $x$, then $P(x) = 0$ everywhere.

Proof:  Consider the first $n$ roots $a_0, ... a_{n-1}$.  Then we can write

$P(x) = c(x - a_0) ... (x - a_{n-1})$ for some suitable constant $c$.  But then the polynomial has another root $a_n$, meaning

$P(x) = c(a_n - a_0)... (a_n - a_{n-1}) = 0$, and since all roots are distinct this implies $c = 0$.

[b]Lemma:[/b]  If a polynomial $P(x)$ is zero for an infinite number of values of $x$, then $P(x) = 0$ everywhere.

Proof:  Suppose the polynomial is of some finite degree $n$.  It follows immediately from the lemma that $P(x) = 0$ everywhere.[/hide]

Interesting thought raised from the first question:

Can anyone characterize the function $P(x) = \lim_{n \to \infty} \frac{ (x-1)(x-2)...(x-n) }{ n!}$?


calculus
integration
algebra
function
domain
abstract algebra
Ring Theory

Let $ R$ be a integral domain with fraction field $ L$, and assume that $ a$ is algebraic over $ L$.
Show that $ S$={$ r \in R : ra$ integral over $ R$}  is a nonzero ideal in $ R$.

How can I prove that for any $ r,s \in S$, $ rs \in S$?
I can't prove just that.

Don't try to prove that, because it isn't the right condition. Show that $ rs\in S$ when $ r\in R$ and $ s\in S$. For that- well, $ rsa$ is an algebraic integer when $ sa$ is. If $ x\equal{}sa$ is a root of $ p(x)\equal{}x^n\plus{}c_{n\minus{}1}x^{n\minus{}1}\plus{}\cdots\plus{}c_0\equal{}0$, $ x\equal{}rsa$ is a root of $ q(x)\equal{}x^n\plus{}rc_{n\minus{}1}x^{n\minus{}1}\plus{}^2c_{n\minus{}2}x^{n\minus{}2}\cdots\plus{}r^nc_0\equal{}0$.

Oh, I am sorry!!
It is not 'for any $ r,s \in S$, $ rs \in S$' but 'for any $ r,s \in S$, $ r\plus{}s \in S$'.

[hide="Hint"]Assuming that everything takes place in a field $ K$ use the criterion that an element $ b\in K$ is integral over $ R$ iff $ b$ is contained in a [b]finitely generated[/b] $ R$-module $ M$ which is contained in $ K$.[/hide]

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=202382]here[/url] for sketches of a constructive proof that sums and products of algebraic integers are algebraic integers. That's plenty here, since we get $ ra\plus{}sa$ an algebraic integer when $ ra$ and $ sa$ are.


Does anybody know about any books that have a lot of easy proof problems with the solutions?

wickedjestr, you've posted this already in the pre-olympiad forum. First off, first do hard problems without proofs and get 80%-90% of them correct. Proofs require problem solving skills that need a lot of experience to achieve. Why do think none of the AMC 10/12's or AIME's require proofs? but, if you still want one then the Problem Book (available at AoPs) by Titu Andreescu.

Hmm... Most books have proof problems.

However, you can write proofs for problems like find N.

Also, try doing hard problems that you can still do and writing proofs for those. Writing a proof for a very easy-to-prove problem is not going to help you as much as writing a proof for, say, an USAMO problem.

If you want help writing a solution, there is an article on Aops for it..

http://www.artofproblemsolving.com/Resources/AoPS_R_A_HowWrite.php

Oh!!! Okay thanks. I just thought that I would need practice with actually proving something, because I really don't know how to.

The Second Part of the [url=http://gauss.math.oakland.edu/mmpc/]Michigan Math Prize Contest (MMPC)[/url] has good problems for people just getting started in writing proofs. They have solutions to most of the past years' contests.


May '95. Why?

P.S. Miyomiyo and I are the same age. How weird...

[quote="davidyko"]This is weird. June '92. I feel so old compared to you two.[/quote]
same

June '93. :D

Apparently I'm neither young nor old...

[quote="miyomiyo"]My sister was born June '93.  How weird....  Why is everybody born June 93?  What day in June were you guys born?[/quote]
10th for me

OK. Also post the day you were born.

May 25.

Not 12 yet....

December 9th.  Not $11\frac{1}{2}$ yet...

This topic has been posted rather recently i think.

To break the chain of young people, Jan 14, 1991.


If $x^4 +ax^3 +2x^2 +bx +1 =0$ has real roots then $a^2+b^2 \geq 8.$

Hmmm... Do you mean that all its roots are real?

In that case :
Let x,y,z,t be the real roots of the polynomial. Thus, it is well-known that
x+y+z+t = -a  (1)
xy+xz+xt+yz+yt+zt = 2   (2)
xyz+xyt+xzt+yzt = -b   (3)
xyzt = 1  (4)

From (1) and (2), since the roots are real, we have :
a 2 - 4 = (x+y+z+t) 2  - 2(xy+xz+xt+yz+yt+zt) = x 2 +y 2 +z 2 +t 2   \geq 0

In another hand, from (3) and (4), we get :
b 2 - 4 = b 2 - 2*(xy+xz+xt+yz+yt+zt) = b 2 - 2*(1/xy + 1/xz + 1/xt + 1/yz + 1/yt + 1/zt) = (1/x+1/y+1/z+1/t) 2 - 2*(1/xy + 1/xz + 1/xt + 1/yz + 1/yt + 1/zt) = 1/x 2  + 1/y 2  + 1/z 2 + 1/t 2   \geq 0

Summing, it leads to the desired result. Note that the inequality may easily be proved to be strict.

Pierre.

Yes,all roots are real, therefore your proof is good.

It seems that the inequality is far from the best possible since, according to my previous message, we have :

a 2 + b 2  - 8 = x 2 +y 2 +z 2 +t 2 + 1/x 2 +1/y 2 +1/z 2 +1/t 2   \geq 2+2+2+2 = 8.

Thus, a 2 +b 2   \geq  16.  (*)

Since the equality case in (*) does not satisfy (2), this is still not the optimal result.

Pierre.

You are right. I copied the problem from the web, but I don't remember
where.

15 th Tournament  of  the  Towns  Autumn  1993


Find the equation of the line such that its distance between $3x-2y-6=0$ and $6x-4y-3=0$ are equal.

[hide]
The slope of theses lines are the same, so they are parallel.  The line with the same distance will be right in the middle, with the same slope.  It's y-intercept will be exactly between the y-intercepts of the two lines.  The y-intercepts are $-3$ and $-\frac 34$, so the y-intercept of the middle line is $-\frac{15}8$.  The equation of the line is $y = \frac 32 x - \frac{15}8 \implies 12x - 8y - 15 = 0$
[/hide]


[quote="Silverfalcon"]A bowling contest consists of several series. Mary got 185 points in her previous series. Now, let's say that her average is between 176 and 177 points (so it's not integer).

Then which of the following points below does she need to score in next series to increase her average to 178?

$ \text{(A)}\ 184 \ \qquad \text{(B)}\ 185 \ \qquad \text{(C)}\ 186 \ \qquad \text{(D)}\ 187 \ \qquad \text{(E)}\ 188$[/quote]
Um, what?

[hide="My solution sketch"]

I let the average (the one between 176 and 177) to be 176.5.

$ \frac{nX \plus{} 185}{n\plus{}1} \equal{} 176.5, \ \frac{nX \plus{} 185 \plus{} P}{n\plus{}2} \equal{} 178$

where $ nX$ is the sum of previous scores and $ P$ is the new score that she must make.

Simplifying two equations give

$ A\plus{}P\minus{}356 \equal{} (178\minus{}A)n$

Further simplifying this gives

$ P \equal{} 1.5n \plus{} 179.5$

Now, for $ n \equal{} 3$ and $ n \equal{} 5$, we have two numbers that belong in the given range (184 and 187). I'm not sure how to go from here though.. ha  :P  [/hide]

Uh, what I meant was, are you sure the problem is correct? What I'm reading it as is she scored 185, and her average is somewhere between 176 and 177. That tells us nothing...

Ah, you're right.

I copied the first part and didn't fix it.

Arr.. I'm just going to lock the problem because I don't feel like fixing it to what I wanted to do.  :D


inequalities
integration
real analysis
real analysis unsolved

Suppose $ f$ is continuous and bounded  on $ \mathbb{R}$ and $ \alpha>0$.
Show that there is a unique $ u$ bounded wich verify : $ \minus{}u'' \plus{} \alpha^2 u \equal{} f$.
If $ f$ is in $ L^p$, show that $ \parallel{}u\parallel{}_p \leq C \times \parallel{}f\parallel{}_p$ with $ C$ constant and $ \parallel{}.\parallel{}_p$ the norm $ L^p$.

I sometime meet the following boundary value problem
\[ \left\{\begin{gathered} \minus{} u'' \plus{} u \equal{} f \hfill \\
u\left( 0 \right) \equal{} \alpha \hfill \\
u\left( 1 \right) \equal{} \beta . \hfill \\
\end{gathered} \right.
\]
If $ f \in L^2(0,1)$ then the existence of $ u \in H^2$ is unique. Moreover if $ f \in C([0,1])$ then $ u \in C^2([0,1])$.

Usually to show existence and uniqueness of the solution to such boundary value problems you use Lax-Milgram theorem.

The uniqueness is clear (just look at the difference of $ 2$ solutions). One can even find an explicit solution. For example, if $ X(t)\equal{}(u(t), u'(t))$ then $ X(t)$ satisfies
$ X'(t) \equal{} A.X(t) \plus{} f(t)$
where $ A$ is the appropriate matrix. The solution of the homogeneous ODE is simply $ X(t) \equal{} e^{t A} . X_0$, and the solution of the non homogeneous ODE is (variation of constant):
$ X(t) \equal{} e^{t A} [X_0 \plus{} \int_0^t e^{s A} f(s) ds]$
From here it should not be very hard to conclude ..


Let $ a,b,c$ be non-negative real numbers and they aren't all equal to zero. Prove that
$ \sum\limits_{cyc}a^3\plus{}3abc\ge \sum\limits_{cyc}ab(a\plus{}b)\plus{}\frac{\sum\limits_{cyc}ab(a\minus{}b)^2}{a\plus{}b\plus{}c}$

[quote="The soul of rock"]Let $ a,b,c$ be non-negative real numbers and they aren't all equal to zero. Prove that
$ \sum\limits_{cyc}a^3 \plus{} 3abc\ge \sum\limits_{cyc}ab(a \plus{} b) \plus{} \frac {\sum\limits_{cyc}ab(a \minus{} b)^2}{a \plus{} b \plus{} c}$[/quote]
It's equivalent to $ \sum_{cyc}(a\minus{}b)^2(a^2\plus{}b^2\plus{}ab\minus{}c^2)\geq0,$ which obviously true.

$ \sum\limits_{cyc}a^3\plus{}3abc\ge \sum\limits_{cyc}ab(a\plus{}b)\plus{}\frac{\sum\limits_{cyc}ab(a\minus{}b)^2}{a\plus{}b\plus{}c}$

$ \Leftrightarrow c(x^2\plus{}xy\plus{}y^2)\plus{}x^2(x\plus{}2y)\;\ge\; \frac{2(x^2\minus{}xy\plus{}y^2)c^2\plus{}(2x^3\plus{}x^2y\plus{}xy^2\plus{}2y^2)c\plus{}xy(x^2\minus{}xy\plus{}y^2)}{2c\plus{}x\plus{}y}$

$ \Leftrightarrow \alpha c^2\plus{}\beta c\plus{}\lambda\ge 0$....

I hope it is right  :)


There are 3 piles of 6 sticks each. What is the smallest number of sticks that can be moved so that the second pile contains twice as many sticks as the first pile, and the third pile contains three times as many sticks as the first pile?

There are $ 3(6)\equal{}18$ sticks total. Call the number of sticks in the first pile $ x$. Then, $ x\plus{}2x\plus{}3x\equal{}18$, and $ x\equal{}3$. Thus, the last pile will need $ 3x\equal{}3(3)\equal{}9$ sticks. By moving $ 6\minus{}3\equal{}\boxed{3}$ sticks from the first pile to the third pile, we accomplish what we set out for.


I tried this code on latex but i dont know whats wrong i will type in the code here without the dollar sign

x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{5\pm\sqrt{-5^2-4(4)(-9)}}{2(4)}
=\dfrac{5\pm\sqrt{25+144}{8}

$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}=\frac{5\pm\sqrt{-5^{2}-4(4)(-9)}}{2(4)}=\frac{5\pm\sqrt{25+144}}{8}$

can you tell me where and put that brace in bolld by just copying my equation using the copy function and pasting then put the bbrace in bold


Yes, I am wrong. Let $D=gcd(C_{n}^{1},...,C_{n}^{n-1})$. Because $C_{n}^{1}=n$ we have $D|n$. If $n=p^{r}k, (p,k)=1$, then $p\not |C_{n}^{p^{r}}$. Therefore $D>1$ if and only if $n=p^{r}$ in this case $ord_{p}(C_{p^{r}}^{p^{r-1}})=1$, therefore $D=p$.


Let $xOy$ be convex angle and $M$ point inside it. Proove that there exists unique point $P$ in the plane of the angle, such that, for any line passing through $M$ and intersecting lines $xO$ and $yO$ (including extensions of rays $xO$ and $yO$) at points $X$ and $Y$, it holds that $\angle XPY \leq \frac \pi 2$.

$MP$ is parallel to $Ox$ and intersect $Oy$ at $P$, $MQ$ is parallel to $Oy$ and intersect $Ox$ at $Q$, $l_1$ passes $P$ and ortho to $Ox$, $l_2$ passes $Q$ and ortho to $Oy$, then $P$ is the intersection of $l_1$ and $l_2$

yes, but you have used $P$ two times :)
but how did you prooved it? i am interested in solutions other than official one. it uses continuity, when line tends to parallel with $Ox$ and $Oy$. if angle when line is "almost parallel" with $Ox$ is $\phi$ then the other is $\pi - \phi$, so, it needs to be right angle. and when we find the point, we proove it using vectors.

so, if your solution is different, could you post it please?


http://www.artofproblemsolving.com/Forum/viewtopic.php?t=218429 didn't inspire many people, here is another one slightly different but which also yields to a nice result:
Find all continuous morphisms from $ \mathrm{GL}_n(\mathbb{R})$ to $ \mathbb{Z}$.

$ GL_n(\mathbb{R})$ has only $ 2$ connected component isn't it ?


I used to write \mathbb R to get $ \mathbb{R}$ but now this command produces
$ \mathbb R$. I'm not too lazy to put {} around R, but a lot of existing posts are now less readable than they used to be. :(


[quote="Inequalities Master"]Let  $ a_1,a_2,a_3,a_4,a_5,a_6 \in\ [\frac {1}{\sqrt {3}},\sqrt {3}]$.Prove that:

$ \frac {a_1 \minus{} a_2}{a_2 \plus{} a_3} \plus{} \frac {a_2 \minus{} a_3}{a_3 \plus{} a_4} \plus{} ... \plus{} \frac {a_6 \minus{} a_1}{a_1 \plus{} a_2}\geq 0$.[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=197705


What product is expected to be obtained when o-diformylbenzene is treated with sodium hydroxide, followed by acidification?

Just totally ignore the post if its wrong
is it benzene 1,2 dicarboxylic acid  :?:
Hypothesis : There is decarboxylation to form benzoic acid  :!:

[quote="VARUNARASU"]is it salicylic acid?[/quote]

But that's Dakin West reaction and you require H2O2 for that  :maybe:

sorry its not salicylic acid. wat i meant was there is an ch2oh group and also a cooNa group ortho to each other which on treatment with with acid forms a cyclic ester

Yes the reaction is a cannizaro reaction and then the formed acid and alcohol will undergo esterification to give lactone.

Do you mean a 3- 6membered cyclic compound with the middle ring being a lactone?  :D

no a five membered cyclic compound with a lactone in the middle .

Yes, that's right, there is a Cannizaro reaction.

[quote="VARUNARASU"][quote="madness"][quote="madness"]Do you mean a 3 membered cyclic compound with the middle ring being a lactone?  :D[/quote]

i meant 3 - 6 membered cyclic compds  :D  with lactone in the middle[/quote]
wat do you mean by 3-6 membered cyclic compound. are you talking abt bicyclic compounds? this compound has a five membered ring adj to benzene ring.[/quote]

I mean isnt there a lactone ring (6 membered) in between 2 benzene rings   :maybe:

no, its an intramolecular canizzaro reaction followed by intramolecular lactonization

[quote="madness"]Has it got something to do with Cannizaro reaction then? if yes then is the answer either di CH2OH benzene or 2 CH2OH benzoic acid?  :D[/quote]

Thats what i meant by 2 CH2OH benzoic acid  :D

[quote="Carcul"]Actually, lactonization is not expected to happen.[/quote]
and why is it not expected to happen? you have acidified the product in the end, and suppose a strong enough acid is used lactonisation must happen?

I am very sorry lactonisation will not occur here because it will create a very large ring and so is unstable (I had read a similar type and got confused :blush: ). But after cannizaro I dont think there is any other step possible

what do you mean a very large ring. a five membered ring is created

no there are 2 benzene rings sandwitching that ring as i have been saying all along  :D

Sorry I thought about intermolecular esterification I think the intramolecular lactonisation is surely 100% correct :coolspeak:


induction
real analysis
real analysis solved

For any n natural number <>0 we consider the sequence (x_n)n>=1 defined like so: x1>0 and x_(n+1)=x_n/(n*(x_n)^3+2).
a)study the convergence of the sequence;
b)compute its limit.

cheers! :D

All x_n > 0 by induction
x_(n+1)/x_n = 1/(n*(x_n)^3+2)  \leq 1/2 <1
x_n is decreasing
x_n converge lim x_n = k  \geq 0

suppose k> 0
from x_(n+1) = x_n/(n*(x_n)^3+2) take the limit
k=k/(lim ( n(x_n)^3+2) ) = k/oo = 0 contradiction

so k = 0


search
abstract algebra
group theory
superior algebra
superior algebra unsolved

Prove that if $ G$ is a group with $ 2009$ elements then the number of solutions of the equation $ x^7 \equal{} e$, where $ e$ is the identity element, is divisible by 7.

There is a general statement (that I cannot prove ) which says that " If $ G$ is a finite group and $ m | |G|$ then the number of solution of the equation $ x^m \equal{} 1$ is divisible by $ m$."

But when $ m$ is prime, this easily follows from the proof of Cauchy's theorem. (have a look [url=http://www.mathlinks.ro/viewtopic.php?search_id=686093762&t=27036]here[/url] at grobber's post )

Another way you can solve the problem in this particular case is to prove that $ |G| \equal{} 2009$ is an abelian group (using Sylow's theorems ), therefore the set of solution of the equation $ x^7$ is a subgroup of $ G$, with the order divisible by $ 7$.

I like the proof using Sylow's theorems said above.

Just for a different proof which also depends on $ p$ being prime though this has the same partitioning flavor as the proof of Cauchy's theorem already mentioned :P

Let $ N \equal{} \{x\in G: x^7 \equal{} e\}.$

Partition $ N \backslash \{e\}$ into 6-sets of the form $ \{a, a^2, ..., a^6\}$ for $ a\in N\backslash \{e\}.$

(If two parts share an element, say $ a^k \equal{} b^l$ for $ k,l\in [6]$, then $ a \equal{} b^{ln}$ for some $ n\in [6]$, meaning the classes of $ a$ and $ b$ are identical.)

Thus $ |N| \equal{} 6k \plus{} 1$.  The only sets of cardinality dividing 2009 that could satisfy a relation are of size $ 7$ or $ 49$.

-evo

i should have said the following instead: the group action by $ \mathbb{Z}_7^{\times}$ on $ G$ defined by $ a\cdot x \equal{} x^a$ has orbits partition which partition $ N$.  all but one are of size 6.  thus $ |N| \equal{} 6k\plus{}1$.  as $ |N|$ divides 2009, 7 must divide $ |N|$.

-evo

[quote="lamphong"]Hey, why don't we try to prove this
[quote="Svejk"]There is a general statement (that I cannot prove ) which says that " If $ G$ is a finite group and $ m | |G|$ then the number of solution of the equation $ x^m \equal{} 1$ is divisible by $ m$."

[/quote][/quote]

If I remember correctly, you show know some character theory.

No character theory is necessary.  For a very easy proof, see Isaacs and Robinson, AMM 99, pg. 352.

Thanks Steve. [url=http://www.math.wisc.edu/~dynerman/content/frobtalk.pdf]Here[/url] is what I've found.


[quote="Lovasz"]Let $a, b, c > 0$ satisfy $a+b+c=1$, then:
$\sum\frac{ab}{1+c} \le \frac{1}{4}$[/quote]

Too.......easy....!

$\frac{ab}{1+c}=\frac{ab}{(a+c)+(b+c)} \le \frac{1}{4}(\frac{ab}{a+c}+\frac{ab}{b+c})$

sum up and done!
 :)


calculus
integration
derivative
trigonometry
calculus computations

I was told that I have five of the questions in the first pdf file below wrong but I'm convinced that I have all of the answers right. The second pdf file contains all of my calculations to the 25 questions - can someone tell me, only, how many questions I have wrong if any? TIA.

http://www.sendspace.com/file/2ckowm

http://www.sendspace.com/file/dhg3lj

It seems to me like several of the choices are wrong.  For example, on number 25, the correct answer is $ \minus{}\cot(x/2)\plus{}C$.  For number 23, it is $ x\plus{}\frac{4}{2\plus{}x}$.  Neither of those choices are there.

Thanks for your feedback. I agree with you that there is no correct answer in the choices for #23 but for #25 I believe choice a) is correct.

[url=http://img387.imageshack.us/i/number25.jpg/][img]http://img387.imageshack.us/img387/274/number25.th.jpg[/img][/url]

Yes, my mistake.  The two are equivalent.  I didn't actually compute the integral, I instead computed the derivative of what you found, so I must have made a small mistake in my calculations.

$ \minus{}\cot(x/2)\plus{}C$ is also correct according to my math software.

They are equivalent, the formula you expressed is just the half-angle formula for cotangent.  Since $ \cot(\alpha\minus{}\beta)\equal{}\frac{\cot{\alpha}\cot{\beta}\plus{}1}{\cot{\beta}\minus{}\cot{\alpha}}$, letting $ \alpha\equal{}\theta$ and $ \beta\equal{}\theta/2$, we get
$ \cot\left(\frac{\theta}{2}\right)\equal{}\frac{\cot\theta\cdot\cot\left(\frac{\theta}{2}\right)\plus{}1}{\cot\left(\frac{\theta}{2}\right)\minus{}\cot\theta}$.  Let $ u\equal{}\cot\theta$ and $ v\equal{}\cot(\theta/2)$ (to shorten the amount of writing I have to do):
$ v\equal{}\frac{uv\plus{}1}{v\minus{}u}\Rightarrow v^{2}\minus{}2uv\minus{}1\equal{}0\Rightarrow v\equal{}\frac{2u\pm\sqrt{4u^{2}\plus{}4}}{2}\equal{}u\pm\sqrt{u^{2}\plus{}1}$.  Recalling that $ cot^{2}{x}\plus{}1\equal{}\csc^{2}{x}$, we get that the above expression is equivalent to:
$ \cot{\theta}\pm\sqrt{\csc^{2}{\theta}}\equal{}\cot{x}\plus{}\csc{\theta}\equal{}\frac{1\plus{}\cos{\theta}}{\sin\theta}$


algebra
polynomial
function
domain
Ring Theory
superior algebra
superior algebra solved

Which of the following rings are principals: $\mathbb{Z}[i]$, $\mathbb{Z}_{2} [X]$, $\mathbb{Z} [X,Y]$ ??? :)

[quote="alekk"]what is $\mathbb{Z}_2[X]$ ?[/quote]

You probably ask what does he mean by $\mathbb Z_2$, because otherwise it is just the polynomial ring over $\mathbb Z_2$ :). In Romania we use $\mathbb Z_2$ for $\mathbb Z/2\mathbb Z=\mathbb F_2$. Generally (internationally), $\mathbb Z_2$ denotes the ring of $2$-adic integers.

$\mathbb Z[\imath]$ is euclidean thus a principal ideal domain.

$\mathbb Z_2=\mathbb Z/2\mathbb Z$ is a field, so $\mathbb Z_2[X]$ is euclidean thus a principal ideal domain.

$\mathbb Z[X,Y]$ is not a principal ideal domain because the ideal $(X,Y)$ cannot be generated by just one element.
Actually not even $\mathbb Z[X]$ is principal. The ideal $(2,X^2+1)$ cannot be generated by one element.

I'd say for that $\mathbb{Z}_{2} [X]$ that the ring of polynomials in one variable over any field is an euclidean domain and thus certainly principal ideal domain


In how many ways can 6 people be seated in a motor car if only 2 can occupy the driver's position?

how many people can seat in the car normally?(what is the capacity of the car)

also, do you mean only two people know how to drive?

120 for each driver 120 * 2 = 240.
I think that's what he meant.

Yeah, I think it would be $\binom 2 1 \cdot 5! = 240$

yes, i think it is a 5 person car, and the answer is $2!5!$


function
combinatorics open
combinatorics

Thank you very much!! :D In fact, I don't have this program (Mathematica), so I didn't think I could find the answer using it :lol:

However, it is possible for certain values of $ \alpha$. For example, $ \alpha = 2$ has the generating function $ (1-4x)^{\frac{-1}{2}}$ as can be verified by the extended binomial theorem.

Hi! You can find the enjoyable about the generating function at following file :D

For $ \alpha\equal{}3$, the generating function is given at: http://www.mathlinks.ro/viewtopic.php?p=422893#422893

[quote=Jutaro]If $\alpha$ is a complex number, is it possible to express the generating function
\[\sum_{n \geq 0}\binom{\alpha n}{n}x^{n}\]
in closed form?[/quote]
Complex $\alpha$ may be tricky, but at least for real $\alpha\ne 0$, we can use [url=https://en.wikipedia.org/wiki/Lagrange_inversion_theorem#Lagrange%E2%80%93B%C3%BCrmann_formula]Lagrange–Bürmann formula[/url].
Let $w(t)$ be the function satisfying $w(t) = t(1+w(t))^{\alpha}$ (in other words, $w(t)$ is the series reversion of $\frac{t}{(1+t)^{\alpha}}$).
Then
$$\sum_{n\geq 0} \binom{\alpha n}{n} x^n = \frac{1}{1-\alpha x (1+w(x))^{\alpha-1}}.$$

See also [url]https://artofproblemsolving.com/community/c6h281653p1847194[/url]


[quote="jonathanchou711"]So what are we trying to find here? 2^infinity power?[/quote]
correct! :D

[quote="jonathanchou711"]..... uh... is that even worth posting?   :maybe:[/quote]

[hide="huh?"]err, i don't know[/hide]


counting
distinguishability
combinatorics proposed
combinatorics

How many ways are there to put $n$ balls in $3$ boxes

(1) if the balls are distinguishable (labelled as $1,2,\cdots n$) and the boxes are also distinguishable (labelled as $A,B,\ and\ C$)?

(2) if the balls are indistinguishable but the boxes are distinguishable (labelled as $A,B,\ and\ C$)?

(3) if the balls are distinguishable (labelled as $1,2,\cdots n$) but the boxes are indistinguishable?

(4) if neither the balls nor the boxes are distinguishable and $n$ is a multiple of $6$ denoted by $6m\ (m\in \mathbb{N})$

Note (for all 4 problems): There are some cases that one or two boxes contain no balls.
                                      These cases count as well.

[hide]1)  Since the boxes are distinguishable the distribution of the balls between the boxes matter and since each of the $n$ numbers(labeled in the balls)  belong to exactly one of the three boxes the number of ways to put the balls is $3^{n}$

2) Since the balls are indistinguishable just matter the quantity of the balls in each box, so the number of ways is the number of ordered solution of the equation $x+y+z=n$ that is $\frac{ (n+1)(n+2)}{2}$

3) In that case the distribution of the balls do not matter. Let $m$ be number of ways to put the balls (obeying case ' 1) ' condition) so that there are no empty box and $l$ be the number of ways to put the balls (again obeying case '1)' condition) so that there are exactly one empty box.Then, the answer to case '3)' is :

$\frac{m}{3!}+\frac{l}{3.2!}+1 = \frac{m+l+6}{6}=\frac{3^{n}+3}{6}=\frac{3^{n-1}+1}{2}$

 4) Here we have to find the number of unordered triples $(x,y,z)$ of naturals numbers that are solutions of the equation $x+y+z=n$. If $x=y=i$ and $z$ ≠ $i$  the number of ordered solution of the equation $2i+z=n=2.(3m)$ is $3(3m)=9m$, because $0 \leq i \leq 3m$, $3$ is the number of permutation of $(i , i , z)$ and the triple $(2m,2m,2m)$ cannot be included. Now if $k$ is the number of ordered triple $(x,y,z)$ that are solutions of the given equation so that $x$ ≠ $y$  ≠ $z$, then, considering case '2)' and including $(2m,2m,2m)$ we have that:

 $k+9m+1=\frac{ (n+1)(n+2)}{2}\implies k=\frac{n^{2}+6n+4}{2}$ the answer to case '4)' is :${\frac{k}{3!}+3m+1}$[/hide]


advanced fields
advanced fields unsolved

Let $ A$ be a commutative ring with unit. Let $ V(I): \equal{}\{\mathfrak{p}\in\operatorname{Spec}A: I\subset \mathfrak{p}\}$. Let $ F$ be a subset of an affine scheme $ \operatorname{Spec}A$. Show that the closure $ \bar{F}$ of $ F$ in the underlying topological space of $ \operatorname{Spec}A$ is equal to $ V(\cap_{\mathfrak{p}\in F}\mathfrak{p})$.

every closed subset of $ \operatorname{Spec} A$ has the form $ V(I)$ for some ideal $ I$. now $ F \subseteq V(I)$ is equivalent to $ I \subseteq \mathfrak{p}$ for all $ \mathfrak{p} \in F$, i.e. $ I \subseteq \cap_{\mathfrak{p} \in F} \mathfrak{p}$. in particular, $ V(\cap_{\mathfrak{p} \in F} \mathfrak{p}) \subseteq V(I)$, and $ V(\cap_{\mathfrak{p} \in F} \mathfrak{p})$ is the smallest closed subset containing $ F$.


calculus
integration
trigonometry
calculus computations

An algebraic note: $ A\cos (\omega x)\plus{}B\sin (\omega x)$ can always be rewritten as $ R\cos(\omega x\minus{}\delta),$ where $ R\cos\delta \equal{} A$ and $ R\sin\delta \equal{} B.$ In this case, that would make $ R\equal{}\sqrt{41}$ and $ \delta\equal{}\arctan\left(\frac{4}5\right).$


calculus
integration
trigonometry
derivative
calculus computations

What method can be used to solve this integral? I first of looked at the problem with integration by parts, but that would give me problems pretty soon.

$\int_0^1x\cdot 3^xdx$

If we say $u=x$, then $u'=1$ and we will have $v'=3^x$ which gives us $v=\frac{3^x}{ln(3)}$, but this $v$ will later give me problems i can't handle with my current knowledge given by the textbook about integration methods, as i will have another integral called $\int_0^1 1\cdot \frac{3^x}{ln(3)}$ that i can't work with.

Switching between the values of $u$ and $v$ will also give me a problem AFAIK.

No, no, you're on the right track. You seem to have no trouble with $3^x$. That $\ln(3)$ is just a constant -- you can move it out of the integral. Then you'll just have to integrate $3^x$ one more time and evaluate at the bounds. :D

So i end up with $\int_a^bk\cdot f(x)$ which doesn't usally give me a problem. I got it :) It is just about finding those small tricks you know works usally and see what trick can be applied in this case. Easier said than done though, unfortunately. Will remember this trick for the next time.

What about this one. Have been trying several things on this one, and i have a feeling it is easier than i think.

$\int_0^\pi e^x\cos(x)$

It says i should use integration by parts twice. I'm pretty sure i can't eliminate the $e^x$, so i thought about $cos(x)$. But then i get (looking at the $\cos(x)$ only):

$\cos(x)-\sin(x)-(-\cos(x))$

And this doesn't eliminate anything.

$2\cos(x)-\sin(x)$

Not even if i go on integrating $-\cos$

$\int e^x\cos x dx=e^x\cos x-\int e^x(-\sin x)dx=e^x\cos x-\left(-e^x\sin x+\int e^x\cos x dx\right)$
$\Longleftrightarrow 2\int e^x\cos x dx=e^x(\cos x+\sin x)$

To give a little more explanation for what he did:

You want to integrate by parts twice not to eliminate the original integral, but to get a second copy of it. If $I$ is the original integral, then you have $I = A + B = A + C - I$, so $2I = A + C$.

This is a common way to deal with integrals involving sine and cosine, since integrating those twice (or the derivative twice) will give you the same thing back.

Ok, i made it finally. Took a bit more time though. I was still having problems with the first one. Two tasks i really learnt a great deal from. But expect that i will drop by for a glas of wine later with some more integration - i don't expect myself to be able to do the rest of the tasks honorstly ;) But ok, will try and leave the integration problem for a day if i get into trouble. That often solves a lot of the problems i encounter.


limit
logarithms
integration
real analysis
real analysis theorems

I was just wondering how the definition of $e=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n}$ follows from the definition of e as the number such that $\ln e=\int_{1}^{e}\frac{1}{t}dt=1$.

With substitution $t=u^{n}$ you get $\int_{1}^{(1+1/n)^{n}}\frac{dt}{t}=n\int_{1}^{1+1/n}\frac{du}{u}$, which tends to $1$ as $n\to\infty$ (by the Mean Value Theorem for integrals).


1.)  The sum of the infinite fraction $ \frac {1}{8} \plus{} \frac {2}{8^2} \plus{} \frac {3}{8^3} \plus{} ...$equals what simple fraction?

2.)  How many integers between 1000 and 9999 are perfect squares and have sums equal to 14?

3.)  You start with 128 cents, and six times bet half of your current amount of money, with probability .5 of winning on each occasion. You win 3 times and lose 3 times in some order. Does the amount of money which you have at the end depend upon the order of wins and losses, and if not, how many cents will you have at the end? Your answer should be either the word "depends" or else the number of cents that you will always have at the end.

For what real number c does the equation $ |x^2 \plus{} 8x \plus{} 12| \equal{} c$ have exactly three solutions?

[hide="1"]Call the sum $ S$. Now, consider $ S\minus{}\frac18S$. This can be written as $ (\frac18\plus{}\frac{2}{8^2}...)\minus{}(\frac{1}{8^2}\plus{}\frac{2}{8^3}...)\equal{}\frac18\plus{}\frac{1}{8^2}...\equal{}\frac{\frac18}{1\minus{}\frac18}\equal{}\frac17$. So, $ \frac78S\equal{}\frac17$, so $ S\equal{}\boxed{\frac{8}{49}}$.[/hide]

[hide="3"]3 times you will lose half your money, and three times you will gain half your current money. In other words, multiply by $ \frac12$ 3 times and $ \frac32$ 3 times. No matter the order, therefore, the end total is $ (\frac12)^3(\frac32)^3\times128\equal{}\boxed{54}$ cents.[/hide]

2 is unclear, and I'm not in the mood for 4.

I believe 2) should read "the sum of whose digits is 14."

[quote="ajai"]
For what real number c does the equation $ |x^2 \plus{} 8x \plus{} 12| \equal{} c$ have exactly three solutions?[/quote]
[hide="4"]
If you look at the shape of $ y\equal{}|x^2 \plus{} 8x \plus{} 12|$, the only way to get a horizontal line intercepting at 3 points is if it is tangent to the vertex. Now, as $ x^2 \plus{} 8x \plus{} 12 \equal{} (x \plus{} 4)^2 \minus{} 4$, normally the vertex would be at $ ( \minus{} 4, \minus{} 4)$, but as it is flipped about the x-axis, it becomes $ ( \minus{} 4,4)$ for $ y\equal{}|x^2 \plus{} 8x \plus{} 12|$. So the horizontal line is $ y \equal{} 4 \Rightarrow \boxed{c \equal{} 4}$[/hide]

[hide="#4"]$ x_{min} \equal{} \minus{} 4\Longrightarrow f( \minus{} 4) \equal{} \minus{} 4$
so, $ \boxed{c \equal{} 4}$[/hide]

Thought so...

[hide="2"]The SOD must be 14, so $ n\equiv5(\mod9)$, and thus $ n\equiv2(\mod3)$. All squares are 0 or 1 mod 3, so there are no solutions.[/hide]


Find the largest positive integer that if you divide it into each of $4201, 6301$ and $7351$ the remainder will be $1$.

[hide]Find the HCF of 4200, 6300, 7350

clearly the solution is [size=150]1050[/size][/hide]

What is the process that you used to calculate the HCF of 4201, 6301, 7351
Other than just reconizing that each is 1 away from a multiple of 1050
Is there acually a algebraic way... :? :)

AMPIS, you can prime factorise then multiply the common prime factors.
6300 = 2 * 3^2 * 5^2 * 7
4200 = 2^3 * 3 * 5 * 7
7350 = 2 * 3 * 5^3 * 7^2

Taking common prime factors 5^2 * 2 * 3 * 7 = 1050.


algebra
polynomial
calculus
derivative
inequalities
real analysis
real analysis unsolved

1) It there exists polynomial $ P(x)$, for which $ P(x) > P''(x)$ and $ P'(x) > P''(x)$ for all $ x$ from $ R$?
2) Show that if $ P(x) \geq P'(x)$ for all x from R then $ P(x) \geq 0$ for all $ x$ from R.

For number 1, no: if $ P'(x) > P''(x)$ for all $ x$, $ P'$ must be of even degree.  Then $ P$ and $ P''$ are of odd degree and the degree of $ P$ is larger.  Then as $ x \to \pm \infty$ (sign depending on the sign of the leading coefficient of $ P$), $ P$ will approach $ -\infty$ faster than $ P''$ will and so will at some point be smaller.

For number 2; Put $ F(x)=\exp(-x)P(x)$. Then the derivative of $ F$ is $ \exp(-x)P'(x)-\exp(-x)P(x) \leq 0$ for all real $ x$. If there was a value $ y$ with $ P(y) <0$ then $ F(y)$ would be negative but this is a contradiction since the size of $ F(x)$ goes to $ 0$ as $ x\top\infty$ - i.e there is a value $ a>y$ with $ F(a)>F(y)$ contradicting the fact that $ F$ decreases. (I've just realised the inequality isn't strict - however assuming that $ F'(x)$ is $ 0$ for sufficiently large $ x$ leads to a contradiction anyway  :) .)

Yes, thank you.Post this problem be bad. :blush:


The 5th Annual Rockdale Math Competition is December 12, 2009.

If you or a team would like to attend, go to [url]http://web.me.com/drcgarner/RMC[/url] and download the registration form and information letter.  This is a tournament with a focus on middle school and junior varsity.  (Up to three 6-person teams from the same school may attend in each division.)  There is also a Varsity division, but only one six-person team is invited, and that invitation is contingent upon bringing a JV team as well.  (Details are in the information letter.)

The registration deadline is November 30.  There is late registration for an extra fee; the late registration deadline is December 7.

However, for AoPS members, I will waive the late registration fee.  Simply write "I saw this at AoPS" at the bottom of the registration form and mail it in by December 7, and your school will only have to pay the standard registration fee.

I hope to see you on December 12!

Hmm, the Rockdale Tourney website doesn't have the questions for tiebreakers.  Are they highly classified, or can I get them from a Rockdaler?  Thanx!

Sorry, the tie-breaker questions are highly classified!  I'm the only one who sees them until they are ready to be used.

Aww...  Oh well... Thanks though. :)

How about the answer key?  :lol:


A particle $ P$ of mass $ m$ is attached to points $ A$ and $ B$, where $ A$ is a distance $ 9a$ vertically above $ B$, by elastic strings, each of which has modulus of elasticity $ 6mg$. The string $ AP$ has natural length $ 6a$ and the string $ BP$ has natural length $ 2a$. Let $ x$ be the distance $ AP$.

The system is released from rest with $ P$ on the vertical line $ AB$ and $ x \equal{} 6a$. Show that the acceleration $ \ddot{x}$ of $ P$ is $ \frac {4g}{a}(7a \minus{} x)$ for $ 6a < x < 7a$ and $ \frac {g}{a}(7a \minus{} x)$ for $ 7a < x < 9a$.

Find the time taken for the particle to reach $ B$.

Modulus of elasticity is defined as the quotient of the stress by the strain. Here, what exactly means $ 6mg$ ?

The question did not specify.. a good guess is $ m$ is the mass of the particle and $ g$ is the acceleration due to gravity.

So, $ 6mg$ must be a force and can't be Young's modulus.

But given that it's a Cambridge STEP question, they would have issued an errata if there was something wrong. Hmm


inequalities
quadratics
symmetry
inequalities proposed

[quote="shaaam"]Given that $ a,b,c \geq 0$. Prove that,
\[ \left(a \plus{} b \plus{} c\right) \cdot \left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right)\geq 4\cdot\frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \plus{} 5
\]
[hide="Challange"]Find the maximal $ k$, so that the following inequality holds:
\[ (a \plus{} b \plus{} c)\cdot \left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right) \geq k \cdot \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \plus{} 9 \minus{} k
\]
I have a feeling that, $ k \equal{} 4$ is best constant. Try to prove or disprove me.
Thanks!  :) [/hide][/quote]
You are right. $ k\equal{}4$ is the best constant!
[hide="proof"]
I've written the general in pqr notation:
$ pq^2 \minus{} kp^2r \minus{} 9qr \plus{} 3kqr \ge 0$.
Assume wlog $ p \equal{} 1$. Then we just have to prove $ f(a,b,c) \equal{} q^2 \minus{} kr \minus{} 9qr \plus{} 3kqr$ with the constraint $ g(a,b,c) \equal{} p \minus{} 1 \equal{} 0$.
It is obviously true at the extremum values: $ f(1,0,0) \equal{} 0$, so we just have to find other local extremas.

Btw. Note that $ \frac {\Delta p}{\Delta a} \equal{} 1$, $ \frac {\Delta q}{\Delta a} \equal{} p \minus{} a$, $ \frac {\Delta r}{\Delta a} \equal{} \frac {r}{a}$.
So $ f_a(a,b,c) \equal{} \frac {\Delta f(a,b,c)}{\Delta a} \equal{} 2(p \minus{} a)q \minus{} k\frac {r}{a} \plus{} (3k \minus{} 9)(pr \minus{} ar \plus{} \frac {qr}{a})$.

Using Lagrange Multipliers, since $ g_a(a,b,c) \equal{} g_b(a,b,c) \equal{} g_c(a,b,c) \equal{} 1$ we have to find $ a,b,c$ such that $ f_a(a,b,c) \equal{} f_b(a,b,c) \equal{} f_c(a,b,c)$. But this implies that two of the variables must be equal. Say $ a \equal{} b$. (It is because $ f_a(a,b,c)$ is only a quadratic, so there is only two solutions to $ f_a(a,b,c) \equal{} k$. Because of symmetry, we have through the pigeon hole principle, that two of them must be equal if a extramum is attained!).

So we only have to prove $ f(a,a,c) \ge 0$. Now forget we assumed $ p \equal{} 1$ :) Instead assume wlog $ a \equal{} 1$. Then we just have to prove $ f(1,1,x) \ge 0$. Then $ p \equal{} 2 \plus{} x$, $ q \equal{} 2x \plus{} 1$, and $ r \equal{} x$.

So we have to find the max $ k$ such that:

$ (2 \plus{} x)(2x \plus{} 1)^2 \minus{} k(2 \plus{} x)^2x \minus{} 9(2x \plus{} 1)x \plus{} 3k(2x \plus{} 1)x \ge 0 \iff$
$ (4 \minus{} k)x^3 \plus{} (2k \minus{} 6)x^2 \minus{} kx \plus{} 2 \ge 0$.

Then it's obvious that if $ k > 4$ then it is wrong for large $ x$. So we just prove it for $ k \equal{} 4$. Then it becomes:
$ 2x^2 \minus{} 4x \plus{} 2 \equal{} 2(x \minus{} 1)^2 \ge 0$. So equality iff $ x \equal{} 1$. That is: $ a \equal{} b \equal{} c$.[/hide]

[quote="shaaam"]Given that $ a,b,c \geq 0$. Prove that,
\[ \left(a \plus{} b \plus{} c\right) \cdot \left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right)\geq 4\cdot\frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \plus{} 5
\]
[/quote]
Let $ a \plus{} b \plus{} c \equal{} 3u,$ $ ab \plus{} ac \plus{} bc \equal{} 3v^2,$  $ abc \equal{} w^3$ and $ u^2 \equal{} tv^2.$
Hence, your inequality is equivalent to
$ \frac {9uv^2}{w^3}\geq\frac {4(3u^2 \minus{} 2v^2)}{v^2} \plus{} 5\Leftrightarrow3uv^4\geq(4u^2 \minus{} v^2)w^3.$
But $ (a \minus{} b)^2(a \minus{} c)^2(b \minus{} c)^2\geq0$ gives $ w^3\leq3uv^2 \minus{} 2u^3 \plus{} 2\sqrt {(u^2 \minus{} v^2)^3}.$
Id est, it remains to prove that $ 3uv^4\geq(4u^2 \minus{} v^2)\left(3uv^2 \minus{} 2u^3 \plus{} 2\sqrt {(u^2 \minus{} v^2)^3}\right)\Leftrightarrow$
$ \Leftrightarrow4u^5 \minus{} 7u^3v^2 \plus{} 3uv^4\geq(4u^2 \minus{} v^2)\sqrt {(u^2 \minus{} v^2)^3}\Leftrightarrow$
$ \Leftrightarrow(4u^2 \minus{} 3v^2)u\geq(4u^2 \minus{} v^2)\sqrt {u^2 \minus{} v^2}\Leftrightarrow(4t \minus{} 3)^2t\geq(4t \minus{} 1)^2(t \minus{} 1)\Leftrightarrow$
$ \Leftrightarrow8t^2 \minus{} 8t \plus{} 1\geq0,$ which is true when $ t\geq1.$

The expanding is better here. It gives $ \sum_{cyc}c^3(a \minus{} b)^2\geq0.$

$ k \equal{} 4$ is the best constant because if $ b \equal{} c \equal{} 1$ we obtain:
$ \left(a \plus{} b \plus{} c\right) \cdot \left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right)\geq k\cdot\frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \plus{} 9 \minus{} k\Leftrightarrow k\leq4 \plus{} \frac {2}{a}$ and
for $ k \equal{} 4$ your inequality is true.

[quote="shaaam"]Given that $ a,b,c \geq 0$. Prove that,
\[ \left(a \plus{} b \plus{} c\right) \cdot \left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right)\geq 4\cdot\frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \plus{} 5
\]
[/quote]

We can also prove it by Muirhead inequality:
Our inequality is equivolent to
 $ (a\plus{}b\plus{}c)(ab\plus{}ac\plus{}bc)^2 \geq 4abc(a^2\plus{}b^2\plus{}c^2)\plus{}5abc(ab\plus{}ac\plus{}bc)$
$ \Leftrightarrow$
$ \sum_{sym}{a^3b^2}\plus{}5\sum_{cyclic}{a^2b^2c}\plus{}\sum_{sym}{a^3bc} \geq 2\sum_{sym}{a^3bc}\plus{}5\sum_{cyclic}{a^2b^2c}$
$ \Leftrightarrow$
$ \sum_{sym}{a^3b^2} \geq \sum_{sym}{a^3bc}$
which is right by Muirhead inequality.

[quote="shaaam"]Given that $ a,b,c \geq 0$. Prove that,
\[ \left(a \plus{} b \plus{} c\right) \cdot \left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right)\geq 4\cdot\frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \plus{} 5
\]
[hide="Challange"]Find the maximal $ k$, so that the following inequality holds:
\[ (a \plus{} b \plus{} c)\cdot \left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right) \geq k \cdot \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \plus{} 9 \minus{} k
\]
I have a feeling that, $ k \equal{} 4$ is best constant. Try to prove or disprove me.
Thanks!  :) [/hide][/quote]
For a simple calcultation, we will set $ a: \equal{} \frac {1}{a},b: \equal{} \frac {1}{b},c: \equal{} \frac {1}{c}$, then our inequality becomes
\[ (a \plus{} b \plus{} c)\left( \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right) \ge \frac {4(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)}{abc(a \plus{} b \plus{} c)} \plus{} 5
\]

\[ (ab \plus{} bc \plus{} ca)(a \plus{} b \plus{} c)^2 \ge 4(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2) \plus{} 5abc(a \plus{} b \plus{} c)
\]

\[ ab(a \minus{} b)^2 \plus{} bc(b \minus{} c)^2 \plus{} ca(c \minus{} a)^2 \ge 0
\]
This proof also shows that the best constant is $ k \equal{} 4$. So, your statement is valid, shaam. :)

Thank you very much everyone. Can_Hang, your substitution is very smart  :lol:
And i have a question to Arqady. Can you please explain step by step how do you get:

$ w^3\leq3uv^2 \minus{} 2u^3 \plus{} 2\sqrt {(u^2 \minus{} v^2)^3}$

Because this bound is very strong and i believe, it has solved so many inequalities.

Thanks.  :)

It is only the following indentity  :D
\[ (a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2\ge 0\]

which is the strongest 3-variables inequatily  :wink:

Dear Zaizai, i was asking how to transform it. Because:

$ (a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2 \Leftrightarrow$
$ \Leftrightarrow \sum_{cyc} (a^4b^2 \plus{} a^4c^2 \plus{} 2a^3b^2c \plus{} 2a^3bc^2) \geq \sum_{cyc} (2a^3b^3 \plus{} 2a^4bc \plus{} 2a^2b^2c^2)$

Please, tell me how to transform into $ w,v$ and $ u$ and

$ w^3\geq3uv^2 \minus{} 2u^3 \minus{} 2\sqrt {(u^2 \minus{} v^2)^3}$

[quote="zaizai-hoang"]It is only the following indentity  :D
\[ (a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2\ge 0
\]
which is the strongest 3-variables inequatily  :wink:[/quote]strongest? how do you prove it?

We can also solve it easily by SOS with $ Sa\equal{} \frac{1}{bc} \minus{} \frac{2}{ab\plus{}bc\plus{}ca}$...
 :)

[quote="shaaam"]
And i have a question to Arqady. Can you please explain step by step how do you get:

$ w^3\leq3uv^2 \minus{} 2u^3 \plus{} 2\sqrt {(u^2 \minus{} v^2)^3}$

Because this bound is very strong and i believe, it has solved so many inequalities.

Thanks.  :)[/quote]
Of cause, shaaam!
Let $ a \plus{} b \plus{} c \equal{} 3u,$ $ ab \plus{} ac \plus{} bc \equal{} 3v^2$ and $ abc \equal{} w^3.$
Hence, $ (a \minus{} b)^2(a \minus{} c)^2(b \minus{} c)^2 \equal{}$
$ \equal{} \sum_{cyc}(a^4b^2 \plus{} a^4c^2 \minus{} 2a^4bc \minus{} 2a^3b^3 \plus{} 2a^3b^2c \plus{} 2a^3c^2b \minus{} 2a^2b^2c^2) \equal{}$
$ \equal{} 27( \minus{} w^6 \plus{} 2(3uv^2 \minus{} 2u^3)w^3 \minus{} 4v^6 \plus{} 3u^2v^4)\geq0.$
It gives the following estimation:
\[ 3uv^2 \minus{} 2u^3 \minus{} 2\sqrt {(u^2 \minus{} v^2)^3}\leq w^3\leq3uv^2 \minus{} 2u^3 \plus{} 2\sqrt {(u^2 \minus{} v^2)^3}
\]

[quote="zaizai-hoang"]It is only the following indentity  :D
\[ (a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2\ge 0
\]
which is the strongest 3-variables inequatily  :wink:[/quote]

How about $ (a \plus{} b \minus{} 2c)^2(b \plus{} c \minus{} 2a)^2(c \plus{} a \minus{} 2b)^2 \ge 0$  :wink:

[quote="Shishkin"]$ 4\frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ac} \plus{} 5 \equal{} 4\frac {(a \plus{} b \plus{} c)^2}{ab \plus{} bc \plus{} ac} \minus{} 3$
$ (a \plus{} b \plus{} c)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}) \plus{} 3 \equal{} \sum_{cyc}(\frac {a}{b} \plus{} \frac {b}{a} \plus{} 2) \equal{} \sum_{cyc}\frac {(a \plus{} b)^2}{ab} \ge 4\frac {(a \plus{} b \plus{} c)^2}{ab \plus{} bc \plus{} ac}$[/quote]
My comment is definitly spam. But I just had to say that your solution is great! Nice job! :thumbup:


Gauss
function
geometry
geometric transformation
reflection
algebra
floor function

I need help on this one

What is the greatest positive integer n such that 3 to the n power is a factor of 200!   :?:

Well there are 4 numbers smaller than 200 which is in the form 3^x (3, 9, 27, and 81) multiplying all of them together will give us 3^10 ((3^1)*(3^2)*(3^3)*(3^4)), so the answer is 10.

I hope I'm right...

guess not :cry:

Well, the problem is pretty much to find the number of 3's in 200!. Refer this this thread for explanation of a very similar problem:

http://www.artofproblemsolving.com/wforum/viewtopic.php?t=893

Yes, I know.
By brute force
3 and 6 each has one 3
9-24 has two
27-78 has three,
and 81-198 has four, so
2+6*2+18*3+40*4=228, I guess.

each multiple of 3 adds one factor of 3, each multiple of 9 adds another, each multiple of 27 adds another, and each multiple of 81 adds another. 81 is the biggest power of 3 that is less than 200, so we can stop there. that should be 97.

p.s. this question was numba 10(?) state mc this year!

More generally: ([ x ] is the gauss function - largest integer less than or equal to x)

If you have a question of the form "find the largest power of prime p that divides into n!", the answer is of the form

[ n / p ] + [ n / p^2 ] + [ n / p^3 ] + ... + [ n / p^m ],

where p^m <= n but p^(m+1) > n.

Variants of this question were surprisingly popular on my finite math book in the chapter dealing with permutations and combinations.

[color=green]Be careful, Osiris -- that formula only works for prime numbers.

Also, you can make the sum infinite, since for bigger powers of k, the floor will just give you 0's.[/color]

Yikes, you're right... (Edited post to reflect that) Thanks!

Hmm...I saw something like that on USAMTS.

Hey JBL, can you explain why that works only for prime numbers?  Shouldn't that work for everything?

For a non-prime, you have to take its largest factor and use that, I believe. That is because as we go go 1, 2, 3, 4, ... there are more occurences of that factor than any of the other ones.

Consider this question:

Find the largest pos. int. n such that 10^n divides into 1000!.

[hide]One might be tempted to use the formula - but that's wrong. The reason is, not only can multiples of 10 (10, 20, 30, ...) make up powers of 10, but products of other numbers (2*5, 4*15, etc... they're not multiples of 10 themselves, but multiplied together they are). So one should write 10 as 2*5. Since there are lots more multiples of 2 than there are multiples of 5 in 100!, one just needs to count the number of multiples of 5 in 100!. The correct expression for this question is then

[ 1000 / 5 ] + [ 1000 / 25 ] + [ 1000 / 125 ] + [ 1000 / 625 ], which comes out to be 200 + 40 + 8 + 1 = 249.[/hide]

Ahhh... Yes yes, I finally realized that I knew that :)

[color=#A000FF]Then, it gets really nasty when you talk about, say, the number of powers of 360, because you need to count the number of 2's divided by 3 and compare that to the number of 3's divided by 2 and compare that to the number of 5's.  Messy.[/color]

Could you explain the floor function that was used to get 97? I know that the numbers round down to the base number, but I don't understand why 200 is divided by 3,9,27,81 and so on..

Here's the way I always explained this kind of problem:

Count the number of numbers that are divisible by 3 in the first 200 numbers.

You get 3, 6, 9, 12, 15, ..., 198. There are [200 / 3] = 66 such numbers.

Now notice that 9, 18, ... are multiples of 3:^2:. We have counted them only ONCE in in the previous iteration, but they have 2 3's so we must count them again: [200 / 9] = 22.

Now notice that 27, 54, ... are multiples of 3:^3:. We have counted them only TWICE in the previous iterations, but they have 3 3's so we must count them again: [200 / 27] = 7.

Now notice that 81 and 162 are multiples of 3^4. As usual, we count those again: [200 / 81] = 2.

There are no multiples of 3:^5: within the first 200 numbers, so we stop here.

Add them up: 66 + 22 + 7 + 2 = 97.


inequalities
number theory
Diophantine equation
number theory unsolved

Show that there are only finitely many triples $ (x,y,z)$ of positive integers satisfying the equation $ abc\equal{}2009(a\plus{}b\plus{}c).$

In general, for positive integers $ n$ and $ k$, prove that Diophantine equation $ \prod_{i\equal{}1}^n x_i \equal{} k\sum_{i\equal{}1}^n x_i$ has a finite number of solutions made of positive integers.

Assume $ 1 \leq x_1 \leq x_2 \leq \cdots \leq x_n$. Then
\[ \frac {1} {k} \equal{} \sum_{i\equal{}1}^{n} \frac {1} {\prod_{j \neq i} x_j} \leq \frac {n} {\prod_{i\equal{}1}^{n\minus{}1} x_i}, \ \textrm{ hence } \  \prod_{i\equal{}1}^{n\minus{}1} x_i \leq nk.\]
Therefore $ 1 \leq x_i \leq nk$, $ 1 \leq i \leq n\minus{}1$. Since $ x_n$ is uniquely determined, the conclusion ensues.

[hide]
If $ a > 2009$, we have the inequality
\[ (b \minus{} 1)(c \minus{} 1) < \left (b \minus{} \frac {2009}{a} \right ) \left (c \minus{} \frac {2009}{a} \right ) \equal{} 2009 \plus{} \frac {2009^2}{a^2} < 2010\]
, which gives finite values of $ b, c$. Similarly, there are a finite number of solutions if $ b > 2009$ or $ c > 2009$, so there are a finite number of solutions total.
[/hide]

orznorz, is that enough to prove that there are finitely many solutions?

i think two answers above are both  :lol: right

Let suppose that $a=max(a,b,c)$. We have $abc=2009(a+b+c)<2009*3*a$, so $bc<2009*3$ ,  which gives finite values of $b,c$.If $bc<2009$ because $a<a+b+c$ , we have $abc<2009(a+b+c)$ , so $bc>2009$. Because $abc=2009(a+b+c)$ , we obtain $a(bc-2009)=2009(b+c)$.So $a=\frac{2009(b+c)}{bc-2009}$.We know that $a$ is a positive integer so we chose only the values for $b,c$ such that $bc-2009 | 2009(b+c) $ but we know that these are finite. So we have a finite value for $a$.


You see a castle in the middle of a small square island. All around the island, there is a 10 foot wide water moat. You are given two boards of equal length in order to cross the river. What is the minimum length of the boards needed in order to cross? (Assume you can't nail them together, all you have is the boards, and assume they are both lines.) Express in simplest form.

The answer's probably no for both of these questions, but just to make sure:
Can you jump any distance?
Can you cut the boards?
Also, I came up with one solution can be better than just throwing a 10.5 foot board over, but it depends on how large the square island is.

Nice question. There seem to be a lot of good question writers on AoPS... MC should come here to hire question-writers...

Umm.... assuming that a board can be held up by the friction of the sides of the moat,...

[hide]
At one of the outer corners throw one board diagonally at a 45 degree angle, and then at the midpoint of this board, throw the second board across to the inner corner. (Hopefully people can understand this.) Let the board's length be $ x$. Draw a line from the inner corner to the outer corner. Because of the 45 degree angles, this length is $ \displaystyle\frac{3x}{2}$. This length is also $ 10\sqrt{2}$, because the width of the moat is $ 10$ feet. Setting these equal and solving yields $ \displaystyle\boxed{ x\equal{}\frac{20\sqrt{2}}{3}}$ feet.

It's probably wrong  :oops: .

[/hide]


I'm learning graphing piecewise functions and step functions in algebra, and I really don't get anything about it at all. My teacher never teaches anything, she expects us to figure everything out on our own. I felt really stupid today because everyone in my class got them perfectly except me. Anyways, how do you graph piecewise functions and step functions? The only thing that I understand about this is that there is this gap between the dots called "points of discontinuity":/

Hi kuhristine. See here: http://www.analyzemath.com/Graphing/piecewise_functions.html

or here: http://www.math.uakron.edu/amc/PreCalculusFoundationsToCalculus/Limits/PiecewiseLimit_Nspire/LinearPiecewiseFunctions_NspireWorkshop.DOC

Hope these help you.


inequalities algebra polynomial probability function

First off, my apologies if this is in the wrong forum. I do not really believe this technique to be really Olympiad level yet.

This really exists for the purpose of expanding Olympiad inequalities (for the unholy purpose of Muirheading them.)

Take, for instance, the following inequality:

$ \frac{1}{a^3 \plus{} b^3 \plus{} abc} \plus{} \frac{1}{a^3 \plus{} c^3 \plus{} abc} \plus{} \frac{1}{b^3 \plus{} c^3 \plus{} abc} \leq \frac{1}{abc}$.

Yes, I am aware that this inequality has been posted multiple times. Yes, I aware this has a very nice solution. But for the sake of example, I shall Muirhead this to demonstrate this technique.

To summarize, I am taking advantage of the fact that the polynomial is symmetric, and I am writing everything out in symmetric mean notation without first expanding everything out  (thus saving a lot of time and paper, also reducing the probability of getting something wrong.)

The first step above is to clear the denominators: we get that we need to show that
$ abc((a^3 \plus{} b^3 \plus{} abc)(a^3 \plus{} c^3 \plus{} abc) \plus{} (a^3 \plus{} b^3 \plus{} abc)(b^3 \plus{} c^3 \plus{} abc) \plus{} (b^3 \plus{} c^3 \plus{} abc)(a^3 \plus{} c^3 \plus{} abc)) \leq (a^3 \plus{} b^3 \plus{} abc)(a^3 \plus{} c^3 \plus{} abc)(b^3 \plus{} c^3 \plus{} abc)$.

Before we proceed, we first note that any symmetric expression can be written as a sum of symmetric sums.

Now, for the expansion. I will describe the technique in this specific case; you may choose to generalize this yourself.

In short, it suffices to consider the term $ abc(a^3 \plus{} b^3 \plus{} abc)(a^3 \plus{} c^3 \plus{} abc)$ to get almost the entire left-hand side; this is the crux of the technique. Suppose that, when expanding, we encounter a term like $ a^7 bc$. Then, since this is symmetric, the terms $ a b^7 c$ and $ abc^7$ must appear as well. Furthermore, if $ n$ is the coefficient of $ a^7 bc$, it is also the coefficient of $ a b^7 c$ and $ n abc^7$. Finding the coefficient of $ a^7 bc$ itself is not very cumbersone (we simply analyze the rest of the terms.) When the coefficient is found, we simply write that part of the sum as $ 3n[7, 1, 1]$; the 3 is present, since $ [7, 1, 1] \equal{} \frac{x^7 yz \plus{} x y^7 z \plus{} xyz^7}{3}$. Similarly, some portion of $ [a, b, c]$ with $ a \geq b \geq c$ and $ a \plus{} b \plus{} c \equal{} 7$ that will appear in the expansion of the left-hand side will appear in the expansion of $ abc(a^3 \plus{} b^3 \plus{} abc)(a^3 \plus{} c^3 \plus{} abc)$.

We apply a similar procedure to the right-hand side, cancel, and Muirhead.

Long story short, given that $ a^x b^y c^z$ appears in the expansion and given its coefficient, we can find the coefficient of $ [x, y, z]$ in the expansion without actually expanding. Repeating this procedure, we almost painlessly expand the inequality.

Very likely, this technique is not original. I just came across it while expanding an inequality the hard way on a plane (yes, I was bored), so I'm sured others have too. I'm completely fine with that. In any case:

1. Does this have a name?
2. Is this technique sufficiently well-known to be used on the Olympiads without showing the work in your expansion? (I.e., using [name of technique], we see that this inequality reduces to this.) I don't want to be using this and then getting major points deducted for not showing work (and I don't want to have to waste my time describing this technique either.)

If it actually is original (woo, but I won't get my hopes up) I'd gladly write up a more rigorous, precise, and general version of this post here.

Any response would be appreciated. Thanks in advance.

$ abc(\sum_{cyc} (a^3 \plus{} b^3 \plus{} abc)(b^3 \plus{} c^3 \plus{} abc)) \le (a^3 \plus{} b^3 \plus{} abc)(a^3 \plus{} c^3 \plus{} abc)(b^3 \plus{} c^3 \plus{} abc)$

Would be another way to write the desired expression. $ \sum_{cyc}$ means that we take the sum that was just written, and then add it to all sums attained by CYCLING the variables.

E.G. If we had 3 variables $ a,b,c$, then $ \sum_{cyc}f(a,b,c)$ would mean $ f(a,b,c) \plus{} f(b,c,a) \plus{} f(c,a,b)$

4 variables, we would have $ \sum_{cyc}f(a,b,c,d) \equal{} f(a,b,c,d) \plus{} f(b,c,d,a) \plus{} f(c,d,a,b) \plus{} f(d,a,b,c)$.

You get the idea, note that $ \sum_{cyc}$ goes over ALL variables even if they aren't in the function under the sum, for example, with 4 variables a,b,c,d, $ \sum_{cyc}\frac {a}{b \plus{} c}$ means $ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} d} \plus{} \frac {c}{d \plus{} a} \plus{} \frac {d}{a \plus{} b}$. It's usually clear in context.

That's $ \sum_{cyc}$. The following rules apply: $ \sum_{cyc} f(a) \plus{} g(a) \equal{} \sum_{cyc}f(a) \plus{} \sum_{cyc}g(a)$, and $ \sum_{cyc}f(a,b,c,d) \equal{} \sum_{cyc}f(b,c,d,a) \equal{} \sum_{cyc}f(c,d,a,b) \equal{} \sum_{cyc}f(d,a,b,c)$ (here I wrote out the case for 4 variables, but you can see how it easily genralizes.

Note however that $ \sum_{cyc}ab^2$ is not the same as $ \sum_{cyc}ac^2$ in three variables (or in any number of variables for that matter), only if one of them has the variables cycled of the other are they equal.

This is essentially what you were saying, just with better notation. Note however that $ \sum_{cyc}$ is pretty useless when it comes to muirhead ($ ab^2 \plus{} bc^2 \plus{} ca^2$ doesn't do much with Muirhead), so we have another notation, $ \sum_{sym}$

$ \sum_{sym}f(a,b,c)$ would be the sum taken over all PERMUTATIONS of $ a,b,c$ that means that we take this to mean $ f(a,b,c) \plus{} f(a,c,b) \plus{} f(b,c,a) \plus{} f(b,a,c) \plus{} f(c,a,b) \plus{} f(c,b,a)$ in the case of three variables, of course it generalizes.

It is critical to note that $ \sum_{sym}a^2b^2 \equal{} 2(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)$ not $ a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2$ because even though we have $ a^2b^2 \equal{} b^2a^2$, we still need them in the symmetric sum.

In the case of the problem, we would have $ abc(\frac {1}{2}\sum_{sym}(a^3 \plus{} b^3 \plus{} abc)(b^3 \plus{} c^3 \plus{} abc))\le (a^3 \plus{} b^3 \plus{} abc)(a^3 \plus{} c^3 \plus{} abc)(b^3 \plus{} c^3 \plus{} abc)$. (We had the factor of $ \frac {1}{2}$ since each term was counted twice in the symmetric sum). Now, when you expand out a symmetric sum, here is the beautiful thing. $ \sum_{sym}a^{p}b^{q}c^{r} \equal{} [p,q,r]$, perfect for muirhead.

So in summary, use $ \sum_{cyc}$ when the variables are cyclically interchanged, then, when you feel the time is right, use your common sense and see how you can combine some of those cyclical sums to symmetric sums (e.g. $ \sum_{cyc}a^2c \plus{} \sum_{cyc}a^2b \equal{} \sum_{sym}a^2b$ when taken over three variables).

Hope this helped.

I'm pretty sure almost everyone knows this trick; however, it will be extremely enlightening for those who don't. Perhaps this could be moved to the Intermediate forum?

Also, regarding work: you can say something like "we multiply everything by such-and-such expression and get such-and-such monstrous equation" without showing all the painstaking work of actually multiplying it out. You should show the steps before and after every merge of terms.

If it's too monstrous, though, you'll get marked down for having a terrible solution.

Thanks for your post Rofler, but it didn't really address any of the questions I had in mind. Nor did I ever mention anything about cyclic sums. :/ I did learn a few things about them (e.g., I'm now better with the notation and the like). And to say the least, I shall certainly take advantage of that while writing down my expansions, as it will undoubtedly save time and space. So even if that was a bit off, thanks anyway.

[quote]Also, regarding work: you can say something like "we multiply everything by such-and-such expression and get such-and-such monstrous equation" without showing all the painstaking work of actually multiplying it out. You should show the steps before and after every merge of terms. [/quote]
I don't quite understand. The purpose of this method was to avoid having to get the actual equation at all (perhaps I wasn't clear enough), and simply jump from the inequality to the expanded version, with everything written as symmetric sums. How would we justify this?


Mary bought a pair of jeans on sale for $ \$24.00$ The sale price of the jeans was the result of a 20\% reduction from the original price. What was the number of dollars in the original price?

Being 20% off is the same as being 80% of the original price, so $ 24\equal{}.8x\Rightarrow x\equal{}30$.


Hi, could someone help me solve the following problem? :)

Consider a circuit consisting of a battery of voltage $E$, a switch $S$ and a parallel-plate capacitor with capacitance $C$. The capacitor consists of two parallel conducing plates of equal area $A$ separated by a distance $d$. After the switch $S$ is closed and the capacitor is fully charged, a conducting plate of thickness $d/3$ and area $A$ is inserted slowly between the plates of the capacitor, and it is kept parallel to the conducting plates of the capacitor.

a) Find the capacitance of the capacitor after the conducting plate is inserted.

b) How much is the increase in the charge stored in the capacitor caused by inserting the conducting plate?

c) How much is the increase in the energy stored in the capacitor caused by inserting the conducting plate?

d) How much work is done by the battery during the insertion of the conducting plate?

e) How much work is done by the force applied to the conducting plate during its insertion?

f) If the switch $S$ is opened and the conducting plate is removed slowly, how much work is done by the force applied to the conducting plate to remove it?

Thanks in advance for your help! :D

[quote="Carcul"]Hint: consider the new system as two capacitors connected in series.[/quote]

Doing so I was able to solve the first 3 items :D but I cannot figure out items d, e, f! Could you help me again, please? :oops:


There's a problem in the AoPS book that I have a hard time understanding.

Here is goes:

A chemist has 80 mL of a solution containing 20% acid.  How many mL must be removed and replaced by pure acid in order to obtain a 40% solution?

Source: Mu Alpha Theta 1992

What answers do you get?

[hide]I got 20. You need to remove 16 ounces of water and in order to do that, just remove 20 ounces of the solution.[/hide]

[hide]I get that you'd have to remove and replace 20 mL of solution. Right now you have 16 mL of acid. You need to have a total of 32 mL of acid. You know that for every 5 mL of liquid you remove, you remove 1 mL of acid, and can replace it with 5 mL of pure acid. I made a chart.

-5 mL of mix = 15 mL of acid
+5 mL of acid= 20 mL of acid
-5 mL of mix = 19 mL of acid
+5 mL of acid= 24 mL of acid
-5 mL of mix = 23 mL of acid
+5 mL of acid = 28 mL of acid
-5 mL of mix = 27 mL of acid
+5 mL of acid = 32 mL of acid

You replace as much acid as you take out 4 times, and now your solution is 40% acid. You have to remove 20 mL of acid.  :)  [/hide]

[hide="Solution"]There are 16mL of acid in the 80mL solution.

$\frac{16 + \frac{4}{5}n}{80} = \frac{2}{5}$

This can be concluded from the fact that for every 1mL removed, the acid increases by $\frac{4}{5}$ mL's.

It is apparent that $n = 20$, after solving.[/hide]

hmm...do you mean the final volume is still $80ml$? Or you just add pure acid to the $80ml$ solution?