counting
distinguishability
pifinity is ze best

How many ways are there to put 5 balls in 2 boxes if the balls are distinguishable but the boxes are not?

you can choose a box and for each ball it can either be in or out. then all the rest of the balls will be in the other box.
so there are 2^5 or 32 possibilities but since we counted everything twice we have to divide be 2 so the answer is 16

Note that this only works because $ 5$ is odd.

Why does it work only because 5 is odd?
I think that it works generally.
We have 2^k subsets, so we have 2^k ways to divide them into different boxes.
however, given S,U-S (U being {1,...,k}) , it appears both as (S,U-S) and (U-S,S) (those are distinct since S does not equal U-S). So, if we divide it by 2, we get the number of ways to divide into identical boxes.

for example, for k=4
2^4 / 2 = 8
as we have
1234,
123,4
124,3
134,2
234,1
12,34
13,24
14,23

Am I wrong?

Nope.   :D  In the problem, if the boxes were distinguishable, it would be 2^5 = 32.  But, since the boxes are not distinguishable, each possibility is equivalent to the reverse, so we divide by 2 to get 16. :yup:

If both me AND Spiritus are wrong, someone correct us and EXPLAIN WHY!

[hide]After you place one of the balls, all the others possible placements give different outcomes, so we want $2^4=\boxed{16}$[/hide]

[hide=Solution]Here's my solution:

There are a total of $2^5=32$ ways to place the balls in general, as each box has two "choices".

We know that there is one case that is repeated twice (all 5 balls in either box), so we have $32-2=30$ other arrangements. For every other case, the contents of each box is different, and so these cases are each counted $2!=2$ times. [hide=Here's an explanation why dividing by two only works in this case, and if there were, say, three boxes, dividing by three would not work.][[quote=saagarjha]Divide by 2 only works when you have 2 boxes.[/quote]
The reason that dividing by two works in this case is because $2!=2$, while $a!=a$ is not true for larger numbers[/hide] Because each of these 30 distinguishable cases are counted twice, we now have $\frac{30}{2}=15$ of these cases.

However, we must remember to add back the $1$ case that represents when all 5 balls are in one box, giving us: $15+1=\boxed{16}$ cases.

[hide=Alcumus Solution]Since the boxes are indistinguishable, there are 3 possibilities for arrangements of the number of balls in each box.

Case 1: 5 balls in one box, 0 in the other box. We must choose 5 balls to go in one box, which can be done in $\binom{5}{5} = 1$ ways.

Case 2: 4 balls in one box, 1 in the other box. We must choose 4 balls to go in one box, which can be done in  $\binom{5}{4} = 5$ ways.

Case 3: 3 balls in one box, 2 in the other box. We must choose 3 balls to go in one box, which can be done in $\binom{5}{3} = 10$ ways.

This gives us a total of $1 + 5 + 10 = \boxed{16}$ arrangements.[/hide][/hide]

[quote=Mathlete2017][hide]Here's my solution:

There are a total of $2^5=32$ ways to place the balls in general, as each box has two "choices".

We know that there is one case that is repeated twice (all 5 balls in either box), so we have $32-2=30$ other arrangements. For every other case, the contents of each box is different, and so these cases are each counted $2!=2$ times. [hide=Here's an explanation why dividing by two only works in this case, and if there were, say, three boxes, dividing by three would not work.][[quote=saagarjha]Divide by 2 only works when you have 2 boxes.[/quote]
The reason that dividing by two works in this case is because $2!=2$, while $a!=a$ is not true for larger numbers[/hide] Because each of these 30 distinguishable cases are counted twice, we now have $\frac{30}{2}=15$ of these cases.

However, we must remember to add back the $1$ case that represents when all 5 balls are in one box, giving us: $15+1=\boxed{16}$ cases.

[hide=Alcumus Solution]Since the boxes are indistinguishable, there are 3 possibilities for arrangements of the number of balls in each box.

Case 1: 5 balls in one box, 0 in the other box. We must choose 5 balls to go in one box, which can be done in $\binom{5}{5} = 1$ ways.

Case 2: 4 balls in one box, 1 in the other box. We must choose 4 balls to go in one box, which can be done in  $\binom{5}{4} = 5$ ways.

Case 3: 3 balls in one box, 2 in the other box. We must choose 3 balls to go in one box, which can be done in $\binom{5}{3} = 10$ ways.

This gives us a total of $1 + 5 + 10 = \boxed{16}$

[hide][/hide] arrangements.[/hide][/quote]
[/hide]

Nice solution! Juts remember to hide your work using hide tags ;) :
[c][hide]SOLUTION IN HERE[/hide][/c]

[hide="Much easier solution"]
Note that it is $2^5$ if the boxes [b]are[/b] distinguishable. Since they are not, we divide by 2 to acheive the result of $\boxed{16}$
[/hide]

[quote=OlympusHero]Instead of using these silly solutions :) (jk) use my handmade formula! If there are x balls going into y boxes and the balls are distinguishable but the boxes are not, there are ((y^x-y)/y!) + 1 ways to do it![/quote]

Sorry, on the note of your formula, is it: $\frac {y^x-y}{y!}+1, \frac {y^x-y}{y}!+1, \frac {y^{x-y}}{y!}+1, \text { or } \frac {y^{x-y}}{y}!+1$

[quote=pog][quote=OlympusHero]This formula is incorrect; it only works in a few cases.[/quote]

Could you elaborate?[/quote]

Why bother. It will be too much work to remember the exact cases.

[quote=P3Tan][hide]16  :rotfl:  :rotfl: [/hide][/quote]

1. Explanation
2. Why the rotfl?


analytic geometry
algebra
polynomial
algebra proposed

A liquid is moving in an infinite pipe. For each molecule if it is at point with coordinate $x$ then after $t$ seconds it will be at a point of $p(t,x)$. Prove that if $p(t,x)$ is a polynomial of $t,x$ then speed of all molecules are equal and constant.

I am not sure but something like this should work :

For $(t,k,x) \in R^{3}, t > k$ we have : \[p(t,x)-p(k,x)= p(t-k,p(t-k,x))\] (Assume the molecule is travelling from point $x$ for  $k$ seconds and then $t-k$ seconds .)
$\longrightarrow : degP(t,x)=1 \longrightarrow P(t,x)=x+at$ where $a \in R$ is fixed.

Real beautiful Problem for the final exam.

Can i know who was the author ?


Sa se arate ca in orice triunghi, exista o pereche de conjugate izogonale $(M_{1},M_{2})$ astfel incat : $OM_{1}\cdot OM_{2}> OI^{2}$.

[hide="parere"]
daca nu am gresit eu la calcule, mi se pare ca daca $BC$ este cea mai mica latura in triunghiul $ABC$, iese ca $OA\cdot OD>OI^{2}$.
apoi aleg $X\rightarrow A$ , cu $X$ pe bisectoare, si atunci $X^{*}\rightarrow D$ (izogonalul), si deci $OX\cdot OX^{*}> OI^{2}$ pentru un $X$ acolo.
nu mai scriu calculele, dar sper sa fie bine
[/hide]


umm..
ch1n353ch3s54a1l, thanks for the answers, but
i need to know HOW to solve them..
can anyone explain? :huh:

[quote="ch1n353ch3s54a1l"][quote="polarpan523"]Please teach me how to solve these problems.

(1) Patricia has 12 dollars more than Rhoda and 15 dollars more than Sarah. Together all three have 87 dollars. How much does Patricia have?

(2) If I start with the number 7 and count by 4's, the following sequence is obtained: 7,11,15,19,23, and so forth. A new sequence is formed when I start with a different number and count by a different number. Suppose the 2nd number of the new sequence is 8 and the 5th number is 17. What is the 10th number of the new sequence?

(3) The entire treasury of the Alpha club, consisting of 240 dollars, was to be divided into equal shares for each club member. When it was discovered that one member was not eligible for a share, the share of each of the remaining members increased by 1 dollar. How many eligible members got a share of the 240 dollars?[/quote]

[hide](1) $P=R+12$ $P=S+15$
$P+R+S=87$
$2R+S=75$
$R+12=S+15$
$S=R-3$
$3R=78$
$R=26$
$P=38$
$S=23$
(2)The common difference is 3. the first term is 5. 5+3(9)=$32$.
(3)$\frac{240}{x}=y$
$\frac{240}{x-1}=y+1$
$\boxed{x=16}$
[/hide][/quote]

quoting myself just for reference.
for the first one, since Patricia has 12 more dollars than rhoda, P=R+12, and doing the same for others, we get$P=R+12$ $P=S+15$
$P+R+S=87$
and then you can substitute so that R+12=S+15 (Because P=P)
and then substitute in the equation P+R+S+87 for one variable and then you can solve from there.
for the second one, fifth term = 17 and second = 8. there are three terms in between so 17-8=9, and 9/3=3, so the common difference is 3. since the first term is 5 and you have to add 3 (9) times, you have 5 + 3(9) = 32
for the third one, the number of people = x so each person gets 240/x which i called y. 240/x=y, and if one less person is there, it is x-1, and each person gets one more, so its y+1. 240/x-1 = y+1. then you can solve those equations simultaneously. I hope this helps

it does help
it helps a bunch
THANK YOU SO MUCH!! :D

polarpan523 wrote:

“(1) Patricia has 12 dollars more than Rhoda and 15 dollars more than Sarah. Together all three have 87 dollars. How much does Patricia have?”

Sarah: x.

Patricia: (x+15)

Rhoda: (x+15)-12= (x+3)

Then: x+x+15+x+3=87

         3x+18=87 => x=23.

So, Sarah has 23, Patricia 38, and Rhoda 26.


Hey guys have you ever heard about Euler's numbers ?

Note for future reference: Saying something about Euler [blank] with no other distinction means we have no idea what you're talking about. Euler probably has a lot of numbers I've never heard of. I've seen $ e$ attributed to him on occasion, and I definitely know $ \gamma$ is named after him.

Please be more specific.

Very true.  Euler's name is attached to more theorems and concepts than any other mathematician, to my knowledge.  Even with the plural, you could be talking about three completely different things:

http://mathworld.wolfram.com/EulerNumber.html (and this link contains two [i]other[/i] possible meanings)
http://mathworld.wolfram.com/EulerianNumber.html
http://mathworld.wolfram.com/EulerCharacteristic.html


modular arithmetic
Divisibility
number theory
IMO Shortlist
Chinese Remainder Theorem

Let $a$, $b$ be positive integers such that $b^n+n$ is a multiple of $a^n+n$ for all positive integers $n$. Prove that $a=b$.

[i]Proposed by Mohsen Jamali, Iran[/i]

This is ISL 2005 and is designed by Mohsen Jamali an Iranian teacher, the designer of IMO 2004 Problem 6

Omid Hatam wrote:
##  This is ISL 2005 and is designed by Mohsen Jamali,
##  an Iranian teacher, the designer of IMO 2004 Problem 6.

To stress it once again:
The problem IMO 2004/6 on "alternating" numbers is just a rewritten
version of the "wobbly" number problem from the IMO 1994 shortlist.

Using it for IMO'2004 was nothing but a bad mistake of the problem
selection committee.

http://www.mathlinks.ro/Forum/resources-1-16-2004-99760.html
http://www.kalva.demon.co.uk/short/soln/sh94n7.html

Let's take a prime $p>b-a$.
Then we take an integer $k>1$ s.t. $(k,p-1)=1$.
We have that $a^k \equiv -x \pmod{p}$. Now we can find two numbers $y,z \geq 0$ s.t. $k+y(p-1) = x+zp = N$.

Now we have $a^N \equiv -N \pmod {p}$ and so $p|a^N+N|b^N+N$ and so $a^N+N \equiv b^N + N \pmod{p}$.

Now we can say that $a^k \equiv b^k \pmod{p}$ and so $a \equiv b \pmod{p}$  (we remember that $(k,p-1)=1$) and so $a=b$ because $p|b-a$ but $b-a<p$

I'm sorry, i don't understand the first part of your solution  :?  Could you explain it to me, please ?

[quote]The problem IMO 2004/6 on "alternating" numbers is just a rewritten
version of the "wobbly" number problem from the IMO 1994 shortlist[/quote]
may be he don't know this
however problem3 may be known before by contestant too

Is there anyone who can explain this problem ? Please can you help ... is there any other solutions ?

Davron

There is an asymptotical solution by Peter Scholze, but it's long and ugly...
(mine is more or less the same as above: look for some big prime $\mod$ that they are equal)

Thank you very much but what is going on there i couldn't understand, i will try my best to get
the solution. Please help me with some points i will ask you :
-which method of problem solving there used ?

Davron

The main ideas (what exactly do you call a method¿) for mine are:

If $p \nmid a,b$ is any prime and we have any $k$ with $p|a^{k}+k$, then $p|a^{k}+k|b^{k}+k \implies p|b^{k}+k$, thus $a^{k}+k \equiv 0 \equiv b^{k}+k \mod p \iff a^{k}\equiv b^{k}\mod p$. If we additionally would have $k \equiv 1 \mod (p-1)$, then Fermats Little Theorem gives $a \equiv a^{k}\equiv b^{k}\equiv b \mod p$.

But for $k \equiv 1 \mod (p-1)$ we have also $a^{k}+k \equiv a+k$, so to get $p|a^{k}+k$ we just need $p|a+k$.
So lets look for some $k \equiv 1 \mod (p-1)$ and $k \equiv-a \mod p$, which can be found by the Chinese Remainder Theorem or directly as $k=(p-1)(a+1)+1$.

Till now we didn't use any special property of $p$ except that it doesn't divide $a,b$. But we know that for any such prime we have $p|(a-b)$ from chossing a suitable $k$ with $p|a^{k}+k$, thus if $a \neq b$ we would get a contradiction by choosing a big prime.

But what about the other cases, what will happen when k doesn't give a remainder 1 up on the division with p .
Thank you ZetaX.

Sincerely Davron

We ignore that cases since we don't need them.

More than if $ P(n)\equal{}xn\plus{}y$ and
$ a^n\plus{}P(n)|b^n\plus{}P(n)$ then $ a\equal{}b$

For any prime $p$, let $n \equiv 1 \pmod {p-1}$ and $n \equiv -a \pmod p$. then $$a^n + n \equiv 0 \pmod p \implies b^n + n \equiv b-a \equiv 0 \pmod p.$$ Taking $p > |b-a|$ yields $a=b$, as required.

We have that \[a^{n}+n \mid b^{n}-a^{n}\], note that we can pick $n$ such that $n \equiv 1 (\mod p-1)$ & $n \equiv -a (\mod p)$, we get that, this prime $p$ divides $a-b$, note that we can pick an arbitrarily large prime which divides $a-b$, which implies $a = b$.

Suppose FTSOC $a\ne b$. Let $p>2$ be a prime not dividing $b-a$.

By CRT, set $n$ such that $n\equiv -a\pmod p$ and $n\equiv 1\pmod{p-1}$.

We have $a^n + n\equiv 0\pmod p$, and $b^n + n\equiv b - a\pmod p$. Thus $b-a\equiv 0\pmod p$, contradiction since $p\nmid b-a$.

[quote=The QuattoMaster 6000][hide="Solution"]
[b]Lemma:[/b] There exists a prime $ p$ so that for every natural, $ l$, we have that there exists an $ n$ so that $ a^n \plus{} n \equiv 0\bmod p^l$.

[i]Proof:[/i] The following is a far more general case of the lemma: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=354533#p354533

Let $ n$ and $ a$ be relatively prime.  Then, $ b^n \plus{} n \equiv 0\bmod (a^n \plus{} n)$, meaning that $ b^n \equiv a^n \bmod (a^n \plus{} n)$.  Then, $ (\frac {a}{b})^n\equiv 1\bmod (a^n \plus{} n)$.  Since $ n$ and $ a$ are coprime, we have that $ a^n \plus{} n$ and $ n$ are coprime.  Let $ p$ be a prime factor of $ a^n \plus{} n$ so that $ p^l\parallel{}(a^n \plus{} n)$.  We see that $ p$ and $ n$ are coprime since $ n$ and $ a$ are coprime.  Let $ k$ be the order of $ \frac {a}{b}\bmod p^l$.  Now, we have that $ (\frac {a}{b})^n \equiv 1\bmod p^l$, so $ k|\gcd (p^{l \minus{} 1}(p \minus{} 1), n) \equal{} \gcd (p \minus{} 1, n)$.  Thus, $ (\frac {a}{b})^{p \minus{} 1}\equiv 1\bmod p^l$. This implies that $ p^l|(b^{p \minus{} 1} \minus{} a^{p \minus{} 1})$.  Yet, by the lemma, we can make $ l$ very large, so $ b^{p \minus{} 1} \equal{} a^{p \minus{} 1}$, meaning that $ a \equal{} b$.   [/hide]

EDIT: Another proof of the lemma:

[hide="Proof"]
We proceed with induction on $ l$.  When $ l \equal{} 1$, we consider $ a^{p \minus{} 1} \plus{} (p \minus{} 1) \equiv 1 \plus{} p \minus{} 1\equiv 0\bmod p$.  Say that we have proven it for $ l$, so $ a^x \plus{} x \equiv kp^l \bmod p^{l \plus{} 1}$ for some integer $ k$.  Then, we consider $ n \equal{} x \plus{} m\varphi (p^{l \plus{} 1})$, where $ m\equiv \frac {k}{1 \minus{} p}\bmod p$.  Notice that
\[ a^{x \plus{} m\varphi (p^{l \plus{} 1})} \plus{} x \plus{} m\varphi (p^{l \plus{} 1}) \equiv a^x \plus{} x \plus{} mp^l(p \minus{} 1)\equiv p^l(k \plus{} m(p \minus{} 1))\equiv 0\bmod p^{l \plus{} 1}
\]
and induction is complete. [/hide][/quote]

[hide=New link]The link for your lemma appears to be broken. Here is a new one for completeness: [url=https://artofproblemsolving.com/community/c6h57713p354533]Brazil MO 2005/6[/url] [/hide]

Almost the same as others.  :o
[i]Solution:[/i] By ``Chinese Remainder Theorem'', there exists a $n$ such that
    \begin{align*}
        n & \equiv -\frac{1}{a} \pmod{p} \\
        n & \equiv -1 \pmod{p-1}.
    \end{align*}
for an arbitrary prime $p$. Call the $n$ which satisfies these congruences as ``constructive''. One can show that $p \mid a^n + n$ for all such constructive $n$ for any prime $p$ with the aid of ``Fermat's Little Theorem''. Clearly
    \[a^n + n \mid b^n + n \iff a^n + n \mid b^n - a^n.\]
Pick a constructive $n$ in the above relation. We would get $p \mid a - b$ for any prime $p$. Since $a-b$ have infinitely many divisors, we must have $a = b$ and we are done. $\blacksquare$

Let $p$ be any prime. Then $p\mid a^n+n\mid b^n-a^n$. Choosing $n=k$ such that $k\equiv 1\mod{p-1}$ and $k\equiv -a\mod p$, gives $a^n+n\equiv 0\mod p$. Therefore,
	\[b^n-a^n\equiv b-a\mod p\]
choosing $p$ large gives $a=b$, as needed.

Obviously $a\le b$. Now consider a prime $p>b+1434$. Now let $n$ be a positive integer such that $n\equiv 1\pmod{p-1}$ and $n\equiv p-a\pmod{p}$. Clearly $p$ divides $a^n+n$. The only way that $p$ divides $b^n+n$ is if
\[p\mid (b+p-a)\rightarrow a=b.\]
QED.

Let $p>b\ge a$ be a prime and let $n\equiv -a\pmod p$ and $n\equiv 1\pmod {p-1}$. The latter means that $x^n\equiv x\pmod p$ for all $a$. Thus, \[a^n+n\equiv a+n\equiv 0\pmod p\] and so $p\mid b^n+n\implies p\mid b-a$. Since $b-a<b<p$, $b-a=0$ as desired.

Oops...
[hide=Solution]
Assume for the sake of contradiction that $a \neq b$. Then, clearly $a < b$, as $a + 1 \mid b+1$, so $a+1\leq b+1$, implying $a\leq b$. Now, take a prime $p > (a+b)^{1434}$. Then, we have that $$a^p + p \mid b^p + p,$$ so $$a^p + p \mid b^p - a^p.$$ Let $g = \gcd(a, b)$, and let $y = \frac{b}{\gcd(a, b)}$, $x = \frac{a}{\gcd(a, b)}$. Then, we have that $$a^p + p \mid b^p - a^p = g^p(y^p-x^p).$$ Note that since $g \mid a$, we have that $$\gcd(a^p+p, g^p) \mid \gcd(a^p + p, a^p) = \gcd(p, a^p) = 1,$$ as $p > (a+b)^{1434} > a$. Thus, we have that $\gcd(a^p+p, g^p)$ itself is equal to $1$, so $$a^p + p \mid y^p-x^p = (y-x)\left(\frac{y^p-x^p}{y-x}\right).$$ Note that $$\frac{y^p-x^p}{y-x} = y^{p-1} + y^{p-2}x + \cdots + yx^{p-2} + x^{p-1},$$ so $\frac{y^p-x^p}{y-x}$ is a positive integer. Now, let $q$ be a prime divisor of $\frac{y^p - x^p}{y-x}$. If $q = p$, then we would have that $p \mid \frac{y^p-x^p}{y-x} \mid y^p - x^p$, so by FLT, since $y^p \equiv y\pmod{p}$ and $x^p \equiv x\pmod{p}$, we would have $p \mid y-x$. However, we know that $p > (a+b)^{1434} \geq (x+y)^{1434} > y-x > 0$, so we cannot have $p \mid y-x$. So $q \neq p$. Then, we have that $$q \mid y^p - x^p,$$ and we cannot have $q\mid x$, since then $q\mid y$, so $q \mid \gcd(x, y) = \gcd\left(\frac{a}{\gcd(a, b)}, \frac{b}{\gcd(a,b)}\right)$, so $q\gcd(a,b)\mid \gcd(a,b)$, absurd (this also shows that $\gcd(x, y) = 1$). Thus, $x$ has an inverse modulo $q$, and so we can divide by $x^p$ to get $$\left(\frac{y}{x}\right)^p \equiv 1\pmod{q}.$$ If we had $$\frac{y}{x} \equiv 1\pmod{q},$$ then we would have $q \mid y-x$, however, since $\gcd(x, y) = 1$ as discussed earlier, we have that by LTE, $$\nu_q\left(\frac{y^p-x^p}{y-x}\right) = \nu_q(y^p-x^p) - \nu_q(y-x) = \nu_q(p) = 0,$$ as $p \neq q$, contradicting the fact that $q$ divides $\frac{y^p-x^p}{y-x}$. Thus, we have that $\text{ord}_q(y/x) \mid p$ and $\text{ord}_q(y/x) \neq 1$, so $\text{ord}_q(y/x) = p$, which must divide $q-1$ by FLT, so $q \equiv 1\pmod{p}$. Thus, every prime divisor of $\frac{y^p-x^p}{y-x}$ is $1\pmod{p}$. Now, note that since $0 < y-x < p$, it follows that every positive divisor of $$y^p - x^p = (y-x)\left(\frac{y^p-x^p}{y-x}\right)$$ is congruent to a positive divisor of $y-x$ modulo $p$, since every prime divisor of $\frac{y^p-x^p}{y-x}$ is $1\pmod{p}$. Thus, $a^p + p \equiv d\pmod{p}$ for some positive divisor $d$ of $y-x$. By FLT, $a^p + p \equiv a\pmod{p}$, so $a\equiv d\pmod{p}$. However, since $0 < a < p$, $0 < d \leq y-x < p$, we have that $a = d$, so $a$ is a divisor of $y-x$. Thus, we have that $$a^p + p \mid a\left(\frac{y-x}{a}\right)\left(\frac{y^p-x^p}{y-x}\right),$$ and $\frac{y-x}{a}$ is a positive integer. Note that $\gcd(a^p + p, a) = \gcd(p, a) = 1$, as $p > a$, so $$a^p + p \mid \left(\frac{y-x}{a}\right)\left(\frac{y^p-x^p}{y-x}\right).$$ By the same logic, $a$ divides $\frac{y-x}{a}$, so we can write $$a^p + p \mid a\left(\frac{y-x}{a^2}\right)\left(\frac{y^p-x^p}{y-x}\right),$$ and $\frac{y-x}{a^2}$ is a positive integer, and we can remove the $a$ since it has gcd $1$ with $a^p + p$, and so $$a^p \mid \left(\frac{y-x}{a^2}\right)\left(\frac{y^p-x^p}{y-x}\right),$$ etc., and keep repeating this process, and we get that $a^k \mid y-x$ for every integer positive integer $k$. This implies that $a = 1$, since $y - x > 0$.

Thus, since $a = 1$, we have that $$n + 1 \mid b^n + n$$ for every positive integer $n$. Note that $b > a = 1$. Then, $$n + 1 \mid (b^n + n) - (n+1) = b^n - 1.$$ Now, for $r\geq 1$, let $n = 2^r - 1$. Then, $$2^r = (2^r-1) + 1 \mid b^{2^r-1} - 1.$$ From taking $r = 1$, we have $2\mid b-1$, so $b$ is odd. Now, if $b\equiv -1\pmod{4},$ then $b^{2^r-1} - 1 \equiv (-1)^{2^r-1} - 1 = -2,$ so $\nu_2(b^{2^r-1}-1) = 1$, but taking $r = 2$ gives a contradicting as $\nu_2(2^2) = 2 > 1$. Thus, we have that $b \equiv 1\pmod{4}$, so we can apply LTE to obtain that $$r = \nu_2(2^r) \leq \nu_2(b^{2^r-1} - 1) = \nu_2(b-1) + \nu_2(2^r-1) = \nu_2(b-1),$$ but taking $r = \nu_2(b-1) + 1$ yields a contradiction. (Note that $b > 1$, so $\nu_2(b-1)$ is finite.)

Thus, we have a contradiction and so we must have had $a = b$, as desired. $\blacksquare$

Condition becomes \[ a^n + n \mid b^n - a^n \] for all positive integers $n$.

Let $p > 2541434198442069459ab$ be a prime. Choose $n=1+(a+1)(p-1).$ FLT gets that $p$ divides $a^n + n$, but also $p$ cannot possibly divide $b-a$ unless $b=a$ due to how stupidly large $p$ was chosen to be.

ISL 2005 N6
Very easy construction problem :)
[hide=simple solution]The problem is equivalent to proving $a^n+n|b^n-a^n$. If we can choose n to get $a^n+n$ to be a multiple of some prime $p>b^{n-1}+...+a^{n-1}>b\ge a$, we would be done since it would be forced that b-a<p is a multiple of p, whence b=a. Indeed, taking mod p $a^n+n\equiv 0\pmod p$ can be obtained by letting $a+n\equiv 0\pmod p$ (suitably choosing an n), and letting $n\equiv 1\pmod {p-1}$. There is a unique solution by CRT (motivated because we can naively try to get rid of the exponent n in order to just have p, which is very easy).[/hide]

@above why such a weird number lol i see 1434 but what else bruh

Rewrite the relation as $a^n+n \mid b^n-a^n$. WLOG $b \geq a$. Take a prime $p > b$ and allow $n \equiv 1 \pmod {p-1}$ and $n \equiv -a \pmod {p}$ which exists by CRT. It follows that $a^n+n \equiv a-a \equiv 0 \pmod p$ so $p \mid a^n+n$. This means that $p \mid b^n-a^n$ but since $b^n \equiv b \pmod {p}$ and $a^n \equiv a \pmod p$ we have $p \mid b-a$ which means $b-a = 0$ or $a=b$ as desired.

[b][color=pink]Claim:[/color][/b] There are infinitely many primes $p$ such that $a^n + n \equiv 0 \pmod{p}$ and $\gcd(n, p-1) = 1$ for some $n$.
[i]Proof.[/i] We show this holds for all odd primes $p$.
Take $n = 1 + (a+1)(p-1)$.
By FLT it follows that $a^{1 + (a+1)(p-1)} + 1 + (a+1)(p-1) \equiv a - a \equiv 0 \pmod{p}.$ $\blacksquare$
Note that this is equivalent to \[ a^n + n \mid b^n - a^n. \]
Taking this $n$, we then get that \[ p \mid a^n + n \mid b^n - a^n \] so $b^n \equiv a^n \pmod{p}$, and since $\gcd(n, p-1) = 1$ it follows that $b \equiv a \pmod{p}$.
Equality follows as there are infinitely many such primes.

We claim $p \mid b-a$ for all primes $p$, which directly implies the desired. Construct a value of $n$ such that
\begin{align*}
n &\equiv 1 \pmod{p-1} \\
n &\equiv -a \pmod p,
\end{align*}
which must exist by CRT. Then $a^n + n \equiv a-a \equiv 0 \pmod{p}$, so
\[0 \equiv b^n + n \equiv b-a \pmod{p}.~\blacksquare\]

[hide = How is this a TST 3]
Take a prime $p >> ab$
Set $n$ such that $n \equiv 1 \pmod{p-1}$ and $n \equiv -a \pmod{p}$
$p \mid a^n + n \mid b^n + n \equiv b-a$
$\therefore b=a$
[/hide]

BTW I believe a very nice solution is possible using limits...
Will upload if I finish it


floor function
function
ceiling function
algebra

how many numbers are divisible by $ 49$ in the set $ {1,2,3,4,\ldots2008}$

I believe that's the greatest integer value function.  $ \left\lfloor\frac{2008}{49} \right\rfloor\equal{}40$.  So I think there are 40 integers less than or equal to 2008 divisible by 49.

o thanks alot i didnt know i could use greatest integer value function( floor function) for finding the divisors. what is ceiling functon used for ?

Uh, please stop posting middle school problems in the intermediate topics forum, binmial_4eva.

In counting, for example, you could say if each bus holds $ x$ and there are $ y$ people, then you would need $ \left\lceil\frac {y}{x}\right\rceil$ buses.


AMC
USA(J)MO
USAMO
AIME
AMC 12
AMC 10
AMC 12 B

Alright, this thread is probably slightly premature considering the AIME II hasn't been administered and nothing can be finalized yet. But since this issue has become so popular, this is a chance for everyone to comment on this year's USAMO selection. If you have questions, post them. And of course all "expert" and official opinions are welcome. :-)

Here's how USAMO selection is generally done:

1. AMC score + 10*AIME score = USAMO index (out of a possible 300)
2. The top (approximately) 160 indices of AMC 12 takers are selected as qualifiers. The lowest index of these is considered "qualifying index".
3. Out of the people making the qualifying index, the lowest AIME score is selected. This AIME score is what we call the "floor value".
4. All sophomores, freshmen, and middle schoolers scoring the AIME floor value or above who did not make the qualifying index are invited to take USAMO.
5. In any state or territory without a USAMO qualifer already, the person with the highest index will be invited to take USAMO.

USAMO selection seems to differ slightly every year. That was how it was done last year and how it will most likely be done this year. However, the original criteria for the second selection was "out of sophomores and frehmen making the AIME floor value" instead of "take all sophomores and freshmen making the AIME floor value".

As you can see from the other threads, the predicted floor value for this year is a 9 (although 8 and 10 are also possibilities). It is also agreed that the qualifying index will probably be around 230.

Edit: Yea, Dave is right. No one can really be sure of how it'll be done except the AMC people. [url=http://www.unl.edu/amc/e-exams/e8-usamo/usamo.html]Here is the USAMO page on the AMC website.[/url] At the bottom, there's the criteria for USAMO selection. :-)

This is not entirely accurate.  I suggest checking the AMC's web site if you want completely accurate criteria.

I read the AMC website, but how is this year different?

you might make the floor value and the AMC 10 thingy does not hurt u at all...

[quote="mkkool64"]I have a question.  I'm a tenth grader who took the AMC12 but not the AMC10(my teacher told me to take the 12A and they don't do the B tests).  Does the AIME floor value for USAMO still count for me?  I got 123 on AMC12 and a 9 on the AIME, so It's possible I'll make the floor but not the index.  On the AMC site it says something about the first 160 people being chosen from the AMC12 and next group being from among people who took the AMC10.  Is it possible I'll miss USAMO because I took AMC12 instead of AMC10?  Or am I just a confused newbie that doesn't know what he's talking about...
Thanks! :D  :D  :D[/quote]

If the AIME floor value is 9 or below, you should make USAMO. If it's not, then taking the AMC 10 wouldn't have helped anyways.

The official statement on the AMC site is that "In advising students who desire to be selected for the USAMO whether to take the AMC 12 level contests or the AMC 10 level contests, it will be to their advantage to take the AMC 12 level contests."

If the index is 222 or less you'll make it

isn't $112 + 10*10 = 212 ?$

we sure can do math here at AoPS

[quote="Philpoleon"]I'm in 11th grade, and new to AoPS. I received a 112 on the AMC 12B and a 10 on the AIME I.[/quote]

You probably wont make it with your USAMO index, but if you live in a state with only a few AIME qualifiers (you can look at the state statistics on the AMC page to find out), you might make it if you have the highest USAMO index.

[quote="h_s_potter2002"]
You probably wont make it with your USAMO index...[/quote]

That may not necessarily be a fair or safe assumption...

so is this saying that the person in wyoming with a 98.5 amc b score (highest in state) will make it to the USAMO without taking the AIME?
*wishes living in wyoming with the wild buffalo

I've looked at the previous years and the way they selected USAMO qualifiers, and it seems that take the top 160 but then have categories for 11th grade and under, 10th grade and uder, etc. Is it different now?

So when Miths says AMC score, does he mean only the AMC 12 score?  Becausefi if this is not the case, then you could take AMC-10 and get some crazy score like 150 but an AIME score barely high enough to allow for a qualifying USAMO index but not high enough to make the floor value of everyone besides himself.  Because of this, if you feel confident of making the AIMEs it might be advantageous to take AMC-10 instead of AMC-12 (AMC-10s are easier).

[quote="bubala"](For example, last year, if you took the AMC 12 in 8th grade and got a 150 but only got a 6 on the AIME, you would make it.  If you took the 10 and got a 150, you'd need to hit the floor value of 7 on the AIME to make USAMO.[/quote]

This is an interesting hypothetical example, but no one in 8th grade took the AMC-12 and got a 150 on it and a 6 on the AIME last year. Based on my limited understanding of the system, I believe that if this situation had occurred, the floor score would have been a 6 (if they decided to follow the system that they have been using the past few years).

The AMC only takes about 80 additional qualifiers in grades 10 and below after they take the top 160 AMC 12 participants.  So if there are significantly more than 80 students in grade 10 or below (who don't get in the top 160) who make the floor, only the top 80 (based on USAMO index) will advance to the USAMO.

My advice:  Don't stress over it.  It's out of your hands.  You can't change the outcome by worrying about it.  You'll know next week.  Enjoy your weekend.  Be happy.

[quote="southside"]did they use this index thing for 10th graders last year?[/quote]Last year I believe that the numbers worked out so that they took everyone in grades 10 and below who made the floor (i.e. they had the right number of students, so they didn't need to resort to the USAMO index).

[quote="bubala"]How come?[/quote] As JGeneson pointed out, since last year's qualifying index was 210, a 150 + 6 would've made the cutoff and the floor would've been 6. If that had happened, a person having 150 on AMC 10 and 6 on AIME would also qualify.

Everyone needs to obtain the floor value in order to qualify (speaking to sophomores and below). If you do not have the floor on AIME, then you don't qualify no matter what you made on whichever AIME you took. 7 was floor so you needed 7.

Now again, remember the floor is set AFTER the top indices have already been taken, not before. Knowing that, a 150 with a low-but-in-qualifying-range AIME score would set the floor.

Two question: what is wrong with [quote]if you're a sophmore or below, it doesn't really matter whether you take the AMC 12 or the AMC 10 because if you get a high enough score on the AIME, then you make USAMO regardless.[/quote]? No matter which AMC you take, it's the AIME that really counts for 10th graders and below (except for the following scenario which is stupid a JGeneson pointed out).

JGeneson: I agree with you that my hypothetical was stupid, and yes, the floor value would become a 6.  But I'm saying that this situation would be the only time it'd help a 10th grader or below get into the USAMO who did only fairly well on the AIME.  If Mr. Someone took the AMC 10 and got a 150 and then a 6 on last year's AIME, he wouldn't make it.  But if Mr. SOmeone took the AMC 12 and got a 150 and then a 6 on last year's AIME, he would make it.  And yes, the floor value would change.  Am I missing something?

[quote="bubala"]Two question: what is wrong with [quote]if you're a sophmore or below, it doesn't really matter whether you take the AMC 12 or the AMC 10 because if you get a high enough score on the AIME, then you make USAMO regardless.[/quote]? No matter which AMC you take, it's the AIME that really counts for 10th graders and below (except for the following scenario which is stupid a JGeneson pointed out).[/quote]

This has been explained many times over, and I think Dave just made it quite clear. We've been lucky that the numbers have worked out well in the past. Just because a sophomore or below makes the AIME floor, it does not guarantee USAMO qualification. If the numbers do not work out (which they may not this year) then the second selection goes to USAMO indices as well.

[quote]
JGeneson: I agree with you that my hypothetical was stupid, and yes, the floor value would become a 6.  But I'm saying that this situation would be the only time it'd help a 10th grader or below get into the USAMO who did only fairly well on the AIME.  If Mr. Someone took the AMC 10 and got a 150 and then a 6 on last year's AIME, he wouldn't make it.  But if Mr. SOmeone took the AMC 12 and got a 150 and then a 6 on last year's AIME, he would make it.  And yes, the floor value would change.  Am I missing something?[/quote]

If you're pointing out the fact that indices with AMC10 scores are not counted in the first selection, you are correct.

[quote="filletwho"]Hmm...if they go to the index, won't ppl who took the AMC 10 have the advantage, having scored higher.[/quote]

I don't think you really have an 'official USAMO index' if you only took the AMC 10. So maybe you would be put at the bottom or something. That would suck. And I put my AMC 10 score on the AIME form, too. Hopefully this won't matter.

"Official USAMO index" is AMC + 10*AIME as defined on the AMC homepage. It just so happens that the indices coming from AMC10 takers are not considered in the first selection, but that doesn't mean you don't have an official index.

As for index selection the second time around, I'm not quite sure what'll happen. Apparently, AMC12 takers have an advantage. What sort of advantage...I'm not sure.

Ugh, the "advantage" better not be that AMC12 takers are all taken and then high scorers on the AMC10 after that...

Yah, I have often wondered about the advatage and only AMC can probably give us definite answer, but i suppose AMC12 could be weighed or something (if more than 80)

[quote="MithsApprentice"]Well... :-)

Rumor has it that AMC12 takers are given an automatic 150. *cough* Of course, that hasn't been tested yet but eh. We'll see...[/quote]

Oh *bleep*


For the class transcripts, next to the Transcript link, would it be possible to list the major points that were discussed in class that day? I find that I do a lot of looking-back when I work on MidTerms and it would be such a time saver if we didn't have to look through the transcript to know what it was about.

Thanks

well on the top they do have the name of the problems we did to skip to them. and usually each problem teaches another concept so its not that hard to find out


Find all triples of positive integers $(a,b,c)$ such that
\[ \Bigl(1+\frac{1}{a}\Bigl)\Bigl(1+\frac{1}{b}\Bigl)\Bigl(1+\frac{1}{c}\Bigl) = 2  \]

A solution to this problem can be found here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=42687

Note that all permutations of enescu's answers work.


AMC
AIME
geometry
circumcircle
ratio
irrational number
area of a triangle

The triangle ABC has AB = 24. The median CE is extended to meet the circumcircle at F. CE = 27, and the median AD = 18. Find area ABF.

Hmm...I can find the area of the triangle, the sides of the triangle, and the circumradius. It can't be far now...

yay i think i solved it. I actually didn't need the circumradius and the sides.

Ok: solution. Triangle BFE is similar to triangle CAE because angle FBE equals angle ECA and angle EFB equals angle CAB. So, that means the ratio of the area of BFE to triangle CAE is the square of the ratio of the length of BE to CE, which is 12/27=4/9. Thus, BFE is 16/81 times the area of CAE. Similarly, we can find that AFE is 16/81 the area of CBE. Adding those two equations together, we find that the area of ABF is 16/81 the area of ABC. Ok, hold that thought now.

If we let G be the centroid of ABC, we know AE=12, EG=9, and AG=12. We can find the area of this isosceles triangle. (i get an irrational number, so I think i made a mistake. oh well...) We multiply that by 6 to get the area of ABC, because the medians of a triangle divide the triangle into six regions of equal area. So now, we can get the area of ABC, and using the work in my last paragraph, we can get the area of ABF.

Someone please check why this doesn't work as I get an irrational number.

When I did this a couple of days ago (havent checked the forums since then), I came out with an irrational answer, which was this
" [color=black]((8 :sqrt: 395) / 3)[/color] ".  This is what I did:

Start spoiler: [color=black]If we make the figure, it comes out to cyclic quad AFBC with triangle ACB inside it.  IF we call the centroid G, then AG is 12 and GD is 6. Along the same lines, CG is 18 and GE is 9. Now because AE is 12, we get an isosceles triangle in AEG. Drawing the altitude down to EG, we get that it equals  (:sqrt: 12^2-(9/2)^2 ), which equals  (:sqrt: 395) / 2.  So that altitude is the altitude from A to side FE in triangle AFE. Now all we need to do is find FE and multiply it by that altitude. THe division by 2 for the area of a triangle is taken away by the fact that we are trying to get two of this specific area, because triangle AFE is equal in area to triangle BFE. We note that tri CEB is similar to tri AEF, so CE/AE = EB/EF,
= 27/12 = 12/FE. x here comes down to 144/27. Multiplying 144/27 by the aformentioned altitude and we get what I had said earlier, which is an irrational number. [/color]Perhaps this question had something more to it, like being equal to [i]k (:sqrt: l) / P and then finding what k + l + p equals?[/i]

Ok, my solution:

[hide]Alright, im assuming everyone's got the diagram done, so:

We know medians cut itself into 1:2 ratio parts. So, letting P be the centroid, we know AP=12, PD=6, CP=18, and PE=9.

Now, since we know that triangles AEF and BEF have the same area, since they have the same base and the same height, so if we could find out the area of one of them, we can find out the area of ABF. Im going to find the area of triangle BEF.

Ok, by power of a point, 12*12=27*EF.

So EF=16/3.

Now, if we could sinBEF, then we know the area.

Notice that triangle APE is isosceles, with legs 12 and base 9. By dropping a height, we easily see that sinPEA is sqrt[495]/24.

So now, area of ABF is 2*1/2*12*16/3*sqrt[495]/24.

That comes out to...8*sqrt[55].

i hope.....[/hide]

Who has the answer and what is it exactly? Talking about the one given as the solution by the AIME people.

[color=cyan]The correct answer to this one on the actual AIME, which is I think what jelyman was asking, was almost certainly either 63 = 8 + 55 or 440 = 8*55.  I think it may have been the latter, but I'm not sure.[/color]

its a nice problem tho

really nice problem

lotsa different ways to solve it

The answer on the actual AIME was 63 (8+55).


how many 8- digits number are which consist from 1,2,3,4, ( every number appear 2 times) such that two neighbor digits are distinct ( example 12314234, 23414231)
the number 12234314 do not satisfies the condition

Hint: Use PIE to count the sequences where at least two identical digits are next to each other

but will be a lot of sets to analyze. Is there any simple solution

if you use PIE correctly, it's really not that bad. there's symmetry. for example, the number of sequences with 2 adjacent 1's is the same as the number of sequences with two adjacent 2's.

can you help me in this problem , because i use
11 _ _ _ _ _ _
_ 11 _ _ _ _ _
_ _ 11 _ _ _ _
_ _ _ 11 _ _ _
_ _ _ _ 11 _ _
_ _ _ _ _ 11 _
_ _ _ _ _ _ 11
then for  2,3,4.
 i started like this , but I can not not reach the solution

Hi;

I have 864 by direct count. This might help to back engineer a method.

it's pretty unlikely the number 864 would lead you to backwards-engineer a solution.

Hi;

Yes, but you could check it against the answer you come up with by analytical means.


trigonometry
geometry
ratio
analytic geometry
graphing lines
slope

What is the relationship between the tangent in trigonometry and the tangent used in euclidean geometry? They have the same names but they mean different things!

The tangent in trig is the ratio between the measure of a side opposite an angle in a right triangle, and the leg that is adjacent to it.  The tangent in geometry is a line that intersects a circle at one point, and is perpendicular to a radius drawn to the point of intersection.

http://www.mathnstuff.com/math/spoken/here/2class/330/gif/pyth2.gif

Have a look at the trigo tan in there. We know that tan = sin/cos. We can see that from that picture, sin/cos = tan/1 (similar triangles). Can you see how it relates?

the first time I noticed that by mistake I was amazed...

Here's what I think.  Tangent lines are very much related to slope, which is basically the tangent of a right triangle formed; this is probably why they are named the same.

Masoud Zargar

EDIT: I suppose that image explains it nicely. :)

definitely. did you make a connection like that once?
I think they are called "blinding flash of the obvious" ;)

Wow!  I never thought of it that way!   :idea:  :idea:

Edit:  When did we get a smilie limit?


How many pairs of positive inetegers $ (a,\ b,\ c,\ d,\ e,\ f)$ with $ a>b>c>d>e>f$ are there such that $ a\plus{}f\equal{}b\plus{}e\equal{}c\plus{}d\equal{}22?$

[hide="Solution"]This can be rephrased as $ (11 \plus{} x,11 \minus{} x,11 \plus{} y,11 \minus{} y,11 \plus{} z,11 \minus{} z)$ with $ 11 > x > y > z > 0$ Because we want them all to be different, there are $ 10*9*8 \equal{} 720$ possibilites for $ (x,y,z)$. Out of every 6 possibilities, only one has the right order ($ x > y > z$), so there are $ \frac {720}{6} \equal{} 120$ possible pairs.[/hide]

For the general equation of $ x_1\plus{}x_2\equal{}22$ with $ x_1>x_2$  and both positive integers there are $ (21\minus{}1)/2\equal{}10$ solutions. Since a $ x_2$ is fixed once $ x_1$ is chosen we just have to look at $ a, b,$ and $ c$ because if we can successfully pick an $ a, b,$ and $ c$ with $ a>b>c$ then we will automatically have the remaining variables fixed and in the correct order. We have 10 possible values that $ a, b,$ and $ c$ can be. Since the order is already fixed once the 3 numbers are picked, there are 10 choose 3 = 120 ways to pick these three variables.


A five-digit number is called a mountain number if the first three digits increase and the last three digits decrease. For example, $ 35,763$ is a mountain number but $ 35,663$ is not. How many five-digit numbers greater than $ 70,000$ are mountain numbers?

The first three digits of that particular mountain number must be $ 789$.  The next number can be from 1-8.  The number after, varies according to the number.

If $ 7891n$, then $ n$ has $ 1$ possibility
If $ 7892n$, then $ n$ has $ 2$ possibilites
If $ 7893n$, then $ n$ has $ 3$ possibilities

We are developing a pattern.

If $ 7898n$, then $ n$ has $ 7$ possibilities

So the answer is $ 1 \plus{} 2 \plus{} 3 \plus{} ... \plus{} 6 \plus{} 7$ or $ \boxed{28}$.

[quote="Mrdavid445"][quote="Jasonion"][quote="isabella2296"]Actually, it would be $ 1 \plus{} 2 \plus{} 3 \plus{} 4 \plus{} \dots \plus{} 7 \plus{} 8 \equal{} \boxed{36}$, because you forgot $ 789n0$.

EDIT: Oops, typo. Fixed.[/quote]
how do you get so many posts????[/quote]
What a great reason to revive this...
She has so many posts cause she makes quality posts that aren't deleted and that she has been on aops for a long time.[/quote]

posts get deleted????


abstract algebra
superior algebra
superior algebra unsolved

Find all automorphisms of $(\mathbb{Z}_{6},+)$.

Okay, the solution I found says: Let $f: \mathbb{Z}_{6}\to\mathbb{Z}_{6}$ a morphism then $f(x+y)=f(x)+f(y)$ for all $x,y\in \mathbb{Z}_{6}$. We have ${f(\hat{0}})=\hat{0}$ and $f(x)=x\cdot f(\hat{1})$. Denote $f(\hat{1})=a\in\mathbb{Z}_{6}$ then $f(x)=ax$ is the form of an automorphism of $\mathbb{Z}_{6}$ [b][color=red]only if $a$ is inversable?[/color][/b]

I have something unclear about this... since $(\mathbb{Z}_{n},+), \ n\in\mathbb{N}$ is an abelian [b]group [/b] I don't understand the "red  condition" part from the solution.

Slizzel, an automorphism is bijective if I'm not wrong. So $a$ must have an inverse. If it didn't, then some elements wouldn't have a preimage.

Right or wrong?

Well $(\mathbb{Z}_{6},+)$ has all elements inversable because is a group.

It doesn't matter. We must have a solution to $ax = b$, where $a,b$ are given !!!

PS: When I say inverse, I say inverse in $\left( \mathbb Z_{6}, \cdot \right)$. [Yeah, I wrote a stupid thing  :blush: But the idea is there]

$(\mathbb{Z}_{6},\cdot)$ is not a group so not all its elements are inversable (only $\hat{1},\hat{5}$) but my problem isn't reffering to multiplicative operation and the condition seems strange. ...

Just to finally settle that thing:

Yes, the red part concerns inverses via multiplication. This should be clear since $ax$ is the product, not the sum, of $a$ and $x$.

The whole problem in abstract terms:
$\mathrm{Aut}(\mathbb{Z}/n\mathbb{Z}) = (\mathbb{Z}/n\mathbb{Z})^{*}$


search
combinatorics unsolved
combinatorics

On an infinite table,a horizontal line is chosen.Checkers may be placed below the lined,
at most one per square.The following move may be performed:A checker may jump over
an adjacent checker(Horizontally&vertically)into an adjacent empty square;the checker
jumped over is then removed.Determine the maximum number of the squares that one checker can go across the chosen line.

I found the following threads by searching for the word "checker":
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1658193305&t=21914
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1658193305&t=169621
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1658193305&t=182741


I see the symbol: ^ many times on other posts. Does anybody know what it means?

Thanks! By the way, where do get the symbol: squared or cubed?  :huh:

I think you mean the Latex, if so.
[hide="I hope I understand"] to write x to the yth power post this code
 [code]$x^{y}$[/code]$x^{y}$
I'm probably misunderstanding what you meant but if this helps
you can check out http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php
[/hide]

[quote="KrazyNerd] Also, anything ^1 equals itself, and anything ^0 equals one.[/quote]
With the exception of zero, of course.

[quote="KrazyNerd"]^ means "to the power of." It's like multiplying something by itself that many times. You can also write it as a little number above and to the right of the number you're doing the operation to.

Two to the third power (two cubed, or 2^3):
$2^{3}= 2*2*2 = 8$

Four to the second power (four squared, or 4^2):
$4^{2}= 4*4 = 16$

Also, anything ^1 equals itself, and anything ^0 equals one.[/quote]

I got you on the part that ^=to the power of, so 2^3=$2^{3}$, but i don't get the part with the "anything ^0 equals one...."  :play_ball:  :starwars:

Then why did not you ask that in the first place?  :huh:
It is by the law of exponents with fractions.
Example $\frac{2^{3}}{2^{3}}= \frac{8}{8}= 1 = 2^{3-3}= \frac{2^{3}}{2^{3}}$.  If things have the same base in a fraction, like with $2$ in this example, you subtract the exponents.
An exception, $0^{0}= \text{U}\not= 1$ , since there is no division by zero.

Why did you write "$0^{0}=U$"? What is U???
You said "^=to the power of". So, $2^{3}$ = 2 to the power of 3. Is that correct? If not, please show me the right one.

U probably means "undefined", as $0^{0}$ is undefined.
and yes, that is correct.


rotation
symmetry
geometry
geometric transformation
trigonometry
calculus
integration

Consider how any two particles move on the line connecting them.  The problem requires no calculus.

BTW, you already have Irodov?

No, I just have found a few problems on the internet.

Also, what do you mean 2 particles moving in a line connecting them?  In this case the line that connects the two particles continuously rotates, because the motion changes direction all the time.  :?:

By symmetry, I think that the triangle shrinks at an equal rate for all three sides, and that because of the change in direction, the triangle rotates continuously.  What we care is the shrinking rate of the triangle, but I don't know how much of the v component is put into the shrinking, and how much into the rotation.  Wow, this problem is confusing.

I can't really describe this quantitatively...

They will all meet in center, because they are constantly forming a equilateral triangle, and it's reducing. As they are $ s\equal{}\frac{a\sqrt3}{3}$ from the center, and the center-moving velocity is $ u\equal{}\frac{v}{\sqrt3}$they will meet in: $ t\equal{}\frac su\equal{}\frac{a}{3v}$

That is not quite correct because the "center-moving velocity" is not constant with time.

That's what I was thinking...  I was like, Irodov can't be [i]that[/i] easy  :D

Hey Genericme, could you read my previous post?  I gave basically all of my ideas, am I on the right track?

(for example, thinking that the 3 lattices of the triangle for equilateral triangles over time that shrink at a constant rate, while the triangle formed by the 3 particles continously rotate.  We ignore the rotation b/c it's motion is not relevant, and calculate the rate of shrinking...)

I do believe moledude has the right idea. The particles always form an equilateral triangle, and the velocity of each is always pointing along one of the sides. So the component of velocity towards the center is $ \sin{60^{\circ}} v \equal{} \frac{\sqrt{3}}{2} v$. And the distance to the center is $ \frac{\sqrt{3}}{3} a$, so the time it takes for them to meet at the center is $ \frac{2 a}{3 v}$.

Another way you can do this is like genericme said: consider the path between two of the particles. One of them will have speed $ v$ along this path towards the other, which itself will have a velocity component of $ \frac{1}{2} v$ towards the first.

Wow! Thanks a bunch.

Arghhh... I was so dumb. :oops:

[quote="integration"]Three particles a, b and c are at the vertices of an equilateral triangle with side length a.  a moves continuously toward b (is aimed at b continuously), and b moves toward c and c moves toward a, all at a speed v.

How much time is required until all 3 particles meet?

My attempt:

I have no idea.  I just set up the vertices and tried to solve using a dx from a causes a d(theta) in direction of c, and so on, but don't know how to progress from that  :oops:[/quote]
I kinda get an equation of a spiral which means these poor guys will never meet or meet at $ t= \infty$ as theres a $ t$ term in the denominator of $ e^{(a \: lot \:  of \: stuff)/t}.$ Thats a long way to go.

My equation (general for a polygon with angle = $ \alpha = \frac{1}{2} \left( \pi - \frac{2\pi}{n}\right)$:
$ r = r_0e^{(-\cot \alpha) \theta}$. And ${ \theta}$ is angle covered by radius of particle throughout its journey.

Can someone tell me what I'm doing wrong?

Call the area of the equilateral triangle A, which is equal to 1/2*a^2*sin60. In an infinitesimal time dt, the triangle shrinks to a smaller equilateral triangle. The total area lost is 3*1/2*vdt*a*sin60. When the points converge, the triangle's area will be 0, and so the original triangle will have lost a total area of A. Equating A = integral of total area lost in time dt from time 0 to t, and then solving for t gives t = a/3v.

Here is an attached image for what i'm envisioning. The larger triangle is the original triangle connecting the 3 points, and the smaller one is the triangle after a time dt. The area lost is equal to 3*the area of one of those thin smaller triangles.


Let $ S$ be a field of characteristic $ p\neq 0$. Let $ T\equal{}S(x^{1/{p^\infty}}) \equal{} limS(x{1/p^n})$. Show that the module of Kahler differentials $ \Omega_{T/S}\equal{}0$.

For this problem I have shown that $ dx\equal{}0$ by looking at $ d((x^{1/p})^p$. However, after this, I am quite stuck. I am trying to show that $ d(x^{1/{p^n}})\equal{}0$, but it seems any trick using derivations to do it doesn't lead anywhere. If anyone could give some insight that would be great. thanks

What about doing the same again¿:
$ x^{\frac{1}{p^n}} \equal{} \left( x^{\frac{1}{p^{n\plus{}1}}} \right)^p$.


Euler
calculus
integration
exterior angle

Hello all,
              Here are the problems ,

             1. The number of different positive integer triplets (x,y,z) satisfying the equations x^2 + y - z = 100 and x + y^2 - z = 124 is

             2.In r ABC, the altitude from A to BC meets BC at D and the altitude from B to CA meets AD at H. If AD = 4, BD = 3, CD = 2 then the length of HD is

             3. A number with 8 digits is a multiple of 73 and also a multiple of 137. The second digit from the left equals 7. Then the 6th digit from the left equals
(A) 1 			(B) 7 		(C) 9 		(D) can be any digit

             4. The rays PX and PY cut off arc AB and CD of a circle with radius 4. If the length of arc CD is 2 times the length of the arc AB and length of CD is 4pi/5  , then the angle APB is

                                         Enjoy them ,
                 Post solutions as soon as possible .
                                                                              Yours
                                                                            Syed Imam

[3] If you haven't been bored enough one day to memorise the factoriations of 1001, 10001, and maybe 100001, then you really should one day :D

Hi,

For (3)

[hide]

73 and 137 are both prime, so if both divide a number, then 73*137 = 10001 will also divide it. Since the number has 8 digits, so it could be 10001*abcd; which means the number is of the form abcdabcd. So, if the 2nd digit from left is 7, then b=7, implies the 6th digit from left is also 7.

[/hide]

About (4), I didn't get it properly. If PX intersects the circle at A and B, then angle APB = 0, right? If your question was to find angle AOB where O is the center, then the answer is [hide]pi/10[/hide]; and if you wanted angle ACB or ADB, then

[hide]ACB = ADB = 1/2 * AOB = pi/20[/hide]

********************************************************************

Sambit

(2) is not really difficult. You call the intersection of BH with AC E. Then because of Pythagoras you have AB=sqrt(3^2+4^2)=5, AC=sqrt(4^2+2^2)=sqrt(20). Because ABC is isosceles you have EC=sqrt(20)/2=sqrt(5). Then you have BE=sqrt
(25-5)=sqrt(20). Because of EC:HD=BE:BD HD=1.5 if I didn't nake a calculation mistake.

Misha

I like number 1.

Sambitnayak, I believe he means that the arcs cut off are the arcs between the two rays, rather than the arcs cut by each ray.  That is:

I think I'm doing something wrong for 1

[hide]im just getting 1.

y^2-x^2+x-y=24

(y-x)(x+y-1)=24 .. x,y=13,12 and all other x,y are too small since z has to be positive?

[/hide]

Hi all,

JBL, Thanks for posting the figure. It helped a lot. But since Euler_India said that the arc AB is half the arc CD, so I guess you mistakenly interchanged AB and CD in your figure.

For (4),

[hide]

Let the center be O. Angle COD = pi/5. (got from length of arc CD and the radius of circle)

Angle AOB = 1/2 * Angle COD = pi/10.

Angle CBD = 1/2 * Angle COD = pi/10.

Angle ACB = 1/2 * Angle AOB = pi/20.

In triangle CPB, exterior angle CBD = angle CPB + angle PCB. But since angle PCB is same as angle ACB, so angle CPB = CBD - ACB = pi/10 - pi/20 = pi/20.

But angle APB is the same as angle CPD = angle CPB, thus angle APB = pi/20.

PS : The explanation seems involved owing to the angle names, but once one sees the figure, its simple enough.

[/hide]

********************************************************************

Sambit

My pleasure with the diagram, and you are correct that I've flipped by letters.

Geomar -- I get the same thing you do for number 1, in addition to 3 integral (not all the variables positive) solutions.


Is possible to use Pstricks in aops? If so, how?
Some programs like GeoGebra can export the figure using the package pstricks. It's usefull when you don't want to type asymptote.

thanks

It's not possible at the moment as, as far as I'm aware, the PSTricks packages aren't supported. The only way to do is it is to create an image and use that.


function geometry rectangle calculus integration

before i start, sorry this is not in latex.   :oops:

let a function,f be defined on the closed interval [0,1].f is continuous and monotonic. f(0)=0 and f(1)=1.
prove that

 f(1/10) + f(2/10) ...... +f(9/10) + f inverse(1/10) +f inverse(2/10) +....... f inverse(9/10) < 99/10

a nice hint would be appreciated even more than a step by step solution.thanks a lot in advance!

i was able to prove that the sum is less than 10 by using the rectangle method of definite integrals, but the further 1/10 eludes me :mad: anyone have any ideas how i could build on this idea?????once again thanks a lot :lol:

Here's a sketch of what's going on. Of course, you will have to formalize this.

Let's look at the equivalent situation

$\sum_{k=1}^{9}[f(x)+f^{-1}(x)] < 99$

[hide="Visually, we have"] that at each integral value of $x$ from $1$ to $9$, find $f(x)$, and draw a horizontal line across to $(x+1,f(x))$, and for each integral value of $f(x)$ from $1$ to $9$, find $x$ and draw a vertical line to $(x,f(x)+1)$. Note that the square with corners $(\{ 0,1 \},\{ 0,1 \})$ cannot be shaded![/hide]

[hide="Then,"] note that the only function that can fully fill in the remaining $99$ squares is a step function, which would have horizontal segments that do not satisfy the condition that it is monotonic.[/hide]


$ 1.$  Solve for $ x$ and $ y$
    $ 5x(1 \plus{} \frac1{x^2 \plus{} y^2}) \equal{} 12$
    $ 5y(1 \minus{} \frac1{x^2 \plus{} y^2}) \equal{} 9$

$ 2.$  Solve for $ x$ and $ y$
    $ xy^2\equal{}15x^2\plus{}17xy\plus{}15y^2$
    $ x^2y\equal{}20x^2\plus{}3y^2$

$ 3.$   Find $ a,b,c$ such that
        $ (1\plus{}\frac1a)(1\plus{}\frac1b)(1\plus{}\frac1c)\equal{}3$

Pls I dont have a clue how to start. They're not to be very lengthy and tedious

I think for question number 3 $ (a,b,c) \equal{} (\frac {1}{2}, \minus{} \frac {1}{2}, \minus{} \frac {1}{2})$ and its permuations satisfy the condition.
For all your problems,do you [b]consider[/b] $ x,y,a,b,c$ to be real numbers?

[quote="new_member"]For all your problems,do you consider $ x,y,a,b,c$  to be real numbers?[/quote]

Yes for all the 3 problems $ a,b,c,x,y,z \in R$

[quote="new_member"]For the third one i found those solutions just by considering the fact that 3 is a prime and if $ a,b,c$are INTEGERS one possible
case for example for the product would be$ 3( \minus{} 1)( \minus{} 1) \equal{} 3 ..$
[/quote]

I couldn't follow that : If the product is written as $ 3( \minus{} 1)( \minus{} 1) \equal{} 3$ then $ a,b,c$ can't be integers :?:

I apologize.
Edited.I deleted the nonsense I had written.
I will try your problems more carefully.I am sorry about that.

For 1 let $ z \equal{} x \plus{} yi$ so adding the first equation and $ i$ times the second equation we get $ 5z \plus{} \frac5z \equal{} 12 \plus{} 9i$.

For 2 we have
\[ xy^2(20x^2 \plus{} 3y^2) \equal{} x^2y(15x^2 \plus{} 17xy \plus{} 15y^2)\implies(x^2 \plus{} y^2)(5x \minus{} y) \equal{} 0.\]

[b]For the third question:[/b] Can't we just simply assign random values to $ a$ and $ b$ such that $ \left(1 \plus{} \frac 1 a\right)\left(1 \plus{} \frac 1 b\right) \ne 3$ and solve for $ c$? It seems like there are infinitely many solutions. Then again, I might have misunderstood the question.


induction algebra function domain inequalities

Problem: Show that $ 2^n<n$ for $ n>3$

Solution: Consider the base case $ n\equal{}4$. We can show that the formula holds because:

$ 2^4<4!$
$ 16<24$

Now we move on to the induction step:

When $ 2^n<n!$, then we can multiply $ n$ to both sides to get:

$ 2^{n}(2)<n!(2)$
$ 2^{n\plus{}1}<n!(2)$

How are you supposed to take $ n!(2)$ and get $ (n\plus{}1)!$?

Comparing $ n! \cdot 2$ to $ (n\plus{}1)!$ is the same thing as comparing $ 2$ to $ n\plus{}1$.  So...

hello, you must show that $ 2n!<(n\plus{}1)! \Leftrightarrow 2n!<n!(n\plus{}1) \Leftrightarrow n>1$ is true.
Sonnhard.

[b]Dr Sonnhard Graubner: [/b]
That can be shown by (after diving $ n!$ from both sides):
$ 2<n\plus{}1$
$ 1<n$

And since $ n>3$, $ 2n!<(n\plus{}1)!$ is always true for any value of $ n$.

[b]t0rajir0u:[/b]

I take it you were hinting to the same thing as Dr Sonnhard Graubner. However, is there a way to prove this is true without having to also prove that $ 2n!<(n\plus{}1)!$ for all the $ n$ in our domain?

Does this work?
I sort of figured that if you have an inequality and you perform an operation on the side with greater value which would, without exception, make its value even greater, then the inequality still holds.

[hide="Proof"]
Prove $ \forall{n}\in\mathbb{Z}$, $ n>3$, that $ 2^n<n!$

[b]Establish a base case, $ n\equal{}4$:[/b]
$ 2^4<4!$ is true because $ 16<24$

[b]Assume that the condition holds for some integer $ k$:[/b]
$ 2^k<k!$ is assumed to be true.

[b]Show that if the condition is true for $ k$, it's true for $ k\plus{}1$[/b]
$ 2^k<k!$ is assumed to be true.
Multiplying both sides by 2 we get: $ 2^{k\plus{}1}<2k!$
We know that $ k>2$ because our condition has declared that $ k$ is an integer greater than 3. Therefore, $ k\plus{}1>2$ as well. Dividing this inequality by 2 gives us: $ \frac{k\plus{}1}{2}>1$.
Because of this fact, we can multiply the right hand side by $ \frac{k\plus{}1}{2}$ and still preserve the inequality.
Thus, we arrive at:$ 2^{k\plus{}1}<\frac{2k!(k\plus{}1)}{2}\implies 2^{k\plus{}1}<(k\plus{}1)!$.

This shows that $ \forall{n}\in\mathbb{Z}$, $ n>3$, $ 2^n<n!$, Q.E.D.
[/hide]

How can you multiply just one side of an inequality (or an equation for that matter) and expect it to hold just because the term you multiply in is greater than one?

i.e. $ 2>1$:  $ 4 < 3$ but $ 4 > 2(3)$

[quote="happyme"]\is there a way to prove this is true without having to also prove that $ 2n! < (n \plus{} 1)!$ for all the $ n$ in our domain?[/quote]
Why would you want to?  It's the only step necessary to make the induction work and it follows immediately.

[quote="happyme"]How can you multiply just one side of an inequality (or an equation for that matter) and expect it to hold just because the term you multiply in is greater than one?[/quote]
Inequality is transitive; that is, if $ a < b$ and $ b < c$ then $ a < c$.  If all the terms involved are positive and we have a constant $ k > 1$, then we know that

$ a < b, b < bk \Rightarrow a < bk.$

Your example is invalid because you've switched some signs.

My logic was more along the lines of: if $ 4>3$ then $ 4k>3$ for all $ k\geq 1$


The follwing sequence uses the Fibonacci sequence on the top row. How do you get to the bottom row. You could add like 8 (the 6th fib number) to 3 (the 4th fib number) to get 11 (the 3rd unnamed sequence number) for example. But find a continuous pattern.

   0|1|1| 2 | 3 | 5 | 8 |13
   5|3|8|11|19|30|49|79

wait is the bottom row related to the top row or are they different patterns?

from what i understand...
yes the bottom row is related to the top row. you apply a certain rule to a number in the top row to get the corresponding number in the second row.

There is no pattern relating the terms specifically in the first row to the terms in the second row (at least I don't think  :D ), but notice that:

The sequence is $5,3,8,11,19,30,49,79...$

5+3=8, 3+8=11, 8+11=19 etc.

It is basically another Fibonacci-type sequence where each term is produced by summing the previous two numbers.  :)

correct! but what I need help on is what to do on the Fib sequence to get the bottom part.

hmm.... i don't see how u could apply 1 rule or function to the same number to get two different results. :huh:
l 1 l 1 l
l 3 l 8 l
or i think that this could mean the rule is based on the position of the number in the sequence, but, no, that can't work. :?:

What I don't get is the 5, 3, when the Fibonacci sequence is always increasing...

Look at the problem, the fibonacci is on top and a different sequence with fibonacci rules is on bottom. Also look at example in question.

I don't think there is a possible way to convert the top sequence to the bottom one. (at least with common formula.)

What do you mean? Its the Fibonacci sequence

look, the answer can be like the 6th fibonacci nmber plus the 4th fibonacci number. One example formula is fib compared to the lukas:
1 1 2 3 5...
2 1 3 4 7...n

The way to find the lukas from fib is by starting at the third number, the number above say n plus the number two numbers to the left would equal n.

I think this problem is making too much confusion

The top sequence is $\frac{(\frac{1+\sqrt{5}}{2})^n-(\frac{1-\sqrt{5}}{2})^n}{\sqrt{5}}$.  The bottom sequence is $\frac{(25+\sqrt{5})(\frac{1+\sqrt{5}}{2})^n+(25-\sqrt{5})(\frac{1-\sqrt{5}}{2})^n}{10}$.  This means that the ratio of the bottom sequence to the top sequence gets closer and closer to $\frac{5\sqrt{5}+1}{2}$ as $n$ increases.

yes, but say you went on a couple more numbers on the fib sequence, what would you do to get the bottom number?

Suppose we start the second-order homogeneous linear recurrence $a_{n+1} = a_n + a_{n-1}$ with two arbitrary values $x, y$.  In this case, $x = 5, y = 3$.  We will define the Fibonacci sequence $F_{0} = 0, F_{1} = 1, F_{n+1} = F_n + F_{n-1}$.  Now let us calculate the values of our other sequence.

$a_0 = x$
$a_1 = y$
$a_2 = x + y$
$a_3 = x + 2y$
$a_4 = 2x + 3y$
$a_5 = 3x + 5y$

Etc.

It is trivial to prove $a_n = a \cdot F_{n-1} + b \cdot F_{n}$.

In this case, we can use the rule $a_n = 5 \cdot F_{n-1} + 3 \cdot F_{n}$ to calculate the second row from the first.  :)

Oh, ok, now I understand the question. :)

You could use the closed-form definitions of the two sequences to define the second sequence in terms of the first, but it's not the recursive answer that vicseb asked for.


complex analysis
complex analysis unsolved

Prove that the two annuli $ A_1 \equal{} \{z \in \mathbb{C} : r_1 < |z| < R_1 \}$ and $ A_2 \equal{} \{z \in \mathbb{C} : r_2 < |z| < R_2 \}$ are conformally equivalent iff $ R_1/r_1 \equal{} R_2/r_2$.


[u][b]Day 1[/b][/u]([i]23/02/[/i][i]2006)[/i]
[b][u]Problem 1[/u][/b]Solve the symtem:
             $\sqrt{x^{2}-2x+6}.log_{3}(6-y)=x$
             $\sqrt{y^{2}-2y+6}.log_{3}(6-z)=y$
             $\sqrt{z^{2}-2z+6}.log_{3}(6-x)=z$
[u][b]Problem 2[/b][/u]Let a convex $ABCD$.$M\in (AB)$,$M\neq A,B$Let $N\neq M,N=(MAC)\cap (MBD)$.Prove that
 1/$N\in$ a fixed circle.
 2/$MN$ through to a fixed point[u][i].Note[/i][/u]$(XYZ)$is the circle through to $X,Y,Z$
[u][b]Problem 3[/b][/u]Let $m,n\in Z,m,n>3$.A board $m\times n$($m$row ,$n$columm).Consider a figure (K)contains 4 squares with 2 squares in a line and 2 squares in  a line (anybody from Vietnam can draws it for me?  :D ).With
1/$m=2004.n=2006$
 2/$m=2005,n=2006$
Can we lay 4  breaks in a K, such that after finite time:the number of breaks in each square is const?
tomorrow,I will post the problem of day 2 when the competion end.  :)

[quote="prowler"]Just ideas:
Problem $1$
First of all $\displaystyle x,y,z\leq 6$, take $\displaystyle f(x)=\frac{x}{\sqrt{x^2-2x+6}}$

$\displaystyle f'(x)=\frac{6-x}{(x^2-2x+6)^{3/2}}$, thus on $(-\infty,6)$ it is increasing one.

WLOG  $x\leq 3$ then
$\displaystyle log_{3}(6-y)=f(x)\leq f(3) =1$,  $y\geq 3$.
$\displaystyle log_{3}(6-z)=f(y)\geq f(3) =1$,  $z\leq 3$.
$\displaystyle log_{3}(6-x)=f(z)\leq f(3) =1$,  $x\geq 3$, contradiction.
Analogously for $x\geq 3$, thus unique solution $x=y=z=3$.

Problem $2$:
1) Let $T={AC}\cap{BD}$ then $C, T, D, N$ are concyclic (angle chasing).
2) Suppose $MN$ intesects $(CTD)$ in point $X$ then $NC, NT, NX, ND$ form $r$-corelative
harmonic fascicle on a fixed circle (again angles), hence $X$ is a fixed point.

Can anybody post exact condition for problem $3$.[/quote]Please, not in this topic! Post the problems in their own topics in the olympiad forum and then the links back here. Thanks.

[u][b]Day 2[/b][/u][u][i](24/02/2006)[/i][/u]
[u][b]Problem 4[/b][/u]Let
 $f(x)=\sqrt{(x+a)(x+b)}-x$with$a,b>0,a\neq b$
Prove that
$\forall s\in (0,1)$always exist one and only one $\alpha >0$such that
            $f(\alpha)=(\frac{a^s+b^s}{2})^{\frac{1}{s}}$

[u][b]Problem 5[/b][/u]Find all $P(x)\in R[x]$such that
          $P(x^2)+x(3P(x)+P(-x))=(P(x))^2+2x^2\forall x\in R$

[u][b]Problem 6[/b][/u]Let $S$is a set of $2006$positive integer number.Consider$T\subseteq S$is a "good" subset if and only if$\forall u,v\in T\rightarrow$ $T$don't contain $u+v$
a)if $S=$  {$1,2,...,2006$}.Prove that $\forall T$is a "good" subset of $S$ then $|T|\le 1003$
b)Prove that $\forall S$contains $2006$positive integer number,we have a$T\subset S$and $|T|=669$

[quote="11235813"][quote="nthd"]

[u][b]Problem 5[/b][/u]Find all $P(x)\in R[x]$such that
          $P(x^2)+x(3P(x)+P(-x))=(P(x))^2+2x^2\forall x\in R$

[/quote]

Is P a polynomial?[/quote]
yes,P is a polynomial

There might be something wrong with problem 3.


A line cuts the perimeter and the area of a triangle in half. Prove it goes through the incentre.

The line $l$ must cut at least two sides of the triangle.

[b]Case 1.[/b] It cuts exactly two sides. (See the attached diagram.)

The intersection points must be inner points of some two sides. WLOG assume $l\cap AC=P$ and $l\cap BC=Q$. Draw the angle bisector from $C$ and let it intersect $l$ in $D$. Draw the perpendiculars $DM$ to $AC$ and $DN$ to $BC$. By the properties of an angle bisector, $DM=DN=d$. Let $2s$ be the perimeter of $\triangle ABC$. Then $CP+CQ=s$ (by the construction) and $[ABC]=rs$ where $r$ is the inradius.

Now we have $[CPQ]=[CPD]+[CQD]=\frac{CP\cdot d}{2}+\frac{CQ\cdot d}{2}=\frac{sd}{2}$

But on the other hand, by the construction $[CPQ]={1\over 2}[ABC]=\frac{rs}{2}.$ Therefore ${rs\over 2}={sd\over 2}\iff r=d$, which means that $D$ is the incenter.

[b]Case 2.[/b] It cuts all three sides.

In that case $l$ must pass through one vertex and intersect the opposite side. Let it pass through $C$ and intersect $AB$ at $M$. Since $[CAM]=[CBM]$, it follows $AM=BM$ because those two triangles share the height. On the other hand, the perimeter condition gives $CA+AM=CB+BM\iff CA-CB=BM-AM=0$, which means that $\triangle ABC$ is isosceles and $CM$ is its symmetrial axis, which necessarily must pass through the incenter.


function
induction
algebra proposed
algebra

[quote="behdad.math.math"]find all functions $ f : R^ \plus{} \longrightarrow R^ \plus{}$ such that:

$ f(\frac {x \plus{} y}{2})$ $ \equal{}$ $ \frac {2f(x)f(y)}{f(x) \plus{} f(y)}$[/quote]

$ f(\frac {x \plus{} y}{2})$ $ \equal{}$ $ \frac {2f(x)f(y)}{f(x) \plus{} f(y)}$

I suppose that $ R^\plus{} \equal{} \{x>0\}$ (in my contry, it's $ \{x\geq 0\}$

Let then $ g(x)\equal{}\frac{1}{f(x)}$. The equation is $ g(\frac{x\plus{}y}{2})\equal{}\frac{g(x)\plus{}g(y)}{2}$

1) $ g(x)$ is a non decreasing function :
Assume we have $ a,b,r\in \mathbb{R}^\plus{}$ such that $ g(a\plus{}b)\equal{}g(a)\minus{}r$
Using $ x\equal{}a$ and $ y\equal{}a\plus{}2b$ in the equation, we get $ g(a\plus{}2b)\equal{}g(a)\minus{}2r$ and, with an immediate induction : $ g(a\plus{}2^nb)\equal{}g(a)\minus{}2^n r$ and so, with $ n$ great enough, $ g(a\plus{}2^nb)<0$, which is impossible.
Q.E.D.

2) $ g(x)$ is a segment on any non empty interval
Let $ a_0,b>0$ and $ \alpha\geq 0$ and $ \beta$ such that $ g(a_0)\equal{}\alpha a_0 \plus{} \beta$ and $ g(a_0\plus{}b)\equal{}\alpha (a_0\plus{}b) \plus{} \beta$
Considering $ a_n^k\equal{}a_0\plus{}\frac{k}{2^n}b$ (with $ n\in\mathbb{N}\cup\{0\}$ and integer $ k$ such that $ 0\leq k\leq 2^n$), it's easy to show with inductions that $ g(a_n^k)\equal{}\alpha a_n^k \plus{} \beta$.

Since $ \{a_n^k\}$ is dense in $ [a,a\plus{}b]$ and since $ g(x)$ is monotonous (non decreasing), we get $ g(x)\equal{}\alpha x \plus{} \beta$ for any $ x\in[a,a\plus{}b]$

3) $ g(x)$ is a line
It's then immediate to get $ g(x)\equal{}\alpha x \plus{} \beta$ for any $ x\in R\plus{}$ (consider any interval containing $ [a,a\plus{}b]$, $ g(x)$ is a segment over him, but is also a segment over $ [a,a\plus{}b]$ and so these two segments belong to the same line).

So, $ f(x)\equal{}\frac{1}{\alpha x \plus{} \beta}$ for any $ \alpha \geq 0$ ($ g(x)$ is non decreasing) and any $ \beta>0$ (or $ \beta\geq 0$ if $ \alpha>0$).

And it's easy to check that these functions (necessary conditions) are solutions.


linear algebra
matrix
linear algebra unsolved

Suppose $A\in M_{m\times n}(\mathbb F)$ and $Rank (A)=r$. Prove that if we choose $r$ independent rows and $r$ independent columns then the intersection of these rows and columns is an invertible $r\times r$ matrix.

The fact the we change two rows (or two colums) will not change the rank of the matrix or anything else. Now suppose the we make such transformation to our matrix such that the first $r$ rows are those independent rows and the first $r$ columns are the independent columns. Let's assume by contradiction that the rows (of the $rxr$ matrix ) are not independent  than:

$c_{1}*Row_{2}+c_{2}*Row_{3}+...+c_{r-1}*Row_{r} = Row_{1}$

for some real constants, not al equal to $0$. But this would at the same time would imply that the first $r$ rows of the original "modificated" matrix are linearly dependent which is false so the rows of the $rxr$ matrix are independent thus the conclusion follows.


Can anyone find a trinomial probem like($x^2+2x+1=0$) that has incredibly high numbers??? I would like it if it wasnt able to be factored by any other way than what most people do(shown below)

$2x^2+x-10=0$
  2x[color=white]````````[/color]5
  1x[color=white]````````[/color]-2

(2x+5)(x-2)=0
x=5/2 or x=2

Does this count?

      $\frac{2x^2}{0}+\frac{393939x}{0}+\frac{4}{0}$

Nice infinities.

i like $\infty$
because it looks like a sideways 8

$x^2 + 669x - 2682022$.  Neat solution, I guarantee.  :)

Y'know, you can generate arbitrarily hard "trinomial problems" for other people by expanding out $(ax + b)(cx + d)$ for any sufficiently large choices of $a, b, c, d$, right?

[hide="So for this problem, I chose..."]
$a = c = 1, b = -1337, d = 2006$.  :D
[/hide]

try using guess and check on my trinomial to see if you can factor it

this is probably one of the hardest trinomials that you can factor.     [hide]$x^2+19997x+99970002$[/hide]
You can factor that. try it.
I SAID TRY IT!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! :cursing:  :cursing:  :cursing:

$(x+9998)(x+9999)$ easy just use vieta's formula

it might have been hard in 2006 when we didn't have all these neat formulas and tricks...


In the figure, ABNM, BCON, CDPO, and DEQP are congruent rectangles. If AM=21, find the length of KN.

CLICK ON IT BECAUSE SOME LINES DO NOT SHOW  :o

[quote="roadnottaken"]The answer is 21/4=5.25.[/quote]

explain?

Because the rectangles are congruent, point k is one-fourth of the way up segment bn.  Therefore, $kn=.25bk=.25am=5.25$.

Similar triangles. MN=1/4 MQ, thus, KN=1/4 EQ =21/4


This was created by Marian Tetiva and Darij Grinberg. :D

[quote="Inequalities Master"]Lat $ a,b,c\geq0$.Prove that:
$ \dfrac{a^3 \minus{} b^3}{a \plus{} b} \plus{} \dfrac{b^3 \minus{} c^3}{b \plus{} c} \plus{} \dfrac{c^3 \minus{} a^3}{c \plus{} a}\leq\dfrac{(a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2}{8}$[/quote]
It's equivalent to $ \sum_{cyc}c(a^2\minus{}b^2\minus{}2ac\plus{}2ab)^2\geq0.$

Very nice, Arqady. See also http://www.mathlinks.ro/Forum/viewtopic.php?t=64860 .

  darij

See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3614025#p3614025

[quote=Inequalities Master]Lat $ a,b,c\geq0$.Prove that:
$ \dfrac{a^3\minus{}b^3}{a\plus{}b}\plus{}\dfrac{b^3\minus{}c^3}{b\plus{}c}\plus{}\dfrac{c^3\minus{}a^3}{c\plus{}a}\leq\dfrac{(a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2}{8}$[/quote]

We have
\[RHS-LHS=\frac{\sum{c(a^2-2ac+b^2)^2}}{4(a+b)(b+c)(c+a)}\ge{0}\]


Let $ a,b,c\in\mathbb{R_{ + }}$ such that $ a + b + c = ab + bc + ca$. Prove the following inequality:
\[ \displaystyle\sum_{cyc}\frac {a}{b + c}\leq \frac{9}{8}(a + b + c) - \frac {15}{8}.
\]

[quote="Maxim Bogdan"]Let $ a,b,c\in\mathbb{R_{ + }}$ such that $ a + b + c = ab + bc + ca$. Prove the following inequality:
\[ \displaystyle\sum_{cyc}\frac {a}{b + c}\leq 9(a + b + c) - \frac {15}{8}.
\]
[/quote]
\[ \frac {a}{{b + c}} + 1 = \frac {{a + b + c}}{{b + c}} = \frac {{ab + bc + ca}}{{b + c}} = a + \frac {{bc}}{{b + c}} \le a + \frac {{b + c}}{4}
\]

\[ \Rightarrow \sum {\frac {a}{{b + c}}} + 3 \le \frac {3}{2}\left( {a + b + c} \right)
\]

Can you post the Romania JBMO TST 2008 problem?

Let us prove the following stronger inequality for $ a,b,c > 0$ such that $ a \plus{} b \plus{} c \equal{} ab \plus{} bc \plus{} ca$,
\[ \sum_{cyc}\frac {a}{b \plus{} c}\le\frac 32(a \plus{} b \plus{} c) \minus{} 3
\]
From the inequality posted [url=http://www.mathlinks.ro/viewtopic.php?p=966147#966147]here[/url], we deduce that
\[ \sum_{cyc}\frac {a}{b \plus{} c}\le\frac {3(a^2 \plus{} b^2 \plus{} c^2)}{2(ab \plus{} bc \plus{} ca)} \equal{} \frac {3(a \plus{} b \plus{} c)^2}{2(ab \plus{} bc \plus{} ca)} \minus{} 3,
\]
which was what we wanted.

Btw, when does equality hold in your problem? :maybe:

EDIT: Oops, beaten! :oops:

Oh, sorry! I modify the inequality! Now equality holds for $ a\equal{}b\equal{}c\equal{}1$

[quote="Maxim Bogdan"]Let $ a,b,c\in\mathbb{R_{ + }}$ such that $ a + b + c = ab + bc + ca$. Prove the following inequality:
\[ \displaystyle\sum_{cyc}\frac {a}{b + c}\leq \frac {9}{8}(a + b + c) - \frac {15}{8}.
\]
[/quote] The same idea
its equal
$ \sum\frac {a}{b + c}\sum ab+ \frac {15}{8}\sum ab\le \frac {9}{8}(a + b + c)^2$
$ \sum\frac {abc}{b + c}+\sum a^2\le \frac{9}{8}\sum a^2+\frac{3}{8}\sum ab$
$ \sum\frac {abc}{b + c}\le \frac{1}{8}\sum a^2+\frac{3}{8}\sum ab$ but $ \frac{abc}{b+c}\le \frac{a(b+c)}{4}$ and we are done

Romania JBMO TST problem: [url]http://www.mathlinks.ro/viewtopic.php?p=1154600#1154600[/url]

[quote="nayel"]Let us prove the following stronger inequality for $ a,b,c > 0$ such that $ a \plus{} b \plus{} c \equal{} ab \plus{} bc \plus{} ca$,
\[ \sum_{cyc}\frac {a}{b \plus{} c}\le\frac 32(a \plus{} b \plus{} c) \minus{} 3
\]
From the inequality posted [url=http://www.artofproblemsolving.com/viewtopic.php?p=966147#966147]here[/url], we deduce that
\[ \sum_{cyc}\frac {a}{b \plus{} c}\le\frac {3(a^2 \plus{} b^2 \plus{} c^2)}{2(ab \plus{} bc \plus{} ca)} \equal{} \frac {3(a \plus{} b \plus{} c)^2}{2(ab \plus{} bc \plus{} ca)} \minus{} 3,
\]
which was what we wanted.

Btw, when does equality hold in your problem? :maybe:

EDIT: Oops, beaten! :oops:[/quote]
\[ \sum_{cyc}\frac {a}{b \plus{} c}\le\frac 32(a \plus{} b \plus{} c) \minus{} 3
\iff \sum_{cyc}\frac {1}{b \plus{} c}\le\frac 32 .\]
for $ a,b,c > 0$ such that $ a \plus{} b \plus{} c \equal{} ab \plus{} bc \plus{} ca$.
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&p=1176002#p1176002]Romanian Junior  2008:[/url]
\[ \sum \frac {1}{b\plus{}c}\geq  \frac {1}{a+b+c}+1.\]
for $ a,b,c \ge 0$ such that $ a \plus{} b \plus{} c \equal{} ab \plus{} bc \plus{} ca$.


inequalities
search
number theory unsolved
number theory

find all answers of below quality in $\mathbb{Z}$
$x^4+x^3+x^2+x+1=y^2$

This was in national maths olympiad a few yrs back.

Bomb


function
algebra unsolved
algebra
functions

Here is an absolutely fantastic problem:
  Given is a sequence $ f_1, f_2,...$ of functions from the real numbers to the same set. Then there exist 2 functions $f,g$ such that any $f_i$ can be expressed as a composition of f and g. I do not know if I should put it in analysis, since the statement is absolutely elementary, but I have no idea of solution.

Just to clarify, is the sequence of functions infinite?

Here's an idea: take $f$ any bijective function $f:R\rightarrow \left(0,1\right) $ and $h\left( x\right) =x+1.$ Define $h_{1}=h\circ f,$ $%
h_{n}=h\circ h_{n-1},$ $h_{n}:R\rightarrow \left( n,n+1\right) .$ Clearly, all $h_{n}$'s are bijective. Define $g:R_{+}-Z$ as follows: if $x\in \left(
n,n+1\right) ,$ then $g\left( x\right) =f_{n}\left( h_{n}^{-1}\left(x\right) \right) .$ We have
\[
f_{n}\left( x\right) =g\left( h_{n}\left( x\right) \right) ,
\]
so
\[
f_{n}=g\circ \underset{n\text{ times}}{\underbrace{h\circ \ldots \circ h}}\circ f,
\]
almost the required result.


Για ποιο k η παρασταση

$ A_k\equal{}\dfrac{19^k\plus{}66^k}{k!}$

ειναι μεγιστη

θα δώσω μία βοήθεια για όσους θέλουν [hide]Προσέξτε πόσες φορές αυξάνει η παράσταση αν όπου κ βάλουμε κ+1 και παρατηρείστε ότι όσο μεγάλο και αν είναι το κ η παρασταση αυξάνει κατα =<65 φορές .[/hide]

Αυτή τη σκέψη την έκανα κι εγώ. Μετά όμως, αν και βρίσκω μέγιστη τιμή δεν μπορώ να βρω το k για την ποία ισχύει. Για να βρεθεί το k πρέπει να έχουμε <=, ενώ με τη σκέψη σου, αν πολλαπλασιάσουμε κατά μέλη βγαίνει <. Μπορείς να δώσεις όλη τη λύση;

Θα δώσω μια λύση που βρήκα:

Παρατηρούμε ότι (παίρνοντας τη διαφορά επόμενος πλην προηγούμενος όρος):
-η παράσταση $ \frac {19^k}{k!}$ αυξάνεται κατά $ \frac {19}{k \plus{} 1} \minus{} 1$ για κάθε μοναδιαία αύξηση του $ k$
-η παράσταση $ \frac {66^k}{k!}$ αυξάνεται κατά $ \frac {66}{k \plus{} 1} \minus{} 1$ για κάθε μοναδιαία αύξηση του $ k$

άρα η παράσταση $ A_k$ για κάθε μοναδιαία αύξηση του $ k$ αυξάνει κατά
$ \frac {19}{k \plus{} 1} \minus{} 1 \plus{} \frac {66}{k \plus{} 1} \minus{} 1 \equal{} \frac {85}{k \plus{} 1} \minus{} 2$.

-για $ k > 41$, η αύξηση είναι αρνητική, άρα η παράσταση $ A_k$ είναι φθίνουσα
-για $ k < 41$, η αύξηση είναι θετική, άρα η παράσταση $ A_k$ είναι αύξουσα
άρα στο $ k \equal{} 41$ η $ A_k$ αλλάζει μονοτονία από αύξουσα σε φθίνουσα (βασικά είναι 41,5 αλλά το k είναι ακέραιος)
άρα το $ k$ είναι μέγιστο, αφού δεν αλλάζει σε άλλο σημείο η μονοτονία

άρα για $ k \equal{} 41$ έχουμε το μέγιστο της $ A_k$

Διορθώστε με, αν κάνω λάθος.

Ξαναδές το γιατί βγαινει μεγιστο για κ=65

Για k=65 είναι το μέγιστο της $ \frac{66^k}{k!}$.
Όχι και της $ \frac{19^k}{k!}$
Βρες μου καλύτερα αντιπαράδειγμα για την απάντηση ή κάποιο λάθος στη λύση.

Βάλε και τη δική σου λύση.

Σόρρυ που πρέπει να πω και αυτό :( . Όχι ότι έχει μεγάλη σημασία αλλά για να μη γεμίζουμε ένα topic με σελίδες από ποστς  :mad: καλύτερα άμα κάποιος έχει ξεχάσει να πει κάτι να κάνει edit στο προηγούμενο ποστ του αντί να γράφει καινούριο γιατί μετά η σελίδα κάνει δέκα ώρες να κατέβει ολόκληρη... :D
Ευχαριστώ...

Μανιέ μπορεις να μου εξηγήσεις που βρίσκεις την αύξηση $ \dfrac{19}{k \plus{} 1} \minus{} 1$ διότι εάν το προσθέσω και αυτο και το αντίστοιχο για το 66 στο αρχικό δεν βγαίνει $ \dfrac{19^{k \plus{} 1} \plus{} 66^{k \plus{} 1}}{(k \plus{} 1)!}$ . Η αύξηση για το 19 είναι $ \dfrac{19^{k}}{k!}*\dfrac{19 \minus{} (k \plus{} 1)}{k \plus{} 1}$ . Τοσο πρέπει να προσθέσω για να βγει και το αντίστοιχο στο 66 .Ακόμα και αν θεωρήσεις οτι  πολλπλασιιάζω στην αρχικη το $ \dfrac{85}{k \plus{} 1} \minus{} 2$ δεν βγαίνει  $ \dfrac{19^{k \plus{} 1} \plus{} 66^{k \plus{} 1}}{(k \plus{} 1)!}$  . Διορθώστε με αν κάνω λάθος.

Δίκιο έχεις.
Πάντως αρκεί να βρεις το k για το οποίο το άθροισμα των μεταβολλών ισούται με 0.

Υπάρχει και άλλη λύση πιστεύω ωραιότερη αν και μοιάζουν λίγο

[quote="bilot4"]Χωρίς παρεξήγηση, αν μπορείτε μη βάζετε posts που δεν προσφέρουν τίποτα , όπως π.χ. "υπάρχει και πιο ωραία λύση" κ.τ.λ. :mad:  Θα ήταν καλύτερο αν θέλετε να βάζετε και τη λύση αυτή τουλάχιστον :)[/quote]

Pes ta re dimitraki....

Δημήτρη, ανέβασε τη λύση σου.

Πραγματικά βγαίνει 65.

Στο $ [1, 18]$ η $ A_k$ είναι αύξουσα.(Προφανώς)
Στο $ [66, \plus{}\infty)$ η $ A_k$ είναι φθίνουσα.(Προφανώς)
Άρα το ολικό μέγιστο θα ταυτίζεται με το μέγιστο στο διάστημα [19, 65].
Επειδή όμως στο διάστημα αυτό η παράσταση $ \frac {19^k}{k!}$ μειώνεται λιγότερο από όσο αυξάνει η $ \frac {66^k}{k!}$ (σχηματίζοντας πηλίκο βγαίνει αμέσως) η $ A_k$ σε αυτό το διάστημα θα είναι αύξουσα, άρα στο διάστημα αυτό θα είναι το μέγιστο στο άνω άκρο, δηλαδή στο k = 65.
Άρα και το ολικό μέγιστο βρίσκεται στο k = 65.

H λύση μου είναι ιδια με τη δική σου μόνο απο άλλο δρόμο τα ίδια ακριβως λαμβάνω υπόψην μου .


If the endpoints of one side of a square are at $ (2,\minus{}3)$ and $ (5,4)$, then how many square units are in the area of the square?

Just plug in the distance formula: $ \sqrt(x\#1-x\#2)^2 + (y\#1-y\#2)^2=\sqrt(9+49)=\sqrt58$.
Therefore, the side of the square is $ \sqrt58$, and the area's $ 58$.

We are given that the endpoints of one side of a square are $(2,-3)$ and $(5,4)$.  The distance between those points is $\sqrt{(2 - 5)^2 + (-3 - 4)^2} = \sqrt{9 + 49} = \sqrt{58}$.  So this is the length of one side of the square.  The area of a square is the square of its side, so $[\sqrt{58}]^2 = \boxed{58}$ square units.


inequalities
function
inequalities unsolved

Let $ \frac{1}{2}\le a_1, a_2, \ldots, a_n \le 5$ be given real numbers and let $ x_1, x_2, \ldots, x_n$ be real numbers satisfying $ 4x_i^2\minus{} 4a_ix_i \plus{} \left(a_i \minus{} 1\right)^2 \le 0$. Prove that \[ \sqrt{\sum_{i\equal{}1}^n\frac{x_i^2}{n}}\le\sum_{i\equal{}1}^n\frac{x_i}{n}\plus{}1\]

Vietnam Comrades
it is Vietnam 1986
please  give your  solutions

Here is a solution.

We can easily find that

$\frac{a_i-\sqrt{2a_i-1}}{2} \leq x_i \leq \frac{a_i+\sqrt{2a_i-1}}{2}$

$\frac{1}{2} \leq a_i \leq 5 \: \Longrightarrow \: \frac{a_i+\sqrt{2a_i-1}}{2} \leq 4$

On the other hand

$ 0 \leq \frac{a_i-\sqrt{2a_i-1}}{2} $

$\Longrightarrow \: 0 \leq x_i \leq 4$ for all $1 \leq i \leq n$

Now we define a function $f$ such that

$f(x_1,x_2,...,x_n)=\sqrt{\frac{{x_1}^2+{x_2}^2+...+{x_n}^2}{n}}-\frac{x_1+x_2+...+x_n}{n}-1$

We will prove that $f(x_1,x_2,...,x_n) \leq 0$ when $0 \leq x_i \leq 4$ for all $1 \leq i \leq n$

$\frac{\partial f^2}{\partial{x_i}^2}= \frac{1}{\sqrt{n}} \cdot \frac{{x_1}^2+{x_2}^2+...+{x_n}^2-{x_i}^2}{({x_1}^2+{x_2}^2+...+{x_n}^2)^{\frac{3}{2}}} \geq 0$

This means that $f$ gets its maximum values at the endpoints, namely

$f(x_1,x_2,...,x_n) \leq f(0,0,...,0,4,4,...,4)$

Let us assume that there are $a$ zeros and $b$ fours.

$ f(0,0,...,0,4,4,...,4) = \sqrt{\frac{16b}{a+b}}-\frac{4b}{a+b}-1$

$\sqrt{\frac{16b}{a+b}}-\frac{4b}{a+b}-1 \leq 0 \: \Longleftrightarrow$

$a+5b \geq \sqrt{16b(a+b)} \: \Longleftrightarrow \:  (a-3b)^2 \geq 0$

Equality holds when $a=3b$

Solution is done.


Michael walks at the rate of 5 feet per second on a long straight path. Trash pails are located every 200 feet along the path. A garbage truck travels at 10 feet per second in the same direction as Michael and stops for 30 seconds at each pail. As Michael passes a pail, he notices the truck ahead of him just leaving the next pail. how many times will Michael and the truck meet?

This particular problem has been thoroughly discussed: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1050534]Click here to view the thread[/url]  :)

Past AMC problems can be found in the "Contests" section -- it's one of the buttons on the horizontal navigation bar (Profile, Private Messages, ...)


\[ \begin{array}{l} if: p = q \Rightarrow (p;q)any/prime/number \Rightarrow {\rm For definiteness/p > q} \\ {\rm p}^{\rm 2} + 7pq = (n - q)(n + q)here/n - natural/number \Rightarrow \frac{{n - q}}{p}or\frac{{n + q}}{p} \in N \\ 1.if\frac{{n - q}}{p} \in N \Rightarrow n = pk + q/here/k > 1 - natural/number/because/ \\ n^2 = p^2 + 7pq + q^2 > (p + q)^2 \Rightarrow p + 7q = pk^2 + 2qk/k \ge 2 \Rightarrow pk^2 + 2pk \ge 8p > p + 7q \Rightarrow \emptyset \\ hence\frac{{n + q}}{p} \in N \Rightarrow n = pk - q \Rightarrow k > 1 \Rightarrow p + 7q = pk^2 - 2qk \Rightarrow p(k^2 - 1) = q(2k + 7)/if/k \ge 4 \\ k^2 - 1 \ge 2k + 7/and/p > q \Rightarrow \emptyset /hence/k = 2/or/k = 3///1)k = 2 \Rightarrow 3p = 11q \Rightarrow p = 11,q = 3 \Rightarrow \\ (3: 11)(11: 3)/2)k = 3 \Rightarrow 8p = 13q \Rightarrow \emptyset \\ \end{array}  \]

[quote="#20002 pwns"][quote="Andreas"][hide]$p^2 + 7pq + q^2 = n^2$.

$5pq = (n - p - q)(n + p + q)$.

Every factor on the right can be $5pq, 1, 5p, q, 5q, p, 5, pq$.

So solutions are $(A, A)$, $(3, 11)$, $(11, 3)$, $(-3, -11)$, $(-11, -3)$.[/hide][/quote]
So this is the first part of my solution, next we have:
$(p-q)^2+9pq=n^2 \implies (n-p+q)(n+p-q)=9pq$
$9pq=n-p+q, 1=n+p-q \implies p_1,q_1$
$3pq=n-p+q, 3=n+p-q \implies p_2,q_2$
$pq=n-p+q, 9=n+p-q \implies p_3,q_3$
$9p=n-p+q, 1q=n+p-q \implies p_4,q_4$
$3p=n-p+q, 3q=n+p-q \implies p_5,q_5$
$p=n-p+q, 9q=n+p-q \implies p_6,q_6$
$9q=n-p+q, 1p=n+p-q \implies p_7,q_7$
$3q=n-p+q, 3p=n+p-q \implies p_8,q_8$
$q=n-p+q, 9p=n+p-q \implies p_9,q_9$
So there should be up to 9 additional solutions $\{ p_i,q_i \}$[/quote]i  think your  answer  is  wrong

[quote="José"]Can you explain where/why?[/quote]

The solutions aren't guaranteed to be distinct, I think.  Try working them all out and see.

I asked Utkirstudios because maybe he can point out where the problem was

[quote="Andreas"][hide]$p^2 + 7pq + q^2 = n^2$.

$5pq = (n - p - q)(n + p + q)$.

Every factor on the right can be $5pq, 1, 5p, q, 5q, p, 5, pq$.

So solutions are $(A, A)$, $(3, 11)$, $(11, 3)$, $(-3, -11)$, $(-11, -3)$.[/hide][/quote]
Umm, can we say $-3$ is a prime?

Of course not sir
Read article about prime

I wanna ask what problem is this???

Well, it's a matter of convention. Personally I tend to answer YES, since the primes in $\mathbb{Z}$ are associated in divisibility $(p,-p)$. The convention is to take the positive representative. But in the ring of Gauss integers $\mathbb{Z}[\textrm{i}]$, where $\pm 1 \pm \textrm{i}$ are primes associated in divisibility, which one is the conventional representative?

What about p=q=2?
it´s going to be:
4+28+4=36, which is the square of 6

[quote=Night_Witch123]All the solutions are: (Can someone confirm?)
$$(p,q) \equiv (p,p),(3,11),(11,3)$$
Also I didn't understand how at #2 arrived at $(-3,-11)$ as a solution...is it a typo??[/quote]

Umm, it has been repeatedly told that yours ar ethe only solutions. #2 has also considered that primes can been negative

Official solution believe in only positive primes BTW

Bt any one tell why there is no other solution

https://youtu.be/mMnDpjcv_3Q solution by pj sir

[quote=Rushil]Find all primes $p$ and $q$ such that $p^2 + 7pq + q^2$ is a perfect square.[/quote]

$$ (p, q) = (p, p), (3,11), (11,3) $$
they are the only solutions


i compete in the state of NM, i know its not the least competitive state. i was wondering which score i'ld need to get in to the top 4. please let me know. thanks :)

yeah k if u could help me too that would be shweet, i live in Oregon......

Depending on the difficulty of the test this year it will probably be mid 30s. Im not sure of the skill level of people in New Mexico (I think NM is New Mexico...) but a mid 30 to high 30 should get you there. It also depends if CD is official or not. If it is, try to get high 30s so you don't lose your position. It should be the same of Oregon.

yeah, i think the countdown round is official in new mexico. so would that change the chances?

I dont know if countdown's official. What grade are you in?, just curious

Hey, no score discussion for state. Read the stickies. :wink:

The scoring system has always been the same as before. Individual score=sprint score+2*target score
Maximum 30 sprint, maximum 8 target. Maximum total is 46.

[quote="mathlovers1731"]do u have any 1995-1999 state practice tests?[/quote]

see the thread in the MC forum called "MATHCOUNTS Resources", something along those lines.

[quote="hongsquare"]7 on the sprint gets you in top 10 in new mexico?! wow here in california you need something like 44[/quote]

As beast as California is, a 44? Thats really high....

No, nobody said aristotlepi is in nm. I don't think a 7 sprint (and even conbined with an 8 target for a total of 23) could get you into into a top 10 in any state except for a very few number. Does anybody know the weakest state? Guam, US Virgin Islands, Okla, NM, Nevada, etc.?

I believe I heard from the Virgin Islands coach last year that they get last place almost every year, though that's an unverified statistic.  I'm pretty sure they got last last year at least.  It was odd to find an area with an even smaller population than my state :D

what about florida? what would you need to make countdown in florida? :?:

I think it's more important just to try to do as well as you can than think about scores.
Basically, I'm saying what xpmath said here:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=263486[/url]

[quote="funtwo"]what about texas?[/quote]

*cough*
r u serious....
and generally states with official cds are tougher...so maybe mid-30s will get u in? (don't really know about NM)

[quote="funtwo"]but why was the cutoff for top 10 like 42 last year[/quote]

It's Texas...

What would Washington be?

I assume it's gonna be pretty high... :(

umm last years team had 3 out of 4 as 7th or 6th graders so the competition is gonna be super tough.............this if for washington btw

[quote="funtwo"]but why was the cutoff for top 10 like 42 last year[/quote]

why do u always assume every1's talking about texas...
i mean texas is awesome but not THAT awesome


Find equivalent ressistance across A and B if all the geven resistances are diff(in magnitude).

Are you saying that they are all different?

Yup,all the resistances are diff in magnitude :lol:

[quote="shadysaysurspammed"]Yup,all the resistances are diff[/quote]

Why don't you just say different instead of diff, that's not so confusing?
Is it that hard to post in correct english?

We've done the general case but the equations are tooooo tough and invlove a lot of variables. Anyways, do you know KIrchoff's laws???

[quote="eminem"]by the way shady , i would be glad if you would ask me whether you can post it or not , it being my question but neway

hey , the concept of the wheatstone bridge would not be applied as you dont know wheter r1/r2= r3/r4 ,otherwise it would be xtremely easy

i asked my tutor , he couldnt get it too , can you please tell the solution anyway  Rushil ;)  , I am Xtremely Eager to know  .........[/quote]

Yo shady,again ur overactin,who  the heck said its ur sum?I got it in H.C.VERMA!!!!!ENOUGH??!!!HOW DARE U SAY I SHOULD ASK U BEFORE POSTIN IT,YO MOD XXXREDDEVILXXX,I GET REAL ANGRY WHEN SOMEONE UNNECESSARILY POSTS STUPID STUFF LIKE THAT AGAINST ME FOR NO REASON.I'LL START GETTIN NASTY OUT HERE.I'M SREIOUS.Sorry if i'm soundin harsh,not in a good mood thnx to eminem.

ANd Arne,whats wrong,u dont get abbreviated internet language out there in ur country :rotfl:  :rotfl: It isnt hard to post "CORRECT" english,i thought u guys could at least get that.NO offence Arne,ur a great guy,highly intelligent,not makin fun of ya,just kiddin :lol:

Now to Rushil,Yeahh kirchhaof's laws (whatever the spellin is_)
Are pretty basic,thoughnone of them are really helpful in this qustion,cmon solve it pleaaasseeeeee.

First of all, please don't post confusing, hard-to-read and unneccessary posts to argue, that's what the PMs are for...

The question of calculating resistance of a network was discussed here a few times. You can always (in a finite network) "put a voltage accross it", denote the potentials at each node, write a system of equations, solve, calculate total current and then just divide to get the resistance (I've even posted a computer program I think). You can do the same here, with resistances of the individual resistors as parameters.

You can also use a little trick, the star-triangle transformation. It takes a while to see that if you have a triangle of resistors in a scheme (see the picture), you can replace it with a star (with appropriately changed resistances, also see the picture). Apply this to your original problem...

[hide="and you get..."]
six resistors like the right-bottom picture, with respective resistances, according to the star-triangle transformation. The total resistance between the desired endpoints is...

$R = \frac{AB}{A + B + C} + \frac{\big(\frac{AC}{A + B + C} + \frac{DC}{D + E + C}\big)\big(\frac{BC}{A + B + C} + \frac{EC}{D + E + C}\big)}{\big(\frac{AC}{A + B + C} + \frac{DC}{D + E + C}\big) + \big(\frac{BC}{A + B + C} + \frac{EC}{D + E + C}\big)} + \frac{DE}{D + E + C}$

which I don't feel like simplifying, but you can do it for yourself ... I'm sure lots of things will cancel out. (And lots won't;-)
[/hide]