[hide]In order for the expression to be defined, it must be greater than or equal to 0, since no log can be negative. Thus

$\log_{2004} (\log_{2003} (\log_{2002} (\log_{2001} x))) \ge 0$
$\log_{2003} (\log_{2002} (\log_{2001} x)) \ge 1$
$\log_{2002} (\log_{2001} x) \ge 2003$
$\log_{2001} x \ge 2002^{2003}$
$x \ge 2001^{2002^{2003}}$

So the answer should be (E).[/hide]

Wroooooong!

Logs can be sooo negative, but the expression UNDER the log must be positive.

Hence

\begin{eqnarray*}\log_{2003}\log_{2002}\log_{2001}x &>& 0\\ \log_{2002}\log_{2001}x &>& 1\\ \log_{2001}x &>& 2002\\ x &>& 2001^{2002}\end{eqnarray*}

The answer is $\boxed{B}.$

[hide]

heh did u ever plug it back in and try it?

so first step we have:

$\frac{\log 2001^{2002}}{\log 2001} = 2002$

then we have:

$\frac{\log 2002}{\log 2002} = 1$

then we have:

$\frac{\log 1}{\log 2003} = 0$

and finally

$\frac{\log 0}{\log 2004} = DOMAIN ERROR$

wat does work is E, in which case after all the steps you end up with $0$

[/hide]

No, Farenhajt is right. I messed up, not sure why, but when I got my answer I wasn't sure about it. Probably cause I've done this problem before.

gotztahbeazn: Note that x > c, so when x = c it's not true.

[quote="=s"]i saw this question on amc 2004 i think[/quote]refer to source above the problem. :)

How do you know that the value has to be greater than 0?

EDIT: Nevermind  :P

Do we have a definitive answer?  I can see it both ways...

Yes, we do.
[hide="The answer is..."] $\boxed{B}$ [/hide]

What chesspro neglected was that log outside can be negative. He assumed it had to greater than or equal to 0.

$\log_{12}0$ is undefined.

$\log_{12}1$ is 0.

So what is $\log_{12}0.5$ ??  Is it defined?

BTW:  I checked the AMC Website, and the answer is indeed $\boxed{B}$.

[quote="mysmartmouth"]$\log_{12}0$ is undefined.

$\log_{12}1$ is 0.

So what is $\log_{12}0.5$ ??  Is it defined?[/quote]
Yes, it is. It's the number $x$ such that
$12^{x}=0.5$
and it's about $x=-0.279$.

Aha!  Then this makes sense.  Because when you plug in E, you get a clearly defined value, yet we want all x GREATER than E, not including E itself.

I think Farenhajt's answer is the most complete, because it clearly shows how the answer is arrived at. Good job  Farenhajt :)


function
floor function
logarithms
continued fraction
topology
real analysis
real analysis unsolved

Prove that the cardinality of $ \mathbb R$ is the same as the cardinality of the set of all injections $ f: \mathbb N\to\mathbb N$.

My favorite proof uses the fact that every element of $ \mathbb{R}$ has an essentially unique continued fraction expansion.  I won't give all the details since this is by far the most important idea.

[hide="using positional notation representation"]We can express each $ x\in (0,1)$ uniquely as $ x \equal{} \sum_{k \equal{} 1}^\infty a_k (x)2^{ \minus{} k}$ where each $ a_k(x)$ is either $ 0$ or $ 1$ and $ \liminf _{k\to \infty} a_k(x) \equal{} 0$.  Given an $ x$, define $ f$ recursively: $ f(1) \equal{} 1$ and $ f(k) \equal{} f(k \minus{} 1) \plus{} 1 \plus{} a_k(x)$ for $ k\geq 2$.  Each $ x$ then defines a unique $ f$, and each such $ f$ is evidently an injection, so $ | \mathcal{F}| \geq|(0,1)| \equal{} |\mathbb{R}|$, where $ \mathcal{F}$ denotes the set of injections on $ \mathbb{N}\to \mathbb{N}$.

We now show $ |\mathcal{F} |\leq|\mathbb{R}|$, which together with the previous result gives $ |\mathcal{F}| \equal{} |\mathbb{R}|$.  Define the function $ d: \mathbb{N}\to \mathbb{N}$, $ d(k) \equal{} \lfloor 1 \plus{} \log_2 k \rfloor$.  Given an $ f\in \mathcal{F}$, recursively define the following two sequences:
\[ p_1 \equal{} d(f(1)), \ \ p_{k \plus{} 1} \equal{} p_k \plus{} d(f(k \plus{} 1)) \plus{} 1 \\
x_1 \equal{} f(1)3^{ \minus{} p_1} \plus{} 2\cdot 3^{ \minus{} p_1 \minus{} 1}, \ \ x_{k \plus{} 1} \equal{} x_k \plus{} f(k \plus{} 1)3^{ \minus{} p_{k \plus{} 1}} \plus{} 2\cdot 3^{ \minus{} p_{k \plus{} 1} \minus{} 1}\]
Essentially we are building a real number in base $ 3$, adding on the base $ 2$ representation of $ f(k)$ to the end of $ x_k$ at each step to produce $ x_{k \plus{} 1}$, and separating succeeding values with a $ 2$.  It's easy to check that $ x_k$ converges to a real number $ x\in (0,1)$, and $ x$ is therefore a unique representation of $ f$, so it follows $ |\mathcal{F} |\leq|\mathbb{R}|$ and hence $ |\mathcal{F}| \equal{} |\mathbb{R}|$.[/hide]

Actually, the continued fraction idea naturally gives you functions from $ \mathbb{N}$ to $ \mathbb{N}$. Making those injective is a slight tweak; the simplest thing to do is to replace a value of $ k$ with a value of the $ k$th natural number not yet used.

Denote by $ \mathbb{N}_{inj}: \equal{} \{f: \mathbb{N}\to \mathbb{N}|f\mbox{ is injective}\}$ the set of such injective functions, and as usual $ \mathbb{N}^{\mathbb{N}}: \equal{} \{f: \mathbb{N}\to \mathbb{N}\}$. By definition, $ \mathbb{N}_{inj}\subset \mathbb{N}^{\mathbb{N}}$, so $ |\mathbb{N}_{inj}|\leq |\mathbb{N}^{\mathbb{N}}|$.

Now define $ \phi: \mathbb{N}^{\mathbb{N}}\to \mathbb{N}_{inj}$ by $ \phi(f) \equal{} g$, where $ g: \mathbb{N}\to \mathbb{N}$ is given by
$ g(n) \equal{} p_1^{f(1)}\cdot\cdot\cdot p_n^{f(n)}$, here $ p_n$ denotes the n'th prime number. By the fundamental theorem of arithmetic $ g$ is injective, so indeed $ \phi(f)\equal{}g$ is in the range $ \mathbb{N}_{inj}$, and also $ \phi$ is injective. Hence $ |\mathbb{N}^{\mathbb{N}}|\leq |\mathbb{N}_{inj}|$. [Another injection could be $ \psi(f)(n) \equal{} f(1) \plus{} ... \plus{} f(n)$.]

It follows that $ |\mathbb{N}_{inj}| \equal{} |\mathbb{N}^{\mathbb{N}}| \equal{} |\mathcal{P}(\mathbb{N})| \equal{} |\mathbb{R}|$.

Note that just proving uncountability is next to trivial. The function $ d\left(f,g\right)\equal{}\begin{cases}
\frac{1}{\min\left\{ n\in\mathbb{N}\mid f\left(n\right)\ne g\left(n\right)\right\} } & f\ne g\\
0 & f\equal{}g\end{cases}$ turns $ \left(\mathbb{N}^{\mathbb{N}},d\right)$ into a complete metric space with no isolated points and the subspace $ \mathbb{N}_{inj}$ is closed. The conclusion follows from the Baire category theorem.

Why would you use the Baire category theorem when diagonalization works just fine?

Another proof: Let $ I(N)$ denote the set of injective maps of $ N$ into $ N$. Let $ q_1$, $ q_2$, ... be the rationals. Map $ x \in R$ to $ \{n : q_n < x\}$. We can regard the last set as a strictly increasing sequence in $ N$,  so this gives us an injective map of $ R$ into $ I(N)$.

To see that $ I(N)$ can be injectively mapped into $ R$, just map all of $ N^N$ injectively into $ R$ via the map $ (n_1, n_2, ...) \rightarrow 1/10^{n_1} \plus{} 1/10^{n_1\plus{}n_2} \plus{} ...$

By Cantor-Bernstein we're done.


How many factors of $70^{11}$ are multiples of $70^7$?

[hide]I am assuming it is 70^11

the prime factorization is 2^11*5^11*7^11

and there must be at least 2^7*5^7*7^7

that is 7-11 inclusive, which gives 5 possibilities

so 5^3

125?[/hide]

[hide]
So $70^{11}=7^{11}5^{11}2^{11}$
$70^7=7^75^52^2$
For the factor of $70^{11}$ to be a multiple of $70^7$, it has to have at least $7$ of each prime factor.
There are 3, and each has 5 ways of working.
$5^3=125$[/hide]


A 900N crate rests on the floor.  How much work is required to move it at a constant speed 6.0 m along the floor against a friction force of 180N.

I know that work equals force times displacement and that force equals mass times acceleration.  If it's at a constant speed, that means the acceleration is 0 but I'm not sure where the friction comes into play.

if you draw a free body diagram, there are 4 forces.  you have the force due to gravity acting down, the normal force acting up, the force you are exerting on it (to the right let's say), and the frictional force acting opposite the motion.  let's call down positive in the y direction and right positive in the x direction.

based on $F = ma$,
$mg = 900 N$
$F_N = -900 N$ because there is no acceleration in the y direction
$F_f = -180 N$
$F = 180 N$ because it must be equal and opposite to the frictional force for a to be 0
$W = Fd = (180 N)(6 m) = 1080 J$


Hi please  I need help to find the range of the following function:

f(x,y)= (e^x - e^y) / (e^x + e^y)

I check the answer in the book . It is [-1,1] but I don't know how they get that.

Thank you for your help.

B

[quote="blahblahblah"]It's easy to make some guesses as to what is going on, but try dividing through by $e^x$

We get

$f(x,y)=\dfrac{1-e^{y-x}}{1+e^{ y-x}}$

Introduce the new variable $t=y-x$; $t$ can assume any value in $\mathbb{R}$.

So $f(t)=\dfrac{1-e^t}{1+e^t}$; this is monotone decreasing, and it goes to $1$ as $t\to-\infty$ and to $-1$ as $t\to\infty$.  So the range is actually $(-1,1)$.

Hope this helps.  By the way, problems of this nature should go in the Calculus Computations and Tutorials section.[/quote]

Thanks . That make more sense now


function
algebra
domain
calculus
derivative
complex analysis
complex analysis theorems

Prove or disprove:

If f is a continuous complex-valued function on a bounded domain D
such that f' exists on a dense set E contained in D then f is analytic
in D.

False. Just take $ f(z)\equal{}g(\text{Re\,}z)$ where $ g$ is any real-valued differentiable function of one real variable whose derivative vanishes on a dense set (say, the inverse function to $ x\plus{}\sum_n c_n(x\minus{}q_n)^{1/3}$ where $ q_n$ is an enumeration of rationals and $ c_n>0$ decay fast enough).


Around 3 seconds for 1-pointers, 5 seconds for 2 pointers, and maybe 8-10 seconds on 3 pointers? I haven't played for a while, so these could be off.

[quote="D3m0n Shad0w"]24 is a widely known game. Object: Try to make 24 by adding, subtracting, multiplying and dividing all 4 numbers to get 24. Use each number once.

How long does it take for you to get an average 1, 2, and 3 pointer?[/quote]

it isn't really hard once you get the hang of it. It still takes like 10 secodns for me to get 3 points though because i'm slow. I've gotten some impossible ones though, where it's impossible to make 24, such as 1,1,1,1 ()not saying that I got that)

[quote="archimedes1"]I'm new to this game.  Which 4 numbers are used?[/quote]

usually when I play I just use regular playing cards.
most people use numbers from 1-10 I think

[quote="Hamster1800"]Here's a challenge for anyone who hasn't already seen it. I guarantee that it is possible:

Make 24 with 8, 8, 3, 3.[/quote]

I have got it, $ \frac{8}{3\minus{}\frac{8}{3}}\equal{}24$. :P

like sudoku, and checkers and unlike chess and otherstuff, i used to be a beast at this. i won 2nd place at my school.

2 years ago: 1 pointers-0-3 seconds(i had some memorized). 2 pointers-3-5 seconds. 3 pointers 10 seconds

now: 1 pointsers- 2-5 seconds. 2 pointers-about 7 seconds 3 pointers-like 15 seconds

ive gotten so much worse

[quote="Hamster1800"]Here's a challenge for anyone who hasn't already seen it. I guarantee that it is possible:

Make 24 with 8, 8, 3, 3.[/quote]

i got another simpler way...

$ 24\equal{}(8: :8)*(3: :3)$

if you can use averages.

[quote="Hamster1800"]Here's a challenge for anyone who hasn't already seen it. I guarantee that it is possible:

Make 24 with 8, 8, 3, 3.[/quote]

Flip one of the threes and put it with the other 3 to make an eight.

If you can do that … how about $ 3\cdot 8\plus{}\frac{3}{\infty}$?  :wink:

thats what i was thinking. we should have a contest. give like 100 different 24 thingys. 1, 2, and 3 pointers and see who gets more right.

Prove that the follwing set of numbers cannot be "made" to 24.(Of course, according to the normal rules mentioned in the previous posts."

[size=200]{3,4,5,6}[/size]

Hmm, for some reason, I can't think of a way to prove that unless by trying at all possibilites?

[quote="luimichael"]Prove that the follwing set of numbers cannot be "made" to 24.

{3,4,5,6}[/quote]
How about I disprove it?

$ 6\cdot(3 \minus{} 4 \plus{} 5) \equal{} 24$

Another solution, if the rules can be bent: $ 6\cdot(3: : 4: : 5) \equal{} 24$

I've always played by the fact that you can never go into negatives or decimals at any point, though.... plus, ratios aren't +-*/ operations.

[quote="Temperal"]I've always played by the fact that you can never go into negatives or decimals at any point...[/quote]
Even if this rule is used (IMO, overly-strict), the expression can be re-written as $ 6\cdot(5 \plus{} 3 \minus{} 4) \equal{} 24$. :)

Yes, I saw that. Here's a pretty hard one: 8 6 3 3

Took me about 10 seconds. :)

[hide="8, 6, 3, 3"]$ 8\cdot\frac {3 \plus{} 6}{3} \equal{} 24$[/hide]


Suppose you had an array x[] with positions from 0 to 3 such that each number in the four parts were random integers from -3 to 3 inclusive.

An example of such an array would be
x[0]= -3
x[1]= -2
x[2]= -1
x[3]=  0
x[4]=  2

So we have the list -3,-2,-1,0, and 2. Then, we print out the number from left to right starting with the zeroth position.

-3 -2 -1 0 2
[b]
How would you write a program that accomplishes/satisfies the following? -->
- if a number to the left of a given number x[n] is negative, then replace x[n] with zero
- any changes to the original array do not affect the above step
- You are not allowed to copy any x[n] of the above array or create another array
[/b]
[hide="My difficulties with this problem"]Examining the example given in this post, we have -3 -2 -1 0 5.

-3 is negative so we swap -2 for zero for -2's original array position.

However, now we have lost the data for -2 and the computer has no way of knowing that -2 was originally to the left of -1 in the array. How can I get around this without creating another array or making copies of the original values of the numbers in my array?[/hide]

I'm not quite sure I understand your restrictions, but how about you just
[hide]work from right to left? And each time you find a negative, replace the number to its right by 0.[/hide]

As tripleM says
[hide]
for(i=4 to 1)
if(a[i-1]<0)
 a[i]=0;[/hide]

this should do the trick :)


inequalities
function
trigonometry
inequalities proposed

Let $ a,b,c$ be nonnegative real numbers with sum $ 1$. Prove that the following inequality holds:
$ \frac{1}{\sqrt{(a^2\plus{}ab\plus{}b^2)(a^2\plus{}ac\plus{}c^2)}}\plus{}\frac{1}{\sqrt{(b^2\plus{}bc\plus{}c^2)(b^2\plus{}ba\plus{}a^2)}}\plus{}\frac{1}{\sqrt{(c^2\plus{}ca\plus{}a^2)(c^2\plus{}cb\plus{}b^2)}} \ge 4\plus{}\frac{8}{\sqrt{3}}.$
:)

[quote="Agr_94_Math"]Indeed a nice inequality Can_Hang but I have already killed it.
Will just post the idea as I dont have much time now. I will post my complete solution later. :
[hide=" HINt"]
Use Cauchy Schwartz by writing the numerator $ 1$as $ (a \plus{} b \plus{} c)^2$ in each term and then  by AM GM $ \sqrt {(a^2 \plus{} ab \plus{} b^2)(a^2 \plus{} ac \plus{} c^2)} \le \frac {(a^2 \plus{} ab \plus{} b^2) \plus{} (a^2 \plus{} ac \plus{} c^2) } {2}$[/hide][/quote]
Can you please post your full solution? I am really curious to see it ;)

[color=purple]
Dear Can,
Here is some of my thinking.
  Denote $ T$ is $ Fermat \minus{} Torricelli$ point of triangle $ XYZ$ (which has two radii $ R,r$) such that $ TX \equal{} a,TY \equal{} b,TZ \equal{} c$.
Let $ YZ \equal{} x,XZ \equal{} y,XY \equal{} z$.
  We have $ 1 \equal{} (a \plus{} b \plus{} c)^2 \equal{} \frac {x^2 \plus{} y^2 \plus{} z^2 \plus{} 4\sqrt {3}S_{XYZ}}{2}$
Note that $ x \equal{} \sqrt {b^2 \plus{} bc \plus{} c^2},y \equal{} \sqrt{c^2 \plus{} ca \plus{} a^2},z \equal{} \sqrt {a^2 \plus{} ab \plus{} b^2}$ and $ \frac {1}{ab} \plus{} \frac {1}{bc} \plus{} \frac {1}{ca} \equal{} \frac {1}{2Rr}$ then your inequality is equivalent to :
\[ \sqrt {3}.\frac {x^2 \plus{} y^2 \plus{} z^2 \plus{} 4\sqrt {3}S_{XYZ}}{2}\ge16(\sqrt {3} \plus{} 2)Rr
\]
I think the last problem can solve by $ Rrs$ and function method.
 Sorry for my spam.
Evarist.
[/color]

[quote="evarist"][color=purple]
Dear Can,
Here is some of my thinking.
  Denote $ T$ is $ Fermat \minus{} Torricelli$ point of triangle $ XYZ$ (which has two radii $ R,r$) such that $ TX \equal{} a,TY \equal{} b,TZ \equal{} c$.
Let $ YZ \equal{} x,XZ \equal{} y,XY \equal{} z$.
  We have $ 1 \equal{} (a \plus{} b \plus{} c)^2 \equal{} \frac {x^2 \plus{} y^2 \plus{} z^2 \plus{} 4\sqrt {3}S_{XYZ}}{2}$
Note that $ x \equal{} \sqrt {b^2 \plus{} bc \plus{} c^2},y \equal{} \sqrt {c^2 \plus{} ca \plus{} a^2},z \equal{} \sqrt {a^2 \plus{} ab \plus{} b^2}$ and $ \frac {1}{ab} \plus{} \frac {1}{bc} \plus{} \frac {1}{ca} \equal{} \frac {1}{2Rr}$ then your inequality is equivalent to :
\[ \sqrt {3}.\frac {x^2 \plus{} y^2 \plus{} z^2 \plus{} 4\sqrt {3}S_{XYZ}}{2}\ge16(\sqrt {3} \plus{} 2)Rr
\]
I think the last problem can solve by $ Rrs$ and function method.
 Sorry for my spam.
Evarist.
[/color][/quote]
I think you forgot the condition for exist Fermat point of $ XYZ$, the condition is $ \max (X,Y,Z) < \equal{} 120^0$, and with this condition your inequality with $ R,r,p$ will be true, but of course the inequality with $ R,r,p$ with condition of triangle is not easy. Also, the identity$ \frac {1}{ab} \plus{} \frac {1}{bc} \plus{} \frac {1}{ca} \equal{} \frac {1}{2Rr}$ is not true, it must be $ \frac {1}{yz} \plus{} \frac {1}{zx} \plus{} \frac {1}{zx} \equal{} \frac {1}{2Rr}$ with your notation. Anyway, thanks for your interests :)
[quote="Agr_94_Math"]Indeed a nice inequality Can_Hang but I have already killed it.
Will just post the idea as I dont have much time now. I will post my complete solution later. :
[hide=" HINt"]
Use Cauchy Schwartz by writing the numerator $ 1$as $ (a \plus{} b \plus{} c)^2$ in each term and then  by AM GM $ \sqrt {(a^2 \plus{} ab \plus{} b^2)(a^2 \plus{} ac \plus{} c^2)} \le \frac {(a^2 \plus{} ab \plus{} b^2) \plus{} (a^2 \plus{} ac \plus{} c^2) } {2}$[/hide][/quote]
Could you please post your full solution. Dont just spam. ;)

[color=purple] Dear Can,
Thank for your correction.I use computer to solve $ Rrs$ inequality.Have you got any different way ?
Can you post your solution for this problem ? I like this problem but I cannot solve it  :)
[/color]


We say that a natural number $n$ is [i]charrua[/i] if it satisfy
simultaneously the next conditions:
- Every digit of $n$ is greater than 1.
- Every time that four digits of $n$ are multiplied, it is obtained a
divisor of $n$

Is it true that for every natural number $k>3$ there exists a [i]charrua[/i] number with [b]exactly[/b]  $k$ digits.

(original: http://www.mathlinks.ro/Forum/viewtopic.php?t=83737 )

there is one easy way to solve (i think that the proposer of the problem for the exam didn't see this), if all digits of the number are 9, we can choose n suficiently big such that $9^{4}|10^{n}-1$.

the oficial solution is more nice, but i can't remember!  :(

ps: this problem is from ibero 2001, not 2004  :)

EDIT: sorry Megus, now i see the 'exactly k digits'. nice problem, i will try!


The Round 4 problems are due March 13.  Assuming they give us about five weeks to do the problems (which has been normal for this year, it seems), I'd guess sometime during the first full week of February.


geometry
rectangle
symmetry
AMC
AIME
ratio
geometric transformation

Rectangle $ABCD$ is given with $AB=63$ and $BC=448.$ Points $E$ and $F$ lie on $AD$ and $BC$ respectively, such that $AE=CF=84.$ The inscribed circle of triangle $BEF$ is tangent to $EF$ at point $P,$ and the inscribed circle of triangle $DEF$ is tangent to $EF$ at point $Q.$ Find $PQ.$

[quote="sapphyre571"][hide="I can not believe I missed this"]
By Pythagorean Theorem, EB=FD=105. Drop in the altitude EH from E to BC and we have a rectangle ABHE, so BH=AE=84. From this, HF=BC-BH-CF=448-84-84=280. By Pythagorean again, EF=287. Let the incircle of BEF be tangent to BE at U, to EF at P, and to BF at V. Let EP=x. By two-tangent theorem, UE=EP=x, so BU=105-x. This means BV=105-x, so VF=364-(105-x)=259+x. Knowing that EP=FQ by symmetry and EP=FQ=x, PQ=259+x-x=$\boxed{259}$
[/hide]
But leave it to me to make a stupid mistake...[/quote]

ya, that's the way i did it. it was one of the few i got right!

[quote="Ignite168"][hide]Draw everything out in millimeters, note that everything can be scaled down by 7, that is your check (AIME writers can always be devious though). Then, you will realize you have to subtract 2*14 from 287 to easily get $\boxed{259}$.

Diagrams coming shortly.[/hide][/quote]

hahahaha, nice!  :)

[quote="worthawholebean"]Ignite seriously drew an insanely accurate diagram like that. It spanned two pages. I was impressed. He constructed all the angle bisectors and everything. It was off by two millimeters and he didn't check, though.[/quote]

It was NOT off by 2 millimeters. You had to subtract double the length of one segment and I measured it wrong, the diagram was perfectly accurate. I said it was 15 millimeters when it was 14 (it should've been obvious as 14 is a multiple of 7 like everything else but 15 wasn't) My stupid mistake, not the diagrams.

Man I'm glad I convinced a friend to put a compass and protractor in between us so that we could both use it when we needed to.

[quote="Ignite168"][quote="mwpl11"][quote="Ignite168"][hide]Draw everything out in millimeters, note that everything can be scaled down by 7, that is your check (AIME writers can always be devious though). Then, you will realize you have to subtract 2*14 from 287 to easily get $\boxed{259}$.

Diagrams coming shortly.[/hide][/quote]

whoa, i tried using a diagram to check my work. it wasn't pretty... i got the answer right, but got 640 from my diagram...[/quote]

How the heck did you get a length that was over twice the length of the diagonal?[/quote]

oh, whoops. wrong problem! just don't listen to me, i'm like going insane. what i said was for #3. i didn't try to draw an accurate diagram for this one...

[quote="samath"]I thought that this problem was really easy, everyone else that I've talked to hasn't seemed to agree though.[/quote]

I think this is one of those problems that required that little "click" to go off in your head. When I was first doing this problem, I spent about 25 minutes finding the lengths of every which segment I could find and tried setting up some sort of weird proportion to find PQ, but once I skipped it and came back to it a while later, I realized within one minute that all I needed was that diagonal minus twice the tangent length.
The solution may not have the typical complexity of a number 9, but I do think the problem itself requires enough thinking and analysis for a number 9.

lol.. If you look closely, you can prove $PQ=BF-BE$, then the answer comes out easily hehe!...

$BF=BC-CF=364$
$BE=\sqrt{BA^{2}+AE^{2}}=105$
$PQ=364-105=259$

Yeah, that's the solution I saw right after the test.  Too bad I spent about 30 minutes trying to figure out why my answer wasn't an integer during the test.  Then I realized that I had drawn the diagram wrong (I thought $AE=BF$ instead of $AE=CF$).  When I saw what the problem actually was, I just skipped it, figuring that it would be harder.  \begin{sarcasm} Luckily, \end{sarcasm} it was way easier.  :oops:

yeah i wasn't intended to do that, but when i was assigning the measurements, I realized the answer is just equal to BC-CF.. saved me a lot of algebra works

I thought it was a good problem - lots of information, but you have to apply it exactly right.

I didn't even see the scaling by 7 thing - that could have saved me some nasty computation.  Bleah.

But what I found weird was that the two hardest geometry problems on the test both required tangents to the incircle.  That is an advanced concept, but very useful.

Note: Tangents to the incircle came in useful in finding cotangents of the half-angles to determing the ratio of similitude of a homothety about the incenter from the vertices $A,B,C$ to the centers of $\omega_{A},\omega_{B},\omega_{C}$ in problem 15.

scaling is a smart way to reduce many messy numbers. I liked #15 too, I found that problem similar to problem 8(i think) in AIME I, which is about two circles are drawn outside of a triangle.

I spent wayyyy too long on this problem because I thought P and Q lay on AD and BC....

That couldn't come out to an integer, no matter how many times I tried it.  Then I re-read the problem and was like.... wow....

Dude #3 was pretty easy... all you have to do is, take the bottom and top of the twisted dreidel thing connect them, and notice that it goes through the center of the square because of symmetry. Then use law of cosines, and BAM, you're done.

someone had an even cooler solution to number 3... but yeah go look at that topic to see it ;)

[quote=coolkid1213][quote=Allan Z]lol.. If you look closely, you can prove $PQ=BF-BE$[/quote]

Had a question about this problem, can someone show how to prove this? Thanks a lot![/quote]

[hide=Proof]
$BF$ can be expressed as $b+c$. $BE$ can be expressed as $a+b$. $EF$ can be expressed as $a+c$.

Notice that we want to solve for $c+a-2a=c-a$ (or $a-+c-2c=a-c$, depending on your diagram)

$BE- BF=(a+b)-(b+c)=a-c$

Sidenote: I didn't notice this  :blush:, so I solved for $a$ and $c$.

This problem is quite easy for its placement... just find the lengths of two tangents (they are each $14$) and subtract from $EF = 287$.


geometry
perpendicular bisector
geometry unsolved

My solution maybe wrong,so somebody check this.
We are defining the length $s$ as $s$ times some standard value.
For instance, $s$ centimetres would be defined as $s$ times $1\text{ cm}$ .
So,
If we draw a line segment $MN$ of length $s\text{ units}$,we produce $MN$ to $X$ such that $NX$ is $1\text{ unit}$
Draw a perpendicular bisector of $NX$,take any point on the perpendicular bisector, say $O$ and with $O$ as centre,$OX$ as radius,construct a circle which passes through $X,N$.
Join $MO$, Draw a line which is perpendicular to $OM$ which meets the circle at point $K,K'$.
Let the length $KA=K'A$ be denoted by $t$.
Now,
Notice two such triangles are possible,
Either $AB>AC \text{ or } AC>AB$
WLOG,$AB>AC$, now, as we have the length $t$, we can draw a line of length $\frac{t}{2}$, this is one of the sides of the triangle(other than the hypotenuse this one is $AC$), and the other one is obtained by trisecting a line of four times the length of $\frac{t}{2}$.
Thus we have found out the length of the two sides of the triangle(other that the hypotenuse) and we can construct the triangle.
Please point out any mistakes.


(i) Show that $ \underline{lim}\ x_{n}\leq \overline{lim}\ x_{n}$ and that $ \underline{lim}\ x_{n}=\overline{lim}\ x_{n}= l$ if and only if $ l=\overline{lim}\ x_{n}$.

(ii) Prove that $ \overline{lim}\ x_{n}+\underline{lim}\ y_{n}\leq \overline{lim}\ /(x_{n}+y_{n})\leq \overline{lim}\ x_{n}+\overline{lim}\ y_{n}$ provided that the right and left sides are not of the form $ \infty-\infty$.

Seems like a typical homework assignment to me. In principle, you may ask questions about homeworks on this forum but you should try your best first and show exactly what you achieved and where you got stuck. We are not here to do routine exercises instead of you ;).

[quote="queue_zniv16"](i) Show that $ \underline{lim}\ x_{n}\leq \overline{lim}\ x_{n}$ and that $ \underline{lim}\ x_{n}=\overline{lim}\ x_{n}= l$ if and only if $ l=\overline{lim}\ x_{n}$.
[/quote]

As fedja said this is a "trivial" homework assigment testing your knowledge of defintions.

I'll just do this one cuz I hate doing people's homework to show you the main idea.

[b]Theorem:[/b] If $ (s_{n})$ is a sequence then $ \liminf s_{n}\leq \limsup s_{n}$.

[b]Proof:[/b] If $ (s_{n})$ is not bounded above then $ \limsup s_{n}=+\infty$ and the proof is complete. If $ (s_{n})$ is not bounded below then $ \liminf s_{n}=-\infty$ and the proof is complete. So it is safe to say $ (s_{n})$ is bounded, in which case $ \limsup$ and $ \liminf$ are [i]real[/i] numbers. By definition $ \limsup s_{n}$ is the limit of the sequence $ \sup \{s_{n}, s_{n+1},...\}= \sup \{s_{i}|i\geq n\}$ and $ \liminf s_{n}$ is the limit of the sequence $ \inf \{s_{n}, s_{n+1},...\}= \sup \{s_{i}|i\geq n\}$.
Confirm that,
$ \sup \{s_{1},s_{2},...\}\geq \sup \{s_{2},s_{3},...\}\geq \sup \{s_{3},s_{4},...\}\geq ...$
And,
$ \inf \{s_{1},s_{2},...,\}\leq \inf \{s_{2},s_{3},...\}\leq \inf \{s_{3},s_{4},...\}\leq ...$

So the superior sequence is a non-increasing sequence with a lower bound $ \inf \{s_{1},s_{2},...,\}$.

So the inferior sequence is a non-decreasing sequence with an upper bound $ \sup \{s_{1},s_{2}.,,,\}$

Hence the limits exists, but since $ \inf \{s_{i},s_{i+1},...\}\leq \sup \{s_{i},s_{i+1},...\}$ we have $ \liminf s_{n}\leq \limsup s_{n}$. Q.E.D.


Euler
function
real analysis
real analysis unsolved

I know $ \sum_{k \equal{} 1}^{\infty}\frac {3^{k} \plus{} k}{k! \plus{} 2} \equal{} 14.82782755....$

Any tips on how can this can be derived?.  I have noticed there are many on this site that are very good at these. Perhaps I can learn

something.  I ran it through Maple to find what it converges to, but I do not know how to do it 'by hand'.

I have tried several typical methods, but they have led me nowhere.

I broke it up into $ \sum_{k \equal{} 1}^{\infty}\frac {3^{k}}{k! \plus{} 2} \plus{} \sum_{k \equal{} 1}^{\infty}\frac {k}{k! \plus{} 2}$ and tried to proceed from

there by listing the terms and seeing of there was a nice pattern. I did not see it, but I am sure it's there.

Perhaps there is an identity I am unaware of that I could use.

$ \sum_{k\equal{}1}^{\infty}\frac{3^{k}}{k!}\equal{}e^{3}\minus{}1$

I don't immediately see a way to produce a closed-form answer, not with that denominator.

$ \sum_{k\equal{}1}^{\infty}\frac{3^k}{k!}\equal{}e^3\minus{}1$ and $ \sum_{k\equal{}1}^{\infty}\frac{k}{k!}\equal{}e.$ But changing the denominator to $ k!\plus{}2$ doesn't help.

For what it's worth: this converges quite quickly, and I agree with your numerical result.

There's really no reason to suspect that that denominator can be dealt with.  It's not hypergeometric, for one thing.

Thank you for your time. I thought maybe there was some clever method I often see on this site.

As an aside what was the method people like Euler used to compute limits to 14 decimal digits and what not? Even for quickly converging series they weren't adding term by term in base 10 by hand, were they?

By hand, or abacus-aided, or the like. Before calculators became common, people needed and developed serious calculation skills.
Also, addition is easy. It's multiplication and division that takes more time.

To get the answer to 8 places, about 20 terms will do.

To improve the rate of convergence, we can start with Kent Merryfield's approximation $ e^3 \plus{} e \minus{} 1$. We get
\[ e^3 \plus{} e \minus{} 1 \minus{} 2\sum_{k \equal{} 1}^\infty \frac {3^k \plus{} k}{k!(k! \plus{} 2)} \approx e^3 \plus{} e \minus{} 1 \minus{} 2\sum_{k \equal{} 1}^{10} \frac {3^k \plus{} k}{k!(k! \plus{} 2)} \equal{} 14.827827547...\]

I did that myself, but took another look at it and decided that

(1) $ e^3\plus{}e\minus{}1$ isn't really closed enough to the final value to deserve the name "approximation", and

(2) The series converges so rapidly anyway that there isn't all that much motivation for accelerating the convergence. You're trading two or three terms on one end for two or three terms on the other.

So after I did what Yury did, I chose not to post it that way.

[quote="Kent Merryfield"]
(2) The series converges so rapidly anyway that there isn't all that much motivation for accelerating the convergence. You're trading two or three terms on one end for two or three terms on the other.[/quote]
I agree. What I propose is another method of computing the sum by hand. People used to have tables of special functions - so it was easy for them to find the value of $ e^3 \plus{} e \minus{} 1$.


Putnam Harvard college MIT Princeton Stanford

How do colleges form the teams for Putnam and ACM ICPC?

They pick from among their strongest competitors.  At schools like Harvard, MIT, Princeton,... I'm sure there is quite a process to determine it.  At schools that are less involved however, I would imagine it is much easier to make the team.

There's not much of a "process" at Princeton. The team is picked based on who did well the previous year and occasionally underclassmen with Olympiad experience from high school. The whole notion of picking a team of three from an entire University is absurd to me though. It renders the team rankings basically meaningless. They should just count the highest three from each school.

In MIT,they just pick the competitors who did the best in the previous year.

At Stanford, the Putnam team is more or less the best scorers of the previous year (although I hear on occasion it is not). The ACM team is chosen based on the results of a local contest held at the beginning of the year.

At Hopkins, we basically just pick who did the best the previous years.

At Waterloo the Putnam team is chosen by the prof who organizes the contest. To the best of my knowledge even the team members don't know who they are until after the contest (I think their strategy in this respect is to make sure no one has unnecessary pressure). As far the ACM goes, I think our teams are chosen based largely on past results and reputation, I know the only 1st year on one of our ACM teams this year was an IOI silver medallist.


quadratics
conics
ellipse
hyperbola
rotation
geometry
geometric transformation

I read something that said that a quadratic relation is in the form $Ax^2+Bxy+Cy^2...$ etc. But then it goes on about circles, ellipses, and hyperbolas, and none of them include the xy term...what does the xy term have to do with the graphs?

It rotates the graphs, and makes them much less fun to play with.

It gives it a angle of rotation of x in this equation:
$\tan{x}=\frac{B}{A-C}$

Then you have to use a transformation matrix on each x and y term to yield an equation which eliminates the xy term.  All in all, its a very ugly process.  It can usually be found ONCE in every precalculus/analytic geometry textbook.  People definitely don't like doing it, its a long process generally.


floor function
ceiling function
induction
algebra
polynomial
binomial theorem
number theory unsolved

let $ n \equal{} \lfloor 4^m \minus{} (2 \plus{} \sqrt {2} )^m \rfloor$ when $ m \nmid 2, m \nmid 3$

Prove that 112|n.

Are you meaning, that $ m\equal{}6k\plus{}1$, or $ m\equal{}6k\minus{}1$? Because for $ m\equal{}4$: $ n\equal{}120$.

May be, it will help you:
$ n \equal{} 4^m \minus{} (2 \plus{} \sqrt {2})^{m} \minus{} (2 \minus{} \sqrt {2})^{m} \equal{} 4^m \minus{} 2(2^m \plus{} C^2_m2^{m \minus{} 1} \plus{} C^4_m2^{m \minus{} 2} \plus{} ... \plus{} C^{m \minus{} 1}_m2^{m \minus{} \frac {m \minus{} 1}{2}}) \equal{} 2^{\frac {m \plus{} 3}{2}}(2^{\frac {3m \minus{} 3}{2}} \minus{} 2^{\frac {m \minus{} 1}{2}} \minus{} C^2_m2^{\frac {m \minus{} 3}{2}} \minus{} ... \minus{} C^{m \minus{} 1}_m)$
For $ m \equal{} 1$: $ n \equal{} 0$, so $ 112|n$. For $ m\geq 5$: $ 2^\frac {m \plus{} 3}{2}|n$, so $ 16|n$. But how we can prove that $ 7|n$?

[hide="Solution"]I am assuming that Kirill's interpretation of the problem is taken, i.e., $ m \equiv \pm 1 \mbox{ mod } 6$.

Define $ a_i \equal{} \lfloor 4^i \minus{} (2 \plus{} \sqrt{2})^i \rfloor$. Clearly, $ a_i \equal{} 4^i \minus{} \lceil (2 \plus{} \sqrt{2})^i \rceil$. Let $ b_i \equal{} (2 \plus{} \sqrt{2})^i \plus{} (2 \minus{} \sqrt{2})^i$. Observe now that by the binomial theorem, $ b_i$ is always an integer.

I claim that $ \lceil (2 \plus{} \sqrt{2})^i \rceil \equal{} b_i$. Since $ 0 < 2 \minus{} \sqrt{2} < 1$, we have $ 0 < (2 \minus{} \sqrt{2})^i < 1$. Thus, $ \lfloor (2 \plus{} \sqrt{2})^i \rfloor < (2 \plus{} \sqrt{2})^i \plus{}(2 \minus{} \sqrt{2})^i < (2 \plus{} \sqrt{2})^i\plus{} 1$. Since $ (2 \plus{} \sqrt{2})^i \plus{}(2 \minus{} \sqrt{2})^i$ is an integer, and there is exactly one integer in the range $ (\lfloor (2 \plus{} \sqrt{2})^i \rfloor, (2 \plus{} \sqrt{2})^i\plus{} 1)$, it follows that $ b_i \equal{} \lceil (2 \plus{} \sqrt{2})^i \rceil$.

We may now readily prove by induction that $ b_0 \equal{} 2$, $ b_1 \equal{} 4$, and $ b_{i \plus{} 1} \equal{} 4b_i \minus{} 2b_i$ (this is because $ 2 \plus{} \sqrt{2}$ and $ 2 \minus{} \sqrt{2}$ are the roots of $ x^2 \minus{} 4x \plus{} 2$, the characteristic polynomial of this recursion.)

So now, we consider $ a_i \equal{} 4^i \minus{} b_i$ mods 16 and 7; if we can show that whenever $ i \equiv \pm 1 \mbox{ mod } 6$, $ a_i \equiv 0 \mbox{ mod } 16$ and $ a_i \equiv 0 \mbox{ mod } 7$, we will be done.

Mod 16 is easy: if $ i \equiv \pm 1 \mbox{ mod } 6$, then $ i > 2$, and thus, $ 4^i \equiv 0 \mbox{ mod 16}$. Thus, $ a_i \equiv \minus{} b_i \mbox{ mod 16}$ so it suffices to show that $ i \equiv \pm 1 \mbox{ mod } 6$ implies $ b_i \mbox{ mod 16}$. To prove this, we merely consider this sequence mod 16. The first few terms are $ 2, 4, 12, 8, 0, 0, 0, \ldots$. Since two terms uniquely determine the rest of the sequence, it follows that $ b_i \equiv 0 \mbox{ mod 16}$ whenever $ i \geq 4$.

For mod 7, we observe that the order of 4 mod 7 is 3. Thus, when $ i \equiv 5 \mbox{ mod 6}$, $ 4^i \equiv 2 \mbox{ mod 7}$ and when $ i \equiv 1 \mbox{ mod 6}$, $ 4^i \equiv 4 \mbox{ mod 7}$.

We now consider $ b_i$ mod 7. We have that the first terms are $ 2, 4, 5, 5, 3, 2, 2, 4$, we see that this sequence repeats with a period of 6, with $ b_{5} \equiv b_{6k \plus{} 5} \equiv 2 \mbox{ mod 7}$, i.e., $ 4^i \minus{} b_i \equiv 0 \mbox{ mod 7}$ whenever $ i \equiv \minus{}1 \mbox{ mod 6}$, and $ b_1 \equiv b_{6k \plus{} 1} \equiv 4 \mbox{ mod 7}$, i.e., $ 4^i \minus{} b_i \equiv 0 \mbox{ mod 7}$ whenever $ i \equiv 1 \mbox{ mod 6}$.

Thus, $ 112 | 4^i \minus{} b_i$, i.e., $ 112 | 4^i \minus{} \lceil (2 \plus{} \sqrt{2})^i \rceil$, as desired. [/hide]


modular arithmetic
number theory unsolved
number theory

p is a given prime. Please prove that if (p^2p+1)/(p^2+1)is divisible by k,then k-1 is divisible by 4p.

Let $ q$ is prime $ q|\frac{p^{2p}\plus{}1}{p^{2}\plus{}1}$

$ \implies$  $ q|p^{4p}\minus{}1$,$ q|p^{q\minus{}1}\minus{}1,(p,q)\equal{}1$

$ \implies$  $ p^{(4p,q\minus{}1)}\equiv 1\pmod{q}$

$ q>2$ Because $ val_{2}(p^{2p}\plus{}1)\equal{}val_{2}(p^{2}\plus{}1)$

$ \implies$  $ q\equiv \pm 1 \pmod{4}$

if $ q\equiv \minus{}1\pmod{4}$  $ \implies$  $ x^{2}\equiv \minus{}1\pmod{q}$ Contradiction!

$ \implies$  $ q\equiv 1\pmod{4}$

if $ (q,p\minus{}1)\equal{}1$  $ \implies$  $ q|p^{4}\minus{}1$

$ (p^{2}\minus{}1)(p^{2}\plus{}1)\equiv 0\pmod{q}$  $ \implies$  $ q|p^{2}\plus{}1$

Because if $ q|p^{2}\minus{}1$  $ \implies$  $ p^{2p}\plus{}1\equiv 2\pmod{q}$  Contradiction!

$ \implies$  $ val_{q}(p^{2}\plus{}1)\equal{}val_{q}(p^{2p}\plus{}1)$ Contradiction!

$ \implies$  $ p|q\minus{}1,4|q\minus{}1$  $ \implies$  $ 4p|q\minus{}1$ for $ p>2$

if $ p\equal{}2$  $ \implies$  $ \frac{p^{2p}\plus{}1}{p^{2}\plus{}1}$ isn't integer QED


inequalities
number theory
relatively prime

Do this without a calculator or brute-force program:

There exists exactly one Pythagorean triplet for which $ a \plus{} b \plus{} c \equal{} 1000$.
Find the product $ abc$.

[hide]Let $ a \equal{} x^2 \minus{} y^2, b \equal{} 2xy, c \equal{} x^2 \plus{} y^2$ for positive integers $ x\ge y$.

From $ a \plus{} b \plus{} c \equal{} 1000$ we deduce that $ 2x^2 \plus{} 2xy \equal{} 1000\implies x(x \plus{} y) \equal{} 500$.

Thus $ (x,y) \equal{} \left(t,\frac {500}{t} \minus{} t\right)$ for any $ t\mid 500$. But $ x\ge y\ge 0$ gives $ t\ge \frac {500}{t} \minus{} t\ge 0$. From the left inequality we get $ 2t^2\ge 500\implies t\ge 5\sqrt {10} > 15$. And from the right one we have $ t^2\le 500\implies t\le 10\sqrt 5 < 25$. Thus $ 15 < t < 25$. Hence the only value for $ t$ left (remember $ t\mid 500$ ) is $ t \equal{} 20$, which gives the pair $ (x,y) \equal{} (20,5)$ and so $ (a,b,c) \equal{} (375,200,425)$. Now to find $ abc$ is just a matter of time. :wink: [/hide]

EDIT: Sorry, my bad. I should've thought of this before. Anyway look at the posts of dtl42 and MellowMelon.

nayel, sorry if this is very obvious, but can all Pythagorean Triples be generated from that formula?

All primitive triples can be generated using that form, as well as a large amount of non-primitive, but not all non primitive triples.

The above poster is correct, meaning that solution gets an answer but doesn't prove its uniqueness.

[hide="If that is wanted"]Let $ a \equal{} x^2 \minus{} y^2, b \equal{} 2xy, c \equal{} x^2 \plus{} y^2$ for relatively prime positive integers $ x,y$ with $ x > y$ and one of $ x,y$ even. (These are the necessary and sufficient conditions for $ a,b,c$ to be a primitive pythagorean triple)

We require $ x(x\plus{}y) \mid 500$, then multiplying the triple by a constant can satisfy the condition. Note that $ (x,x\plus{}y) \equal{} 1$. Therefore one of $ x,x\plus{}y$ is a power of $ 2$ and the other is a power of $ 5$, since each divides $ 500$. Now we note that $ x$ cannot be a power of $ 5$. If it were, then $ x\plus{}y$ is a power of $ 2$. Since $ x\plus{}y > 1$, $ x\plus{}y$ is even, and since $ x$ is odd, $ y$ must also be odd, contradicting the assumption that one of $ x,y$ is even. Thus $ x$ is a power of $ 2$. Then since $ x \mid 500$, $ x$ can only be $ 1,2,4$. It's easy to see that $ x \equal{} 1,2$ fails, since $ x > y$ leaves few choices. So $ x \equal{} 4$. The only $ y < x$ that makes $ x\plus{}y$ a power of $ 5$ is $ y \equal{} 1$. So $ a \equal{} 15, b \equal{} 8, c \equal{} 17$ with this solution. Multiplying each by $ 25$ gives nayel's answer, and it is unique.[/hide]


Each of the numbers x1,x2,x3,...,xn  equals 1 or -1, and  x1x2x3x4 + x2x3x4x5 + x3x4x5x6 + ... +xn-3xn-2xn-1xn + xn-2xn-1xnx1 + xn-1xnx1x2 + xnx1x2x3 =0   Prove that n  is divisible by 4.

[hide="solution"]
I think this is well known

We have every element as $ \minus{} 1$ or $ 1$, and their sum, $ S \equal{} 0$. This implies $ S\equiv 0 \mod 4$

Now suppose we take one element that is $ \minus{} 1$ and change it to $ 1$. Since this term appears in $ 4$ streams of numbers $ S$ remains constant $ \mod 4$

Now if we change all elements to $ 1$ then $ S \equal{} n$, but $ S\equiv 0 \mod 4 \Longrightarrow n\equiv 0 \mod 4\qquad Q.E.D$
[/hide]


find all of (x,y,n) that  (x+y)(1+xy)=2^n


Rep123max wrote:How would you do #5: Point P is 9 units from the center of a circle of radius 15. How many different chords of the circle contain P and have integer lengths?

Wouldnt all chords that satisfy that have length 24, how could you figure out how many chords there are?

Weird, I just did this problem with my students today...

[hide]

So you've already seen that you can get 24 when the chord is perpendicular to the line connecting P to the center.  That's one way that works, but you can also get chords of different lengths:  if you take the diameter through P, that's also a chord, and it has length 30.  But you can also get stuff with lengths in between 24 and 30 if you tilt the chord at different angles -- this should be enough for you to figure out the problem.

[/hide]

You draw 2 concentric circles one with radius 15 and other one with radius 9. there is a triangle with  length 9 and 15 and x. The length of the chord =y. (y-x)x=6(18+6)=144 and x is between 6 and 24, and x is a factor of 144, the possible values are 8,9,12,16,18, so 5*2+1=11

When two chords intersect inside a circle, then the product of the segments of one chord equals the product of the segments of other chord. So whenever a chord going through the circle with radius 9, you could always draw a diameter of the big circle with part of it length 6. So x(y-x)=6(24)=144. I made a mistake because x doesn't have to be an integer, but the equation will be x^2-yx+144=0, where y^2-4(1)(144) :ge: 0, so minimum value of y is 24, the maximum is the length of diameter 30. My idea is really not that good because it makes the problem too complicated. All you have to do is think that smallest length of the chord is when it's tangent to the smaller circle, that's when the length of the chord is 24.

Yep.  I really like this problem because there's all sorts of fancy chord stuff you can play with on this, but the solution is really simple and intuitive.

[hide]Draw the radius through P and the center. Through P, draw the chord that is perpendicular to that radius. This chord has length 24 if you use the Pythagorean Theory. This is the shortest chord. The longest chord would be the diameter of length 30. There are only 1 of each of these chords. But there are two chords for lengths 25, 26, 27, 28, and 29. So this gives 12 chords total.[/hide]


Find all $ f: \mathbb{R}\to\mathbb{R}$ such that
\[ f(6x(1\minus{}f(x)))\plus{}f(6x^2)\equal{}f(3xf(y))\plus{}(2\minus{}y)f(3x)\] for all $ x,y\in\mathbb{R}$.

[quote="jedaihan"]Find all $ f: \mathbb{R}\to\mathbb{R}$ such that
$ f(6x(1 \minus{} f(x))) \plus{} f(6x^2) \equal{} f(3xf(y)) \plus{} (2 \minus{} y)f(3x)$ for all $ x,y\in\mathbb{R}$

own[/quote]

Let $ P(x,y)$ be the assertion $ f(6x(1\minus{}f(x)))\plus{}f(6x^2)\equal{}f(3xf(y))\plus{}(2\minus{}y)f(3x)$

$ f(x)\equal{}0$ $ \forall x$ is a solution. We'll from now consider $ \exists c$ such that $ f(c)\neq 0$

Consider then $ f(a)\equal{}f(b)$. Comparing $ P(\frac c3,a)$ and $ P(\frac c3,b)$, we get $ a\equal{}b$ and $ f(x)$ is injective.
$ P(0,0)$ $ \implies$ $ f(0)\equal{}0$

$ P(\frac 13,x)$ $ \implies$ $ f(f(x))\equal{}xf(1)\plus{}f(2(1\minus{}f(\frac 13)))\plus{}f(\frac 23)\minus{}2f(1)$ $ \equal{}xf(1)\plus{}u$
Setting $ x\equal{}0$ in the above equation gives $ u\equal{}0$ and so $ f(f(x))\equal{}xf(1)$
Setting $ x\equal{}1$ in the above equation and using injectivity gives $ f(1)\equal{}1$ and so $ f(f(x))\equal{}x$ and so $ f(x)$ is a bijection.

Subtracting $ P(\frac x3,f(y))$ from $ P(\frac x3,0)$ gives $ f(xy)\equal{}f(x)f(y)$

Using this property for $ x\neq 0$, $ P(x,y)$ may be simplified as $ f(1\minus{}f(x))\plus{}f(x)\equal{}\frac 2{f(2)}$
Using $ x\equal{}0$ in the above equation gives $ f(2)\equal{}2$ and so $ f(1\minus{}f(x))\equal{}1\minus{}f(x)$

And so, since $ f(x)$ is bijective (so $ 1\minus{}f(x)$ is surjective) : $ f(x)\equal{}x$ $ \forall x$ and it is easy to verify that this indeed is a solution

So two solutions :
$ f(x)\equal{}0$ $ \forall x$
$ f(x)\equal{}x$ $ \forall x$


Round Four March 1994

8. Find all possible values for $c$ if the numbers $a$, $b$, and $c$ satisfy \[ a^2 + b^2 + c^2 = 16064,  \] where $a \le b \le c$ are positive integers.

[hide]
examining the equation in mod 4, we see that $a, b, c$ must be even, so we set $a = 2x$, $b = 2y$, $c = 2z$ and divide out the 4 to get $x^2 + y^2 + z^2 = 4016$.  repeat the same process until we got $p^2 + q^2 + r^2 = 251$.  Examine the equation mod 4 again, we see that $p, q, r$ must be odd, plug in $p = 2d+1$, $q = 2e+1$, $r = 2f+1$, subtract out the three 1's and divide out 4 we get $d^2 + d + e^2 + e + f^2 + f = 62$.  Factoring we get $d(d+1) + e(e+1) + f(f+1) = 62$.  Each of these terms are twice the triangular number, so we list out the first few terms and guess...they are:2, 6, 12, 20 30.  See that 12, 20, 30 works, which implies that $d = 3$, $e = 4$, and $f = 5$.  Substitute all the way back up we get $a = 56$, $b = 72$, and $c = 88$.  QED.
[/hide]

There's one more possibility for $d,e,f$ (keep in mind that $d\leqslant e\leqslant f$, so the numbers can be equal).

[hide]$2+30+30=62$[/hide]


calculus
derivative
calculus computations

What is the derivative of $f(x)=e^{bx}$?

I ***THINK*** that it is $be^{bx}$

However I don't believe calculus problems are allowed in this section

Yes, that is correct, and yes, this is the wrong forum for this question  :P.

[hide="Explanation of the previously-given solution"]

By the chain rule, $\frac{d}{dx}f(g(x)) = f'(g(x))*g'(x)$. Thus, in this case, let $f(x)=e^{x}$ and $g(x) = bx$. Obviously, $g'(x) = b$, and $f'(x) = e^{x}$. So, using our formula, we get that $\frac{d}{dx}(e^{bx}) = e^{bx}* b = be^{bx}$.[/hide]


Today, I couldn't go on AoPS in the morning and now. All websites worked fine, but my computer said Internet connection was lost when I typed artofproblemsolving.com in my address bar. However, I easily fixed this by changing my IP address. Now, I'm thinking, is my IP address like, banned or something from AoPS?

Even though I renewed my IP address, Alcumus still doesn't work. When I click on it, nothing happens.

Even though it's very unlikely my IP address is banned or something similar, I am still inquiring as to whether it is or if there's a way to solve the glitch, bug, etc.

I think the AoPS server went down for awhile - it was the same for me.  Haven't tried Alcumus though.

You changed your IP address?  From the way you phrased that, it sounds like you're talking about your internal IP on your local network.  That would not affect your ability or inability to access the site, since it can only see your external IP.


I still don't understand exactly what mathematical research is. I know that research in other scientific disciplines involves doing experiments and finding a relationship by using the results of the experiments. But with math, there is no experimentation (is there?). What would a typical day be like for a person doing research in some area of math? If someone could clarify this for me, that would be great.

Research in mathematics tends to involve proving formerly unproven conjectures. You might work in large groups and throw out wild ideas, and perhaps some of those ideas will be useful. You can think of it sort of like a higher-level ARML team or power round.

So essentially, it is pretty much just discussion and thinking, right?

So wat do you suggest as some good research topics?

A friend of mine recently did a research project on square permutations. (That is, permutations that are the composition of two identical permutations.) Come up with a problem you think is challenging, find some other people to work with you, and then see what you can come up wtih.

If the idea of mathematical research seems odd, think of it as investigation instead.  You might not be physically carrying out experiments (though, with computers, sometimes you are), but you are testing whether various ideas bear fruit within the realm of an interesting problem.


[b]1)[/b] Solve the equation $ \displaystyle\left[\frac{2x\plus{}1}{3}\right]\equal{}\left[\frac{3x\minus{}1}{2}\right]$

[b]2)[/b] Prove that $ 2[x]\leq[2x],$ for all $ x>0$

[b]3)[/b] Solve the equation $ [x[x]]\equal{}1$

[quote="nickolas"][b]1)[/b] Solve the equation $ \displaystyle\left[\frac {2x \plus{} 1}{3}\right] \equal{} \left[\frac {3x \minus{} 1}{2}\right]$

[b]2)[/b] Prove that $ 2[x]\leq[2x],$ for all $ x > 0$

[b]3)[/b] Solve the equation $ [x[x]] \equal{} 1$[/quote]

[hide="1)"][b]1)[/b] $ \frac {3x \minus{} 1}{2} < \frac {2x \plus{} 1}{3} \plus{} 1$ gives $ x > 2.2$
$ \frac {2x \plus{} 1}{3} < \frac {3x \minus{} 1}{2} \plus{} 1 \Rightarrow x > \minus{} \frac {1}{5}$

So $ 1.8 > \frac {2x \plus{} 1}{3} > 0.2$.

[b]Case 1:[/b] $ \displaystyle\left[\frac {2x \plus{} 1}{3}\right] \equal{} 0 \Leftrightarrow 1 > x \ge \minus{} \frac {1}{2}$
, then $ \left[\frac {3x \minus{} 1}{2}\right] \equal{} 0 \Leftrightarrow 1 > x \ge \frac {1}{3}$,
So $ 1 > x \ge \frac {1}{3}$ is solution for this case.

[b]Case 2:[/b] $ \displaystyle\left[\frac {2x \plus{} 1}{3}\right] \equal{} 1 \Leftrightarrow \frac {5}{2} > x \ge 1$
$ \displaystyle\left[\frac {2x \plus{} 1}{3}\right] \equal{} 1 \Leftrightarrow \frac {5}{3} > x \ge 1$
So $ \frac {5}{3} > x \ge 1$ is the solution for this case[/hide]

[hide="2)"][b]2)[/b] If $ \frac {1}{2} > \{x \} \ge 0$, then $ 2[x] \equal{} [2x]$
If $ 1 > \{x \} \ge \frac {1}{2}$, then $ [2x] \equal{} 2[x] \plus{} 1$.[/hide]

[hide="3)"][b]3)[/b]The question $ \Leftrightarrow 2 > x[x] \ge 1$
Clearly $ 1 > x \ge 0$ is impossible or $ [x] \equal{} 0$.

[b]Case 1:[/b] $ x$ is positive, then $ 2 > x[x] \ge [x]^2$ gives $ [x] \equal{} 1 \Rightarrow 2 > x \ge 1$ is the solution for this case.
[b]Case 2:[/b] $ 0 > x \ge \minus{} 1$ is negative, then $ \minus{} 2 < x \le \minus{} 1$ gives $ x \equal{} \minus{} 1$ as solution for this case.
[b]Case 3:[/b] $ \minus{} 1 > x$, then $ x[x] \ge x( \minus{} 2) > 2$, so no solution for this case.[/hide]


algebra
polynomial
Euler
inequalities
algebra solved

Let f(x) be a polynomial with real coefficients and degree greater or equal to 1.Show that for every real number c>o,there is a positive integer n satisfying the following condition: If the polynomial  P(x) of degree greater or equal to n with real coefficients and leading coefficient equal to 1. Then the number of integers x for which Modulus f(P(x)) \leq c is not
greater than the degree of P(x).

Hello, Euler_India!

I have long proof.

Let c>0.

If |f(y)|<c then |y|<Kc where Kc is some finite number. Let C=max(2,Kc).

Now we will prove following statement.
There is a positive integer n satisfying the following condition: If the polynomial P(x) of degree greater or equal to n with real coefficients and leading coefficient equal to 1 then the number of integers x for which |P(x)|<C is not greater than the degree of P(x).

Let n0=20*C.

Consider P(x), deg(P(x))=n>n0. If P(x) has non-real root a then conjugate complex number a' also root of P(x). Replacing a and a' with Re(a) we obtain new polynomial P'(x) and |P'(x)|<|P(x)| ==> set {x | |P(x)|<C} is smaller then {x | |P'(x)|<C}. So we can WLOG suppose P(x) has real roots only, denote them a1,...,an, P(x)=(x-a1)...(x-an).

Let t1, ..., tn+1 - distinct integers. Suppose |P(ti|<C for all i=1,2,...,n+1.
It is clear there is tk such that |tk-ai| \geq 1/2 for all i=1,2,...,n. Suppose WLOG |tk-ai|<=2C, i=1,2,...,s and |tk-ai|>2C, i=s+1,...,n. We have C>|P(tk)|>(1/2)s(2C)n-s ==> Cn-s-1 < 22s-n ==> n-s-1 < 2s-n ==> s>=2n/3. Segment [tk-2C-2,tk+2C+2] contains 4C+5 integer numbers ==> there n-4C-5 of ti (i=1,2,...,n+1) outside [tk-2C+2,tk+2C+2] and only n/3 of ai (i=1,...,n) ==> within such ti exists tp such that |tp-ai|>=1 for all i=(s+1),...,n ==> C>|P(tp)|>2s>=22n/3>C !!!!

P.S. Some inequalities denote order of values only (for example, s>=2n/3) and haven't exact meaning.


What is the sum of all integers less than 100 which have exactly 12 divisors?

[quote="mathnerd314"][hide]
They all have prime factorizations with exponents that are:

1, 1, 2
OR
2, 3
[/hide]

But I don't know where to go from there.[/quote]
[hide="hint"]Another possibility is $1$ and $5$. You know what the primes' exponents are, and you know the primes can't be very big (or else you'd get way over $100$), so for each of the three cases see how many ways there are to do it so the number is less than $100$.[/hide][hide="my answer"]$96+72+60+90+84$, whatever that is.[/hide]

Correct AntonioMainenti! Use the prime factorizations (this problem came up on Mathcounts once, something like this).

[hide]The answer is 402. [/hide]


[quote="SplashD"]Express $ \sum_{i=0}^{n}\frac{2^{i}}{1+x^{2^{i}}}$ in closed form, preferably without using induction.[/quote]
I think you will ultimately have to use induction (or an equivalent telescopic sum), since it's an expression for a general integer $ n$.

[hide="Hint"]$ \frac{2^{i}}{1+x^{2^{i}}}+\frac{2^{i}}{1-x^{2^{i}}}=\frac{2^{i+1}}{1-x^{2^{i+1}}}$[/hide]
[hide="Induction"]By induction we can prove that
$ \frac{1}{1-x}+\sum_{i=0}^{n}\frac{2^{i}}{1+x^{2^{i}}}=\frac{2^{n+1}}{1-x^{2^{n+1}}}$,
(it's true for $ n=0$, and if it's true for $ n-1$, using the hint it's also true for $ n$)
so
$ \sum_{i=0}^{n}\frac{2^{i}}{1+x^{2^{i}}}=\frac{2^{n+1}}{1-x^{2^{n+1}}}-\frac{1}{1-x}$[/hide]
[hide="Telescopic sum"]Rewriting the hint
$ \frac{2^{i}}{1+x^{2^{i}}}=\frac{2^{i+1}}{1-x^{2^{i+1}}}-\frac{2^{i}}{1-x^{2^{i}}}$,
so by a telescopic sum
$ \sum_{i=0}^{n}\frac{2^{i}}{1+x^{2^{i}}}=\frac{2^{n+1}}{1-x^{2^{n+1}}}-\frac{1}{1-x}$[/hide]

not sure if it's rigorous at all, but [hide="hint"]write it in terms of infinite series: $ \frac{1}{1+x}=1-x+x^{2}-x^{3}+\cdots$[/hide]EDIT: oh.. but i like lordWings' way  :surrender:

I just noticed I should have considered separately the case $ x=1$. [hide]In that case it's just the sum of a geometric series:
$ \sum_{i=0}^{n}{2^{i-1}}=\frac{2^{n}-2^{-1}}{2-1}=2^{n}-\frac{1}{2}$[/hide]


pigeonhole principle
probability and stats

Hello Everybody,

I had a question about the power of the probabilistic method. I have not seen much problems this method can solve and in fact have only begun looking at some of the problems you can solve using it.

What seems strange to me is that I kind of feel like it is just Pigeonhole Principle in disguise.
(Now before you mention it, I know i must be wrong - but still I feel compelled to ask)

Is it just me or is there indeed a connection between these two techniques?

Please let me know. I will keep studying this method too but a bit of 'push' will be extremely helpful.

Thanks and Have a nice day
-Akash Kumar

There is a connection between the Probabilistic Method and the Pigeonhole Principle.  For example, you can prove the Pigeonhole Principle via the Probabilistic Method.  Namely, if you have $ n \plus{} 1$ pigeons in $ n$ holes, then the average hole has $ \frac{n \plus{} 1}{n}$ pigeons.  So some hole has at least $ \frac{n \plus{} 1}{n}$ pigeons, which is more than $ 1$.  Since the hole has more than $ 1$ pigeon, it has at least $ 2$ pigeons.

The Probabilistic Method can often be viewed as a generalized Pigeonhole Principle.

By the way, I posted a bunch of problems that can be solved with the Probabilistic Method:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23507[/url].

Hello Ravi,

Thanks for your reply. Makes things a bit clear to me now.

Its neat that Probabilistic Method seems to be atleast as powerful as the pigeonhole principle.

I understand that the reverse direction (that is pigeonhole being as "strong" as the Probabilistic method) must be false, but owing to my limited understanding of the method at the moment, i cannot see it clear enough.

Could you lend me a hand again with this?

Thanks in advance
-Akash

BTW, thanks for telling me about the problem set on Probabilistic Method. I will definetly go through that.
Have a good day

Well, for example, if you have $ 2n \plus{} 1$ pigeons in $ n$ holes, some hole has at least $ 3$ pigeons.  That's trivial by the Probabilistic Method, and can be viewed as a generalized Pigeonhole Principle, but it's not obvious how to prove it with just the ordinary Pigeonhole Principle.

Oh..Okay I see your point. Now its time for me to start studying this method. But just as an aside - I would like to assume as I go onwards that Probabilistic Method better be more powerful than the 'generalized pigeon hole principle'

Is that a right statement to make?

I sure hope so. Will try coming up with an example on my own and if in the meantime you find some free time, please let me know how it actually compares against Generalized Pigeonhole

Bye
-Akash

Well, I have seen some books refer to the Generalized Pigeonhole Principle as the statement that if you have $ kn \plus{} 1$ pigeons in $ n$ holes, then some hole has at least $ k \plus{} 1$ pigeons.  I don't see how to apply just that principle to solve many of the problems in my Probabilistic Method handout.  So I guess I would say that the Probabilistic Method is stronger than the Generalized Pigeonhole Principle.

Okay, thats neat enough for me!

Thanks for your help.

Good day Man!
-Akash


algebra
polynomial
number theory
least common multiple
number theory unsolved

Let $ P(x)$ be a polynomial with rational coefficients.Suppose that for any integer $ n$ ,$ P(n)$ is an integer.Prove that for any distinct integer $ m$ and $ n$ then:
            $ lcm(1;2;...;deg(P)).\frac{P(m)\minus{}P(n)}{m\minus{}n}$ is an integer.

[quote="math10"]Let $ P(x)$ be a polynomial with rational coefficients.Suppose that for any integer $ n$ ,$ P(n)$ is an integer.Prove that for any distinct integer $ m$ and $ n$ then:
            $ lcm(1;2;...;deg(P)).\frac {P(m) \minus{} P(n)}{m \minus{} n}$ is an integer.[/quote]
Good problem! Here is my solution :
The first ,we will prove following lemma :
[b][i]Lemma 1 [/i][/b] Let P is a polynomial with real coefficients , and P obtain integer values at n+1 consecutive integer points iff  P takes integer value for all integers $ x$ .
The second part is obvious so we only need to prove part 1 . Suppose P takes integer values at n consecutive points 1,...,k-1,k,k+1(problem said that this polynomial takes integer values at h+1,...,h+n and we can assume h=0 ,if not we consider polynomial $ P(x \minus{} h)$ ) , we can rewrite P in the form :
\[ P(x) \equal{} a_k(x \minus{} 1)...(x \minus{} k) \plus{} .. \plus{} a_1(x \minus{} 1) \plus{} a_0
\]
Now we only need to consider $ P(1),...,P(k \plus{} 1)$ ,it is not hard to show that $ \frac {a_i}{i!}$ be integers for all $ 1\leq i\leq k$ . But we know that product of k consecutive integers be a multiple of $ k!$ so P takes integer values for all integers n ,more than we can find $ c_1,..,c_k$ such that :
\[ P(x) \equal{} \sum_{i \equal{} 0}^k c_i {x\choose k}
\]
By this lemma we only need to consider $ P(x) \equal{} {x \choose k}$
Let $ h \equal{} m \minus{} n$ , we will show that for all m and h then
\[ f(m) \equal{} lcm \{1,..,k\} \frac {P(m \plus{} h) \minus{} P(m)}{h}
\]
are integer numbers . Let consider this as a polynomial of $ m$ . For all $ i \equal{} 1,...,k$ we have
\[ f(i) \equal{} \frac {lcm\{1,..,k\}}{i} .{{m \plus{} i}\choose k}
\]
which obvious be integers .
Lemma 2 :f(k+1) is an integer number . We use above trick again ,consider it as a polynomial of h : $ g(h) \equal{} f(k \plus{} 1)$ . It is obvious that $ g( \minus{} 1),...,g( \minus{} k)$ are integers , and we must prove $ g( \minus{} k \minus{} 1)$ is also an integer , but it is very obvious (we only need to consider two cases , h be a power of prime or not ) . Hence by lemma 1 ,for all k g takes integer value and so $ f(k \plus{} 1)$ is integer . Apply lemma 1 again , f takes integer values for all integers k and our proof completes here .
Note : It is a complex solution , and I'm waiting for a new solution that more simple :)

You were wrong.His name is Nguyễn Thọ Tùng.

oh, I also listen to name him.
he is really good.

[quote="math10"]thank you very much.
can you introduce yourself?
I think you are a teacher of DHSP.[/quote]

And you,Cuong.
I think you are a student in Lam Son Thanh Hoa


Hi every one,
I have a very very silly question. I cannot prove why:
(Au,v) = (u,ATv); AT is the Transpose of A
Please help me!

In general, $\langle u,v\rangle=u^{T}v$. Rewrite the two expressions using this. What do you know about the transpose of a product?

Both Sayeh's original question and jmerry's hint presume that we are dealing with real matrices and the real inner product. A suitable modification of the statement could be found for complex matrices and vectors.


What is the value of the reciprocal of $ \frac{\minus{}2}{5}$? Express your answer as a decimal to the nearest tenth

the reciprocal would be $ \frac{5}{\minus{}2}$ when simplified it is [b]-2.5[/b]

The reciprocal of $ \frac{x}{y}$ is $ \frac{y}{x}$.


factorial
function
algebra
domain
floor function

There is this problem I've been trying to figure out how to do for a long time

What is the smallest factorial greater than a googol?

You can do it with a calculator, which is what I did when given it for a summer packet, but I still haven't figured out how to do it without a calculator. Can someone help? Or at least tell me if you were supposed to use a calculator. Thanks.

i don't really know, but it'll be pretty cool if math had a new notation: $ !^{ \minus{} 1}$. like the inverse operation of factorial, u know (just like sin has $ sin^{ \minus{} 1}$ as its inverse function, u no). like if i were to find $ 24!^{ \minus{} 1}$, it would be 4. and then it gets into calcs and all that good stuff. :)

edit: @AIME15: probably "ERR:SYNTAX" or something. yeah, u would have to do a bunch of trial/error if that is made. probably something like the floor function inverse operation of factorial.  :|

That wouldn't work out, because factorials are product of integers, so what if we had a number like 36?  What would 36!^-1 be?

You can use the inverse of the gamma function, an extension to factorials that actually takes negative numbers (besides negative integers, those are undefined) and complex numbers.

Although that takes calculus.

[quote="1-1 is 3"]There is a way to do it by "hand", but it is less math and more memorization.

[hide]If you have a TI-83/84, you might have noticed that the factorial function tops out at 69!. At 70! the calculator gives an overflow error. And the calculator's overflow is one google. Therefore, $69!<googol<70!$. The answer is $\boxed{{70}}$[/hide]

I would call this the cheesy method.[/quote]

Lol.  That works.

The inverse of the gamma function would have to have domain approximately $\left[.8856,\infty\right)$, from what's presented [url=http://en.wikipedia.org/wiki/Particular_values_of_the_Gamma_function]here[/url].


[quote="MrMagician"]make 3_3_3_3_3 =100.[/quote]

Can the blanks be, well, blank? If not, I don't believe there is a solution, although I may be missing something obvious.

I agree, I've tried everything and it doesn't seem to work.

Me too, it doesn't seem to work.  If there were four blanks, it could be $ 3\cdot33\plus{}3/3$.  Maybe its a typo?

Well, division must be used at least once, because $ 100 \equiv 1$ modulo $ 3$. This leads to having three $ 3$s and one $ 1$, which can not yield a number greater than $ 100$ if we act under the assumption that we cannot concatenate numbers.

Assume that the problem disregards order of operations and moves from left to right.
Then, at least $ 3$ of the blanks must possess multiplication signs.  To see this, consider the case with only two multiplication signs: in that case the largest possible number is $ 3\plus{}3\plus{}3*3*3\equal{}81<100$. Further, all $ 4$ of the blanks cannot possess multiplication signs, as $ 3*3*3*3\neq100$.

Thus, exactly $ 3$ of the blanks have multiplication signs.
From here, we can see that no solution can work.

Assume that the problem does follow order of operations.
Then, the largest possibility with $ 3$ multiplication signs does not work ($ 3*3*3*3\plus{}3\equal{}86<100$). Since the only possibility with $ 4$ multiplication signs does not work ($ 3*3*3*3*3\equal{}243\neq100$), we can conclude no set of operators works.


inequalities
function
inequalities proposed

It reminds me a same problem which comes from China MO 1989 and there the problem was

Let $x_1,...,x_n>0$ such that $\sum_{i=1}^n x_i=1$. Prove that

$\sum_{i=1}^n {\frac{x_i}{\sqrt{1-x_i}}}\geq\frac{ \sum_ {i=1}^n {\sqrt x_i}}{\sqrt{n-1}}$

where $n>1$

let $a_i=\sqrt{1-x_i}$ for $i=1,2,\cdots,n$
then $x_i=1-a_i^2$ and $\sum_{i=1}^n a_i=n-1$
we have to prove $\sum_{i=1}^n \frac{1}{a_i}-\sum_{i=1}^n a_i \geq \sqrt{\frac{n}{n-1}}$ (*)
but we have $\sum_{i=1}^n a_i \leq \sqrt{n(n-1)}$  (1)
and use caughy inequality:
$(\sum_{i=1}^n \frac{1}{a_i} )^2(\sum_{i=1}^n a_i^2 ) \geq \frac{1}{n}(\sum_{i=1}^n \frac{1}{a_i} )^2(\sum_{i=1}^n a_i)^2 \geq n^3$
so $\sum_{i=1}^n \frac{1}{a_i} \geq \sqrt{\frac{n^3}{n-1}}$ (2)
by (1) and (2) we get (*) is right.
and it is stronger than China MO 1989

this is one of my favorites inequalities :D

WLOG suppose that
$\forall \ m>n, \ \ \ x_m\geq x_n \\\Rightarrow \frac{1}{\sqrt{1-x_m}}\geq\frac{1}{\sqrt{1-x_n}}$,
then by chebyschev and cauchy-schwarz we have that
$\sum \frac{x_k}{\sqrt{1-x_k}}\geq\frac{\sum x_k \cdot \sum \frac{1}{\sqrt{1-x_k}}}{n} = \frac{\sum \frac{1}{\sqrt{1-x_k}}}{n}\geq \frac{1}{n}\cdot\frac{n^2}{\sum \sqrt{1-x_k}} \geq\frac{n}{\sqrt{n\cdot\sum (1-x_k )}}=\frac{n}{\sqrt{n\cdot(n-1)}} = \frac{\sqrt n}{\sqrt{n-1}}$
and we're done!

[quote="e.lopes"]the function $f(x) = \frac{x}{\sqrt{1-x}}$ is convex, so, the inequality follow by jensen.  :)[/quote]
Should without jensen ;)


algebra
polynomial
complex numbers
algebra unsolved

Suppose $ x,y,z \in \mathbb{C}$,then solve the system of equation:
$ x\plus{}y\plus{}z\equal{}3$
$ x^2\plus{}y^2\plus{}z^2\equal{}3$
$ x^5\plus{}y^5\plus{}z^5\equal{}3$

[quote="greenvert"]
\[ 9 \equal{} (x \plus{} y \plus{} z)^2 \equal{} x^2 \plus{} y^2 \plus{} z^2 \plus{} 2(xy \plus{} yz \plus{} zx) \equal{} 3 \plus{} 2(xy \plus{} yz \plus{} zx)\Rightarrow xy \plus{} yz \plus{} zx \equal{} 3
\]
so:
\[ 0 \equal{} x^2 \plus{} y^2 \plus{} z^2 \minus{} xy \minus{} yz \minus{} zx \equal{} \frac {1}{2}((x \minus{} y)^2 \plus{} (y \minus{} z)^2 \plus{} (z \minus{} x)^2)\Rightarrow x \equal{} y \equal{} z \equal{} 1
\]
[/quote]
Recall that $ x,y,z$ are complex numbers, hence the last implication is not valid.
Instead, consider $ P(t)\equal{}t^3\minus{}at^2\plus{}bt\minus{}c$ the polynomial with roots $ x,y,z$. Using Viete's relations we can easily obtain $ a\equal{}3,b\equal{}3,c\equal{}1$, therefore $ P(t)\equal{}(t\minus{}1)^3$, so $ x\equal{}y\equal{}z\equal{}1$.

The sequence $ s_n \equal{} x^n \plus{} y^n \plus{} z^n$ satisfies the recurrence

$ s_{n\plus{}3} \equal{} (x \plus{} y \plus{} z) s_{n\plus{}2} \minus{} (xy \plus{} yz \plus{} zx) s_{n\plus{}1} \plus{} xyz s_n$.

We can compute $ xy \plus{} yz \plus{} zx \equal{} 3$ and so $ s_3 \equal{} 3xyz$.  Let $ xyz \equal{} p$.  Then

$ s_4 \equal{} 9p \minus{} 9 \plus{} 3p \equal{} 12p \minus{} 9$
$ s_5 \equal{} (36p\minus{}27) \minus{} 9p \plus{} 3p \equal{} 30p \minus{} 27 \equal{} 3 \implies p \equal{} 1$.

Fun.
The Viete's relation, I just searched that stuff. Seems quite useful, as just shown :D


In a closed plane, there are n lines. How many figures are formed from these n lines?
I tried to make a formula:
$ \dfrac{n(n\plus{}1)}{2}\plus{}1$
I tried it on 1, 2, 3, 4, and 5 lines, but I still am not sure if it is right.
Thanks.

[quote="gogators"]In a closed plane, there are n lines. How many figures are formed from these n lines?
I tried to make a formula:
$ \dfrac{n(n \plus{} 1)}{2} \plus{} 1$
I tried it on 1, 2, 3, 4, and 5 lines, but I still am not sure if it is right.
Thanks.[/quote]

[hide="solution?"]
$ a_{1}\equal{}2$
$ a_{n}\equal{}a_{n\minus{}1}\plus{}n$
[/hide]

@gogators For the most part, you are correct. They are a few cases against it ($ 3$ lines concurrent, for example). Btw, nice avatar, I used to live in Gainesville ... :D

But it does work for 3...

I forgot to add that every line intersects every other line.

If that is the case, my formula works for 3 concurrent lines.

:)

"Concurrent" means that $ 3$ lines intersect a common point. (You may already know this, but I'll throw it out anyway). Also, if any $ 3$ lines are concurrent in your drawing with $ n$ lines, the formula simply is $ \frac{n(n\plus{}1)}{2}$.(This also assumes that no $ 2$ lines are parallel).


I'm on page 34 on volume 2 (top).

I just have a small question that i'm dealing with since the last 15 minutes. how does angle D=$ \frac{\beta\plus{}\theta}{2}$ since D isn't even on the circumference?

Draw segments EF and AC.  Note that angle FED + angle EFD = 180 - angle D.  Angle FED intercepts arc FC and angle EFD intercepts arc EA.  So Angle D = 180 - (angle FED + angle EFD) = 180 - (arc FC + arc EA)/2.

Now just write arc FC + arc EA in terms of Arc theta and Arc beta.  That is, arc FC + arc EA = 360 - Arc theta - Arc beta.  Plug this expression for FC + EA into the expression for Angle D and you get the step they took.


Let $ a,b,c$ be positive numbers such that: $ a\plus{}b\plus{}c\equal{}1$.Prove that:
$ \frac{a}{b^2\plus{}b}\plus{}\frac{b}{c^2\plus{}c}\plus{}\frac{c}{a^2\plus{}a} \ge \frac{36(a^2\plus{}b^2\plus{}c^2)}{ab\plus{}bc\plus{}ca\plus{}5}$

No one interested in this inequality ? :)

[quote="tuantam1lan"][quote="quykhtn-qa1"]Let $ a,b,c$ be positive numbers such that: $ a \plus{} b \plus{} c \equal{} 1$.Prove that:
$ \frac {a}{b^2 \plus{} b} \plus{} \frac {b}{c^2 \plus{} c} \plus{} \frac {c}{a^2 \plus{} a} \ge \frac {36(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca \plus{} 5}$[/quote]
hello,i can prove your problem by one result of hungkhtn and vacs is
if $ a \plus{} b \plus{} c \equal{} 1$, $ a,b,c$ be positive numbers then $ a^2b \plus{} b^2c \plus{} c^2a \plus{} abc \le \frac {4}{27}$
uses cauchuy-schawrs we have:
$ \sum{\frac {a}{b^2 \plus{} b}} \ge \frac {(a \plus{} b \plus{} c)^2}{\sum{ab} \plus{} \sum{b^2a}} \ge \frac {36(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca \plus{} 5}$
   let $ q \equal{} \sum{ab}$,$ r \equal{} abc$
we have
$ 36\sum{a^2}(\sum{ab^2} \plus{} \sum ab) \ge 36\sum{a^2}(\frac {4}{27} \minus{} abc \plus{} \sum{ab}) \ge \sum{ab} \plus{} 5$
<=>$ q \plus{} 5 \ge 36(1 \minus{} 2q)(\frac {4}{27} \minus{} r \plus{} q)$(1)
becase $ 1 \minus{} 2q \ge 1 \minus{} \frac {2}{3} \ge 0$so uses schur third degree we have$ r \ge \frac {1 \minus{} 4q}{9}$
supposing (1)<=>$ q \plus{} 5 \ge 36(1 \minus{} 2q)(\frac {4}{27} \minus{} \frac {4q \minus{} 1}{9} \plus{} q)$<=>$ (1 \minus{} 3q)(40q \plus{} 13) \ge 0$
equality when $ a \equal{} b \equal{} c \equal{} \frac {1}{3}$[/quote]

You are wrong.
$ r \ge \frac{4q\minus{}1}{9}$With:$ q \ge \frac{1}{4}$

[quote="caubetoanhoc94"][quote="tuantam1lan"][quote="quykhtn-qa1"]Let $ a,b,c$ be positive numbers such that: $ a \plus{} b \plus{} c \equal{} 1$.Prove that:
$ \frac {a}{b^2 \plus{} b} \plus{} \frac {b}{c^2 \plus{} c} \plus{} \frac {c}{a^2 \plus{} a} \ge \frac {36(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca \plus{} 5}$[/quote]
hello,i can prove your problem by one result of hungkhtn and vacs is
if $ a \plus{} b \plus{} c \equal{} 1$, $ a,b,c$ be positive numbers then $ a^2b \plus{} b^2c \plus{} c^2a \plus{} abc \le \frac {4}{27}$
uses cauchuy-schawrs we have:
$ \sum{\frac {a}{b^2 \plus{} b}} \ge \frac {(a \plus{} b \plus{} c)^2}{\sum{ab} \plus{} \sum{b^2a}} \ge \frac {36(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca \plus{} 5}$
   let $ q \equal{} \sum{ab}$,$ r \equal{} abc$
we have
$ 36\sum{a^2}(\sum{ab^2} \plus{} \sum ab) \ge 36\sum{a^2}(\frac {4}{27} \minus{} abc \plus{} \sum{ab}) \ge \sum{ab} \plus{} 5$
<=>$ q \plus{} 5 \ge 36(1 \minus{} 2q)(\frac {4}{27} \minus{} r \plus{} q)$(1)
becase $ 1 \minus{} 2q \ge 1 \minus{} \frac {2}{3} \ge 0$so uses schur third degree we have$ r \ge \frac {4q \minus{} 1}{9}$
supposing (1)<=>$ q \plus{} 5 \ge 36(1 \minus{} 2q)(\frac {4}{27} \minus{} \frac {4q \minus{} 1}{9} \plus{} q)$<=>$ (1 \minus{} 3q)(40q \plus{} 13) \ge 0$
equality when $ a \equal{} b \equal{} c \equal{} \frac {1}{3}$[/quote]

You are wrong.
$ r \ge \frac {4q \minus{} 1}{9}$With:$ q \ge \frac {1}{4}$[/quote]
i'm sorry
i Type is wrong
i repaired it
thanks you very much, because you readed and show my mistake

[quote="tuantam1lan"][quote="quykhtn-qa1"]Let $ a,b,c$ be positive numbers such that: $ a \plus{} b \plus{} c \equal{} 1$.Prove that:
$ \frac {a}{b^2 \plus{} b} \plus{} \frac {b}{c^2 \plus{} c} \plus{} \frac {c}{a^2 \plus{} a} \ge \frac {36(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca \plus{} 5}$[/quote]
hello,i can prove your problem by one result of hungkhtn and vacs is
if $ a \plus{} b \plus{} c \equal{} 1$, $ a,b,c$ be positive numbers then $ a^2b \plus{} b^2c \plus{} c^2a \plus{} abc \le \frac {4}{27}$
uses cauchuy-schawrs we have:
$ \sum{\frac {a}{b^2 \plus{} b}} \ge \frac {(a \plus{} b \plus{} c)^2}{\sum{ab} \plus{} \sum{b^2a}} \ge \frac {36(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca \plus{} 5}$
   let $ q \equal{} \sum{ab}$,$ r \equal{} abc$
we have
$ 36\sum{a^2}(\sum{ab^2} \plus{} \sum ab) \ge 36\sum{a^2}(\frac {4}{27} \minus{} abc \plus{} \sum{ab}) \ge \sum{ab} \plus{} 5$
<=>$ q \plus{} 5 \ge 36(1 \minus{} 2q)(\frac {4}{27} \minus{} r \plus{} q)$(1)
becase $ 1 \minus{} 2q \ge 1 \minus{} \frac {2}{3} \ge 0$so uses schur third degree we have$ r \ge \frac {1 \minus{} 4q}{9}$
supposing (1)<=>$ q \plus{} 5 \ge 36(1 \minus{} 2q)(\frac {4}{27} \minus{} \frac {4q \minus{} 1}{9} \plus{} q)$<=>$ (1 \minus{} 3q)(40q \plus{} 13) \ge 0$
equality when $ a \equal{} b \equal{} c \equal{} \frac {1}{3}$[/quote]
You can edited it?Good luck :)

[quote="quykhtn-qa1"]Let $ a,b,c$ be positive numbers such that: $ a \plus{} b \plus{} c \equal{} 1$.Prove that:
$ \frac {a}{b^2 \plus{} b} \plus{} \frac {b}{c^2 \plus{} c} \plus{} \frac {c}{a^2 \plus{} a} \ge \frac {36(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca \plus{} 5}$[/quote]
This inequality very nice.I hope you can find an nice solution :)

[quote="quykhtn-qa1"]Let $ a,b,c$ be positive numbers such that: $ a \plus{} b \plus{} c \equal{} 1$.Prove that:
$ \frac {a}{b^2 \plus{} b} \plus{} \frac {b}{c^2 \plus{} c} \plus{} \frac {c}{a^2 \plus{} a} \ge \frac {36(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca \plus{} 5}$[/quote]
my solution:
$ \frac {a}{b^2 \plus{} b} \plus{} \frac {b}{c^2 \plus{} c} \plus{} \frac {c}{a^2 \plus{} a} \ge \frac{3}{4}\left( {\frac{a}{b} \plus{} \frac{b}{c} \plus{} \frac{c}{a}} \right) \ge \frac {36(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca \plus{} 5}$

[quote="Toan_VN_LC"][quote="quykhtn-qa1"]Let $ a,b,c$ be positive numbers such that: $ a \plus{} b \plus{} c \equal{} 1$.Prove that:
$ \frac {a}{b^2 \plus{} b} \plus{} \frac {b}{c^2 \plus{} c} \plus{} \frac {c}{a^2 \plus{} a} \ge \frac {36(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca \plus{} 5}$[/quote]
my solution:
$ \frac {a}{b^2 \plus{} b} \plus{} \frac {b}{c^2 \plus{} c} \plus{} \frac {c}{a^2 \plus{} a} \ge \frac {3}{4}\left( {\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}} \right) \ge \frac {36(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca \plus{} 5}$[/quote]
How about the stronger, Toan  :)
$ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a} \geq\ \frac{30(a^2\plus{}b^2\plus{}c^2)}{3(a\plus{}b\plus{}c)^2\plus{}ab\plus{}bc\plus{}ca}$

[quote="nguoivn"][quote="Toan_VN_LC"][quote="quykhtn-qa1"]Let $ a,b,c$ be positive numbers such that: $ a \plus{} b \plus{} c \equal{} 1$.Prove that:
$ \frac {a}{b^2 \plus{} b} \plus{} \frac {b}{c^2 \plus{} c} \plus{} \frac {c}{a^2 \plus{} a} \ge \frac {36(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca \plus{} 5}$[/quote]
my solution:
$ \frac {a}{b^2 \plus{} b} \plus{} \frac {b}{c^2 \plus{} c} \plus{} \frac {c}{a^2 \plus{} a} \ge \frac {3}{4}\left( {\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}} \right) \ge \frac {36(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca \plus{} 5}$[/quote]
How about the stronger, Toan  :)
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \geq\ \frac {30(a^2 \plus{} b^2 \plus{} c^2)}{3(a \plus{} b \plus{} c)^2 \plus{} ab \plus{} bc \plus{} ca}$[/quote]
I think we need to use the lemma:
$ 2\left( {\frac{a}{b} \plus{} \frac{b}{c} \plus{} \frac{c}{a}} \right) \plus{} 1 \ge \frac{{21({a^2} \plus{} {b^2} \plus{} {c^2})}}{{{{(a \plus{} b \plus{} c)}^2}}}$ :)

Yes, that's it. I also used this lenma to creat the stronger  :)
With this lenma, I had had an easy proof by Am-Gm and CS :)


[size=125]
مجموعه ای اعدادی طبیعی است که جمع اعداد هر دو زیرمجموعه ی متفاوت این مجموعه متمایز باشند نشان دهید $S=\{a_1,a_2,\dots,a_n\}$
\[ \frac 1{a_1}+\dots \frac1{a_n}<2  \]
[/size]

mishe begi masale male kojast?

mer30 ;)

fekr konam masaleye jalebi bashe.chon hameye soalaye mosene jamali ziba hastand. :D

mer30 omid.

yadam bashe mogheye class azash be porsam ;)

Trivial.

Let a1<a2<...<an.
For every set b1<...<bj with the property given in the problem let M(b1,...,bj) denote the smallest natural number bigger than
bj that we can take as bj+1 without contradicting the condition.
We use induction to show that 1/a1  + 1/a2 +... + 1/an + 1/an =<2
 Note that if an= an-1 + x then we require that x can not be written as : x= X - Y where X and Y are sums of elements of two subsets
of {a1,...,an-2}. Therefore x>= M(a1,...,an-2). Thus an>= an-1 + M(a1,...,an-2)
One can see that to maximize 1/a1 + 1/a2 +... + 1/an
the best choice would be to take an-1 = M(a1,...,an-2) and an = 2an-1
Hence 1/a1 + ... + 1/an  + 1/an < 1/a1 + ... + 1/an-2 + 1/an-1 + 1/2an-1 + 1/2an-1 = 1/a1 +... + 1/an-1 + 1/an-1 <2
by induction hypothesis.

to make induction complete we need to show it for n=2:
Indeed 1/a1 + 1/a2 + 1/a2 <= 1 + 1/2 + 1/2 <=2

-Ali


What is the simplified value of $ \frac{10! \plus{} 11! \plus{} 12!}{10! \plus{} 11!}$?

Factoring out a 10! on both the numerator and denominator, the problem becomes easy:
$ \frac {10!(1 \plus{} 11 \plus{} 12*11)}{10!(1 \plus{} 11)}$

$ \equal{} \frac {12 \plus{} 12*11}{12}$

$ \equal{} \frac {12(1\plus{} 11)}{12}$

$ \equal{} \boxed{12}$


Define/(x) = |x|  + [x] , where [x] is the greatest integer less
than or equal to x. What is the value of 2/(-7i) + 2/(1.5)? Express
your answer in terms of K.

Can you also graph this? Ive always wondered if you can graph imaginaries on a carteseian plane...

You certainly can graph imaginaries on the [b]Complex Plane[/b]. It is like the Cartesian plane, except the real number part is the x-axis and the imaginary is the y-axis.

neat! but how do u solve the problem? And can anyone send me a pic of a complex graph of that function?

Hmm...where'd you get this problem?

Also, Imaginary Numbers aren't MC level.

[quote="AIME15"]There is no "absolute value" of $ i$.  Also, what's K?[/quote]

Actually, if $ z\equal{}a\plus{}bi$, then $ |z|\equal{}\sqrt{a^2\plus{}b^2}$

Allright, then what's the floor value of an imaginary number?

The correct problem is

13. Define $ f(x)\equal{}|x|\plus{}[x]$, where $ [x]$ is the greatest integer less than or equal to $ x$. What is the value of $ 2f(\minus{}\pi)\plus{}2f(1.5)$? Express your answer in terms of $ \pi$.

I have no idea where $ K$ came from, or the other random stuff. :wink:

[quote="math154"]The correct problem is

13. Define $ f(x) \equal{} |x| \plus{} [x]$, where $ [x]$ is the greatest integer less than or equal to $ x$. What is the value of $ 2f( \minus{} \pi) \plus{} 2f(1.5)$? Express your answer in terms of $ \pi$.

I have no idea where $ K$ came from, or the other random stuff. :wink:[/quote]

First we evaluate $ f(\minus{}\pi)$.  This equals $ |\minus{}\pi|\plus{}[\minus{}\pi]\equal{}\pi\minus{}4$.  Multiply by $ 2$ to get $ 2\pi\minus{}8$.
Then we evaluate $ f(1.5)$.  This equals $ |1.5|\plus{}[1.5]\equal{}2.5$.  Multiply by $ 2$ to get $ 5$.

Adding these two, we get $ \boxed{2\pi\minus{}3}$.

Well then, the answer would be:

$ 2 (\pi \minus{} 4) \plus{} 2 (1.5 \plus{} 1) \equal{} 2\pi \minus{} 8 \plus{} 3 \plus{} 2 \equal{} 2\pi \minus{} 3 \equal{} \boxed{\minus{}3 \plus{} 2\pi}$

EDIT: Beaten.


[quote="usaha"]What is the probablility that if u draw 5 cards from a 52 card deck, that they can be aranged into a straight flush (5 cards in chronological order with common difference one and all of the same suit) (the Ace can be used as the lowest or highest card in the straight flush)?[/quote]
[hide]The straight flush can be any one of the $10$ arrangements
$(\mbox{A},2,3,4,5), (2,3,4,5,6), \dots, (10,\mbox{J,Q,K,A})$
for any of the $4$ suits, so the total number of ways to get a straight flush is $10\times 4=40$. We divide this by ${52\choose 5}$, which is the number of ways to choose any 5 cards from the deck, and our answer comes out to be
$\frac{40}{{52\choose 5}}=\boxed{\frac{1}{64974}}$, which is pretty low.  :wink: [/hide]

[hide]its 1/64974.  there are 40 total scenerios (A-5, 2-6,...10-A), thats 10 but u have to times 4 because there are 4 suits u can get the flush part on. So it is (1/52)(1/51)(1/50)(1/49)(1/48) *5!*40 as there are 5!(120) ways to get the 5 cards because u said arranged into a straight flush (so 2,A,4,3,5 all spades will work) and if u mulitply out its 1/64974.[/hide]

[size=75][color=darkred]Fixed your hide tags and the typo.
-nebula42[/color][/size]

[hide]For one suit, you can have (A,2,3,4,5) all the way up to (10,J,Q,K,A). That's 10 straights and since there are 4 suits, there are 40 distinct straight flushes. There are a total of 52C5 combinations of 5 cards, so the answer is (40)/(52C5)=$\boxed{\frac{1}{64974}}$
[/hide]

[quote="johnz"][hide]its 1/64975.  there are 40 total scenerios (A-5, 2-6,...10-A), thats 10 but u have to times 4 because there are 4 suits u can get the flush part on. So it is (1/52)(1/51)(1/50)(1/49)(1/48) *5!*40 as there are 5!(120) ways to get the 5 cards because u said arranged into a straight flush (so 2,A,4,3,5 all spades will work) and if u mulitply out its 1/64975.[/hide]

[size=75][color=darkred]Fixed your hide tags.
-nebula42[/color][/size][/quote]
Was that a typo? When I multiplied it out I got
[hide]$\frac{1}{64974}$, while you got $\frac{1}{64975}$, so that really looks like a typo to me.  :wink: [/hide]

no it was a typo but wehn i tried to fix it it didnt fix.  i initally didnt hide it right and i had a typo so i tried to fix both, but when i clicked submit, all it said is "your hide tags have been fixed by nebula", so i dont know what i did wrong

[quote="SplashD"][quote="nebula42"][quote="SplashD"]Extension:

If you draw a hand of five cards from a deck of 52 cards in poker, what is the probability that it will beat a set of 7s with Ace and King kicker?[/quote]
Does that mean the set is $(7,7,7,A,K)$?
I don't know that much about poker...  :roll:  :P[/quote]

Yeah that does, and note you still have a full deck of cards so there're no split pots.[/quote]
But then there are many different complicated cases you have to consider, so this is well above MathCounts level.

Techinically, 10,J,Q,K,A isn't a straight flush. It's a royal flush, but whatever.

I'll change the problem a little.

You draw five cards from a standard deck of 52 cards. What the probability that you have a straight, flush, full house, 4 of a kind, straight flush or a royal flush?

[quote="Klebian"][quote="bpms"][quote="usaha"]What is the probablility that if u draw 5 cards from a 52 card deck, that they can be aranged into a straight flush (5 cards in chronological order with common difference one and all of the same suit) (the Ace can be used as the lowest or highest card in the straight flush)?[/quote]
Umm. Don't use leetspeak :ninja:[/quote]
Typing "u" instead of "you" once doesn't really make it leetspeak...[/quote]
i agree. Chatspeak is this: Liek omg dood j00 just g0+s h4xor3d!!!11


geometry
circumcircle
perpendicular bisector
geometry unsolved

Let $ O$ be the circumcenter and $ N$ the center of the nine-point circle of a triangle $ ABC$. Let $ N'$  be the point such that
$ \angle N'BA \equal{} \angle NBC$ and $ \angle N'AB \equal{} \angle NAC$.The perpendicular bisector of $ OA$ meets $ BC$ in $ A'$ .Points  $ B'$ and $ C'$ are deﬁned analogously. Prove that $ A',B',C'$ line on a line $ l$ that is perpendicular to $ ON'$


calculus
integration
function
derivative
calculus computations

When is it OK to take the derivate or integral of both sides of some equality? i.e do we always end up with an equality after we take the derivative/integral of both sides?

When we take derivate/integral of both sides, do we lose any pieces of information?

What makes taking derivative/integral of both sides different from, say, multiplying both sides by 2?

The first caveat: you say "both sides of an equality." Is this a conditional equation like $3x+2=12,$ only true for some particular values of the variables involved? Or is it an identity like $x^{2}-a^{2}=(x+a)(x-a)$ which is true for all values of the variables?

You differentiate functions and you integrate functions; you don't differentiate or integrate points. That means that you cannot differentiate or integrate conditional equations in any meaningful way, but you can differentiate or integrate identities freely, subject only to the issues of whether the derivative or integral exist (and subject to my second caveat). Let's put it this way: if both sides are the same function, then they have the same derivative.

But the issue of needing both sides to be the same function and not just the same number is what's different than just multiplying by 2.

The second caveat: beginning calculus students invariably say "integral" when they really mean "antiderivative" or "primitive." A statement like $\int x^{2}\,dx=\frac13x^{3}+C$ is [b]not[/b] a statement about an integral - it is a statement about an antiderivative. And, as indicated by that "$+C,$" an antiderivative is not a function, it is an infinite family of functions. If you try to take the antiderivative of both sides of some identity, you're at the mercy of the result only being true up to the addition of some constant. If you restrict "integral" to mean just that, then you're on safe ground.

For example: if $f(x)=g(x)$ for all $x\in[0,a],$ then for all $x\in[0,a],$

$\int_{0}^{x}f(t)\,dt=\int_{0}^{x}g(t)\,dt.$

(Note my use of a "dummy variable.")


Split from inactive thread. Moving to Support.

In the future, please read the forum guidelines. :-) Thanks.

[quote="math92"]somethings wrong with all the avatars...nobody's is showing up... :([/quote]

I see the avatars and nobody else is reporting this problem, so it's almost surely local.  You might try deleting your cookies, rebooting, or any of the usual processes that restore robustness to your computer.

You should also check your [url=http://www.mathlinks.ro/Forum/profile.php?mode=editprofile]Profile[/url] to see if you Allowed Avatars to be displayed.


limit
trigonometry
calculus
logarithms
calculus computations

Find and prove the following limits:

i) $ \lim_{n\rightarrow \infty} ((n\plus{}n^{1/2})^{1/2}\minus{}(n\minus{}n^{1/2})^{1/2})$
ii) $ \lim_{n\rightarrow \infty} \displaystyle\sum_{i\equal{}2}^{\infty}\log(1\minus{}\frac{1}{k^2})$

On the calculus test I gave last Thursday, I had a limit question very similar to (i), and I asked whether $ \sum_{n\equal{}2}^{\infty}\ln\left(1\plus{}\frac1{n^2}\right)$ converges (but not what its value was.)

They weren't easy questions.

The minus sign does make things easier, particularly when looking for exact values; $ \log\left(1\minus{}\frac1{k^2}\right)\equal{}\log\left(1\plus{}\frac1k\right)\plus{}\log\left(1\minus{}\frac1k\right)\equal{}\log\frac{k\plus{}1}{k}\minus{}\log\frac{k}{k\minus{}1}$
The series thus telescopes to $ \lim_{n\to\infty}\left(\log\frac{n\plus{}1}{n}\minus{}\log\frac{2}{1}\right)\equal{}\minus{}\log 2$.

This is all working under the assumption that in the original problem, $ \infty \equal{} n$ and $ k \equal{} i$, right?


algorithm
number theory
Euclidean algorithm
\/closed

i looked back at the first class of the introductory number theory class, and i found
out that the euclidean algorithm is not explained as fully as it should be.

MCrawford (20:20:00)
(1152, 1392) = (1152, 240) = (912, 240) = (672, 240)
= (432, 240) = (192, 240) = (192, 48) = (144, 48)
= (96, 48) = (48, 48) = 48.

This could be shortened to:
(1152, 1392) = (1152, 240) = (192, 240) = (192, 48) = 48

MCrawford failed to point out that (r, sr + q) = (r, q)
instead of just (r, r + q) = (r, q)

Is this is a problem? I mean the way MCrawford did it simply includes *every* step. It may seem unnecessary and perhaps the formula you propose works more efficiently, but the method described seems to be more "comprehensive," in a sense. I would suspect that a student could figure out the more efficient way through what MCrawford showed. Just my thoughts.

EDIT: not sure if this should go here, either, since it's talking about specific info from a class.

That's how Mr. Crawford began with the Euclidian Algorithm, but I believe in the next class he explained the "Extended Euclidian Algorithm" which shortcuts just as your method does.  This also should not be in this forum, and should if anything be sent directly to him, not posted to everyone on AoPS.

Yes, we do cover the fact that steps can be cut out of the algorithm.

We don't mind the posting of excerpts from the class for the sake of addressing the classes themselves, though we would appreciate it if nobody posted more extensive portions of the classes or use them as direct instruction.


LaTeX
modular arithmetic
number theory
relatively prime

If $ p$ and $ 8p^2 \plus{}1$ are relatively prime ,then $ 8p^2 \plus{}1$ is prime

In my book there is a solution,By the way I will try to write it in LaTeX  :maybe:

This is my solution from the book : If $ p \equal{} 3k \plus{} 1$ or $ p \equal{} 3k \plus{} 2$ then $ p^2 \equal{} 3q \plus{} 1 \equal{} > 8p^2 \plus{} 1 \equal{} 3t$.From here we see $ p \equal{} 3$ and $ 8p^2 \plus{} 1 \equal{} 73$ but in that case $ 8p^2 \minus{} 1 \equal{} 71$ which is prime.

I really don't understand this solution and I think the problem should be : If $ p$ and $ 8p^2 \plus{} 1$ are relatively prime , then one of the number $ 8p^2 \plus{} 1$ or $ 8p^2 \minus{} 1$  is prime .                   :(

That is, indeed, the correct phrasing of the problem. The first step means that if $ p$ is either 1 or 2 $ \pmod 3$, then $ p^2$ will be $ 1 \pmod 3$, leading to the conclusion that $ 8p^2\plus{}1$ is $ 0\pmod 3$ if $ p$ isn't divisible by $ 3$. See where you can go from there.

Note that...

[hide]p and $ 8p^2\plus{}1$ are always relatively prime.[/hide]

[quote="enndb0x"][quote="pythag011"]Note that...

[hide]p and $ 8p^2 \plus{} 1$ are always relatively prime.[/hide][/quote]

hmm ..Is that a lemma ?
 can you prove it ?[/quote]

note that $ p$ divides $ 8p^2$, so $ p$ does not divide $ 8p^2\plus{}1$

[quote="enndb0x"]If $ p$ and $ 8p^2 \plus{} 1$ are relatively prime ,then $ 8p^2 \plus{} 1$ is prime[/quote]
In competition this problem would be a time waster. As has been shown by several counterexamples already it is not true that $ 8p^2 \plus{} 1$ is prime if $ 8p^2 \plus{} 1$ and $ p$ are relatively prime.

\[ \textrm{Consider }8p^2 \plus{} 1. \\ \\ \textrm{If }p\textrm{ and }3\textrm{ are relatively prime then }p^2\equiv 1\pmod{3}\textrm{ by Fermat's Little Theorem.}\\ \\ \textrm{It follows that }8p^2 \plus{} 1\equiv 8(1) \plus{}1\equiv 9\equiv 0\pmod{3}. \\ \\ \textrm{If } 8p^2 \plus{} 1\equiv 0\pmod{3}\textrm{ then } 3|(8p^2 \plus{}1),\text{ and if }3|(8p^2 \plus{} 1)\textrm{ then }\\ 8p^2 \plus{} 1\textrm{ is not prime.}\]

[quote="Temperal"]That is, indeed, the correct phrasing of the problem. The first step means that if $ p$ is either 1 or 2 $ \pmod 3$, then $ p^2$ will be $ 1 \pmod 3$, leading to the conclusion that $ 8p^2 \plus{} 1$ is $ 0\pmod 3$ if $ p$ isn't divisible by $ 3$. See where you can go from there.[/quote]
so it becomes 72t + 1 or 72t -1 is prime. How do you prove something is prime?

One problem I remember from the USSR olympiad problem book was "If $ p$ and $ 8p^2 \plus{} 1$ are both prime, then $ 8p^2 \minus{} 1$ is also prime." Maybe that's what was meant.

[quote]so it becomes 72t + 1 or 72t -1 is prime. How do you prove something is prime?[/quote]
In general it is very difficult to do, which is why questions like whether there are infinitely many primes of the form $ n^2 \plus{} 1$ are still open. One way to check if $ n$ is prime in a general manner (if you know some things about the number) is analyze the structure of $ \mathbb{Z} \mod n$; every element is invertible in it if and only if $ n$ is prime, for instance. (that's not what you use in this problem though; this one is a silly trick)

[quote="Temperal"]You two might consider reading the thread before posting.[/quote]

I did read the thread before I posted my comments. Granted, I missed enndb0x's correction of the problem statement, but that's irrelevant. My remarks are still valid with respect to the problem as originally stated.

As pertains to problem solving, especially where competition is concerned, it is just as important to know when a problem cannot be solved as it is to know the proper approach to use in solving it.

Well, as previously shown:

[hide]if p is not a multiple of 3, 8p^2+1 is a multiple of 3. Note that it can't equal 3, so it can't be prime. so p = 3, which solves the problem.[/hide]