calculus integration logarithms ratio AMC USA(J)MO USAMO

The increasing geometric sequence $x_{0},x_{1},x_{2},\ldots$ consists entirely of integral powers of $3.$ Given that \[\sum_{n=0}^{7}\log_{3}(x_{n}) = 308\qquad\text{and}\qquad 56 \leq \log_{3}\left ( \sum_{n=0}^{7}x_{n}\right ) \leq 57,\] find $\log_{3}(x_{14}).$

i need help with this one too. this is what i had so far...

[hide]
if you express the sequence in the form $3^{a}, 3^{b}, 3^{c},\ldots$, then you know a, b, c, ... are in an arithmetic sequence.

the first part basically tells you that the sum of the first eight powers ($a+b+c+...+h$) is 308, right?

the second part tells you that the product of those eight powers is between 56 and 57 inclusive i think.

that's where i got stuck. i don't see how that's possible.[/hide]

[quote="shake9991"]i liked this problem![/quote]

Same here :)

But it definetely was to easy for a number 12.

I would agree.... the first equation told you the sum of an arithmetic series and the second one told you one of the terms... makes it into a simple algebra problem

[quote="sapphyre571"][hide="I can not believe I missed this one too"]
Since the series $x_{0},x_{1},\ldots$ is geometric and consists of only powers of 3, let the series be $3^{a}, 3^{a+k},3^{a+2k},\ldots$.

The first sum equation tells us that $a+(a+k)+...(a+7k)=8a+28k=308$ because $x_{7}$ is the [b]eighth[/b] term, not the seventh :wallbash:

The second sum tells us that these numbers have a sum betweemn $3^{56}$ and $3^{57}$. Let's say $a+7k<56$. if this is true, there is no way to possibly satisfy the second sum, and if $a+7k>57$, there is no way to satisfy the second sum. So $a+7k=56$.

We now have
$8a+28k=308$ and $a+7k=56$. Multiplying the second equation by 8 and subtracting the first yields $28k=140\rightarrow{k}=5$

Our target is $a+14k=56+7(5)=\boxed{091}$
[/hide]
But leave it to me to miss the $x_{0}$ part and probably miss the floor because of it.[/quote]

wow. I actually did miss the floor because of it.

[quote="sapphyre571"]
The second sum tells us that these numbers have a sum betweemn $ 3^{56}$ and $ 3^{57}$. Let's say $ a \plus{} 7k < 56$. if this is true, there is no way to possibly satisfy the second sum, and if $ a \plus{} 7k > 57$, there is no way to satisfy the second sum. So $ a \plus{} 7k \equal{} 56$..[/quote]

What is the "second sum" you are referring to? Can someone elaborate this part of the solution? It seems to be the key to solving this problem.

Tachyon:
$ 56\leq\log_{3} (\sum_{n=0}^{7}x_{n} )\leq 57$
Exponentiating with a base of 3 yields:
$ 3^{56}\leq \sum_{n=0}^{7}x_n  \leq 3^{57}$
Try it now.

I thought of that specific part this way:
[hide]
As in sapphyre's post, let $ x_0 \equal{} 3^a$, and the common ratio be $ k$.
$ \log_{3}\left (\sum_{n \equal{} 0}^{7}x_{n}\right ) \equal{} \log_{3}(3^a \plus{} 3^{a \plus{} k} \plus{} ... \plus{} 3^{a \plus{} 7k})$

$ \log_{3}(3^a \plus{} 3^{a \plus{} k} \plus{} ... \plus{} 3^{a \plus{} 7k}) \equal{} a \plus{} 7k \plus{} \log_{3}(3^{ \minus{} 7k} \plus{} 3^{ \minus{} 6k} \plus{} ... \plus{} 3^{ \minus{} k} \plus{} 1)$

Since $ k$ has to be positive, $ 3^{ \minus{} 7k} \plus{} 3^{ \minus{} 6k} \plus{} ... \plus{} 3^{ \minus{} k} < \frac {1}{2}$. Therefore,

$ 0 < \log_{3}(3^{ \minus{} 7k} \plus{} 3^{ \minus{} 6k} \plus{} ... \plus{} 3^{ \minus{} k} \plus{} 1) < 1$,
so the only integral value of $ a \plus{} 7k$ that will place the sum between the right values is $ 56$. The solution follows from there.
[/hide]

I wish the average #12 was this easy. That would make USAMO placement a reasonable goal for me...

[quote="pkothari13"]The increasing geometric sequence $ x_{0},x_{1},x_{2},\ldots$ consists entirely of integral powers of $ 3.$ Given that

$ \sum_{n \equal{} 0}^{7}\log_{3}(x_{n}) \equal{} 308$ and $ 56 \leq \log_{3}\left ( \sum_{n \equal{} 0}^{7}x_{n}\right ) \leq 57,$

find $ \log_{3}(x_{14}).$[/quote]

Denote $ x_0, x_1, x_2 \dots x^7$ as $ 3^p, 3^{p \plus{} q} \dots$

Then, $ p \plus{} (p \plus{} q) \plus{} \dots \plus{} (p \plus{} 7q) \equal{} 308 \implies 8p \plus{} 28q \equal{} 308 \text{ \minus{} Equation I}$.

Also, $ 3^{56} \leq \sum_{n \equal{} 0}^{7}x_{n} \leq 3^{57}.$  (This is obtained from raising 3 to the power of both sides of the inequality.)

This means, $ 3^{56} \leq 3^p(1 \plus{} 3^q \plus{} 3^{2q} \plus{} \dots \plus{} 3^{7q}) \leq 3^{57} \text{ \minus{} Equation II.}$

The problem tells us that the powers of 3 are integral.  We will consider the last power, namely $ 3^{a \plus{} 7k}.$

By $ \text{Equation II}$, we find that the last power cannot be 57 or greater nor less than 56.  Therefore, the only integral power of $ p \plus{} 7q$ that is remotely possible is $ 3^{p \plus{} 7q} \equal{} 3^{56}.$

Taking $ \log_{3}$ of both sides, we find that $ p \plus{} 7q \equal{} 56.$

Now we have:
$ 8p \plus{} 28q \equal{} 308$ along with $ p \plus{} 7q \equal{} 56.$  Solving for $ q$ and $ p$, we get that $ p \equal{} 21$ and that $ q \equal{} 5.$

Therefore, $ a_{14}$, namely $ p \plus{} 14q \equal{} 21 \plus{} 14(5) \equal{} \boxed{091.}$

[quote="rpond"]

$ \log_{3}(3^a \plus{} 3^{a \plus{} k} \plus{} ... \plus{} 3^{a \plus{} 7k}) \equal{} a \plus{} 7k \plus{} \log_{3}(3^{ \minus{} 7k} \plus{} 3^{ \minus{} 6k} \plus{} ... \plus{} 3^{ \minus{} k} \plus{} 1)$

[/quote]

Nice solution, however I'm confused on how you got the above equation. Thanks :).

Just factor:  \[ 3^a + 3^{a+k} + \cdots + 3^{a+7k} = 3^{a+7k}(3^{-7k} + 3^{-6k} + \cdots + 1), \] then take the base-3 logarithms of both sides.

@Maxima, I liked your solution but there’s just one question I still have- why can’t p+7q be less than or greater than 56? I get that it can’t be equal to 57 because then that would not satisfy the RHS of the inequality, but why can’t it be greater than or less than 56? If the last term had a power of 3 that was less than 56, there’s still 6 more terms to go! So maybe, just maybe, their sum might still be bigger than or equal to 3^56? How do we know this won’t happen? Also, I don’t understand why p+7q can’t be more than 56…how can we be sure that there won’t be any geometric sequence that are powers of 3 that satisfy all the given constraints but has a power of 3 that’s greater than 56?

The first equation gives us $x_0x_1 \dots x_7 = 3^{308}$. Since all $x_0, x_1, \dots$ are power of three we get $x_0 = 3^r$. So this means $3^{8r + 28s} = 3^{308}$ which means $2r + 7s = 77$. Since $r, s$ are integers we know that $2r = 7(11 - s)$ which implies $s$ is odd. The second equation gives $3^{56} \le x_1 + x_2+ \dots + x_7 \le 3^{57}$.  Replacing with $r ,s$ we know that $3^{56} \le 3^{r} + 3^{r+s} + \dots + 3^{r + 7s} \le 3^{57}$. Using the geometric series formula we get $3^{56} \le \frac{3^r(3^{8s} - 1)}{3^s - 1 } \le 3^{57}.$ Now we can just casework on $s$.

Case 1: $s = 1$
This means $r = 35$. So we have $3^{21} \le \frac{3^8 - 1}{3-1} \le 3^{22}$. Clearly this isn't true.

Case 2: $s = 3$
This means $r = 28$. So we have $3^{28} \le \frac{3^{24} - 1}{3^3 -1} \le 3^{29}$. Clearly this can't be true.

Case 3: $s = 5$
This means $r = 21$. So we have $3^{35} \le \frac{3^{40} - 1}{3^5 - 1} \le 3^{36}$. Clearly $\frac{3^{40} - 1}{3^5 - 1} \le 3^{36}$. Multiplying by $3^5 - 1$ we get $3^{40} - 3^{35} \le 3^{40} - 1$ and this is clearly true. Thus our answer is $21 + 14(5) = 091$.


search
combinatorics proposed
combinatorics

Several knights participated in tournament .Knights fought with each other.We will call $ A$ a king,if for every knight $ B$, either $ A$ win $ B$,or there exist $ C$,such that $ A$ won $ C$ and $ C$ won $ B$.Prove that
1.There exist a king
2.If each knight has lost at least one battle,then there exist at least three kings.

read the hidden section in my post here:

[url]http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1779208502&t=159625[/url]

for the first part.

and for the second part use the following lemma:

[b]let $ T$ be a tournament having no vertex with indegree $ 0$.prove that if $ x$ is a king in $ T$,then $ T$ has another king in $ N^\minus{} (x)$[/b]

(here $ N^\minus{} (x)$ means the in-neighborhood of vertex $ x$,i.e. the set $ \{y\in V(G): y\to x\}$)

[quote="BaBaK Ghalebi"]read the hidden section in my post here:

[url]http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1779208502&t=159625[/url]

for the first part.

and for the second part use the following lemma:

[b]let $ T$ be a tournament having no vertex with indegree $ 0$.prove that if $ x$ is a king in $ T$,then $ T$ has another king in $ N^ \minus{} (x)$[/b]

(here $ N^ \minus{} (x)$ means the in-neighborhood of vertex $ x$,i.e. the set $ \{y\in V(G): y\to x\}$)[/quote]
Ok,i agree that the first part is very easy,it is just enough to consider the vertex with maximum degree and the second part is very simple too...but please post a complete proof,for those who didn't solve this problem.

ok,the proof of the first part is given in the link above,for the second part,first of all we will prove the lemma I mentioned in my last post:

[b]lemma.[/b]let $ T$ be a tournament having no vertex with indegree $ 0$.now let $ x$ be a king in $ T$,then $ T$ has another king in $ N^ \minus{} (x)$.

the proof of this lemma can be done in a similar way as part 1.because note that the vertices of $ T$ can be partitioned into three subsets $ \{x\},N^ \plus{} (x),N^ \minus{} (x)$.
now note that the set $ N^ \minus{} (x)$ is a tournament itself,so it has a king,let it be $ v$.
now we claim that $ v$ is also a king in $ T$.
note that $ v\in N^ \minus{} (x)$ hence there is an edge in $ T$ from $ v$ to $ x$.
and also for every $ y\in N^ \plus{} (x)$ consider the path $ vxy$ which is a path with length $ 2$ from $ v$ to $ y$.so every vertex in $ N^ \plus{} (x)$ can be reached from $ v$ by a path with length at most $ 2$,also $ v$ was a king in $ N^ \minus{} (x)$,so every vertex in $ N^ \minus{} (x)$ can be reached from $ v$ by a path with length at most 2.
hence $ v$ is a king in $ T$,as wanted.

now use this lemma another time for $ v$.so there exists a vertex $ u$ in $ N^ \minus{} (v)$ which is a king in $ T$.now note that we had $ v\in N^ \minus{} (x)$,hence $ u\neq x$,thus there are at least three different kings (x,v,u) in $ T$.

All parts have appeared before and are linked to from http://www.artofproblemsolving.com/Forum/viewtopic.php?t=162646


logarithms
inequalities
inequalities unsolved

Show that for any positive reals $ x,y,z,t$ is true that :   $ (x^xy^yz^zt^t)^4 \ge (xyzt)^{(x\plus{}y\plus{}z\plus{}t)}$.

[quote="Adriana N."]Show that for any positive reals $ x,y,z,t$ is true that :   $ (x^xy^yz^zt^t)^4 \ge (xyzt)^{(x \plus{} y \plus{} z \plus{} t)}$.[/quote]
$ \Leftrightarrow x\ln x\plus{}y\ln y \plus{} z \ln z \plus{}t\ln t \ge \frac{1}{4}(x\plus{}y\plus{}z\plus{}t)(\ln x \plus{} \ln y \plus{} \ln z \plus{}\ln t )$ - true by Chebyshev

nice...Why I dindn't think it?!        :maybe:  :maybe:


number theory
least common multiple
prime factorization

What is the least common multiple of the numbers 1332 and
888?

Factoring the numbers, we get $ 6\times222$ and $ 4\times222$. Now we just need to find $ LCM(4,6)$ which is $ 12$. The answer is thus $ 12\times222\equal{}\boxed{2664}$

ill answer this in a series of steps

[hide="ANSWER"]1.our best chance of success is to first is to factor the #'s 1332 and 888

2.when factoring these #'s we get :$ 6\times222 $ and $ 4\times222 $

3.now that we have simplified our question we can just now find the LCM  (6,4)

4.doing so, we get an answer of 12

5.and back to the prime factorization we know the #'s 222 repeat themselves so we multiply 222*12=2664

6.[hide]2664[/hide][/hide]

[hide="."][size=200][color=#BF0000]PLEASE RATE MY POST THANK YOU SO MUCH!![/color][/size][/hide]

[size=50][b][i][u]please be fair when you're rating my post and please PM me if there is a better way to explain this thank you so much!![/u][/i][/b][/size]


geometry
rectangle
geometric transformation
rotation
function
analytic geometry
combinatorics proposed

Give a one-line argument showing that it is impossible to dissect a $1 \times 2$ rectangle into finitely many smaller rectangles and reassemble the pieces (rotations are allowed) to form a $\sqrt{2} \times \sqrt{2}$ rectangle.

 ;)

--Vesselin

Let $f: \mathbb R\to\mathbb R$ be an additive function which vanishes precisely on $\mathbb Q$, and to each rectangle $\mathcal R$ with axes parallel to the coordinate axes and sides $x,y$ assign $T(\mathcal R)=f(x)f(y)$. Clearly, $T$ is additive, and it vanishes on a $1\times 2$ rectangle, but not on a $\sqrt 2\times\sqrt 2$ one.

These are $3$ lines, but now we're just splitting hairs :).


geometry
function
trigonometry
inequalities
circumcircle
cyclic quadrilateral
geometry unsolved

$A,B,C,D,E,F$ are on the circle $O$ counterclockwise. $BD$ is a diameter of O and it lies at right angles to $CF$. Also, $CF, BE, AD$ are concurrent. The $M$ is a foot of perpendicular of $B$ on $AC$, and $N$ is a foot of perpendicular of $D$ on $CE$. Prove that
\[S[MNC]\leq\frac{S[ACEF]}{2}\]
(Function $S$ is the area function.)

Quadrilateral ACEF is cyclic, hence $\angle F = 180^\circ-\angle C,$ $\sin F = \sin C.$ The quadrilateral area is

$S(ACEF) = \frac{1}{2}\ CA \cdot CE\ \sin C+\frac{1}{2}\ FE \cdot FA\ \sin F =$

$= \frac{1}{2}\ (CA \cdot CE+FE \cdot FA) \sin C$

On the other hand, area of the triangle $\triangle MCN$ is

$S(MCN) = \frac{1}{2}\ CM \cdot CN\ \sin C$

The proposed inequality $S(MCN) < \frac{1}{2}\ S(ACEF)$ is actually sharp. Thus we have to show that

$(?)\ \ \ 2\ CM \cdot CN < CA \cdot CE+FE \cdot FA$

Since $CF \perp BD,$ B, D are midpoints of the circumcircle arcs CF containing A, E, respectively. By the problem conditions, C lies on the opposite side of the circumcircle diameter BD than A, E, F, hence $CA > CB = FB > FA$ and $CE > CD = FD > FE.$ Since in addition $BM \perp CA,\ DN \perp CE,$ by Archimedes theorem (see the problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=115281]External bisector[/url]), M cuts in half the polygonal path CAF, while N cuts in half the polygonal path CEF. Thus $2\ CM = CA+FA,$ $2\ CN = CE+FE.$ Substituting this to the equation in question,

$(?)\ \ \ (CA+FA) \cdot (CE+FE) < 2\ (CA \cdot CE+FE \cdot FA)$

As a result, we no longer have to consider the triangle $\triangle MCN.$ Multiplying the left side out and canceling equal terms on both sides, this becomes

$(?)\ \ \ CA \cdot FE+CE \cdot FA < CA \cdot CE+FE \cdot FA$

Rearranging and factoring,

$(?)\ \ \ (CA-FA)(CE-FE) > 0$

Since $CA > FA,\ CE > FE$ (see above), the last inequality trivially holds.

[b]Comment:[/b] The concurrence of CF, BE, DA is not necessary and it was not used. Since B, D are midpoints of the circumcircle arcs CF containing A, E, respectively, AD bisects $\angle A,$ while EB bisects $\angle E$ of the cyclic quadrilateral ACEF. Assuming CF, BE, DA concur at P, we have $\frac{CA}{FA}= \frac{CP}{FP}= \frac{CE}{FE}$ and it follows that $MN \parallel AE.$ But again, it is not necessary and it was not used.


vector
geometry
parallelogram
analytic geometry
graphing lines
slope

This is problem 158 chapter 10 vol 2 aops.
Three vertices of parallelogram PQRS are p(-3,-2), q(1,-5), r(9,1) with P and Q diagonally opposite. What is the S coordinate?

First, P and Q being diagonally opposite, does that mean they are opposite angles of the parallelogram?

If my above question is yes, then i dont see how the book solution (i think it said (5,4), but i dont have the solution guide with me.) is wrong i think.

This is what i did:
I set it up as two circles, one
(x+3)^2 + (y+2)^2 = 100.
The other is (x-1)^2 + (y+5)^2 = 153. The first one is centered at P and the second at Q and this would determine the intersection line of the two.

Solving, I get -4x + 3y = 20. Now I need to find another equation of a line to find where point S is. I know that from point Q, there is going to be a 1/4 slope till point S because the slope will be equal to the slope of R to P.
y = x/4 + b.   Plugging Q in, -5 = 1/4 + b, and b = -21/4.
so y = x/4 -21/4, and -x + 4y = -21.
Solving both equations found, i get the intersection at (-11,-8). First, I see no reason at all for this answer to be positive like the book says. And second, I see no reason why I would be wrong.
Help is appreciated, and I thank you in advance.

Gah - you caught us on a typo  :oops: .  Kind of surprised no one has mentioned before.  Problem (and first line of solution) should have had P and R diagonally opposite.

(Yet another reason we're glad we started this forum...  Can clear up typos...)

Go back through the problem (as stated in the book) and do it using the vector approach (as described in the solutions, but for the case of P and R being diagonally opposite).  It's much faster.  You'll know when you've done it right.


Find the Difference Quotient of f, that is, find
f(x + h) - f(x)/h, where h does not = 0, for each function. Be sure to simplify.

NOTE: This is a precalculus question. Please, do not answer the question using calculus concepts.

(1) f(x) = -3x + 1

(2) f(x) = 1/(x + 3)

Post some progress or an attempt first, then we'll help you. We can help with understanding concepts related to homework, but we won't do it.


AMC
USA(J)MO
USAMO
trigonometry
calculus
geometry
linear algebra

What type of calculator restrictions are there for the USAMO and the IMO??? I want to start looking more seriously into these tests and I want to know what instruments I can use.

wait, are you even allowed calculators on the USAMO?
i thought you weren't...

that would be an answer :\, i didn't state there was.. for all I know, you aren't allowed to use one, this is what I am trying to figure out.

No calculators.

[quote]No notes, books, headphones, slide rules, mathematical tables, calculators or calculator watches are allowed during the exam. [/quote]

(from http://www.unl.edu/amc/d-publication/d-1-pubarchive/2002-3pub/03usamo/html/04-03usamo-text.html )

You can not even use calculators on the AIME.(Although you probably already knew that)

No calculators are allowed for the IMO ... :D

I think the only famous competitions where calculators allowed are AMC and part of MC...

Can you use a calculator in the AMC ? Didn't know that :?

Do some of you know the TI 200 ?
I won a TI 200 with our national olympiad and it's really an incredible thing :D

Is it a Voyager 200?

You can use the 89 on the AMC

I hadn't heard of the TI-200 until you mentioned it, Arne. Would you mind telling us about it?

It's a "voyager" indeed :D There are lots of interesting programs on it such as mathcad, derive, cabri, geometer's sketchpad, ... It has a special calculus & linear algebra program. ... and lots of memory !

[color=cyan]So it's really a computer, not a calculator?  QWERTY keyboard and such?[/color]

[url=http://education.ti.com/us/product/tech/89/features/features.html]Explanation of Voyage 200[/url]
[url=http://epsstore.ti.com/webs/catlist.asp?deptid=609&catid=902]$200@TI site[/url]

[url=http://www.artofproblemsolving.com/wforum/viewtopic.php?p=14418#14418]My attempt at making a post entirely linked :) (Even the emoticon's linked)[/url]

[quote="Tare"]Yes you can. You can use any calculator permitted in SAT, so you could use TI-83 se, which is pretty useful when it comes to logs and trigs and such.
I'm not sure about 89 though.[/quote]

For AMC,

89 = yes
Voyager/200 = no

If I remember correctly (yeah, I sound like a really old, old person...), the Voyager isn't allowed because it has a QWERTY keypad.  This is how they distinguish between "calculator" and "computer" - that is, you can't use a laptop, either.

But the '89 is allowed.  I believe the AMC allows the same calculators as the SAT in order to be consistent.

Voyage 200 does have a lot of memory for programs, but the bad thing is... It's probably not allowed in most competitions. They won't let you use anything with a qwerty keyboard. Voyage 200 has that. Most of the apps that runs on the voyage 200 will run on TI-89. So I suggest getting a TI-89. It's very helpful. I also heard some HP graphing calculators is even more powerful then a TI-89.

Really? Calculators aren't allowed in IMO? I can see how calculators are problems. For example if they want you to calculate the digits in for example 299!; you can enter it on your TI-89 and get the exact digits.

Well IMO's are like proofs (and correct me if I'm wrong) and hard geometry problems so calculation might be not required at all/required very little.
Anyway if a person is good enough to go to the IMO I'll bet that anyone of them could do long division :)

That isn't necessarily true. There are myriads of anecdotes about the greatest mathematicians in history who were unable to do basic arithmetic. If you're proving difficult theorems, you might not need to know basic arithmetic.


modular arithmetic
induction
number theory

Okay I just learned this cool modular arithmetic trick from ernie on the post titled remainder, and I want to see if I'm doing this correctly. If you know how to solve the problem, then please tell me if I did it right.

Problem;
What is the remainder when [125^(625)]/6

My Solution;

125^625 (mod 6) = [125^(104 + 1)] (mod 6)
125^105 (mod 6) = [125^(17 + 3)] (mod 6)
125^20 (mod 6) = [125^(3 + 2)](mod 6)
125^5 (mod 6) = 5^15 (mod 6) = [5^(2 + 3)] (mod 6)
5^5 (mod 6) = 3125 (mod 6)
5 (mod 6)

So the remainder is 5.

That's correct.
[hide="A faster way"]$ 125^{625} \pmod 6 \equiv (\minus{}1)^{625}\pmod 6 \equiv \minus{}1\pmod 6 \equiv \boxed{5}\pmod 6$.[/hide]

[hide="This also works, I think?"]
\[ 625\equiv{1}\pmod{6}\\
125\equiv{5}\pmod{6}\\
5^1\equiv{\fbox{5}}\pmod{6}\\
\]
But I'm new to modular arithmetic so I'm not sure.
Anyway, RisingFlame's solution is probably the easiest one.
[/hide]

[quote="grn_trtle"][hide="This also works, I think?"]
\[ 625\equiv{1}\pmod{6} \\
125\equiv{5}\pmod{6} \\
5^1\equiv{\fbox{5}}\pmod{6} \\
\]
But I'm new to modular arithmetic so I'm not sure.
Anyway, RisingFlame's solution is probably the easiest one.
[/hide][/quote]
it's not right.

[quote="Brut3Forc3"][quote="grn_trtle"][hide="This also works, I think?"]
\[ 625\equiv{1}\pmod{6} \\
125\equiv{5}\pmod{6} \\
5^1\equiv{\fbox{5}}\pmod{6} \\
\]
But I'm new to modular arithmetic so I'm not sure.
Anyway, RisingFlame's solution is probably the easiest one.
[/hide][/quote]
No, that doesn't work. Sorry.
Basically, you can't take the mod off of an exponent...
There are ways to do this, but it uses Fermat's little theorem.[/quote]

:lol: Didn't think so.
What about this:
\[ 125\equiv{5}\pmod{15} \\
5^{625}\equiv{25^{312}*5}\pmod{15} \\
25\equiv{10}\pmod{15} \\
10^{312}\equiv{10}\pmod{15} \\
10*5\equiv{50}\equiv{\fbox{5}}\pmod{15}
\]
Which sort of uses my newly invented conjecture that $ 10^n\equiv{10}\pmod{15}, {n\in{\mathbb{Z}}}$  :o

[quote="grn_trtle"]
Which sort of uses my newly invented conjecture that $ 10^n\equiv{5}\pmod{15}, {n\in{\mathbb{Z}}}$  :o[/quote]
:) Did you check $ n \equal{} 2$?

Sorry, sorry, typo.
I meant $ 10^n\equiv{10}\pmod{15}$, not $ \equiv{5}$

How can we prove that $ 10^n \minus{}10 \equiv 0 \mod 15$?
[hide="hint"]Consider that it must be divisible by both 5 and 3.[/hide]

I'm going to try it without looking at the hint :)
It suffices to show that $ 15|(10^n - 10), \forall{n}\in\mathbb{Z}^ +$
Let $ P(n) = 10^n - 10$
$ P(1) = 10 - 10 = 0$, and $ 15|P(1)$ because everything divides 0.
$ P(2) = 100 - 10 = 90$, and $ 15*6 = 90$, so this is true.
And, making the assumption that $ P(k)$ is true, we can show that $ P(k)$ implies $ P(k + 1)$
\begin{eqnarray*} P(k + 1) & = & 10^{k + 1} - 10 \\
& = & 10(10^k - 1) \\
& = & 10(P(k) + 9) \\
& = & 10*P(k) + 90 \end{eqnarray*}
Now, because we assumed that $ 15|P(k)$, we know that $ 15|(10*P(k))$. We also know that $ 15|90$ by considering $ P(2)$. Adding a multiple of 15 to another multiple of 15 results in a multiple of 15.

Thus, by induction, $ 15|(10^n-10), \forall{n}\in{\mathbb{Z}^ + }$, Q.E.D.

(And yes, I'm sure there's a much more elegant way to do it than induction.)

It's correct, but you used $ P(n)$ both as a number and as a statement.
(in the line "And, making the assumption that $ P(n)$ is true...)

Thanks for pointing that out :). Taken care of.

i think this is simpler.
powers of 10 give remaindre 10 u can observe.
so, 10^n = 10(mod 15)
implies, 10^n-10=0(mod 15)
hence, proved.

[quote="analytic"]
powers of 10 give remaindre 10 u can observe.[/quote]

I think archimedes1 was trying to get me to prove that $ 10^n\equiv{10}\pmod{15}$.
To prove that $ 10^n\minus{}10\equiv{0}\pmod{15}$ you must prove that $ 10^n\equiv{10}\pmod{15}$... you can't just "observe" or it wouldn't be a proof so much as a hypothesis.

remainder 10^1 =10(mod 15)
remainder 10^2= 10*10=100=10(mod 15)
this chain will continue till infinity.just like u have non terminating recurring decimals.

Yes, the premise behind my proof by induction was to show that that chain continues infinitely.

What about this for a proof?

[hide]We need to show that $ 10^n - 10 \equiv 0 \mod 15$

Since $ 15 = 3 \times 5$ , and 3 and 5 are both prime, it is sufficient to show that $ 10^n - 10 \equiv 0 \mod 3$ and $ 10^n - 10 \equiv 0 \mod 5$.

$ 10^n - 10 \equiv 1^n - 1 \mod 3$
$ 10^n - 10 \equiv 0 \mod 3$

$ 10^n - 10 \equiv 0^n - 0 \mod 5$
$ 10^n - 10 \equiv 0 \mod 5$

Therefore, $ 10^n - 10 \equiv 0 \mod 15$ and $ 10^n \equiv 10 \mod 15$.

But about the $ {n\in{\mathbb{Z}}}$, I'm not too sure...[/hide]

And Wickedestjr, can you post the link with the modular arithmetic trick?
Thanks

Nice proof  :lol: I would have done something more creative if I hadn't been so tired.
$ n\in\mathbb{Z}$ simply means $ n$ in the set of all integers. (Basically, defines $ n$ as an integer).

N is the set of Natural numbers.
Z is the set of Integers.
Q is the set of Rational Numbers.
R is the set of Real Numbers.

For those that don't know.

[quote="Poincare"][quote="RisingFlame"]That's correct.
[hide="A faster way"]$ 125^{625} \pmod 6 \equiv ( \minus{} 1)^{625}\pmod 6 \equiv \minus{} 1\pmod 6 \equiv \boxed{5}\pmod 6$.[/hide][/quote]
i am not very good with modular arithmetic so, could you please explain your solution?[/quote]
divide 125 by 6 , y'll get 5 or -1 in terms of 6. replace 125 by -1 and rest is simple.

[quote="ernie"][quote="mathemonster"]Wickedestrjr will you teach me that method of Ernies? Now well be even because ive helped you.[/quote]

Why learn it from Wickedestjr, when you could learn it from ernie himself!

Ok, sample problem.

What is $ 6^{25432} \text{ mod } 7$?

First, you take $ 25432$, and divide it by $ 7$, to get $ 3633 \frac {1}{7}$.  Take $ 3633 \plus{} 1$.

What is $ 6^{3634} \text{ mod } 7$?

Take $ 3634$ and divide it by $ 7$, to get $ 519 \frac {1}{7}$.  Take $ 519 \plus{} 1$.

What is $ 6^{520} \text{ mod } 6$?

Take $ 520$ and divide it by $ 7$, to get $ 74 \frac {2}{7}$.  Take $ 74 \plus{} 2$.

What is $ 6^{74} \text{ mod } 7$?

Take $ 74$ and divide it by $ 7$, to get $ 10 \frac {4}{7}$.  Take $ 10 \plus{} 4$.

What is $ 6^{14} \text{ mod } 7$?

Take $ 14$ and divide it by $ 7$, to get $ 2 \frac {0}{7}$.  Take $ 2 \plus{} 0$.

What is $ 6^2 \text{ mod } 7$?

Since it is obviously cannot be simplified any more, take $ 36 \text{ mod } 7$, to get the answer of $ \boxed{1}$.

There are many easier ways. xD[/quote]

Would if it is something like;

x^y (mod z), when they are all very large numbers like 37, 81, and 93.

Solve this;

37^81 (mod 93) That procedure doesn't always work.

Why does the ernie method work?
Yes it can't be used on $ 37^{81}\pmod{93}$
how would u solve this one?

This is exactly what I couldn't figure out in my other modular arithmetic post.


I know I may break some rules when I post this post,so if necessaryly medortor can delete it.
It is  a solution to crux 2950,the problem appears in crux 2005.4 again,it is unsolved now.
But when I sent my solution to the editor,he said he can't understand what I wrote.I am so sorry,and may there is something unclear.I don't what can I do with it.So I post it here,I hope somebody can help me :)

I'm sorry but I don't understand even the first line of the problem: let $ABC$ be a triangle whose largest angle does not exceed , for , Can you please write more clearly what you are going to say.

Sorry,I forget "$\frac{2\pi}3$ there,now I've edited my file.
Thank you

Hi,you have a nice solution. :)
Did you think about generalizations? :P
Compare $\sum sin\frac{A}{2}.sin\frac{B}{2}$ and $x+y.\sum cos\frac{A}{2}.cos\frac{B}{2}$


function
calculus
derivative
limit
absolute value
real analysis
real analysis unsolved

Let $f: R^2 \to R$ be defined by:

[img]http://img211.imageshack.us/img211/3874/image0102lr.gif[/img]

Show that $f$ is not differentiable at $(0,0)$

How would you do this question? I am having trouble dealing with the $|y|$ and showing that the function is indeed not differentiable at $(0,0)$

The partial derivatives are zero, so if it were differentiable the derivative would be zero and $\frac{f(x,y)}{\|(x,y)\|}=\frac{x|y|}{x^2+y^2}$ would tend to zero at zero.

The absolute value is just there so that the directional derivatives exist.

I am quite confused about what you wrote, sorry. So the function IS differentiable at 0? Or were you just suggesting to use a proof by contradiction showing that $f$ is not differentiable at 0?

$\frac{\partial f}{\partial y}$ and $\frac{\partial f}{\partial x}$ exist and are zero at $x=0$.  So if $f$ were differentiable, then we would have to have

$\lim_{\|(x,y)\|\to 0} \frac{f(x,y)-f(0,0)}{\|(x,y)\|}=0$

But this is not true (take $x=y$ and let $x,y\to 0$)...


arithmetic sequence
geometric sequence
geometric series
arithmetic series

Would someone please refer some sites or ebooks where I can read information about sequences in general and about geometic and arithmetic progressions? My portuguese textbooks are too cold, if I make me understand...

Thank you...  :lol:

Always start at Wikipedia or MathWorld. :D
[url=http://en.wikipedia.org/wiki/Arithmetic_progression]Arithmetic Progression - Wikipedia[/url]
[url=http://en.wikipedia.org/wiki/Geometric_progression]Geometric Progression - Wikipedia[/url]
[url=http://mathworld.wolfram.com/GeometricSeries.html]Geometric Series - MathWorld[/url]
[url=http://mathworld.wolfram.com/ArithmeticSeries.html]Arithmetic Series - MathWorld[/url]


[quote="aznness"]Phelpedo, who's your brother?

Alex is right, i did exaggerate a bit. but hey, we are still going to own you guys! And whats with Texas. Just because bush won the election doesnt mean texas owns the world. CAlifornia has the most electoral votes. EAT THAT!![/quote]

Please note that bush is not capitalized

[quote="NoSoupForYou"]Does 'bush' have to be capitalized?[/quote]

Joseph Chu was capitalized... while bush wasn't... does that tell you something?

[quote="NoSoupForYou"]not really[/quote]

Joseph Chu>bush

[quote="NoSoupForYou"]$x>bush \text{ where } [x|x\text{ is any human, some animals, and most inanimate objects}]$[/quote]
nice one :D
wow a lot of people here dont' like bush, myself included.


Let $ x_n$ such that
$ x_n\equal{}\sum\limits_{k\equal{}1}^{n} \frac{1}{(k!)^2}$
Prove that a) this sequence has lim ( it is easy)
                b) This lim is not a rational number (it is hard)
Help help

a) note that the sequence is increasing and is bounded(for example by $ e$). this implies that ti has a limit.
b) suppose it is equal to $ x\equal{}\frac{p}{q}$
$ x(q!)^2\equal{}n\plus{}\left(\frac{1}{(q\plus{}1)^2}\plus{}\frac{1}{(q\plus{}1)^2(q\plus{}2)^2}\plus{}\cdots\right)$
where $ n\in \mathbb{Z}$
but $ 0<x(q!)^2\minus{}n\le \sum_{i\equal{}1}^{\infty}\frac{1}{(q\plus{}1)^{2i}}\equal{}\frac{1}{q(q\plus{}2)}<1$
which is impossible  :wink:


What is the greatest number of pieces into which a circular pizza can be cut using five straight edge-to-edge cuts?

0 cuts: 1 piece
1 cut: 2 pieces
2 cuts: 4 pieces
3 cuts: 7 pieces
4 cuts: 11 pieces
5 cuts: 16 pieces

Or, in general, it's the nth triangular number plus 1.

Answer is 16.

This is Wikipedia's proof. If you want to learn more about this sequence, just google Lazy Caterer's Sequence!

[quote=Wikipedia: Lazy Caterer's Sequence] When a circle is cut $n$ times to produce the maximum number of pieces, represented as $p = f(n)$, the $n$th cut must be considered; the number of pieces before the last cut is $f(n-1)$, while the number of pieces added by the last cut is $n$.

To obtain the maximum number of pieces, the $n$th cut line should cross all the other previous cut lines inside the circle, but not cross any intersection of previous cut lines. Thus, the $n$th line itself is cut in $n-1$ places, and into $n$ line segments. Each segment divides one piece of the $(n-1)$-cut pancake into $2$ parts, adding exactly $n$ to the number of pieces. The new line can't have any more segments since it can only cross each previous line once. A cut line can always cross over all previous cut lines, as rotating the knife at a small angle around a point that is not an existing intersection will, if the angle is small enough, intersect all the previous lines including the last one added.

Thus, the total number of pieces after n cuts is

$f(n)=n+f(n-1).\,$
This recurrence relation can be solved. If $f(n-1)$ is expanded one term the relation becomes

$f(n)=n+(n-1)+f(n-2).\,$
Expansion of the term $f(n-2)$ can continue until the last term is reduced to $f(0)$, thus,

$f(n)=n+(n-1)+(n-2)+\cdots+1+f(0).\,$
Since $f(0)=1$, because there is one piece before any cuts are made, this can be rewritten as

$f(n)=1+(1+2+3+\cdots+n).\,$
This can be simplified, using the formula for the sum of an arithmetic progression:

$f(n)=1+\frac{n(n+1)}{2}=\frac{n^2+n+2}{2}.$ [/quote]


The separator inside the box designates the median in a box-and-whisker plot, in this case, $ \boxed{8}$.


trigonometry
inequalities
triangle inequality
inequalities unsolved

[quote="warut_suk"]Let $A,B,C$ be angles of a triangle, prove that
$-1 < \frac{\sin{A}}{\sin{B}}+\frac{\sin{B}}{\sin{C}}+\frac{\sin{C}}{\sin{A}}-\frac{\sin{B}}{\sin{A}}-\frac{\sin{C}}{\sin{B}}-\frac{\sin{A}}{\sin{C}} <1$[/quote]

Let a, b, c be the sides of the triangle. Then, by the sine law, $\frac{\sin A}{\sin B}=\frac{a}{b}$, $\frac{\sin B}{\sin C}=\frac{b}{c}$, $\frac{\sin C}{\sin A}=\frac{c}{a}$, $\frac{\sin B}{\sin A}=\frac{b}{a}$, $\frac{\sin C}{\sin B}=\frac{c}{b}$ and $\frac{\sin A}{\sin C}=\frac{a}{c}$. Hence, your inequality becomes

$-1<\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-\frac{b}{a}-\frac{c}{b}-\frac{a}{c}<1$.

This is equivalent to

$\left|\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-\frac{b}{a}-\frac{c}{b}-\frac{a}{c}\right|<1$.

But this is clear, since

$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-\frac{b}{a}-\frac{c}{b}-\frac{a}{c}=\frac{\left(b-c\right)\left(c-a\right)\left(a-b\right)}{abc}$,

and thus

$\left|\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-\frac{b}{a}-\frac{c}{b}-\frac{a}{c}\right|=\left|\frac{\left(b-c\right)\left(c-a\right)\left(a-b\right)}{abc}\right|$
$=\frac{\left|b-c\right|\cdot\left|c-a\right|\cdot\left|a-b\right|}{abc}$
$<\frac{a\cdot b\cdot c}{abc}$         (since |b - c| < a, |c - a| < b and |a - b| < c by the triangle inequality)
$=1$.

  Darij


Cho $ a,b,c \in [0,1]$  and $ a\plus{}b\plus{}c\equal{}1$.Prove that !
$ \sum \frac{1}{1\plus{}x^2} \ge \frac{27}{10}$

[quote="bigbang195"]Cho a,b,c \in [0,1]  and a+b+c=1.Prove that !
\sum \frac{1}{1+x^2} \ge \frac{27}{10}[/quote]

Translated into English from Vietnamese.

Let $ a,\ b,\ c$ be real numbers such that $ a,b,c \in [0,1]$ and $ a\plus{}b\plus{}c\equal{}1$.

Prove that $ \frac{1}{1\plus{}a^2}\plus{}\frac{1}{1\plus{}b^2}\plus{}\frac{1}{1\plus{}c^2}\geq \frac{27}{10}.$

[quote="kuing"][quote="kunny"][quote="bigbang195"]Cho a,b,c \in [0,1]  and a+b+c=1.Prove that !
\sum \frac{1}{1+x^2} \ge \frac{27}{10}[/quote]

Translated into English from Vietnamese.

Let $ a,\ b,\ c$ be real numbers such that $ a,b,c \in [0,1]$ and $ a \plus{} b \plus{} c \equal{} 1$.

Prove that $ \frac {1}{1 \plus{} a^2} \plus{} \frac {1}{1 \plus{} b^2} \plus{} \frac {1}{1 \plus{} c^2}\geq \frac {27}{10}.$[/quote]

then use tangent line method can easy to prove that

$ \frac {1}{1 \plus{} a^2} \plus{} \frac {1}{1 \plus{} b^2} \plus{} \frac {1}{1 \plus{} c^2}\le \frac {27}{10}.$[/quote]

oh yes $ \le$

[quote="kuing"][quote="kunny"][quote="bigbang195"]Cho a,b,c \in [0,1]  and a+b+c=1.Prove that !
\sum \frac{1}{1+x^2} \ge \frac{27}{10}[/quote]

Translated into English from Vietnamese.

Let $ a,\ b,\ c$ be real numbers such that $ a,b,c \in [0,1]$ and $ a \plus{} b \plus{} c \equal{} 1$.

Prove that $ \frac {1}{1 \plus{} a^2} \plus{} \frac {1}{1 \plus{} b^2} \plus{} \frac {1}{1 \plus{} c^2}\geq \frac {27}{10}.$[/quote]

then use tangent line method can easy to prove that

$ \frac {1}{1 \plus{} a^2} \plus{} \frac {1}{1 \plus{} b^2} \plus{} \frac {1}{1 \plus{} c^2}\le \frac {27}{10}.$[/quote]
easy to show that:
$ \frac{1}{1\plus{}{x}^{2}}\leq \frac{27}{50}\left(2\minus{}x \right)$
$ \frac{1}{1\plus{}{y}^{2}}\leq \frac{27}{50}\left(2\minus{}y \right)$
$ \frac{1}{1\plus{}{z}^{2}}\leq \frac{27}{50}\left(2\minus{}z \right)$
Add them up,we have Q.E.D :)

[quote="bigbang195"]Cho $ a,b,c \in [0,1]$  and $ a + b + c = 1$.Prove that !
$ \sum \frac {1}{1 + x^2} \ge \frac {27}{10}$[/quote]

$ \frac {5}{2} = \sum_{cyclic} \frac {2 - x}{2}\le \sum_{cyclic} \frac {1}{1 + x^2} \le \sum_{cyclic} \frac { - 27x + 54}{50} = \frac {27}{10}$

$ RHS$ The proof is completed. Equality holds if and only if $ x=y=z=\frac{1}{3}$

$ LHS$ The proof is completed. Equality holds if $ {x=y=0\\z=1\\Cyclic}$


A charity sells 140 benefit tickets for a total of 2001 dollars. Some tickets sell for full price (which is a whole number amount), and the rest sell for half price. How much money is raised by the full-price tickets?

[hide]

Let a be the price of a full price ticket, and let x be the number of full price tickets sold.

We're therefore trying to find ax.

$140-x=y$
 $ax + a/2 * y = 2001$
 $ax + a/2 * (140-x) = 2001$
$ax + 70a -ax/2 = 2001$
$ax/2 + 70a = 2001$
 $ax = 4002 - 140a$ [/hide]

Eh... can't see what to do after there. What should I do, and/or where did I mess up?

Try to rewrite the equation so it says

[hide]$px+0.5p(140-x)=2001$, where p is the price of the full price ticket, x is the amount of full price payers, and that can be factored to get $p(0.5x+70)=2001$. Remember that p has to be a whole number.... [/hide]

Does the half price have to be a whole number because 2001 was factorable by 29,23,and 3

[hide]
$p$=price of ticket
$x$=full price tickets
Since 2001 factors into 3x23x29, $(x+\frac{140-x}{2})p$ p has to be either 3,6,23,46,29, or 58.  If all of the tickets were full-price, and it cost 6 for a full-price ticket, the amount paid would be 1040.  That eliminates 3 and 6.  If all of the tickets were half-price, and it cost 46 for a full-price ticket, the amount paid would be 3220, too much.  That eliminates 46 and 58.  We are left with 23 and 29.  140 half-price tickets at $\frac{29}{2}$ dollars each would make 2030 dollars, too much.  We are left with 23 dollars for each full-priced ticket.

$(x+\frac{140-x}{2})23=2001$
$x+\frac{140-x}{2}=87$
$x+140=174$
$x=34$ <-full priced tickets

Multiply by price of ticket:
$34\cdot 23=782$

So 782 dollars.
[/hide]


The rent-a-stall horse barn has stalls for 1000 horses. Forty percent of the stalls are for ponies. On Tuesday, there were 200 ponies and a bunch of quarter horses at the horse barn. The horse barn was 75 percent full.
How many quarter horses were in the stalls?

[hide="answer"]
There were 550 quarter horses.
20% of 1000 = 200
55% of 1000 = 550[/hide]


linear algebra
matrix
complex numbers
linear algebra unsolved

If $ A\in M_{n}(\mathbb{R})$ and $ A^T\equal{}A$

Then there exists $ P \in M_{n}(\mathbb{R})$,  $ PP^T\equal{}I_n$

such that: $ P^TAP\equal{}\begin{pmatrix}
\lambda_1 &  &  & \\
 & \lambda_2 &  & \\
 &  &  ...& \\
 &  &  & \lambda_n
\end{pmatrix}$

[color=red]My problem is:[/color]

If $ A \in M_{n}(\mathbb{K})$ and $ A^T\equal{}A$

can we have some similar conclusion?

Your proposition would imply that if $ A$ is symmetric then $ A$ is diagonalizable.
It is false over the complex numbers because any complex matrix is similar to a symmetric matrix.


pigeonhole principle
function
calculus
integration

Source: Past MMPC Part 2 Exams
Don't hesitate to answer if you know; there will be lots of problems. Solutions are, of course, necessary.

2002 1)
a) Show that for every positive integer m>1 there are positive integers x and y such that x^2 - y^2 = m^3.
b) Find all pairs of positive integers (x,y) such that x^6 = y^2 +127

2002 5)
Each of the first 13 letters of the alphabet is written on the back of a card and the 13 cards are placed in a row in the order
A,B,C,D,E,F,G,H,I,J,K,L,M
The cards are then turned over so that the letters are face down. The cards are rearranged and again placed in a row, but of course they may be in a different order. They are rearranged and placed in a row a second time and both rearrangements were performed exactly the same way. When the cards are turned over the letters are in the order
B,M,A,H,G,C,F,E,D,L,I,K,J
What was the order of the letters after the cards were rearranged the first time?

2000 5)
The Olive View Airline in the remote country of Kuklafrania has decided to use the following rule to establish its air routes: If A and B are two distinct cities, then there is to be an air route connecting A with B either if there is no city closer to A than B or if there is no city closer to B than A. No further routes will be permitted. Distances between Kuklafranian cities are never equal. Prove that no city will be connected by air routes to more than ve other cities.

1998 1)
An organization decides to raise funds by holding a $60 a plate dinner. They get prices from two caterers. The first caterer charges $50 a plate. The second caterer charges according to the following schedule: $500 set-up fee plus $40 a plate for up to and including
61 plates, and $2500 log (p/4) for p > 61 plates.
a) For what number of plates N does it become at least as cheap to use the second caterer as the first?
b) Let N be the number you found in a). For what number of plates X is the second caterer's price exactly double the price for N plates?
c) Let X be the number you found in b). When X people appear for the dinner, how much profit does the organization raise for itself by using the second caterer?

1. [hide]a) x=(m+m:^2:)/2, y=(m:^2:-m)/2. Note that those will indeed be integers as m and m:^2: have the same parity. b)(x:^3:-y)(x:^3:+y)=127 (prime), so x:^3:-y=1 and x:^3:+y=127=> x:^3:=64, y=63 => (x,y)=(4,63)[/hide]

3. [hide]Assume that city A is connected to six cities, named B, C, D, E, F, and G. Connect A to these 6 cities. By pigeonhole, one of the angles between two cities and A (such as BAC, CAD, etc.) will be :le:60. Let the other two cities at the endpoints of this angle be B and C wlog. Then either ABC>60 or ABC<60. ABC cannot be equal to 60 b/c then the triangle would be equilateral, violating the assumption that all distances between cities are different. In the first case above, there should not be a road between A and C b/c AB<AC and BC<AC. The second case is dealt with similarly.[/hide]

4. [hide]a)let's call the first caterer's price f(n) (n is the number of plates) and the second caterer's price g(n). Clearly h(n)=g(n)-f(n) is a decreasing function of n, so as h(61)=-110<0, we don't have to worry about that freaky log part of g(n) right now. anyway, we want the minimal n for which 500+40n:le:50n=>500:le:10n=>n:ge:50, so the smallest integral n is 50.

b)ok, we want the x for which g(x)=2*(2500)=5000. g(61)=2940, so as g(n) is an increasing function of n, we do have to worry about that freaky log part. okay, if y=x-61, g(x)=2940+2500 log (y/4). if this is to equal 5000, 2940+2500 log (y/4)=5000=> log (y/4)=.824 => y/4=10^(.824)=6.668 => y=27.672 okay so the smallest integral x is thus 89.

c) we want 60*89-g(89)=5340-2940-2500log(7)=2400=$287.25[/hide]

2. [hide]1 a 3 b 6 c 7 d 5 e 8 f 4 || g 9 h 11 i 12 j 10 k 13 l 2 m. in the above string of letters and numbers, the numbers refer to the places of the letters in the stack of cards, while the letters are variables. for example, the first three characters represent that the card in the first position is mapped to something in the "a"th position, which will then be mapped to the third position. Notice that we have 26 characters up there, but we only have cycles of 13 characters. Note also that the double vertical lines separate the two sets of 13 characters. Thus, they must be the same cycle, i.e. 1=g, a=9, 3=h, etc. This makes the cycle as follows:

1 9 3 11 6 12 7 10 5 13 8 2 4, meaning that ABCDEFGHIJKLM is mapped to DHIBJKLMAGCFE. You can indeed check that after another reiteration of this cycle, DHIBJKLMAGCFE is mapped to BMAHGCFEDLIKJ[/hide]

fun contest. when do we get a second round?

How did you think to do that for number 2? That is a very interesting method. Have you seen that used somewhere before?

no, it just seemed logical to do it that way. I am not entirely sure that it is rigorous, as I never explicitly stated (proved) that the cycles could not be of length <13, which would obviously detract from the method, but besides that little thing, it just seems logical to write out the sequence like that.

Second round will be up as soon as I get the problems together. I'm starting now.


In a tri. ABC, ang.B = 90 deg. ;  ang.A = 30  and if the hypotenuse AC = 10 cm then find the other angles.

im trying to prevent myself from thinking that this is a trick question.

[hide]sum of angles in triangle = 180
90 + 30 + x = 180
x = 60[/hide]

60. This barely qualifies as a math problem, certainly not one for competitions.

i am sorry ......the question is find the remaining sides... :(

This is a famous "30-60-90" triangle, whose ratio of side lengths is $1: \sqrt{3}: 2$. To prove this, consider an equilateral triangle, and cut it into two 30-60-90 triangles. (You need to use Pythagorean Theorem to finish the proof.)

So the remaining side lengths are $AB=5\sqrt{3}$ and $BC=5$.

I think this thread should be moved to middle school forum.


limit
real analysis
real analysis unsolved

$x_1=\frac 13\ ,\ x_2=\frac 12\ ,\ x_3=1\ ,\ x_4=\ \frac 10\ ?$ You are cunning !
In my opinion this sequence hasn't sense. Can you think that $x_n=\frac{1}{4-n},\ \left(\forall\right)\ n\in N^*,\ n\ne 4\ !$ Maybe possible, because from a sequence we can eliminate a finite number of terms. In this case, $x_n\nearrow 0\ .$

What do you see, Virgil? I see a perfectly normal decreasing positive sequence $x_{n+1}=x_n\cdot(1-x_n)$, which clearly converges to zero.

Just drawing lines $y=x$ and $y=x(1-x)$ and bouncing from one to another gives an answer in those cases :)


Ben rolls four fair 20-sided dice, and each of the dice has faces numbered from 1 to 20. What is the probability that exactly two of the dice show an even number?

[quote=SlurpBurp]Shouldn't it be $\frac{4!}{2!2!} \cdot \frac{1}{16}$?[/quote]

That's the same thing as the post above you just in different notation.

[quote=kittenwarrior][quote=SlurpBurp]Shouldn't it be $\frac{4!}{2!2!} \cdot \frac{1}{16}$?[/quote]

$\binom42=\frac{4!}{2!(4-2)!}.$[/quote]

FTFY

@above lol, thanks! I don't know why I even typed a $3\dots $ :P

[hide="Sol"]
It is $\left(\frac12\right)^4=\frac{1}{16}$. However, they can be rearranged in $\frac{4!}{2!2!}=6$ ways. So it is $\frac{6}{16}=\boxed{\frac38}$
[/hide]

Wow we can literally just let them be fair coins because it only cares about parity.

[hide=pog // 896 posts]
Note that each die has a probability of $\dfrac{1}{2}$ of getting an even number or odd number, so we have $\left(\dfrac12\right)^2 \cdot \left(\dfrac12\right)^2 = \dfrac{1}{16}.$ Also, we must choose $2$ of the die to even, so we have $\dfrac{4 \cdot 3}{2}$ ways (choosing die $1$ and $2$ is the same as choosing die $2$ and $1.$) Thus we have $6 \cdot \dfrac{1}{16} = \dfrac{6}{16} = \boxed{\dfrac{3}{8}}.$
[/hide]

[hide=Solution]The number of ways to choose the dice that are even is $\binom{4}{2} = 6$. The rest must be odd.
Hence, the probability that Ben rolls an even number is
$6 \cdot (\frac{1}{2})^2 \cdot (\frac{1}{2})^2 = \frac{6}{16} = \boxed{\frac{3}{8}}$.[/hide]


[img]http://i42.tinypic.com/2n3fy0.jpg[/img]

Is there a mistake in the solution?

The line $ V_y\equal{}U_x$. Should that not be $ V_y\equal{}\frac{\minus{}1}{y}{U_x}$ and the proceed from there?

thanks,
bos1234

no, that's true, why should it be $ v_y \equal{} \frac { \minus{} 1}{y}u_x$? I think you've confused with the polar version of Cauchy-Riemann equations.


Find all triples of prime numbers $ (p,q,r)$ such that $ p^q\plus{}p^r$ is a perfect square.

$ q < r$   $ \implies$   $ p^q \plus{} p^r \equal{} p^q.(1 \plus{} p^{q \minus{} r}) \equal{} a^2$
$ q > 2$   $ \implies$   $ p| 1 \plus{} p^{q \minus{} r}$  contradiction!
$ q \equal{} 2$   $ \implies$ $ p^2 \plus{} p^r \equal{} a^2$ $ \implies$ $ (a \minus{} p)(a \plus{} p) \equal{} p^r, a \minus{} p \equal{} p^x,a \plus{} p \equal{} p^y,2p \equal{} p^y \minus{} p^x$
$ x > 1$   $ \implies$   $ p|2$ $ \implies$ $ p \equal{} 2,2^y \minus{} 2^x \equal{} 4,y \equal{} 3,x \equal{} 2, r \equal{} 5,q \equal{} 2,p \equal{} 2$
$ x \equal{} 1$    $ \implies$   $ a \equal{} 2p$  $ \implies$ $ 3p \equal{} p^y$ $ \implies$ $ p \equal{} 3. a \equal{} 6, a^2 \minus{} p^2 \equal{} p^r$
$ 36 \minus{} 9 \equal{} 3^3$ $ \implies$ $ r \equal{} 3,p \equal{} 3,q \equal{} 2$
$ x \equal{} 0$    $ \implies$  $ a \equal{} p \plus{} 1,2p \plus{} 1 \equal{} p^y$  Contradiction!
$ q \equal{} r$     $ \implies$  $ 2p^q \equal{} a^2,p \equal{} 2,q > 2 (2,q,q)$


pick it

Its the rules we'll play with for the MARCH tournament

wats with 2 of the same names?
R u all crazy?

personally, i really don't care (both seem fine to me)

but seeing as verizon's on my team, i'll support his rules

GO BOYZ OF DARKNESS!!!!!

I voted verizonwires.  Otherwise we will have no countdowns which means no fun. :(

Wait a second.  The normal round was for the first round.  I am thinking about a way to have a countdown for the second round and then a long normal game for the last round, which will be loads of fun!
Thats why I said in my Private Message that I sent to everyone, that those rules [b]only[/b] apply to the first round.  It will get better and better each round making people want to win!

Uhh, wasn't that what I originally said on my very first post?  I'm pretty sure there was a reason we decided not to do it that way :D .
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=187175

dunno cause when I read your rules, I was under the impression that all that would be done in EACH round. Isn't that why you only had two types of competition? Otherwise you would've had three. Either way, it was a bit unclear and verizonwires/minicon restated the rules so it was crystal. At least in a relative sense XD

let's just go with minicon's - further changes will only confuse everyone.


A swarm of evil bunnies that were being breeded by Drunken_Math contains 1000 evil bunnies.

The evil bunnies can do this: Every minute, each evil bunny can bite one person, and that person will instantly turn into an evil bunny, but that human who just turned into an evil bunny can't bite anyone until the next minute. For example, if one evil bunny bites someone at 12:38, the person who just turned into an evil bunny can't bite anyone until 12:39.

Another thing about the evil bunnies, after an evil bunny bites 100 people, it shrivels up and turns into a human.

If my swarm of 1000 bunnies attack the earth, in how long will the entire Earth be made completely into evil bunnies (round to the nearest day)? Assume the earth has 10 billion people.

PS: I tried to do this problem and it got really messy!

:rotfl: Haha! This is a really well made problem. I'll post my solution if nobody else posts.

:o this problem is crazy!  I could solve it if you gave me like 3 hours of adding.  I bet you can make it algebra though where did you come up with this?  Please show us how 4everwise :D  It has a pattern 2,4,8,16,32,64,128,256,512,1024,2048,4096,8192 ok i know how to do it! Yay.  2^100 is more than a billion.  It will take the bunnies 30 minutes to destroy the world, if you do not count the time it will take to travel all over the world.   :D


Support
abstract algebra
superior algebra
superior algebra unsolved

I don't have a very deep understanding of the concepts $ Ass$,$ Supp$,etc.. (yet). Can someone clear up this line.

If $ I$ is an ideal of $ A$ then $ Supp_A(A/I)$ is the set of prime ideals containing $ I$.

So we look for $ p \in Spec(A)$ such that the localization of the $ A$-module $ A/I$ is non-zero

so $ (A/I)_p$ as $ A_p$-module may not be zero

equivalently

$ (A_p/IA_p)$ as $ A_p$-module may not be zero

this is non-zero as long $ p \supset I$

else if $ p \subset I$, then $ IA_p \equal{} A_p$ and $ A_p/IA_p \equal{} 0$

I've been learning these things way to fast. Apparently I now pay the price for this. I have a very chaotic understanding of these abstract concepts.

Does the above make any sense ?

the first part is ok. but in the second part you are very inaccurate and you assume that every two sets are comparable by inclusion ..

$ I A_p \equal{} A_p$ is equivalent to: there is $ i \in I$ and $ s \in A \backslash p$ such that $ i/s \equal{} 1$ in $ A_p$, i.e. there is $ t \in A \backslash p$ such that $ t(i \minus{} s) \equal{} 0$. it follows $ i \minus{} s \in p$ and then $ i \notin p$. conversely, every $ i \in I \backslash p$ yields $ 1 \equal{} i/i \in I A_p$. thus: $ supp_A(A/I) \equal{} \{p \in Spec(A) : A_p \neq I A_p\} \equal{} V(I)$.

Is this why the name "prime divisor" is used? I never read any motivation actually

So $ P$ is a prime divisor of $ I$ if $ P \in Ass(A/I)$.

Because $ Ass_A(A/I) \subset Supp_A(A/I) \equal{} V(I)$, the elements in $ Ass_A(A/I)$ are prime ideals containing $ I$ too.

Then we can use the same terminology as in $ \bf Z$,  i.e. if $ (a) \subset (b)$ one writes $ b \mid a$

so $ P \in Ass(A/I)$ means $ I \subset P$ means "$ P$ divides $ I$" means "$ P$ is a prime divisor of $ I$"

An ideal $ J$ divides an ideal $ I$ means $ J\supseteq I$. Unfortunately, "prime divisor" is not equivalent to prime + divisor. Any minimal prime over some ideal $ I$ (i.e., a minimal element w.r.t. inclusion of the set of primes containing $ I$) is a prime divisor, but there may or may not be more. I am not quite sure whether this answers your question, or what your question was in the first place. owk


geometry
perimeter
geometric transformation
reflection
area of a triangle
Heron\u0027s formula
inequalities theorems

How to show that if $n\geq 3$ is a given integer and $s>0$ is also given, then there is an $n$-gon with perimeter $s$ which has a maximal area? Furthermore, show that the maximal area is achieved if and only if the polygon is regular.

Okay, I have heard a sketch of the proof: First you have to prove that the polygon is convex. This is easy, since we can always find two points in a non-convex polygon where the line segment goes also outside the polygon. Then we just reflect the outer part to get a bigger polygon with the same perimeter. Then we prove that if the polygon has different length of sides, we can step by step change the polygon so that finally the polygon is regular. By Heron's formula one can show that the area won't get smaller. From the isoperimetric theorem, it follows that a polygon with fixed side lengths, the largest one is cyclic. Finally, by compactness argument one can show that the largest polygon exists.

But I think that my formulation is not correct. In my opinion, there are non-regular cyclic quadrilaterals with the same area and the same perimeter as a regular one.

[quote="Albanian Eagle"]No you are correct the maximal polygon is cycliq and has all equal sides so it will be regular :wink:[/quote]
What part of my sketch is not true?

you don't have any mistake in your sketch,
you have actually shown that in the maximal polygon all sides are equal(by mixing the sides+Heron formula), also that it will be cyclic so it must be regular. Therefor your statement was correct since the beginning :)


I need a solution using vectors to this problem, BAEDC is a pentagon, with vector BC equal and parallel to vector ED. X Y and Z are midpoints of respective vectors; BA, AE and DC respectively. Prove that XYZC is a parallelogram.

Why you don't want a solution of pure geometry?  :mad:

In the triangle $ABE$, the segment $XY$ is joining the midpoints of $AB,AE$, so $XY \parallel BE$ and $XY=\frac{BE}{2}$

But $BCDE$ is a parallelogram, so $CD \parallel = BE \Rightarrow  CZ \parallel = XY\Rightarrow CZYX$ parallelogram

($\parallel =$ means parallel and equal)

Ok here is a solution with vectors:

$\overline{BC}=\overline{ED}\Rightarrow BCDE$ parallelogram $\Rightarrow \overline{BE}=\overline{CD}$

$\overline{BE}=\overline{BA}+\overline{AE}=2\overline{XA}+2\overline{AY} = 2(\overline{XA}+\overline{AY}) = 2\overline{XY}$

$\overline{CD}=2\overline{CZ}$

$\overline{BE}=\overline{CD}\Rightarrow \overline{CZ}=\overline{XY}\Rightarrow XYZC$ parallelogram


geometric sequence
number theory
modular arithmetic

Hi, im not sure how to approach this problem.  Any help would be appreciated, thanks.

For how many natural numbers n =<12 is 4^n + 5^n + 6^n + 7^n a multiple of 11

Other questions that have been bugging me: why is it that in most textbooks the sum  of a geometric sequence is shown as a(1-r^n)/(1-r)?  Since this is used for |r|>1, 1-r^n will be negative, as will 1-r.  So why isnt -1 factored out so it is a(r^n-1)/(r-1)?
Also, if a question asks for the sum of the 1st n powers of 2, the answer is 2^(n+1)-1.  This means that it goes from 2^0...2^n.  Why does it start at 2^0 instead of 2^1?  Thanks.

The formula for the summation of a geometric sequence can be written as a(1-r^n)/(1-r) or a(r^n-1)/(r-1); they're just different forms of the same formula.

For the powers of two, I assume that n starts from 0 since it is the smallest natural number.

"For how many natural numbers n =<12 is 4^n + 5^n + 6^n + 7^n a multiple of 11"

This problem can probably be solved with modular arithmetic.  Using the fact that a number is divisible by 11 iff the alternate sum of its digits is divisible by 11 (i.e. ex. 121 :equiv: 1-2+1 (mod 11) :equiv: 0 (mod 11).

-interesting_move

I realize that they're different forms of the same equation, but what's the purpose of having the first form if you're going to get a negative numerator and denominator and you're going to have to cancel them?  In most textbooks it's written the first way, but the 2nd way seems more practical to me.
If you count to 20 you start at 0?
In order to solve this problem by using the divisibility rules for 11, you have to actually sum the numbers, so it becomes tedious after n=2, unless if there is a pattern or something, but im not sure if we can find a pattern by calculating for n=1,2?

wow, how systematic hgmm76.

-interesting_move

[quote="interesting_move"]For the powers of two, I assume that n starts from 0 since it is the smallest natural number.[/quote]

No, 0 is not the smallest natural number.  It isn't a natural number.

I quote from The Big Book (AoPS #1, pg. 99): "We start with the natural numbers: 1,2,3,..."

It's easy to confuse natural numbers and whole numbers.  One way to remember it is this: for whole numbers, think "hole".  When you think "hole" you think of the number that looks like a hole - 0!  So whole numbers include 0, as well as the positive integers.  Whereas natural numbers don't.

[quote="interesting_move"]"For how many natural numbers n =<12 is 4^n + 5^n + 6^n + 7^n a multiple of 11"

This problem can probably be solved with modular arithmetic.  Using the fact that a number is divisible by 11 iff the alternate sum of its digits is divisible by 11 (i.e. ex. 121 :equiv: 1-2+1 (mod 11) :equiv: 0 (mod 11).[/quote]

Yes, you said you could use modular arithmetic.  But you don't specify how - you just informed hgmm76 about the sum/difference of digits of multiples of 11.  How is [i]saying[/i] that you could use modular arithmetic going to help him?

[color=cyan]Look at the cycles of things mod 11.  For example:
4^0 :equiv: 1 (11)
4^1 :equiv: 4 (11)
4^2 = 16 :equiv: 5 (11)
4^3 = 4*4^2 :equiv: 4*5 = 20 :equiv: 9 (11)
4^4 :equiv: 3
4^5 :equiv: 1
and then it has to repeat.  You can do the same with the others.  Also, you might note that
4 :equiv: -7 (11) and 5 :equiv: -6 (11), so that [i]all[/i] odd powers will work, since the four and seven will cancel each otehr, as will the 5 and 6.  The even powers, you will have to check by hand or some other clever method.  Also, it has to be cyclic every tenth power, by Fermat's Little Theorem.

As for the other question, you start with 2^0 because then the formula looks nice; you use 1-r^n because the most interesting thing geometric sequences can do happens when |r| < 1, when this will be positive.  I mean, those aren't real answers, but the real answer is "by convention." Someone started doing it, and everyone else picked up on it.  That's all.[/color]

[quote="mathfanatic"]

No, 0 is not the smallest natural number.  It isn't a natural number.

[/quote]

There is no convention on if to include 0 to the natural numbers. Sometimes one includes sometimes not. So to avoid ambiguousness I usually write Z+ to denote the positive integers and Z* the nonnegative integers.

Here's something I just discovered that applies to this problem:

Let x and y be positive integers such that x+y=11

Then for all odd positive n, x^n + y^n mod 11 :equiv: 0

That is, x^n + y^n is a multiple of 11.

I haven't been able to prove this, but it seems to work.  Any ideas on a proof?  It would probably involve modular arithmetic of some sort.

In this problem, we have 4^n + 5^n + 6^n + 7^n
which can be written as (4^n + 7^n) + (5^n + 6^n).

These two pairs are examples of x^n + y^n such that x + y = 11 (4+7=5+6 = 11); hence the above statement applies and for odd n, 4^n + 5^n + 6^n + 7^n is a multiple of 11.  This is for n = 1, 3, 5, 7, 9, 11, ...; that is, [b]6[/b] values for n :le: 12.

Sure. Rewrite y as 11-x, or -x (mod 11). Then x^n+(-x)^n=0 if n is odd because (-x)^n=(-1)^n*x^n=-x^n, so adding them gives x^n + - x^n = 0.

Here's some modular arithmetic that tells something about the previous statement that for x+y=11 and odd n, x^n + y^n mod 11 :equiv: 0

(n=1)
4 mod 11 = 4
7 mod 11 = 7
4+7 = 11

(n=2) (Even power; shouldn't work)
16 mod 11 = 5
49 mod 11 = 5
5 + 5 ≠ 11

(n=3)
64 mod 11 = 9
343 mod 11 = 2
9+2 = 11

(n=4) (shouldn't work)
256 mod 11 = 3
2401 mod 11 = 3
3 + 3 ≠ 11

(n=5)
1024 mod 11 = 1
16807 mod 11 = 10
1 + 10 = 11

(n=6)
4096 mod 11 = 4
117649 mod 11 = 4
4 + 4 ≠ 11

Another startling discovery (again, I can't prove it but it'll definitely have something to do with modular arithmetic) is:

For x+y=11 and even n, x^n mod 11 = y^n mod 11

Any ideas?  This confirms our suspicion that even powers of n won't ever work, because x^n mod 11 + y^n mod 11 will always produce an even mod 11 number (not 11)

Again, that's right. Let us write x=2k. Then x^n+(-x)^n=x^(2k)+(-x)^2k=(x^2)^k+(x^2)^k=2(x^2)^k, and since x^2 is not congruent to 0 (mod 11), 2(x^2)^k will be twice a number not congruent to 0 (mod 11), so it will not be divisible by 11. (Specifically, 11/2:equiv: 0 (mod 11), if you are used to seeing fractions with modulos.)

Oh!  After reading ComplexZeta's proof of my first statement, I can prove the second one.

Again, by writing y as 11-x, we have:
y^n mod 11 = (11-x)^n mod 11 = (-x)^n mod 11
Since n is EVEN, (-x)^n = x^n
So y^n mod 11 = x^n mod 11.

This is why the mod 11 of x^n and y^n for even n is equal.

Here's as summary of the weird 11-things:
For all positive integers such that x + y = 11,
If n is odd, then x^n + y^n mod 11 = 0
If n is even, then x^n mod 11 = y^n mod 11

Ah, we can generalize this statement for all numbers (since ComplexZeta's proof can be applied to other values other than 11; i.e. if x+y=6, then you could write y as 6-x, then apply the same logic for mod 6)

Suppose x + y = Z

Then for odd n, (x^n + y^n) mod Z :equiv: 0

For even n, x^n mod Z = y^n mod Z

For instance, if x + y = 21,

Then if n is 5, x^5 + y^5 mod 21 :equiv: 0

If n is 6, x^5 mod 21 :equiv: y^5 mod 21

A problem similar to hgmm76's original one is this:

For how many values n:le:100 is 2^n + 3^n + 6^n + 7^n divisible by 9?

[hide]The answer is, of course, 50, since for all the odd ns, 2^n + 9^n and 3^n + 6^n will be divisble by 9.  For all the even ns, 2^n mod 9 :equiv: 7^n mod 9, etcetera.[/hide]

So that's the easy way to do problems like this: by knowing this special formula or discovering it through experimentation.[/hide]

I don't quite see your logic for your problem, mathfanatic. Your answer is right though. But here's how I did it. We use the results from before to write for even n 2(2^n+3^n) (mod 9). 3^n will be divisible by 9, so we are left with 2*2^n (mod 9), to which there are clearly no zero solutions.

Here is the original problem:

[quote]For how many natural numbers n =<12 is 4^n + 5^n + 6^n + 7^n a multiple of 11?[/quote]

Here's how I solved it.

I noticed that 4 + 7 = 5 + 6 = 11
Then I noticed that 4^3 + 7^3 mod 11 :equiv: 5^3 + 6^3 mod 11:equiv: 0
Similarly, 4^5 + 7^5 mod 11 :equiv: 5^5 + 6^5 mod 11:equiv: 0
And 4^7 + 7^7 mod 11 :equiv: 5^7 + 6^7 mod 11:equiv: 0

So I figured out that if x + y = 11 (just as 4&7, 5&6 do), then x^n + y^n mod 11 :equiv: 0 for odd n only.

Eventually, I expanded this to read:

[quote]Suppose x + y = Z
Then for odd n, (x^n + y^n) mod Z :equiv: 0
For even n, x^n mod Z :equiv: y^n mod Z [/quote]

You can apply this rule to the new problem, "For how many values n:le:100 is 2^n + 3^n + 6^n + 7^n divisible by 9?"

In this case, 2+7 = 3+6 = 9.  If we split the terms up into:
2^n + 7^n
and 3^n + 6^n
We find that each one works with the above rule.  In the first pair, 2^n + 7^n mod 9 :equiv: 0 for all odd n.  In the second pair, 3^n + 6^n mod 9 :equiv: 0 for all odd n.  Because 2^n + 3^n + 6^n + 7^n is just the sum of two things that are each divisible by 9 when n is odd, the four-term sum is also divisible by 9 when n is odd.

Hence, there are 50 values for n that are less than 100 which satisfy the condition that 2^n + 3^n + 6^n + 7^n is divisible by 9.


Simply the following multiplication:
$(x+z+y)^-1(x^-1+z^-1+y^-1)(xy+yz+xz)[(xy)^-1+(yz)^-1+(zx)^-1]$

Simply the following:
$(\sqrt[6]{27}-\sqrt{6\frac{3}{4}})^2$

When$x^5,x+\frac{1}{x},$ $and$ $1+\frac{2}{x}+\frac{3}{x^2}$
are multiplied, what is the degree of the resulting polynomial?

1.[hide]
$\displaystyle\frac{1}{x+y+z}(\frac{1}{x}+\frac1y+\frac1z)(xy+yz+xz)(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz})$
$\displaystyle\frac{1}{x+y+z}(\frac{yz+xz+xy}{xyz})(xy+yz+xz)\frac{xyz^2+x^2yz+y^2xz}{x^2y^2z^2}$
$\displaystyle\frac{1}{x+y+z}(\frac{yz+xz+xy}{xyz})(xy+yz+xz)\frac{(xyz)(x+y+z)}{x^2y^2z^2}$
Now we cancel:
$\displaystyle\frac{(xy+yz+xz)^2}{x^2y^2z^2}\Rightarrow (\frac{xy+yz+xz}{xyz})^2$
[/hide]

2. [hide]
We'll start with simplification of the powers:
$\displaystyle(\sqrt[6]{3^3}-\sqrt{\frac{3^3}{2^2}}\Rightarrow (3^{\frac12}-\frac{3^{\frac32}}{2})^2$

Let y be $3^{\frac12}$. Then we have the equation $\displaystyle(\frac{y-y^3}{2})^2\Rightarrow (\frac{y(y-1)(y+1)}{2})^2$. Now we expand:  $\displaystyle\frac{y^2(y^2-1)^2}{4}=\frac{3(3-1)^2}{4}=\frac{3(4)}{4}=\box{3}$
[/hide]

Billy

You guys made 2 a lot harder than it needed to be...
$\sqrt[6]{3^3}-\sqrt{\frac{3^3}{2^2}}= \sqrt 3 - \frac{3\sqrt{3}}{2} = \frac{-\sqrt{3}}{2}$.  Square to get $\frac 3 4$

[quote="solafidefarms"]
2. [hide]
We'll start with simplification of the powers:
$\displaystyle(\sqrt[6]{3^3}-\sqrt{\frac{3^3}{2^2}}\Rightarrow (3^{\frac12}-\frac{3^{\frac32}}{2})^2$

Let y be $3^{\frac12}$. Then we have the equation $\displaystyle(\frac{y-y^3}{2})^2\Rightarrow (\frac{y(y-1)(y+1)}{2})^2$. Now we expand:  $\displaystyle\frac{y^2(y^2-1)^2}{4}=\frac{3(3-1)^2}{4}=\frac{3(4)}{4}=\box{3}$
[/hide]

Billy[/quote]

There is a problem in the section that states $\displaystyle(\frac{y-y^3}{2})^2$; it is supposed to be $\displaystyle(3^{\frac12}-\frac{3^{\frac32}}{2})^2\Rightarrow (y-\frac{y^3}{2})^2=(\frac{2y-y^3}{2})^2$...

In any case, this problem is easier than the amount of work you've put into it.

P.S.  I have currently typed my first Latex code by finding a pattern in other people's codes; hence, I may have some typos in there too.

Masoud Zargar

[hide="3"]We want the highest degree term.  So $x^5\cdot x\cdot 1=x^6$, so the answer is $6$.[/hide]


£1000 is placed in a bank account at the start of each year, beginning in the year 2000.
the account pays 5% interest per year at the end of each year.
how much is the account after 15 years?

[hide]He'll have 2078.93 in the bank account after 15 years.

1000 (1+5%)^15

Interest= Principal x Rate.

So after 15 years, he'll have 2078. 93 (a profit of 1078.93!).[/hide]

Is it simple interest or compound interest?

Assuming compound interest, the way I read your problem is that we have 15 total payments (one at the start of each year), but we valuate one year after the last payment. Giving : [hide]$1000(1.05)^{15}+1000(1.05)^{14}+...+1000(1.05)^{1}=1000(1.05) \frac{(1.05^{15}-1)}{(1.05-1)}=22657.49$[/hide]

what I would do is make a chart...

if it takes too long..

I'll choose the longer way! :D

make a graph! :P


The following two statements are easy to "sense" that they're true, but I found it hard to write a formal proof.

1) Let $f: R^{+}\rightarrow R^{+}$ be a strictly decreasing function. Show that there exists $x$ such that $f(x)=x$

2) Let $f: R^{+}\rightarrow R^{+}$ be a strictly increasing function. Show that $\lim_{x\rightarrow a^{-}}f(x)$ and $\lim_{x\rightarrow a^{+}}f(x)$ exist for all $a\in R^{+}$

1) I'm assuming it's continuous (it's not implied by strictly decreasing, right?). Consider $g(x) = f(x)-x$. Since $\lim_{x \rightarrow-\infty}g(x) = \infty$ and $\lim_{x \rightarrow \infty}g(x) =-\infty$ so by the intermediate value theorem there exists an $x_{0}$ such that $g(x_{0}) = 0 \Rightarrow f(x_{0}) = x_{0}$.

2) Let f be strictly increasing.  Fix a in R+.  Let xn be a increasing sequence converging to a-.  Then f(xn) is an increasing sequence in R+ bounded above by f(a) thus it coverges in R+ to the limit.  The limit exists since one sequence implies all sequences will converge, easy in reals.

Without the assumption of continuity, #1 is false. Example: $f(x)=\begin{cases}1-x&x\le0\\-1-x&x>0\end{cases}$

[quote="Jumbler"][quote="paladin8"] Since $\lim_{x \rightarrow-\infty}g(x) = \infty$ [/quote]

But I defined $f: R^{+}\to R^{+}$...[/quote]

Oops. Sorry. Then you just have $g(0) = f(0) > 0$ instead.


so right now there are 40 candles, and we want to add $x$ white so that white will be half
for white to be half, it has to = the rest.
So since there are 40 right now, we must add 40 white.
So 40 white candles is half of the package, which means 80 total.

[quote="helpmex55"]The president of the company has decided to use a new packaging scheme. she wants to have the same number of red and blue candies as are in the original package (10 red, 30 blue) but now she wants to make the same number of white candies 50% of the total number of candies in the package.

-how many white candies would need to be added to make then 50% of the total mixture?
-Explain
-How many candies would NOW be put in each package?

THANK YOU :roll:[/quote]

[hide="te prueba es muy facil"]I guessed and checked. Seemed faster that way...

Original$=10 R, 30 B$

$a)$ Add $40$ white candies. You get:
                     $10 R+30 B+40 W=80 RBW$, where the white candy is 50% of total
$b)$ It would be a total of 80 candies[/hide]


How do you have braces facing up with text under it like this?

111...111
Then you have $\}$ opening up
and under it says 100 1s

[quote="stevem"]Do you mean like this $\underbrace{111...111}_{100}$ which is given by \underbrace{111...111}_{100}?[/quote]

Yeah, thanks.


Prove that the sum of an infinite convergent geometrical series of rational numbers is rational.

[hide="Quick Attempt"]
The sum of an infinite geometric series is $ \frac {x}{1 \minus{} r}$.  Since the statement says the terms of the series are rational, that makes the first term rational.  For the rest of the terms to be rational the rate, $ r$ , must be rational.  This makes the [i]fraction[/i] a rational number.
[/hide]

Nice job, though maybe you should explicitly mention closure on the operation of multiplication in the field of rationals.  :)

Yeah, that would make it a bit more complete.  :P


[quote="akimatsu"]solve for x:
\[{{3\log_{2}x+8}\over{2\log_{2}x+2}}+{{5\log_{2}x-4}\over{\log_{2}x}}= 6 \]
[/quote]
[hide]Let's simplify this by eliminating the fractions.
$\frac{3\log_{2}x+8}{2\log_{2}x+2}+5-4\log_{x}2=6$
$3\log_{2}x+8+10\log_{2}x+10-8-8\log_{x}2=12\log_{2}x+12$
By simplyfying and applying the substitution $\log_{2}x=a$ We get $a-2-\frac{8}{a}=0$
$a^{2}-8a-2=0$
$a=4+3\sqrt{2}$
So $x=2^{3\sqrt{2}+4}$[/hide]

[quote="bpms"]$a-2-\frac{8}{a}=0\implies a^{2}-8a-2=0$[/quote]

Do recheck this step, please.

That makes much more sense as teh answer...


[hide="Solution"]
From AM-GM,

$ a(x_k^2 \plus{} 1)\ge2ax_k\implies ax_k^2 \plus{} a^2x_k \plus{} x_k \plus{} a\ge a^2x_k \plus{} 2ax_k \plus{} x_k$

$ \implies(x_k \plus{} a)(ax_k \plus{} 1)\ge(a \plus{} 1)^2x_k\implies\prod_{k \equal{} 1}^{2008}(x_k \plus{} a)(ax_k \plus{} 1)\ge\prod_{k \equal{} 1}^{2008}(a \plus{} 1)^2x_k$.

It is required by multiplying the given equations for

$ \prod_{k \equal{} 1}^{2008}(x_k \plus{} a)(ax_k \plus{} 1) \equal{} \prod_{k \equal{} 1}^{2008}(a \plus{} 1)^2x_k$, so by the equality condition of AM-GM,

$ x_k \equal{} \boxed1$.
[/hide]

Equality occurs when $ x_k^2\equal{}1$ so couldn't $ x_k$ be $ \minus{}1$, too? But nice solution anyways...

If you plug in $ x_k\equal{}\minus{}1$, it doesn't work. But it's not a valid candidate in the first place because it's really AM-GM on $ x_k$ and $ 1$ ($ (x_k)^2\plus{}1\ge2(x_k)(1)$).

[quote="math154"]If you plug in $ x_k \equal{} \minus{} 1$, it doesn't work. But it's not a valid candidate in the first place because it's really AM-GM on $ x_k$ and $ 1$ ($ (x_k)^2 \plus{} 1\ge2(x_k)(1)$).[/quote]
Shouldn't it be $ x_k^2 \plus{} 1\ge 2\sqrt {x_k^2*1} \equal{} 2|x_k|$?
EDIT: hmm actually we require $ x_k\equal{}|x_k|$ in order to factor that out...

Yeah, sorry; I only solved it for nonnegative $ x$. Now I'm not really sure how to solve it...

[hide="Straightforward"]Expand the LHS's, sum up all the equations, and denote $ S_2: \equal{} \sum x_k^2, S_1: \equal{} \sum x_k$ to get

$ aS_2 \plus{} (a^2 \plus{} 1)S_1 \plus{} 2008a \equal{} (a \plus{} 1)^2S_1\iff S_2 \minus{} 2S_1 \plus{} 2008 \equal{} 0$

However, $ S_2 \minus{} 2S_1 \plus{} 2008 \equal{} \sum(x_k \minus{} 1)^2$, hence the only solution is $ x_1 \equal{} x_2 \equal{} \dots \equal{} x_{2008} \equal{} 1$

It actually works for $ a<0$ too, hence $ a\neq 0$ is sufficient.[/hide]


$x,y,z$ are positive prove that
$x.(x-z)^2+y(y-z)^2\ge (x-z)(y-z)(x+y-z)$

:rotfl:

$x\left(x-z\right)^2+y\left(y-z\right)^2-\left(x-z\right)\left(y-z\right)\left(x+y-z\right)$
$=x^3+y^3+z^3+3xyz-y^2z-z^2y-z^2x-x^2z-x^2y-y^2x\geq 0$

by the Schur inequality; thus,

$x\left(x-z\right)^2+y\left(y-z\right)^2\geq\left(x-z\right)\left(y-z\right)\left(x+y-z\right)$.

  Darij

i wrote that it is easy inequality  :D  ;)


Charlene's calculator displays the digits 0, 1, 6, 8 and 9 so that, when the calculator is held upside-down, these digits appear to be 0, 1, 9, 8 and 6 respectively. How many three-digit numbers look the same upside-down and right-side-up when using these five possible digits? (A three-digit string that begins with either 0 or 00 is not considered a three-digit number.)

The numbers must be of the form $ \overline{aba}$, where $ a,b\neq6,9$ and $ a\neq0$, or $ \overline{6a9}$ or $ \overline{9a6}$, where $ a\neq6,9$, for a total of $ 2\cdot3 + 2\cdot3 = \boxed{12}$.

9 and 6 look different when upside-down.


rotation
combinatorial geometry
geometry solved
geometry

In the plane, n circles of unit radius are drawn with different centers. Of course, overlapping circles partly cover each other's circumferences. A given circle could be so overlaid that any uncovered parts of its circumference are all quite small; that is, it might have no sizable uncovered arcs at all. However, this can't be true of every circle; prove that some circle must have a continuously uncovered arc which is at least 1/n th of its cirumference.

Clearly, we have to assume that the circles are pairwise distincts...
So let C be the convex hull of the set of the n circles. It is the union of segments and circular arcs from the circles. As we rotate to 360 when travelling along the boundary of C, it follows that the total length of the circular arcs used in C is 2 \pi . Since there are n circles, there are no more than n circular arcs so that the average length of these circular arcs is at least 2 \pi /n, and we are done.

Pierre.


If $ [a,b][c,d] \equal{} [0,0]$, prove that $ ac\minus{}bd \equal{} 0$ and $ bc\plus{}ad\equal{}0$, and hence that $ (a^{2}\plus{}b^{2})(c^{2}\plus{}d^{2})\equal{}0$. Deduce that either $ [a,b] \equal{} [0,0]$ or $ [c,d]\equal{}[0,0]$

How to do the second part and third part?

Hmm, do you mean $ [a,b] \equal{} a\plus{}bi$? [hide="Then"] $ 0 \equal{} [0,0] \equal{} [a,b][c,d] \equal{} (a\plus{}bi)(c\plus{}di) \equal{} (ac\minus{}bd)\plus{}(ad\plus{}bc)i$. So $ ac\minus{}bd \equal{}\Re([a,b][c,d]) \equal{} 0$ and $ ad\plus{}bc \equal{}\Im([a,b][c,d]) \equal{} 0$. Now one remarks that $ (a^{2}\plus{}b^{2})(c^{2}\plus{}d^{2}) \equal{} (ac\minus{}bd)^{2}\plus{}(ad\plus{}bc)^{2}\equal{} 0$. And this means $ 0 \equal{} |a\plus{}bi|\cdot|c\plus{}di|$, so either $ |a\plus{}bi| \equal{} 0$, i.e. $ [a,b] \equal{} [0,0]$ or $ |c\plus{}di| \equal{} 0$, i.e. $ [c,d] \equal{} [0,0]$.[/hide]

ok thanks i understand now

yes thats what i meant $ [a,b] \equal{}\Rightarrow  a\plus{}bi$


trigonometry
inequalities
LaTeX
geometry
geometric transformation
inequalities proposed

Let ABC be a triangle .Prove that
    $\frac{1]{cos\frac{B-C}{2}cos\frac{C-A}{2}cos\frac{A-B}{2}} + 1 \geq \frac{3}{cosA+cosB+cosC}$
    I posted this problem a long time ago, but finally it wasn't completely proved.
At that time, my inequality was for acute triangle.tranminhhoang, who is my friend, did tell me this one was also right for any triangle. So can anyone help me?
Any solution you can, even if very very "buffalic" :D

1/{cos(B-C)/2cos(C-A)/2cos(B-C)/2} +1 >= 3/{cosA+cosB+cosC}.
  I 'm afraid you can't see my latex.  :?

Here is the LaTeX:

[quote="treegoner"]Let ABC be a triangle. Prove that
$\displaystyle \frac{1}{\cos\frac{B-C}{2}\cos\frac{C-A}{2}\cos\frac{A-B}{2}} + 1 \geq \frac{3}{\cos A+\cos B+\cos C}$[/quote]

You can always go back to edit your original post instead of double posting.

OK, I did a translation, no idea whether it's helpful or not.

\[
\frac{(abc)^2}{(a+b)(b+c)(c+a)(s-a)(s-b)(s-c)} + 1 \geq \frac{3abc}{abc + 4(s-a)(s-b)(s-c)}
\]

Let $x=s-a ,y =s-b, z=s-c$, then $a=y+z, b= z+x, c=x+y$, and it's translated into:
[quote] If $x,y,z>0$, then
\[
\frac{[(x+y)(y+z)(z+x)]^2}{(x+y+2z)(x+2y+z)(2x+y+z)xyz} + 1 \geq \frac{3(x+y)(y+z)(z+x)}{(x+y)(y+z)(z+x) + 4xyz}.
\]
[/quote]

[quote="treegoner"]Let ABC be a triangle. Prove that
$\displaystyle \frac{1}{\cos\frac{B-C}{2}\cos\frac{C-A}{2}\cos\frac{A-B}{2}} + 1 \geq \frac{3}{\cos A+\cos B+\cos C}$[/quote]
We can easily prove these two equalities
${\\cos\frac{B-C}{2}\cos\frac{C-A}{2}\cos\frac{A-B}{2}} = \frac{p^2 +2Rr + r^2}{8R^2} $
$\cos A+\cos B+\cos C = 1 + \frac{r}{R} $
So treegoner 's inequality is equivalent to this one
$\displaystyle \frac{8R^2(R+r)}{2R-r} -r^2-2Rr \geq \ p^2 $

Because treegoner did prove for acute triangles, now I prove for obstute triangles.
[b]Lemma[/b] : for obstute triangles we have $\ p \leq\ 2R+r $

Apply this lemma, all we have to do now is to prove the inequality
$\displaystyle \frac{8R^2(R+r)}{2R-r} -r^2-2Rr \geq \ (2R+r)^2 $

But this one is equivalent to $\ 4R^2 \geq\ 4R^2 - r^2 $ (obviously true)
So treegoner 's inequality is true for all triangles.
Also use this method we can prove for acute triangles but we must use another lemma which is rather complicated. Therefore, treegoner should post his solution for this case.
After all, I hope my solution is not very buffalic  :D

Hey, but I don't even know how to prove the inequality for acute triangle.
 Or I should try proving it one more time. Maybe I have luck this time. :)
 But can you send your solution in this topic, tranminhhoang ? :cool:

Hm treegoner, I really do not want to post such a complex solution.

The lemma which I talked about is that
[b]Lemma[/b] For all triangles we have
$ p^2 \leq \ 2R^2 + 10Rr - r^2 +2(R-2r)\sqrt{R(R-2r)} $

Use this lemma we have to prove
$\displaystyle \frac{8R^2(R+r)}{2R-r} -r^2-2Rr \geq \ 2R^2 + 10Rr - r^2 +2(R-2r)\sqrt{R(R-2r)} $

Denote $ t= \frac{R}{r} $ the inequality becomes
$ (t-2)t(2t-3) \geq \ (t-2)(2t-1)\sqrt{t(t-2)} $

This one is easy because $ t \geq \ 2 $

The lemma is very strong and the proof is too long for me to write. So... I must stop here  :D

Don't worry, tranminhhoang. I know this lemma. Nice solution.  :)


quadratics
function
floor function
algebra
algebra unsolved

Let $ f(x)\equal{}a_1 x^2 \plus{}b_1 x \plus{} c_1$ $ g(x)\equal{}a_2 x^2 \plus{}b_2x \plus{}c_2$ ,  $ a_1 \neq  0$  ,$ a_2  \neq 0$ quadratic functions such that  $ [f(x)]\equal{}[g(x)]$  ,$ x \in R$.Prove that $ f(x)\equal{}g(x)$

___________________________
Azerbaijan Land of the Fire   :ninja:  :roll:

do you mean $ \lfloor f(x) \rfloor \equal{} \lfloor g(x) \rfloor$?

where $ x\leq [x] < x\plus{}1$   :ninja:

[quote="Pirkuliyev Rovsen"]where $ x\leq [x] < x \plus{} 1$   :ninja:[/quote]

this isn't well-defined. I think what you mean to say is greatest integer function, i.e. floor function.

[quote="Pirkuliyev Rovsen"]coefficients are real !!![/quote]

yes, and?

We may assume $ a_1,a_2>0$

Then $ f,g \rightarrow \infty$ as $ x\rightarrow \infty$.

Hence f(x) and g(x) take on infinitely many integer values i.e. f(x) = g(x) at infinitely many values of x

Since $ f(x)\minus{}g(x)$ is of degree utmost 2 we have $ f(x)\minus{}g(x) \equiv 0$.

Hence $ f(x) \equal{} g(x)$


Prove that for all natural n, $\frac{1}{2n}<$ {$n\sqrt{7}$} $<1-\frac{1}{6n}$

Is it hard? or just very easy.i  cannot solve it  :(

One part is solved here: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=74032[/url]


Prove that the equation

$x^{2}+x+1=py$ has integer solutions $(x,y)$ for infinitely many primes $p$.

Here's a hint

[hide]Another interpretation of the problem is that inifnitely number of primes occur for among the factors for numbers in the form of $x^{2}+x+1$.[/hide]

[hide]
Suppose there are only finite $p$ such that $p|(x^{2}+x+1)$ for some integer $x$. Let the set of all such $p$ be $S = \{p_{1}, p_{2}, \ldots, p_{k}\}$. Then take

$x = p_{1}p_{2}\cdots p_{k}\Rightarrow x^{2}+x+1 = p_{1}p_{2}\cdots p_{k}(p_{1}p_{2}\cdots p_{k}+1)+1$.

This, however, means that $x^{2}+x+1$ is not divisible by $p_{1}, p_{2}, \ldots, p_{k}$ which implies that it is divisible by some other prime $p_{k+1}$, contradicting the fact that $S$ was the set of all primes dividing $x^{2}+x+1$ for any integer $x$. Hence $S$ is infinite. [/hide]


calculus integration inequalities symmetry

1)Find the number of $non \ negative$ integral solution to:-

$2x+y+z \leq 19$

2)Find the number of $non \ negative$ integral solution to:-

$x+y+z=18$,where $x<y<z$

 :)

[quote="shadysaysurspammed"]1)Find $non \ negative$ integral solution to:-

$2x+y+z \leq 19$

2)Find $non \ negative$ integral solution to:-

$x+y+z=18$,where $x<y<z$

 :)[/quote]

I think he want the number of non-negative integral triplets.  :?

[hide="1"]
$x$ can assume all values from $1$ to $8$. So for $y=1$ we have $16$ different values of $z$, for $y=2$ we have $15$ and so on

$\frac{16*17}{2}$ cases. Expanding this reasoning for all possible values of $x$ we get that

the number of triplets is $\frac{16*17}{2}+\frac{14*15}{2}+\frac{12*13}{2}+...+\frac{2*3}{2}=444$ triplets[/hide]

Now generalise for both  :D(for the first one,take $x+y+z\leq n$)

[hide="1 generalized"]
the minimum value of $x$ is $1$. Then, if $y=1$ we have $n-2$ possible values of $z$. For $y=2$ we have $n-1$ possible values and so on until $y=n-2$. So the number of triplets is $\frac{(n-2)(n-1)}{2}$. The same reasoning when $x=2$ but $y_{max}=n-3$.
The number of triplets that satisfy the inequality $x+y+z \leq n$ is $\displaystyle \sum_{i=1}^{n-2}\frac{(n-i)(n-i-1)}{2}$[/hide]

is it right? :oops:

[quote="hydro"][hide="2"]
the maximum value for $x$ is $5$, because if $x=6$ the minimum value for $y$ is $7$ but then $z$ should be equal to $5<y$. Now, if $x=1$ we have to consider all possible couples $(y,z)$ s.t. $y+z=17$ with $y<z$. We have $8-1$ cases. expand this method to all possible values of $x$ to get number of triplets $=7+5+4+2+1=19$
[/hide][/quote]

Agreed.  But note that the problem says non-negative integers.  I believe that means $\ x,y,z$ can be equal to $\ 0$ so considering the case where $\ x=0$ then we add on another $\ 8$ solutions.

[quote="hydro"]1 generalized
the minimum value of $x$ is $1$. Then, if $y=1$ we have $n-2$ possible values of $z$. For $y=2$ we have $n-1$ possible values and so on until $y=n-2$. So the number of triplets is $\frac{(n-2)(n-1)}{2}$. The same reasoning when $x=2$ but $y_{max}=n-3$.
The number of triplets that satisfy the inequality $x+y+z \leq n$ is $\displaystyle \sum_{i=1}^{n-2}\frac{(n-i)(n-i-1)}{2}$

is it right? :oops:[/quote]

I have no idea :D Could someone confirm that? The generalisation of the second one should be easy...

I'm getting a different answer for number 1.  I think there are $\ 515$ triplets.

Ok, heres how I figure for the first one.  $\ 2x + y + z \leq 19$, so $\ 2x + y + z + k =19$ where $\ k$ is a non-negative integer.  This equation has $\ \binom{21}{3}$ triplets $\ (2x,y,z,k)$.  This triplet will have a corresponding $\ (x,y,z,k)$ solution if $\ 2x$ is even.  By symmetry this should happen half the time so the answer is $\ \binom{21}{3} * \frac{1}{2} = 515$


Three sides of a quadrilateral have the same length, and they include angles measuring 60 and 70 degrees. What is the measure of the largest angle in the quadrilateral?

the answer to the question is 70 dergress

that is impossible, since the full quadrilateral must have 360 degrees, thus the final two angles have a combined 360-(60+70)=230 degrees, so one must be at least 115 degrees.

I don't know what's more ridiculous. The fact that he defended his answer in another thread might trump his claiming that the answer is 70 without any explanation in this thread.

Clearly the answer cannot be 70.

If 70 were the LARGEST angle in the triangle, then even 70*3 + 60 = 270 ==|== 360.

Just basically examine this as if the sides were paired and tangents to a circle. This leads to the fact that a quad can only have angles 60 and 70 if and only if the other two angles are equal. If it were otherwise, the tangents, meaning the sides would be unequal.

[hide]
Let $ABCD$ be the quadrilateral, with $\angle A=60^\circ$ and $\angle B=70^\circ$. Drop perpendiculars from $D$ and $C$ to $AB$, meeting at $P$ and $Q$, respectively. WLOG we can set $AB=1$. Then $AP=\frac{1}{2}$, $PD=\sin 60^\circ$, $QB=\sin 20^\circ$, $QC=\sin 70^\circ$.

Now let $X$ be the point on $QC$ such that $DX\perp QC$. Then $PQXD$ is a rectangle, so $DX=PQ=\frac{1}{2}-\sin 20^\circ=\sin 30^\circ-\sin 20^\circ$ and $XC=\sin 70^\circ-\sin 60^\circ$. We now compute the measure of $\angle XDC$:
\begin{eqnarray*}\tan \angle XDC &=&\frac{\sin 70^\circ-\sin 60^\circ}{\sin 30^\circ-\sin 20^\circ}\\ &=&\frac{2\cos 65^\circ \sin 5^\circ}{2\cos 25^\circ \sin 5^\circ}\\ &=&\frac{\sin 25^\circ }{\cos 25^\circ }\\ &=& \tan 25^\circ \\ \end{eqnarray*}
It follows that $\angle XDC$ is 25 degrees, so $\angle ADC$ measures 30+90+25=145 degrees, and $\angle DCB$ measures 85 degrees. This makes the largest measure $\boxed{145^\circ }$.
[/hide]

Denote $ABCD$, where $AB=BC=CD$, $\angle B=60$, $\angle C=70$, clearly, $ABC$ is equilateral, so $\angle BCA=60$, and $AC=DC$, then $\angle ACD=70-60=10$, so $\angle CAD=\angle ADC=85$, it follows that the largest angle is $60+85=145$


integration
inequalities
logarithms
algebra
partial fractions
calculus
calculus computations

Write
$\frac{1}{x(x+1)\cdots(x+n)}$
as
$\frac{A_0}{x} + \frac{A_1}{x+1} + \cdots + \frac{A_n}{x+n}$ (**)

To find $A_k$, we multiply (**) by $x+k$ and it equals to $\frac{1}{x(x+1)\cdots(x+k-1)(x+k+1)\cdots(x+n)}$.
Let $x=-k$, we obtain that $A_k = \frac{1}{(-1)^k k!(n-k)!}$.

$\frac1{n\cdot(n+1)!}<I_n<\frac1{n\cdot n!}$

This inequality follows from $\sum_{k=1}^{\infty}\frac1{k(k+1)\cdots(k+n)}= \frac1{n\cdot n!}.$

Unfortunately, the telescoping series computation that was so effective in calculating the sum is less effective with the integral.

liyi's partial fractions coefficients can be written $A_k=\frac{(-1)^k}{n!} {n\choose k}.$

Pursuing the calculation that way, we obtain the following:

$I_1=\ln 2$

$I_2=\frac12\ln\left(\frac{2^2}{1\cdot3}\right)$

$I_3=\frac16\ln\left(\frac{2^3\cdot4}{1\cdot3^3}\right)$

$I_4=\frac1{24}\ln\left(\frac{2^4\cdot4^4}{1\cdot3^6\cdot5}\right)$

In general, $I_n=\frac1{n!}\sum_{k=0}^n(-1)^{k+1}{n\choose k}\ln(1+k).$


demontrer que l'equation $x^5 +ax^4 + bx^3 + cx^2 +dx + e$ nadmet pas des solutions  reels si : $2a^5 < 5b$

This should be $2a^2 < 5b$ or something?


let $ a,b$ and $ b$ be positive real numbers.
prove that: $ 6(a^{3}\plus{}b^{3}\plus{}c^{3})\geq\ (a\plus{}b\plus{}c)(a^{2}\plus{}b^{2}\plus{}c^{2})$


HAPPY NEW YEAR!!!

DON't SPEND IT ON AOPS!!!

TO INFINITY AND BEYOND!! THE FUTURE!!!! :lol:

thank you !
i am so suprised that two of you post in our Sub_forum.any way ,thanks again.
long time absent in mathlinks.
Cuong

[quote="Mircea Lascu"]Happy New Year !

Mircea Lascu[/quote]

Thank you!  :wink:


For what values of n and k is C(n,k) odd? Find as simple and elegant a criterion as possible.

More generally, given a prime p, find a simple and elegant descrip-
tion of the largest power of p dividing C(n,k)

[quote="Qruni"]GREAT PROBLEM RELATED :
Simplify :
2^0C(n,0)+2^1C(n,1)+2^2C(n,2)+....+2^nC(n,n)[/quote]
[hide]It's just the binomial expansion of $ (2+1)^{n}$, and is thus $ 3^{n}$.[/hide]

[b]Solution.[/b]

[b]Theorem (Lucas).[/b] If $ n=\overline{n_{m}n_{m-1}\hdots n_{0}}_{p}$ for $ p$ prime (p-th base representation of n), and $ k=k_{0}+k_{1}p+\hdots+k_{m}p^{m},\ k_{i}\in[0,p)\cap\mathbb{Z}$, then
\[ \binom{n}{k}\equiv \prod_{i=0}^{m}\binom{n_{i}}{k_{i}}\pmod p. \]
[b]Informal Proof.[/b]
\[ \\ \begin{*eqnarray}f(x)&\equiv& (1+x)^{n}\equiv (1+x)^{n_{0}}[(1+x)^{p}]^{n_{1}}[(1+x)^{p^{2}}]^{n_{2}}\hdots[(1+x)^{p^{m}}]^{n_{m}}\\ &\equiv& (1+x)^{n_{0}}\hdots (1+x^{p^{m}})^{n_{m}}\pmod p\end{*eqnarray}\]
$ [x^{k}]f(x)=[x^{k_{0}}(x^{p}){k_{1}}\hdots (x^{p^{m}})^{k_{m}}]f(x)\implies\binom{n}{k}\equiv \prod_{i=0}^{m}\binom{n_{i}}{k_{i}}\pmod p$. $ \Box$

As a result, $ \binom{n}{k}\equiv 1\pmod 2$ iff all of the digits of the base-2 representation of $ k$ are less than or equal to their correspoding (by index) digits in the base-2 representation of $ n$. $ \Box$

For your second question, use the well known Legendre's formula.


Find all real numbers that satisfy the equation $|x+3|-|x-1|=x+1$. (Note: $|a| = a$ if $a\ge 0$; $|a|=-a$ if $a<0$.)

This is trivial.  Just consider 3 cases, and you to get 3 solutions.

EDIT: The above is the correct solution.


analytic geometry
geometry
geometric transformation
reflection
parallelogram
geometry proposed

[b][size=100][color=DarkBlue]Let $ \bigtriangleup A'BC,\ \bigtriangleup B'AC,\ \bigtriangleup C'AB$ be, three similar isosceles triangles with bases the side-segments $ BC,\ AC,\ AB$ respectively of a given triangle $ \bigtriangleup ABC,$ erected outwardly to it. Prove that the line segments $ AA',\ BB'\ CC',$ where $ A',\ B',\ C'$ are the midpoints of the segments $ EF,\ DF,\ DE$ respectively, are concurrent at one point.[/color][/size][/b]

Kostas Vittas.

Let $ \phi$ be the angle base of these isosceles triangles. Using Conway's formula, the barycentric coordinates of $D,E,F$ are:

$ D \equiv ( \minus{} a^2: S_C \plus{} S_{\phi}: S_B \plus{} S_{\phi})$

$ E \equiv (S_C \plus{} S_{\phi}: \minus{} b^2: S_A \plus{} S_{\phi})$

$ F \equiv (S_B \plus{} S_{\phi}: S_A \plus{} S_{\phi}: \minus{} c^2)$

Since the sum of the coordinates of $D,E,F$ equals $ 2S_{\phi},$ we compute easily the coordinates of the midpoints of segments $DE,EF,FD$ as:

$ A' \equiv (a^2 \plus{} 2S_{\phi}: S_{\phi} \minus{} S_C: S_{\phi} \minus{} S_B)$

$ B' \equiv (S_{\phi} \minus{} S_C: 2S_{\phi} \plus{} b^2: S_{\phi} \minus{} S_A)$

$ C' \equiv (S_{\phi} \minus{} S_B: S_{\phi} \minus{} S_A: 2S_{\phi} \plus{} c^2)$

Therefore, $ AA',BB',CC'$ concur at the inner Kiepert perspector relative to $\phi$

$K \equiv  \left (\frac {1}{ S_A \minus{} S_{\phi}}: \frac {1}{ S_B \minus{} S_{\phi}}: \frac {1}{ S_C \minus{} S_{\phi}} \right )$

Reflect $\triangle BCD$ about sideline $ BC$ into $\triangle BCD'.$ Notice that $ \angle FBD' \equal{} \angle ABC,$ since $ \angle FBD'$ is obtained by rotating $\angle ABC$ about $ B$ with angle +$\phi.$ Further,  $ \triangle BFD'$ and $ \triangle BAC$ are similar with similarity coefficient $\frac{_{ BF}}{^{BA}}.$ But $ \frac {_{BF}}{^{BA}} \equal{} \frac {_{AE}}{^{AC}}$  $ \Longrightarrow$ $ FD' \equal{} AE$ and analogously $ ED' \equal{} AF,$ which implies that $ AFD'E$ is a parallelogram $\Longrightarrow$ Line $ AA'$ goes through $ D'.$ By similar reasoning, we conclude that $ AA',BB',CC'$ concur at the inner Kiepert perspector relative to $\phi.$


geometry
geometric transformation
reflection
analytic geometry
perimeter
perpendicular bisector
puzzles

say you have Rx=0: (x,y)--> (-x,y)
and Ry=-x: (x,y)--> (-y,-x)

what is the composition of the above two reflections:
Ry=-x o Rx=0.
What single transformation will accomplish the same thing as the composition of these two reflection?

Rx=0: (x,y)--> (-x,y)
This means that the reflection of the point (x,y) over line x=0 (i.e. y axis) is (-x,y).

Ry=-x: (x,y)--> (-y,-x)
This means that the reflection of the point (x,y) over line y=-x is (-y,-x).

The composition of the above two reflections: Ry=-x o Rx=0 is (x,y)--> (-y,x)

That's all I could do.

This isn't so bad if you think about it.

Let $R_1: (x,y)\longrightarrow(-x,y)$,
and let $R_2: (x,y)\longrightarrow(-x,-y)$

The composition $R_1+R_2$ Will map $(x,y)\longrightarrow(-x,y)\longrightarrow(x,-y)$

Remember that if $P$ reflects across a line to $Q$, then the reflection line is the perpendicular bisector of $PQ$ (draw the diagram in your head to convince yourself). So we are looking for the perpendicular bisector of the line that goes through $(x,y)$ and $(x,-y)$.

The midpoint of this line is quite obviously $(x,0)$. It is also easy to notice that since they have the same $x$-coordinates, they two points are on the same vertical line, meaning that their perpendicular bisector must be horizontal. The horizontal line that goes through $(x,0)$ is $f(x)=0$.

Also note that the reflections can be done in either order. $R_1+R_2=R_2+R_1$.

Here's an problem I saw recently regarding reflections:
You have a point $P$ on the line $f(x)=x$, and a point $Q$ on the line $f(x)=0$. You have another point $R$ inbetween the these two lines. $P$, $Q$, and $R$ are all in the first quadrant. Defining $R$ as $(a,b)$, find points $P$ and $Q$ in terms of $a$ and $b$ such that the perimeter of triangle $\Delta PQR$ is a minimum.

[quote="gopherhole112"]Also note that the reflections can be done in either order. $R_1+R_2=R_2+R_1$.[/quote]

Be careful, though -- this isn't a general truth, but works in this case just because the lines of reflection are perpendicular to each other.


geometry
3D geometry
pyramid
Pythagorean Theorem

Two regular square pyramids have all edges 12cm in length. The pyramids have parallel bases and parallel edges, and each has a vertex at the center of the other pyramid's base. What is the total number of cubic centimeters in the volume of the solid of intersection of the two pyramids? Express your answer in simplest radical form.

[quote="dragon96"]If you "chop" this combined figure in half, then there will be a tiny pyraimd inside of a frustrum. The base of the tiny pyramid is 1/4 the area of one of the original pyramids, while the height is 1/2 the original pyramid. THerefore, 1/2 of the intersection thing is 1/8 of one pyramid. So the entire intersection is 2 * 1/8 = 1/4 of the original pyramid's volume.

Using Pythagorean Theorem and stuff, we get the height of one pyramid is $ 6\sqrt{2}$. (I can't really show how, but it's not too hard) Therefore, the entire pyramid's volume is $ \frac{(12)^2\cdot 6\sqrt{2}}{3}\equal{}288\sqrt{3}$. Thus, the volume of the intersection is $ 72\sqrt{3}$[/quote]

How to find the intersection, any picture or so?
I don't see why you have $\sqrt{3}$ from the first thing

[quote="professordad"]Try to picture two pyramids stacked on top of each other, just that one pyramid's vertex is at the other's center.[/quote]

Is the correct answer $72\sqrt{2}$ than?
( I got that if I understand your explaining)

yes, the answers say that 72 \sqrt 2 is the answer.


I don't know how to write matrices in latex, sorry, but let

"B = Matrix([[1, 3, 4, 1, 3], [3, 9, 8, 6, 5], [5, 8, 7, 5, 2], [7, 7, 6, 3, 9], [9, 6, 5, 7, 7]])" (Maple notation).

Given that 13413, 39865, 58752, 77639, 96577 are all divisible by 17, show that the determinant of B is divisble by 17. As a hint it is given that $13413 = 1\cdot 10^4 + 3\cdot10^3 + 4\cdot10^2 + 1\cdot10 + 3$.

Replace the fifth column $C_5$ with $10^4C_1+10^3C_2+10^2C_3+10C_4+C_5$. This doesn't change the determinant, and the new fifth column is a multiple of 17.


I'm going to india in a few days and I want to buy a few books from there. Anywhere from algebra, geometry, etc. to analysis and linear algebra. Any recommendations? Thin, and light books are very welcome since I can convince my mom easier that they won't be a hassle on the plane :)

Well [b]Challenges and Thrills in Pre - College Mathematics[/b] is a very nice book. The Authors are B.J.Venkatachala, K.N.Ranganathan, C.R.Pranesachar.
Also [b]An Excursion In Mathematics[/b] is very good.
[b]Problem Primer For the Olympiad[/b] is also a nice book.


analytic geometry
vector
linear algebra
linear algebra unsolved

Let $E$ be a vectorial space with dimension $n$ and $\{v_1,\ldots,v_n\}$ its base, and let $a_1,\ldots,a_n$ be real numbers, not all zero. Take

$F=\{x=x_1v_1+\ldots+x_nv_n \in E|a_1x_1+\ldots+a_nx_n=0\}$

How can I show that the dimension of $F$ is $n-1$?

thanks

Intuitively, since the $a_{k}$ are not all $0$ and because of the definition of $F$, each element of $F$ has one of its coordinates entirely determined by its $n-1$ other coordinates.

I feel retarded i couldn't see that. I thought of it as a given vector $(a_1,\ldots,a_n)$, and F as the subspace containing all perpendicular vectors to $(a_1,\ldots,a_n)$. For exemple, in R^3, it should be a plane passing through the origin containing the vectors perpendicular to $(a_1,a_2,a_3)$, and I knew that it had to loose one dimension...too geometric, less algebric  :P