MATHCOUNTS articles calculus probability

I saw an interesting poster in my math class:  "Be An Actuary."  I asked around, and I found out what an actuary does.  Basically, an actuary manages [i]risk[/i].  I can't explain very well, but it seems that the job of an actuary has been ranked as one of the top 5 jobs consistently.  So check out [url=http://www.beanactuary.org/about/whatis.cfm]this website[/url], as I did, and see if... actuarying (:D) is right for you!

:) I heard about Actuary when I was in Grade 7. One day, I went to a bookstore with my mum. By the way, that was in China. We saw a book named "the most popular careers in 21 century". Mum read it and found one called Actuary. And then she knew what I was going to do in the future. Well, I did not really care about that for a long time. Until last summer, I was going to apply for a university. We started to talk about my major. I don't remember how we recalled Actuary. Actually, I am quite interested in science, such as bio, chem and even physics. But math works better with me. Pure math definitely "no." Actuarial science seems very fun. By the way, I am in Canada now. I will go to Actuarial Science in University of Waterloo this September. I hope I will like my university and the major.

Please keep us updated about actuaries, as it seems like something I'd really want to look into.  I have PLENTY of time, though.  I still have to win MathCounts next year, and then score a perfect SAT score, if that's still possible out of 2400...

[quote="yif man12"]You don't have to get a perfect SAT or win Mathcounts to be an actuary. That's silly nonsense.

If you saw that poster, I'm pretty sure you've heard of the 9 (?) actuary tests. Those are way more important than an SAT scofre.[/quote]
Actuary is a type of position that insurance company hire math people to do. Not really that high level of math. The pay is average in my opinion. What they do is to plug in the factors (age, sex, years of work in the company, retirement age, how many more years to go before retirement, contribution to the retirement account....) to determine the level in a defined benefit plan for retirement. There are basically two types of retirement plans: defined benefit (employer bear the investment risks) and defiend contribution (employee bear the investment risks). I knew those cuz I just finished a 21-months evening classes in CFP (certified financial planner). CFP has nothing to do with my real job. I did it purely for my personal info.

Why is 9 actuary tests? I think it's 7. In waterloo, they usually finish the first four when they are taking bechelor. And then, the other three will be finished during the next couple of years while they are working.

I wasn't sure about the number. I knew there were quite a few though, and 9 seemed like a fitting number. But it may very well be 7. In fact, now that you mention it, 7 seems right. Yeah, probably 7 tests.

[quote="yif man12"]You don't have to get a perfect SAT or win Mathcounts to be an actuary. That's silly nonsense.

If you saw that poster, I'm pretty sure you've heard of the 9 (?) actuary tests. Those are way more important than an SAT scofre.[/quote]

Can't anyone take a joke?  Or maybe its just my jokes...

[quote="qvc122"]Actuary is a type of position that insurance company hire math people to do. Not really that high level of math. The pay is average in my opinion. [/quote]
An actuarial career has consistently ranked high in both salary and job satisfaction (see  [url=http://www.beanactuary.org/counselors/best_job.cfm]article[/url]).  There are two types of actuary: casualty and life.  Both require passing a series of exams for promotion.  Casualty actuaries, for example, must pass 9 exams in order to be certified as a "fellow".  It is not unusual to pass one exam per year.  Salaries for casualty fellows start around 100,000 and go up to 300,000+ (see [url=http://www.actuary-recruiter.com/salarysurvey.html]stats[/url] for "FCAS", which stands for "Fellow Casualty Actuarial Society").  Few technical careers can offer such high salaries.

The Bureau of Labor Statistics site lists 69,970 as the median salary for actuaries, with the top 10% earning at least 137,650.  Compare to salaries for mathematicians (76,470 / 112,780), electrical engineers (69,930 / 103,860) and financial analysts (57,100 / 108,060).  Unlike some industries, where the highest paid positions are reserved for sales and marketing types, in insurance actuaries can rise to the highest positions.

It's true you don't need an extensive math background to be an actuary.  Calculus, probability, statistics, and economics at the college level are all you need, along with an interest in business.  The rest of the training takes place on the job.

[quote="qvc122"]What they do is to plug in the factors (age, sex, years of work in the company, retirement age, how many more years to go before retirement, contribution to the retirement account....) to determine the level in a defined benefit plan for retirement. [/quote]
Not quite. Actuaries primarily analyze and manage risk.  Few actuaries work on retirement plans.  The ones that do are the ones who create the plans; they're not the ones who plug in numbers.

I have 3 Ivy League PhD friends who have worked in insurance.  One specializes in risk management, one models the effects of natural disasters on the economy, and the third worked for many years as a casualty actuary.

[quote]I will go to Actuarial Science in University of Waterloo this September. I hope I will like my university and the major.[/quote]

I'm going to Waterloo this fall too! I'm also writing the SOA/CAS Probability exam this May 24th.

I worked as an actuarial technician while in college and the work was fairly easy and paid well.  If you're interested in a career that allows you to make good money and that doesn't cost you 60-80 hours per week, it's a great option.  I recommend it to bright math students whose interest in a healthy family life exceeds their ambition in taking over the world.  You're more likely to take over some niche of the world as a programmer than as an actuary.

I was wondering, can business majors become actuaries?

It doesn't matter what your college major is.  The best way to show your interest in an actuarial career is to pass the first couple of actuarial exams (see [url=http://www.beanactuary.org/exams/exam_info.cfm]description[/url]) while still in college.  The first few exams are the same for both casualty and life actuaries, I believe.  The first exam covers probability, the second covers financial math.  Study guides are available.

[quote="hello"]But I mean is it related to Business? If I am majoring in Accounting and Economics, will I be able to ba an actuary too after graduating?[/quote]
Yes.

Hi,

I have been an actuary for 30 years.  We are not theoretical mathematicians, but rather we are people who use our mathematics to solve business problems in insurance and related fields.  You don't have to be an actuarial science major and you don't even have to be a math major, but you have to have studied enough college math to compete with the math majors on the actuarial exams.  You do have to be someone who wants to work in the world of business, and insurance in particular, and who can tolerate that the problems we solve have no single correct answer (for example which of a hundred applicants should we insure) and whose ultimate answer in fact will not be known for many years (for example, did we charge enough premium for the medical malpractice liability insurance we sold this year, after all the lawsuits that we covered are finally settled).

The actuarial societies around the world offer actuarial exams leading to the designation of "Fellow."  This word is derived from the word "fellowship" and is not intended to be sexist.  Although there are many female Fellows, there seems to be no push to change the word.

Not every beginning actuary passes all the exams to become a Fellow.  But those of us who do generally get a pretty good income and interesting work assignments.  Some experienced actuaries go into non-actuarial work, including becoming presidents of their companies.  I don't think any Fellow, regardless of what work he or she is doing, regrets becoming an actuary.

Here is a web page you can visit called Advice to Beginning Actuarial Students on http://users.aol.com/fcas/advice.html  which answers a lot of questions about actuarial careers and refers you to additional resources.

Best wishes,

[quote="joeorez"]Here is a web page you can visit called Advice to Beginning Actuarial Students on http://users.aol.com/fcas/advice.html  which answers a lot of questions about actuarial careers and refers you to additional resources.[/quote]
Very interesting site! Thanks.

[quote="wild octagon"] I have 3 Ivy League PhD friends who have worked in insurance.  One specializes in risk management, one models the effects of natural disasters on the economy, and the third worked for many years as a casualty actuary.[/quote]

There is a difference between working in the insurance industry, and being an actuary.
When a high-level technical PhD works in insurance it is typically not as an actuary.  The ones who do end up as actuaries often do so as a result of having children during grad school, or looking for something lower-risk but still OK-paying as compared to other types of technical work.

I had a long posting in an earlier thread about actuaries, which was lost during the system crash.  If time allows I will write more about this, but for now suffice it to say that qvc122 and MCrawford's remarks are right on target, IMO.  Actuary is certainly a good job in many respects, but those job satisfaction surveys are significantly overblown.

At my university in Bergen you can be an actuary after 5 years study.
You need a lot of statistic-courses, but no one of them are especially challenging (compare to pure mathematic-courses). You also need some calculus, analysis, and some computer-programming. It is very easy to get a job as an actuary in Norway. You start earning
about 400 000 NOK (about 60 000 USD).

It is an exciting job, but I like pure math too much, so I will not choose that direction..

[quote="fleeting_guest"]
There is a difference between working in the insurance industry, and being an actuary.
When a high-level technical PhD works in insurance it is typically not as an actuary.  The ones who do end up as actuaries often do so as a result of having children during grad school, or looking for something lower-risk but still OK-paying as compared to other types of technical work.

I had a long posting in an earlier thread about actuaries, which was lost during the system crash.  [b]If time allows I will write more about this[/b].......[/quote]

I am very interested if you have time to write more about actuary. Not that I want to become one, but just for the information.

i have wanted to be an actuary but not many colleges have actural sciences as a major. is it really necessary to have this as a major or will a major in mathematic be ok?

Mathematics or statistics are both fine majors for someone who desires to be an actuary. Many universities that don't have a specific major in actuarial sciences DO have an actuarial track for math majors, with specific courses that prepare for work as an actuary.

I never took an actuarial course and worked full time as an actuarial assistant (and took actuarial exams) while in college.  So long as you have a strong statistics background, you can work through the actuarial track on the job.  Companies even provide leave time to study for the exams.

The plus side of doing the actuarial track in college is getting the higher starting salary due to having more exams under your belt.  However, such a distinction is temporary, and a better math and computing education might make you a more attractive employee in the long run.

What kind of background in Business is required for entry-level Actuarial positions? Is this not very important if you're a competent math major?

At this point I have just been exposed to proof-based math that are presented using a "pure" theoretical approach. The emphasis is on the math and not on the applications of the math. However, before I graduate I will have taken a class in Statistics, a class in Probability, and several Economics classes. Is there a significant amount of material besides this required for the first few tests and a chance for a starting actuarial position?

You don't need classes beyond math if you have a good general understanding of finance and economics.  But if you don't feel that you naturally see how incentives work, I would recommend taking finance/econ courses or actuarial courses.

[quote="joeorez"]Hi,

As a practicing actuary, I see a lot of resumes.

You don't have to have an actuarial major to be an actuary.

But if you do, then:

1. I would expect you to know something about insurance when I interview you.  I wouldn't expect this of a math major, but an actuarial major who can't discuss anything about insurance (how about insurance current events?) is fooling him- or herself.

2. I would expect you to be reasonably familiar with the actuarial exam system, have passed one or two exams, and (assuming you are North American) have taken the appropriate courses towards Validation by Educational Experience (VEE).  If you don't know what VEE is, ask your professors.

3. I would expect that you have had at least one actuarial science professor who is him- or herself a credentialed actuary, and therefore keeps up with actuarial research and with educational requirements like VEE.  An actuarial science department with no credentialed actuaries is bogus.

I see graduates with actuarial degrees who can't do the above, and I snicker at what value their major is compared to a math major.

Comments?[/quote]

That sums it up pretty well folks.

i keep seeing that there are different types of actuaries, however, it is never mentioned what the difference between them is.
I am wondering if someone could list the different kinds of actuaries and how they differ from the other in the list.
I at this point i know that i want to be an actuary but an not sure which kind i am most suited for.
any help would be greatly appreciated.

[quote]I am a grade 12 student in Ontario. I want to see how easy/difficult the first exam is and also if like acturial science or not.
What I want to know is that what book should I read (as a high school student) which will give me a good idea of what is expected on the first exam and it will prepare me for the first exam?[/quote]

Can someone please post a reply to this question?

See

http://www.beanactuary.org

Or visit the CAS or SOA websites to see what the first exams cover, and what some recommended texts are.

At the high school level, it probably makes sense to focus on learning calculus, probability, and economics.  The exams are not designed for high school students, but you can certainly prepare for them now if you are mathematically advanced.


floor function
search
number theory unsolved
number theory

This one took me a little less than a day's work last week to figure out :P
But I know someone here can do it in 5 minutes.

Question from my book states it as : Show that if $ f(n) \equal{} \frac{\left(1\plus{}\sqrt{8n\minus{}7}\right)}{2}$ then the $ n$th term of the sequence
$ 1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,...$ is given by $ \lfloor{f(n)}\rfloor$.

$ 1 \plus{} \sqrt{8n \minus{} 7} \equal{} 2m \implies$
$ 8n \minus{} 7 \equal{} (2m \minus{} 1)^2 \equal{} 4m^2 \minus{} 4m \plus{} 1 \implies$
$ n \equal{} \frac{m^2 \minus{} m}{2} \plus{} 1$;

in other words, $ \lfloor f(n) \rfloor$ increments whenever $ n$ is one greater than a triangular number, exactly the observed pattern.

That solution is a lot more cool than mine. I derived the $ f(n)$ given instead ><
Thanks again t0rajir0u !

Do you have any good practice questions like this one ?

[url=http://www.amazon.com/Problem-Solving-Strategies-Problem-Books-Mathematics/dp/0387982191]Engel[/url] covers sequences of this form briefly.  You might find some problems if you search for "Engel" on the forums, although you might just get a lot of references to Cauchy-Schwarz in Engel form.


I was playing around with the factorizations for differences of squares, cubes, etc. and found this:

[tex]
a^n-b^n = (a-b)\left(\sum_{i=0}^{n-1} a^{n-1-i}b^i\right)
[/tex]

Is this correct?  Also, I tried to prove it and I want to know if I made any mistakes (I'm not very good at avoiding stupid mistakes).  I put my proof at [url]http://home.comcast.net/~carlosrd/nths.pdf[/url]  I didn't know if this should be in Getting Started or Intermediate (I'm in-between these two difficulties, so I don't know where to post these things).

Very nice. Though you can just expand the right side:

[tex]\displaystyle ~~ (a - b)(a^{n-1}+a^{n-2}b+\ldots+ab^{n-2}+b^{n-1})[/tex]

[tex]\displaystyle =(a^n + a^{n-1}b + \ldots + a^2b^{n-2} + ab^{n-1})[/tex]
[tex]\displaystyle - (a^{n-1}b + a^{n-2}b^2 + \ldots + ab^{n-1} + b^{n})[/tex]

[tex]\displaystyle = a^n - b^n +[/tex]
[tex]\displaystyle (a^{n-1}b + \ldots + a^2b^{n-2} + ab^{n-1} - a^{n-1}b - a^{n-2}b^2 - \ldots - ab^{n-1})[/tex]

The stuff inside the brackets all cancel out, so we're left with [tex]a^n - b^n[/tex].

Can you figure out a similar formula for [tex]a^n + b^n[/tex]?

Thank you!  I did way more work than I needed to.   :D  I'll try to find that other formula.

[tex]a^n + b^n = (a+b)^n - \sum_{i=1}^{n-1} {n\choose i}a^{n-i}b^i[/tex]

Not exactly similar to the previous formula, but it works.

I was looking for something along the lines of (a + b)(......). That is, a factorisation of an + bn.

(Note: n must be odd for this factorisation to work - can you see why?)

Oh!  I found that factorization a while ago - it is

[tex](a+b)(\sum_{i=0}^{n-1} (-1)^i a^{n-1-i} b^i)[/tex]

I didn't post that one because I thought you wanted one that worked for all integers n.

Ah - sorry, I didn't make myself clear then. Anyway, that's right.


algebra
polynomial
Vieta
modular arithmetic

Na eksetastei an iparxei timi tou $d\in\mathbb{Z}$ tetoia wste to poliwnimo $P_{d}(x)=x^{3}-87x^{2}+181x+d$, na exei treis akeraies rizes.

Alexandros

Έστω ότι το πολυώνυμο έχει $3$ ακέραιες ρίζες $\alpha ,\, \beta, \, \gamma$ .

Από τους τύπους του Vieta έχουμε :

$\\ \begin{array}{cccc}\alpha+\beta+\gamma &=&87&(1)\\\ \alpha \beta+\beta \gamma+\gamma\alpha &=&181&(2) \\\ \alpha \beta \gamma &=&-d&(3)\end{array}$

Από την $(1), \,\, \alpha+\beta+\gamma \equiv 0 \mod3$ άρα τα υπόλοιπα των $\alpha ,\, \beta, \, \gamma , \, \mod3$

είναι $\{(1,1,1),(0,0,0),(2,2,2),(1,2,0)\}$ και οι μεταθέσεις του τελευταίου .

Από την $(2)$ είναι $\alpha \beta+\beta \gamma+\gamma\alpha \equiv 1 \mod3$ .

Αλλά κανένας από τους συνδυασμούς των υπολοίπων δεν δίνει υπόλοιπο  $1$

άρα δέν υπάρχουν ακέραιοι $\alpha ,\, \beta, \, \gamma$ ώστε να ικανοποιείται η $(3)$ .

Συγχωρήστε με αν κάνω κάποιο λάθος  :roll:

Apolitws swsta Ramanujan :)

Mporeis kai anti $\pmod{3}$ na doulepseis kai $\pmod{8}$ (pou doulepsa egw giati pisteua otan to eida oti itan diskolotero) me katallilo tropo, alla einai pio diskolo tote kai den kerdizeis kati perissotero apo ton diko sou tropo.

Alexandros

αρκει να αποδειξουμε οτι το πολυωνυμο αυτο δεν γραφεται ως γινομενο τρειων ακεραιων πολυωνυμων πρωτου βαθμου δηλ x-c  οπου c ριζα του πολυωνυμου?

Nai Kwsta! Opou c riza tou poliwnimou kai c akeraios arithmos. H lisi tou Ramanujan parapanw nomizw kaliptei plirws to zitoumeno tou provlimatos kai einai poli apli (kai aplousteri apo tin diki mou opws proanefera parapanw).

Alexandros


Is there a way to find how many digits a certain expression has, like $ 30!$ :?:

I forgot exactly how the problem goes, but it was a crazy problem that asked how many kilometers long would a certain factorial be  :lol:

Thanks for the help.

With a non-programmable calculator, your best bet is Stirling's Approximation.  We have

$ n!\approx\sqrt{2\pi n}\left(\frac{n}{e}\right)^{n}$

To find the number of digits, we take the base-$ 10$ logarithm and add $ 1$.  In other words,

$ d \equal{}\lfloor\log\sqrt{2\pi n}\plus{}n\log\frac{n}{e}\plus{}1\rfloor$

The difficulty in this problem lies in the fact that large factorials cannot be directly computed on a calculator, as they will overflow.

So if I wanted to find the amount of digits of $ 100$!, then I would do this, right?

$ d\equal{}log\sqrt{2*\pi*100}\plus{}100\;log\;\frac{100}{e}\plus{}1$

$ d\equal{}\frac{1}2\;log\;200\pi\plus{}100\;log\;36.7879\plus{}1$

$ d\equal{}\frac{1}2\;log\;628.3185\plus{}156.5705\plus{}1$

$ d\equal{}1.399\plus{}156.5705\plus{}1$

$ d\approx159\;digits$

Wow that's nice. Thanks for your help.


geometry
3D geometry
tetrahedron
IMO 1972
IMO

Let our planes be $\pi_1,\pi_2,\pi_3,\pi_4$, which we assume to be parallel to the $xy$-plane, listed in the increasing order of their $z$-coordinates. First take a plane $\pi$ orthogonal to $\pi_i$, which cuts $\pi_1,\pi_2,\pi_3$ along three lines $d_1,d_2,d_3$. On these three lines, take three vertices $A_1,A_2,A_3$ respectively of an equilateral triangle (it is well-known that this is possible; in fact, the problem here is the $3$-dimensional version of this), and then complete the two regular tetrahedra $A_1A_2A_3P_1,A_1A_2A_3P_2$ having $A_1A_2A_3$ as one of their faces. Both $P_i$ lie below $\pi_4$.

Now take another plane $\pi$ and repeat the construction above. If $\pi$ makes a small enough angle with the $\pi_i$'s, one of the $P_i$'s we get this time must lie above $\pi_4$. Now, if we move the initial position of $\pi$ towards the new one continuously and record the $z$-coordinates of $P_1,P_2$, these will be continuous functions of the angle that $\pi$ makes with $\pi_i$, and for one of the points $P_1,P_2$ the $z$-coordinate will move continuously from being smaller than that of $\pi_4$ to being larger than it, meaning that at some point, one of the points $P_1,P_2$ will lie on $\pi_4$, and this is what we want.

[quote="orl"]Given four distinct parallel planes, prove that there exists a regular tetrahedron with a vertex on each plane.[/quote]

Here is another solution.

Let's denote the (directed) distance between two parallel planes p and p' by d (p; p'), and the (directed) distance between two parallel lines g and g' by d (g; g'). (Directed distances are defined as follows: If $p_1$, $p_2$, $p_3$, ... is a family of parallel planes in space, then we choose a unit vector $\overrightarrow{v}_p$ perpendicular to all of these planes (there are two such unit vectors, and we have to choose one of them), and then, by the [i]directed distance[/i] between two of these planes $p_i$ and $p_j$, we denote the real number k such that the translation with translation vector $k\cdot\overrightarrow{v}_p$ maps the plane $p_i$ to the plane $p_j$. Similarly, we define the directed distance between two of a family of parallel lines in a plane.)

The problem can be rewritten as follows: Given four distinct parallel planes $p_1$, $p_2$, $p_3$, $p_4$ in space, prove that there exists a regular tetrahedron XYZW such that $X\in p_1$, $Y\in p_2$, $Z\in p_3$, $W\in p_4$.

In order to do this, it is enough to find a regular tetrahedron ABCD somewhere in space and four parallel planes $q_1$, $q_2$, $q_3$, $q_4$ such that $A\in q_1$, $B\in q_2$, $C\in q_3$, $D\in q_4$ and $d\left(q_1;\;q_2\right): d\left(q_2;\;q_3\right): d\left(q_3;\;q_4\right)=d\left(p_1;\;p_2\right): d\left(p_2;\;p_3\right): d\left(p_3;\;p_4\right)$. In fact, once we have found such a tetrahedron ABCD and such planes $q_1$, $q_2$, $q_3$, $q_4$, then, because of $p_1\parallel p_2\parallel p_3\parallel p_4$, $q_1\parallel q_2\parallel q_3\parallel q_4$ and $d\left(q_1;\;q_2\right): d\left(q_2;\;q_3\right): d\left(q_3;\;q_4\right)=d\left(p_1;\;p_2\right): d\left(p_2;\;p_3\right): d\left(p_3;\;p_4\right)$, there exists a similitude transformation which maps the planes $q_1$, $q_2$, $q_3$, $q_4$ to the planes $p_1$, $p_2$, $p_3$, $p_4$; this similitude transformation will then obviously map the regular tetrahedron ABCD with $A\in q_1$, $B\in q_2$, $C\in q_3$, $D\in q_4$ to a regular tetrahedron XYZW with $X\in p_1$, $Y\in p_2$, $Z\in p_3$, $W\in p_4$; hence, the existence of such a tetrahedron XYZW will be proven, and the problem will be solved.

So consider a regular tetrahedron ABCD lying arbitrarily in space; we try to find four parallel planes $q_1$, $q_2$, $q_3$, $q_4$ such that $A\in q_1$, $B\in q_2$, $C\in q_3$, $D\in q_4$ and $d\left(q_1;\;q_2\right): d\left(q_2;\;q_3\right): d\left(q_3;\;q_4\right)=d\left(p_1;\;p_2\right): d\left(p_2;\;p_3\right): d\left(p_3;\;p_4\right)$.

In fact, we start working in the plane ABC. Let T be the point on the line AC such that $AT: TC=d\left(p_1;\;p_2\right): d\left(p_2;\;p_3\right)$ (where the segments AT and TC are directed). Let $g_2$ be the line BT, and let $g_1$ and $g_3$ be the parallels to the line $g_2=BT$ through the points A and C, respectively. Then, the lines $g_1$, $g_2$, $g_3$ are parallel and, by Thales, $d\left(g_1;\;g_2\right): d\left(g_2;\;g_3\right)=AT: TC$. Thus, $d\left(g_1;\;g_2\right): d\left(g_2;\;g_3\right)=d\left(p_1;\;p_2\right): d\left(p_2;\;p_3\right)$. Now, denote by $g_4$ the line in the plane ABC which is parallel to the lines $g_1$, $g_2$, $g_3$ and satisfies $d\left(g_1;\;g_2\right): d\left(g_2;\;g_3\right): d\left(g_3;\;g_4\right)=d\left(p_1;\;p_2\right): d\left(p_2;\;p_3\right): d\left(p_3;\;p_4\right)$.

Now, let $q_4$ be the plane passing through the line $g_4$ and the point D. Let $q_1$, $q_2$, $q_3$ be the planes parallel to $q_4$ and passing through the lines $g_1$, $g_2$, $g_3$, respectively (of course, we can construct such planes since the lines $g_1$, $g_2$, $g_3$ are parallel to $g_4$). Thus, we have found four parallel planes $q_1$, $q_2$, $q_3$, $q_4$ such that $A\in q_1$, $B\in q_2$, $C\in q_3$, $D\in q_4$, and these planes obviously satisfy $d\left(q_1;\;q_2\right): d\left(q_2;\;q_3\right): d\left(q_3;\;q_4\right)=d\left(g_1;\;g_2\right): d\left(g_2;\;g_3\right): d\left(g_3;\;g_4\right)$. Since $d\left(g_1;\;g_2\right): d\left(g_2;\;g_3\right): d\left(g_3;\;g_4\right)=d\left(p_1;\;p_2\right): d\left(p_2;\;p_3\right): d\left(p_3;\;p_4\right)$, we thus have $d\left(q_1;\;q_2\right): d\left(q_2;\;q_3\right): d\left(q_3;\;q_4\right)=d\left(p_1;\;p_2\right): d\left(p_2;\;p_3\right): d\left(p_3;\;p_4\right)$. Hence, according to the above, the problem is solved.

  Darij

How can I find new plane π in grobber's solution it isn't described? :)  :-D


What is the units digit of the product of 5 consecutive whole numbers?

[hide="Solution"]Note that, given any two consecutive whole numbers, one must be even. This is obvious; two consecutive whole numbers cannot be odd, for their difference must be $ 1$, yet that is an odd number.

We can generalize. Given any $ n$ consecutive whole numbers, one must be divisible by $ n$. The proof of this is similar; if $ n$ consecutive numbers do not contain one number divisible by $ n$, by the pigeonhole principle, two numbers must be equivalent modulo $ n$, yet their difference is strictly less than $ n$; they are equal. Contradiction, therefore, the original claim is true.

Therefore, our product is divisible by $ 10$, so the units digit is $ \boxed{0}$.[/hide]

Simpler:
Any set of 5 consecutive integers contains one multiple of 5 and at least two multiples of 2. The product can be written as $ 2^25k\equal{}20k$ for some whole number value of k, which obviously ends in a 0.

Even simpler:
Any set of 5 consecutive integers contains one multiple of 5 and a multiple of 2. This make the product 2*5k=10k for some k, which means that the product must always end in 0

The product of 'n' consecutive nos. is divisible by n!

so this is divisible by 5! = 120

so its unit's digit is 0


function
real analysis
real analysis unsolved

Consider a real valued function f(x) satisfying $ 2f(xy)\equal{}(f(x))^y  \plus{}(f(y))^x$ for all x,y $ \in$ R and f(1)=a where a $ \neq$1.Prove that $ (a\minus{}1)\sum_{i\equal{}1}^{n}{f(i)}\equal{}a^{n\plus{}1}\minus{}a$

take $ xy \equal{} i \in \mathbb{N}$.

$ 2f(i) \equal{} f(i)^1\plus{}f(1)^i \implies f(i) \equal{} f(1)^i \equal{} a^i$

$ (a\minus{}1)\sum_{i\equal{}1}^{n}{f(i)}\equal{}(a\minus{}1)\sum_{i\equal{}1}^{n}{a^i} \equal{} a^{n\plus{}1}\minus{}a$


Let $A$ be closed subset of a plane and let $S$ be a closed circle which contains $A$ such that for every closed circle $S'$ if $A\subseteq S'$ then $S\subseteq S'.$ Prove that every inner point of $S$ is a midpoint of some segment with endpoints in $A.$

By "circle", I guess you actually mean "disk".

We can show that $A$ contains the boundary $\partial S$ of $S$, and the desired conclusion follows immediately from this.

Suppose $x\in\partial S\setminus A$. Because $A$ is closed, there is an open neighborhood $U$ of $x$ in the plane which is not contained in $S$. Let $a\ne b\in U\cap\partial S$ such that the intersection of one of the hyperplanes bounded by $ab$ with $S$ is completely contained in $U$ (a drawing might make things clearer). Now let $D_{n}$ be a sequence of open disks whose boundaries pass through $a,b$, with radii going to $\infty$, and such that the intersection of $D_{n}$ with the mentioned hyperplane bounded by $ab$ goes to $\emptyset$.

Since $\bigcup_{n}D_{n}$ is the other hyperplane bounded by $ab$, and this hyperplane clearly contains $A$, and since $D_{n}$ are open while $A$ is compact, there is a $D_{n}$ which contains $A$. The closure of $D_{n}$ is a closed disk which contains $A$ but does not contain $S$, contradicting the hypothesis.


For any positive real numbers $a,b$ and $c$, prove that
\[ \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+a+b+c \ge \frac{6(a^2+b^2+c^2)}{a+b+c} \]

What does SOS means?How to solve this one?

SOS means Sum of Squares. (there are two threads dealing with it : one by Darij Grinberg, and the other one by Hungkhtn)

And yeah, SOS kills this one! :)

Jin,If you want the solution,i will post (by use SOS method) :)  :)

nhat, If so the following proof is easier. ;)
$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+a+b+c \ge \frac{6(a^2+b^2+c^2)}{a+b+c}\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(a^4c+a^3c^2+2a^2b^2c-4a^3bc)\geq0.$
Let $a=min\{a,b,c\}$ and $b=a+x,$ $c=a+y,$ where $x\geq0,$ $y\geq0.$
Hence, our inequality equivalent to the following:
$2(x^2-xy+y^2)a^3+(x^3-3x^2y+6xy^2+y^3)a^2+$
$+(x^4-4x^3y+5x^2y^2+2xy^3+y^4)a+x^2y^3+xy^4\geq0,$ which obviously true.

[quote="nhat"]OK,Here is my solution :)
$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+a+b+c \geq 6\frac{a^2+b^2+c^2}{a+b+c} \Longleftrightarrow \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}-3\frac{a^2+b^2+c^2}{a+b+c} \geq 3\frac{a^2+b^2+c^2}{a+b+c}-(a+b+c) \Longleftrightarrow \sum a^2(\frac{1}{b}-\frac{3}{a+b+c}) \geq \frac{1}{a+b+c}(\sum (b-c)^2) \Longleftrightarrow \sum a^2(\frac{(a-b+b-c)}{b(a+b+c)}) \geq \frac{1}{a+b+c} (\sum (b-c)^2) \Longleftrightarrow \sum (b-c)(\frac{b^2}{c}-\frac{a^2}{b}) \geq \sum (b-c)^2 \Longleftrightarrow \sum (b-c) \frac{(b^3-a^2c)}{bc} \geq \sum (b-c)^2 \Longleftrightarrow \sum (b-c) \frac{(b^3-c^3+c^3-a^2c)}{bc} \geq \sum (b-c)^2 \Longleftrightarrow \sum (b-c)^2\frac{b^2+bc+c^2}{bc}+\sum (b-c)(c-a) \frac{c(a+c)}{bc} \geq \sum (b-c)^2$(1)
We have:$2\sum (b-c)(c-a) \frac{a+c}{b}=\sum ((a-b)^2-(a-c)^2-(b-c)^2)\frac{a+c}{b}=-\sum (b-c)^2 (\frac{a+c}{b}+\frac{b+c}{a}-\frac{b+a}{c})$
So that (1) $\Longleftrightarrow 2\sum (b-c)^2\frac{b^2+c^2}{bc} \geq \sum (b-c)^2(\frac{a+c}{b}+\frac{b+c}{a}-\frac{a+b}{c}) \Longleftrightarrow \sum (b-c)^2(2\frac{(b^2+c^2)}{bc}+\frac{a+b}{c}-\frac{a+c}{b}-\frac{b+c}{a}) \geq 0$(2)
Put:$S_{a}=2\frac{(b^2+c^2)}{bc}+\frac{a+b}{c}-\frac{a+c}{b}-\frac{b+c}{a}$
      $S_{b}=2\frac{(a^2+c^2)}{ac}+\frac{c+b}{a}-\frac{a+c}{b}-\frac{a+b}{c}$
      $S_{c}=2\frac{(a^2+b^2)}{ab}+\frac{c+a}{b}-\frac{b+c}{a}-\frac{a+b}{c}$
From here you can check yourseft two statement:$S_{a}+S_{b}+S_{c} \geq 0$ and $S_{a}S_{b}+S_{b}S_{c}+S_{c}S_{a} \geq 0$ (it's too long ,so i don't post the solution here)
After this we apply the criterion of SOS,we easily get $\sum S_{a} (b-c)^2 \geq 0$
Hence the problem is proved  :D  :D[/quote]

It's too long. SOS is more clearly.
\[ \sum_{cyc} (\frac{a^2}{b}+b-2a) \ge 2(\frac{3(a^2+b^2+c^2)}{a+b+c}-a-b-c)  \]
\[ \Leftrightarrow\sum_{cyc} \frac{(a-b)^2}{b} \ge 2\sum_{sym}\frac{(a-b)^2}{a+b+c}  \]
\[ \Leftrightarrow \sum_{cyc} (a-b)^2(\frac{1}{b}-\frac{2}{a+b+c}) \ge 0  \]
Easy to see that we only need to check the case $a\ge b \ge c$. So $S_c \ge 0,S_a \ge 0$. Easy to prove
\[ S_a+2S_b,S_c+2S_b \ge 0  \]

About SOS, i think the best details is here

http://www.mathlinks.ro/Forum/topic-80127.html

Yes [b]hungkhtn[/b], this was also my solution, and this is why I said "SOS kills it", although, when I've seen [b]nhat[/b]'s solution, I've beginning to wonder whether my solution was wrong.. :)

And yeah, your post is really helpful w.r.t. the SOS technique.

hungkhtn, whether you can explain case $a-b\geq b-c,$ $a\geq b+c,$ $a\geq b\geq c$? ;)  Thanks.

[quote="hungkhtn"]There's a misktake typo in that and I just correct it. In case $a-b \le b-c$ we have $a-c \le 2(b-c)$.[/quote]
Sorry, hungkhtn. How to prove our inequality for $a\geq b\geq c$ and $a-b\geq b-c$? ;)
[quote="hungkhtn"]
Easy to see that we only need to check the case $a\ge b \ge c$. So $S_c \ge 0,S_a \ge 0$. Easy to prove
\[ S_b+2S_a,S_c+2S_a \ge 0  \][/quote]
I think it proves nothing. :)
Maybe you mean to $S_a+2S_b\geq0,S_c+2S_b\geq0,$ but the last is wrong for $a\geq b\geq c.$  ;)

Sorry, I'm in a hurry so the solution maybe wrong, but I think it's not to hard. I will check now and I promise to give a true solution after tomorrow.

The first part of my proof (the rewriting part) is the same as hungkhtn's and the second part is the same as nhat's, for it's very easy (as arqady noted above) to verify that $S_a+S_b+S_c\ge 0$ and $S_aS_b+S_bS_c+S_cS_a\ge 0$ (and we don't need to split in cases).

[quote="arqady"]Dear hungkhtn!
You allege that enough to prove $S_a+S_b+S_c\geq0$ and $S_aS_b+S_aS_c+S_bS_c\geq0.$
But $S_a+S_b+S_c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{6}{a+b+c}\geq0$ and $S_aS_b+S_aS_c+S_bS_c\geq0\Leftrightarrow$
$\Leftrightarrow a^3+b^3+c^3-a^2b-a^2c-b^2a-b^2c-c^2a-c^2b+6abc\geq0,$
 which obvuously true. :)
But why this is enough? Thank you.[/quote]
You can see here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=80127&start=20

:(  Yes, do i don't need to think more. :D

[quote="ductrung"]For any positive real numbers $ a,b$ and $ c$, prove that
\[ \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a} \plus{} a \plus{} b \plus{} c \ge \frac {6(a^2 \plus{} b^2 \plus{} c^2)}{a \plus{} b \plus{} c}
\]
[/quote]
<=>$ \sum_{cyc}(\frac{a^2}{b}\minus{}2a\plus{}b) \geq 6(\frac{a^2\plus{}b^2\plus{}c^2}{a\plus{}b\plus{}c}\minus{}\frac{a\plus{}b\plus{}c}{3})$
<=>$ \sum_{cyc} \frac{(a\minus{}b)^2}{b} \geq \frac{2}{a\plus{}b\plus{}c} \sum_{cyc} (a\minus{}b)^2$
or $ (b\minus{}c)^2S_a\plus{}S_b(c\minus{}a)^2\plus{}S_c(a\minus{}b)^2 \geq 0$
with $ S_a\equal{}\frac{a\plus{}b}{c}\minus{}1;S_b\equal{}\frac{b\plus{}c}{a}\minus{}1;S_c\equal{}\frac{c\plus{}a}{b}\minus{}1$
Note that $ S_b\plus{}S_c\equal{}\frac{(a\minus{}b)^2}{ab}\plus{}c(\frac{1}{a}\plus{}\frac{1}{b}) \geq 0$
and $ S_aS_b\plus{}S_bS_c\plus{}S_cS_a \equal{}3\plus{}\frac{\sum_{cyc} a^3\plus{}3abc\minus{}\sum_{cyc}ab(a\plus{}b)}{abc} \geq 3$(Schur)
Done!

Similar problem:
Let $ a;b;c$ be positive real numbers.Prove that:
$ \frac{ab^2}{c^2}\plus{}\frac{bc^2}{a^2}\plus{}\frac{ca^2}{b^2} \plus{}a\plus{}b\plus{}c \geq \frac{6(a^2\plus{}b^2\plus{}c^2)}{a\plus{}b\plus{}c}$

[quote="ductrung"]For any positive real numbers $ a,b$ and $ c$, prove that
\[ \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a} \plus{} a \plus{} b \plus{} c \ge \frac {6(a^2 \plus{} b^2 \plus{} c^2)}{a \plus{} b \plus{} c}
\]
[/quote]
$ \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a} \plus{} a \plus{} b \plus{} c \ge \frac {6(a^2 \plus{} b^2 \plus{} c^2)}{a \plus{} b \plus{} c}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a^4c\plus{}a^3c^2\plus{}2a^2b^2c\minus{}4a^3bc)\geq0\Leftrightarrow\sum_{cyc}a(b^2\plus{}ac\minus{}2bc)^2\geq0.$ :)

Have you a solution without SOS, Ductrung?

Nice proof, Arqady. :)

[quote="arqady"]I think, ductrung means the following reasoning:
$ \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a} \plus{} a \plus{} b \plus{} c \ge 2\sqrt {(a^2 \plus{} b^2 \plus{} c^2)\left(\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\right)}\geq\frac {6(a^2 \plus{} b^2 \plus{} c^2)}{a \plus{} b \plus{} c}.$[/quote]
Actually, I found the weaker version first (which I solved by SOS at the time) and then the stronger version later. I solved the second ([url]http://www.mathlinks.ro/Forum/viewtopic.php?t=87192[/url]) and trivially "generalized" it as:
Let $ k,m$ and $ n$ be real numbers. Then for any positive real numbers $ a,b$ and $ c$,
\[ \sum_{cyc}\left(\frac{a^{n\plus{}m\plus{}k}}{b^m}\plus{}a^nc^k\right)\ge 2\sqrt{\sum_{cyc} a^{n\plus{}m}\sum_{cyc}\frac{a^{k\plus{}n}c^k}{b^m}} \]
which can be regarded as a consequence of Titu-Dospinescu's inequality
([url]http://reflections.awesomemath.org/2008_1/unexpected_ineq.pdf[/url])
But what I found rewarding (to me) is that while comparing the weaker and the stronger version I discovered this one [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=120859[/url] which I again solved by SOS.

@manlio: sorry that I didn't see you post earlier.
cheers

[quote="arqady"]Thank you, ductrung!
[quote="ductrung"]For any positive real numbers $ a,b$ and $ c$,
\[ (a \plus{} b \plus{} c)^{2}(ab^{2} \plus{} bc^{2} \plus{} ca^{2}) \ge 9abc(a^{2} \plus{} b^{2} \plus{} c^{2})
\]
[/quote]
What do you think about the next  stronger inequality:
For any positive real numbers $ a,$ $ b$ and $ c$  prove that:
\[ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\ge \sqrt {\frac {13(a^{2} \plus{} b^{2} \plus{} c^{2})}{ab \plus{} bc \plus{} ca} \minus{} 4}
\]
[/quote]
Nice, arqady  :wink:  What do you think about the next  stronger inequality:
Problem: [zaizai-hoang]
\[ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\ge \sqrt {\frac {143(a^{2} \plus{} b^{2} \plus{} c^{2})}{10(ab \plus{} bc \plus{} ca)} \minus{} \frac{53}{10}}
\]

[quote="zaizai-hoang"][quote="arqady"]Thank you, ductrung!
[quote="ductrung"]For any positive real numbers $ a,b$ and $ c$,
\[ (a + b + c)^{2}(ab^{2} + bc^{2} + ca^{2}) \ge 9abc(a^{2} + b^{2} + c^{2})
\]
[/quote]
What do you think about the next  stronger inequality:
For any positive real numbers $ a,$ $ b$ and $ c$  prove that:
\[ \frac {a}{b} + \frac {b}{c} + \frac {c}{a}\ge \sqrt {\frac {13(a^{2} + b^{2} + c^{2})}{ab + bc + ca} - 4}
\]
[/quote]
Nice, arqady  :wink:  What do you think about the next  stronger inequality:
Problem: [zaizai-hoang]
\[ \frac {a}{b} + \frac {b}{c} + \frac {c}{a}\ge \sqrt {\frac {143(a^{2} + b^{2} + c^{2})}{10(ab + bc + ca)} - \frac {53}{10}}
\]
[/quote]
The strongest for this type is
\[{ \sum \frac {a}{b} \ge\sqrt{ \frac {k(a^2 + b^2 + c^2)}{ab + bc + ca} + 9 - k}}
\]
with $ k = 3 + 3\sqrt [3]{4} + 6\sqrt [3]{2}.$
@hoang: May I see your solution for your inequality? I think it is nice. :)

[hide="@hoang"]Cam on em[/hide]

[quote="can_hang2007"][/quote][quote="zaizai-hoang"][quote="arqady"]Thank you, ductrung!
[quote="ductrung"]For any positive real numbers $ a,b$ and $ c$,
\[ (a + b + c)^{2}(ab^{2} + bc^{2} + ca^{2}) \ge 9abc(a^{2} + b^{2} + c^{2})
\]
[/quote]
What do you think about the next  stronger inequality:
For any positive real numbers $ a,$ $ b$ and $ c$  prove that:
\[ \frac {a}{b} + \frac {b}{c} + \frac {c}{a}\ge \sqrt {\frac {13(a^{2} + b^{2} + c^{2})}{ab + bc + ca} - 4}
\]
[/quote]
Nice, arqady  :wink:  What do you think about the next  stronger inequality:
Problem: [zaizai-hoang]
\[ \frac {a}{b} + \frac {b}{c} + \frac {c}{a}\ge \sqrt {\frac {143(a^{2} + b^{2} + c^{2})}{10(ab + bc + ca)} - \frac {53}{10}}
\]
[/quote][quote="can_hang2007"]
The strongest for this type is
\[{ \sum \frac {a}{b} \ge\sqrt {\frac {k(a^2 + b^2 + c^2)}{ab + bc + ca} + 9 - k}}
\]
with $ k = 3 + 3\sqrt [3]{4} + 6\sqrt [3]{2}.$
@hoang: May I see your solution for your inequality? I think it is nice. :)

[hide="@hoang"]Cam on em[/hide][/quote]
I used to a powerful lemma  :blush:
$ \frac{a}{b}+\frac{b}{c}+\frac{c}{a} \ge 3\sqrt[5]{\left(\frac{a^2+b^2+c^2}{ab+bc+ca}\right)^4}$

So we will prove that:
$ f(x)=3\sqrt[5]{x^4}-\sqrt{\frac{143}{10}x-\frac{53}{10}}\ge 0,\forall x\ge 1$

which $ x=\frac{a^2+b^2+c^2}{ab+bc+ca}$
Easy to check it out  :wink: I don't think my solution is nice  :wink:

[hide]mấy chuyện linh tinh trên VMF anh thông cảm nhá, mà ko hiểu sao lão suppermember lại nhận xét linh tinh như vậy :([/hide]

[quote="zaizai-hoang"]
I used to a powerful lemma  :blush:
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \ge 3\sqrt [5]{\left(\frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}\right)^4}$
[hide]mấy chuyện linh tinh trên VMF anh thông cảm nhá, mà ko hiểu sao lão suppermember lại nhận xét linh tinh như vậy :([/hide][/quote]
This inequality is not true, Hoang. Try $ a\equal{}3.5,b\equal{}1.5,c\equal{}1$. :)

[quote="zaizai-hoang"][quote="can_hang2007"][quote="zaizai-hoang"]
I used to a powerful lemma  :blush:
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \ge 3\sqrt [5]{\left(\frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}\right)^4}$
[hide]mấy chuyện linh tinh trên VMF anh thông cảm nhá, mà ko hiểu sao lão suppermember lại nhận xét linh tinh như vậy :([/hide][/quote]
This inequality is not true, Hoang. Try $ a \equal{} 3.5,b \equal{} 1.5,c \equal{} 1$. :)[/quote]

Sorry for my hateness  :blush: I saw the lemma in MnF but I don't know it is wrong ...
The strongest form of this one is:
Let $ a,b,c\ge 0$. Find the best constant such that:
\[ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \ge 3\left(\frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}\right)^k
\]
I am finding the best constant for the moment!
I think we can solve the inequality by above lemma. I will solve late. I believe it work well  :)

[hide]@anh Cẩn: em thấy cái bổ đề này bên MnF, em chưa thử chứng minh nhưng cũng nghĩ là nó đúng vì anh tanpham nói vậy trong topic đó ! Em sẽ thử tìm hằng số k rồi post lời giải sớm nhất có thể  :wink: hix sorry vì ko check kỹ bổ đề [/hide][/quote]
My calculation gives $ k \le 0.73$, Hoang. :)

[quote="arqady"][quote="can_hang2007"]
The strongest for this type is
\[{ \sum \frac {a}{b} \ge\sqrt {\frac {k(a^2 + b^2 + c^2)}{ab + bc + ca} + 9 - k}}
\]
with $ k = 3 + 3\sqrt [3]{4} + 6\sqrt [3]{2}.$
[/quote]
Please, give me link, can_hang2007 ( or someone that knows ), for  proofs of the similar results.
I know the link was posted in our forum but I lost it. Thank you very much![/quote]
Maybe you ask this one? http://www.mathlinks.ro/viewtopic.php?t=202347
Forgive me, I know little about English. :(

[quote=arqady][quote="manlio"]Have you a solution without SOS, Ductrung?
[/quote]
I think, ductrung means the following reasoning:
$ \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a} \plus{} a \plus{} b \plus{} c \ge 2\sqrt{(a^2\plus{}b^2\plus{}c^2)\left(\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\right)}\geq\frac {6(a^2 \plus{} b^2 \plus{} c^2)}{a \plus{} b \plus{} c}.$[/quote]

[img]http://s5.sinaimg.cn/middle/006ptkjAzy75tT93jtGd4&690[/img]


quadratics
Diophantine equation
number theory proposed
number theory

Prove that there are infinitely many positive integers $ n$ such that $ n!$ is not divisible by $ n^2 \plus{} 1$.

Additional problem: Prove that there are infinitely many positive integers $ n$ such that $ n!$ is divisible by $ n^2 \plus{} 1$.

For the first part, we see that for each prime $ p$ of the form $ 4k \plus{} 1$, $ \minus{} 1$ is a quadratic residue modulo $ p$, so there is some positive $ n$ such that $ n^{2} \plus{} 1 \vdots p$. Choose the smallest $ n$ so $ n < p$
Therefore obviously we have $ n!$ can't divisible by $ p$ so $ n^{2} \plus{} 1$
While there are infinitely many such prime $ p$, hence we have the result

The second part is a fascinating application of Pell equation
First apply the properties of Pell equations one can prove
$ x^{2} \minus{} 5y^{2} \equal{} \minus{} 1$ has infinitely many solution in $ Z^{ \plus{} }$
Hence root $ x,y$ of this must satisfy $ x > 2y$ so $ x! \vdots y^{2}$
When $ x \rightarrow \infty$ eassy to obtain $ x! \vdots 5y^{2}$
The result was proved

BTW, the second problem is on [url=http://www.math.uu.nl/people/beukers/getaltheorie/pen0795.pdf]PEN[/url], problem A 26.


[quote="13375P34K43V312"]Find the sum of all five digit numbers using the numbers 1, 3, 4, 5, 9 exactly once.

I keep getting 5866608, but the answer key says 1199976. Can someone post a solution so I see where I went wrong?

Thanks[/quote]
[hide]isn't it just
$(13459+95431) \cdot 30$[/hide]

Is that Gaussian pairing? I think I've seen it before

BPMS do you mean it is 108890 x 30=3266700?

[quote="13375P34K43V312"]Find the sum of all five digit numbers using the numbers 1, 3, 4, 5, 9 exactly once.

I keep getting 5866608, but the answer key says 1199976. Can someone post a solution so I see where I went wrong?

Thanks[/quote]

I just did that test. Remember they have to be even.

The answer is 11110(9+5+3+1)*6+24(4)=1199976.

pgpatel, that is not correct, since you won't have a complement for every number in the set.

[quote="Walk Around The River"][quote="frost13"]LEET your answer would be right if it was all numbers, but the test question does say EVEN.[/quote]

yeah, i took this test at a practice 2 weeks ago and did the exact same thing. i didnt see the even.[/quote]

that's coincidental,  me and Ubemaya both made the same mistakes on our practice tests!!!! :lol:


Let x and y be integers such that

$((x!)!)!=((y^{2})!)!$

Find, with proof, the number of ordered pairs $(x,y)$ that satisfy this equation.

[quote="Altheman"]since we have the factorial $f: \mathbb{Z}^+\rightarrow \mathbb{Z}^+$, then it is clearly one to one, thus an inverse exists, and we apply that 2 times to get $x!=y^2$, i think the only solutions are $x=0,1$ but i don't know how to prove it rigiriously, perhaps induction[/quote]

Strictly speaking, factorial isn't $1-1$ coz $0!=1!=1$

My idea of the proof is based on the possibility of finding a prime $p$ such that $p\leqslant x$ and $x<2p$ for any $x\geqslant 2$. Can you help on that?

[hide]
x!=y^2
I think the only solutions are 0,1 0,-1 1,1 and 1,-1 but I am not sure how to prove it.[/hide]

Farenhajt, I think I had the same idea, but I also looked for a prime $p$ such that $x\geq{3p} and x\leq{4p}$ and so on. I'll look into it some more


Just a simple question. Are times on the message board now referring to EDT when they simply say Eastern Time?

Yes EDT stands for Eastern Daylight Time

So if the class starts at, say 7:30 Eastern Time, is that with or without DST?

It would be nice if someone could get me a UTC equivalent for the posted class times.

7:30 Eastern time means 7:30 in NYC, whether daylight savings or not.


Have a LaTeX code for a Mathcounts Test?
Or do they use something else to type them up?
And if they don't use LaTeX, can someone teach me how to make lines? Since using ____ in LaTeX causes errors.

\rule[ [i]lift[/i]]{[i]width[/i]}{[i]height[/i]}
The optional parameter lift is how far above or below the text baseline the line should be
For example:
No lift $ x\rule{3cm}{0.1mm}x$
lift=1mm $ y\rule[1mm]{3cm}{0.1mm}y$
lift=-1mm $ z\rule[\minus{}1mm]{3cm}{0.1mm}z$

If you want to fill out a complete line so you get left ________ right then use \hrulefill instead.


(D.Shnol) Two opposite angles of a convex quadrilateral with perpendicular diagonals are equal. Prove that a circle can be inscribed in this quadrilateral.

[hide]Suppose $ ABCD$ be the given quadrilateral and $ \angle ABC\equal{}\angle CDA$. Fix points $ A,B,C$. So $ D$ is on the line which is perpendicular to line $ AC$. Note that the point $ D$ for which $ \angle CDA\equal{}\angle ABC$ is unique. Note also that the point $ D$ for which $ ABCD$ is a kite satisfies $ \angle CDA\equal{}\angle ABC$. Hence $ ABCD$ must be a kite, which follows that $ AB\plus{}CD\equal{}BC\plus{}AD$. It is well-known that a circle can be inscribed in such quadrilateral.[/hide]


solve this equation
$arctan\ x+ arctan\ 2x =\frac{\pi}{4}$

Let $x=\tan\alpha, 2x=\tan\beta$ where $\alpha +\beta=\frac{\pi}{4}$
Then $3x=\tan\alpha +\tan\beta=\tan(\alpha +\beta)(1-\tan\alpha\tan\beta)=1-2x^2$
$2x^2+3x-1=0$
$x=\frac{-3+\sqrt{17}}{4}$ or $\frac{-3-\sqrt{17}}{4}$(rejected after checking)

[hide="Another way..."]
$\arctan x+ \arctan 2x =\frac{\pi}{4}$
$\tan(\arctan x+ \arctan 2x) = \tan\left(\frac{\pi}{4}\right)$
$\frac{\tan(\arctan x) + \tan(\arctan 2x)}{1 - \tan(\arctan x)\cdot\tan(\arctan 2x)} = 1$
$\frac{x + 2x}{1 - 2x^2} = 1$
$3x = 1 - 2x^2$
$2x^2 + 3x - 1 = 0$
...
[/hide]


pigeonhole principle
combinatorics unsolved
combinatorics

Fix $G$.  The sum of the elements of $G$ is clearly $< 1000$, and there are $2^{10}-2>1000$ distinct subsets of $G$ containing between $1$ and $9$ elements, inclusive.  Clearly two distinct nonempty subsets have the same sum, by the Pigeonhole Principle.  If they contain elements in common, remove them from both sets.  Clearly this does not leave us with one or more empty sets, since they were specified to be distinct, and thus we have our $S,T$, which are clearly disjoint, nonempty, and have the same sum.

If we suppose the opposite then for each 2 nonempty disjoint subsets S and T of G we have |S| <> |T| where |X| = card(X).
Results also S <> G and T <> G.
But there are N = 2^10 - 2 = 1022 such subsets of G [without empty set and G itself].
If these subsets are Si[i=1;N] and bi = |Si|, wlog suppose 1 <= b1 < ... < bN. It follows that :
1022 = N <= bN <= |Y| = 864, where Y = {100, 99, ... 92} is the largest subset of F with 9 elements [all bi has at most 9 elements].
So we get a contradiction.

[quote="silouan"]What do you mean with this |S| <> |T| where |X| = card(X). Expain to me the symbols please and be more analitical to your solution.[/quote]

[quote="silouan"]Let F = {1, 2, . . . , 100} and let G be any 10-element subset of F. Prove
that there exist two disjoint nonempty subsets S and T of G with the
same sum of elements.[/quote]

A much harder problem is to prove the same thing when $|G| = 9$.

a <> b means a ≠ b and card(X) means number of elements of set X


If anyone knows about PDA's(push-down automata) and CFG's(Context-free grammars), I need help with these following problems:

1. Consider the following language:
a*b*c* − {(a^i)(b^i)(c^i) | i >= 0}

• Give a CFG that generates this set
• Give a PDA that recognizes this set

2.Give a PDA for the following language:
{x ∈ {0, 1}* | count(0, x) <= 2 * count(1, x)}

Also, does anybody know what Chomsky Normal Form is? I also need to do the 2 problems:

Give a grammar in Chomsky Normal Form for the language
{a^(i)b^(2i)c^(j) | i, j >=1 }

Give a PDA that recognizes the language of the following grammar:
A -> aA | aE | bAA
E -> e | aEb | EE
where e is the empty string

Context-free grammar is in Chomsky Normal Form if(f) all the rules are in form

A -> BC (one nonterminal to two nonterminals) or
A -> a (one nonterminal to one terminal) or
S -> e (starting nonterminal to empty string).


abstract algebra
group theory
superior algebra
superior algebra unsolved

[quote="grobber"]Another one of those so-called "quickies" :):

Prove that every uncountable abelian group has a torsion quotient group with the same cardinality.[/quote]
Perhaps i'm not understand what is means "torsion quotient group". Consider $<\mathbb{R},+>$ the torsian subgroup is $\{0\}$.

EDIT. I think I understand now, you mean the factor group. $|\mathbb{R}/\{0\}|=|\mathbb{R}|$

"torsion quotient group" simply means "quotient group which is torsion" :). More precisely, you have to prove that every uncountable abelian group $G$ has a subgroup $H$ such that:

(i) $|G/H|=|G|$, and
(ii) every element in $G/H$ has finite order.

If $G=\mathbb R$ then $H$ cannot be $\{0\}$, because then (ii) doesn't hold.

Let $A$ be the initial group. $|A|\ge\aleph_{0}$.
Let $I\supseteq A$ be an  injective hull for $A$. Then:
(1) $|A| = |I|$,
(2) $T(A)\subseteq T(I)$ ($T(\cdot)$ denotes torsion subgroup),
(3) $T(I)$ is an injective hull for $T(A)$,
(4) $I=J\oplus T(I)$, and $J=\bigoplus_{j\in K}e_{j}\mathbb Q$.

There exist $n_{j}\in\mathbb N$ and $a_{j}\in A\setminus 0$ such that $a_{j}=e_{j}\cdot n_{j}$ for all $j\in K$. Consider subgroup $B$ in $A$ generated by $\{a_{j}\cdot 2 | j\in K\}$. We have:
(a) $A/B$ is a torsion group,
(b) $B\cap T(A)=0$,
(c) $A/B$ contains direct sum of $|K|$ 2-element subgroups (direct sum of $(a_{j}\mathbb Z+B)/B$, $j\in K$).

So $|A/B|\ge max(|T(A)|, |K|)$. But $|A|=|I|=|J|\cdot|T(I)|=|K|\cdot|T(A)|=max(|T(A)|, |K|)$. Thus $|A/B|=|A|$.


i say a letter, someone else says a letter, and whoever spells a word loses, the word has to be 4+ letters, and making a word shorter than that, doesn't count, you submit the letters previous to you, adding your letter to the end. for example, i start wit h, someone else says e, they would put in as he, someone else says m, putting it in as hem,, now the next person doesn't think that is part of a word, so they challenge, someone then has to say a word with those letters in it, for instance, hemoglobin, then say what it means. the challenger if proven wrong, loses. all words have to be english words.

so i start with [b]L[/b]

if that is not clear, you can ask what i meant.

Dilberto :)

Uhhh... can we restart this game?  Pretty obvious why.

[quote="EggyLv.999"][b]lesbo[/b][/quote]
pwned.
do i win?

Clarification on the rules:

The word starting with "LESB" was fine. Eggy posted "LESBO", which does NOT start an english word (according to my dictionary), but nobody challenged. Had someone challenged, Eggy would have lost.

You can bluff!

LIke if the word so far is "impersonat" you continue "impersonats" and hope that the next player doesn't realize, snagging the next player, not you.

[b]AS[/b]

That is against the rules.
Ignoring fishythefish's post, [b]asi[/b].

The point is to force another to make a word.

(I think surmisably could work as well, but would also result in a loss)

[b]j[/b]

Wait. You just lost. :(
New word!
[b]w[/b]

[b]walt[/b] No, please do not fail and say "waltz".

btw, I think Yoshi won last round because jest is a world.


The oldest tests I 've seen on the internet are 2001. If you can tell me where to find 2000 tests, I would be pleased.

I do have tests from 1994, that I got from my teacher, not the internet. Some of those chapter sprint round questions were so easy that a dog could do it up-side down with its eyes closed.  :)

[quote="purplefairy"]Does anyone know where I can get older versions (before 2000) of state mathcounts tests? I can't find it anywhere on the Internet. :? [/quote]
There is only one way to get the older tests, and that is only if your school was good enough to win the states. If your school ever won the states, your teacher will probably have copies of all (or most) of the Chapters, States, and Nationals. There absolutely is no legal way to get the old States online... Mathcounts doesn't want the older problems to be circulated (probably because they may reuse and edit problems from the older ones)

[quote="yif man12"]No. Mathcounts used to sell the old tests.[/quote]
Ya.. but we're talking about getting them now... but you do have a good point...
[quote="WJS"]Where'd ya even find tests from 2000/2001.[/quote]
You have to buy them from mathcounts for like $15-20 each

Is there any other way to get the 2000 state tests besides buying them? I have some old nationals I downloaded but I can't find any 2000 states.

trading is one way... im not sure whether it is legal tho...

FFR, please don't awaken 2 year old topics. 2nd of all, there is a topic titled MATHCOUNTS Resources HERE. Look there on yashar.biz (2nd or 3rd post) It doesn't have 2000, though. Just google search. States are possible to find sometimes.


vector
inequalities
linear algebra
matrix

How can I do this exercise?

Draw diagrams in $ R^2$ of the solutions sets of the vector inequalities Ax ≤ b for the give matrices A end vectors b.

$ A = \begin{bmatrix}{1}&{1}\\{0}&{-1}\end{bmatrix},  b= \left[{\begin{array}{ccc}{1}\\{0}\end{array}\right]$

$ A = \begin{bmatrix}{2}&{4}\\{4}&{2}\end{bmatrix}, b=  \left[{\begin{array}{ccc}{3}\\{3}\end{array}\right]$

$ \begin{pmatrix} 1 & 2 \\ 0 & -1 \\ -1 & 0 \end{pmatrix}, b = \left[{\begin{array}{ccc}{5}\\{0}\\{0}\end{array}\right]$

$ A = \begin{pmatrix} 1 & 2 \\ 1 & 3 \\ -1 & 0 \\ 0 & -1 \end{pmatrix}, b = \left[{\begin{array}{ccc}{3}\\{5}\\{0}\\{0}\end{array}\right]$

Thank you very much in advance

These are systems of linear inequalities; just draw each of those lines, note which side we want, and take the intersection of the regions.

Hi! Thank you very much for your help, I could do:

Now I doubt this one:

[b]The fitness sets of Levins provide an interesting application of the idea of convex sets in $ R^n$. A genotype of a particular species may have different "fitness",  $ W_1$ and $ W_2,$ in two different environments. Suppose that all genotypes of the species correspond to the points to the convex set $ W_1^2 \plus{} W_2^2 \leq{10}$, [$ 3W_2 \minus{} W_1 \geq{0},$ and $ 3W_1 \minus{} W_2 \geq{0}$. Which genotype (point of this set) maximizes $ W_1?, W_2, W_1 \plus{} W_2?$

(These are the fittest genotypes in the first environment, the second environment, and a mixed environment, respectively).[/b]

I guess we are only talking of drawing the intersections, and the body set the maximum given the sense of inequality.

I thank you if you help me,

many health


Έχουμε άπειρα άτομα φυλακισμένα (λέμε τώρα...) Στο μέτωπο κάθε ενός από αυτούς γράφουμε μία λέξη. Όλοι μπορούν να δούν όλες τις λέξεις εκτός από αυτήν που είναι γραμμένη στο μέτωπό τους. Όλοι προσπαθούν ταυτόχρονα να μαντέψουν την λέξη που είναι γραμμένη στο μέτωπό τους. Αν υπάρξουν άπειρα άτομα που μαντέψουν λάθος θα τους εκτελέσουμε όλους. Αλλιώς θα τους αφήσουμε όλους ελεύθερους.

Φυσικά οποιαδήποτε επικονωνία μεταξύ τους απαγορεύεται. Μπορούν όμως το προηγούμενο βράδυ να συμφωνήσουν σε κάποια στρατηγική που θα ακολουθήσουν.

Τι λέτε. Μπορούν να γλυτώσουν;

[quote="Demetres"]Έχουμε άπειρα άτομα φυλακισμένα (λέμε τώρα...)[/quote]
Πλακα πλακα ετσι οπως ειναι τα 'σοφρωνιστικα μας ιδρυματα' καπως ετσι ειναι τα πραγματα.... :mad:


I believe it's around April Fool's Day (No kidding)

It's within a week I think. Well they announce USAMO qualys by the 28th so it cant be too long from now.

Well according to the teacher's manula, March 28th. However, I was wondering if confirmation by mail or stats would be released before that date.

[quote="SPiKEzZ"]When did we give them our email addresses?[/quote]

Haha...going off on a tangent here...

I see how easy it is for identity fraud criminals to do their job now. We give out personal information without a second thought and then forget about doing so. Eh...not to embarrass you or anything spikezz...hehe...I just thought it was kinda funny. ^-^

[quote="MithsApprentice"][quote="DPopov"]Does anyone know when the results will be sent to qualifiers and also statistics on the website now that the AIME2 has been released.[/quote]

Generally AIME statistics are released a few hours before USAMO qualifiers are. There will be a link from the homepage on March 28th to the list of this year's USAMO qualifiers. If you see yourself on there, you made it and congrats. You should receive an email within a few days congratulating you and giving you your student ID. Your coach will also receive notification via snail mail.[/quote]

What's a few hours?

The amount of time between AIME statistics release and USAMO qualifiers?

(A few) $\in [1, 3]$ I think, which probably means no, they will not come out today.  They will come out tomorrow, a little earlier than the USAMO.  Of course I'm assuming MithsApprentice is right.


Show that $ (n \minus{} 1)^{n \plus{} 1} > n^n > (n \plus{} 1)^{n \minus{} 1}$ for integers $ n > 4$.

nope! try n=4
3^5 4^4 5^2
243 256 125

n=3 doesnt work either
it may be true for n>4

The problem is true for $ n > 4$.

[hide="Hint"]The function $ \left( 1\plus{}\frac{1}{n} \right) ^n$ is bounded by $ e$, at least for values of $ n$ being considered.[/hide]


1. Some individuals with Down Syndrome are the result of chromosomal translocation in which one copy of chromosome 21 becomes attached to chromosome 14. How does this lead to Down Syndrome?

2. Two genes of a flower, one controlling blue (B) versus white (b) petals and the other controlling round (R) versus oval (r) stamens, are linked and are 10 map units apart. You cross a homozygous blue-oval plant with a homozygous white-round plant. The resulting progeny are crossed with homozygous white-oval plants, and 1000 progeny are obtained. How many plants of each of four phenotypes do you expect?

#2 is a dihybrid problem, I believe.

For #1, read the section on Down Syndrome in your biology book. It should have the answer.

#1 Down syndrome is caused by 3 copies of Chromsome 21.  It doesn't matter how the **** the 3rd copy gets there. I think.

#2 The progeny of the first cross are all $C^{B,r},C^{b,R}$.  Thus, the relevant crosses are:
-90% of the time (no recombination): $(C^{B,r},C^{b,R})\times (C^{b,r},C^{b,r})$
-10% of the time (with recombination): $(C^{b,r},C^{B,R})\times (C^{b,r},C^{b,r})$

Thus, out of 1000 plants, you'd expect 450 blue/oval, 450 white/round, 50 white/oval, 50 blue/round. $\omega00\tau$

[quote="236factorial"]1. Some individuals with Down Syndrome are the result of chromosomal translocation in which one copy of chromosome 21 becomes attached to chromosome 14. How does this lead to Down Syndrome?
[/quote]

During meiosis and formation of gametes in one parent, the above event occurs. This will
result in ¼ of the gametes containing a copy of 21 and a copy of 14 with the translocated 21. This is the equivalent of 2 copies of 21. If this gamete is involved in the formation of a zygote, the resulting organism will have 1 copy of 21 from the other parent and the one normal copy of 21 plus the “second” copy from that same parent attached to 14. The result is a triple set of genes from 21 – creating Down’s syndrome in the child.
¼ of the gametes from the above parent will be missing chromosome 21 entirely, receiving no 21 and a normal copy of 14 during meiosis. This gamete will create a lethal 2N-1 condition in a zygote. There will be only 1 copy of 21 from other parent.
¼ of the gametes will be normal – getting a normal copy of 14 and a normal copy of 21.
¼ of the gametes will carry the gene equivalent of 1 copy of 14 and its translocated 21.
Th individual inheriting the genes in this gamete will be a “ translocation carrier” passing the possibility of familial Downs on to other generations.

i hope thats enough; i never unterstood how much you should write at biology; thats how i got a 99/100 to my national scool exams :)


Let $T_1=B_1F\cap BK,T_2=B_2E\cap BK$. Since $B_1F,B_2E$ have fixed directions, it means that $KT_1,KT_2$ depend linearly on the length $KB_1=KB_2$. Since they are equal when the circle has radius $r=IK$ and when it has radius $IB$, it means that they are always equal, so $T_1=T_2$.

[quote="cauchyguy"]Let $ABC$ be a triangle whose incircle touches side $AC$ at $K$. Pick a concentric circle $S$ which intersects all the sides of the triangle. Let $E$ and $F$ be the intersection points on $AB$ and $BC$, respectively, which are closer to $B$. Let $B_1$ and $B_2$ be the intersection points on $AC$ so that $B_1$ is closer to $A$. Also, let $P$ be the intersection point of $B_2 E$ and $B_1 F$. Prove that $B$, $K$, and $P$ are collinear.[/quote]

There is also a very easy canonical solution:

The incircle of triangle ABC and the circle S are concentric; let I be their common center. Also, additionally to the point K where the incircle of triangle ABC touches the side CA, define the points L and J where this incircle touches the sides AB and BC. Then, $IJ\perp BC$, $IK\perp CA$ and $IL\perp AB$. Hence, the triangles ILE, IJF, $IKB_1$ and $IKB_2$ are all right-angled. These right-angled triangles all have the same hypotenuse (in fact, $IE=IF=IB_1=IB_2$, since the points E, F, $B_1$, $B_2$ all lie on the circle S which is centered at I) and one equal catet (in fact, IL = IJ = IK = IK, since the points L, J, K all lie on the incircle of triangle ABC which is centered at I). Thus, these triangles are congruent; hence, $LE=JF=KB_1=KB_2$.

Now, since the two tangents from a point to a circle are equal in length, we have AL = AK. Comparing this with $LE=KB_2$, we get $\frac{AL}{LE}=\frac{AK}{KB_2}$. Thus, $LK\parallel EB_2$. In other words, LK || EP. Similarly, JK || FP. Finally, since the two tangents from a point to a circle are equal in length, we have BL = BJ, what, compared with LE = JF, yields $\frac{BL}{LE}=\frac{BJ}{JF}$ and thus LJ || EF. Hence, altogether, we have LK || EP, JK || FP and LJ || EF. Thus, the triangles LKJ and EPF are homothetic; consequently, the lines LE, KP and JF concur at one point. In other words, the line KP must pass through the point $LE\cap JF$. But $LE\cap JF=B$; hence, the line KP must pass through the point B. In other words, the points B, K and P are collinear.

  Darij


A jar contains two red marbles, three blue marbles, and four green
marbles. Nikisha draws one marble from the jar, and then Tom
draws a marble from those remaining. What is the probability
that Nikisha draws a green marble and Tom draws a blue marble?
Express your answer as a common fraction.

The probability that Nikisha will draw a green marble is $ \frac {4}{9}$.  Then , the probability that Tom will draw a blue marble is $ \frac {3}{8}$.  Multiply them, to get an answer of $ \boxed{\frac {1}{6}}$.

[color=blue][size=75]Fixed LaTeX. Please preview before submitting to ensure all LaTeX is written properly. -Isabella2296[/size][/color]


Let $ a,b,c > 0$ , $ a \plus{} b \plus{} c \equal{} 3$ . Prove that :
$ \sum_{cyc} (3a^2 \minus{} 1)(3b^2 \minus{} 1) \leq 12$
I'm sorry

[quote="conan_naruto236"]Let $ a,b,c > 0$ , $ a \plus{} b \plus{} c \equal{} 3$ . Prove that :
$ \sum_{cyc} (3x^2 \minus{} 1)(3y^2 \minus{} 1) \leq 12$[/quote]
It's wrong. Try $ x \equal{} y \equal{} z \equal{} 5.$ :wink:  :D  :)
If you have meant to prove that $ \sum_{cyc} (3a^2 \minus{} 1)(3b^2 \minus{} 1) \leq 12$
then try $ a \equal{} b$ and $ c\rightarrow0^ \plus{} .$ :wink:


Hello all, I'm working on algebra for pre-calculas. Here are two problems that are extremely simple but I keep getting the wrong solution according to the back of the book.

Here is the 1st problem.

45. 5(a-1)43 = 28

my solution
5(a-1)43 = 28
5a-5(43) = 28
5a-215    = 28
5a           = 243
a           = 48.6
              {48.6}

books solution
{6}

and the 2nd problem

77. 7 - 3(k - 6) = 2 - k

      7 - 3k + 18 = 2 - k
     -3k + k      = -7 - 18 + 2
     -2k            = -23
     - k            = 23/2
                     {23/2}

books solution
{6}

Can someone please show solution to these two problems


[quote="shadysaysurspammed"]If $a+b=2$
Find the minimum value of:-
$a^4+b^4-a^3-b^3$(For a,b reals.)(I hope this question is determinable)[/quote]

[hide="My solution"]

$A^4+B^4-A^3-B^3=A^3*(A-1)+B^3*(B-1)$.

Then, if $A<0$, then, $A^3$ is negative, but $A-1$ is also negative, so $A^3*(A-1)$ is positive.Then, if $A>0$, $A^3$ is positive, and $A-1$ is too, so $A^3*(A-1)$ is positive.

The same happens to $B$. So, $A^3*(A-1)+B^3*(B-1)$ is always non-negative.

Then, the smallest non-negative number is 0. If $A=B=1$, then the equation is $\boxed {0}$, the smallest possible value.[/hide]

I don't feel like posting the whole thing because the expansion and stuff gets long, but say a=2-b and plug that in for a. Then take the derivative and set it equal to 0. You also get b=1. Not a very elegant solution though  :(

[quote="Virgil Nicula"]$E\equiv (a^4+b^4)-(a^3+b^3)\ ;\ P\equiv ab$$\Longrightarrow a^4+b^4=2P^2-16P+16\ ,\ a^3+b^3=8-6P\ .$
$4ab\le (a+b)^2\Longrightarrow P\le1\ ;\ a+b=2\Longrightarrow E=2(P-1)(P-4)\ge0\ .$
Therefore, $\boxed {\ E\ge 0\ \mathrm {and}\ E=0\Longleftrightarrow P=1\Longleftrightarrow a=b=1\ }\ .$

[b]Remark.[/b] $S_n\equiv a^n+b^n\ ,\ n\in N\Longleftrightarrow S_0=2\ ,\ S_{n+2}=S\cdot S_{n+1}-P\cdot S_n\ ,\ n\in N\ .$[/quote]

cOULD YOU PLEASE GIVE ME A GENERAL OUTLINE FOR THIS SOLUTION,PLEASE XPLAIN FROM THE FIRST STEP OF CONGREUNCE PLEASE(Oh shoot,sorry for the caps,sorry)

[b]Remark.[/b] $S_n\equiv a^n+b^n\ ,\ n\in N\Longleftrightarrow S_0=2\ ,\ S_{n+2}=S\cdot S_{n+1}-P\cdot S_n\ ,\ n\in N\ .$

[quote]cOULD YOU PLEASE GIVE ME A GENERAL OUTLINE FOR THIS SOLUTION,PLEASE XPLAIN FROM THE FIRST STEP OF CONGREUNCE PLEASE(Oh shoot,sorry for the caps,sorry)[/quote]

$(x-x_1)(x-x_2)\equiv \boxed {x^2-Sx+P=0}$, where $S=x_1+x_2$ and $P=x_1x_2$ $\Longrightarrow$
$x^2_1-Sx_1+P=0$ and $x^2_2-Sx_2+P=0\ ,\ n\in N\Longrightarrow$
$x_1^{n+2}-Sx^{n+1}_1+Px^n_1=0$ and $x^{n+2}_2-Sx^{n+1}_2+Px^n_2=0$ $\Longrightarrow$
$\left(x^{n+2}_1+x^{n+2}_2\right)-S\left(x^{n+1}_1+x^{n+1}_2\right)+P\left(x^n_1+x^n_2\right)=0\Longrightarrow$
$\boxed {\ S_0=2\ ,\ S_1=S\ ;\ \left(\forall\right)\ n\in N\ ,\ S_{n+2}-S\cdot S_{n+1}+P\cdot S_n=0\ }\ .$

$\boxed {\ S_0=2\ \ \wedge\ \ S_1=S\ }$
$\boxed {S_2}=S\cdot S_1-P\cdot S_0=\boxed {S^2-2P}$
$\boxed {S_3}=S\cdot S_2-P\cdot S_1=S(S^2-2P)-PS=\boxed {S^3-3PS}$
$\boxed {S_4}=S\cdot S_3-P\cdot S_2=S(S^3-3PS)-P(S^2-2P)=\boxed {S^4-4PS^2+2P^2}$
$\boxed {S_5}=S\cdot S_4-P\cdot S_3=S(S^4-4PS^2+2P^2)-P(S^3-3PS)=\boxed {S^5-5PS^3+5P^2S}$ \[ \mathrm {\ a\ .\ s\ .\ o\ .\ }  \]

[quote="Farenhajt"][quote="José"]Then, if $A>0$... $A-1$ is positive[/quote]

This is very wrong :([/quote]

What´s your solution then, man???

[quote="Farenhajt"]This is very wrong :([/quote]

And, please, try to be more POLITE.

[quote="Farenhajt"]I'm just asking what if $A={1\over 2}$? Then $A^3(A-1)$ is negative.

PS. I'm sorry that you consider me to be impolite for pointing out the mistakes in your reasoning.[/quote]

No problem, then.


If $x,y$ be two positive rationals such that $x^5 + y^5 = 2x^2y^2$
then prove that $1-xy$ is a square of a rational number.


advanced fields
advanced fields unsolved

[quote="Xevarion"]First, here is [url=http://en.wikipedia.org/wiki/Surreal_number]Wikipedia's[/url] article about surreal numbers, which is a decent introduction and overview, though I'd recommend Conway's book [u]On Numbers and Games[/u] if you have that available to you.

I am wondering something about the surreal numbers: is the Axiom of Choice used implicitly somewhere in the construction? If so, where? If not, who else thinks that's pretty amazing?

I can't find where, but there are some parts I don't understand fully (like when you have $\{A | B\}$ and (at least one of) $A$ and $B$ are $P(\mathbb{R})$ in cardinality -- maybe actually the cardinality doesn't matter but I'm a noob, or also how you decide what is the "simplest" number between the LHS and the RHS, which Conway doesn't seem inclined to explain).[/quote]

Every surreal number has an ordinal birthday.  0 = { | } is born on day 0; 1 = {0 | } and -1 = { | 0} (and * = {0 | 0}, if you're doing Games instead of Numbers, although I don't reccomend it) have birthday 1; 2 = {1 | }, -2 = { | -1}, 1/2 = {0 | 1} and -1/2 = {-1 | 0} have birthday 2, etc.  Taking finite birthdays gets you the dyadic rationals (those with denominators $2^{n}$).  But then at birthday $\omega$ you get all sorts of new numbers (like $\omega$ = {0, 1, 2, ... | }), and so on.  One surreal is simpler than another if it was born earlier.

I've seen that stuff.... The definitions of the ordinal numbers are fine by me. But to be honest, I don't know how we know $\{0\ |\ 1\}= \frac{1}{2}$ until we define addition and check.

Also, I'm still kinda amazed by the structure of the surreal numbers and I want to know if the Axiom of Choice is involved in it. Maybe you meant to answer that and I just didn't get it. But I still sorta feel like a lot of stuff is hidden behind the notation...

It is a non-trivial theorem that addition and multiplication of surreals correspond to addition and multiplication of reals and so is not "immediately obvious" that {0 | 1} = 1/2, except to the extent that the "simplest thing between 0 and 1" certainly [i]sounds[/i] like it should be 1/2.  After you get that addition and multiplication work for the dyadic rationals, though, the fact that it works for all reals is (to me) exactly as natural as the fact that Dedekind cuts work for generating all the reals.

As for AxChoice, my initial instinct is that the answer is "no, it's not used in the construction."  But, I'm not fully confident of that at all.  Is Choice necessary for the construction of the ordinals?  If not, I think it's likely not necessary for the construction of the surreals.


Let $P$ be an interior point of triangle $ABC$ and extend lines from the vertices through $P$ to the opposite sides.  Let $a$, $b$, $c$, and $d$ denote the lengths of the segments indicated in the figure.  Find the product $abc$ if $a + b + c = 43$ and $d = 3$.

[asy]
size(200);
defaultpen(fontsize(10));
pair A=origin, B=(14,0), C=(9,12), D=midpoint(B--C), E=midpoint(A--C), F=midpoint(A--B), P=centroid(A,B,C);
draw(D--A--B--C--A^^B--E^^C--F);
dot(A^^B^^C^^P);
label("$a$", P--A, dir(-90)*dir(P--A));
label("$b$", P--B, dir(90)*dir(P--B));
label("$c$", P--C, dir(90)*dir(P--C));
label("$d$", P--D, dir(90)*dir(P--D));
label("$d$", P--E, dir(-90)*dir(P--E));
label("$d$", P--F, dir(-90)*dir(P--F));
label("$A$", A, SW);
label("$B$", B, SE);
label("$C$", C, N);
label("$P$", P, 1.8*dir(285));[/asy]

[hide="Hint"]Area ratios should do the trick.[/hide]

[hide]$\frac{3}{3+a}+\frac{3}{3+b}+\frac{3}{3+c}=1$ by a form of ceva, simplifying gives, $9*\sum a+54=abc\implies abc=441$[/hide]

Edit by 4everwise: Please hide your solution.  ;)

Sorry for bumping this back up, but can someone please show me a more complete solution? I'm not really good at this geometry stuff.
Thanks!

[hide]
Call the cevians AD, BE, and CF.

$ \frac{d}{a+d}=\frac{[PBD]}{[ABD]}$

$ \frac{d}{a+d}=\frac{[PCD]}{[ACD]}$

$ \frac{d}{a+d}=\frac{[PBD]+[PCD]}{[ABD]+[ACD]}=\frac{[PBC]}{[ABC]}$

Similarily,
$ \frac{d}{b+d}=\frac{[PCA]}{[ABC]}$

$ \frac{d}{c+d}=\frac{[PAB]}{[ABC]}$

Then,
$ \frac{d}{a+d}+\frac{d}{b+d}+\frac{d}{c+d}=\frac{[PBC]}{[ABC]}+\frac{[PCA]}{[ABC]}+\frac{[PAB]}{[ABC]}=\frac{[ABC]}{[ABC]}=1$

This identity, $ \frac{d}{a+d}+\frac{d}{b+d}+\frac{d}{c+d}=1$ is the "form of ceva" that Altheman used.
[/hide]

[hide="Another Solution Using Mass Points"]
Call the cevians $AX, BY, CZ$.

Let the mass of $A$ be $3$ and the mass of $X$ be $a$.  Then we have $M(P)=3+a$.  From that, we get $M(B)=(3+a)\left(\frac{3}{3+b}\right)$ and $M(Y)=(3+a)\left(\frac{b}{3+b}\right)$.

Then we find $M(C)=M(X)=M(B)=a-(3+a)\left(\frac{3}{3+b}\right)$ and $M(F)=M(A)+M(B)=3+(3+a)\left(\frac{3}{3+b}\right)$.  Then we can write the equation
\[c\left[a-(3+a)\left(\frac{3}{3+b}\right)\right]=3\left[3+(3+a)\left(\frac{3}{3+b}\right)\right]\]
The rest is just algebra and we get $abc=\boxed{441}$.
[/hide]

[u]See[/u] [b]lemma[/b] from [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=150&t=75942&p=2691135#p2691135][color=darkred][b]here[/b][/color][/url].

$\boxed{\prod \frac ad=2+\sum \frac ad}$ ([u][b]Euler[/b] relation[/u]) $\iff abc=2d^3+d^2(a+b+c)\iff$ $abc=2\cdot 3^3+3^2\cdot 43\implies \boxed{abc=441}$ .

[quote="tjhance"]This identity, $ \frac{d}{a+d}+\frac{d}{b+d}+\frac{d}{c+d}=1$ is the "form of ceva" that Altheman used.[/quote]

How do you get that?

[hide="Proof of the Form of Ceva"]
Consulting the figure, we see that \[\frac{d}{a+d}=\frac{[PBC]}{[ABC]}\]
Similarly, \[\frac{d}{b+d}=\frac{[PAC]}{[ABC]}\] and \[\frac{d}{c+d}=\frac{[PAB]}{[ABC]}\]
Adding these together produces the desired result.
[/hide]

Sorry, I still dont understand. I thought Ceva's theorem only applies to the sides.

I don't know why it's called a form of Ceva; that's for the etymologists. The "form of Ceva" is indeed true, but I don't see how it relates to or is a form of Ceva. It doesn't really establish a necessary or sufficient condition for concurrency. It does not have any meaning without the concurrency.

Of course, a more general form of it (assuming that the $d$'s are not equal) would be \[\frac{d}{a+d}+\frac{e}{b+e}+\frac{f}{c+f}=1\] where $e$ and $f$ are collinear with $b$ and $c$, respectively.

Trivialized by a simple theorem.

[hide=Solution]Applying Gergonne's Theorem, we get $\frac{3}{3+c}+\frac{3}{3+a}+\frac{3}{3+b}=1$. Expanding gives $abc=9(a+b+c+6)=9 \cdot 49 = \boxed{441}$.[/hide]


induction
inequalities
inequalities unsolved

For $a_i,b_i,c_i>0,i=1,2,\cdots,n$
Prove:
\[ \sum_{i=1}^n(a_i+b_i+c_i)\sum_{i=1}^n \frac{b_ic_i+c_ia_i+a_ib_i}{a_i+b_i+c_i} \sum_{i=1}^n \frac{a_ib_ic_i}{b_ic_i+c_ia_i+a_ib_i} \le \sum_{i=1}^n a_i \sum_{i=1}^n b_i \sum_{i=1}^n c_i  \]

[quote="zhaobin"]For $a_i,b_i,c_i>0,i=1,2,\cdots,n$
Prove:
\[ \sum_{i=1}^n(a_i+b_i+c_i)\sum_{i=1}^n \frac{b_ic_i+c_ia_i+a_ib_i}{a_i+b_i+c_i} \sum_{i=1}^n \frac{a_i+b_i+c_i}{b_ic_i+c_ia_i+a_ib_i} \le \sum_{i=1}^n a_i \sum_{i=1}^n b_i \sum_{i=1}^n c_i  \][/quote]
do you want to say $\[ \sum_{i=1}^n(a_i+b_i+c_i)\sum_{i=1}^n \frac{b_ic_i+c_ia_i+a_ib_i}{a_i+b_i+c_i} \sum_{i=1}^n \frac{a_ib_ic_i}{b_ic_i+c_ia_i+a_ib_i} \le \sum_{i=1}^n a_i \sum_{i=1}^n b_i \sum_{i=1}^n c_i$
to this problem I have a solution:
Lemma 1:$\sum {(a_i+b_i)}\sum\frac{a_ib_i}{a_i+b_i}=<\sum a_i\sum b_i$
use induction we can prove it.
Lemma 2:$\sum_{i=1}^n(a_i+b_i+c_i)\sum_{i=1}^n \frac{b_ic_i+c_ia_i+a_ib_i}{a_i+b_i+c_i}\leq \sum a_i\sum b_i+\sum c_i\sum b_i+\sum a_i\sum c_i$
use lemma 1
we have $\sum ((a_i+b_i)+c_i)\sum \frac{(a_i+b_i)c_i)}{a_i+b_i+c_i}\leq (\sum a_i+\sum b_i)\sum c_i$
we have  another two similar Ineq
and sum the three Ineq we obtain:$\sum_{i=1}^n(a_i+b_i+c_i)\sum_{i=1}^n \frac{b_ic_i+c_ia_i+a_ib_i}{a_i+b_i+c_i}\leq \sum a_i\sum b_i+\sum c_i\sum b_i+\sum a_i\sum c_i$
Lemma 3 $(\sum a_i\sum b_i+\sum c_i\sum b_i+\sum a_i\sum c_i)\sum_{i=1}^n \frac{a_ib_ic_i}{b_ic_i+c_ia_i+a_ib_i}\leq \sum a_i\sum b_i\sum c_i$
also use lemma 1
Let $t_i=\frac{c_ib_i}{b_i+c_i}$
we have$\sum {(a_i+t_i)}\sum\frac{a_it_i}{a_i+t_i}\leq \sum a_i\sum t_i$
and $\sum {(c_i+b_i)}\sum\frac{c_ib_i}{c_i+b_i}\leq \sum c_i\sum b_i$

is the general inequality (not 4 sequences, but m sequences) true?

And I don't understand the proof of lemma 3.

[quote="Soarer"][quote="Hawk Tiger"]To zhaobin ,I found i can't use indcution to prove the similar Lemma for 4 vairiables with ease.
could you show your solution to this two Inequalities?[/quote]

do you mind explaining how did you prove your lemma 3?[/quote]
$\sum {(a_i+t_i)}\sum\frac{a_it_i}{a_i+t_i}\leq \sum a_i\sum t_i$
product $\frac{\sum{i=1}^nb_i \sum{i=1}^nc_i}{\sum_{i=1}^n t_i}$ both side.
Is it OK?

To Hawk Tiger,We may write four lemmas to prove 4 groups varialbes inequality,which one you can't prove? ;)

[quote="Hawk Tiger"]to zhaobin ,I can't prove the third lemma with the same method,but use induction,i think i maybe can prove it[/quote]
Can you write the lemma you can't prove?
PS:I don't know how to use induction on the inequality. :blush:

[quote="Hawk Tiger"][quote="zhaobin"]Thank you for your nice solution. :)

We may also can prove:
 \[ \sum_{i=1}^n(a_i+b_i+c_i+d_i) \sum_{i=1}^n \frac{b_ic_i+c_ia_i+a_ib_i+d_ia_i+d_ib_i+d_ic_i}{a_i+b_i+c_i+d_i} \sum_{i=1}^n \frac{a_ib_ic_i+b_ic_id_i+c_id_ia_i+d_ia_ib_i}{b_ic_i+c_ia_i+a_ib_i+d_ia_i+d_ib_i+d_ic_i}  \] \[ \sum_{i=1}^n \frac{a_ib_ic_id_i}{a_ib_ic_i+b_ic_id_i+c_id_ia_i+d_ia_ib_i} \le \sum_{i=1}^n a_i \sum_{i=1}^n b_i \sum_{i=1}^n c_i \sum_{i=1}^n d_i  \]
do you want to have a try? :)[/quote]
I can't prove$\sum_{i=1}^n \frac{a_ib_ic_i+b_ic_id_i+c_id_ia_i+d_ia_ib_i}{b_ic_i+c_ia_i+a_ib_i+d_ia_i+d_ib_i+d_ic_i} \leq \frac{\sum a_i\sumb_i\sum c_i+\sum b_ic_id_i+\sum b_i\sum a_i\sum d_i+\sum c_i\sum d_i\sum a_i }{\sum a_i\sumb_i+\sum b_ic_i+\sum b_ii\sum d_i+\sum c_i\sum d_i+\sum a_i\sum c_i+\sum c_i\sum a_i$
the so-called induction is when one can prove when $n=2$ then you can prove it when $n\geq 2$
by the way ,how do you post such a long fomula?I need to spent a lot of time in posting,even more than thinking.[/quote]
you can quote my post and see :)

I still don't know how you can use induction.
How can you prove the $n+1$ use $n$?

p.s but I haven't got a nice solution when $n=2$,either.
My english is poor,is 80% leant by myself.
I aslo know little about computer.

[quote="Hawk Tiger"]\sum_{i=1}^n \frac{a_ib_ic_i+b_ic_id_i+c_id_ia_i+d_ia_ib_i}{b_ic_i+c_ia_i+a_ib_i+d_ia_i+d_ib_i+d_ic_i} \leq \frac{\sum a_i\sumb_i\sum c_i+\sum b_ic_id_i+\sum b_i\sum a_i\sum d_i+\sum c_i*\sum d_i*\sum a_i}{\sum a_i\sumb_i+\sum b_i\sum c_i+\sum b_ii\sum d_i+\sum c_i\sum d_i+\sum a_i\sum c_i+\sum c_i\sum a_i
can you read this ?
I don't find even one error,but "Error 6"
and if $n=K+1$,we first use the n-variables inequality for $a_1,\cdots,a_n,b_1,b_n,c_1,c_n$
and $let a=\sum_{i=1}^n {a_i}$b,c,d,is similar
then use the two variables ineq for $a,b,c,d,a_{n+1},\cdots$[/quote]
1.May it should be  "\sum_{i=1}^n \frac{a_ib_ic_i+b_ic_id_i+c_id_ia_i+d_ia_ib_i}{b_ic_i+c_ia_i+a_ib_i+d_ia_i+d_ib_i+d_ic_i} \leq \frac{\sum a_i\sumb_i\sum c_i+\sum b_ic_id_i+\sum b_i\sum a_i\sum d_i+\sum c_i*\sum d_i*\sum a_i}{\sum a_i\sumb_i+\sum b_i\sum c_i+\sum b_ii\sum d_i+\sum c_i\sum d_i+\sum a_i\sum c_i+\sum c_i\sum a_i}"
May they are too long,you should cut them to two part.

2.For this inequality,you should just let $t_i=\frac{a_ib_i+b_ic_i+c_ia_i}{a_i+b_i+c_i}$,You will find a nice solution. :)

3.I still don't know how can you use induction,can you express it more clearly or send me a pm with chinese ;)

Thank you

[quote="Soarer"]for n = 2, you only need to use cauchy schwarz.

$\sum (a_i+b_i) \sum \frac{a_i b_i}{a_i+b_i} \leq \sum a_i \sum b_i \\ \Leftrightarrow \frac{1}{4}(\sum a_i + \sum b_i)^2 - \sum a_i \sum b_i \le \frac{1}{4}(\sum a_i + \sum b_i)^2 - \sum (a_i+b_i) \sum \frac{a_i b_i}{a_i+b_i} \\ \Leftrightarrow (\sum a_i - b_i)^2 \leq \sum (a_i+b_i) \sum \frac{(a_i-b_i)^2}{a_i+b_i}$[/quote]
This is the solution just for $m=2$ not n.
And my solution is a little different from yours.
$\sum (a_i+b_i) \sum \frac{a_i b_i}{a_i+b_i} \leq \sum a_i \sum b_i \\ \Leftrightarrow (\sum a_i + \sum b_i)^2 - 2\sum a_i \sum b_i \le (\sum a_i + \sum b_i)^2 - \sum (a_i+b_i) \sum \frac{2a_i b_i}{a_i+b_i} \\ \Leftrightarrow (\sum a_i)^2+(\sum b_i)^2 \leq \sum (a_i+b_i) \sum \frac{a^2_i+b_i^2}{a_i+b_i}$ :)

Finally, it was posted before, http://www.mathlinks.ro/Forum/viewtopic.php?t=29341&view=next#.


calculus
integration
modular arithmetic
algebra
polynomial
algebra unsolved

Find a necessary and sufficient condition on the natural number $ n$ for the equation
\[ x^n \plus{} (2 \plus{} x)^n \plus{} (2 \minus{} x)^n \equal{} 0
\]
to have a integral root.

[hide="Solution"]The necessary and sufficient condition is that $ n \equal{} 1$.

Note that $ x,2 \plus{} x$, and $ 2 \minus{} x$ cannot equal zero simultaneously. Hence, if $ n$ is an even number, the left hand side of the given equation is positive. So $ n$ is an odd number. Now we can rewrite the equation as
\[ x^n \plus{} (2 \plus{} x)^n \equal{} (x \minus{} 2)^n.
\]
If $ x$ is non negative, then the left hand side is greater than the right hand side. So $ x$ is negative. Let $ a \equal{} \minus{} x$. The equation becomes
\[ \minus{} a^n \plus{} (2 \minus{} a)^n \plus{} (2 \plus{} a)^n \equal{} 0,
\]
i.e.
\[ (a \plus{} 2)^n \equal{} (a \minus{} 2)^n \plus{} a^n.
\]
By Fermat's Last Theorem, if $ n > 2$, this equation has no solutions in non-zero integers. If one of them equals zero, it is easy to see that the equation does not have any solution. Hence $ n\le2$. So the only possible value of $ n$ is 1, which is a sufficient condition for the given equation to have an integral root, that is $ x \equal{} \minus{} 4$.[/hide]

Fermat's Last Theorem wasn't proven when this problem was proposed, so there is almost certainly another solution.

[i]Let $ P(x) \equal{} x^n \plus{} (2 \plus{} x)^n \plus{} (2 \minus{} x)^n, n \in \mathbb{N}$: we know that $ \alpha \in \mathbb{Z}$ exists s.t. $ P(\alpha) \equal{} 0$.[/i]

If $ n \equiv 0 \pmod 2$ then $ LHS > 0$ since we cannot have $ x \equal{} 2 \plus{} x \equal{} 2 \minus{} x \equal{} 0$, so let $ N \equal{} 2n \plus{} 1, n \ge 0$.

Note that if $ m \equiv 1 \pmod 2$ then $ P(m) \equiv 1 \pmod 2$, so we can define $ x \equal{} 2X$.

So $ P(x) \equal{} 0$ iff $ P(X) \equal{} X^N \plus{} (1 \plus{} X)^N \plus{} (1 \minus{} X)^N \equal{} 0$, but $ P(X)$ is a monic polinomial with last term 2, so we can affirm that $ |\alpha| \equal{} 2$.

Then if $ \alpha \equal{} 2 \implies 2^N \plus{} 3^N \equal{} 1$, absurd and if $ \alpha \equal{} \minus{} 2 \implies 3^N \equal{} 2^N \plus{} 1$ that has the unique solution $ N \equal{} 1$.  :wink:

N cannot be odd number greater than 1 ,else given equation contradicts fermat's last theorem .
N cannot be even ofcourse as ,the each term will have to be individually zero ,which is not possible .
Hence N can only be equal to 1


Well I think it is going to be really very hard problem. Any one come up with solution?

We have N vertices. You have to choose maximum possible vertices such that every vertex have at least K edge from them to the other selected vertices.

For example:
1 and 2
 1 and 3
 1 and 5
 1 and 6
 2 and 5
 2 and 6
 3 and 4
 3 and 5
 4 and 6
 4 and 7
 5 and 6
 7 and 8
(edge is undirected)
are the edges. Then if K=2 then the answer would be {1,..,6} and if K=3 then {1,2,5,6} but if K>=4 then {}.

N<3000

I have no IDEA!!!

Suppose you store your graph in an adjacency list. Then start scanning the graph row by row. Since the edges are undirected, we can consider each edge counted for two vertices. Create a list called "Counter list" of size N and type integer to store the number of edges connected to each vertex. Then set your k for instance to 2 and start scanning. Each time you see a 1, you increament the counter for the two vertices which are connected to this edge and so on. At the end scan your counter list and report every vertex with counter ==2 or k in general.
This solution takes O(N^2).


Prove that there are no continuous functions $f: \mathbb R \to \mathbb R$ satisfying the functional equation
\[f(1+f(x))=2-3x \quad \forall x \in \mathbb R.\]

There are no function (without continiosly).
Because 2-3x is bijective f(x) must be bijective (for any y exist $ x \equal{} (2\minus{}y)/3, z \equal{} 1\plus{}f(x)$ suth that $ f(z) \equal{} y$, and $ f(x) \equal{} f(y)\to f(1\plus{}f(x)) \equal{} 3\minus{}2x \equal{} f(1\plus{}f(y)) \equal{} 3\minus{}2y\to x \equal{} y$).
From $ f(3\minus{}3x) \equal{} f(1\plus{}2\minus{}3x) \equal{} f(1\plus{}f(1\plus{}f(x))) \equal{} 2\minus{}3(1\plus{}f(x)) \equal{}\minus{}3f(x)\minus{}1$ we get $ f(3/4) \equal{}\minus{}1/4$ .

[quote="ekaragoz"]Prove that there are no continuous function satisfying the functional equation below. (f:R-R)

f(1+f(x))=2-3x[/quote]

$ f(x)$ is clearly bijective, so, if $ f(x)$ is continuous, $ f(x)$ must be stricty monotonous (since bijective and continuous), so $ f(1\plus{}f(x))$ would be stricty increasing, which is wrong since $ f(1\plus{}f(x))\equal{}2\minus{}3x$, strictly decreasing.

So $ f(x)$ can't be continuous.

Suppose that $ f(x)$ is strictly increasing, then if $ y>x$,

$ f(1\plus{}f(y)) \equal{} 2\minus{}3y < f(1\plus{}f(x))$

But since $ f$ is strictly increasing,

$ 1\plus{}f(y) > 1\plus{}f(x)$
$ f(1\plus{}f(y)) > f(1\plus{}f(x))$, but $ f(1\plus{}f(y))<f(1\plus{}f(x))$. Contradiction.

The other way follows similarly.

[quote="MellowMelon"]pco, can you elaborate more on how you can conclude that $ f(x)$ is strictly increasing? I got the rest of the solution.[/quote]

$ g(x)\equal{}1\plus{}f(x)$ is strictly monotonous. So $ g(g(x))$ is stricty increasing. So $ g(g(x))\minus{}1\equal{}f(1\plus{}f(x))$ is strictly increasing.

I found mistake in my solution.
Let $ g(x)\equal{}\frac{1}{4}\plus{}f(\frac{3}{4}\plus{}x)$, then equation equavalent to $ g(g(x))\equal{}\minus{}3x$ and it had infinetely many (non continiosly) solutions.


게시? 제목?는 수학 올림픽스 ?고 쓰여있는?..
'한국 수학 올림피아드', '국제 올림피아드' ?고 하지 않나요?
올림픽? 알아?; 올림픽스는 처? 들어보는?.. ?/

yes, it should be,

수학 올림피아드

( PM to Velantin )


Im having problems with quoting someone and when I put the quote it says quote and not the name of the person im quoting :furious:

[quote="Name"][code]
[quote="Name"]Quoted.[/quote]
[/code]
[/quote]

Yes.  More generally, see my advice [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=173255]here[/url].

in the quote tag, just put = after "quote" in the first quote tag and then put in quotation marks what you want to say.
For example, [code][quote="Nerd_of_the_Ages"]in the quote tag, just put = after "quote" in the first quote tag and then put in quotation marks what you want to say.[/code]
For example, [\quote]  (except put a / instead of a \) produces [quote="Nerd_of_the_Ages"]in the quote tag, just put = after "quote" in the first quote tag and then put in quotation marks what you want to say.
For example, [/quote]


MATHCOUNTS
analytic geometry
FTW
function
inequalities
integration

Last year, we helped Kevin win the Relly Award for Best Junior Achiever.

In so doing, we had an impact on the Regis & Kelly Show's programming: you'll see that this year, there are 3 contest winners among the nominees.  Click here to cast your vote, and to help give MATHCOUNTS some much-deserved publicity.  Vote early and vote often!

[i]Edit: I've removed the link; voting is closed/[/i]

Darryl Anyday!!!!!! Does Darryl have an Aops account?

Hmm. I recall from last year:

VOTE, DELETE YOUR COOKIES, VOTE AGAIN, DELETE YOUR COOKIES ect. ect.

I'm not so sure you have to delete your cookies this time around...

am i the only one voting for demi lovato here??  =]
LOL i'll vote for both

Optimally, code something up using the java Robot class and let it run while you're at school. (But randomly distribute your votes between Sameer Mishra and Darryl Wu, and put a random time interval between the voting, so it doesn't look suspicious ;))

Bias that random voting towards Darryl, please ;)  The Spelling Bee surely doesn't need more publicity.

i voted about 20 times now...you can probably tell im bored.

Can someone please upload some code to make a bot to keep voting over over and over again in java? :D :D

Get your parents, get your friends, get everybody.

did my share of voting for darryl, but what? they get an award called the golden stool award? golden stool? thanks regis and kelly...?

I voted. I'll probably make a macro with C later...

does anyone have any idea how to do the bot thing? :maybe:

no one for demi??

i'm actually surprised

it might be now, but it wasn't last evening :D

so if voting is over, who won? which website can we check to find out the winner?

can someone watch for me? i don't get tv... :)

Watch them track IP addresses and realize that darryl got voted like only by 100 differnt people acctually instead of 35000. But I voted 32 times so I did my share.

dang it, should have refreshed my ip everytime i did it! NOW DARYLL WON"T WIN!!!!!!! :noo: :mad:  :wallbash_red:

We just hafta make sure they don't find this thread :D

lol if they find this thread we fail and rrusczyk is in trouble for encouraging rigging a vote.

Quick! Everyone edit your posts to talk about Lagrange Multipliers! We can confuse the Regis and Kelly people with amazing greek letters! I'll give an example:

Let $ U(g, b) \equal{} 25g^2 \plus{} 12b$ be the utility function relating happiness in terms of gizzards and babies. (alternatively, use flamethrowers and hand grenades.)
You have $ \$1000$ to purchase gizzards and babies. Gizzards cost $ \$15$, babies cost $ \$4$. Maximize happiness.

Solution: add some weird looking $ \lambda$s to the mix and you've got gizzards, babies, and lambdas. You MUST be happy now.

[hide="Solution"]
Assume for the sake of contradiction that $ \sqrt {\lambda} \neq g\omega \plus{} b^2$ (where $ \omega$ is the ordinal constant). Then we have
\[ 2\lambda \plus{} 2\gamma \plus{} 2\zeta \equal{} \frac {\lambda^2 \plus{} \gamma^2 \plus{} \zeta^2}2
\]
But this is clearly impossible, by AM-GM. Therefore, the original equation must constitute happiness for all values of $ \lambda$.[/hide]

[hide="Alternative Solution"]
We have $ 15g \plus{} 4b \equal{} 1000$, so $ b \equal{} \frac {1000 \minus{} 15g}4$. Plugging in, we see that $ U(g) \equal{} 25g^2 \minus{} 45g \plus{} 3000$. Note that the critical point is the MINIMUM, and we want to find the MAXIMUM (within the range $ g, b > 0$). Therefore, we test the endpoints, where $ g \equal{} 0$ and $ b \equal{} 0$.

$ g \equal{} 0$: $ 25(0)^2 \plus{} 12 \cdot 250 \equal{} 3000$
$ b \equal{} 0$: $ 25\left(\frac {1000}{15}\right)^2 \plus{} 0 \approx 111111$.

We can't buy fractions of gizzards, however, and we want to buy as many gizzards as possible, so we set $ g \equal{} 66$ and $ b \equal{} 2$, for a happiness value of 108924 (with $ \$2$ left over! :D)[/hide]

Erm.  Did you see the thread title? Help MATHCOUNTS: Vote for Darryl

um AIME15 read that page 4 at teh bottom and you'll understand

[quote="Yongyi781"][hide="Solution"]
Assume for the sake of contradiction that $ \sqrt {\lambda} \neq g\omega + b^2$ (where $ \omega$ is the ordinal constant). Then we have
\[ 2\lambda + 2\gamma + 2\zeta = \frac {\lambda^2 + \gamma^2 + \zeta^2}2
\]
But this is clearly impossible, by AM-GM. Therefore, the original equation must constitute happiness for all values of $ \lambda$.[/hide]

[hide="Alternative Solution"]
We have $ 15g + 4b = 1000$, so $ b = \frac {1000 - 15g}4$. Plugging in, we see that $ U(g) = 25g^2 - 45g + 3000$. Note that the critical point is the MINIMUM, and we want to find the MAXIMUM (within the range $ g, b > 0$). Therefore, we test the endpoints, where $ g = 0$ and $ b = 0$.

$ g = 0$: $ 25(0)^2 + 12 \cdot 250 = 3000$
$ b = 0$: $ 25\left(\frac {1000}{15}\right)^2 + 0 \approx 111111$.

We can't buy fractions of gizzards, however, and we want to buy as many gizzards as possible, so we set $ g = 66$ and $ b = 2$, for a happiness value of 108924 (with $ \$2$ left over! :D)[/hide][/quote]

Yes but you are forgetting the $ \phi$ constant which we use to set up the obvious inequality $ \frac{x^2+\lambda}{72}\geq\theta$ in the case where $ \int{9\theta^\circ-7=43}$ silly yongyi!

[quote="mewto55555"]Yes but you are forgetting the $ \phi$ constant which we use to set up the obvious inequality $ \frac {x^2 + \lambda}{72}\geq\theta$ in the case where $ \int{9\theta^\circ - 7 = 43}$ silly yongyi![/quote]

I think you may mean $ \int_0^\infty {(9\theta^\circ - 7)}\,d\theta = 43$ :D
Never forget the $ d\theta$'s or the bounds :D


who's here that's going to hartford chapter but will make it to states?
-jorian

How can anybody be sure they'll make it to states?

Why are there two topics with the same titel?

two different topics, so let me rephrase

who is 95+% sure they will make states?
-jorian

If you're reasonably good, you can be sure of making state in CT.

[quote="Phelpedo"]Obviously, he improved between chapter and state  :wink:  Like 40 people make it every year from each chapter, so...[/quote]

Only 24 advanced from ours and if your school wasn't good like mine, it became 12.

[quote="Ignite168"]#90th at Nats this year got #15 at his Chapter so I would say you shouldn't be that sure.[/quote]

he actually got 92 at nats, but he's from my school and i can match him this year
-jorian

i don't know if that was a dis or what?
-jorian

It's just that you keep talking about how he's not all that good, but I don't think 2nd at states and 90th at nats is bad for a 7th grader...

nononono i don't think u understood me

he's amazing at math but it feels to me that he's not as good as you guys, and still did better than most
-jorian


Let $ a;b;c$ be positive real numbers such that $ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\leq 1$.Find the smallest possible value of $ P=[a+b]+[b+c]+[c+a]$ where $ [x]$ is the largest integers which is smaller than $ x$

You've already post this problem here:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=159281[/url]

Do not double post.  Find the right forum, and then post only once in that forum.


Prove that for every $ n \in \mathbb{N}^*$ exists a multiple of $ n$, having sum of digits equal to $ n$.

Let n= (2^a). (5^b). L where L is not divisible by 2 or 5.

Let k be a positive integer such that L|10^k -1

Take N= 10^(a+b)* [10^k + 10^(k^2) +10^(k^3)+....+ 10^(k^n) ]

We have n|N and S(N) =n

I saw this problem in the Andrei Negut's book: "Problems for the mathematical Olympiads".  :!:

Can't we just multiply the number by 10?
$ 10n$ is a multiple of $ n$ and has the same sum of digits.

you need the sum of digits to be equal to n. not the sum of digits to be equal to the sum of digits of n.

for example when n=12 you need a multiple that is divisible by 12 and with sum of digits=12
10*12=120 but the sum of digits of 120 is 3.

instead a proper multiple in this case would be 444. Do you understand?

Can anyone solve this or has it become forgotten?


quadratics
algebra
polynomial
number theory proposed
number theory

[quote="Christine"]Prove that for every irrational real a there exist irrational numbers b, b' such that a + b, ab' are rational but a + b', ab are irrational.[/quote]

1) if $a$ is root of no quadratic polynomial with rational coefficients, then you can choose any rationals $p$ and $q$ and write $b=p-a$ and $b'=\frac{q}{a}$. Then :
$b$ and $b'$ are irrational numbers
$a+b=p$ is rational
$ab'=q$ is rational
$a+b'=a+\frac{q}{a}$ is irrational, else $a$ would be root of $a^{2}-s*a+q=0$ for some rational $s$
$ab=a(p-a)$ is irrational, else $a$ would be root of $a^{2}-p*a+s=0$ for some rational $s$.

2) if $a$ is root of some quadratic polynomial with rational coefficients, then $a=u+\sqrt{v}$ for some rationals $u$ and $v$ such that $\sqrt{v}$ is irrational.
Then let $b=u+1-\sqrt{v}$ and $b'=\frac{u^{2}+1}{u^{2}-v}(u-\sqrt{v})$
$b$ and $b'$ are irrational numbers
$a+b=2u+1$ is rational
$ab'=u^{2}+1$ is rational
$a+b'=(1+\frac{u^{2}+1}{u^{2}-v})u-\frac{v+1}{u^{2}-v}\sqrt{v}$ is irrationnal
$ab=u^{2}+u-v+\sqrt{v}$ is irrational

Fix any 2-elements sets of rational numbers $A$ and $B$ and let $a$ be any irrational number. Take any $p\in A$ and $q\in B$.
Then for $b=p-a$ and $b'=\frac{q}{a}$ the half of requirements is satisfied.
To satisfy the another one we only have to change if needed our choise of $p$ and $q$.
It is possible because the difference between old and new values of $ab=a(p-a)$ (similary of $a+b'=a+\frac{q}{a}$ ) is irrational  if we change $p$ (or $q$ ) by another element of $A$ (or $B$).

[quote="Christine"]Prove that for every irrational real a there exist irrational numbers b, b' such that a + b, ab' are rational but a + b', ab are irrational.[/quote]

Was this from the Red MOP Homework?

See http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=134466 .  It is essentially the same solution as pco's.

for more solutions, see [url=http://www.mathlinks.ro/viewtopic.php?p=191619#191619]here[/url] :wink: