Solve these two recursions:

1. $ a(n,k) \plus{} a(n,k \plus{} 1) \equal{} a(n \plus{} 1,k \plus{} 1); \ n, k \in \mathbb{W}, a(0,0) \equal{} c \in \mathbb{R}$
Find $ a(n,k) \equal{} f(n,k,c)$

2. $ a(n \plus{} 1,k) \plus{} a(n \plus{} 1,k \plus{} 1) \equal{} a(n,k); \ n, k \in \{0, 1, 2, 3, ..., N\}$
Find $ a(0,0) \equal{} f(a(N \plus{} 1,0),a(N \plus{} 1,2),a(N \plus{} 1,3),..., a(N \plus{} 1,N \plus{} 1))$

A quick question: for the first problem, what does it mean for $ n, k \in \mathbb{W}$?

Also, I think that the first question needs another seed value.. Having a seed for a(0,0) is good, but you need one for a(0,1) as well. I might be completely wrong though.

$ \mathbb{W} \equal{} \text{the set of whole numbers} \equal{} \mathbb{N}_0 \equal{} \mathbb{N} \cup \{0\}$

Nowadays we don't use $ \mathbb{W}$ nor the term "whole numbers" that often; I usually use "nonnegative integers" in place of "whole numbers."

I'm not sure if $ a(0,1)$ is required, but you can just leave it as it is in your closed form.  Could you give me a reason for having $ a(0,1)$ in the closed form?

How does one work out $ a(1,1)$ without a second base case of $ a(0,1)$?
Otherwise, I think I've seen this problem before.

Where I REMEMBER this recursion coming from is a problem where you're supposed to cut an infinite watermelon into the maximal number of pieces in $ k$ dimensions using $ n$ cuts. I think it's a fairly famous problem.

Really?  Where I remember that recursion is [url=http://en.wikipedia.org/wiki/Pascal_rule]Pascal's identity[/url].  :)

Oh. Yep. That's pretty much the solution :)

I feel stupid now :P


[quote="mmrproblem"]Solve the equation   1.2^x+2.3^x+3.4^x+4.5^x+5.6^x=11^(5-x).13^(2x-7)[/quote]
solving in R or N? or...?

I checked your equation using MathCad. Unique root is something like 8.25...

where does this equation come from excactly and what's the sense of solving it ?

[quote="mmrproblem"]what about the solution other sets(Z  or N)?[/quote]

if there's no solution in R, there won't be one in Z or N^^


induction
real analysis
real analysis unsolved

Is it true that every uncountable set has a subset which is not only uncountable, but also has an uncountable compliment in the original set? I suspect that it is, but I can't think of a slick proof for it.

Offhand, one way to show this would be to generalize the construction of a Cantor set, but that appears to require imposing a topology.

The [url=http://en.wikipedia.org/wiki/Ordinal_number]first uncountable ordinal[/url] ought to give a counterexample.

[b]Edit.[/b] No it doesn't. In fact, the set of successor ordinals and the set of limit ordinals are both uncountable subsets of the first uncountable ordinal. That gives an alternative proof to that of Kent Merryfield below. In fact, given any uncountable set you can embed the first uncountable ordinal in it by a process of transfinite induction.

(Transfinite induction is equivalent to the axiom of choice, and I would be very surprised if this result can be proved without using AC.)

Suppose $ S$ is uncountable.

Consider all sets of the form $ P\equal{}\{(x_{\alpha},y_{\alpha})\mid \alpha\in A\}$ such that $ x_{\alpha}$ and $ y_{\alpha}$ are members of $ S$ and no element of $ X$ appears more than once among the $ x_{\alpha}$ or $ y_{\alpha}.$

Consider the collection $ \mathfrak{P}$ of all such sets $ P.$ Partially order $ \mathfrak{P}$ by inclusion. The union of any chain of sets in $ \mathfrak{P}$ is itself a set in $ \mathfrak{P}.$ Apply Zorn's Lemma: there must exist a maximal set.

Consider such a maximal set $ P.$ We claim that the set of points that appear either as $ x_{\alpha}$ or $ y_{\alpha}$ can omit at most one element of $ S.$ If it omitted two elements, we could form another ordered pair and $ P$ wouldn't be maximal. Let $ X\equal{}\{x_{\alpha}\mid \alpha\in A\}$ and $ Y\equal{}\{y_{\alpha}\mid \alpha\in A\}.$ By our construction, $ X\cap Y\equal{}\emptyset$ and either $ S\equal{}X\cup Y$ or $ S\equal{}X\cup Y\cup\{s\}.$ $ X$ and $ Y$ have the same cardinality. If $ X$ and $ Y$ were countable, then $ S$ would be countable. Since we were assuming that  $ S$ is uncountable, we must have both $ X$ and $ Y$ uncountable, and we have the desired partition.

I don't see any way to do this without the Axiom of Choice.

Another way (except the previous one, which is fine too): for any infinite set A
$ |A| \equal{} |A \times A|$, and then in the product $ A \times A$ it is easy to find such a set
(e.g. $ \{p\} \times A$ for some $ p \in A$). Then use the bijection between $ A$ and $ A \times A$ to pull it back to $ A$ itself. Again, this uses AC (for the first fact).


geometry
incenter
ratio
angle bisector
geometry solved

A line passing through the incenter of the triangle $ ABC $ cuts the sides $ AB , AC $ respectively at $ M , N $ . Prove that :
    $ \frac{BM\cdot CN}{AM\cdot AN} \leq \frac{BC^2}{4\cdot AB\cdot AC} $

[quote="levi"]$(*)b\cdot \frac{MB}{MA}+c\cdot \frac{NC}{NA}=a$
[b]Remark.[/b] The relation (*) is wellknown ! I don't know address in Forum where you can find the relation (*). [/quote]
Well, sometimes it's easier to post a proof than to find it on the Forum. Let me state a problem from the Romanian Math Olympiad 1984, proposed by prof. Ioan Tomescu (a world top expert in combinatorics, who also served as a leader of the Romanian IMO team several years):
Let $\alpha ,\beta $ be positive real numbers and let $M,N$ be variable points on the sides $AB,AC$ of triangle $ABC$ such that
\[\alpha \cdot \frac{MB}{MA} +\beta \cdot \frac {NC}{NA}=1.
\]
Prove that the line $MN$ passes through a fixed point.

Now, if $\alpha =\beta =1$ the fixed point is the centroid, while if $\alpha=\frac ba$ and $\beta =\frac ca$, the fixed point is the incenter.

[quote="levi"]$(*)\ \ \underline{\overline{\left| b\cdot \frac{MB}{MA}+c\cdot \frac{NC}{NA}=a\right| }}\Longrightarrow a\ge 2\cdot \sqrt {bc\cdot \frac{MB}{MA}\cdot \frac{NC}{NA}}\Longrightarrow \frac{a^2}{4bc}\ge \frac{MB}{MA}\cdot \frac{NC}{NA}.$

[b]Remark.[/b] The relation (*) is wellknown ! I don't know address in Forum where you can find the relation (*). You ask sir Darij Grinberg-manager of the this section.[/quote]

Thanks. Never thought I was a manager.  :?  :oops:  :)

I can't recall the relation (*) being explicitely proven here, but [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=20289]the transversal theorem[/url] provides a generalization and is more or less equivalent to the generalization posted by Enescu, but more constructive (it actually identifies the fixed point).

Here is how to derive the relation (*) from the transversal theorem:

Let I be the incenter of triangle ABC, so that the line MN passes through the point I, and let the line AI meet the side BC of triangle ABC at a point X. Since I is the incenter of triangle ABC, the line AI is the angle bisector of the angle CAB, and since the angle bisector of an angle of a triangle divides the opposite side in the ratio of the adjacent sides, we have $\frac{BX}{XC}=\frac{c}{b}$; thus, $\frac{BC}{XC}=\frac{BX+XC}{XC}=\frac{BX}{XC}+1=\frac{c}{b}+1=\frac{b+c}{b}$, so that $\frac{XC}{BC}=\frac{b}{b+c}$. Similarly, $\frac{BX}{BC}=\frac{c}{b+c}$. Also, since I is the incenter of triangle ABC, the line BI is the angle bisector of the angle ABC; speaking from the viewpoint of triangle ABX, this means that the line BI is the angle bisector of the angle ABX. Thus, since the angle bisector of an angle of a triangle divides the opposite side in the ratio of the adjacent sides, $\frac{XI}{IA}=\frac{BX}{AB}$, so $\frac{XI}{IA}=\frac{BX}{c}=\frac{BX}{BC}\cdot\frac{BC}{c}=\frac{c}{b+c}\cdot\frac{BC}{c}=\frac{BC}{b+c}=\frac{a}{b+c}$.

Now, applying the transversal theorem to the collinear points B, C, X, the point A, and the collinear points M, N, I on the lines AB, AC, AX, respectively, we get

$BX\cdot\frac{CN}{NA}+XC\cdot\frac{BM}{MA}+CB\cdot\frac{XI}{IA}=0$

(with directed segments). Thus,

$BX\cdot\frac{CN}{NA}+XC\cdot\frac{BM}{MA}=-CB\cdot\frac{XI}{IA}=BC\cdot\frac{XI}{IA}$.

Dividing by BC, we transform this into

$\frac{BX}{BC}\cdot\frac{CN}{NA}+\frac{XC}{BC}\cdot\frac{BM}{MA}=\frac{XI}{IA}$.

Using the ratios computed above, this becomes

$\frac{c}{b+c}\cdot\frac{CN}{NA}+\frac{b}{b+c}\cdot\frac{BM}{MA}=\frac{a}{b+c}$,

and thus $c\cdot\frac{CN}{NA}+b\cdot\frac{BM}{MA}=a$. This is exactly the required identity (*).

  darij


Prove that for all positive reals a b c d
a^4b + b^4c + c^4d + d^4a >=  abcd(a + b + c + d)

Why not? They are not hard to create, most of them have nice proofs, ...

For the one you posted, see http://www.mathlinks.ro/Forum/viewtopic.php?t=45218 .

  Darij

thank you Darij  :)  sorry but i have a question: hardsouls http://www.mathlinks.ro/Forum/viewtopic.php?t=4521 is nice but how can i find the coefficients for the similar problems ?


AMC
USA(J)MO
USAMO
number theory unsolved
number theory

[quote="greatestmaths"]let a and b are natural numbers such that a>b. a and b in N
prove that
                        (a^2+b^2)/(a^2-b^2) is in N[/quote]

That's obviously false : take for example $ a\equal{}2$ and $ b\equal{}1$

I think he post find (a,b) but this problem have solved.

This problem (or whatever the original / intended problem is) is not USAMO 2007...


Let $ a,b,c$ be nonnegative real numbers. Prove that
\[ \frac{a}{\sqrt{a\plus{}b}}\plus{}\frac{b}{\sqrt{b\plus{}c}}\plus{}\frac{c}{\sqrt{c\plus{}a}} \le \frac{5}{4}\sqrt{a\plus{}b\plus{}c}\]
:)

I want to see a proof using classical mixing variable not entirely mixing variable. :)

I think we can solve this one by the following way:
Assume that $ a\plus{}b\plus{}c\equal{}4$
The first case $ c\ge b\ge a$. Using Cauchy-Schawrz inequality we have:
\[ (\sum \frac{a}{\sqrt{a\plus{}b}})^2\le (a\plus{}b\plus{}c)(\sum \frac{a}{a\plus{}b})\]
We will prove that:
\[ \sum \frac{a}{a\plus{}b} <\frac{25}{4}\]
It is very easy to prove ;)

The second case $ a\ge b\ge c$ we will prove that $ f(a,b,c)\le f(a\plus{}c,b,0)$  and $ f(a,b,c)\le f(a\plus{}c/2,b\plus{}c/2,0)$.
This is the base idea to prove this one  :)

[quote="zaizai-hoang"]I think we can solve this one by the following way:
Assume that $ a \plus{} b \plus{} c \equal{} 4$
The first case $ c\ge b\ge a$. Using Cauchy-Schawrz inequality we have:
\[ (\sum \frac {a}{\sqrt {a \plus{} b}})^2\le (a \plus{} b \plus{} c)(\sum \frac {a}{a \plus{} b})
\]
We will prove that:
\[ \sum \frac {a}{a \plus{} b} < \frac {25}{4}
\]
[/quote]

But i think this part is not ture.

Are you sure about your opinion?  :wink: We always have:
\[ \sum \frac{a}{a\plus{}b}\le \frac{3}{2}\le \frac{25}{4},\forall c\ge b\ge a\]

But my hint is not all things what can help us prove this one. There is a very hard solution by Classical MV method of canhang_2007 :wink:

[quote="zaizai-hoang"]Are you sure about your opinion?  :wink: We always have:
\[ \sum \frac {a}{a + b}\le \frac {3}{2}\le \frac {25}{4},\forall c\ge b\ge a
\]
But my hint is not all things what can help us prove this one. There is a very hard solution by Classical MV method of canhang_2007 :wink:[/quote]

\[ \sum \frac {a}{a + b}\le \frac {3}{2}\le \frac {25}{4},\forall c\ge b\ge a
Then?
\[
(\sum \frac {a}{\sqrt {a + b}})^2\le (a + b + c)(\sum \frac {a}{a + b})
\]
?
I don't know how to use your hint to slove the original inequality.
 :(


Is it possible to give all solutions  $(r,s,t)\in \mathbb{N}^3$ of $1+3^r+3^s=7^t$ ?

I think you can easily adapt my proof ([url=http://www.mathlinks.ro/Forum/viewtopic.php?t=25448]here[/url]) by studying $v_3(7^t-1)$
Try it! :D

[quote="blang"]you can easily adapt my proof by studying $v_3(7^t-1)$Try it! :D[/quote]
I tried but I failed :( Can you give me some details :blush:

In fact I think we can extend this method to solve the following equation in $(r,s,t)\in \mathbb{N}$:
$1+p^r+p^s=n^t$  where $p$ is prime and $n\in \mathbb{N}$ such that: $n \equiv 1[p]$ and $n \not \equiv 1[p^2]$.
(try to calculate $v_p(n^t-1)$ with $v_p(t)$...)


function
algebra
polynomial
modular arithmetic
algebra unsolved

For every $n\in N$, prove that $[(\frac{7+\sqrt37}{2})^{n}]$ is multiple of $3$.
Where $[x]=n/n\in Z;n\leq x<n+1$

Let $a_{n}=\left({7+\sqrt{37}\over 2}\right)^{n}$. Observe the sequence $b_{n}=\left({7+\sqrt{37}\over 2}\right)^{n}+\left({7-\sqrt{37}\over 2}\right)^{n}$. By the theory of recursive equations, the characteristic polynomial for this sequence is $\left(t-{7+\sqrt{37}\over 2}\right)\left(t-{7-\sqrt{37}\over 2}\right)=t^{2}-7t+3$, which gives $b_{n+2}-7b_{n+1}+3b_{n}=0$ for $n\geqslant 1$. Since $b_{1}=7, b_{2}=43$, all $b_{n}$ are integer, and the equation also gives $b_{n+2}\equiv b_{n+1}\pmod{3}\implies (\forall n\in\mathbb{N})b_{n}\equiv 1\pmod{3}$

Now $0<{7-\sqrt{37}\over 2}<1$, which gives $0<\left({7-\sqrt{37}\over 2}\right)^{n}<1\implies b_{n}-1<a_{n}<b_{n}$. But that yields $[a_{n}]=b_{n}-1\equiv 0\pmod{3}$. QED


calculus
derivative
calculus computations

A lamp-post, $ 4.5m$ high, is $ 4.2m$ away from a straight footpath. A boy, $ 1.5m$ tall, walks along the path at the rate of $ 1.8m/s$. Find the rate at which his shadow is shortening when he has $ 14.4m$ more to cover in order to arrive at the point on the path nearest to the lamp-post.

Hello
Why r u posting a question whose answer is given in the book itself?Also the solution is so much clear to understand. :wink:
Thank u.

That was for alternate solutions. The method proposed in the book is too clumsy.


$ P_{n}Q_{n} \equal{} 1, Q_{n} \equal{} Q_{0} \plus{} \sum _{k \equal{} 0} ^{n \minus{} 1} P_{k}$ for all $ n \in N_{0}, Q_{0} \equal{} \alpha$
Following properties are given.
1. $ P_{n}$ is decreasing and positive sequence
2. $ Q_{n} \succ \sqrt {\alpha^2 \plus{} 2n}$
3. $ Q_{n} \prec \sqrt {\alpha^2 \plus{} 2n \plus{} \alpha^{ \minus{} 2} \plus{} \frac {1}{2}ln(\frac {2n}{\alpha^2} \plus{} 1)}$
Then, find minimum value of n that
$ P_{n} \prec \frac {1}{2e}$

shouldn't it be:[quote]$ P_{n}Q_{n} \equal{} 1, Q_{n} \equal{} Q_{0} \plus{} \sum _{k \equal{} 0} ^{n \minus{} 1} P_{k}$ for all $ n \in N_{0}, Q_{0} \equal{} \alpha$
Following properties are given.
1. $ P_{n}$ is decreasing and positive sequence
2. $ Q_{n} \ge \sqrt {\alpha^2 \plus{} 2n}$
3. $ Q_{n} \le \sqrt {\alpha^2 \plus{} 2n \plus{} \alpha^{ \minus{} 2} \plus{} \frac {1}{2}ln(\frac {2n}{\alpha^2} \plus{} 1)}$
Then, find minimum value of n that
$ P_{n} \le \frac {1}{2e}$[/quote] :?:

ok but i think it isn't important. It is for natural number when it was written in paper.
Thank you^^


[quote="andyciup"]Let $a,b,c$ be real numbers satisfying the condition $a^{2}+b^{2}+c^{2}=9$.
Prove that the following inequality occurs: \[a^{3}+b^{3}+c^{3}+2(a+b+c)+\frac{a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}}{3}\leq 35\] When does the equality occur?[/quote]
$a^{3}+b^{3}+c^{3}+2(a+b+c)+\frac{a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}}{3}\leq 35\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}\left(a^{3}+2a+\frac{27-a^{4}}{6}\right)\leq35\Leftrightarrow\sum_{cyc}(a^{4}-6a^{3}-12a+43)\geq0\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(a^{4}-6a^{3}-12a+43+13(a^{2}-3))\geq0\Leftrightarrow\sum_{cyc}(a-2)^{2}(a-1)^{2}\geq0.$
Equality holds when $(a,b,c)\in\{(2,2,1),(2,1,2),(1,2,2)\}.$ :)

Exactly!That is also the way I solved it :wink:
Are there other solutions?

$\Leftrightarrow\sum_{cyc}\left(a^{3}+2a+\frac{27-a^{4}}{6}\right)\leq35$

how did you get this?


SuMAC
summer program
PROMYS
geometry
Stanford
college
geometric transformation

I'm curious to see how your SAT scores were. Now, only answer if you did preparations by yourself (That is, if you just took it and got 2400, I don't want to know..). Also, reveal how many hours and # of books you went through to achieve that score.

This means a lot for my summer. Thanks.

2140, no studying...

My other standardized test scores are alot better and they are also no studying, so I guess I'm just not that great at the SAT.

2340 - three practice tests 2 weeks before and vocab practice and grammar practice through sparknotes book. I also took it twice in middle school. I went through more essays than actual practice tests.
1390- eigth grade, no practice.

2320, flipped through practice test the day before. I don't think it hurt that I had good math and english teachers, though.

By the way, if you are planning to spend your entire summer (or some non-trivial portion) preparing for a test, I would recommend not doing so.

This one girl I know got a 2380, but she's in an SAT prep class and probably put in a lot of outside study time as well.  I always thought spending a serious amount of time for this kind of test was, well, there are [i]much[/i] better things a kid could do with his time.

btw, 2100, went over some critical reading and writing mc, no math prep. otherwise, same as elemennop's comments.

I'd agree that spending a long time during the summer to prep for SAT is a bad idea. Basically, the marginal benefit of doing really well on the SAT is really not that great compared to doing something cool like a mathcamp(mathcamp, sumac, promys, etc). There's not that much you'll learn from studying for the SAT besides vocab and uh...how to take the SAT. Whereas the alternative teaches you problem solving which carries across in many areas. Besides, if you are doing this for college, "spent summer studying for SAT" is not nearly as impressive as "learned new math/pursued an interest in a subject I really cared about".

The PSAT is probably more important anyways, if you are planning to apply to summer programs like RSI or SSP.

Yeah, I'm planning on studying extensively in other areas so that the SAT comes naturally.  There should be no need to study (other than getting used to writing those essays in 25 min).

[quote="brady00n"]Yeah, I'm planning on studying extensively in other areas so that the SAT comes naturally.[/quote]

Yeah. Except for the detail that I don't call it "studying," that's basically what I advise. For sure, to enjoy the college experience a student should be able to read thoughtfully, understand SAT I math thoroughly (which is not usually a problem for someone who hangs out on this forum), and write a few clear paragraphs in short time span. So there is nothing about being ready for college that will make a student worse on the SAT I. But I dread students spending too much time on SAT I preparation, because there are so many better things to do with your time.

The best test-prep-as-such is still to take GENUINE old SAT I tests, such as those found in the College Board publications, and take those according to the official time limits. Then check your work, THINKING about each wrong answer you get, until you have an idea how you would notice and choose the right answer the next time. The College Board [url=http://www.amazon.com/gp/product/0874477182/]Official SAT Study Guide[/url] book contains enough practice tests for anyone. The tricky thing is to get a useful critique of your essay writing--but that is also the least important, as many colleges are ignoring the writing section of the new SAT I completely as they make admission decisions.

Other than that, the usual advice applies: READ, READ, READ, and READ. Read about something you like. Definitely read about math. Read about something fun. Read books that are supposedly good for you, but mostly read what you've always wanted to read. Read a lot, and you'll do better on all sections of the SAT I.

Well this is a problem with AoPS.. No, not a problem but conflict with me...

People in AoPS are extremely talented people and moreover, most people in here have good teachers. I DON"T. I go to rural school and here is how my English teachers are (at least from 9-10).

9th grade teacher: She just became a teacher after graduating from Messiah College but on my final, she gave me 88. I didn't think my writing was [i]that[/i] bad.

10th grade teacher: This teacher was definitely not the best one there is because she emphasized one quantity > quality. But, on my midterm, I got 100 (my first one). If I was really bad in 9th grade, I don't understand how I would possibly got 100. It's obvious that there's something wrong with my 9th grade teacher as well.

I'm not saying that I should get good grades for each one but this is the situation I'm having. I don't have problem with math section.. I can handle that.. But reading section and writing section.. I think good schools have students preparing for it but it's not that easy when English is your second language and you happen to have bad teachers.. Plus, if you want to score at least 2200 and got avg. on CR and Writing on PSAT.  :(

[quote]READ, READ, READ, and READ. Read about something you like. Definitely read about math. Read about something fun. Read books that are supposedly good for you, but mostly read what you've always wanted to read. Read a lot, and you'll do better on all sections of the SAT I.[/quote]

I remember reading this tip back long time ago too.. Okay, so how many books you mean READ? Your tip is good but it's so broad. I don't think it is possible to seriously raise SAT score soley by reading. I have one summer and that's it.

the ppl who are reading the essays...they have to read hundreds, so i've heard that as long as you have a solid first papragraph and lots of quantity, you can get a good score.  true or not?

i'm planning to take the SAT in eight  grade, so n e advice (more in the writing and all the language stuff areas) would be helpful.  especially since i am not going to be counting on any teachers to teach me n e thing.

and really, i've read enough on this forum about people who have said 'oh no studying, 2400".  maybe some tips from people who arent all[i] that[/i] smart?

@falcon: I wouldn't immediately assume that the 88 is the teacher's fault. Obviously I don't know the full situation, but I've found that in general teachers that grade harshly are an excellent oppurtunity to learn (for example I think my two best english teachers ever were last year, a teacher who refused to give A's unless the writing was really good, and this year a teacher where the class average on a 25 point essay is a 16). Also I think you are missing the point about reading when you ask "how much?" The point is to read something you [i]enjoy[/i]. Then the quantity will take care of itself. As for SAT stuff, I would suggest learning grammar. If you know grammar it is difficult to do poorly on the SAT grammar questions, which make up a lot of the test.

@star: SAT doesn't measure intelligence, it measures the amount of fundamental reading, writing, and mathematics skills you have picked up in high school. Again, the best way to develop those skills is to [i]use them continuously[/i] via reading, writing, and doing math.

I don't know which way I can argue as one of those people who has spent WAY too much time/money on SAT reading/writing prep (no doubt thanks to my parents), but I got a 2370. To be fair, I did take the December SAT, which IMO seems to be considerably easier to do well on. I think my SAT Prep was just a way to force me to read and write, because honestly I would never do that on my own time (go math!). In the end I'm just happy I have a decent SAT score to make me at least seem more well-rounded.

While I wouldn't recommend devoting several years to SAT Prep, definitely check out the grammar section - this one can kill you if you aren't familiar with it. Other than that, it's just what you know put into a test. Of course taking a few practice tests is helpful, too. A friend of mine didn't do any prep until the month before and she got a 2210, so not bad. Admittedly, she is a lot better than me at reading/writing so I have to say that my prep wasn't completely useless (if all you care about is SAT score).

I am not a bit worried at all about SAT. I never took the PSAT (poor dumb me I missed the deadline), and recently I took a practice test just to see what the SAT looks like. It turned out that, in the end, reading was 410, 800 math, and writing got me scratching my head. I never prepared for SAT before and I have no plans to prepare for it. Well maybe because I can use TOEFL to replace the Verbal sections of SAT :D

[quote="star99"]JSteinhardt:  you said high school knowledge is what SAT is.  but i'm going to take it in eighth grade, so the teachers i have wont be preparing me for that.  and its mainly the grammar and writing and all, because math isn't that hard for me.  so any books, online courses for the summer would be helpful[/quote]

Why take it in the 8th grade? Why not wait until at least the end of Freshman year so that you'll have had a year of high school education? There really is no rush, and even if you have some outstanding reason to take it, I don't think people (colleges) will really care how if you don't do well on the SATs in the 8th grade. Heck, a 1000 means that you did better than the average college-bound senior.

Don't spend your summer studying for the SAT, [i]especially[/i] not as a rising 8th grader. There are so many better things you could be doing with your time. If you want to practice something, at least practice [i]real problem-solving skills[/i]. Your underclassmen years are years of [i]oppurtunities[/i]. Don't waste them. Do something you enjoy instead.

If you're [i]really, really[/i] intent on preparing for this, then subscribe to CollegeBoard's SAT Question of the Day. You'll only be dedicating approximately 90 seconds each day to SAT preparation, and it gets you very used to the format of the test, teaches you stuff, etc.

[quote="Iversonfan2005"][quote="mysmartmouth"]I got a 2180 in 7th grade. [/quote]

I hate you sean. For people like me who are inept at taking the SAT, you should try the ACT. It's MUCH easier. Reasoning VS. Knowledge.[/quote]
Is there any kind of disadvantage for taking ACT instead of SAT?
I know that some colleges want ACT scores, but since it's much easier than SAT.....

[quote="Elemennop"][quote="mysmartmouth"]I got a 2180 in 7th grade. [/quote]

Wow, you got a higher score than I did when I took it as a junior.

That makes me feel bad.[/quote]

Wow, my AMC 12 score is a [b]factor [/b]of your AMC score.  That makes me feel bad.  :heli:

[quote="kimby_102"]
Is there any kind of disadvantage for taking ACT instead of SAT?
I know that some colleges want ACT scores, but since it's much easier than SAT.....[/quote]

I wouldn't say that the ACT is easier than the SAT in general.  If you have good reasoning skills, the ACT is easier.  If you have good study skills, SAT is easier.  Since we are all AoPSers, most of us will have a better ACT score than SAT score.

As far as colleges go, from what I've heard, SAT is more of an east coast thing; ACT is more popular on the west.  Although that's definitely something you want to look in to for your specific college(s), I would say that it can't hurt to take both tests.

[quote="bleumoose"]Actually, SAT's more Northeast and West Coast; ACT's more in the Southeast. Most schools accept both, check with the school (I think CollegeBoard has that info on their website, too).[/quote]

That's strange.  I live in SC, and I don't know anyone who took (primarily) the ACT.  However, out in Nebraska, I have a couple of friends who took ONLY the ACT.  Weird.

Huh, according to my memory ACT is primarily in the Midwestern states.

To a first approximation, the SAT is preferred along the coasts, and the ACT inland. But I grew up inland (in Minnesota) and took only the SAT. Looking at the list of available testing sites for each test in states that interest you will give some clue as to which test is more popular in which state.

As JSteinhardt said earlier, do SAT Question of the Day. It e-mails you a question, and you spend 30 (not 90) seconds each day doing an SAT question. Starting from freshman-sophmore year like I am, I should garner up a relatively vast amount of experience with the SAT. And practicing something for a little bit over a large amount of time is a very good way to get good at something, is it not?

Also, learn Latin :p. Well, I'm learning Latin because I like learning other languages and I'll enjoy it, but heck, doing better on the SAT verbal isn't a bad thing ;).

Finally, read, read, read. As with most other activites, the more you do it, the better and more used to it you get, and if you get good at reading, you'll only do better on the reading sections of the verbal SAT.

Eh, I've always thought that I'm going to get a 1800 on the SAT: 800 math, 800 verbal, 200 essay... Frankly, I cannot write an essay in *15 minutes*. Seriously, how do they expect anyone to write a [i]well-developed[/i] and [i]thoughtful [/i]essay in [i]15 minutes[/i]?

Regarding the ability to artificially inflate your score, consider this scenario: you're a well-qualified student who would do very well at Harvard and you're trying to get into it. However, being possibly the most prestigious school in the USA, they simply have too many (overly-qualified) applicants. Wouldn't you want to do everything you can to get in?

It's actually 25 minutes.

I did okay on the essay (9).  Basically what I did was spend a few minutes thinking of examples to support my position on the topic, then I just wrote.

I don't think the essay has to be well-polished or flow extremely well to get a decent grade - as long as you use lots of details, good sentence structure, and emphasize how the stuff supports your position/topic, you should be able to pull off a decent score, like a 9 or 10.

I realized that my essay score (7) is lagging behind my other scores as well.  The problem is that I don't have enough practice.  It's INCREDIBLY hard to motivate myself to write essays on my own time, and it wouldn't help anyway, because I can't critique myself well.  I go to public school, and it SUCKS.  (All I learned in ELA last year was the Ding Dong Theory.  I thought it was our teacher BSing, but I googled it, and it apparently is an "archaic, no longer believed" theory on the origins of language.  Great!)  We haven't written a paper all year, let alone a 25 minute essay.  Any tips for improving the essay score?  And don't say keep practicing, because I'm not that self-motivated.  :D

My SAT II scores (from the June test):

Math II: 800
Chemistry: 800
US History: 800

My SAT I scores (from the December test):

Math: 760 (unfortunately, 2 silly mistakes)
Writing: 800
Reading: 710

Cumulative: 2270

For the SAT II, I was already taking AP courses in the subjects I took the tests on (AP Calc, AP Chem, and AP US Hist.), and thus no extreme preparation was necessary. (Of course, I took a few practice tests from each subject to prepare in the few weeks before the test). The SAT I was a wholly different story, especially in terms of the Reading and Writing sections. In my sophomore year, my scores on these sections were consistently in the low-to-mid 600's. In order to remedy this, I practiced the SAT English sections thoroughly throughout the summer, and studied vocabulary lists in my free time. Therefore, by the end of the summer, I was able to raise my SAT Reading and Writing scores to the upper 600's (for reading) to the lower 700's (for writing). Then, once my junior year started, my AP course load required me to read advanced-level texts extensively (this was especially true for AP History). This obviously played a major role in greatly increasing my Reading and Writing skills, as manifested in the fact that when I resumed SAT practice tests in early October, I suddenly discovered that I was receiving scores consistently in the 700's for both English sections (with Writing scores often reaching the high 700's). Consequently, my cumulative scores shot up to the 2250-2350 range and stayed there with every practice test I took henceforth. In the official College Board practice test, I actually received a 2370 (800 M, 800 W, 770 R) :) With such scores on practice tests, I was able to take the October PSAT and December SAT I with confidence, receiving a 225 on the PSAT and the aforementioned scores on the SAT I.

[quote="mysmartmouth"]I got a 2180 in 7th grade.  The only thing I can really recommend is to read a LOT.  I didn't study for SAT, but I was constantly reading, sometimes getting disciplined in class for reading instead of doing an assignment.  My advice to middle school AoPSers:  For most of you guys, middle school is trivial.  During class, sit in the back and read a good book.  Reading is the best way to pick up vocabulary and spelling naturally.[/quote]

Yeah, I don't like you.  1980 in 7th grade. (800 Math, 590 for reeding/riting  :P )

I took the practice one.  As for classes, since MC nationals, I just play calculator games during Social Studies and Band.  Talk to your lab partner during science.  Relax so you can study math and stuff at home, it's more useful than learning stuff in school.  As for tests, just cram in the few minutes before.

SAT 2 scores in June:
Math II C: 800
Physics: 800
Chemistry: 800

[quote="kyyuanmathcount"]
Yeah, I don't like you.  1980 in 7th grade. (800 Math, 590 for reeding/riting  :P )

I took the practice one.  As for classes, since MC nationals, I just play calculator games during Social Studies and Band.  Talk to your lab partner during science.  Relax so you can study math and stuff at home, it's more useful than learning stuff in school.  As for tests, just cram in the few minutes before.[/quote]

Allen, 1980 isn't bad for a 7th grader. As an 8th grader I kind of bombed every section and got a 2080 (760 Math, 710 Writing, 610 Reading). Other than the previously given advice, is there anyone here who has perhaps gotten an 800 on the Reading who could possibly give me some advice on that.

I also say that the ACT is easier since as an 8th grader I received a 33 (33 in everything except English, 34). However, only the Midwestern schools really care about it I think.

[quote="life=tennis+math"]

I also say that the ACT is easier since as an 8th grader I received a 33 (33 in everything except English, 34). However, only the Midwestern schools really care about it I think.[/quote]

most colleges regardless of the region they're in accept it. :wink:


inequalities
floor function
number theory
prime numbers
number theory proposed

Let $p_{n}$be the nth prime:$p_{1}=2,p_{2}=3,p_{3}=5\dots$\\
Show that $\sum_{n=1}^\infty{\frac{1}{p_{n}}}$ is not bounded.

I hope this problem haven't been posted. :)

You can find a proof By ERDOS in the book : \[\text{Proofs from the BOOK}\] By Martin Augner and Gunter M.Ziegler.

Every $k \in \mathbb{N}$ can be written as $k=t \cdot s^{2}$ with $t$ not divisible by a square $>1$. This gives the inequality \[\sum_{k=1}^{n}\frac{1}{k}\leq \prod_{ \substack{ p \in \mathbb{P}\\ p \leq n}}\left(1+\frac{1}{p}\right) \sum_{s=1}^{n}{\frac{1}{s^{2}}}.\] Since $\frac{1}{s^{2}}\leq \frac{1}{s(s-1)}= \frac{1}{s-1}-\frac{1}{s}$ for $s \geq 2$ we get $\sum_{s=1}^{n}{\frac{1}{s^{2}}}\leq 1+\sum_{s=2}^{n}\frac{1}{s-1}-\frac{1}{s}= 2-\frac{1}{n}\leq 2$.
Together with the easy to verify property $1+x \leq e^{x}$ for all $x \in \mathbb{R}$ this yields \[\sum_{k=1}^{n}\frac{1}{k}\leq \prod_{ \substack{ p \in \mathbb{P}\\ p \leq n}}\left(1+\frac{1}{p}\right) \sum_{s=1}^{n}{\frac{1}{s^{2}}}\leq 2 \prod_{ \substack{ p \in \mathbb{P}\\ p \leq n}}\left(1+\frac{1}{p}\right) \leq 2 \cdot \prod_{ \substack{ p \in \mathbb{P}\\ p \leq n}}e^\frac{1}{p}= 2 \cdot e^{\sum_{}\frac{1}{p}}\] where the last sum also runs over all primes $p \leq n$.
To show that $\sum_{ p \in \mathbb{P}}\frac{1}{p}$ diverges, it now suffices to show that $\sum_{k=1}^\infty \frac{1}{k}$ diverges. But the latter one is a well known property, shown by the following since for all $n \geq 2^{m}$ we have: \[\sum_{k=1}^{n}\frac{1}{k}\geq \sum_{k=1}^{2^{m}-1}\frac{1}{k}= \sum_{s=0}^{m-1}\sum_{k=2^{s}}^{2^{s+1}-1}\frac{1}{k}\geq \sum_{s=0}^{m-1}\sum_{k=2^{s}}^{2^{s+1}-1}\frac{1}{2^{s+1}}= \sum_{s=0}^{m-1}\frac{1}{2}= \frac{m}{2},\] giving the divergence because $m$ can be chosen arbitrary large.

Thank you very much ,Zetax.the first step of the proof is really wonderful.
To lomos_lupin:Can I get the book on the net?

I am not sure if its available online, Its highly unlikely
Anyway here is the proof ,Completely elementary ,By Paul Erdos :

[color=blue]Problem.[/color]Let $P_{1},P_{2},...$ be sequance of the prime numbers then \[\sum_{i=1}^{\infty}\frac{1}{P_{i}}\] is not bounded.

[color=blue]Proof.[/color]Assume the contrary to our claim.
Then there exist a natural number like $k$ such that \[\sum_{i>k}^{\infty}\frac{1}{P_{i}}< \frac{1}{2}\] Lets call $p_{1},...,p_{k}$ the small and $p_{k+1},p_{k+2},....$  the big prime numbers.

For any natural $N$ we have : \[\sum_{i>k}^{\infty}\frac{N}{P_{i}}< \frac{N}{2}\] Let $N_{b}$ be the number of natural numbers like $n\leq N$ that are devisible by at least one big prime number and $N_{s}$ be the number of natural numbers like $n \leq N$ that have only small prime factors.We will show \[N_{b}+N_{s}< N\] Which is our desirade contradiction ,cause clearly we have $N_{b}+N_{s}=N$

Note that we have \[(1) N_{b}\leq \sum_{i>k}^{\infty}\lfloor \frac{N}{P_{i}}\rfloor < \frac{N}{2}\] Now we will estimate $N_{s}$.Consider any $n \leq N$ that has only small prime factor and write it like $n=a_{n}b_{n}^{2}$ (This part is in zetax solution too ) were $a_{n}$ is free square.
So $a_{n}$ can be the multipying of some different small prime numbers ,so we have $2^{k}$ different $a_{n}$ .also $b_{n}\leq \sqrt{n}\leq \sqrt{N}$ therefore \[N_{s}\leq 2^{k}\sqrt{N}\] because $(1)$ holds for every $N$ ,its suffice to find a natural number that \[2^{k}\sqrt{N}\leq \frac{N}{2}\longrightarrow 2^{k+1}\leq \sqrt{N}\] $N=2^{2k+2}$ is such a Number .Proof Complete.

My hat is off to this Elegant,Mysterious Solution.

Thank you very much.Yes ,it is really nice. :)


Find a 3 digit positive integer N such that when 113744 is divided by N, there is a remainder of 119, and when 109417 is divided by N, there is a remainder of 292.

My attempt to solve this was to solve: $ xN\plus{}119 \equal{} 113744, yN\plus{}292\equal{} 109417$

I got y= 873, x=909, and N=125. However the problem is that for one of them we have a remainder of 292, which is larger than 125 :(

After the system you wrote down in the previous post, $ (x\minus{}y)N\plus{}(119\minus{}292) \equal{} 113744\minus{}109417$, i.e. $ (x\minus{}y)N \equal{} 2^{2}*3^{2}*5^{3}$. So it's enough to choose $ N$ being any integral positive divisor of $ 2^{2}*3^{2}*5^{3}$ such that $ 292 < N < 1000$. This is just the case when $ N \equal{} 3\cdot 125 \equal{} 375$.

I can not believe I didn't see that >.< I lost so many marks :( O well, thanks

[quote="Gib Z"]
Find a $ 3$ digit positive integer $ N$ such that when $ 113744$ is divided by $ N$, there is a remainder of $ 119$, and when $ 109417$ is divided by $ N$, there is a remainder of $ 292$.
[/quote]

We can write down this as:

$ 113744\equiv 119\bmod N$ and $ 109417\equiv 292\bmod N$

So, for the first one, $ 113625\equiv 0\bmod N$, and for the second one, $ 109125\equiv 0\bmod N$

Gathering boths:

$ 113625\minus{}109125\equiv 0\bmod N\implies 4500\equiv 0\bmod N$

Since $ 4500\equal{}2^{2}\cdot 3^{2}\cdot 5^{3}$:

$ N\equal{}5^{3}\cdot 2^{2}\equal{}500$ and $ N\equal{}5^{3}\cdot 3\equal{}375$

But 500 doesn't seem to leave remainders of 119 and 292 when you divide those numbers :(


number theory
permutation
counting
combinatorics
IMO
IMO 2005
IMO Shortlist

Solution:
[hide]
${a_n}$ satisfies the conditions if and only if ${a_n-a_1}$ does.
Therefore we can assume that $a_1=0$

First of all, $|a_n|<n$
Otherwise, $a_n = 0$ mod $a_{a_n}$

Claim: $a_{k+1}$ is either the smallest positive number not in ${a_1, a_2, ..., a_k}$ or the largest negative number not in this set.
Proof by induction: the induction hypothesis implies that $a_1, a_2, ..., a_k$ are consecutive. Let m and M be the smallest and largest numbers in this consecutive set, respectively. It is clear that $a_{k+1}=m-1=M+1$ mod $k+1$. But since $|a_n|<n$, $a_{k+1}=m-1$ or $M+1$

This proves that if ${a_n}$ contains an infinite number of positive and negative numbers, it must contain each integer exactly once.[/hide]
That seems very simple so it's probably wrong  :blush:

It seems to be a correct solution and a very easy problem indeed.

Indeed easy problem.I have the same idea(may be the one possible).

My solution was slightly more complicated than the above one, but it was basically like, WLOG $a_1 \ge 0$, if $a_k$ is the first term of the sequence that's negative, it's easy to show $a_k = -1$, and $a_1, a_2, \ldots, a_{k-1}$ is a permutation of $0, 1, \ldots, k-2$, and then if $a_{k_2}$ is the second term of the sequence that's negative, then once again it's straightforward to show $a_{k_2} = -2$, and the terms between $a_{k_2}$ and $a_k$ are the positive integers from $k-1$ to whatever the difference was, and so on.

This seemed like quite a straightforward problem, much easier than many of the problems we got at MOP, so hopefully that bodes well for the USA :).

[quote="jarod"][quote="Valentin Vornicu"]Let $a_1,a_2,\ldots$ be a sequence of integers with infinitely many positive and negative terms. Suppose that for every positive integer $n$ the numbers $a_1,a_2,\ldots,a_n$ leave $n$ different remainders upon division by $n$.

Prove that every integer occurs exactly once in the sequence $a_1,a_2,\ldots$.[/quote]

  Indeed , I couldn't understand both this problem and schulmannerism's solution  :roll:

 For every positive integer $n$ the numbers $a_1,a_2,\ldots,a_n$ leave $n$ different remainders upon division by $n$.  So , we can understand that : There doesn't exist 2 numbers $ a_i, a_j $ in n numbers $ a_1,a_2 ,...a_n $ such that : $ a_i \equiv a_j $ mod n . So the problem is obviously true  :huh:[/quote]

It is true, and even obviously, but I do not understand your arguments. You prove that all numbers are different, but why each integer is a member of a sequence?

it seems to me that it's going to be a really easy IMO.. at least if I didn't make any mistakes, but I think it's really an easy problem.

excuse me for not typing in tex, but i'll correct it in short time..
[hide]first we see that there cannot be two equal elements. suppose they are a(n) and a(m), n<m, then they would give the same residue mod m, a contradiction.
then observe d=|a(n)-a(1)|<=n-1 (*),
 because otherwise when looking the first d elements mod d a(n) and a(1) give the same residues.
lemma. a(1), a(2), .. a(n) are n consecutive integers.
proof. by induction, base is trivial, |a(2)-a(1)|<=1 by (*)
let's assume that {a(1),.., a(n)}={x+1, .., x+n} for some integer x (**).
from (*)  and (**) follows that x-n<a(n+1)<x+2n+1, and from the condition in the problem a(n+1) should give the same residue as x mod n+1, so a(n+1)= x or x+n+1, which proofs the claim.
now, since there is infintely many negative and positive numbers in the sequence, it's trivial to see that every number must appear in the sequence. (e.g. we can take such M that the first M elements of the sequence contain at least N negative and at least N positive numbers, then the numbers N and -N must occur in the first M elements.) [/hide]

I took a look at others' solutions after typing, so I guess I should of done it before typing :/

I have a solution :
 Consider $ (a_1,a_2,...,a_n) $ . Suppose $ max($ a_1,a_2,..a_n $) -  min($ a_1,a_2,..a_n $) \geq n $
 So $ max($ a_1,a_2,..a_n $) \equiv  min( $ a_1,a_2,..a_n $ ) $ mod k >n  : contradiction !
 So $ max( $ a_1,a_2,..a_n $ )  -  min($ a_1,a_2,..a_n $) = n-1 $
 Then $ |a_1| , |a_2| ,..., |a_n| $ are consective integers : Q.E.D

[quote="euclid_gemoetry"]schulmannerism:

I don't think you prove that [b]each integer occurs exactly once in the sequence[/b], which is required by the problem.

You only prove that [b]if the sequence contains every integer, then every integer occurs exactly once[/b].

I think these two claims are different.[/quote]

No, I think he proved it right.  His claim implies that each a1,...,ak is consecutive. Also that at each point, you can only "attach" one more number either from the left side (largest negative number not in the set) or from the right side (smallest positive number).  Now if m>0 is not in the sequence, it means that after a while, you can only attach from the left, which means you have finitely many positive numbers. Contradiction. Same argument for m<0.

You must just prove that the function $f : Z \rightarrow (a_n)$ is bijective (injective and surjective).

Injective :

suppose that $a_i=a_j$ for some $i$ and $j$ such that $i<j$. Then for every $m$ such that $m>j$, we have the sequence :
$a_1,a_2,....,a_i,....a_j,....,a_m $
But this sequence contains $a_i$ and $a_j$ such that $a_i=a_j (mod m)$. And it's absurb.

Surjective:

by induction : let the sequence $a_1,a_2,...,a_n$ which contains n integers. We will construct $a_{n+1}$. $a_{n+1}$ is the missing remainder in $a_1,a_2,...,a_n$. Thus, there is an infinite of values for $a_{n+1}$

[quote="cezar lupu"]Which country proposed this problem?[/quote]
The Netherlands!!! My lovely home country...
I was participant at this year's IMO and I solved this problem, I found it the easiest problem, but I liked it..

:?: There is very easy number theory on IMO 2005

[quote=djmathman]@above: this doesn't work!  The given statement only holds when $n$ is finite; while $n$ can be arbitrarily large, you cannot take $n = \infty$.  (Figuring out how to patch this is the whole point of the problem!)[/quote]

I was just saying its so obvious the problem statement is true, i never said that was my solution

uhh

How is it "obvious" that the problem statement is true?

Obviously $a_i\neq a_j$ for $i\neq j.$ Clearly shifting of each term of sequence on constant integer doesn't affect on problem, so WLOG $a_1=0.$ Since $m\nmid a_n$ for every $m>n>1$ it follows that $|a_n|<n.$ Now, since sequence contains infinitely many of both positive and negative terms we may end with the following

[b]Claim.[/b] For every $k\in \mathbb Z^+$ multiset $S_k=\bigcup_{i=1}^k a_i$ is a sequence (in some order) of consecutive integers.
[i]Proof.[/i] Induct on $k.$ Base case $k=1$ is trivial. Next, for $k\geq 1$ define $m=\min S_k,M=\max S_k.$ Then it's clear that $$a_{k+1}\equiv m-1\equiv M+1 \pmod{k+1}\implies a_{k+1}\in \left \{ m-1,M+1\right \} \quad \Box$$

If there exist two numbers in the sequence $a_{1}, a_{2}, \ldots, a_{n}$ that were alike, then they would have the same remainder upon division by $n$, thus every two number in the sequence must be different.

Suppose that some integer $N$ exists, such that $a_{i} \neq N$ for all $i=1,2,3, \ldots$. Let $a_{1}$ be given. There are two cases to consider depending on whether $a_{1} <N$or $a_{1}>N$. We only have to study the first case since the hunt after a contradiction of the latter is very similar.

Suppose that $a_{1}<N$. Since the sequence contains infinitely many positive integers there must exist a smallest $n>1$ such that $a_{n}>N$ (Well-ordering theorem), that is $a_{i}<N$ for all $i=1,2, \ldots, n-1$. Since the sequence $a_{1},a_{2}, \ldots , a_{n}$ have different remainders upon division by $n$, there exist exactly one $i$ such that $a_{i} \equiv N \pmod n$.

If $a_{n} \equiv N \pmod n$, then by definition there exists an integer $\ell$ such that $a_{n} = N + \ell n$. Since $N<a_{n}=N+\ell n$ we have $\ell \geq 1$. This implies $a_{n}=N+\ell n > a_{1} + n$, that is $a_{n}-a_{1}>n$ which implies that $a_{n} \equiv a_{1} \pmod{a_{n}-a_{1}}$, thus $a_{n}$ and $a_{1}$ having the same remainder upon division by $a_{n}-a_{1}$, contradiction.

This means that there exists an $1\leq i < n$ such that $a_{i} \equiv N \pmod n$, which, again, implies that there exists an integer $\ell$ such that $a_{i}= N + \ell n$. Since $N>a_{i}=N+ \ell n$ one must have $\ell \leq -1$, and so $a_{i} = N + \ell n \leq N-n$. This implies $a_{n} - a_{i}>N-(N-n)=n$, that is $a_{n} \equiv a_{i} \pmod{a_{n}-a_{i}}$, another contradiction.

This contradiction proves the claim.

[hide=Solution]
We begin with a claim.
[b]Main Claim:[/b] For all $n\geq 1$, $(a_1, a_2, \cdots, a_n)$ is a permutation of $(x+1, x+2, \cdots, x+n)$ for some integer $x$.
[b]Proof:[/b] Induct on $n$. For $n = 1$, take $x = a_1 - 1$. Now, assume thet for some $k\geq 1$, $(a_1, a_2, \cdots, a_k)$ is a permutation of $(x+1, x+2, \cdots, x+k)$ for some integer $x$. Then, $a_1, a_2, \cdots, a_k, a_{k+1}$ must have distinct residues modulo $k+1$, so $a_{k+1}\neq a_i$ for any $1\leq i\leq k$. Thus, we either have $a_{k+1} > x+k$ or $a_{k+1} < x+1$.

If $a_{k+1} > x+k$, then if we have $a_{k+1} = x+k+1$, we are done, since $(a_1, a_2, \cdots, a_{k+1})$ is now a permutation of $(x+1, x+2, \cdots, x + (k+1))$. Else, $a_{k+1} = x+r$ for some $r\geq k+2$. Then, note that $$a_{k+1} \equiv x + r \equiv x+1 \equiv a_j \pmod{r-1},$$ for some $1\leq j\leq k$. Since $r\geq k+2$, $r-1\geq k+1$, so this contradicts the fact that $a_1, a_2, \cdots, a_{r-1}$ produce distinct remainders modulo $r-1$.

If $a_{k+1} < x+1$, then if we have $a_{k+1} = x$, we are done, since $(a_1, a_2, \cdots, a_{k+1})$ is now a permutation of $(x, x+1, \cdots, x+k) = ((x-1) + 1, (x-1) + 2, \cdots, (x-1) + (k+1))$. Else, $a_{k+1} = x - r$ for some $r\geq 1$. Then, note that $$a_{k+1} \equiv x - r \equiv x + k \equiv a_j\pmod{k+r},$$ for some $1\leq j\leq k$. Since $r\geq 1$, $k+r\geq k+1$, so this contradicts the fact that $a_1, a_2, \cdots, a_{k+r}$ produce distinct remainders modulo $k+r$.

Thus, our inductive step is complete and the claim is true. $\square$

Now, let $M$ be an integer. Clearly, $M$ cannot appear more than once in $\{a_n\}_{n\geq 1}$, since if $a_i = a_j = M$, then $a_1, a_2, \cdots, a_j$ cannot have distinct remainders modulo $j$ since $a_i$ and $a_j$ would produce the same remainder modulo $j$.

Thus, it suffices to show that $M$ must appear in $\{a_n\}_{n\geq 1}$ at least once. Assume the contrary. Let $a_1 = t$.

We cannot have $t = M$, since $M$ does not appear in $\{a_n\}_{n\geq 1}$, and $t = a_1$.

If $t < M$, then I claim that $a_k < M$ for any $k\geq 1$. To show this, assume the contrary. Then, there exists a $j\geq 1$ with $a_j\geq M$, so $a_j > M$ (as $M$ does not appear in $\{a_n\}_{n\geq 1}$). Then, $a_1, a_2, \cdots, a_j$ are just $j$ consecutive integers by our claim. However, since $t < M < a_j$, we must have $M$ appearing in $a_1, a_2, \cdots, a_j$, since $t$ and $a_j$ appear in it, so every integer in between appears in it. Thus, $a_k < M$ for $k\geq 1$. However, this is absurd, since $\{a_n\}_{n\geq 1}$ has infinitely many positive terms, and no two terms are equal as discussed above, so $\{a_n\}_{n\geq 1}$ must be unbounded above (since if $a_n \leq B$ for some $B\geq 1$ for all $n\geq 1$, then there are at most $B$ positive terms, since each of $1, 2, \cdots, B$ can be achieved at most once by $\{a_n\}_{n\geq 1}$).

If $t > M$, then I claim that $a_k > M$ for any $k\geq 1$. To show this, assume the contrary. Then, there exists a $j\geq 1$ with $a_j \leq M$, so $a_j < M$ (as $M$ does not appear in $\{a_n\}_{n\geq 1}$). Then, $a_1, a_2, \cdots, a_j$ are just $j$ consecutive integers by our claim. However, since $a_j < M < t$, we must have $M$ appearing in $a_1, a_2, \cdots, a_j$, since $a_j$ and $t$ appear in it, so every integer in between appears in it. Thus, $a_k > M$ for $k\geq 1$. However, this is absurd, since $\{a_n\}_{n\geq 1}$ has infinitely many negative terms, and no two terms are equal as discussed above, so $\{a_n\}_{n\geq 1}$ must be unbounded below (since if $a_n \geq -B$ for some $B\geq 1$ for all $n\geq 1$, then there are at most $B$ negative terms, since each of $-1, -2, \cdots, -B$ can be achieved by at most once by $\{a_n\}_{n\geq 1}$).

Thus, we have a contradiction, so $M$ appears in $\{a_n\}_{n\geq 1}$ at least once, so we're done. (Since we showed that $M$ must appear at most once, and this holds for all $M\in \mathbb{Z}$). $\blacksquare$

[b][color=#f00]Claim 1[/color][/b]

    $\mid a_j-a_i\mid=t$, so $t\ge\max\{i;j\}$

Assume that $t<\max\{i;j\}$, so now if we get the first $t$ terms of the sequence we must have that $a_i\not\equiv a_j\pmod{t}$, which is obviously incorrect.

[b][color=#f00]Claim 2[/color][/b]

    $a_k$ is either the least positive number not in the sequence already, or the largest negative number not in the sequence already.

We will prove this by induction, $k=1$ is obvious.

Let $a_1,a_2,\ldots,a_k$ be with the given claim. That means that they are a sequence of consecutive numbers. So they can be written as:

\[\{a_1,a_2,\ldots,a_k\}=\{x,x+1,x+2\ldots,x+k-1\}\]

So it is obvious that

\[a_{k+1}\equiv x+k\pmod{k+1}\]

If $a_{k+1}\le x-k-2$ we will see that $\mid a_{k+1}-a_1 \mid\ge k+2>k+1$, whic is impossible by the first Claim.

If $a_{k+1}\ge x+2k$, then $\mid a_{k+1}-a_1 \mid\ge k+1$ and again contradiction by the first claim.

So given that we have infinitely many negative and positive numbers, it means that every number is in the sequence exactly once.

[b][color=#f00]Claim 1[/color][/b]

    $\mid a_j-a_i\mid=t$, so $t\ge\max\{i;j\}$

Assume that $t<\max\{i;j\}$, so now if we get the first $t$ terms of the sequence we must have that $a_i\not\equiv a_j\pmod{t}$, which is obviously incorrect.

[b][color=#f00]Claim 2[/color][/b]

    $a_k$ is either the least positive number not in the sequence already, or the largest negative number not in the sequence already.

We will prove this by induction, $k=1$ is obvious.

Let $a_1,a_2,\ldots,a_k$ be with the given claim. That means that they are a sequence of consecutive numbers. So they can be written as:

\[\{a_1,a_2,\ldots,a_k\}=\{x,x+1,x+2\ldots,x+k-1\}\]

So it is obvious that

\[a_{k+1}\equiv x+k\pmod{k+1}\]

If $a_{k+1}\le x-k-2$ we will see that $\mid a_{k+1}-a_1 \mid\ge k+2>k+1$, which is impossible by the first Claim.

If $a_{k+1}\ge x+2k$, then $\mid a_{k+1}-a_1 \mid\ge k+1$ and again contradiction by the first claim.

So given that we have infinitely many negative and positive numbers, it means that every number is in the sequence exactly once.

Solved With [b]sman96[/b] and [b]GeNuAlCo_pi[/b]
Solution:
Claim $1$: $a_i\neq a_j$ for any $i\neq j$
Proof: FTSOC, assume $a_i=a_j$ for $i \neq j$, then take a sufficient large $n$, in that case, having $a_i=a_j$ means $a_1,a_2,...,a_n$ is not a complete residue system mod n.Hence,Contradiction!!!
Claim $2$: $a_1,a_2,...,a_n$ is a permutation of consecutive $n$ numbers.
Proof: Now, from $a_1,a_2,...,a_n$, assume $a_i$ is the minimum and $a_j$ is the maximum. Now, $a_j-a_i < n$, FTSOC, assume it isn't true. Then , let $d=a_j-a_i$, Now, $a_i \equiv a_j \pmod d$, but since, $d\geq n$. This means $a_1,a_2,...,a_d$ is not a complete residue system. Hence, Contradiction!!!
Now,In $a_1,a_2,...,a_n$, any two number has difference at most $n-1$ and any two numbers are different, which automatically implies that $a_1,a_2,...,a_n$ are a permutation of n consecutive numbers.

Now, Since there are infinitely many positive as well as negative numbers in the sequence, this means that if $N$ is a big enough positive number in the sequence and $M$ is a small enough negative number in the sequence, then the sequence eventually covers every single number between $M$ and $N$. Henceforth, Every single integer exists on the sequence exactly once. $\blacksquare$

wow uh it seemed really intimidating but actually really easy lol, unless im fakesolving

First, note that all terms must be distinct, because if there's an $i>j$ s.t. $a_i=a_j$, taking mod n>i makes two same residues. Now, we claim that in any sequence $a_1,a_2,...,a_k$, these numbers are consecutive in some order. Assume not, so there are $l,m$ s.t. $|a_l-a_m|=d\ge k$ (since otherwise minimum minus maximum is at most k-1, and there are only k numbers so they would be consecutive). Then taking mod d we get $a_l\equiv a_m\pmod {d\ge k}$, contradiction. We can extend this for sufficiently large k infinitely. $\blacksquare$

In fact, this sequence is always some permutation of a set of consecutive integers.

We use induction to prove this; for our base case, consider when $n=1$; a single integer is definitely a permutation of a set of consecutive integers.

For the inductive step, if $a_1, a_2, \dots, a_n$ are $n$ consecutive integers, consider $a_{n+1}$. If $a_{n+1}$ differs from $a_1$ by some integer $x > n$,  then the remainders when $a_1, a_2, \dots, a_x$ are divided by $x$ won't all be distinct. In particular, $x \neq n+1$ since then $a_1 \equiv a_{n+1} \pmod{n+1}$.

Similarly, $a_{n+1}$ differs from $a_n$ by no more than $n$. Furthermore, $a_{n+1}$ obviously can't equal any of $a_1, a_2, \dots, a_n$, hence either $a_{n+1} = \min (a_1, \dots, a_n) -1$ or $a_{n+1} = \max (a_1, \dots, a_n) + 1$.

Combining this with the condition that there are infinitely many positive and negative integers, we clearly can't skip over any integers.

Actually, we have the following characterization of $\{a_i\}$:

[color=#f00][b]Claim:[/b][/color] $a_1$, $a_2$, $\dots$, $a_n$ is a permutation of some $n$ consecutive numbers. (This solves the problem by the infinitely many positive/negative condition.)

[i]Proof.[/i] Induct - base case $n=1$ is vacuous. Clearly none of the $a_i$ are equal (otherwise take modulo sufficiently large to obtain a contradiction). Note that $\lvert \max(a_1, a_2, \dots, a_n) - \min(a_1, a_2, \dots, a_n) \rvert \le n -1$, otherwise consider $\pmod{\max(a_1, a_2, \dots, a_n) - \min(a_1, a_2, \dots, a_n)}$ - this finishes.

We will show by induction that $a_i$ equals one of $\min{(a_1,a_2,\dots,a_{i-1})}-1$ and $\max{(a_1,a_2,\dots,a_{i-1})}+1.$ For the base case we have $i=2.$ If $|a_2-a_1|=k \ne 1,$ then we would have $a_1 \equiv a_2 \pmod{k},$ contradiction. For the inductive step, if $a_i < \min{(a_1,a_2,\dots,a_{i-1})}-1$ then we have $a_i \equiv \max{(a_1,a_2,\dots,a_{i-1})} \pmod{\max{(a_1,a_2,\dots,a_{i-1})}-a_i},$ and $\max{(a_1,a_2,\dots,a_{i-1})}-a_i > i,$ impossible. Similarly $a_i > \max{(a_1,a_2,\dots,a_{i-1})}$ is impossible, and our inductive hypothesis is equivalent to every $k$ with $\min{(a_1,a_2,\dots,a_{i-1})} \le k \le \max{(a_1,a_2,\dots,a_{i-1})}$ being an $a_j$ for some $j,$ so our induction is complete.

It is clear that each positive integer appears at most once. Notice that our claim above implies that $\max{(a_1,a_2,\dots,a_{i-1})}$ and $\min{(a_1,a_2,\dots,a_{i-1})}$ can only change by at most one when going from $i$ to $i+1,$ so if $x > a_1$ appears, then all $y$ with $a_1 \le y \le x$ appear, and similarly if $x < a_1$ appears then all $y$ with $x \le y \le a_1$ appear. Since the sequence has infinitely many positive and negative terms, this gives that every integer appears at least once, by taking an arbitrary large or small $x,$ so we are done.

We claim that $a_1$, $a_2$, \dots, $a_n$ must be $n$ consecutive numbers which obviously is the problem.

We prove this by induction. Our base case is pretty obvious. Now assume FTSOC that $a_1$, $a_2$, \dots, $a_{n-1}$ are consecutive numbers and $a_n$ is not a part of that block. WLOG let $a_{n-1}$ be the greatest number and WLOG let $a_n$ be such that $a_n > a_{n-1}+1$. Also, WLOG let $a_1$ be the smallest number. We can see that we have $a_n -a_1 \geq n$. Thus we consider that when we divide by $a_n-a_1$ we will not have distinct remainders for $a_n$ and $a_1$. $\blacksquare$

We'll prove two claims about the sequence, which when combined imply the desired result.
[b][color=#f00]Claim 1:[/color][/b] All elements in the sequence are distinct.
[i]Proof.[/i] Assume otherwise; pick any $i < j$ such that $a_i = a_j$. Then, the integers $a_1, a_2, \cdots, a_j$ must have different remainders when divided by $j$, a contradiction, as $a_i$ is also among them. $\blacksquare$
[b][color=#f00]Claim 2:[/color][/b] We have $|a_i - a_j| \le n - 1$ for all $1\le i, j \le n$.
[i]Proof.[/i] Assume otherwise, and let $a_i - a_j = k \ge n$. Then, the integers $a_1, a_2, \cdots, a_k$ must have different remainders when divided by $k$, but since $k \ge n \ge i, j$, $a_i$ and $a_j$ are also among these numbers, which gives a contradiction. $\blacksquare$
So, $a_1, a_2, \cdots, a_n$ are $n$ consecutive numbers in some order for all $n$, since they're all distinct, and the maximum difference between two of them doesn't exceed $n - 1$. Thus, $a_1, a_2, \cdots, a_n$ contain all the integers in the range $[l, r]$ for some $l, r$ with $r - l + 1 = n$. Also, since the next integer $a_{n + 1}$ can't lie in the range $[l, r]$ since all the numbers are distinct, and it can't be $< l - 1$, or $>r + 1$, since we don't skip any number either, this means that $a_{n + 1} = l - 1,$ or $a_{n + 1} = r + 1$. So, either $l$ gets decremented by $1$, or $r$ gets incremented by $1$. It suffices to show that each of these operations happens infinitely many times. Assume that only one of them occurs an infinite number of times, and the other occurs only a finite number of times; WLOG $l$ gets decremented only finitely many times. Then, if the initial value of $l$ was $k$, and it gets decremented $d$ times, any number $x < k - d$ doesn't appear in the sequence at all, a contradiction, since there are infinitely many negative numbers in the sequence. So, each operation occurs an infinite number of times, and we're done.


i might move to boston. are there any good schools in mathcounts or just math in general around there?

Move to Lexington. (Make sure you're rich; the housing is hellishly expensive I heard)
The schools are amazing.
It's scary.

You'd really have ot move to lexington. Most private schools aren't great in MathCounts. Come to Roxbury Latin though!

Don't forget about the insane influx of Asians in Lexington (24.2% at Clarke last year).

But taxes are also pretty insane, they've raised taxes about 60% in 8 years

u should move to Acton...

Lexington is OP at math here...

 :D

Well the team has already been decided so idk if you can still get on if u move here...

i dont think ill be able to move to anywhere far from boston because my mom got a job there and thats the main reason why we are moving.

and im suposedly moving next year or the year after

[quote="abacadaea"]or just math in general around there?[/quote]

i MIGHT move this year.

Well you'd better move in the next week if you even want a chance at any not terrible team around here then. My school's team will be final tomorrow...

Anyway, lexington works, PA works, but those aren't all that close to boston. Well, that's subjective, but...anyway, my school is [i]in[/i] boston, but it's not that east to get into after 9th grade. They accept ~12 for 9th grade, but then at max 1 for 10th and 11th. I'm not sure about the 7th grade enrollment policies-they take 42, but i came this year, so I'm not sure about how stringent they are. Yeah, we need more math people.


Prove that the equation $ x^3\plus{}y^3\plus{}1\equal{}z^3$ has infinitly many solutions in natural numbers.

If $ x^n\plus{}y^n\plus{}d\equal{}z^n, n$ is odd prime, try to prove $ n|d$.

[url]http://www.mathlinks.ro/viewtopic.php?t=199597[/url]

As I see noone solve it, so I'll post a solution soon :)

Yes, you can write my equation in the form $ (\minus{}x)^3\plus{}(\minus{}y)^3\plus{}1\equal{}z^3$ :)

[quote="isomorphism"][quote="Taoli"]If $ x^n \plus{} y^n \plus{} d \equal{} z^n, n$ is odd prime, try to prove $ n|d$.[/quote]

But the original poster claims there are infinitely many solutions, and 3 does not divide 1.What are you trying to say?

[quote]Could someone please post a solution? I'm really interested in finding out an answer.[/quote]

Hey [b]t0rajir0u[/b], 2 questions:

1) Did you find at least one solution? I could not even do that.... I am losing confidence in the statement  :(

2)What does t0rajir0u mean?[/quote]

It is a hint that I will have a try on for a similar problem, not for the right topic. I will try to say something not misunderstanding.

[quote="hsiljak"]I know that the questions were adressed to to [b]t0rajir0u[/b], but here's the answer to your first question, [b]isomorphism[/b]:

$ (6,8,9)$[/quote]

Oh, sorry, I thought [b]isomorphism[/b] is not a nickname  :D

[quote="hsiljak"]I know that the questions were adressed to to [b]t0rajir0u[/b], but here's the answer to your first question, [b]isomorphism[/b]:

$ (6,8,9)$[/quote]

Thanks mate :D

I was doing trial and error and I never reached up to that :(

[quote]It is a hint that I will have a try on for a similar problem, not for the right topic. I will try to say something not misunderstanding.[/quote]

Ok. So how will you prove it?

That implys $ x\plus{}y\minus{}z \equiv \minus{}d \bmod n$. Oh I correct it:

Try to solve the case $ n|d$.

This solution is not mine. I saw this in some number theory book.

[hide] $ x \equal{} 9t^4\minus{}3t, y\equal{}9t^3\minus{}1, z\equal{}9t^4$ [/hide]


Solve that equations:
1)  3x-[x]=8

2) [x+[x]+2]=7

[ ] denotes greatest integer function.

[hide="Solution"]1. Denote $\alpha=x-[x].$ By the definition of $[.]$ function, $0\leq\alpha<1.$

$3([x]+\alpha)-[x]=8$
$2[x]+3\alpha=8$

It follows that $3\alpha$ must be an integer, hence $\alpha=0$ or $\alpha={1\over 3}$ or $\alpha={2\over 3}$. Substituting the first value we get $[x]=4$, which gives the solution $x=[x]+\alpha=4.$ Substituting the second value we get $[x]={7\over 2}$ which is impossible, and the third value gives $[x]=3$ which yields $x=[x]+\alpha={11\over 3}.$ Therefore, the solutions are $x_1=4, x_2={11\over 3}.$

2. If $n$ is an integer, then $[y+n]=[y]+n.$ Hence $[x+[x]+2]=[x]+[x]+2=2[x]+2.$ But $2[x]+2=7$ gives $[x]={5\over 2}$, which is impossible. Hence, the equation has no solution.[/hide]

[hide]
2) $7 \leq x + [x] + 2 < 8$
$5 \leq x + [x] < 6$

Now, if $x < 2$, then $x + [x] < 4$
$2 \leq x < 3$, then $4 \leq x + [x] < 5$
If $3 \leq x$, then $x+ [x] \geq 6$
So no solutions.[/hide]

apparently you are more quickly than me ...  :P

But Farenhajt, I don't think your proof for 1) is complete... What about $\frac{11}{3}$?
Apart from that, nice   :D

Oh, apparently I can't add  :D $3+{2\over 3}\neq {10\over 3}$. Corrected now.

$\blacktriangleright\ 1^{\circ}.\ 3x-[x]=8\Longleftrightarrow [x]=p\in Z\ \left(\ x=\frac{p+8}{3}\ \right)\ ,\ p\le \frac{p+8}{3}<p+1\Longleftrightarrow$
$p\in \left(\frac 52 ,8\right]\cap Z\Longleftrightarrow p\in \{3,4\}\Longleftrightarrow x\in \left\{\frac{11}{3},4\right\}\ .$
$\blacktriangleright\ 2^{\circ}.$ For any $x\in R$ and for any $z\in Z$ there are the following relations: $[x+z]=[x]+z$ and $\{x+z\}=\{x\}$, i.e. $T=1$ is the smallest  period for the function  $\{ x\}$.
[hide="For Farenhajt."] What is more aired, your or my solution ?[/hide]

[hide="For Virgil Nicula"]What does "aired" mean?[/hide]

To expose to the air, to weather, to ventilate, to renew the air in a room, to aerate, to be refreshed, to take an airing. It is enough ? Now understood you ? Shorty, it is means
\[ \mathrm {\ WHAT\ IS\ MORE\ CLEARLY\ } \]

I believe the question you want is, "Which is clearer?"  Since there is a choice between several known options, the appropriate question word is "which" (as in "which of these"), not "what."  There is no thing which is "clearly," (an adverb), so you want to say, "Which is more clear?"  However, we have a word for "more clear," namely "clearer."

The word "aired" is not a synonym of "clear" in English.


$h(t) = c - (d -4t)^2$

At time $t = 0$, a ball was thrown upward from an initial height of 6 feet. Until the ball hit the ground, its height, in feet, after $t$ seconds was given by the function $h$ above, in which $c$ and $d$ are positive constants. If the ball reached its maximum height of 106 feet at time $t = 2.5$, what was the height, in feet, of the ball at time $t = 1$

I tried to solve this problem but I'm not sure If I did it correctly.

$106 = c - (d - 4 (2.5))^2$
$106 = c - d^2 + 100$
$6 = c - d^2$

sove for x when t = 1

$x = 6 + (- 4 * 1)^2$
$x = 22$

what do you guys think?

[quote="pawfood"]$h(t) = c - (d -4t)^2$
At time $t = 0$, a ball was thrown upward from an initial height of 6 feet. Until the ball hit the ground, its height, in feet, after $t$ seconds was given by the function $h$ above, in which $c$ and $d$ are positive constants. If the ball reached its maximum height of 106 feet at time $t = 2.5$, what was the height, in feet, of the ball at time $t = 1$[/quote]Well, since $h(t)$ is a quadratic and 6 is it's vertical shift, $c-d^2=6$ and by taking the derivative, $h'(t)=8d-32t$ and when the ball is at a maximum, the derivative must be 0, so $8d-32 \times 2.5 = 0 \Rightarrow d=10$. Since $c-d^2=6$, plugging in d gives $c=106$ and then plugging d, c, and $t=1$ into the original equation, $106-(10-4)^2 \Rightarrow 106-36 \Rightarrow \boxed{70}$. Think this is right, but you might want to double check it.

yes your answer is correct but I still don't understand why my answer isint. If someone can explain it to me that will be nice.

[quote="pawfood"]sove for x when t = 1
$x = 6 + (- 4 * 1)^2$
$x = 22$[/quote]How did you get here? Because this is where you wen't wrong, since $c \neq 6$ and $d \neq -4$.

Yes, Hawkage, this could easily be done with calculus, but lets use middle school techniques here.

Look at the original problem: $h(t) = c - (d - 4t)^2$.  The problem says that it starts at an initial height of six feet. So substituting zero into the above equation, we have

$h(0) = c - d^2 = 6$

We plug in t = 2.5, h(t) = 106:

$106 = c - (d - 4*2.5)^2$

[b]Now pay attention to this next part, because many students mess up here[/b]

$106 = c - (d-10)(d-10) = c - d^2 + 20d - 100$
$206 = c - d^2 + 20d$

Since $c - d^2$ = 6,

$206 = 6 + 20d$
$d = 10$
$c = 106$

Plugging this into the original equation with $t=1$:

$h(1) = 106 - (10 - 4*1)^2 = 106 - 36 = \hbox{70}$.

It should be pointed out that extra information was given in this problem, you don't even need to know that t=2 was a maximum of the function.  However, this also lets you compute a quick calculus solution, as hawkage did, or you can also use the equation $x=\frac{-b}{2a}$ in an expanded quadratic to find more information.

Since $c-d^2 = 6$,

$106 = 6 + 20d + 100$
$d = 0$
$c = 6$

Now we plug in $t=1$ to the equation:

$h(1) = 6 - (0 - 4)^2$
$h(1) = 6 - 16 = -10$[/b]

[quote]
Now pay attention to this next part, because many students mess up here
[/quote]

damn, I totally missed that binomial. Thanks a lot for the explanation


How many distinct phrases of three words can be made using all the letters in "Math Contest" once (each word must have at least 1 letter in it)?

Edit: the previous one had too many so this one is simpler.

Do you just use logic for this or is there some kind of formula?

This one seems a bit hard for classroom math.  You should move it to high school basics.

do all of the words have to make sense, and does the phrase have to make sense...i.e can you make a phrase like ath testcon m?
or like CA test month... and can you vary the order, as CA month test, Month Test CA, CA test month...sorry about these words, i just made them up.

? I didn't think it was too hard. :/

The words don't have to make sense. And the 3 you listed would be 3 separate phrases.

Um, ok then

My solution:
[hide]
There are 11 letters in math contest, and the only letter that repeats is t (3 times)

So there are 11!/3! ways to arrange the letters in math contest.

Then since you want the phrase to be 3 words, you have to insert 2 spaces. There are 10 places to put the first space and 9 to put the second. But the spaces are the same so its a combination: (10*9)/2!. The total # of 3 word phrases  = (11!*10*9)/(3!*2!) or [b]299,376,000[/b] I believe.[/hide]

That's not right; your method would allow the two spaces to be next to each other.

There should be $10*8/2$ ways for the spaces to go.

No, for the two spaces to be right next to each other one would have to do $\frac{10 \cdot 10}{2}$, right?

um?

10 spaces for the first one to go

MATHCONT_EST

9 spaces for the second

1) M_ATHCONT_EST
2) MA_THCONT_EST
3) MAT_HCONT_EST
4) MATH_CONT_EST
5) MATHC_ONT_EST
6) MATHCO_NT_EST
7) MATHCON_T_EST
8) MATHCONT_E_ST
9) MATHCONT_ES_T

so... yeah.

ra: I counted that in:

So there are 11!/3! ways to arrange the letters in math contest. (there are 3 T's, so 11!/3!)

[hide]
so the solution is
$\frac{11!}{3!}\cdot \frac{10 \cdot 9}{2}= \fbox{299376000}$
[/hide]


He is probably concerned with his score being at the mercy of a person who just happens to be a mean grader, or two mean graders.

All of the graders for any particular problem will get together and agree on a grading standard.  Each individual problem will be read by only one person.  The vast majority of solutions will fit into the grading standard; however, the graders are encouraged to discuss "unusual" solutions with each other to arrive at a consensus score.

There shouldn't be any "mean" or "nice" graders -- our goal is a consistent scoring process.  It's not perfect, though, so there will be an appeal process (which will be described at the time when the Round 1 scores are released in early November).

I think he might be concerned with the grading speed by asking how many graders.

I also want to find out: When can we expect the grading to finish?

umm i just realized, i made a mistake where i was solving for a variable that was half the value of the solution as i stated in the problem, found it and just concluded i was done!! i forgot to multiply be two and give the answer!!!! :o  :o  :maybe:  :mad:
how many points off should i expect???

A trivial error, you should most likely lose at most one point (for having the incorrect answer), as long as your solution is sound.

[quote="AIME15"]How many graders will grade each set of solutions?[/quote]
For the grading session this Friday that graded problems 2 and 3, we had 18 graders. That made 9 graders per problem. The other problems were graded at other sessions, so I don't know how many graders worked on them.

Erin Schram
USAMTS grader

[quote="BOGTRO"]Does this mean that the solutions have been graded?[/quote]
Yes, they are graded. But the score sheets (and the solutions themselves) must be shipped back to California from Maryland in order to record the scores.

We are scanning in the grades and comments as we speak.  As it is late in the day and we are a bit less than halfway through, it will not be done today.  It will hopefully be done tomorrow, and if not, then almost certainly on Monday.


algebra
polynomial
superior algebra
superior algebra unsolved

prove that every a+b(sqr(2)) with b not equal to 0 is a root of some second-degree polynomial with rational coefficients.

See $ a \plus{} b\sqrt {2}$ and $ a \minus{} b\sqrt {2}$ , we have equation $ x^2 \minus{} 2ax \plus{} a^2 \minus{} 2b^2\equal{}0$.  :lol:

Note that if $ x_1\plus{}x_2\equal{}S$ and $ x_1x_2\equal{}P$ then $ x_1,x_2$ are roots of the equation $ x^2\minus{}Sx\plus{}P\equal{}0$.

[quote="stf123"]i dont see how a+b(sqr(2)) is a root of that polynomial.[/quote]

And there is this secret trick. $ r$ is a root of $ P(x)\equal{}0$ if and only if $ P(r)\equal{}0$. Some people would call this the definition of the term root, but what do they know?


tell right or wrong?
The question is to prove that sigma n is multiplicative.
I think it is quite easy to prove this through the definition of multiplicative function .For any value of n, sigma n is not zero as d(n) is a multiplicative function.
Also we can write that sigma n = (sigma n) * (sigma 1)

Hi rationalist.

These results are proved in "Multiplicative Functions" which can be found at [url]http://wdjoyner.com/teach/book/node21.html[/url]

Hope this helps.


inequalities
logarithms
integration
real analysis
real analysis unsolved


Sorry, but Lassiter is booked for Duke, which happens to be on the same day, unfortunately.


Define a sequence of positive integers as follows: $ a_{1}\equal{}2$ and $ a_{n}\equal{}\sum_{k\equal{}1}^{n\minus{}1}ka_{k}$ for all $ n\ge 2$.  Compute $ a_{1337}$.

[quote="lingomaniac88"]Define a sequence of positive integers as follows: $ a_{1}= 2$ and $ a_{n}=\sum_{k = 1}^{n-1}ka_{k}$ for all $ n\ge 2$.  Compute $ a_{1337}$.[/quote]

[hide="Hint"]\begin{eqnarray*}a_{n}&=&\sum_{k = 1}^{n-1}ka_{k}\\ &=& (n-1)a_{n-1}+\sum_{k = 1}^{n-2}ka_{k}\\ &=& (n-1)a_{n-1}+a_{n-1}\\ &=& n a_{n-1}\end{eqnarray*}

Now, try to find a closed form for $ a_{n}$[/hide]
[hide="Solution"]From the hint, $ a_{n}= n a_{n-1}$. So:

\begin{eqnarray*}a_{n}&=& n a_{n-1}\\ &=& n(n-1)a_{n-2}\\ &=&n(n-1)(n-2) a_{n-3}\\ &=&\cdots\\ &=&n(n-1)(n-2)\cdots(3) a_{2}\\ &=& n!\end{eqnarray*}

This doesn't work for $ n=1$, but it works for everything else. So $ a_{1337}=\boxed{1337!}$[/hide]


Prove that the equation

$x^{4}+y^{4}+z^{4}-2y^{2}z^{2}-2z^{2}x^{2}-2x^{2}y^{2}= 24$

Has no solution in integers $x, y, z$.

According to your rewritten equation, those two's in the initial one are not supposed to be there. Alternatively, if the two's ARE supposed to be there, then your rewritten equation is wrong.

...Right.  The $2$s are, indeed, supposed to be there.   :oops:

Since all the exponents in the equation are even, we can assume $x,y,z\geqslant 0$, and because it's symmetric, WLOG we can assume $x\geqslant y\geqslant z$. Rewrite it thus:

\begin{eqnarray*}&& (x^{2}+y^{2}-z^{2})^{2}-4x^{2}y^{2}=24\\ &\iff& (x^{2}-2xy+y^{2}-z^{2})(x^{2}+2xy+y^{2}-z^{2})=24\\ &\iff& [(x-y)^{2}-z^{2}][(x+y)^{2}-z^{2}]=24\\ &\iff& (x-y-z)(x-y+z)(x+y-z)(x+y+z)=24\end{eqnarray*}

Our starting assumptions give $x-y+z>0, x+y-z>0, x+y+z>0$, hence $x-y-z>0$. Assume any of the two factors are equal:

$x-y-z=x-y+z\iff z=0\iff (x-y)^{2}(x+y)^{2}=24$ which is impossible as $24$ isn't a perfect square. Similarly for other pairs where only one term has the sign switched.

$x-y+z=x+y-z\iff y=z\iff x^{2}(x-2y)(x+2y)=24$, hence $x=1$ or $x=2$, but neither gives a natural solution for $y$.

$x-y-z=x+y+z\iff y+z=0\iff y=0\land z=0\iff x^{4}=24$, which is impossible.

Therefore the factors are pairwise distinct. Since the product of four different natural numbers can't be less than $1\cdot 2\cdot 3\cdot 4=24$, we conclude that

\begin{eqnarray*}x-y-z &=& 1\\ x-y+z &=& 2\\ x+y-z &=& 3\\ x+y+z &=& 4\end{eqnarray*}

but the solution to this system is $({5\over 2},1,{1\over 2})$, hence non-integer. Thus is proven that the initial equation has no solution in $\mathbb{Z}$.


A pizza with a diamater of $ 12$ inches is divided into four parts as shown. The central angles for the two larger congruent slices each measure $ 20$ degrees more than the central angles for each of the two smaller congruent slices. What is the measure, in degrees, of a central angle for one of the smaller slices?

[asy]draw(Circle((0,0),20));
draw((16,12)--(-16,-12));
draw((-16,12)--(16,-12));[/asy]

[hide="Solution"] Let $ x$ be the central angle of one of the smaller slices. The central angle of one of the larger slices would be $ x\plus{}20$. A circle has $ 360^{\circ}$, so
$ 2(x)\plus{}2(x\plus{}20)\equal{}360$
$ 4x\plus{}40\equal{}360$
$ x\equal{}80^{\circ}$
[/hide]


For given positive integers $ n$ and $ N$, let $ P_n$ be the set of all polynomials $ f(x) \equal{} a_0 \plus{} a_1x \plus{}... \plus{} a_nx^n$ with integer coefficients such that:
(i) $ |a_j | \leq N$ for $ j \equal{} 0, 1, . . . , n$;
(ii) The set $ \{j | a_j \equal{} N\}$ has at most two elements.
Find the number of elements of the set $ \{f(2N) | f(x) \in P_n\}$.

Could you explain more about that ,specially the last line ?

For every $ f(x)\in P_n$, you look at the value of $ f(2N)$. You see how many different values you have in total.


geometry
rectangle
perpendicular bisector

From AoPS Volume I:

The perpendicular bisectors of two of the sides of triangle $ ABC$ intersect the third side at the same point.  Prove that the triangle is right-angled.

I drew something out in paint  :D but I can't get it to copy onto here....

Anyways, this my proof is completely different than the book's... so I'd like to know if there are any holes in it.

First, I drew $ \bigtriangleup ABC$ then I took the perpendicular bisectors of sides $ AC$ and $ BC$ and let them meet at point $ D$.

This gave us 3 triangles - $ \bigtriangleup BDE$, $ \bigtriangleup EDC$, $ \bigtriangleup BAD$.

We know that $ \bigtriangleup BDE$ $ \cong$ $ \bigtriangleup EDC$ by $ SAS$

Then I drew a line from point $ D$ to side $ AB$ - point $ F$

This line forms rectangle $ FDBE$, thus $ FD$ $ =$ $ BE$

So we know that $ \bigtriangleup BFD$ $ \cong$ $ \bigtriangleup BD$ $ \cong$ $ \bigtriangleup EDC$ by $ SAS$

Similarly, we know that $ \bigtriangleup FAD$ $ \cong$ $ \bigtriangleup BFD$

Thus all four triangles are congruent

So we can find the area of the large $ \bigtriangleup ABC$ by adding the areas of the four smaller triangles...

Which gives us $ \frac{1}{2}(4)(xy)$ $ \Rightarrow$ $ 2xy$

Where $ x$ and $ y$ are the legs

The area of $ \bigtriangleup ABC$ is given by: $ K_t = \frac{1}{2}(2x)(2y) Sin B$

Thus angle $ B$ must equal 90 degrees

I think you overcomplicated the problem. In fact I'm not sure I understand your proof.

We know that two of the bisectors meet at a point on the third side. However, we know that all the perpendicular bisectors in a triangle concur, and the only point on the third side which the perpendicular bisector of this side passes through is the midpoint. Thus the perpendicular bisectors concur at the midpoint of this side. Call this point $ M$. Then $ M$ is the circumcentre of $ ABC$.

This in turn means that the segment which $ M$ lies on is the diameter of the circumcircle. By the theorem of Thales, the angle opposite this side is a right angle.


I've taken a very general chemistry course before but I'm looking for a text that I'm really going to be able to teach myself the in-depth concepts from. Amazon just gives tons of results and many questionable reviews. I'm looking for something more like a textbook.

I second that, assuming you are looking for General Chemistry (as opposed to chemistry in vague generality, as your other course seems to have been).

hey people wich is the best book in chemistry for olympoad learning and for a student of class X1

I'd definitely recommend Atkins as well; I used it for Olympiad prep as well as Carey's Organic Chemistry.  I came very close too....special honors (Top 50) :(

carey's organic chemistry is not a good book imho.
use anything else (wade is good)

Well...I only read the first few chapters of it and it gave me the knowledge to pretty much answer every single organic related question on the National exam this year....so, if you are just looking to have a fairly basic knowledge of orgo for the olympiad, I suppose it'd do the job (plus the orgo sections in Atkins).

once you get to the fun fun orgo cramming at the study camp, carey fails (imo, once again), even though that is what is given.


quadratics
algebra
real analysis
real analysis unsolved

How do you differentiate:

$y=x+\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}$

The first step is to find the value: $\sqrt{y}=y-x$. That's a quadratic equation for $\sqrt{y}$; you want the positive root. Square it, and you have an ordinary algebraic function.


If $ a <  b$, is it true that for positive integer $ n$, $ na \minus{} \lfloor na \rfloor < nb \minus{} \lfloor nb \rfloor$?

What happens if $ a$ is a little less than $ \frac{1}{n}$, and $ b$ is a little greater than $ \frac{1}{n}$?


MATHCOUNTS AMC AIME probability geometry topology LaTeX

Anyone who wants to start another Jeopardy is welcome to do so here. I can't come up with all the problems but I really would like to participate (I couldn't sign up for the other one). Anyone at all?

I will.

Warning: This one is going to be a lot harder...

I'll join too, but can it be 3 players for a more authentic feel? And also, like 1 day time limits.

I would join, but pythag gonna make IMO questions which are way above my level...

ah..who cares..i join

[quote="xpmath"] ARGH NO PYTAHG SHOULD NOT WRITE PROBLEMS[/quote]
 :yup: /me agrees.

[quote="pythag011"][quote="xpmath"]I'll join, as long as this starts soon. ARGH NO PYTAHG SHOULD NOT WRITE PROBLEMS. Sorry, dunno what happened.[/quote]

Well, I can write mathcounts level problems. I just find mathcounts level problems really really boring.

Difficulty:

100- Throwaway question. (i.e. what is 200+346 + 784?)
200- Chapter level sprint medium question.
300- State level sprint medium question.
400- State hard/ National Medium/ AMC 12 Medium
500- National Hard/ AMC 12 Hard/ AIME Medium.[/quote]

Sounds good....instead of National hard, can you just have National..

[quote="1=2"]there are usually 8 categories. They could be Algebra, Counting, Probability, Geometry, Logic, Arithmetic(i'll make some of them hard :diablo:), Graph Theory, and Topology.[/quote]

Topology?

My categories... are going to be normal until the last round. Last round, the categories are not like that...

How about:

Geometry
Algebra
Counting
Probability
Common Sense  :rotfl:

First round:

Algebra
Combinatorics
Geometry
Number Theory
Arithmetic

I'll join but send a PM or something so I remember about this.

for anyone who is interested, I'm making one more of these requests in the High school basics forum.

ok.. so... are we be going to go on with the tournament or not??

Yes, I'll post results for that question.

ARG, sorry, rather busy with some stuff, should be better now...

dafjbomb and Twin Prime Conjecture got the probelm correct.

[quote="akhil0422"]I will join!!!!!!!![/quote]

Err....the games started...

Poincare, please choose the next probelm.

No!!!! Me and BOGTRO need you to help us.....[hide...]but I guess akhil will help us too. [/hide]

I'm not playing; if I do, everyone else will pity Wickedestjr and BOGTRO. :(

[b]quit UNLESS pythag submits answers for ernie's tourney[/b]


[quote="pianoforte"]Martian bus tickets have six-digit numbers, from 000001 to 999999. Martians think that the ticket is lucky if the sum of the first three digits is equal to the sum of the last three digits. Prove that the total sum of all 6-digit numbers which appear on the lucky tickets is divisible by 13.[/quote]
[hide="a feeling"]1001 is divisible by 13[/hide]

[hide]

Any number abcabc has this property and is divisible by 1001, thus making it divisible by 13.

If the number abcdef has that property, than so does defabc.

Those numbers can be expressed as:
$100000a+10000b+1000c+100d+10e+f$
$100000d+10000e+1000f+100a+10b+c$
Summing them, we get
$100100a+10010b+1001c+100100d+10010e+1001f$, which is divisible by 1001, meaning it is divisible by 13.

Thus the sum of all possible numbers is divisible by 13.

[/hide]


I hope I won't make too much mistakes in my text because I have quite small wisdom about math terms in english.
So I have a exercise to do in home but I absolutely don't know how to do it.
Prove that jf (F) is a linear function which isn't a constant function and x1,x2,x3 are numbers that x1<x2<x3 then triangle whom vertex are points (x1,f(x1)), (x2,f(x2)),(x2,f(x1)) is similar to triangle whom vertex are points (x2,f(x2)),(x3,f(x3)),(x3,f(x2)).
Please tell me if I write something wrong or something is incomprehensible, I will try to repeair it.

[edit] No, one know? Did I explain something wrong?[edit]

It might help if u just use the variables a, b, and c instead of x1, x2, and x3. I can't help u with the actual problem though, sorry.

The proof is simple.  Both triangles are right, and aligned to the axes, and both have a hypotenuse with the same slope.  Both of these claims are easily verified.

fireemblem13 you can use a b c and tell that a is x1 b is x2... it doesn't matter.
t0rajir0u (hm it is quite hard to write your nick :lol: so please appreciate this :lol:) It will be easier if you teel how to claim it or where i can find any informations that can help me.
But thanks. I have math today so i will try to do something with it, and for about 9 hours (OMG I have stupid day 9 hours in school  :() I will write correct and completly answer.

Knowing that they are right (and aligned with the axes), you could show that the two legs have equal ratios by showing that $(x3-x2)/(x2-x1)=[f(x3)-f(x2)]/[f(x2)-f(x1)]$

the numbers are supposed to be subscripts....yeah....  :? anybody know how to you do those in Latex?


inequalities unsolved
inequalities
algebra

Let $ a,b,c$ be positive numbers such that $ abc\equal{}1$. Prove that

$ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\minus{}\frac{3}{a\plus{}b\plus{}c}\geq 2(\frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{1}{c^2})\frac{1}{a^2\plus{}b^2\plus{}c^2}$

[quote="jbmorgan"]Let $ a,b,c$ be positive numbers such that $ abc \equal{} 1$. Prove that

$ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \minus{} \frac {3}{a \plus{} b \plus{} c}\geq 2(\frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2})\frac {1}{a^2 \plus{} b^2 \plus{} c^2}$[/quote]
$ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \minus{} \frac {3}{a \plus{} b \plus{} c}\geq 2\left(\frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2}\right)\frac {1}{a^2 \plus{} b^2 \plus{} c^2}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}a^2\sum_{cyc}(a^2b\plus{}a^2c)\geq2\sum_{cyc}a\sum_{cyc}a^2b^2\Leftrightarrow\sum_{cyc}ab(a\plus{}b)(a\minus{}b)^2\geq0.$

[quote=jbmorgan]Let $ a,b,c$ be positive numbers such that $ abc\equal{}1$. Prove  that
$$ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{3}{a+b+c}\geq 2\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\frac{1}{a^2+b^2+c^2}$$[/quote]
[url=https://artofproblemsolving.com/community/c6h126552p717919]Komal, November 2004[/url]
Let $ a,b,c$ be positive numbers such that $ abc=1$. Prove that
$$ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{3}{a+b+c}\leq 2\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\frac{1}{ab+bc+ca}$$


Let $ \mathcal{P}(\mathbb{N})$ be the set of all PMFs on $ \mathbb{N} \equal{} \{1,2,3,...\}$. Is the following set closed w.r.to the total variation metric $ \parallel P \minus{} Q\parallel : \equal{} \sum_{x\in \mathbb{N}}|P(x) \minus{} Q(x)|$?

$ E \equal{} \{P\in \mathcal{P}(\mathbb{N}): \sum_{x\in \mathbb{N}}x^2P(x) \ge K\}$ where $ K < \infty$.
Note: convergence w.r.to the total variation metric is the same as pointwise convergence.

what about the sequence $ P_k \equal{}(1\minus{}\frac{10}{k^2})\delta_0 \plus{} \frac{10}{k^2}\delta_{k} \in E_{K\equal{}10}$.

Thanks alekk. So for every $ K<\infty$, the set $ E$ is not closed.


calculus
derivative
geometry
3D geometry
calculus computations

find f'(1) if f(x)=x^(1/3) using only the defintion of the derivative. Note having a lot of trouble with the algebra due to the cube root


The summer has passed quite nicely, with those of us who were 8th graders at Nats glorying in our success and enjoying finally being smart enough to do pretty much every problem. But soon school will start again, and you will be 9th graders. Since you will be more than Mathcounts age at that time, please stop posting answers to all but the hardest and participating in the mock competitions. Note that this is by grade, not age.

If you could, please post when your school starts again; I have no idea when, since I'm homeschooled.

Billy

U.S. schools will start in this month. I think some schools already started while some schools will start in next week or week after that.  :)

I definitely agree to this forum because if you are done with Mathcounts, move onto Getting Started and Intermediate Forum, etc. You have to keep moving and develop your math skills.

Regards,

As another moderator of this Forum, I agree completely with Billy.  If you are not MATHCOUNTS eligibile, you should leave the problems in this Forum for those who are still in Middle School.

Furthermore, since this is the start of the school year and the start of the [b]coaching phase[/b] of the program, the typical problem posted in this Forum should be early worksheets or low-numbered questions from early competition rounds.

The competition phase of MATHCOUNTS doesn't begin until January or February.  In general, I would discourage contests at this time of the year.  We are coaching at this point, and students still have five months to learn the material for the School and Chapter Competitions.  A poor result on a contest may discourage a student who has great potential.    [b]Contests and challenges that are at the State or National level are completely inappropriate at this time of the year.[/b]  Please save up your great contest questions for use when the official MATHCOUNTS contest season begins.  If you need higher-level challenges, please look to the Getting Started and Intermediate Forums.

While State and National-level problems aren't utterly forbidden, there should be few of them and there should be a good reason you are asking such questions at this stage in your preparation.

Let's all remember that this is not a Forum solely for the elite.  This is a Forum for all MATHCOUNTS students.  Let's all work together to make this forum inviting to new MATHCOUNTS students, but still interesting to experienced MATHCOUNTS students.  If you have seen a challenge problem before, please leave it for the less experienced students.  If you think a challenge problem is easy, please leave it for the less experienced students.

In general, you should not be answering any challenges you think are "easy".  However, there are some exceptions.  If no progress has been made towards a problem after a few days, you might want to offer some guidance or a hint.  Do you have a better way to solve a problem?  Do you think you have a clearer explanation?  Do you have a unique insight?  Can you explain a problem in a different way?  Then please share that with the less experienced students.  The idea is to learn from each other, not to demonstrate your superior problem-solving skills.

Remember that no MATHCOUNTS problem is "easy".  If you think a challenge problem is "easy", you shouldn't be the first to answer it.

I have been posting some difficult national problems, and from what you say, I should stop. Knowing that only easier problems can be posted, should I post my national problems (because I have a bundle of questions) in the high school "getting started" forum?

I'm not a moderator, but here's what I think:

Probably. "Harder Mathcounts Nationals problems may also be posted in the Getting Started forum" is my paraphrase of something in an announcement in the aforementioned forum.

It's fine to post some of the harder MATHCOUNTS problems in the MATHCOUNTS forum so long as the forum maintains the educational spirit of the entire MATHCOUNTS curriculum.

It's also fine to post some of the harder ones in Getting Started as they are also appropriate to that forum.

Ahhh, I'm going to be in ninth grade in eleven months or so...

got to use this wonderful opportunity to spam up the MC forums and get the admins angry *cough* LEARN

[quote="JesusFreak197"]Are we still allowed to help others when they're trying to figure out how to solve a problem, or should we still leave that to those in middle school?[/quote]

If really no one gets it, or there are some serious errors in discussion, then speak up.


calculus
integration
calculus computations

okay so i guess i ran into more problems....

estimate the definite integral?

1. antiderivative from 0-4 of  -(3x^(3/2) + x^(1/2)) dx

2. antiderivative from 4-9 of   (4sqrt(r) - 3r(sqrt(r)) dr

3. antiderivative from 1-4 of   -3/(2p+1)^2     dp

4. antiderivative from 2-3 of  (3x^-2  -  x^-4)  dx

5. antiderivative from 0-3 of  m^2(4m^3  +  2)^3   dM

help pleasee[/img]

If you just want to confirm your answers, go to [url=http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=calculus&s2=integrate&s3=advanced]this[/url] site.

#2 was the only one I actually checked and it is correct.

[quote="jacksmith18"]ok so for #1 i got    -656/15

and for #2 i got    -3038/15

are these correct??[/quote]
You are correct.
Result of #4 is $ \frac{305}{648}$