Which round was harder, Round 1 or Round 2?

On round 1, numbers 1 and 3 were slightly easier, and I found 2, 4, and 5 to be harder, so round 2 was much easier.

Round 2 was probably a bit easier. 4/2 was quite hard, though.

[quote="lotrgreengrapes7926"]For Round 2:
#1 was harder.
#2 was much easier.
#3 was easier.
#4: I found the answer after about 2 hours.  :lol: But it took me a lot longer to prove it.  :roll: So, harder.
#5 was easier.
I used number theory for both number 1s. And both number 5s were relatively easy once you saw the solution. Question 4 took me the longest for both rounds.[/quote]

Agreed.

#3, though... about the same, I'd say.

Round 1 was definitely harder though.


inequalities
induction
number theory
prime numbers
number theory unsolved

A sequence $ \left(p_n\right)$ of natural numbers satisfies:
[b]1.[/b] The numbers $ p_1$ and $ p_2$ are primes.
[b]2.[/b] For every $ n\geq 3$, the number $ p_n$ is the greatest prime divisor of the number $ p_{n\minus{}1}\plus{}p_{n\minus{}2}\plus{}2000$.
Prove that the sequence $ \left(p_n\right)$ is bounded.

[quote="warut_suk"]p_n is the greatest divisor of the number p_(n-1)+p_(n-2)+2000[/quote]
greatest prime divisor?

__________
blang

I think it is Poland Olympiad 2000, Round 3, Problem 3 - if I am not mistaken :)

I haven't solved it yet. It seems that it is hard...

[quote="Arne"]I think it is Poland Olympiad 2000, Round 3, Problem 3 - if I am not mistaken :)[/quote]

exactly Arne - this comes from 51 Polish Olympiad - here's the link to the orignal source if somebody want it:

http://www.om.edu.pl/zadania/om/om51_3.pdf

Not in this particular document, but the solution is in here:

http://www.om.edu.pl/zadania/om/om51_calosc_r.pdf

on a page 43

First, we will prove that for an arbitrary $n \geq 3$ true is following inequality:

(1)   $p_n \leq max(p_{n-2},p_{n-1})+2002$

Indeed, if one of the numbers $p_{n-2},p_{n-1}$ is equal 2, then by definition, number $p_n$ is prime divisor of number $max(p_{n-2},p_{n-1})+2002$. Inequality (1) is then proved. On the other hand, if both $p_{n-2},p_{n-1}$ are odd, number $p_n$ is the greatest prime divisor of even number $p_{n-2}+p_{n-1}+2000$, hence it is also prime divisor of number:

$\frac{p_{n-2}+p_{n-1}+2000}{2}=\frac{p_{n-2}+p_{n-1}}{2}+1000 \leq max(p_{n-2},p_{n-1})+1000$

So inequality (1) is fulfilled in this case either.
Let $M=max(p_1,p_2) 2003!+2$. We will prove by induction that for arbitrary natural number $n \geq 1$ inequality $p_n < M$ is true.
Obviously $p_1<M$ and $p_2<M$
Suppose, for arbitrary number $n \geq 3$ are true following inequalities: $p_{n-2}<M$ and $p_{n-1}<M$. Then, by (1) we have $p_n < M+2002$. Because numbers $M,M+1,M+2,...,M+2001$ are not primes (divisible by 2,3,4....,2003 respectively), so we are given $p_n < M$. What ends inductional proof and solving of a whole problem.

Remark:
Above solution is made on showing that sequence $(p_n)$ can't "jump over" a large hole in prime numbers and on pointing respectively many non-prime numbers.

P.S. I hope you like it Myth    :lol:  :lol:  :lol:


superior algebra
superior algebra unsolved

I'm new to category theory (this is a simple way of saying "I know nothing but the axioms") and I have a question.

We say that $ P$ is a product of $ (A_i)_{i \in I}$ with a set of morphisms $ (\pi_i : P \longrightarrow A_i)_{i \in I}$ called projections if for all $ X$ and all morphisms $ (g_i : x \longrightarrow A_i)_{i \in I}$ exists only one $ f : X \longrightarrow P$ such that $ \pi_if=g_i$ for all $ i$

The question is:
Suppose that there are some products $ P$ and $ P'$ of ${ (A_i)_{i \in I}}$. Prove that $ P$ is isomorphic to $ P'$

A "sketch": if you can't fill in the details (after trying!) I can be a bit more rigorous :D.

The projections $ \pi : P \to A_i$ give a unique map $ f : P \to P'$. The projections $ \pi' : P' \to A_i$ give a unique map $ g : P' \to P$.  Then there is a map $ gf : P \to P$ commuting with the projections. But then $ gf \equal{} id_P$, since $ id_P : P \to P$ is a map induced by the projections $ \pi : P \to A_i$ into the product $ P$; such a map is unique. Likewise $ fg \equal{} id_{P'}$, so $ f$ and $ g$ are isos.

Well, you're taking the $ P$ to be the product and $ X\equal{}P'$ and then $ P'$ to be the product and $ X\equal{}P$, so you get $ \pi_i \equal{} \pi_i fg$ and $ \pi'_i \equal{} \pi'_igf$, but how do you conclude that $ gf\equal{}id_P$ and $ fg\equal{}id_{P'}$?

Because by the definition of product there is a [i]unique[/i] map $ h : P \to P$ such that $ \pi_i \equal{} \pi_i h$ for every $ i$. Since $ \pi_i \equal{} \pi_i (id_P)$ and $ \pi_i \equal{} \pi_i (gf)$, it follows that $ gf \equal{} id_P$.

Oh, of course. I was wondering the uniqueness would appear.
Just forgot that the composition of 2 unique things is another unique thing :P

Thank you


What is the expected value of the roll of a standard 6-sided die?

Well, think about it this way. If you had flipped a coin where 1 side was marked 1 and the other side marked 2, then the expected probability would be 1.5 because it is the expected average of the flips. I other words, since there is a 50% chance of flipping heads or tails in this example, .5(2)*.5(1)=1.5. So expected value does not necessarily have to be something that you can get by flipping the coin or tossing the die.

Another way to do it is add $1$ and $6$ and then divide by $2$.

[quote=PiesAreSquared]This doesn't make any sense to me, since you can't get 3.5...[/quote]

Yes, but expected value is defined as the average of all of the possible outcomes, to put it simply.


Four workers can build a stage in $ 10$ eight-hour days.  At the same rate, how many workers will it take to build the stage in $ 5$ eight-hour days?

If you take 1/2 the time to do something, you take twice the amount of workers (assuming each does the task at the same speed).  Thus, it is 8.

[hide="The Long Way :P"]It takes $ 4$ workers $ 80$ hours to make a stage.  However, they inversely vary.

$ 80 \cdot 4 \equal{} 40x$

$ x \equal{} \boxed{8}$[/hide]


topology
real analysis
real analysis unsolved

Could some one show me how to prove that one-point compactification of $ \mathbb{Z}_\plus{}\equal{}\{ n\in\mathbb{Z} : n>0\}$ is $ \{ n\in\mathbb{Z} : n>0\}\cup \{0\}\subset \mathbb{R}$? For me it is almost obvious . thanks

The space $ \{n \in \mathbb{Z} : n > 0 \} \cup \{0\}$ is not compact. Did you mean $ \{1/n : n \in \mathbb{Z}^\plus{}\} \cup \{0\}$?

To show that $ Y$ is the one-point compactification of $ X$, you need to show that $ Y \setminus X$ is a single point, that $ X$ is dense in $ Y$, and that $ Y$ is compact.

sorry, yes I mean :  one-point compactification of $ \mathbb{Z}_ \plus{} \equal{} \{ n\in\mathbb{Z} : n > 0\}$ is $ \{\frac {1}{n}\ : n > 0\}\cup \{0\}\subset \mathbb{R}$?

Sorry, I have a question :

if I let $ X\equal{}\{\frac{1}{n} : n\in\mathbb{Z}_\plus{}\}$ then I can prove that one-point compactification of $ X$ is $ Y\equal{} \{1/n : n\in\mathbb{Z}^{\plus{}}\}\cup\{0\}$ then can I say that one-point compactification of $ \mathbb{Z}_{\plus{}}\equal{}\{ n\in\mathbb{Z}: n > 0\}$ is HOMEOMORPHISM to $ Y$ above ? thanks

Well, we only care about things up to homeomorphism usually anyway (often only up to homotopy). You can show that $ X$ is homeomorphic to $ \mathbb{Z}^\plus{}$, so you might as well just work with $ X$.


If ABC is a triangle, c is an angle, R is a circumradius, prove:

Area(ABC) > R*(sin C)^2

This can't possibly be true.

[Suppose it were true, then we would have ab > c for any triangle.]

Please disregard my first post. Unfortunately, I made a very stupid mistake somewhere.


Prove x,y,z>=0 with xy+yz+zx=1 that
1/x+yz  +1/y+xz +1/z+xy>=3
Send me please complete and really nice solution
Thanks!

Let $ x,y,z \geq 0$ such that $ xy\plus{}yz\plus{}zx\equal{}1$ Prove that
  $ \frac{1}{x\plus{}yz}\plus{} \frac{1}{y\plus{}zx}\plus{} \frac{1}{z\plus{}xy} \geq 3$
It is easy!


linear algebra
matrix
algebra
polynomial
arithmetic series
number theory

I learned something new by reading your post. obviously the proof has not come to be yet. I don't think it's very complicated though.

[quote="prototypex24"]Does anyone know the proof for the sum of the first n squares
(i.e. $1^2+2^2+3^2+...+n^2=\frac {n(n+1)(2n+1)} {6}$)[/quote]
Once you know this equation, you can prove it by mathmetical induction.

yea proving it is not that hard, it is easy with induction. but to come up with such a formula without knowing it in advance, that is the hard part. i learned a russian method of finding such a formula if we know the formula for the sum of the first $n$ numbers...

here we go....

$(n+1)^3=n^3+3n^2+3n+1$
$n^3=(n-1)^3+3(n-1)^2+3(n-1)+1$
$\text{and so on...}$
$3^3=2^3+3\cdot 2^2+3\cdot 2+1$
$2^3=1^3+3\cdot 1^2+3\cdot 1+1$

add of the equations (there are $n$ of them) and we get
\[(n+1)^3=1^3+3(1^2+2^2+3^2+\cdots+(n-1)^2+n^2)+3(1+2+3+\cdots+(n-1)+n)+n\cdot 1\]
keep the sum of squares to one side because that is what we are looking for...
\[1^2+2^2+3^2+\cdots+(n-1)^2+n^2=\frac13((n+1)^3-1^3-3(1+2+3+\cdots+(n-1)+n)-n\]
but we already know the sum of the first $n$ numbers, which is $\frac{n(n+1)}2$, and so...
\[1^2+2^2+3^2+\cdots+(n-1)^2+n^2=\frac13((n+1)^3-1^3-3\cdot\frac{n(n+1)}2-n\]
\[1^2+2^2+3^2+\cdots+(n-1)^2+n^2=\frac{2n^3+6n^2+6n-3n^2-3n-2n}6\]
\[1^2+2^2+3^2+\cdots+(n-1)^2+n^2=\frac{2n^3+3n^2+n}6=\frac{n(n+1)(2n+1)}6\]

hmm i am also stummped by this proving of standard formulaes.

u see sum of n terms is just the arithmetic series, n/2(n+1)

then, the proving of sum of n^3 is

1^3 = 1                               = 1^2                    = (1)^2
1^3 + 2^3 = 9                    = 3^2                     = (1+2)^2
1^3 + 2^3 + 3+3 = 36        = 6^2                    = (1+2+3)^2

and so on.. so therefore, the sum of n^3 is [AP formula]^2 => [(n/2)(n+1)]^2

since there are proofs for n and n^3 standard formulaes, what is the proof of n^2?

[edit] ->  i mean proof as in where it originated, not by induction. thanks for time!

[/code]

Chrnox, did you read the previous posts before replying? there are 2 proofs already!!!

Just what I wanted. Thanks Palytoxin and chess64!

You can actually motivate the proof in several ways.

AKA: Assume we don't know the formula. Find it.

You can do it in the aformetioned ways.

Another way of looking at it:

Squares are the sums of consecutive odd numbers. Find the sum of squares as the sum of the sum of odd numbers.. This is one occasion where you may see a sum of a sum.

Sorry corrected to odd numbers - I had too much number theory that day...

Hmm,
4 cannot be expressed as the sum of consecutive primes, can it?

can 9 and 16 be expressed as the sum of consecutive primes?

conjecture: prove that every square cannot be expressed as the sum of consecutive primes

hmm, nice conjecture, i should post this in number theory section :D  :pilot:

edit: hmm probably my conjecture is going to be false thou but hehe what a nice attempt

new challenge: if you believe it is false, find the smallest square expressible as the sum of consecutive prime :!:

There was a math jam (a while ago) about how you can use $\sum$ notation to derive the sum of the first n kth powers. However, if you have a calculator with you, there is a much quicker way (it would be extremely tedious deriving the sum of the first n 5th powers using sigma notation). It is obvious that the sum of the first n natural numbers is of degree two. Similarly, you can convince yourself that the sum of the first n kth powers is of degree k+1. This means that it can be written in the form $An^{k+1}+Bn^k+Cn^{k-1}+................$ To find the coefficients, simply plug in the required number of values and use rref (in matrix, math) to solve the system.

the coefficients are called Bell numbers but i think it's too difficult for me right now, not an easy topic (i found out this from a book called "Concrete Mathematics" by D.E. Knuth, the founder of $\text{\TeX}$

[quote="Palytoxin"]i have to say that the sum of fifth powers is wrong, if you plug in $n=1$, you will say "what the heck?"  :D

and it's strange that you got the sum of sixth powers correct...  :huh:[/quote]

sorry... the $-5n^2$ should've been $+5n^2$
let me edit that :)

Psycho, that's even better because you know, -1 is a root of that polynomial factor and so... you know what i mean :D

[hide="huh? what am i talking about? click me to find out"]factor it :P[/hide]

well yeah :)
i always thought it was strange that $(n+1)$ could be factored out everywhere, except with the 5th powers  :?

now it'll turn into $1^5+2^5+3^5+...+n^5=\frac{n^2(n+1)^2(2n^2+2n-1)}{12}$

(let me edit that :P )

theres a nice visual proof in the visual proofs gallery


Let $ a,$ $ b$ and $ c$ are positive numbers such that $ abc\equal{}1.$ Prove that
\[ \frac{1}{a^2\plus{}b\plus{}c}\plus{}\frac{1}{b^2\plus{}a\plus{}c}\plus{}\frac{1}{c^2\plus{}a\plus{}b}\leq1\]

The above inequality is still true for $ a\plus{}b\plus{}c \ge 3$.
Cauchy-Schwarz inequality shows that
$ \sum \frac{1}{a^2\plus{}b\plus{}c} \le \sum \frac{1\plus{}b\plus{}c}{(a\plus{}b\plus{}c)^2} \equal{}\frac{3\plus{}2(a\plus{}b\plus{}c)}{(a\plus{}b\plus{}c)^2} \le 1$

[quote="arqady"][quote="ehku"]The above inequality is still true for $ a \plus{} b \plus{} c \ge 3$.
Cauchy-Schwarz inequality shows that
$ \sum \frac {1}{a^2 \plus{} b \plus{} c} \le \sum \frac {1 \plus{} b \plus{} c}{(a \plus{} b \plus{} c)^2}$[/quote]
I don't see it. Explain pleas. Thank you.[/quote]
$ (a^2\plus{}b\plus{}c)(1\plus{}b\plus{}c) \ge (a\plus{}b\plus{}c)^2$

[quote="ehku"][quote="arqady"][quote="ehku"]The above inequality is still true for $ a \plus{} b \plus{} c \ge 3$.
Cauchy-Schwarz inequality shows that
$ \sum \frac {1}{a^2 \plus{} b \plus{} c} \le \sum \frac {1 \plus{} b \plus{} c}{(a \plus{} b \plus{} c)^2}$[/quote]
I don't see it. Explain pleas. Thank you.[/quote]
$ (a^2 \plus{} b \plus{} c)(1 \plus{} b \plus{} c) \ge (a \plus{} b \plus{} c)^2$[/quote]
Wow! Thank you very much!

[quote="arqady"]The following stronger inequality is true too:
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a \plus{} b \plus{} c\geq3.$ Prove that
\[ \frac {1}{a^2 \minus{} a \plus{} 3} \plus{} \frac {1}{b^2 \minus{} b \plus{} 3} \plus{} \frac {1}{c^2 \minus{} c \plus{} 3}\leq1
\]
[/quote]
Of course.
If $ a,b,c \le 3$, using the inequality $ \frac {1}{x^2 \minus{} x \plus{} 3} \le \frac{4\minus{}x}{9} \ \forall x \le 3$ we have
\[ \sum \frac {1}{a^2 \minus{} a \plus{} 3} \le \sum \frac{4\minus{}a}{9} \leq 1
\]
If $ a > 3$, then
\[ \frac {1}{a^2 \minus{} a \plus{} 3} \plus{} \frac {1}{b^2 \minus{} b \plus{} 3} \plus{} \frac {1}{c^2 \minus{} c \plus{} 3} < \frac{1}{9} \plus{} \frac{4}{11}\plus{}\frac{4}{11} < 1
\]


[quote="anirudh"]A landowner in Missouri employed 10 laborers to plant a total of 10,000 seedlings. If
each laborer planted one seedling every 15 seconds, in how many minutes were all
10,000 seedlings planted?[/quote]

[hide]
$\frac{10000}{60/15) = 2500$
2500/10 = 250?
[/hide]

[quote="ccy"][quote="anirudh"]A landowner in Missouri employed 10 laborers to plant a total of 10,000 seedlings. If
each laborer planted one seedling every 15 seconds, in how many minutes were all
10,000 seedlings planted?[/quote]

[hide]
$\frac{10000}{\frac{60}{15}} = 2500$
2500/10 = 250?
[/hide][/quote]

[hide]That means each laborer has to plant 1000 seedling. So it takes 15000 seconds or [color=indigo]250 minutes[/color].
[/hide]

[quote="math92"][quote="anirudh"]ummm/ tjere s a [rpn;e, with your $\LaTeX$[/quote]
your writing is messed up[/quote]

His hand was shifted over one place (look on your keyboard). It was supposed to be
"ummm, there's a problem with your Latex."


hi

apparently phi (n ) +sigma (n ) is greater than or equal to 2*n

equality occurs only when n=1 or n is prime

can anyone prove this? or is this a difficult theorem?
are there any upper bounds known?

thx

Actually, it is very easy. All you need is that if $p_1,...,p_k$ are the prime factors of $n$, then $\sigma(n)\geq (1+\frac{1}{p_1})...(1+\frac{1}{p_k})n$. Then just divide by $n$ and use Bernoulli inequality. For upper bounds, consult any decent book in number theory and you will find bounds for $\sigma(n)$.


integration
logarithms
inequalities
calculus
IMC
college contests

Prove that
\[ \int^1_0 \int^1_0 \frac { dx \ dy }{ \frac 1x + |\log y| -1 } \leq 1 . \]

It's sweet! :)

We write it as $\displaystyle \int^1_0\int^1_0\frac 1{\frac 1x+\ln \frac 1y-1}dxdy$. We now change both variables to get $\displaystyle \int^{\infty}_1\int^{\infty}_1\frac 1{x^2y^2(x+\ln y-1)}dxdy$. By using the well-known inequality $\ln y\ge \frac{y-1}y$, this is less than or equal to $\displaystyle \int^{\infty}_1\int^{\infty}_1 \frac 1{x^2y(xy-1)}dxdy$, and guess what this last integral is :). If you guessed $1$ you were right (it's trivial to compute, so I won't do it here).


Seven math teachers decided to celebrate National Secretary's Day by buying a gift for their secretary for each day of the work week. Mark spent $ \$3.50$ on Monday, Tess spent $ \$3.25$ on Tuesday, Will spent $ \$4.25$ on Wednesday, Time spent  $ \$5.25$ on Thursday, and Felicia spent  $ \$4.75$ on Friday. Sam and Sally each agreed to pay one-seventh of the total cost. In dollars, how much does Sam owe?


calculus
integration
real analysis
limit
advanced fields
advanced fields unsolved

answer true or false and justify?

1) f belongs to L'(R) implies lim f(x)=0 as n goes to infinity?

2) lim f(x)=infinty as x goes to infinity implies f does not belong L'(R)?

i think both are wrong? is that true

(1) False. Consider $f=\chi_U$, where $\displaystyle U=\bigcup_{n=1}^{\infty} [2n-1, 2n-1+\frac{1}{2^{n-1}}]$

(2) True. Suppose $f\in L^1(\mathbb{R})$. Then so is $|f|$ by definition of Lebesgue integrability. WLOG we may assume $f$ is nonnegative. By Monotone Convergence Theorem, $\lim_{n\to\infty} \int f\chi_{[-n, n]}=\int f$. That in particular means $\lim_{n\to\infty} \int_n^{\infty} f=0$. But by hypothesis, we have that given $M>0$, there exists $N\in\mathbb{N}$ such that $f(x)>M$ for all $x>N$. Thus $\int_n^{\infty} f\geq \int_n^{n+1} f\geq M$ for all $n>N$, which is a contradiction.


number theory
greatest common divisor
function
relatively prime

Prove that for all $ n$, there exist $ a,b$ such that $ \gcd (a \plus{} u,b \plus{} v)\neq 1$ for all $ 1\leq u,v\leq n$.

[hide="This seems really obvious"]
consider that if $ a\plus{}u$ and $ b\plus{}v$ are even, then their gcd would be $ 2$ or greater, so for whatever $ u$ or $ v$, designate the values of $ a$ and $ b$ accordingly
[/hide]

Well, if $ b \plus{} v$ is even then $ b \plus{} v \plus{} 1$ is odd.

[hide="Solution"]Take any $ n^{2}$ distinct primes and put them in an $ n\times n$ array.  Then we choose $ a$ and $ b$ such that all the primes in the first row divide $ a\plus{}1$, all the primes in the second row divide $ a\plus{}2$, etc., and all the primes in the first column divide $ b\plus{}1$, all the primes in the second column divide $ b\plus{}2$, etc.  Finding $ a$ here is essentially solving a system of linear congruences in relatively prime mods (since all our primes are distinct), which has a solution by CRT.  The same goes for finding $ b$.

Then $ a\plus{}u$ and $ b\plus{}v$ are both divisible by the prime in the $ u$th row and $ v$th column of our array, so their GCD is not 1.[/hide]


limit
logarithms
real analysis
real analysis unsolved

A friend of mine proposed the following problem to me, it is a good one:

Let $(x_{n})$ be a sequence of positive real numbers such that $x_{n+m}\leq x_{m}x_{n}$. Show that $\lim_{n\to \infty}(x_{n})^\frac1n$ exists.

... or $\frac1ny_{n}$ diverges to $-\infty$.

In the original problem, a zero limit is possible.

[quote="Kent Merryfield"]I'd prefer to take logarithms and work this equivalent problem:

Suppose $(y_{n})$ is a sequence of real numbers such that $y_{n+m}\le y_{n}+y_{m}.$ Prove that $\lim \frac1ny_{n}$ exists.[/quote]

Nice :)

$y_{n}=\log x_{n}\Rightarrow x_{n}= e^{y_{n}}$ so that

$\lim_{n\rightarrow\infty}(x_{n})^{\frac{1}{n}}= \lim_{n\rightarrow\infty}\left( e^{y_{n}}\right)^{\frac{1}{n}}= \lim_{n\rightarrow\infty}e^{\frac{y_{n}}{n}}= \lim_{n\rightarrow\infty}e^{\lim_{n\rightarrow\infty}\frac{y_{n}}{n}}$

where the last step holds by continuity.

Hard.But the "sup lim-inf lim" method applies.

[quote="Kent Merryfield"]I'd prefer to take logarithms and work this equivalent problem:

Suppose $(y_{n})$ is a sequence of real numbers such that $y_{n+m}\le y_{n}+y_{m}.$ Prove that $\lim \frac1ny_{n}$ exists.[/quote]

Well, this gives $\frac{y_{n}}{n}$ an upper bound: since $y_{n+m}\leq y_{n}+y_{m},$ we have $y_{n}\leq y_{n-1}+y_{1}\leq y_{n-2}+2y_{1}\leq\cdots\leq ny_{1}$ hence $\frac{y_{n}}{n}\leq y_{1}$.

Why go through so much trouble? $x_{n}\leq{x_{1}}^{n}$.. so $\lim x_{n}^{1/n}\leq x_{1}$

[quote="ffrogg"]In point of fact,the limit of $\frac{y_{n}}{n}$ is its lower bound .We can employ the "sup lim-inf lim" method to prove the result.Do you agree?[/quote]

Right. I'd like to complete it here.
Suppose $\liminf y_{n}/n = a$, then for any $\epsilon > 0$, there exists $N$ such that $y_{N}/N < a+\epsilon$.

Each integer can be written in the form of $qN+r$, hence for $n>N$,
$y_{n}= y_{qN+r}\leq y_{N}+y_{(q-1)N+r}\leq \cdots \leq qy_{N}+y_{r}\leq qy_{N}+ry_{1}\leq qy_{N}+Ny_{1}$.
Hence \[\frac{y_{n}}{n}\leq \frac{qy_{N}}{n}+\frac{Ny_{1}}{n}\leq \frac{y_{N}}{N}+\frac{Ny_{1}}{n}\leq a+\epsilon+\frac{Ny_{1}}{n}\] whence \[\limsup \frac{y_{n}}{n}\leq a+\epsilon.\] From the arbitrariness of $\epsilon$, we have \[\limsup \frac{y_{n}}{n}\leq a = \liminf \frac{y_{n}}{n}\] Q.E.D


A non-zero fraction's denominator is increased by 1 and its numerator is decreased by $ 50\%$.  If the resulting fraction is $ \frac{1}{3}$ of the original fraction, what is the original fraction's denominator?

$ \frac{1}{3}\cdot\frac{x}{y}\equal{}\frac{1}{2}\cdot\frac{x}{y\plus{}1}\Rightarrow \frac{1}{3y}\equal{}\frac{1}{2(y\plus{}1)} \Rightarrow 3y\equal{}2(y\plus{}1) \Rightarrow y\equal{}2$


Thanks - that's a lot clearer!

I hope I have understood the modulus steps correctly:

f(2^2006) = 7 mod 9 since 2006 = 6*668 + 2 and for all n = 3k + 2, f(2^n) = 7 mod 9.   (The powers of two cycle though 4, 7, 1, 4, 7, 1...)

Then f(2^n) + 2 is a multiple of 9 hence the sum of its digits is divisible by 9.  So S(f(2^n)) = 9j - 2  f(f(2^n)) = 81j^2 - 18j + 4 = 4 mod9.

Did you pick mod9 specifically because the question was about the sum of digits and because a multiple of 9 always has a digit sum divisible by 9?
What other moduli have useful features in these large exponent problems?
Thanks again.

The post above is supposed to follow the square of sum of digits post.
Apologies.


probability AMC AIME Alcumus USA(J)MO USAMO geometry

Hey,

For all of you guys who have ACoPS, how well does it treat counting and probability compared to Interm. C&P?
Which one goes more in depth? Which one did you learn more from? Finally, which one is most applicable to tough AIME counting problems (those are the ones that make me only get 6 or 7 AIME).

The reason I'm asking this is because I am just finishing AoPS 2, and have realized that the counting section spends at most 1.5 pages on each topic.

Would one recommend buying Interm. C&P, reviewing Intro (I have that one), or buying ACoPS?

Thanks a lot guys!

I used Alcumus and Intro to C&P, and I went from being barely able to solve a medium AMC 12 counting problem to being able to solve most AIME counting problems, and even one or two usamo questions.  Intermediate C&P, though I don't have it, should go more in depth for AIME level problems (because the whole book is about C&P, and ACoPS is about many topics), but I'm not sure.  Intro to C&P has many of the tools needed to solve tough AIME problems though, especially the last chapter, which focuses on tough problems.

What's ACoPS?

Is that like an Advanced series: like Advanced Algebra, Advanced Geometry, Advanced Number Theory, Advanced Counting and Probability?

Is there only one ACoPS book available right now?

ACoPS=art and craft of problem solving, by paul zeitz.  It's sorta like "volume 3", if there was one

So then is that mostly good for olympiad level contests like Putnam, USAMO, and USAMTS?

Not really.  Up to hard AIME, easy USAMTS/USAMO I think.

Then what books would be good for advanced USAMTS?

i has mostly hard AIME usamo problems in it. I have the book.

Where can you order ACoPS? Will it ever be on the AoPS Bookstore?

I recomend Intermediate C+P for a few reasons
1) i own both books. Ive read throught the combinatorics chapter of ACoPS, and it covers about as much stuff as half of C+P.
2) C+P goes into more depth usually.
3) ACoPS does not have solutions with it(appearently, there is one out there, but its very hard to find one)
4) Since you are only trying to get higher AIME scores, C+P is better. ACoPS has olympiad/proof problems which wont help you much. When you get to that level, you might as well buy PSS by Engel.

#3: Yes, there is a solution. But it's restricted to teachers and very very few students. I can trust that it will not be revealed under any circumstances by anyone. So ACoPS essentially is a book without solutions.

[quote="AIME15"]Just wondering: Why aren't the solutions available to students!?!?!?[/quote]

The textbook is sometimes used as a part of an official college course. So in order to prevent cheating, the author decided to release the solution to only few selected people.

sorta depends how you are doing on the AIME. If you are getting 6-8s and think you just need more practice, don't buy ACoPs, but for me, I need to get done with Volume 2 first. And, gauss1181, USAMTS problems are about AIME level (maybe just a wee bit harder) so you can do volume 2 (that's about what you need for USAMTS) and AIME problems (it would be good if you just do the whole test, but sometimes you may want to skip the first 1 or 2. )

[quote="#H34N1"][quote="abacadaea"]When you get to that level, you might as well buy PSS by Engel.[/quote]

It is highly recommended to do ACoPS before Engel. Doing ACoPS first helped me a LOT.

As for solutions, look around in the Books forum. I think Phelepdo released his own solution manual.
And the official solution manual doesn't even contain solutions to most of the problems! It has completed solutions for the ones for which there are hints in the hints document.[/quote]

When i was buying a book, i was deciding between ACoPS and PSS. When i read PSS excerpts on books.google.com, i was kinda intimidated, so i bought ACoPS. Like, 2 weeks after my ACoPS, i wish i had bought engel, so yeah.

he only got a 17 or something on USAMTS so I doubt it. ACoPS as Putnam Problems in it. Also the new edition has geom.

A question so I don't have to start a new topic. I'm taking the combinatorics and geometry sections this year at the Princeton competition at the end of January. I'm also doing AMC 12 in February. I'll probably be able to finish one book before Princeton and another before the AMC. Which two should I do? I already finished AOPS I and I have AOPS II now. Should I get Intro to Counting and Probability instead and do AOPS II before the AMC/AIME?

AoPs II takes about 6 months. You may finish I to C&P before princeton if you don't do all the problems. That probably won't prep you fully for Div A combo though. Im about to finish V2, and thinking about only starting another book after I make USAMO.

[quote="pythag011"][quote="Poincare"]Not really Volume 3 because it sorta goes over the same stuff as AoPs Volume 2 at times but has problems that harder.[/quote]

The problems are the whole point of ACoPs, in my opinion. The problems are rather helpful....

C+P is more deep, though.

Engel is good, but it focuses more on non-enumerative combinatorics, in my opinion (Like coloring, invariants, not counting how many ways.)

Intermediate C+P should be enough for AIME, but you might want to buy Enumerative Combinatorics by Richard Stanley. http://math.mit.edu/~rstan/ec/[/quote]

For anyone who might buy Enumerative Combinatorics be warned that its a graduate level introduction to the subject. Don't expect to be able to complete the entire book during high school (that's my experience at least, although I still have time to complete it).


[quote="fjwxcsl"]Assume that
\[ S_{k}(n) = \sum_{i = 1}^{n}{{x_{i}}^{k}}( x_{i}\ge{0}).\]
$ A.$Let $ S_5(3) = 3,$ prove that  $ S_{10}(3)\ge{S_3(3);}$

$ B.$Let $ S_5(4) = 4,$ prove that  $ S_{10}(4)\ge{S_3(4);}$

$ C.$Let $ S_5(n) = n,$ prove (or disprove) that  $ S_{10}(n)\ge{S_3(n).}$[/quote]

I'll prove the generalization :

We use AM-GM's Inequality as follows :

$ x_i^{10} \geq 2x_i^5 - 1$ and $ x_i^3 \leq \frac {3x_i^5 + 2}{5}$

It follows that $ S_{10}(n) \geq n$ and $ {S_3(n) \leq n}$,the conclusion follows immediately.


The points $ (1,1)$, $ (1,3)$, $ (3,3)$, $ (3,1)$ are the midpoints of a square $ S$. How many square units are in the area of $ S$?

calculate area of this square formed by midpoints, then multiply by 2 for big square(square in square thingy)
so you know side = 2
so area = $ 2^2\equal{}4$
now $ 4*2\equal{}8$
so area of big square is $ 8$


It is maybe a silly question, but at the moment I dont know if it is true or false?!

Let $ f: U\to V$ is a $ 1 : 1$ and onto function ($ U,V$ are open subset of $ R^n$). Given that $ f,f^{ \minus{} 1}$ are continuous, is this true that there exists one constant $ K > 0$ such that
\[ \parallel{}f(x) \minus{} f(y)\parallel{}\ge K\parallel{}x \minus{} y\parallel{}
\]
for every $ x,y\in U$.

Thanks for your help!

Certainly not. Just take $ n\equal{}1$, $ U\equal{}V\equal{}(\minus{}1,1)$, $ f(x)\equal{}x^3$.

Sorry :( I am thinking at a very complicated proof but forgetting the easiest case,


floor function
logarithms
trigonometry
geometry
trig identities
Law of Sines

1.  A circular track has a diameter of 2 miles.  A man attempts to run 10 laps with a constant speed of 10 miles per hour.  Rounded to the nearest foot, how far will the man be from the starting line 30 minutes after he has begun?
[hide]My answer: 6775.  I don't see why the answer should be 6320.[/hide]

2.  Let k=1...1, where k is 10001 digits long and in base 2.  Find $\lfloor \log_5{k} \rfloor$.

Hmm... i got 6775ft also for number 1... but I think the question may be referring to the distance between the man and the starting point in a staight line...

yes, thanks, I calculated the straight line distance and it is approximately 6320  :)

ah ok i got it  :D

[hide] knowing that the circumference of the circle is 2(pi) and that in half an hour the man covers 5 miles, one can establish the following:

2(pi) = 360
5      =  x

By cross-multiplying, one sees that x is equal to 286.4788976 degrees... but to use this one can subtract that value from 360 to obtain 73.52110243 degrees. As you can see this angle is the interior angle of the circle that forms an arc with enpoints on the starting point of where the man began and where the man ended at after 30 min of running. Drawing two points from the center of the circle to both endpoints of the arc and by drawing a strainght line that connects those two points one can see that you have an isosceles triangle with two sides being a length of 1.

Then by subtracting 73.etc from 180 and dividing by 2 one obtains the the measure of the other two angles of the triangle (53.23944879 degrees). Then by using the law of sines one gets the following:

sin(73.etc)     sin(53.etc)
_________= _________

   x                          1

By cross multiplying and solving for x gives you 1.196944288 miles
Multiplying by 5280 gives you 6320 ft when rounded. :D [/hide]

Heh its supposed to be sin(53.etc) over 1... iuno why it did that lol

So if 10001 is also in base two, then 10001 is 17, and our number would be 2^17 - 1, which is not even divisible by 5, but I suppose you can use a calculator.

Yeah, for #2, you just need to find some $k$ such that $5^k$ is the greatest power of $5$ less than or equal to $2^{10001}-1$ or $2^{17}-1$, depending on how you interpret it. I'm pretty sure you would have a calculator during this test.

For two... k is 10001 DIGITS LONG, not 10001.  Since all the digits are 1s, $k = 2^{10002} - 1$.  This is an insanely large number, and I'm not sure how I'd go about finding the largest integer $x$ such that $5^x < k$.  And I don't think a calculator would be too helpful... the highest power of two my calculator can do is like 300 or something.  Maybe TI-89 or TI-92s can do higher, but those are almost never allowed on any tests.

Ohh, I didn't see the floor sign.  And btw it should be 2^10001 and not 2^10002.  I don't know what to do from there though.

\[\log_5k=\log_5(2^{10001}-1)\]

From here we can see that the one makes almost no difference, thus:

\[\log_5(2^{10001}-1) \approx \log_5(2^{10001})\]
\[=10001 \cdot \log_5 2 \approx 10001 \cdot .43068 \approx \fbox{43,068}\]

No, it's 10002.  Consider a base 2 number $1111$.  There are 4 1s.  In base 10, it is $8+4+2+1 = 15 = 2^5 - 1$.  Therefore a  base two number with only 1s and has $n$ digits has a base ten value of $2^{n+1} - 1$.
Since k has all 1s and has 10001 digits, $k = 2^{10002} - 1$.

By the way, nice solution DPopov... not sure how I missed one of the most basic log properties.  However, when I plugged it in my calculator, my answer was $4307$.

thanks for the responses :)

nice solution DPopov, only $\displaystyle 10001 \cdot .43068 \approx \fbox{4306.8}$, not $43,068$ :D


A cube has volume 624 cubic cm. Consider the following tetrahedron whose vertices are one vertex of the cube and the midpoints of the three edges of the cube that are connected to that vertex. Find the number of cubic cm in the volume of this tetrahedron.

[hide="Answer"]13[/hide]

Hello!:)
SORRY- I DON'T HAVE LATEX
[hide]You have to know that the volume of the tetrahedron is
[color=red]BASE[/color][1/3][color=blue]HEIGHT[/color]
If the volume of hte cube is 624 then the side length of the cube is (624^1/3), since the 3 vertices of the tetrahedron is on the midpoint of each side fo the cube, the side lengths of the tetrahedron are (78^1/3). the area of the base is (78^2/3)/2 and the height is 78^1/3
[color=red][(78^2/3)/2][/color][1/3][color=blue][78^1/3][/color]
=(78/2)/3
=13[/hide]

Or, there are 48 little tetrahedrons in the cube, so the answer is $ \frac{624}{48}\equal{}13$.

[quote="alanchou"]Or, there are 48 little tetrahedrons in the cube, so the answer is $ \frac {624}{48} \equal{} 13$.[/quote]
wait, how do you know there are 48 tetrahedrons in the cube?

[quote="alanchou"]Don't ask...

Consider what the sides of the cube look like if you draw in the edges of the tetrahedron that go along the faces of the cube.[/quote]

So is this something that I should just kind of memorize...?

Actually, I think andrewliu's method works better.  It takes a while to find that there are 48 little tetrahedrons.  Actually, I'm assuming that the cube can be dissected into those tetrahedrons, which I'm not sure is true or not.


A sequence $a_{n}$ is defined by $a_{1}=1$ , $a_{2}=12$ and $a_{3}=20$ and $a_{n+3}=2a_{n+2}+2a_{n+1}-a_{n}$  with $n\geq 0$. Prove that for any n the number $1+4a_{n}a_{n+1}$  is a perfect square

First of all,

[quote="silouan"]Noone has a solution ??[/quote]

The correct form is "no one" (unlike "anyone" for example). I don't think that a word a like "noone" exists (although "noon = middle of the day" exists). "none" also exists.

Second of all, by trial-and-error, $1+4 a_{n}a_{n+1}= \left( a_{n+1}+a_{n}-a_{n-1}\right)^{2}, \, \forall n \geq 2$. I'm too lazy to prove my statement :( Induction might work.

EDIT: Just plug $a_{n}= 3 (-1)^{n}+\frac{19+9 \sqrt 5}{15+7 \sqrt 5}\cdot \left( \frac{3+\sqrt 5}2 \right)^{n}+\frac{1+3 \sqrt 5}{2 \sqrt 5}\cdot \left( \frac{3-\sqrt 5}2 \right)^{n}$ into the identity we want to prove (I'd like to thank Mathematica for solving a system with 3 variables, composed of three equations).


let a,b,c is three side of a triangle
prove $(a^{2}+b^{2}+c^{2})^{2}\geq 3(a^{3}b+b^{3}c+c^{3}a)$

It's in my book with four solutions, but VASC creates it. You're also wrong when remenberring that $a,b,c$ are three side-lengths of a triangle, because it's not necessary.

Yes! I'm wrong ! $a,b,c$ are three side-lengths of a triangle.Thankyou very much   :)  :blush:


This isn't dire information or anything, but what is the exact relationship between artofproblemsolving.com and mathlinks.ro?

The best I can tell is that it is more or less of a mirror to art of problem solving, meaning that any action on these forums are copied over. However, posts from mathlinks.ro are [i]not[/i] copied back to AoPS (so best I can tell). Other than the different appearances and portals, is there any other difference between the two sites?

Also, it seems to me for one reason or another than Valentin is more involved than the other administrators at mathlinks, (partially because of the ".ro" part :)), but it is difficult to tell since all the actions are copied over to mathlinks.

Just curios to know. Thanks.

They both merged in like 2004. before that, they used to be different Websites.

Nothing is copied. It is the same site, with a different skin / url. Since all the people before the merge have gotten used with the old url, we did not change it. Your observations are not real :)

And yes I am more involved partly because I created MathLinks and partly because I manage/write almost all the website's php/html/css code.

If this is the case, how come I saw some topics on mathlinks but not on artofproblemsolving. The particular topic I am talking about is found here [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=117369[/url], but is not in the corresponding spot on AoPS.

EDIT: Nevermind, I found I was one sub-forum too high, which was why I could not find it. Sorry  :) .


A positive integer n is called perfect number if the sum of all its positive divisors,excluding n itself,equals n.Let  p be a prime number and
   $ M=2^{p-1}(2^p-1)+1,N=2^{p-1}(2^p-1)-1$
Prove that M and N aren't perfect number

Why don't you post your solution? Anyway the problem is interesting!

Ok,Sailor. I will post my solution if I complete this problem.For N, it's is not hard and I solved it but M... :( (I was stupid these days ago :blush: and I hope today is nice day and I wil complete it).

You shouldn't be so hard with yourself, it's just a math problem ...

Use if $ n \equal{} 4k \minus{} 1, \sigma(n) \equiv 0 (mod 4)$
And same way for $ n\equal{}3k\minus{}1, \sigma(n) \equiv 0 (mod 3)$

It is easy to prove it :lol:


function
superior algebra
superior algebra unsolved

1) Why doesn't the proof of Cayley's theorem show that every cancellation semigroup can be imbedded in a group ?

2) Let $ G$ be a finite group and $ x_1,x_2,...,x_p$ ($ p$ is a prime) be distinct elements of $ G$ s.t.  $ x_1x_2\equal{}x_2x_3\equal{}...\equal{}x_px_1$. Prove that $ 2p$ divides $ |G|$.

a) the proof only shows that every cancellative semigroup M can be embedded into {injective functions M -> M}. in general, M cannot be embedded into a group. look at
http://www.digizeitschriften.de/index.php?id=loader&tx_jkDigiTools_pi1[IDDOC]=364852
b) probably some group action and the orbit formula will help ;-).

sorry , the first question is  nonsens  :oops: , but what about  the second one , it's intresting, isn't it ?


calculus
integration
Ring Theory
superior algebra
superior algebra solved

show in any commutative ring with identity any maximal ideal is prime.

Let $R$ be our commutative unitary ring.
Let $\mathcal{M}$ be a maximal ideal of $R$.
The quotient ring $R/\mathcal{M}$ is a field so an integral ring, so $\mathcal{M}$ is prime.


AMC
AIME
search
function
modular arithmetic

Ninety-four bricks, each measuring $4''\times10''\times19'',$ are to stacked one on top of another to form a tower 94 bricks tall.  Each brick can be oriented so it contribues $4''$ or $10''$ or $19''$ to the total height of the tower.  How many differnt tower heights can be achieved using all 94 of the bricks?

[hide="Answer"]Let a represent the number of blocks with 4'' vertical
b represent the number of blocks with 10'' vertical
c represent the number of blocks with 19'' vertical
So $a+b+c=94$ with $a, b, c \in Z$ and $a, b, c \ge 0$
Plugging in $a=0$ there are 95 combinations for $b+c=94$ from $b= 0 \to 94$.
Plugging in $a=1$ there are 94 combinations for $b+c=93$ from $b= 0 \to 93$.
Thiss will continue to count down to $a=94$ with 1 combination, $b+c=0$.
So the number of combinations is $\displaystyle \sum^{95}_{i=1} i$ which by Gaussian pairing yields $\frac{95(96)}{2}$ which gives $\boxed{4560}$ combinations, or tower heights.[/hide]

Hokkage, AIME answers are integers from 0-999 inclusive  :D

The problem with your argument is you ignore different arrangements of $a, b, c$ that yield the same height.

I thought it seemed a little easy. Ok, well, ignore my post yet again. Maybe I should sit back and watch for a while eh? I've been screwing up on problem after problem. Maybe it could branch off of my method and make a way to tell exclusions for similar heights?

hmm... since no one gave the correct solution, here it is :)

[hide="solution"]
Let a be the number of bricks contributing 4', b be the number of bricks contributing 10', and c be the number of bricks contributing 19'.
we have the following equations:
$a+b+c=94$
$4a+10b+19c=N$ (we want to find the number of possible values of $n$)
notice: $4a+10b+19c=4(a+b+c)+6b+15c=376+3(2b+5c)=N$
so now we have to find the number of values $2b+5c$ takes considering that : $0\le b+c\le 94$
notice that the maximum will be $470$ and min would be $0$, but this doesn't mean we have 471 hights because we cant hit some of the  heights (i.e for example we cant have  height of 1 or 3)
so now start from 94 5's (470), then replace 5's with either nothing or with 2's to get smaller numbers.
If we replace m 5's with nothing and n with 2's our new total is 470 - 5m - 3n. Now we just count how many values 5m+3n cant hit, which will correspond to a high end value we can't reach.
according to The coin exchange problem of Froebenius
 (also called the Chicken Mc nugget thm) the numbers we can't hit are: 1,2,4,7
so this means In addition to not being able to hit 1 or 3 with 2b + 5c, we also can't hit 469, 468, 466, or 463. which imples that the number of possible values for $N$ is $471-6=\boxed{465}$[/hide]

what is the chicken mcnugget theorem?  :huh:

You might want to try the search function :)

Hm well I still don't get the argument from the "other side" which amirhtlusa and white_horse_king reffered to. If you have 81 bricks contributing 4", 12 bricks contributing 10", and 1 contributing 19", the total height will be 463. I can also get 466 with (82,10,2). It shouldn't be too hard to find values for the other ones as well... :?

The way I solved it, we are ignoring the $x$ value and focusing on the values that $2y + 5z$ cannot attain in the equation I derived. Of course, that value of $463$ can be attained in reality, but in our method, $463$ refers to something different.

Oh, right, sorry. But still, why do you need to do it from the other side as well?

[hide] There's nothing wrong with translating all the heights by $ \minus{}4$, so we obtain blocks with height $ 0, 6$, and $ 15$. Using $ 94$ blocks, we want to know all possible heights we can reach. This is equivalent to finding all heights we can reach using blocks of heights $ 6$ and $ 15$, since we can use $ 0$ height blocks for the rest. Obviously, we can only obtain heights divisible by $ 3$. Now we consider cases.

Height = $ 15k$. We simply use $ k$ blocks of height $ 15$. $ k$ can range from $ 0$ to $ 94$, giving $ 95$ possibilities.

Height = $ 15k\plus{}3$. We use $ 3$ blocks of height $ 6$ and then $ k\minus{}1$ blocks of height $ 15$. $ k$ can range from $ 1$ to $ 92$, giving $ 92$ possibilities.

Height = $ 15k \plus{} 6$. We use $ k$ blocks of height $ 15$, then $ 1$ block of height $ 6$. $ k$ can range from $ 0$ to $ 93$, for $ 94$ possibilities.

Height = $ 15k \plus{} 9$. We use $ k\minus{}1$ blocks of height $ 15$, then $ 4$ blocks of height $ 6$. $ k$ can range from $ 1$ to $ 90$, for $ 90$ possibilities.

Height = $ 15k \plus{} 12$. We use $ k$ blocks of height $ 1,5$< then $ 2$ blocks of height $ 6$. $ k$ can range from $ 0$ to $ 92$, for $ 93$ possibilities.

Adding these up, we get 465. [/hide]

[hide="Another Solution"]We want the number of possible values of $ 4a\plus{}10b\plus{}19c$, where $ a,b,c$ are nonnegative integers and $ a\plus{}b\plus{}c\equal{}94$.  To do this, let's figure out when a height can be reached in multiple ways.  If a height can be reached in multiple ways, then we can exchange some bricks for the same number of different bricks and still get the same height.  We can assume that we don't get back any bricks of the same type as the bricks we got rid of.  Then the only possible type of exchange is to trade 4's and 19's for 10's or the other way around.  This gives us the equation $ 4x\plus{}19y\equal{}10(x\plus{}y)$.  Solving this, we get $ 2x\equal{}3y$, so the minimum solution is $ x\equal{}3$ and $ y\equal{}2$.  This corresponds to trading three 4's and two 19's for five 10's.

Since we can make this exchange as much as we want, given any arrangement of bricks, we can keep getting rid of our 10's until we have less than five.  Therefore if we restrict $ b<5$, then $ 4a\plus{}10b\plus{}19c$ takes on all of its possible values with no repetition.  Now we want the number of solutions to $ a\plus{}b\plus{}c\equal{}94$ where $ b<5$, which is $ 95\plus{}94\plus{}93\plus{}92\plus{}91\equal{}465$.[/hide]

[hide="Solution"]

$4x+10y+19z$ is the sum we have, with $x+y+z=94$.  $4x+4y+4z=4*94$.  Therefore, the sum is the same as $4*94+6y+15z=4*94+3(2y+5z)$.  Ignoring the $4*94$ as it has no significance in what we are trying to calculate, we find all possible values of $3(2y+5z)$ when $0\le y+z\le 94$.  We look at numbers of the form $2y+5z$ when $0\le y+z\le 94$ as the factor of $3$ (again) has no real big significance. All sums from $0$ to $5*94$ are going to be what we are analyzing.

Note that $1$ is impossible, $3$ is impossible, but all sums greater than this are possible looking at the lower end.  Also, when $y+z=94$, then we get $2y+2z=188$, therefore we get all numbers of the form $188+3z$ working, therefore all numbers of the form $2\pmod{3}$.  Also, when $y+z=93$, then we get $2y+2z=186$, therefore we get $186+3z$, up to $186+3*93=465$ that are of the form $0\pmod{3}$ working.  Therefore, since we are looking at $470$, the one that doesn't work is $468$ doesn't work.  Also, when $y+z=92$, then we get $2y+2z=184$.  Therefore, we get looking at $184+3z$, with maximum $184+3*92=460$ working of the form $1\pmod{3}$, which means $463$ and $466$ and $499$ don't work.

Therefore, we get $6$ values not working, so the answer is $5*94+1-6=\boxed{465}$ (using complimentary counting).  [/hide]

[quote=white_horse_king88][hide]The answer is $465$.

We have $x+y+z=94$ and $4x+10y+19z = N$.

Substituting, $6y + 15z = N - 376 \Rightarrow 2y + 5z = \frac{N - 376}{3}$.

Letting $M = \frac{N - 376}{3}$,

We know from determining limits of the solutions that $0 \le M \le 470$.

And so $2y + 5z = M$.

Using the ChickenMcNugget analysis of this problem, we have $1$ and $3$ as bad values. But another problem is at the other end.

So we approach the McNugget backwards with $3n + 5m = M'$, starting at $470$ to see that the smallest value $M$ can't take is $463$.

So looking closer, $463, 466, 468, 469$ don't work.
So $471 - 6 = 465$.[/hide][/quote]
"So we approach the McNugget backwards with $3n + 5m = M'$, starting at $470$ to see that the smallest..." Why did you take $3n+5m=M'$, when we started with  $2y + 5z = M$. I don't really understand that step.


x,y,z positive real numbers such that [b] 8<= xyz.[/b]
prove that:     [b]x+y+z <= xyz-2.[/b]


algebra
polynomial
algorithm
number theory
Diophantine equation
number theory open

Given a polynomial $P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\ldots+a_{1}x+a_{0}$, where $a_{i}$ are integers, $a_{n}\not=0$.

Find all integers $x,y$ such that $P(x)=P(y)$.

[hide="some notes"]This is very hard..
related to:
http://www.mathlinks.ro/Forum/viewtopic.php?t=103637
http://www.mathlinks.ro/Forum/viewtopic.php?t=103641
[/hide]

$P(x)-P(y) = 0$
$(x-y)[a_{n}(x^{n-1}+\cdots+y^{n-1})+a_{n-1}(x^{n-2}+\cdots+y^{n-2})+\cdots+a_{1}] = 0$

Clearly $x = y$ is a solution...
But what can we conlude from the second factor ???
Almost impossible to be solved...  :P

[quote="Krzysztof Dorobisz"]I think it's possible only in some special cases. Actually, the tenth Hilbert's problem was connected with finding a universal algorithm for solving Diophantine equations and it was shown by Yuri Matiyasevich in 1970 that such an algorithm doesn't exist...[/quote]

How do you define "Diophantine equation" at all??

See http://mathworld.wolfram.com/DiophantineEquation.html .
As good as every property is equivalent to a diophantine equation to be (not) solvable, e.g. the axiom of choice.


I invite everyone on this forum (middle schoolers included, but you'll compete in the 9/10 division - don't worry, you'll probably have a good chance at winning) to attend the first every 2006 Bellevue BATH Competition hosted by Bellevue High School; it will take place on Saturday, May 20th. The description is attached as a Word file, including who to contact (me) and how to register (with me). It's going to be the best competition ever and will close off a fine year of math competitions! I look forward to seeing ALL of you there.

[quote="amirhtlusa"]with that penalty...i think you guys are trying to sabetoge ppl...

i can see the scores from now:

10th place: -20
9th place: -18
....

2nd place: +1
1st place: +48 (obviously: adam :D)

 :roll:

the relays is like the ARML relays right? (thats probably why you guys are "requiring" 4-person teams)[/quote]

Similar to ARML relays, yeah. You pass an answer to the next person and they use it.

Come on, everyone. You KNOW you want to come...

[quote="xxreddevilzxx"]are all the special events not going to be revealed until the day of the competition?[/quote]

Well there's only one, but yes, we will reveal it on the day of the competition. So if you want to find out, show up!

I have to say, that's the....most original competition I have ever been to :)

That's because Bellevue is super cool and I'm special.


limit
inequalities
Gauss
function
calculus
integration
floor function

$ \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i \equal{} 1}^n {\frac{1}
{n}} \left[ {\sqrt {4i/n} } \right]$
where $ \left[ x \right]$ $ \equal{}$ the greatest integer $ \leq x$

1.Make an inequality by the difinition of Gauss function.

2. Use Riemman Integral.

3. Use Squeezing theorem.

can u please elaborate
I am cluelees about Gauss function and riemann integral

Are all the problems that you have posted so far your home work? :wink:

Nope , just to see how differently the mathlinkers do the problems .I have already done most of them . And again my exams are over so no school for 3 weeks and no one to give homework..And i take no coaching.

All right, :)  what are the values of $ [\pi],\ [3\minus{}\sqrt{2}]?$

@kunny:I know what a greatest integer function is,alright: :huh:
$ [\pi]\equal{}3$
$ [3\minus{}\sqrt{2}]\equal{}1$
Is it a joke, if yes then I don't find it funny. :wink:

Let $ [x]\equal{}m$, find the inequality in which $ x$ should satisy in terms of $ m$.


In how many ways can I choose 2 cooks on a backpacking trip with 8 people?

This is simply $ \dbinom82\equal{}\frac{8\cdot7}2\equal{}4\cdot7\equal{}\boxed{28}$.


Oι πραγματικοί αριθμοί $ a, b, c, x, y, z$ ικανοποιούν $ a \ge b \ge c > 0$
and $ x \ge y \ge z > 0$. Νδο

$ \frac{a^2x^2}{(by \plus{} cz)(bz \plus{} cy)}\plus{} \frac{b^2y^2}{(cz \plus{} ax)(cx \plus{} az)}\plus{}\frac{c^2z^2}{(ax \plus{} by)(ay \plus{} bx)} \ge \frac{3}{4}$

2000 Korean  Mathematical Olympiad , Problem 6


Hi , every one.

Thanks for veiw my post.

PRBLEM : 4 (5,1,0) + 4(2,2,2) > 5(4,1,1) +3(3,3,0)

for all positive x,,y,z equality holds iff x=y=z

NOTICE : IT's A VERY DIFFICULT.  :D  :D

Can you write detail this problem?

[2,2,2]=abc or 6abc?

 :(

ALL sums are symmetric.

So,     (2,2,2)= 6(xyz)^2

[quote="11235813"]Hi , every one.

Thanks for veiw my post.

PRBLEM : 4 (5,1,0) + 4(2,2,2) > 5(4,4,1) +3(3,3,0)

for all positive x,,y,z equality holds iff x=y=z

NOTICE : IT's A VERY DIFFICULT.  :D  :D[/quote]

are you sure with the (4,4,1) ? the degree is 9 while others only 6 .

[quote="11235813"]
PRBLEM : 4 (5,1,0) + 4(2,2,2) > 5(4,1,1) +3(3,3,0)
for all positive x,,y,z equality holds iff x=y=z
[/quote]
If the sign of inequality is strong, equality doesn't hold ;)

I don't think it's true, you have a proof?

[quote="shyong"]
are you sure with the (4,4,1) ? the degree is 9 while others only 6 .[/quote]
Read condition of the problem more attentively and you see (4,1,1) ;)

Pavel kozlov, just read the following post :
[quote="11235813"]I've edited  :D  :D[/quote]
and [quote]Last edited by 11235813 on Thu Feb 16, 2006 10:26 am; edited 1 time in total[/quote]. :)
But, I don't think this inequality holds..

Ha,ha.!

Thank you for read.

But it's true ineq.

I still don't think so. I guess taking some close x,y,z (but not all equal) will make it false, since after SOS the ineq is something like $\sum s_z (x-y)^2 \ge 0$, however, as x,y,z tends to be the same, $s_z \leq 0$, same for $s_x$ and $s_y$

[quote="arqady"]
11235813, what do you mean? Thank you.[/quote]

I mean this ineq is hard without calculus.

I solved by exactly your way

[quote="Gibbenergy"][quote="11235813"]Please solve this problem another way.

It will be a good challenging. ;)[/quote]
 :D Here is one-line solution for your problem after 3 hours of calculation:
Let $x \geq y \geq z$
Denote $a1=2x^2y+2x^2z+xyz+6xy^2-3z^3+9y^3-3y^2z$
            $a2=11y^3-10y^2z-9z^3+12yz^2$
             $a3=-2z^4+3z^3y+12y^2z^2-12zy^3+3y^4$
             $R=y(y-z)^2(y^3-4y^2z+3yz^2+4z^3)$
(Look at Argardy sols!)
$F=(x-y)^3a1+y(x-y)^2a2+(x-y)(y-z)a3+ R \geq 0$  Done! :P[/quote]

Your proof means "imposible is nothing"  Wonderful solution :!:  :first:

This inequality, however, can be solve quite short by apply $ABC$ method, which just consider two case:
$i) a=b=1$
$ii) a=0$
I will post the solution for these cases later  :) .

Who delete Gibbenergy 's post and Why :mad:

[quote="Anh Cuong"]
I will post the solution for these cases later  :) .[/quote]

When?

[quote="Gibbenergy"]I deleted it by myself. I think I am too vain as I put my statement there. Because as reading ABC of Anh Cuong, I think it is a very effective method. And also I read about SOS too. In my opinion, my technique isn't that strong and also I should put more thought for it.
   There are zillion of genius in Mathlinks. I should study more rather than speaking :lol:[/quote]

Do you mind telling me what ABC method is? I can't read a single bit of Vietnamese.......  :(

[quote="Soarer"][

Do you mind telling me what ABC method is? I can't read a single bit of Vietnamese.......  :([/quote]

Hi Soar !
   It is the link to ABC technique: http://mathnfriend.net/index.php?showtopic=900
   Click in the .pdf file.
   It is Vietnamese page so you have to guess it through the context and diagram. I don't translate it because I didn't ask the permission of the author yet and I guess Anh Cuong will translate it soon because he understands it  thoughtful!
  Hope it help you! :)

Actually, we can't see this file unless we register..
So, I guess we'll have to wait for Anh Cuong... :)

Hello? Anh Cuong

Are you sleeping?

Please write ABC method for us ;)

[quote="hungkhtn"]Here's his idea. Because now I don't sure it's actually true.

For all $a,b,c$ exists $x,y,z,x',y',z'$ st
a+b+C=x+y+z=x'+y'+z'
ab+bc+ca=xy+yz+zx=x'y'+y'z'+z'x'
$xyz \ge abc \ge x'y'z'$

And in three numbers x,y,z exists 2 number equal.

(THis idea came from the problem

For $a+b+c=1,ab+bc+ca=m, |m| \e 3$ then find max and min of $abc$. )[/quote]

So How to deal with?


k is a positive integer and t is a real number.
The function f(n) is defined for every integer n not smaller than 0 and not greater than k.
f(n) satisfies:
f(0)=f(k)=0;
for 0<n<k, f(n)=tf(n-1)+(1-t)f(n+1)+1.
Find the function f(n).

The solution for t=0.5 is f(n)=n(k-n), as can be verified easily.
Can somebody find f(n) for t different than 0.5?

Warning, for $t=1$ there is'nt solution !
For $t \notin$

Sorry, latex mistrake in my previous post !

Warning, for $t=1$ there is'nt solution !
For $t \notin \{0.5,1\}$ the solution is $f_n=A(\frac t{1-t})^n + B+ \frac n{1-2t}$
You can find $A,B$ by solving the system $f_0=0 , f_k=0$
 :cool:

And now, what if instead of f(0)=0 we have f(0)=f(1)+1, with all the other conditions unchanged?

For t=0.5 the solution is f(n)=k^2-n^2. What is f(n) for other values of t?