Time for an old IMO problem. Number 4, 1999. If you've seen the solution before please don't post..

Find all pairs (n,p) or positive integers such that:
p is prime
n<=2p
(p-1)^n + 1 is divisible by n^(p-1).

Please post any ideas, small cases, anything even if it seems irrelevant (OK, not tooooo irrelevant ;))

I think what I will do is post a hint or two every so often until we solve it. I'll start after someone posts some trivial solutions first :).

uhh let's see...trivial cases are nice and easy!...for p=2, the only solutions for n are 1 and 2. so after this we can assume p is odd. for n=1, all values of p work...for n=2, only p=2, for n=3, p=3, and for n=4 i dont think there are any values of p. please check my arithmetic =)

also, by fermat's little theorem, n^(p-1) = (i'm using the equal sign to mean "is equivalent/congruent to") 1 (mod p).
thus we must find all m, n, and p for which mp + 1 | (p-1)^n + 1.
i dunno if that really helps but it seems to fit nicely.

[quote="froufrou4189"]uhh let's see...trivial cases are nice and easy!...for p=2, the only solutions for n are 1 and 2. so after this we can assume p is odd. for n=1, all values of p work...for n=2, only p=2, for n=3, p=3, and for n=4 i dont think there are any values of p. please check my arithmetic =)[/quote]

Woohoo! Thanks. Exactly what I was hoping someone would post :)
So we have the solutions (1,p) for all primes p, and (2,2), (3,3).
From now on we can assume p is an odd prime :)

What can we assume about n?

[quote]
also, by fermat's little theorem, n^(p-1) = (i'm using the equal sign to mean "is equivalent/congruent to") 1 (mod p).
thus we must find all m, n, and p for which mp + 1 | (p-1)^n + 1.
i dunno if that really helps but it seems to fit nicely.[/quote]

That would probably make it a bit too general. n^(p-1) is a lot more specific than just any mp + 1. However fermats little theorem is the right idea (which you could probably tell from the question).

I played around with this in calc a little today and here's what i got.

n cannot be even:

Assume n=2m.

Then (2m)^(p-1) | ((p-1)^2)^m + 1. But since p is odd, the rhs is odd, and the lhs is even. An even number cannot divide an odd number, so n must be even.

Now assume a is the smallest prime divisor of n, and b=n/a (a,b,n are odd).

I'm too lazy to prove this now, but after using fermat's little theorem, i got that

(p-1)^b + 1 = 0 (mod a),

from which it follows that p = 0 (mod a), 3 =< a =< p.

From this, we get that (p-1)^n + 1 = 0 (mod a).

Is that right/useful?

That's all i had time to do...any more hints would be appreciated =).

[/hide]

All right, continuing from my last post (I think I finally got it!)

Since p = 0 (mod a), p is prime and therefore p=a.

Now we have

(ab)^(a-1) | (a-1)^(ab) + 1. But since n=< 2p, and b is odd, b=1, and a=n=p.

So <i>now</i> we have

p^(p-1) | (p-1)^p + 1

Multiplying the rhs out and factoring out a p^2, we get that

p^(p-3) | (some expression divisible by p) + 1

clearly, the rhs is relatively prime to the lhs, so we have no solutions if p is a odd prime and p>3.

Therefore, the only solutions are (1,p), (2,2) and (3,3)

Yay! Did I miss anything?

That looks pretty good for me.

Its the proving p is divisible by a (and thus equal a) that was the tricky bit.. in the solution I have, this is how they did it:

[hide]

We still have a is the smallest divisor of n.

We also know that (p-1)^k :equiv: 1 (mod a) for some k. (eg squaring the initial condition shows k=2n works.) So let k be the smallest such positive integer. Now c|2n from what I just had before. Also, c|a-1 by FLT. So c|gcd(2n,q-1). But we know gcd(n,a-1)=1 by the way we chose a. So c=2 (c cannot equal 1 as (p-1)^n :equiv: -1 (mod a)), and thus p-1 :equiv: -1 (mod a), and that proves it.

[/hide]

Ok I know this is an easy question, but if I don't ask, I won't know.

what does the symbol "|" mean between two expressions?

a | b means b is divisible by a. And the same symbol with a slash (like this \) through it, means b is not divisible by a.

btw, couldn't we just prove that p = 0 (mod a) by saying that since (p-1)^b = -1 (mod a), then

(p-1)^b = -1^b (mod a) (since b is odd)

and then that

(p-1)^b = (a-1)^b (mod a), and therefore p = a (mod a) ?

There's prolly something wrong with this but I don't know what.

so then *either* p = a (mod a) or (something else)...so all we have to do is prove that (something else) is not possible or something like that....i'll see.

so p <i> could </i> be equivalent to a =)

[quote="TripleM"]I don't think you can just cancel off powers like that - for example, if x^3 :equiv: y^3 (mod a) then you have (x-y)(x^2 + xy + y^2) :equiv: 0 (mod a) and so *either* x :equiv: y or x^2 + xy + y^2 :equiv: 0 (mod a)..[/quote]

[color=cyan]Not even -- x-y could be congruent to a/2 and (x^2 + xy + y^2) could be congruent to 2 . . . anything resembling division (and taking roots is just like division, but more complicated) will be difficult to do in mods.  All you've proven is that p [i]might[/i] be congruent to 0.  That is, you didn't disprove what you were trying to prove :).[/color]

True, but in this case a was a prime, so you must have one of them divisible by a.


probability
geometry
3D geometry
sphere
integration

What is the probability that 17 points on surface of the  sphere lie on the same hemisphere?
Thx

If u cant solve it comepletely, at least give me some help...

Here is an idea, let the first 2 points be A and B, and the center of the sphere be O, let $\angle AOB=\theta$, now find the surface area that would lie on a hemisphere in terms of $\theta$ then divide that by the total surface area to get $P(\theta)$ do $\left(\int_{0}^{180}f(\theta)\right)^{15}$, the only problem with this is that the chaces of different $\theta$'s are different, but i have to go now

[quote="scorpius119"]Unless i really made an error visualizing the problem...

[hide="Solution"]First look at the first two randomly selected points. The probability that they are the same or on the same diameter is 0 (it's like the probability of picking 1 out of any randomly selected real number), so these two points uniquely determine a great circle on the sphere.

The third point will determine which hemisphere the other 14 points lie on, because the probability that this point lies on the great circle defined by the first two points is again 0. For each point, there is probability of 1/2 that the point is on the same hemisphere as the third point. So the final probability is $\boxed{\frac{1}{2^{14}}}$.[/hide][/quote]

However, the first three points don't uniquely define a hemisphere...

You only counted the one where the first two points lie on the edge of the hemisphere, which has 0 probability...

I would bump this problem to a higher forum...it's really tough.

Anyway, if want to know how to do it, look at #20 on the 1999 BAMM: http://usfca.edu/math/bamm/toi.html


:) 无意间发现这个网站，发现这里有好多高手！
我是山东潍坊的，现在理科实验班就读，高一（实初四）年级。
希望和大家作个朋友！

不知道大家都读了那些书，作了那些题？望不吝赐教！

http://post.baidu.com/f?kw=%CA%FD%D1%A7

那儿有个吧主叫“富里哀”的，（我用的就是他的英文名），已经是数学博士了，
哎~~~~~什么时候咱也~~

欢迎大家去看看！

[quote="Fourier"]不知道大家都读了那些书，作了那些题？望不吝赐教！[/quote]
我也很想知道阿，请高手们指点一下，最好我们成立一个数学小组就好了
我今年考得极差，明年的目标是冬令营

FOURIER...
我也是今天才看到的...
这里的人似乎不是一般的牛...
有N个AIME拿12以上的...
另外建议你去下奥数之家.
也是牛人多的地方.
另外建议你多去看下用英语讨论的那些人的帖..
比中国学生还猛


I'm trying to figure these out...I could use the help. I know the answers if you need them.

 The first one is: The octane number of 87 on gasoline means that it will fight engine "knock" as effectively as a reference fuel that is 87% isooctane, a type of gas. Suppose you want to fill an empty 18 gallon tank with some 87 octane gasoline and some 93 octane fuel to produce a mixture that is 89 octane. How much of each type of gasoline must you use?

Second one: A police officer has chosen a high-yield stock fund that earns 8% annual simple interest for part of a 6000 investment. The remaining portion is used to purchase a preferred stock that earns 11% annual simple interest. How much should be invested in each account so that the amount earned on the 8% account is twice the amount earned on the 11% account. I can do most of them but these are stumping me...I know I was getting close

I think I got the 1st one. Could u tell me the answer so I can check it?

you shouldnt be afraid of posting a wrong solution, as long as it isnt junk...

anyway since the percentage that we want (89%) is 2 times closer to 87 than 93, we use 2 times as much *err* 87-octane gas. Weighting is a very useful tool if you want to solve ratio problems quickly

 so we let a be the amount of 93-octane gas and we use 2a 87-octane gas and this should total 18 so a+2a=18 tells us a=6 and you should use 6 gallons of 87-octane gas and 12 gallons of 93-octane gas

ehh the wording on #2 is freaking me out... maybe u could check it with the answer u have

so the amount invested in the accounts combined is 6000 so we can say we invest $ \$a$ (ugh hate doing this for dollar signs)  for the 8% account and $ \$6000\minus{}a$ for the 11% one.

we want the amount earned on the 8% account to be twice the amount eearned on the 11 % account (i think)
so this tells us $ .08a\equal{}2(.11(6000\minus{}a))$ expand...
$ .08a\equal{}1320\minus{}.22a$
$ .3a\equal{}1320$
$ a\equal{}4400$

so we invest 4400 on the 8% account and 1600 on the 11% account.

1.Nice ratio for number 1 stevenmeow.  Here's a more general answer:
[hide]Let $ x$ = the 87 octane and $ y$= the 93 octane gas.  We know that $ x \plus{} y \equal{} 18$.  We can also find the relative "octane parts" per gallon by setting $ .87x \plus{} .93y \equal{} .89*18$, where the $ .89$ comes from the percentage of the final mixture and the 18 comes from the total gallons we have.  We can then solve for x and y to find $ x \equal{} 12, y \equal{} 6$.[/hide]

2.[hide]Let $ x$ = high yield stock amount and $ y$= preferred stock.  We have that $ x \plus{} y \equal{} 6000$.  We also know that $ .08x \equal{} 2*.11y$.  Solving yields $ x \equal{} 4400, y \equal{} 1600$.[/hide]

Grr...someone beat me to the problems. :mad:

You guys are right! Thanks a lot. I was close but it was giving me problems...I'll have to go back and look at the first one...I kept getting 10.99999...LOL :lol:

 That's pretty good...You guys must be a lot further in your studies then me... :(

I just went through about 30 or so story problems and those were the worst...It was in an Algebra book in the linear systems section....

Weighted averages can start to annoy even the best of us, if you do them a lot.  :wink:  (We've been through weighted averages too, and in my opinion, it isn't one of the more enjoyable parts of math.)

Anyways, if you have an incorrect answer, it's always a good idea to post your solution, so we can help you find out what went wrong.


Uh, did you forget about what happened to the last proofs topic?
There are no proofs in MATHCOUNTS.

[quote="abacadaea"]prove that there is no greatest number.

prove $ n(n+1)(2n+1)$ is divisible by 6.[/quote]

$ n(n+1)(2n+1) = n(2n^{2}+3n+1) = 2n^{3}+3n^{2}+n$

$ 2n^{3}| 2$ while $ 3n^{2}| 3$ and $ n | 1$ (not sure if the notation is right, $ n | 8$ means "$ n$ is divisible by $ 8$ or $ n/8 =$ an integer")

let $ 2n^{3}= 2k$, $ 3n^{2}= 3k$ and $ n = k$ for any integer k.

$ 2k+3k+k = 6k$ substitute backwards to get:

$ 2n^{3}+3n^{2}+n = 6k$

$ \frac{2n^{3}+3n^{2}+n}{6}= k$

Since $ k$ is an integer,  $ n(n+1)(2n+1)$ must therefore be divisible by $ 6$.

Prime factorization , the proof is correct  :) , but $ a|b$ means $ b$ is divisble by $ a$.

Er, actually, it isn't.

[quote="Prime factorization"]$ 2n^{3}| 2$ while $ 3n^{2}| 3$ and $ n | 1$ (not sure if the notation is right, $ n | 8$ means "$ n$ is divisible by $ 8$ or $ n/8 =$ an integer")[/quote]

It should be the other way around.  $ a | b$ is to be read "$ a$ divides $ b$."

[quote="Prime factorization"]let $ 2n^{3}= 2k$, $ 3n^{2}= 3k$ and $ n = k$ for any integer k.[/quote]

This is problematic.  You're basically assuming $ n^{3}\equiv n^{2}\equiv n \bmod 6$ to do this.  You would have to assign three different integers, and then you would get the useless expression $ 2a+3b+c$.  Now, if you wanted to approach the problem in a modular way,

[hide="Then..."] Either $ n$ or $ n+1$ will be even, so $ 2 | n(n+1)(2n+1)$.  We only need to check divisibility by three.  But

$ n(n+1)(2n+1) \equiv n(n+1)(-n+1) \equiv-(n-1)n(n+1)$

And among three consecutive integers, at least one is divisible by three.  Hence $ 2, 3 | n(n+1)(2n+1) \Leftrightarrow 6 | n(n+1)(2n+1)$. [/hide]

Of course, you could always do the fun and indirect

[hide="Method 3"] It's not hard to prove inductively that

$ \frac{n(n+1)(2n+1)}{6}= 1^{2}+2^{2}+...+n^{2}$

Which is clearly an integer. [/hide]

$ 1^{2}+2^{2}+3^{2}+\dots+n^{2}=\frac{n(n+1)(2n+1)}{6}$.

EDIT: sorry i didn't notice that t0rajir0u already posted this.

torajirou, all I was trying to say was that the coefficients 2,3 and 1 were the important thing in considering whether the expression was divisible by 6 or not. In this case, the value of n is irrelevant.

Just out of curiosity, could this statement somehow be used to solve the problem? $ LCM(1,2,3) = LCM(2,3) = 6$.

[quote="Prime factorization"]torajirou, all I was trying to say was that the coefficients 2,3 and 1 were the important thing in considering whether the expression was divisible by 6 or not. In this case, the value of n is irrelevant.[/quote]

Er... no, they're not.  For example, the polynomial $ 2n^{4}+3n^{2}+n$ doesn't have the desired property; when $ n = 2$ the expression evaluates to $ 46$.

Letting $ 2n^{3}= 2k, 3n^{2}= 3k, n = k$ implies that $ k = n = n^{2}= n^{3}$ (at least, $ \bmod 6$), which is plainly not true in the general case.

I'm sorry, you are indeed correct  :oops:

[quote="mdk"][quote="t0rajir0u"][hide="1"] This depends rather a lot on what set of axioms you're using.  This is an axiom in Peano arithmetic, for example.  [/hide]
[hide="2"] $ \frac{n(n+1)(2n+1)}{6}={n+2 \choose 3}+{n+1 \choose 3}$ [/hide][/quote]

just curious,how did you get that expression for 2?[/quote]
Kunny and t0rajir0u see things most of us don't...

[quote="mdk"][quote="t0rajir0u"][hide="1"] This depends rather a lot on what set of axioms you're using.  This is an axiom in Peano arithmetic, for example.  [/hide]
[hide="2"] $ \frac{n(n+1)(2n+1)}{6}={n+2 \choose 3}+{n+1 \choose 3}$ [/hide][/quote]

just curious,how did you get that expression for 2?[/quote]

$ 2n+1 = (n+2)+(n-1)$.  :)

Part of the motivation was looking for strings of three consecutive integers, which are clearly divisible by $ 6$.  There is also a second motivation based on

[hide="A nice proof of the sum-of-squares formula"] Specifically, we use Hockeystick:

$ \sum_{k=1}^{n}k^{2}= \sum_{k=1}^{n}{k \choose 2}+{k+1 \choose 2}={n+1 \choose 3}+{n+2 \choose 3}$ [/hide]

[quote="abacadaea"]has anyone tried the first one yet
[hide="my solution"] keep it nice and simple. asume the largest number was $ n$. then $ n+1$ would be greater than $ n$, the greatest number. those two statements contradict each other. therefore there is no greatest number. [/hide][/quote]

Again, this is an axiom in Peano arithmetic and does not require proof.

[quote="t0rajir0u"][hide="A nice proof of the sum-of-squares formula"] Specifically, we use Hockeystick:

$ \sum_{k=1}^{n}k^{2}= \sum_{k=1}^{n}{k \choose 2}+{k+1 \choose 2}={n+1 \choose 3}+{n+2 \choose 3}$ [/hide][/quote]

Why didn't you complete this proof? Also, the syntax is slightly off:

$ \sum_{k=1}^{n}k^{2}= \sum_{k=1}^{n}{k \choose 2}+{k+1 \choose 2}$

is not true, but I'm guessing you meant:

$ \sum_{k=1}^{n}k^{2}= \sum_{k=1}^{n}[{k \choose 2}+{k+1 \choose 2}]$

Which is true, but there is no explanation of why.

[hide="My Solution"]It assumes that the sum of two consecutive triangular numbers is the appropriate square number.

$ \frac{n(n-1)}{2}+\frac{n(n+1)}{2}$

$ =\frac{n^{2}-n+n^{2}+n}{2}$

$ =\frac{2n^{2}}{2}$

$ =n^{2}$

So then since $ {k \choose 2}=\frac{k(k-1)}{2}$ (because it's the $ (k-1)$th triangular number), and $ {k+1 \choose 2}=\frac{k(k+1)}{2}$ (because it's the $ (k+1-1)$th triangular number). I'll provide a proof for this here, but I'm guessing it's a recognised property so I'll hide it:

[hide="triangular/pascal proof"]

$ {n \choose 2}$

$ =\frac{n!}{2!(n-2)!}$

$ =\frac{1}{2}*\frac{n!}{(n-2)!}$

$ =\frac{n(n-1)}{2}$

[/hide]

So then you could say:

$ {k \choose 2}+{k+1 \choose 2}=k^{2}$

and so

$ \sum_{k=1}^{n}[{k \choose 2}+{k+1 \choose 2}]=\sum_{k=1}^{n}k^{2}$

so then you could say that, due to hockey stick theorem:

$ \sum_{k=1}^{n}{k \choose 2}={n+1 \choose 3}$ and

$ \sum_{k=1}^{n}{k+1 \choose 2}={n+2 \choose 3}$

so

$ \sum_{k=1}^{n}k^{2}= \sum_{k=1}^{n}[{k \choose 2}+{k+1 \choose 2}]={n+1 \choose 3}+{n+2 \choose 3}$

but you should also show that $ {n+1 \choose 3}+{n+2 \choose 3}= \frac{1}{6}n(n+1)(2n+1)$... to do this I'll first find $ {n \choose 3}$:

$ {n \choose 3}$

$ =\frac{n!}{3!(n-3)!}$

$ =\frac{1}{6}*\frac{n!}{(n-3)!}$

$ =\frac{1}{6}n(n-1)(n-2)$

so then $ {n+1\choose 3}$ becomes:

$ \frac{1}{6}(n+1)n(n-1)$

$ =\frac{1}{6}n(n+1)(n-1)$

and $ {n+2\choose 3}$ becomes:

$ \frac{1}{6}(n+2)(n+1)n$

$ =\frac{1}{6}n(n+1)(n+2)$

and to combine:

$ {n+1\choose 3}+{n+2\choose 3}$

$ =\frac{1}{6}n(n+1)(n-1)+\frac{1}{6}n(n+1)(n+2)$

$ =\frac{1}{6}[n(n+1)(n-1)+n(n+1)(n+2)]$

$ =\frac{1}{6}[n(n+1)(n-1+n+2)]$

$ =\frac{1}{6}n(n+1)(2n+1)$

Therefore:

$ 1^{2}+2^{2}+3^{2}+...+n^{2}= \frac{1}{6}n(n+1)(2n+1)$

[/hide]

Sorry, I have a hobby of finding one-line solutions to problems.  I keep the most illuminating steps, and usually the rest is just algebra.  :)


Sorry guys, this is my first time here... I am in desperate need of help on this... I am trying to find the intersection of a system of equations... I can graph both lines just fine... I then hit "F5" and go to "Intersection"..... It then asks me for a "1st curve" and a "2nd curve".... What does that mean?!?!?! I spent waaaay too much on this calculator for me to not beable to use it in my class... I tried reading the manual but NOTHING  :?  :? .....

Anyway, after the 1st curve, 2nd curve thing.... It asks me for "Lower Bound" and "Upper Bound"... However, everytime I hit "enter" an error message comes up that says "Error: BOUND".... What the hell does this mean???? I have no clue... I have been out of school for 5 years in the Marines and have really lost any knowelage of the math that I had already taken.... Please help!!!!

Never mind guys..... I looked online and found a tutorial to help me through it... .Thanks anyway though :)

For the Ti-89, does anyone know how to make it do implicit differentiation?


One day, 45 frogs were captured from a pond, marked and then returned to the pond. The next day, 40 frogs were observed in the pond, of which 10 had been marked the previous day. Assuming the marked frogs are equally distributed among all the frogs in the pond, what is the best estimate of how many frogs live in the pond?

Our best bet would be that 10/40=1/4 of all frogs are marked. Thus, 1/4 of all frogs are 45, so there are going to be 45x5=180 total.

Approximately $ \frac{10}{40}\equal{}\frac{1}{4}$ of the frogs are marked.  Thus, we do $ 4 \times 45\equal{}\boxed{180}$.

edit: beaten

[hide="this is my answer"]1. there were 40 frogs observed and 10 of them were marked so we get a fraction 10/40=1/4

2.45 of them were marked at the beginning so we get a equation of 45*5 we get the 5 from the difference

3.so our answer is $\boxed{180}$[/hide]

[b][i][u]please rate my post it is greatly apprecited ! [=[/u][/i][/b]

please be fair though thanks![=

:stretcher: :stretcher: :stretcher:
 :stretcher:


I make water rockets which are launched with only water and a pressurized gas. You wouldn't think they would go high, but some have been built which can reach 1000 feet. :o


modular arithmetic
number theory
relatively prime
combinatorics proposed
combinatorics

Let $ n$ be an integer, $ n\geq 2$. Find all sets $ A$ with $ n$ integer elements such that the sum of any nonempty subset of $ A$ is not divisible by $ n\plus{}1$.

[hide="Solution"]Let $ A \equal{} \{a_1,\ldots,a_n\}$ be a set. We shall prove that $ a_2 \equiv a_3 \pmod {n \plus{} 1}$, and by rearranging the indexes, we will show that all the elements of $ A$ are congruent to some $ p \pmod {n \plus{} 1}$, such  that $ \gcd(p,n \plus{} 1) \equal{} 1$. Obviously if that is the case, any subset of $ A$ has at most $ n$ elements, so its sum will be $ kp$, with $ 0 < k < n \plus{} 1$, so not divisible by $ n \plus{} 1$.

Consider the sums $ s_1 \equal{} a_1$, $ s_2 \equal{} a_1 \plus{} a_2$, $ \ldots$, $ s_n \equal{} a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n$. Obviously none of these sums can be equal $ \pmod {n \plus{} 1}$ (because if they would taking the difference $ s_i \minus{} s_j \equal{} a_{j \plus{} 1} \plus{} \cdots \plus{} a_i \equiv 0\pmod{n \plus{} 1}$ we get a contradiction. Therefore they cover all the residues $ 1,2,\ldots,n$, modulo $ n \plus{} 1$. Consider now $ a_1 \plus{} a_3$. This is non zero, so it must be equal to some $ s_i$. But if $ i > 3$, then $ s_i \minus{} (a_1 \plus{} a_3) \equal{} a_2 \plus{} a_4 \plus{} \cdots \plus{} a_i$, which cannot be 0. If $ i \equal{} 3$, then $ a_2 \equiv 0\pmod {n \plus{} 1}$ contradiction, if $ i \equal{} 1$, then $ a_3\equiv 0\pmod {n \plus{} 1}$ contradiction, so the only possibility is $ a_1 \plus{} a_2 \equiv a_1 \plus{} a_3$, so $ a_2 \equiv a_3$. Similarly we obtain that all the elements in the set are congruent modulo $ n \plus{} 1$, and we are done.[/hide]

Let A= ( a1,a2,a3,……..,an )

Consider the sums
a 1
a 1+a2
a 1+a2+a3
.
.
.
.
a1+a2+a3+……+an

All these sums must have different non –zero remainders on division by n+1.
Because if we assume two sums to have the same remainder on division by n+1 , then their difference, which is also a subset of A, would have to be divisible by n+1.

Similarly, considering the sums
a 2
a 1+a2
a 1+a2+a3
.
.
.
.
a1+a2+a3+……+an

it could be proved that a1= a2 (mod n+1)

Similarly, it could be proved that all members of A belong to the same residue class modulo n+1.
Now let the remainder of these numbers on division by n+1 be r

If gcd(n+1,r)=d>1, then there exists a positive integer s<n+1 such that sd=n+1 implying that n+1 divides sr which implies that the sum of any s members is divisible by n+1.

Hence, A is such that all of its elements produce the same remainders on division by n+1 and this remainder is relatively prime to n+1.

Also , if r were relatively prime to n+1 , then none of the numbers sr ; s<n+1 are divisible by n+1

Sorry about the notations, i couldn't use subscripts

Sorry to revive this old thread, but the problem has made some reappearance(s), and a full clear solution is now found at [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=417536&p=2353838#p2353838[/url].
Moreover, the original statement of the problem was using a multiset of $n$ integers (which thus were not necessarily required be distinct), with exactly the same conclusion, since just divisibility (and not equality) is involved.


Initially I put this problem up on the middle school Classroom Math section, but nobody got  it, so I've moved it here.   I'm in eighth grade and I love these old, high school Math League problems.  Here's one that I liked:

Let there be acute triangle ∆ABC, such that AE, BF, and CD intersect at point P (each of these are lines to the opposite side of ∆ABC from each angle). If the area of ∆AFP is 40, the area of ∆CFP is 30, and the area of ∆CEP is 35, what is the area of ∆BPD? After I did this, I thought it would be a worthwhile problem to put up. I will post a hint in about 5 days if nobody gets it, and will eventually post the answer as well as my solutions and the book's solutions.

We use the concept of area ratios here; if the height of two triangles are equal, then the ratio of the areas of the triangles to their bases are equal (and vice versa for bases and heights). Let $ x \equal{} [APD]$, $ y \equal{} [BPD]$, and $ z \equal{} [BPE]$.  Since $ \triangle BPC$ and $ \triangle FPC$ share the same height,
\[ \frac{BP}{z \plus{} 35} \equal{} \frac{FP}{30}\]
Also, we notice that $ \triangle BPA$ and $ \triangle FPA$ have these sides and also these two triangles share the same height, so we have
\[ \frac{x\plus{}y}{z\plus{}35} \equal{} \frac{40}{30} \Longrightarrow x\plus{}y \equal{} \frac{4}{3}(z\plus{}35)\]
Also, $ \frac{[APB]}{[BPE]} \equal{} \frac{[APC]}{[EPC]} \Longrightarrow \frac{x\plus{}y}{z} \equal{} 2$. Combining this and the above yield that $ \frac{2}{3}z \equal{} \frac{140}{3} \Longrightarrow z \equal{} 70$.

Finally $ \frac{[APC]}{[APD]} \equal{} \frac{[BPC]}{[BPD]} \Longrightarrow \frac{x}{70} \equal{} \frac{y}{105} \Longrightarrow \frac{2}{3}y \equal{} x$. Combining with above ($ x\plus{}y\equal{}140$), we get $ \frac{5}{3}y \equal{} 140 \Longrightarrow [BPD] \equal{} \boxed{84}$. Hope I did that right..


$ \sqrt{\frac{x\minus{}2}{1\minus{}2x}}>1$

$ \frac{x\minus{}2}{1\minus{}2x}>1$

$ x\minus{}2<1\minus{}2x$

$ 3x<3$

$ \boxed{x<1}$

But if we plug in 0, we get $ \sqrt{\minus{}2}>1$. So I am not sure if the first one has any solutions.

I think mods should move this thread to middle school math...

EDIT: Sorry if this is spam!!

[quote="PowerOfPi"]$ \sqrt {\frac {x \minus{} 2}{1 \minus{} 2x}} > 1$

$ \frac {x \minus{} 2}{1 \minus{} 2x} > 1$

$ x \minus{} 2 < 1 \minus{} 2x$

$ 3x < 3$

$ \boxed{x < 1}$

But if we plug in 0, we get $ \sqrt { \minus{} 2} > 1$. So I am not sure if the first one has any solutions.[/quote]

There are solutions. The thing is that you can't multiply by $ 1 \minus{} 2x$ because we do not know whether it is positive or not.

$ \frac {x \minus{} 2}{1 \minus{} 2x} > 1$
$ \frac {x \minus{} 2}{1 \minus{} 2x} \minus{} 1 > 0$
$ \frac {x \minus{} 2 \minus{} (1 \minus{} 2x)}{1 \minus{} 2x} > 0$
$ \frac {3x \minus{} 3}{1 \minus{} 2x} > 0$
To make this true, either the numerator and denominator are both positive or both negative.

$ 3x \minus{} 3 > 0$
$ x > 1$
$ 1 \minus{} 2x > 0$
$ x < \frac {1}{2}$
There are no cases where they are both positive. Now to check if they are both negative.

$ 3x \minus{} 3 < 0$
$ x < 1$
$ 1 \minus{} 2x < 0$
$ x > \frac {1}{2}$
$ \frac {1}{2} < x < 1$
So the solution interval for $ x$ is $ \boxed{(\frac {1}{2},1)}$.
Also something important to notice is the denominator cannot equal 0, but in this problem it is not important since this value wasn't in the solution set anyways.
$ 1 \minus{} 2x \neq 0$
$ x \neq \frac {1}{2}$

I don't really think this is middle school math. (Average middle school math anyways.) And that isn't spam, because you are letting someone know they posted in the wrong forum.

@PowerOfPi, you got invalid solutions because the first step you did was square both sides, which is inappropriate when working with inequalities with unrestricted variables.
[hide="(a)"]
We start by collecting terms on one side.

$ \sqrt {\frac {x \minus{} 2}{1 \minus{} 2x}} > 1 \implies \frac {\sqrt {x \minus{} 2} \minus{} \sqrt {1 \minus{} 2x}}{\sqrt {1 \minus{} 2x}} > 0\implies \frac {\sqrt {(x \minus{} 2)(1 \minus{} 2x)} \plus{} 2x \minus{} 1}{1 \minus{} 2x} > 0.$

The domain restriction is $ x \in \left[\frac {1}{2},2\right]$. Therefore, we know that the numerator will always be positive. Now in order for the entire fraction to be greater than zero, we must have

$ 1 \minus{} 2x > 0 \implies x < \frac {1}{2}.$

Clearly, that is not possible with our domain restriction, so $ \boxed{x \in \emptyset}$.
[/hide]
[hide="(b)"]
We can just set up a bunch of cases for this problem.

For $ x \ge 2$, we have

$ |x^2 \minus{} 3x \plus{} 2| \equal{} |(x \minus{} 2)(x \minus{} 1)| \equal{} (x \minus{} 2)(x \minus{} 1) \equal{} x(x \minus{} 2) \implies (x \minus{} 2)( \minus{} 1) \equal{} 0 \implies x \equal{} 2.$

For $ 1 \le x < 2$, we have

$ |x^2 \minus{} 3x \plus{} 2| \equal{} |(x \minus{} 2)(x \minus{} 1)| \equal{} \minus{} (x \minus{} 2)(x \minus{} 1) \equal{} x(x \minus{} 2) \implies (x \minus{} 2)(2x \minus{} 1) \equal{} 0 \implies x \equal{} 2, \frac {1}{2}.$
Clearly,  $ x \equal{} 2$ is extraneous.

For $ 0 \le x < 1$, we have

$ |x^2 \minus{} 3x \plus{} 2| \equal{} |(x \minus{} 2)(x \minus{} 1)| \equal{} ( \minus{} 1)^2(x \minus{} 2)(x \minus{} 1) \equal{} x(x \minus{} 2) \implies (x \minus{} 2)( \minus{} 1) \equal{} 0 \implies x \equal{} 2.$

Clearly, $ x \equal{} 2$ is extraneous due to the restriction on $ x$.

For $ \minus{} 1 \le x < 0$, we have

$ |x^2 \plus{} 3x \plus{} 2| \equal{} |(x \plus{} 2)(x \plus{} 1)| \equal{} x^2 \plus{} 3x \plus{} 2 \equal{} x^2 \minus{} 2x \implies 5x \equal{} \minus{} 2 \implies x \equal{} \minus{} \frac {2}{5}.$

For $ \minus{} 2 \le x < \minus{} 1$, we have

$ |x^2 \plus{} 3x \plus{} 2| \equal{} |(x \plus{} 2)(x \plus{} 1)| \equal{} \minus{} x^2 \minus{} 3x \minus{} 2 \equal{} x^2 \minus{} 2x \implies 2x^2 \plus{} x \plus{} 2 \equal{} 0.$

This equation has no real solutions.

For $ x < \minus{} 2$, we have

$ |x^2 \plus{} 3x \plus{} 2| \equal{} |(x \plus{} 2)(x \plus{} 1)| \equal{} x^2 \plus{} 3x \plus{} 2 \equal{} x^2 \minus{} 2x \implies 5x \equal{} \minus{} 2\implies x \equal{} \minus{} \frac {2}{5}.$

This is clearly an extraneous root.

Therefore, $ \boxed{x \in \left\{ \minus{} \frac {2}{5},\frac {1}{2},2\right\}}$.
[/hide]
[hide="(c)"]
We will use casework. Suppose $ x^2 \minus{} 6 \ge 0$, we have

$ x^2 \minus{} 6 > 4x \plus{} 1 \implies x^2 \minus{} 4x \minus{} 7 > 0 \implies x \in \left(2 \plus{} \sqrt {11}, \infty\right).$

If $ x^2 \minus{} 6 < 0$, we have

$ \minus{} x^2 \plus{} 6 > 4x \plus{} 1 \implies x^2 \plus{} 4x \minus{} 5 < 0 \implies (x \plus{} 5)(x \minus{} 1) < 0 \implies x \in \left( \minus{} 5,0\right).$

Therefore, our solution set is $ \boxed{x \in \left( \minus{} 5,0\right) \cup \left(2 \plus{} \sqrt {11}, \infty\right)}$.
[/hide]

This is not middle school math.  It's much closer to "Intermediate Topics" than it is to middle school math.

[quote="sunehra"]
@PowerOfPi, you got invalid solutions because the first step you did was square both sides, which is inappropriate when working with inequalities with unrestricted variables.[/quote]

In general this is true, but in this case it is okay. The process of squaring is reversible in this problem. $ \sqrt {y} > 1 \iff y > 1$

modularmarc101's solution for problem (a) is (now) correct.

sunehra, you have a flaw in your solution at this point:
[quote="sunehra"]
The domain restriction is $ x\in\left[\frac {1}{2},2\right]$. Therefore, we know that the numerator will always be positive.[/quote]
This is not true.

Also sunehra, one of your solutions for problem (b) is incorrect.
And one of your intervals for problem (c) is incorrect.

For a visualization of the solutions to each of these problems, try here:
[url=http://www68.wolframalpha.com/input/?i=sqrt%28%28x-2%29%2F%281-2x%29%29%3E1]Answer to (a)[/url]
[url=http://www68.wolframalpha.com/input/?i=%7Cx%5E2-3%7Cx%7C%2B2%7C%3Dx%5E2-2x]Answer to (b)[/url]
[url=http://www68.wolframalpha.com/input/?i=%7Cx%5E2-6%7C%3E4x%2B1]Answer to (c)[/url]


linear algebra
matrix
linear algebra unsolved

Prove that in $\mathbb{R}^{3}$, a plane and a line not parallel to it must meet in exactly one point.

write the equations of the plane and the line
not parallel means that the rank of matrix corresponding to equations is 3.


Each square of a chessboard contains 0. It is allowed to select any two squares connected by knight move and to increase numbers in them by 1. Is it possible to obtain the chessboard filled with integers $1, 2, \ldots, 64$?

Consider a "knights tour" of the chessboard.  ([url=http://mathworld.wolfram.com/KnightsTour.html]Knights Tour Definition on Mathworld[/url])  Label the squares of the knights tour [1, 3, 4, 2], [5, 7, 8, 6], [9, 11, 12, 10], ...  where each set of 4 is the previous set where each element is increased by 4.

For each set of 4 squares, it is clear how one can add to get those values.  (1-3, 3-4, 3-4, 4-2, 4-2).

So it is possible.


Asymptote
trigonometry
logarithms
geometry
3D geometry
sphere
parameterization

Hello,

I'm trying to draw a pseudosphere in Asymptote. I would like to get something like in the attachment (in black and white, with axis).

I've tried to take tractrix curve (parametric equation $ (\pi/2,\pi \minus{} 0.01) \ni t \to (\cos{t} \plus{} \log{(|\tan(t/2)|)},\sin{t})$) and then draw revolution surface, but I couldn't get the result I want.

I would be grateful for any help.

[code]
size(0,200);
size3(200,IgnoreAspect);
import solids;

currentprojection=perspective(0,-5,10);

triple f(real t)
{
  return (sin(t),0,cos(t)+log(abs(tan(t/2))));
}
path3 h=graph(f,pi/2,pi-0.01);
revolution a=revolution(h,X);
draw(surface(a),green);
[/code]

Which is similar to your picture

In the latest Asymptote 1.77, splined parametric surfaces were implemented
and you can see the pseudo-sphere using this new functionnality

http://asymptote.sourceforge.net/gallery/trumpet.pdf
(Adobe Reader only for PRC)

O.G.


superior algebra
superior algebra solved

I'm studying the actions of a group. How to define the natural action of $ GL_n(\mathbb{R})$ if $ \mathbb{R}^n$? And how acts the group of orthogonal matrices $ O_n(\mathbb{R})$ on the space $ \mathbb{R}^n$?

$ GL_n(\mathbb{R}) \times \mathbb{R}^n \to \mathbb{R}^n: (A,x) \mapsto Ax$.


Let $n \epsilon N$. Show that $n^{4}+4$ is prime only if $n=1$

My Solution:

Check. n =1  ---> 5 which is prime

since $n \epsilon N$, for any n>1, $n^{4}+4$ is allways even.

Even numbers cannot be prime

hence, it works only for 1.

Is this correct? Does anybody have a more rigours argument?

$n^{4}+4$ is not even if n is odd.

[hide]
by euler's totient theorem:

$n^{4}\equiv 1 \mod 5$
$n^{4}+4\equiv 5 \mod 5$
$n^{4}+4\equiv 0 \mod 5$

though this doesn't work if n is divisible by 5... i'm stuck there.
[/hide]

[quote="tjhance"]could you please explain how you found it could be factored like that? i know that u can prove it my expanding that, but how did u fid it could be factored as such?[/quote]

Sophie Germain factorization

$a^{4}+4b^{4}= (a^{2}+2b^{2}+2ab)(a^{2}+2b^{2}-2ab)$

[quote="tjhance"]could you please explain how you found it could be factored like that?[/quote]

[hide="Well..."] Observe that $(n^{2}+2)^{2}= n^{4}+4n^{2}+4$.  Therefore

$n^{4}+4 = (n^{2}+2)^{2}-(2n)^{2}$ [/hide]

both $n^{2}+2n+2$ and $n^{2}-2n+2$ are greater than 1, so $n^{4}+4$ has more than one factor

edit: for all $n > 1$

[quote="iamagenius"][quote="t0rajir0u"][hide="One line"] $n^{4}+4 = (n^{2}+2n+2)(n^{2}-2n+2)$ [/hide][/quote]
sorry if this is a stupid question but how does this show the expression is prime only for $n=1$?[/quote]

$1^{4}+4 = 5$ and $5$ is prime!


Suppose you have a polynomial $ f(x)$, and you want to find the remainder when this is divided by $ (x\minus{}a)(x\minus{}b)$ for a given value of $ a,b$. How do I use synthetic division (no long division) to do this problem? You may plug in your own polynomial, $ a,b$ to illustrate your method.

I remember there was an AHSME problem that asks you to do precisely this.

See http://en.wikipedia.org/wiki/Synthetic_division for an in-depth look at synthetic division.

Let $ f(x) \equal{} (x\minus{}a)(x\minus{}b)g(x) \plus{} [cx \plus{} d]$. When $ x\equal{}a$, we have $ f(a) \equal{} ca \plus{} d$, and when $ x\equal{}b$, we have $ f(b) \equal{} cb \plus{} d$. As $ a,b,f(a),f(b)$ are known, this is a two equation linear system which is easy to solve for $ c,d$.


topology
advanced fields
advanced fields unsolved

Show that every locally compact hausdorff space is a baire space

We want to show that the intersection of a sequence of open dense sets $(G_n)_n$ is also dense.

Take an open set $U_0$. Since $G_1$ is dense, there is a non-empty open set $B_1\subset G_1\cap U_0$. Because of the fact that our space is locally compact Hausdorff, we conclude that there is an open set $U_1$ whose closure is compact and contained in $B_1$. Now, since $G_2$ is dense, we can find a non-empty open set $B_2\subset G_2\cap U_1$, and we consider a non-empty open set $U_2$ whose closure is compact and contained in $B_2$. I think it's clear how to proceed. The intersection of all the $G_n$'s contains the intersection of all the $B_n$'s, which, in turn, contains the intersection of the closures of all the $U_n$'s, but these closures form a descending chain of compact sets and thus have non-empty intersection.


Evaluate $ I \equal{} \oint_{C} \frac {z \minus{} 2}{z \minus{} i} \ dz$ where $ C$ is the positively oriented circle $ |z| \equal{} 2$. Can we just use the Cauchy integral formula and get $ I \equal{} 2 \pi i (i \minus{} 2)$. Or do we need to do the following:

 $ I \equal{} \int_{|z| \equal{} 2} \frac {z \minus{} 2}{z \minus{} i} \ dz$  $ \equal{} \int_{|z| \equal{} 2} \frac {z \minus{} i \plus{} i \minus{} 2}{z \minus{} i} \ dz$ $ \equal{} \int_{|z| \equal{} 2} 1 \cdot dz \plus{} (i \minus{} 2) \int_{|z| \equal{} 2} \frac {1}{z \minus{} i} \ dz$

So we invoke both the Cauchy-Goursat Theorem and the Cauchy Integral Formula. Or is just using the Cauchy Integral Formula enough?

I'm confused. Why do you have $ 2\pi i *(i \minus{} 1)$? Isn't the residue $ i \minus{} 2$?

Yes, it's correct. What would be wrong with the Cauchy Integral Formula?


A friend gave me the following problem :
Let $(a_n)$ be a sequense of positive reals such that $a_n \rightarrow 0$,$n \rightarrow \infty$ and the series $\sum_{n=1}^{\infty}a_n$ diverges. Prove that for any distinct positive reals $a$,$b$ with $0\leq a\leq b$ there exists integers $n$,$m$ such that $a\leq S_n - S_m \leq b$, where $S_n  =  \sum_{i=1}^n a_i$ .
I found a proof 2 lines long , Can you try it ?

P.S How can i write $a$ <$b$ ? NOT $a\leq b$.

Just take $m$ so that $a_k<b-a,\ \forall k\ge m$, and then take $n$ to be the first $k$ for which $a_m+a_{m+1}+\ldots+a_k>a$.

P.S. You can simply put the < between the dollar signs and you'll get $<$. You could have tried that.. :)

That's mine, too! In fact this idea is used to prove Riemann's theorem in non-absulute convergant series.
But  there is no reason for considering that:
[quote]Just take  so that  $a_k<b-a,\ \forall k\ge m$[/quote]

[quote]P.S. You can simply put the < between the dollar signs and you'll get . You could have tried that.. [/quote]
 :blush:  :blush:


For non-negative real $a,b,c,d$ such that

$a^2+b^2+c^2+d^2=1$ , prove that

$a^3(b+c+d)+b^3(a+c+d)+c^3(a+b+d)+d^3(a+b+c)\leq\frac{3}{4}$

Remark : I conjecture that for $n$ variable $a_1,a_2,\cdots a_n$ such that $\sum a_i^2=1$ then

$\sum_{sym}a_1^3a_2\leq\frac{n-1}{n}$

is this true ?

 :)

[quote="arqady"][quote="shyong"]For non-negative real $a,b,c,d$ such that

$a^2+b^2+c^2+d^2=1$ , prove that

$a^3(b+c+d)+b^3(a+c+d)+c^3(a+b+d)+d^3(a+b+c)\leq\frac{3}{4}$
[/quote]
But it is equivalent to $\sum_{sym}(a-b)^4\geq0.$[/quote]

yes arqady , what do you think about the general one ? Is it true ? I dont have any proof yet :)

[quote="arqady"]But it is equivalent to $\sum_{sym}(a-b)^4\geq0.$[/quote]
If find it was $\sum_{sym}(a+b-ab)(a-b)^2 \ge 0$!?!, obviously.

[quote="Lovasz"][quote="arqady"]But it is equivalent to $\sum_{sym}(a-b)^4\geq0.$[/quote]
If find it was $\sum_{sym}(a+b-ab)(a-b)^2 \ge 0$ !?![/quote]
What do you mean? I don't understand.

[quote="Lovasz"]Homogezing and gathering give $\sum_{sym}(a+b-ab)(a-b)^2 \ge 0.$[/quote]
Sorry, where homogezing?

Ah, thanks, it's not. That seems I forgot something, maybe at home.

[quote="arqady"][quote="shyong"]

Remark : I conjecture that for $n$ variable $a_1,a_2,\cdots a_n$ such that $\sum a_i^2=1$ then

$\sum_{sym}a_1^3a_2\leq\frac{n-1}{n}$

is this true ?

 :)[/quote]
I think $\sum_{sym}a_1^3a_2\leq\frac{n-1}{n}\Leftrightarrow\sum_{i\neq j}(a_i-a_j)^2(a_i^2-(n-2)a_ia_j+a_j^2)\geq0,$
 which wrong for $n-2>2.$[/quote]

thank you arqady , and yes , the general one is wrong  :(


A lot of people know the myth of Midas...

So is it a Greek myth(in the greek section of the mythology book), Arabian(Watch Disney's "Aladdin and the King of Thieves"), or English?

Midas is from Greek mythology. According to one story, he was given his power by the Greek god Dionysus after finding Silenus, Dionysus's foster father. Here's [url=http://www.pantheon.org/articles/m/midas.html]a link[/url] with more information if you want it.

The wine god was called bracchus. I know because I once acted bracchus in a classroom play while in the sixth.

Actually, Bacchus is the Roman name for Dionysus.


AMC
USA(J)MO
USAMO
induction
number theory

Ok, we can discuss stuff now. How'd everyone do?

I did 1, 4, and 6. Hopefully partial marks for 2,5.

I'll probably get about 16 marks between 1, 2, 4, and 5.

I got 1, and prolly partial for 3 and 4.
I'd be satisfied with a 14.

lol...i did 1, 2, the lesser half of 3(where one of the classes has fewer than n sets), 4, used induction to prove 5...almost got it...but found out i messed up the induction step in the end, and the very lesser half of 6(the equal part)...i think thats pretty good for a first timer... :D

is the marking for usamo very harsh?????i heard all u could get is 7,6 or 1,0...grrrrr...i tried last years usamo and i thought i was gonna get a 0...i think this year was fairly easy... :|

I wrote unofficially and only fully solved 1 and 4.  My work for 2 and 5 might be worth 1-3 points.  I found this harder than last year.

[quote="rem"][quote="Elyot"]USAMO, where 80% of the problems are extremely ad-hoc, meaning that you just stare at something completely new having no idea what to do with it.[/quote]
It was 100% ad-hoc this year. So I got killed by usamo.

I should have done question 2...

I just hope i will get selected for imo team.....[/quote]
wut?????this year's was relatively easier...although it took me like 15 minutes to figure out wut 4 means...last year's was a beast...it scared me and i had a thought that i would get a 0 this year...

btw...do u get advantage for imo team if u did well on usamo?...cuz they r both 9 hour 6 problem type... :huh: ???

#1 I got a 7 for sure, i wrote everything down very neatly.
#2 I'm not so sure... some sketchy stuff (I couldn't prove what the size of the biggest equilateral triangle that doesn't hit a grid point is, etc.). Probably 6, I'm not sure how hard they mark it - maybe 5
#3 I got 0. Nothing at all :S i stared at it for over an hour and couldn't do anything.
#4 I solved, and my explanation is clear, but perhaps not as organized as my #1. Probably 7, i dunno how hard they mark it...
#5 I did with induction, and got to the point where i have to prove something like x^6-x^5+x^4-x^3+x^2-x+1 is never prime where x=7^(7^n). I couldn't do that... I'm not sure how much my induction is worth. Somewhere in the range of 2-4 marks.
#6 I did only the equality case :) probably 1 or 2 marks.
total = 21 to 26 i think
yay i passed :D

[quote="aasdfg1234"]if you win can you write the TST, if u win the TST can you be on USA team if you are a Canadian???[/quote]

No. But, if you're good enough to make the U.S. team, then you will have surely made the Canadian team.

[quote="probability1.01"]Hmmm so will I be right? No Canadian winners? We shall see :o[/quote]

Probably not. However, most of the best problem solvers don't post here. However I believe that last year Lin Fei got Honourable Mention and he wrote again this year, so we'll see. Dimitri Dziabenko might also make it.

can we discuss the details of the problems now? I want to know how you can prove 8025 is the max number for #4.

Cynthia totally pwned this year. Congrats!
I believe she will go to IMO and pwn it too.

rofl
i'll see how cynthia reacts when she is invited to the imo

Congrats Alex and Cynthia!
I can see an overwhelming team this year!  :clap:  :flex:

[quote="cynthia1991"]I am already super excited with my USAMO result. I think I might laugh to death if I ever got invited to IMO...But I think I still have a long way to go to be good enough for contests like IMO.[/quote]

c'mon man! do some trainings!  ur too lazy
if u want..i can give u problems to do

Cynthia, you are too modest. You are one one phone call away from IMO. :D

I am sure you will also do good on IMO.

lol thats true..she said the same thing about USAMO and how she wouldnt do good..and she got Honourable Mention..

aww scores are out i only got 11 gahhh i overestimated quite  some bit lol

Good job Alex! we are proud of you at WCI
Our school is so bad at math this year... you shouldn't have left :(

haha I got 14... I overestimated my score so much :blush:

9, worst score so far.
I thought my #3 was rigorous, but the official solution is quite different.


Recently, Ive been attempting a lot of problems from the mathlinks sections, but last week I picked up an aime and started going through it. I found that the same problems that I used to get the "2-liner" solutions for Ive now been just trying to pound through and apply the same techniques as always, no application of creativity or ingenuity. Its like the harder problems have dulled my creative blade, so to speak. I had the same feeling once before, just before the aime this year, when I had practiced on about 10 aime's in 2 weeks. It feels like a cloud is over my mind.

Does anyone else get this? Specifically, I mean the creative curtailing.

This is a very good question. I think the loss of creativity can come from being overly concerned about getting a technically flawless answer. My wife the piano teacher occasionally encounters students who worry so much about not making technical mistakes that they lose their musicality in playing. Similarly, when my son A+MATH was taking his first geometry course last school year (the JHU CTY distance learning course, using the Jurgensen-Brown-Jurgensen textbook), it seemed to me that he LOST insight into SAT geometry problems as the course progressed, because the course exercises were too fussy about steps to the solutions, but not deep enough to really stimulate his thinking. He fared better on the SAT after beginning to take the EPGY geometry course, which has more interesting and more thoughtful problems.

What you need is what many Russian and Chinese math textbooks are so good at providing: problems that are "easy" for you from the point of view of what curriculum prerequisites they have, but "hard" in that they demand thinking and creativity. Probably if you drop back one level in AMC-type contests and review old problems in a test level that you ought to know already, you can practice rebuilding your spark of insight. Check your solutions to the official solutions, and see if you can come up with alternative solutions. Reading the Russian textbook series like the Gelfand series, or any Chinese-language textbooks you can find, might help too. (The [url=http://www.singaporemath.com/products_math.htm#Math%207th%20-%2012th%20grades]textbooks from Singapore[/url] are in English, but largely in the Chinee manner of presentation.)

Hope this helps! I'll be interested to see what other participants recommend.

Look at http://www.kalva.demon.co.uk - all soviet union math competitions [ASU]. The first several olympiads are just what you need. Solutions to many problems are awaliable.


or you could give spammers a max number of posts that they can post per day and they can't post any more than their daily limit. that will teach them to think about each post before making it and hopefully their posts will become much more meaningful

Instead of having to decide who is a "spammer" and thus should have a max post count, it might be better to allow moderators to mark a post as "Spam." This tag would be visible to everyone, so they would know what kinds of things are getting marked as spam. When a single user starts getting too many posts marked as spam, more and more serious restrictions on that user's activities come into effect.

Unfortunately, this would require a fair amount of work to implement, I think.

I actually don't think that it would take too much more work. It is fairly obvious which posts are spam, and if there is just a little button that mods could press to mark things as spam, than there should be no problem.

What is the big deal about post counts anyways? Is there a special prize for getting 1000 posts or something stupid like that?

[quote="mustafa"] It is fairly obvious which posts are spam, and if there is just a little button that mods could press to mark things as spam, than there should be no problem. [/quote]
I meant it would be a fair amount of work to implement the system, not to run it once it's implemented. The forum code would have to be modified so taht this extra button was there and did all the stuff we wanted, etc.... There's probably already a long list of stuff to do, also.

[quote="thatin"]Why not allowing users to decide if something is spam?
How about a spam-indicator of each post?[/quote]
I considered that, but I was concerned that a group of very active users would be able to keep each other's posts "non-spam" by always voting positively for each other, and possibly even vandalize other people's good posts by marking them spam when they're not. That's not to say I think our community has such a large malicious element, but you have to think ahead (so that you don't have to change things again later) and I just feel like the user-moderated version has potential problems.
Look at Wikipedia -- most of the time, it's great, but occasionally there are some really serious problems. Remember the Colbert Report incident? The Steve Irwin incident? When I was writing a paper about the internet last year, I also remember reading about an incident where some minor politician or lawyer or something had a page which contained allegations of his involvement with the Kennedy assassination or something, and it had stood uncorrected for months (the guy actually had nothing to do with it). So there are flaws in the system.

[quote="Xevarion"]When I was writing a paper about the internet last year, I also remember reading about an incident where some minor politician or lawyer or something had a page which contained allegations of his involvement with the Kennedy assassination or something, and it had stood uncorrected for months (the guy actually had nothing to do with it). So there are flaws in the system.[/quote]
I think he was actually the editor of a local newspaper, in Tennessee?  The relevant articles I've found are from the New York Times, December 4, 2005: "Rewriting history: snared in the web of a Wikipedia liar," and December 11, 2005: "A little sleuthing unmasks writer of Wikipedia prank," both by Katharine Q. Seelye. (You need to pay to view them, or to be a Times Select member, or have Lexis-Nexis access.  So, no links or complete text from me.)

Of course you are right Xevarion. And majority is not always right. So how about a user's spam-indicator + moderator's spam indicator? Certainly it is a bit complex... :)

[quote="chess64"]It's not hard to implement. I could do it easily, so Mr. Vornicu should have no problem doing it.[/quote]

In principle, it should be easy, but have you looked at the mess of code that's phpBB?  Of course, Mr. Vornicu's probably used to it with all the modifications he's made...

I see that you people have a point here, but can I make something clear? I don't give a ... about post count.
If you think I spam sometimes, can't you ask me to stop nicely before making a topic that is directed against me?  :maybe:

It's not directly only against you. Even if you assume that ccy meant that the two links he posted referred to the only spammers on the forum, you're still only 50% of the target of his post. But a lot of people occasionally post totally meaningless worthless stuff, and there are a few people who regularly post trash and never anything useful (I'm not sure if you're one of those since i don't usually read the forums you post in). There's nothing wrong with informality and friendliness and fun in math -- that's why this is such a lively community. But it starts to become a problem when people are trying to have discussions about math and 75% of the posts in a thread don't have any actual math content.

One thing I hope to implement in the next few months (this depends in part on when we can get Val's visa work finished so he can join us in CA and start working full-time) is a post/thread rating system.  I hope we might even be able to wrap slashdot-like filters around it.  Regardless, it will encourage (I hope) more thoughtful posting, as these ratings might be used to create user ratings, as well.  I welcome your thoughts on the matter.

I am sorry, but my account was actually recently overtaken by a few hackers, but I changed the password, because some posts in some forums were actually ones that I didn't post, so you won't see spam from me any more.

[quote="ccy"]It seems that ALL of the forums started to get spam after posts in G&FF were removed from the post count.  In particular, I'm getting annoyed by post-count-freaks like anirudh and i like pie that post like [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=633890#633890]replies like this[/url] and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=634008#634008]this[/url] that ARE OF ABSOLUTELY NO CONTRIBUTION TO THE TOPIC.  I suggest that the "Posts" be removed from the side of the page when reading a message and that G&FF posts be added back to the post count.  That way, spammers can have a place to spam as they wish and not affect the quality of the math-related forums on the board.

-ccy[/quote]

i don't believe they are spamming. they are just posting. if you want to see spam go to luridphantom.com/forum and look up each thread... that spam makes these posts look liek good posts

I'm against post count in G&FF.
And I still don't see the anounced stronger rules there and in the rest of the forum, as bubkas new signature thread shows. They all started to spam again, and nothing was done against spamming users, just threads/posts got deleted.

I suggest making a pillory:
Everyone that spams gets marked there. If one spams more than three times in a month, he gets banned for a week or so and his name is put on a list (visible for anybody).
The next ban will be longer, then permanent.

If the first people are put there, I think most others will stop spamming.

@rating of posts/threads: I don't think it is a good idea, it just makes some competition between the "older" users, but will surely be neglected by those that spam.

[quote="ZetaX"]@rating of posts/threads: I don't think it is a good idea, it just makes some competition between the "older" users, but will surely be neglected by those that spam.[/quote]
On bash, whenever someone re-submits an old quote, the new copy gets voted down really quickly. If people vote down repeat posts of problems, it might help deal with the way that the forum gets clogged by multiple posts of the same problem. (Maybe not though.)

A high amount of repeated topics comes from new users, so I doubt it would help.

[quote="ZetaX"]

@rating of posts/threads: I don't think it is a good idea, it just makes some competition between the "older" users, but will surely be neglected by those that spam.[/quote]

Imagine this were coupled with filtering, so you didn't have to wade through garbage.  Then would it be worthwhile?

gff doenst have vote count for a good reasons

for example, tof isnt anythin but numbers, and adding tat to ur post count... BAD IDEA

That's actually what I was thinking of when I made the comment about repeat posts.

edit: I was answering rrusczyk's comment about filtering.

I'm just wondering, how did we come to define "spam" as "posts from regular users that are unhelpful or contribute to clutter"?

We do occasionally get some [i]real[/i] spam here: commercial advertising; links to inappropriate outside websites; things utterly irrelevant to the purposes of this site. There are enough moderators and administrators with their eyes open that that stuff disappears pretty quickly. Judged against that, unhelpful or cluttering posts are much lesser offenses.

I'll admit that I frequent mostly the more mature portions of the site, but I tend to take a more tolerant attitude towards what people post. So the problem has been posted before? If it's truly been a while, what's the real harm in posting it again? The problem is new to someone. (And yes, harazi or Moubinool or someone else who's been here a long time and has a good memory may know what to look for and link to - but it may only be a few people like that for whom the problem doesn't seem new.) If the problem was just posted 3 days ago, or has been posted so often that everyone is sick of it - well, that's different. But we don't have to be so quick to dismiss things on "that's been done" grounds.

Sure, posters do many things that can annoy, for most of which there may be no automatic answer. Some, like double-posting or being impatient ("Why has no one answered this? It's been up for an hour!") should draw responses that direct the poster toward our standards of appropriate behavior - and mostly the new ones do get the message. Others, like vaguely-stated or poorly-written mathematics? That's the perpetual struggle for all of us.

On the "College Playground" forums, I'd like for the "Computations and Tutorials" sections to be tolerant of and welcoming for newcomers who don't know the rules. But - this is a still a place of mathematical rigor. Errors should be corrected; posters shouldn't automatically expect a pat on the back. (See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=112297]here[/url] for an example of "Not exactly a pat on the back." But I wasn't saying anything I wouldn't have been willing to say to a student sitting in my office.)

[quote="akalra1"]for example, tof isnt anythin but numbers, and adding tat to ur post count... BAD IDEA[/quote]

Keep in mind that G&FF was actually pretty spam-free for a while there. A major reason that it went to post-count-free was because some people thought that completely pointless, bogus, and otherwise nonsensical threads needed a home. And now that it is post free, we allow threads that aren't "anythin but numbers". Point is, however, that the argument isn't really for v. against having these threads count as posts; rather it's about allowing these threads at all.

And to stay in topic...somewhat...I do believe a rating system would be good.

what types of threads were there? and how do you get rid of post count?

the admins probably used some phpbb tool...some modification to the forumsoftware

yeah its possible to kill post count... i've seen it done. just edit the site thing in the phpbb admin panel or something. also admins can change other's post counts.

[quote="junggi"]guess what? we have a new(and worse, imo) spammer in town(*cough moogra cough*).[/quote]
He's worse than anirudh.

[quote="moogra"]too bad. im posting again

and if you delete, im gonna spam even more

plus give me the week 2 problems. i want to do them[/quote]

From [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=112276[/url].

:furious:

[quote="moogra"]guess who made me post more: junggi[/quote]
Who cares what made you do it, the point is: you spam.

And did junggi make you post that, too?

[quote="moogra"]guess who made me post more: junggi[/quote]
dude, wtf?

[quote="moogra"] :lol: ok so? i spam on useless boards. not on godo ones[/quote]
um I'm pretty sure mathcounts=good, moems=good, HSB=good. and i dont really want to see bay area forum spammed

[quote="junggi"][quote="moogra"] :lol: ok so? i spam on useless boards. not on godo ones[/quote]
um I'm pretty sure mathcounts=good, moems=good, HSB=good. and i dont really want to see bay area forum spammed[/quote]
Yeah, I really don't think those are classified as "useless."

[quote="moogra"]:lol: ok so? i spam on useless boards.[/quote]
And you're happy about it...

[quote="moogra"]i actually post solutions there. i post useless stuff on bay area because no one cares about it.[/quote]
Don't post useless stuff [i]anywhere.[/i] And some people [i]do[/i] care about Bay Area, even if you don't give a - about it.

moogra explain [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=113067]this[/url],
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=642502&search_author=moogra&show_results=posts&sort_dir=desc&return_chars=200#642502]this[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=112825]this[/url]......
i could go on forever you know

only the first one and that was an accident. i didn't know i posted it already and if you keep posting, that'll just make me reply for a high post count, because this is getting lame. :mad:

[quote="moogra"]only the first one and that was an accident. i didn't know i posted it already[/quote]
???
in case you didn't know, posting solutions [i]identical [/i]to the ones already posted, are considered to be spam. and since ALLLLLLL of your "solutions" were already posted before, you rarely make[i] any[/i] good posts.

I think that rating is a good and useful idea. It will help everyone to see only the best topics if he/she has not plenty time.

omg, this is really starting to get annoying. :mad: we never had this problem before. now everyone is like always "spamming" and we're always talking about how to get rid of "spam". why don't the mods just delete the "spam" posts when they see them? where are the mods? i think we need new mods maybe? or maybe we just need more mods?

I think this discussion has degenerated into a shouting match, so I'm going to lock it up.

We will discuss the situation later this week and perhaps issue new guidelines to the moderators.


function
real analysis
calculus
integration
logarithms
real analysis unsolved

$ \left(1\right)$ If $ S\subset\mathbb{R}$ is open and $ f : \mathbb{R}^n\rightarrow\mathbb{R}$ is a measurable function, then $ f^{\minus{}1}(S)$ is a measurable set.

$ \left(2\right)$ Give an example of a measurable function $ f$ on $ [0, 1]$ such that for every open interval $ I$ the function $ f_{\chi_I}$ is not Lebesgue integrable.

(1) Write $ S$ as disjoint union of open intervals and use the fact that inverse images play well with set operations.

(2) Let $ (r_n)$ be some enumeration of rational numbers. The function $ g(x)\equal{}\sum_n 2^{\minus{}n} \frac{e^{\minus{}|x\minus{}r_n|}}{\sqrt{|x\minus{}r_n|}}$ is integrable on $ \mathbb R$, hence finite a.e. However, $ f(x)\equal{}g(x)^2$ is not integrable on any nontrivial interval.

[quote="mlok"](1) Write $ S$ as disjoint union of open intervals and use the fact that inverse images play well with set operations.
[/quote]
Could you please be more specific. As far as I know "$ f$ - measurable function" means
\[ \forall R > 0,\ [f]_R(x) \equal{} \left\{\begin{array}{ll}f(x) & \parallel{}x\parallel{}\leq R\ \&\ |f(x)|\leq R \\
0 & otherwise \\
\end{array}\right.
\]
is L-integrable and a set is measurable if it's characteristic function is measurable ... so ... I can't relate very well to your saying.

[quote="mlok"](2) Let $ (r_n)$ be some enumeration of rational numbers. The function $ g(x) \equal{} \sum_n 2^{ \minus{} n} \frac {e^{ \minus{} |x \minus{} r_n|}}{\sqrt {|x \minus{} r_n|}}$ is integrable on $ \mathbb R$, hence finite a.e. However, $ f(x) \equal{} g(x)^2$ is not integrable on any nontrivial interval.[/quote]
I understand that $ g(x): [0,1]\rightarrow\mathbb{R}$ as a series converges almost everywhere (i.e. on $ [0,1]\setminus\mathbb{Q}$) but why exactly is this $ g$ L-integrable and why its square (and not g itself) is not L-integrable on any non-trivial interval I can't quite see ...

I will think more about it but if somebody could make it clearer I would appreciate it.

Edit: At $ (2)$ I see now why $ g$ is L-integrable  :blush: but why is it that $ g^2$ is not on any non-trivial interval is still beyond me ...

That's a pretty convoluted way to define measurable. There are much simpler definitions of measurable function in terms of the inverse images of half-infinite intervals being measurable.

As for the second example: when asked to integrate a sum, always consider the information gained by interchanging sum and integral. (For sums of nonnegative functions, this is justified by the Monotone Convergence Theoerem.)

As for the lack of square integrability: the whole sum is greater than any single term. What about the integral of the square of a single term?

[quote="Kent Merryfield"]That's a pretty convoluted way to define measurable. There are much simpler definitions of measurable function in terms of the inverse images of half-infinite intervals being measurable.[/quote]

I will get better informed but still I don't get how the problem would be solved by that sole remark mlok gave.

[quote="Kent Merryfield"]As for the second example: when asked to integrate a sum, always consider the information gained by interchanging sum and integral. (For sums of nonnegative functions, this is justified by the Monotone Convergence Theoerem.)

As for the lack of square integrability: the whole sum is greater than any single term. What about the integral of the square of a single term?[/quote]
I see now. That integral would involve a natural logarithm therefore leading to a contradiction.

Concerning (1) - can you prove that the special case $ S\equal{}(a,b)$ ? If you can do that, the rest follows from the fact that a countable union of measurable sets is measurable.


real analysis
function
real analysis unsolved

Lebesgue decomposition decomposes a sigma finite measure into a singular measure + absolutely continuous measure.

I remember there is a similar result on monotone increasing function. (= singular function f' = 0 a.e. + absolutely continuous function)

Is the latter result a special case of the first one? Is there any connection between the f'=0 a.e. property on $ \mathbb{R}$ with the one for measure? I don't quite get what singular measure mean yet.

The latter result is almost a special case of the former. A function constructed from a measure $ \mu$ either by $ f(x)\equal{}\mu((\minus{}\infty,x))$ or $ f(x)\equal{}\mu((\minus{}\infty,x])$ is not just nondecreasing, but also continuous from one of two sides (left- or right-). Monotone functions like ($ f(x)\equal{}0$ for $ x<0$, $ f(0)\equal{}1/3$, $ f(x)\equal{}1$ for $ x>0$) are not naturally associated with a measure.

Anyway, a measure $ \mu$ is singular with respect to Lebesgue measure if and only if the associated function $ f$ satisfies $ f'\equal{}0$ a.e.


Why is that native English speakers pronounce "Beijing" as something like "Bayzhing" where "zh" is like "s" in "measure" rather than like "j" in "jam", as I would think to be more natural? Using "j" as in "jam" is also more accurate and closer to the Mandarin pronunciation.

Hmm...

[hide]And even us Chinese-Americans will go "Bayzzzzzzzhhhhhhing" unless we're saying it the Chinese way with tones and all...[/hide]

(IPA: Why /beɪˈʒiŋ/ instead of the more obvious and natural /beɪˈdʒiŋ/?)

I read in an article a few weeks ago that some people thought that Bei-jing sounded like an ethnic slur so they pronounced it Bei-zhing.

I can give two references:

For the true pronunciation of "Beijing" in Chinese, see this Language Log post:
[url]http://languagelog.ldc.upenn.edu/nll/?p=502[/url]

For a discussion of how the English pronunciation thereof varies, see the article referenced therein, Bosat Man, "Backhill/Peking/Beijing", Sino-Platonic Papers #19, 1990:
[url]http://www.sino-platonic.org/complete/spp019_peking_beijing.pdf[/url]

Thanks JRav and Osud.  :)

@JRav: "Some people"? Does that account for every single English-speaking person I have ever heard say "Beijing"?

@Osud: Very interesting articles. I was attracted after pronouncing 比较 bi3jiao4 to myself and finding that I had said it more like bi3yao4, just as the article predicted. It seems that the article points more to something like IPA /beɪˈjiŋ/ as opposed to /beɪˈʒiŋ/, as to what a foreigner might say trying to imitate Beijingers.

btw what is the convention for the use of // and [] again?

I am not really very well informed in these matters... is there some sort of linguistic connection between approximant [j] and the fricative [ʒ]? Looking on an IPA chart, they don't seem to be very close. And yet with the Spanish "ll" having those very two sounds as possible realizations, and now this... hmm... If it were the case that they were related, then my question would be answered. Without this connection, I still see no very good reason as to why people say "bayzhing".

As for the second article, it seems a little outdated (I don't think I've ever heard any English-speaking person say "Peking"), but still interesting - especially the part about the historical connection between Sinitic and Indo-European languages. Thanks for bringing back fond memories of my conlanging days (wow I'm speaking like some old guy now), without which the latter article would be rather inaccessible to me.

[quote]btw what is the convention for the use of // and [] again?[/quote]

[] for phonetic transcriptions-- the exact record of what would come out of someone's mouth.  // for phonemic representations-- the concept of the sounds in that word that exists in someone's head, minus any automatic sound rules for that language.  So, for an English speaker, the phonemic representation of "cat" would be  /kæt/, whereas a broad phonetic transcription of that word would be [ˈkæt], because English has a sound rule which aspirates all syllable-initial unvoiced stops.  Does that makes sense?  If not I can give more examples.

[quote]I am not really very well informed in these matters... is there some sort of linguistic connection between approximant [j] and the fricative [ʒ]? [/quote]

In terms of phonological feature theory, [j] and [ʒ] only differ by two values, with [j] being [-cons, +son] and [ʒ] vice versa.-- basically the difference between being an approximant and a fricative.  Place of articulation is largely the same-- they're both palatal-- though [j] can also have coronal features.  Try it out yourself-- if you're tongue is closer to your palate, you'll get turbulent airflow and the fricative [ʒ], but if your tongue is slightly lower you'll get smooth airflow and the approximant [j].

[quote]As for the second article, it seems a little outdated (I don't think I've ever heard any English-speaking person say "Peking"), but still interesting - especially the part about the historical connection between Sinitic and Indo-European languages. Thanks for bringing back fond memories of my conlanging days (wow I'm speaking like some old guy now), without which the latter article would be rather inaccessible to me.[/quote]

You're welcome!  But-- you've never heard anyone say "Peking duck"?  Obsolete toponyms tend to hang around in odd places.  Dalmatians, Siamese cats, Ceylon porcelain/Ceylon blue sapphires...


Let $ x,y,z \in \mathbb R$ such that $ x\plus{}y\plus{}z\equal{}xy\plus{}yz\plus{}zx$
Now,Find the minimum value of this expression $ A\equal{}\frac{x}{1\plus{}x^2}\plus{}\frac{y}{1\plus{}y^2}\plus{}\frac{z}{1\plus{}z^2}$
I think the result is $ \minus{}1$,Am I wrong?


The $ 2007\ \text{AMC}\ 12$ contests will be scored by awarding $ 6$ points for each correct response, $ 0$ points for each incorrect response, and $ 1.5$ points for each problem left unanswered. After looking over the $ 25$ problems, Sarah has decided to attempt the first $ 22$ and leave the last three unanswered. How many of the first $ 22$ problems must she solve correctly in order to score at least $ 100$ points?

$ \textbf{(A)}\ 13 \qquad \textbf{(B)}\ 14 \qquad \textbf{(C)}\ 15 \qquad \textbf{(D)}\ 16 \qquad \textbf{(E)}\ 17$

[hide]she needs 100 points.

You can subtract 4.5 from the 3 blanks she is going to leave.

100-4.5=95.5

16x6 is the smallest that goes over 96.

so 16 is the answer[/hide]

[hide="Solution"]Think about like this. If you calculate the score when you get x, to find the score of x + a, all you do is add 6a. Trying 13 yields a score of 82.5. Therefore, 16 will give you 100.5. The answer is [b]16[/b].[/hide]


Point P is randomly picked from within square ABCD. What is the probability that AP + BP < DP?

I think this was posted somewhere else, anyone remember?

Yes it sounds familiar. I tried a solution, it's incomplete and I don't know if someone else has a simpler solution.

Let the coordinates be: A = (0,1) B = (1,1) C = (1,0), D = (0,0). Point p has coordinates (x,y).
Then you can find the lengths of the segments.
AP = $ \sqrt{y^2 + (1-x)^2} $
BP = $ \sqrt{(1-y)^2 + (1-x)^2} $
and DP = $ \sqrt{y^2 + x^2}. $

So if AP + BP < DP, then
$ \sqrt{y^2 + (1-x)^2} + \sqrt{(1-y)^2 + (1-x)^2} < \sqrt{y^2 + x^2}. $

It looks as though if you square that stuff will go...

This turned out to be a rather nasty problem. I posted a solution to it once before on here. Try a search for it.

By the way, where did this problem come from?

Heh, yeah I posted this here a while ago.  http://www.artofproblemsolving.com/Forum/viewtopic.php?t=29997

(if you're wondering, the other question it is referring to is the race probability one)

and to answer gauss202's question, he made it up.

Oddly enough, your not the first one to have said that!!! Thanks!


It's a problem from M&Y magazine!No solution for it!  :(

It's obvios!But i think it's not easy :)

If this ineq is true, I'd like to see a solution.
Straightforward application of Cauchy is useless.

Really nice solution :)
But how can you prove the last line (you wrote using Cauchy, but I can't see it)
I must work to improve my Cauchy-using skill ;)


geometry
geometric transformation
reflection

Who's going to science bowl this year?

First time for our school, BVW, KS.

piano, you're *so* lucky to be going to NSB. It is totally a BLAST; don't pay any attention to what mysticterminator says about it being smelly. I'm not actually going this year, but I wanted to go back there so badly. Unfortunately we came up short in the last 45 secs in the semifinals at regionals  :mad: .

IMO, NSB is actually better than national Mathcounts. It's 5% competition and 95% fun, *and* you get to meet people who you normally wouldn't. NSB was actually my first time to Washington, D.C., and the tour of the monuments at night by the reflection pool was absolutely incredible. That's part of the reason why I wanted to go back badly, plus that I wished I had talked to more people and not stuck to the team so much (and asked the Secretary of Energy something).

Do you (or does anyone) know what happened with the fuel car competition this year? Before, they had one and even gave out around 9,000 dollars as prizes. And is there still a civility award this year? (Think "iPod," think "trip," and help out Sue Ellen :lol: .)

Our school team (which does not include me, because I am lame) is going to nationals. Lexington High School, MA.

[quote="pianoforte"]That's too bad Emton... I'll take your advice to heart.

I think, they added the Division Team Challenge, where you, uh, do short science activities for points that do count (as tie breakers? idk) in the actual competition... something like that.

Sue Ellen from Arthur?[/quote]

I'm pretty sure they had the Division Team Challenge before, except they called it the "Interactive Science Discovery (?)" (it was the "ISD" in any case). They used it as a tiebreaker, and one of the divisions was super competitive and had to use the ISD to break a three-way (?) tie for 1st place (Thomas Jefferson, E.O. Smith, and one other team (was it Poudre?)). If I remember correctly, TJ didn't even qualify for the double-e round because of a lower ISD score! (I was actually predicting that they would win.)

piano--I'm not sure what/where Arthur is, but you'll know Sue Ellen as soon as you get there. Is that tomorrow, BTW?

In any case, have a wonderful time  :lol: !

The show Arthur, based on the children books by Mark Brown? Arthur the aardvark? With Buster, Francine, The Brain, Binky, Prunella, DW, and Pal? Should I ramble more?

Thanks!

Oh, duh  :oops: . Yeah, I haven't seen an episode of that show for several years now, I think.


calculus
integration
real analysis
real analysis unsolved

[quote="jmerry"]Stop repeating your titles. How are people supposed to tell apart your three first-page topics titled "summation" or your four titled "integral"? These titles are too generic and give no real hint at the content.[/quote]

 :rotfl:

you are right, OK, i will try and categorize the topic(s) more from now on...  :)


Let $ x,y,z$ be positive real numbers, then prove:

$ 4 \left (\frac{x^2}{y}\plus{}\frac{y^2}{z}\plus{}\frac{z^2}{x} \right )\plus{}2(x^3\plus{}y^3\plus{}z^3) \ge 5(xz\plus{}yz\plus{}xy)$.

Nice proof !I don't understand why stay a day posted whitout to be solved! Was easy...! :lol:


function
number theory
prime factorization
prime numbers

First, I'll put the question that needs concept understanding in algebra II.

1. Prove that y=x+1 is inverse of y=x-1 while y=x-4 is not inverse of y=x+2 in terms of x.

Factoring.

2. How many factors are in 559 872?

Wonderful. Correct.

Especially for number one, one rule on inverse of function is that if f(x) is inverse of g(x), then f(g(x)) should be equal to x  and g(f(x)) should be equal to.

So, just addition to that:

[tex]f(g(x) = f(x+1)[/tex]
[tex]= (x+1)-1[/tex]
[tex]= x[/tex]

[tex]g(f(x) = g(x-1)[/tex]
[tex]= (x-1)+1[/tex]
[tex]= x[/tex]

I got a question for number 2.  I understand the concept and how to solve those kind of problems from now on but what happens when we get a number like 286488.  I prime factored it and I got [tex](2^3)(3^3)(3979)[/tex].  How would we find the number of factors in this case.

Also, another question is how can we tell if a number is prime or not prime?

[quote="kool_dudy"]Also, another question is how can we tell if a number is prime or not prime?[/quote]

Not easily.  By hand, the only good way is to check every prime up to its square root.

And I thought there was some sort of way to determine if a number is a prime or not.

There are lots of ways, including the one I said.  However, there is no "nice" way to do it by hand other than what I described.

As for your question about how to find the number of factors of a number, given its prime factorization, try to find a pattern.  Rep showed how to do it for any number that is the product of a power of 2 and a power of 3.  Try to figure out how many factors each of the following numbers has, and generalize:

(1) 20 = 2^2*5
(2) 30 = 2*3*5
(3) 120 = 2^3*3*5
(4) 180 = 2^2*3^2*5
(5) 210 = 2*3*5*7

For 20, there will be 6 factors.
For 30, there will be 8 factors.
For 120, there will be 16 factors.

Now I see a pattern...when everwhere a prime factor of 5 comes up, you multiply the # of prime factors you found before by 2.

Am I correct?

Edit : Hmm...I tried the same thing for 180 but it didn't work so something is wrong, I think....

Okay, that's a start.  How about 900 = 2^2*3^2*5^2?  (Also, don't forget about 210.)  Also, how about 105 = 3*5*7, and 63 = 3^2*7?

I understand how you got that got to that but that's not what the generalized form is.  To help you even more try ones that don't have a five or try ones that have more than one five.  Like these  16,  75  ,  35.  Just keep experimenting until you think you have a pattern that you can apply no matter what the factors are.

Better yet, what constitutes a factor?  Try finding factors systematically.

I'm not too sure about 900 for now but ...

For 105, there exists 8 factors
For 63, there exists 8 factors

I can't really see anything this time.  I can't see a pattern with 7.

I believe you mis-counted with 63.

Here's a question to think about:  imagine we have a number, like, I dunno, 30 = 2*3*5.  Now, imagine that we have another number, like, say 33269 = 17*19*103.  I claim that without knowing necessarily how many factors either of them has, that I know that they have the same number of factors.  Can you prove this?  How about in general for two numbers, with all their prime factors expressed in the same powers, but different specific primes.  (For example, 2^32*3^17*19 and 5^32*11^17*7.)

Oh right...there should have been 6 factors for 63.  Now, 7 seems to be like 5.

Anyways, I have very poor skills in proving and it probably shows here that I can't prove if any two given numbers have the same amount of factors.

Hint:

[hide]A factor is something that can divide a number.  Let's take 63 as an example.  63 = 3*3*7.  Now 27 is not a factor because it doesn't divide 63.  Nothing "times" it will make 63.  Why?  Well, if n*27 = 63, then n*3 = 7, and there are no more 3's in the prime factorization to cancel with the 3 on the left side.  So if k is a factor of 63, then n*k = 63.  n = 63/k must be an integer.  What can 63 divide out with some still to spare?  Let's look at the prime factorization... it's 3*3*7.  63 can afford to divide two 3's and a 7 or part of that (e.g., only one 3 and a 7).  You're well on your way now.[/hide]

If you can find the generalized form of the factorization it will be easy to prove :D

I'm not too sure what generalized form means and thanks for the hint

Generalized means that it can be applied to anything.  You have been giving rules for specific cases like the one with fives.  General means that given any number with its factorization you can find the number of factors no  matter the value of its prime factors.

[quote="kool_dudy"]I got a question for number 2.  I understand the concept and how to solve those kind of problems from now on but what happens when we get a number like 286488.  I prime factored it and I got [tex](2^3)(3^3)(3979)[/tex].  How would we find the number of factors in this case.

[/quote]

Hmm... What's the special about 286488?

....random number

anyways...i've been testing a lot of factors at school today and I came up with results that 7 is indeed similar to 5

[quote="JBL"]Here's a question to think about:  imagine we have a number, like, I dunno, 30 = 2*3*5.  Now, imagine that we have another number, like, say 33269 = 17*19*103.  I claim that without knowing necessarily how many factors either of them has, that I know that they have the same number of factors.  Can you prove this?  How about in general for two numbers, with all their prime factors expressed in the same powers, but different specific primes.  (For example, 2^32*3^17*19 and 5^32*11^17*7.)[/quote]

Work with this.  Start with the first part -- can you show that 30 and 33269 have the same number of factors?

Also, you say 7 is similar to 5 in that multiplying a number with no factors of 5 (or 7) by 5 (or 7) doubles the number of factors.  Try the same thing with 11, 13.  Tell me what you see.

We're going to try to use the pattern to find the general result.  And we're basing the pattern on prime factorization.

So let's take the numbers 66 and 78 for example

66 = 2*3*11
78 = 2*3*13

66 should have 8 factors
78 also has 8 factors.

My conclusion is 11 and 13 act as a 4 when counting factors.

So I kind of figure out something.  It's starting to look more like multiplying by 2s but I'm going to experiment with 23 and 29.

JBL, I'll answer your first question later.  Thx for this help, btw.

[quote]My conclusion is 11 and 13 act as a 4 when counting factors. [/quote]

Not quite.

Think about this.

2*3*5 has 8 factors. 2*3*19 has 8 factors. What do you think about 2*3*p? How many factors should it have?

Should p always be something to multiply as a 2?

[quote="Silverfalcon"]No offense, but why we here try to find the pattern? I mean, I think the best way to do is getting prime numbers, rather than trying to find the pattern.. I'm just saying this because finding pattern isn't always easy.. More likely, it can be very difficult.. One time, I tried to find the rule on remainder when divided by 7 and just failed..[/quote]

Basically what JBL said answers your question.  In addition once you learn this general result it will be very valuable.  I have seen it on many math competitions.  We could just tell you what it is but it will be much more valuable and memorable if you find it yourself.

[quote="joml88"]I have seen [the issue of how many factors a number has] on many math competitions.  We could just tell you what it is but it will be much more valuable and memorable if you find it yourself.[/quote]

I have seen that issue on many math competitions too. It came up in the very first math competition that my son ever participated in, which is why I am very sure that this is a legitimate Getting Started level problem. Understanding why the general rule is as it is is quite important, and I invite all the younger problem-solvers to think about it as this thread continues. I promise I won't give away the answer (which I took care to memorize in my first year of coaching our math team) while all of you discuss this.

The first question:

30=1,2,3,5,6,10,15,30
33269=1,17,19,103,323,1751,1957,33269

They have the same number of factors...

As for what I have been trying...

let's take 138 and 174

138=2*3*23 --- Factors : 1,2,3,6,23,46,66,138
174=2*3*29 --- Factors : 1,2,3,6,29,58,87,174

They have same number of factors and they have 8 factors each.

Would this mean that for any prime number without any exponents equal to 2 when calculating for the number of factors?

Yeah. I'm not sure whether this answers your question or not but here it goes:

15=3*5=3 has value of 1 and 5 has value of 1 then = 2*2 = 4
Let's check:

15 = 1,3,5,15

Let's try another one:

102 = 17*3*2 = 17,3,2 has value of 1, meaning there's only one of each then = 2*2*2 = 8
Let's check:

102 = 1, 2, 3, 17, 6, 34, 51, 102

See how that works?

Hope this helps!

Rather than answering what it means, whenever you calculate the number of factors, no exponent number has superscript 1. So, value of 1 means 1.

So, 4 has value of 2 because it's actually 2^2

Oh ok I get it now.

I was getting something like that as well because let's say...

2^2*3^2*19

so then since 2 and 3 both have values of 2...

it would mean 3*3*2 factors...I think...

[quote="beta"]What we want to show is a generally how many factors does[tex] n=a^bc^d...x^y[/tex] have if a,c,...x are all primes. b,d...y are all positive integers.

I think the best way to approach this is combinatorically.
If k is a factor of n, so [tex]k=a^Ac^C...x^X[/tex]
with [tex]b\geq A\geq0, d\geq C\geq0,...y\geq X\geq0[/tex]

I will leave it for you to figure out the rest.[/quote]

I will just finish it.  The total number of value of A are,0,1,2,...b thus b+1.
The total number that possible value of C are 0,1,2,...d, thus d+1
...
The total number of value of X are 0,1,2,...y, thus y+1
The total number of factor of n is (a+1)(b+1)...(y+1)

Check. 30=(3^1)(5^1)(2^1)
by the formula, (1+1)(1+1)(1+1)=8 as expected.

k, lets see...If there is only one of each factor, i.e. 2 * 3  * 5 * 7, then all the different factors are the different products of combinations of the factors, so we have the nth row on the Pascal's row. The sum of these is easy: 2^n. So if there's only one of each prime factor, the number of total factors is 2^n.

Now for when there isn't just one of each...? To conjecture: 2^p where p is the sum of the powers of the prime factorization??

Not quite, but you're on the right track.  Just keep thinking in the same way but maybe start from scratch again using powers > 1.  If you want a hint, you can maybe look at a few previous messages.


Similar with the topic [url=http://www.mathlinks.ro/viewtopic.php?t=286322]AGEb[/url]
[quote="mr.danh"]$ M_aE_b\parallel CH$, so the perpendicular line $ d$ from $ A$ to $ M_aE_b$ is just the sideline $ AB$. Draw the line $ d'$ symmetric to $ d$ wrt $ M_a$. Let $ K$ such that $ M_a$ is the midpoint of $ E_bK$. See that $ KC\parallel BE_b$, so $ KC\perp AC$. Hence $ C$ is just the intersection of $ d'$ and the circle of diameter $ AK$.[/quote]


Let {p1, p2, . . . , pk} be the set of first k prime numbers.
And Let S be the set of integers such that k(n; pi) = 0 for i > k and
k(n; pj) = 0 or 1 for j is less than or equal to k.

Prove the equality of

sum of 1/n for each n is element of S = product of (pj+1)/pj for j = 1 to k.

K: N x p ----> N

k(a,p)= { 0 if p does not divide "a"
{ 1 if p divides a but p^2 does not divide "a"
{ n if p^n divides a but p^n+1 not divide "a"

so k(a;p) = max{n is element of N | p^n divides "a"}

well then it's obvious:

$ \sum_{n\in S}\frac{1}{n}=\sum_{1\leq i_1<\cdots < i_r\leq k}\frac{1}{p_{i_1}\cdots p_{i_r}} = \prod_{j=1}^k (1+\frac{1}{p_j})=\prod_{j=1}^k \frac{p_j+1}{p_j}$


[quote="lingomaniac88"]From [i]Whose Line Is It Anyway?[/i], it's SCENES FROM A HAT!  I gave a scene, and now we try to come up with as many ideas for it as we can.  In this case, it's "How celebrities end their proofs."

I'll start:

Yakov Smirnoff: "But in Soviet Russia, proof writes YOU!"[/quote]

Guess who:
"And if there ain't any, then mah name ain't Dubya!"

Ummmmmmmm... Is it Bush?? I'm bad at these... :blush:

Steve Martin: "And if I'm wrong, well excuuuuuuuuuse me!"


Let p>3 a prime number. Show that 43 divides k=7^p-6^p-1.

Solution:

there exists l from N so that p=6l+1 or p=6l+5. Now we do (mod 43)
7^5==-6, 6^5==-7, 7^6==1, 6^6==1 for any l natural, 6^(6l)==7^(6l)==1.
a) if p=6l+1 then 7^p-6^p-1==7-6-1==0
b) if p=6l+5 then 7^p-6^p-1==(-6)-(-7)-1==0
So 43 divides k for any p prime >3.

hope it is correct! cheers!

Enough is enough!!!  :mad:  :ewpu:  :bomb:

Stop this testing around!

I wrote you a PM.
And mathubu, who is posting from the same place as you, two of them and also one E-Mail.
None of the PMs has ever been read. Neither has the mail been answered.
I will also write you a E-mail now.
Tell me whether you and mathubu are the same person and stop this testing at random threads!

Let this be the last and very last warning!

PS to all others: I will delete this and the previous post in 7 days.


algebra
polynomial
Rational Root Theorem

By the Rational Root Theorem, we find that $ 2x \minus{} 1$ is a factor, which has root $ \boxed{\frac12}$. Dividing, we get $ 4x^2 \minus{} 2x \minus{} 1$, which has solutions $ x \equal{} \boxed{\frac14\pm\frac{\sqrt5}4}$.


Hi everyone! I'm a high school student. Perform badly in school math exam. However, I'm very interested in math. I want to learn, study and solve IMO questions. What can I do? How can I do it? Any suggestions and advice is always welcome. Thank you.

Obtain and read AoPS Volumes 1 and 2 would be a good start.


Source: 360 PMC
Find the maximal value of the expression
$E=\sin x_{1}\cos x_{2}+\sin x_{2}\cos x_{3}+\cdot\cdot\cdot+\sin x_{n}\cos x_{1}$, when $x_{1},x_{2},...,x_{n}$ are real numbers.

[quote="amcavoy"][hide]$\frac{n}{2}$[/hide][/quote]
yes, u got the correct answer.
do you mind posting your solution?

[hide="Solution"]Well, it's easy to show that $\sin x\cos x$ is maximized when $x=\frac{\pi}{4}$.  Therefore we have \[\sum_{k=1}^{n}\sin\frac{\pi}{4}\cos\frac{\pi}{4}=\boxed{\frac{n}{2}}.\] [/hide]

Edit:  Changed $\sin x\cos y$ to $\sin x\cos x$.

[quote="amcavoy"]Well, it's easy to show that $\sin x\cos y$ is maximized when $x=y=\frac{\pi}{4}$[/quote]

This is a grave error. What if $x={\pi\over 2}$ and $y=0$?

Prove the result with induction over $n$ now that you've guessed it..

I should have said $\sin x\cos x$ is maximized at $x=\frac{\pi}{4}$.  That is the only thing I was concerned with because as you can see if the first term is equal to $1$, then the second will be equal to $0$ (the result will be the same, but I just chose $x_{1}=x_{2}=\cdots =x_{n}=\frac{\pi}{4}$).

Of course, I was just explaining my reasoning.  The answer seems obvious, but I'll try to think of a good reason / proof of why.  :)


The release date for Harry Potter and the Half Blood Prince is to be announced by english publishers within 24 hours!!! hoorah!!!

[url=http://www.mugglenet.com/viewer/?image_location=hbpdate.gif]PROOF![/url]

UM, not really. I don't like how borish and simplistic it is. Sure, entertainment, but not much beyond that.

Sooo excited!!!  Love Harry Potter!!!

I do have to say that the movies however ruined some of the fun of creating your own characters and envisioning them...but oh well.

Since the 5th book came out, I'm not really such a fan of Harry Potter anymore.  Hopefully, the 6th book won't be as long.  It feels like except for the 3rd book, the longer the book gets, the more boring it is.

The release date is July 16, 2005! We have been informed also that the HBP is "a bit shorter than the Order of the Phoenix." hooray!

:( Book 6 will be shorter? NOOOOO! The longer they are the better! Including OotP, you...you blasphemers!
A-HEM Anyways, I'm very, very glad about this. That's not too long to wait...compared to what I was afraid the date would be, anyway. (Read: 2006) I don't need to hope it'll be good, because I KNOW it'll be good. The definition of Harry Potter books involves the word good. *Is happy*

[quote="LynnelleYe"][quote="SirErnest"][quote="A+MATH"]
Okay maybe its not like that, but you get my point you can't write 6 books on the same old topic.[/quote]
We can't but J.K. Rowling can.[/quote]
AMEN[/quote] That's a good joke...

[quote="A+MATH"]Harry Potter is surprisingly repeatitive here is the way the books are: Go some where, see new monster thing, go in secret passage, find another secret passage, see a monster, meet new chracter, go some where, see new monster thing, go in secret passage, find another secret passage, see a monster, meet new chracter Go some where, see new monster thing, go in secret passage, find another secret passage, see a monster, meet new chracter go some where, see new monster thing, go in secret passage, find another secret passage, see a monster, meet new chracter.

Okay maybe its not like that, but you get my point you can't write 6 books on the same old topic.[/quote]

Nonsense; that adds to the campiness of it.  It's like going to the same summer camp several years in a row--the familiarity only makes you love it more (provided the previous years were enjoyable, of course).  Why do you think the Indiana Jones flicks were and are so popular?  Or a more extreme example, the James Bond films?  Because they follow a similar, familiar formula but are given a kick of originality that puts a whole new spin on it.

[quote="Jeffreywang"][/quote][quote="LynnelleYe"][quote="SirErnest"][quote="A+MATH"]
Okay maybe its not like that, but you get my point you can't write 6 books on the same old topic.[/quote]
We can't but J.K. Rowling can.[/quote]
AMEN[/quote][quote="Jeffreywang"] That's a good joke...[/quote]
What's a good joke?

yaaaay!!! i luv the hp books 2. but another persons gonna die in this book. dumbledore i think....just a gess. but i lyked the 4th book better than the 5th one.....hopefully this one will be betta

luv HP, but it kinda died when the sixth one didn't come out 4 that long. It seems really long to me b/c I started reading them when the 4th one came out, so I didn't have to wait long 4 the other books.

U sure that's the title 4 the sixth book? had to do a webpage on it, and it says it was called something else

I thought it would never come out, afraid that she was going to quit writing it.... but nvm.... some1 said that she MIGHT write a book after the seventh book about how they lived as an adult... would luv that!

Predictions (and hints from JK Rowling) said that Ron might die in the sixth OR seventh book :(  :?

[quote="LuCy4EvA"] U sure that's the title 4 the sixth book? had to do a webpage on it, and it says it was called something else[/quote]

100% positive it's harry potter and the halfblood prince.
[url]http://www.jkrowling.com/textonly/news_view.cfm?id=82[/url]
that's her official website's announcement of the release date, which uses the title HP&THBP.

also, it's been stated by one of her english publishers that the HBP is less than 38 chapters.


I just wanted to clarify if returning students needed to take the admission test and apply again or if they can just send their cheques and tell Titu or something.

website says:

"Returning students need not resubmit letters of recommendation or the application fee."

Well first they didn't say we didn't have to take the test again and second when do we have to tell them we are coming?

NO seriously guys do returning students have to take the test again and if not how and when should we notify them that we are coming  or do we just send them a check or something? I mean surely they can't just make room if like 50 people suddenly decided they want to come back after they already went through the applications.

Hmmm...I'm pretty sure that returning students have to do the admissions test over again.  It doesn't really count for admission though more like class placement.

The question was answered last year:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=998930#998930[/url].