trigonometry
number theory proposed
number theory

Find the integers $ x,y$ that satisfay the following equation $ 2x^4\plus{}1\equal{}y^2$

$ 2x^4\equal{}y^2\minus{}1\equal{}(y\minus{}1)(y\plus{}1)$. Obviosly $ (y\minus{}1,y\plus{}1)\equal{}1 \ or \ 2$. If $ y\equal{}\pm 1$, then $ x\equal{}0$.
If $ |y|>1$, then $ y$ odd, therefore $ (y\minus{}1,y\plus{}1)\equal{}2$, therefore may be only $ y\minus{}1\equal{}\pm 2,y\plus{}1\equal{}\pm x^4$ or $ y\minus{}1\equal{}\pm x^4,y\plus{}1\equal{}\pm 2$.
In first case $ x^4\equal{}\pm 4$ (no solutions ), in second $ x^4\equal{}\minus{}4$.

@rachid:
note that if $ d\mid y \minus{} 1$ and $ d\mid y \plus{} 1$ then $ d\mid y\plus{}1 \minus{} (y \minus{} 1)\Rightarrow d\mid 2\Rightarrow d \equal{} 1,2$ thus $ \gcd (y \plus{} 1,y \minus{} 1) \equal{} 1\textrm{ or }2$
@Rust:
your argument is wrong,let $ x \equal{} 2^{a}r$ where $ r$ is an odd number,then the case $ y \minus{} 1 \equal{} 2r^4,y \plus{} 1 \equal{} 2^a$ is valid,which you didn't solve the problem for,so your solution is not complete at all...

[quote="BaBaK Ghalebi"]@rachid:
@Rust:
your argument is wrong,let $ x \equal{} 2^{a}r$ where $ r$ is an odd number,then the case $ y \minus{} 1 \equal{} 2r^4,y \plus{} 1 \equal{} 2^a$ is valid,which you didn't solve the problem for,so your solution is not complete at all...[/quote]
Yes. Let x and y natural (if (x,y) - solution, then ($ \pm x,\pm y$) solution too).
We get, that $ x\equal{}x_1x_2, (x_1,x_2)\equal{}1$ and
1. $ y\minus{}1\equal{}2x_1^4,y\plus{}1\equal{}x_2^4$ or
2. $ y\minus{}1\equal{}x_1^4,y\plus{}1\equal{}2x_2^4$.
In first case $ x_2^4\equal{}2x_1^4\plus{}2\equal{}2 \ or 4\mod 32$ - contradition for four degre.
In the second case $ x_1^4\equal{}2x_2^4\minus{}2\equal{}30 \ or \ 31\mod 32$ -contradition.

[quote="Rust"]In the second case $ x_1^4 \equal{} 2x_2^4 \minus{} 2 \equal{} 30 \ or \ 31\mod 32$ -contradition.[/quote]
Don't you mean 0 or 30 mod 32?  I don't see the contradiction.  But we can use descent:

$ 2\left(\frac{x_1}2\right)^4 \equal{} \left(\frac{x_2^2\minus{}1}2\right)\left(\frac{x_2^2\plus{}1}2\right)\equal{}\left(2\tilde x^4\right)\left(\tilde y^2\right)$,

and $ \tilde y^2 \equal{} 2\tilde x^4\plus{}1$, with $ \tilde y \equal{} \sqrt{\frac{x_2^2\plus{}1}2}\equal{}\sqrt{\frac{\sqrt{\frac{y\plus{}1}2}\plus{}1}2} < y$ if $ y>1$.

Yes. Let $ x_1^2\equal{}2a,x_2\equal{}b^2$, then $ 4a^2\minus{}2b^2\equal{}\minus{}2$, with had solution $ 2a_n\plus{}b_n\sqrt 2\equal{}\sqrt 2(3\plus{}2\sqrt 2)^n$.
Obviosly $ (2a_n,b_n)\equal{}1$ and $ 2a_{2n}\plus{}\sqrt 2 b_{2n}\equal{}\frac{1}{\sqrt 2}(2a_n\plus{}\sqrt 2 b_n)^2$, therefore $ a_{2n}\equal{}2a_nb_n, b_{2n}\equal{}b_n^2\plus{}2a_n^2$.
\[ b_n\equal{}\sum_{i\equal{}0}^{[n/2]}C_n^{2i}3^{n\minus{}2i}2^{3i}, \ a_n\equal{}\sum_{i\equal{}1}^{[(n\plus{}1)/2]}C_n^{2i\minus{}1}3^{n\plus{}1\minus{}2i}2^{3i\minus{}2}.\]
If n is odd $ b_n\equal{}\minus{}1\mod 4$ is not square.
If exist $ n\equal{}2^km$, suth that $ 2a_n$ is square, then $ 2a_n\equal{}4a_{n/2}b_{n/2}$ square, therefore $ a_{n/2}$ and $ b_{n/2}$ are squre (because ($ 2a_m,b_m)\equal{}1 \ \forall m$.
Therefore $ 2a_{n/4}$ and $ b_{n/4}$ are square, ... $ 2^{k\%2}a_{n*2^{\minus{}k}}$ and $ b_{n*2^{\minus{}k}}$ are square too.
We get $ (x_1^2)^2\minus{}2(x_2^2)^2\equal{}\minus{}2$ or $ x_1^2\plus{}\sqrt 2 x_2^2\equal{}\sqrt 2(3\plus{}2\sqrt 2)^n$.
It give, that there are not natural solutions of equation $ x^4\minus{}2y^2\equal{}\minus{}2$.

just a little comment on the problem,if you put THE "$ \plus{}1$" on the other side,it turns into a very hard problem:

$ 2x^4\equal{}y^2\plus{}1$

(notice the "+1")

[quote="BaBaK Ghalebi"]$ 2x^4 \equal{} y^2 \plus{} 1$[/quote]
We have

$ x^4 \equal{} \frac {(y \plus{} i)(y \minus{} i)}{(1 \plus{} i)(1 \minus{} i)} \equal{} (u \plus{} 1 \minus{} iu)(u \plus{} 1 \plus{} iu),$

where $ u \equal{} \frac {y \minus{} 1}2$.  We then have

$ u \plus{} 1 \plus{} iu \equal{} (a \plus{} bi)^4$

for some integers $ a,b$, giving

$ a^4 \minus{} 6a^2 b^2 \plus{} b^4 \equal{} 1 \plus{} 4a^3b \minus{} 4ab^3.$

I'm not sure what to do with that, but Mathematica assures me $ (a,b) \equal{} (0,\pm 1)$ or $ \pm(2,3)$ are the only solutions with, WLOG, $ a$ even and $ b$ odd.  Then $ x \equal{} a^2 \plus{} b^2$ is either 1 or 13, giving $ y \equal{} 1$ or $ y \equal{} 239$.

In any case there are only finitely many solutions by Siegel (from the original equation) or Thue (from the above homogeneous equation).

Or (sketch): writing $ a \plus{} bi \equal{} re^{i\theta}$, we have $ \tan4\theta \equal{} \frac u{u \plus{} 1}$, so $ \left|\frac ba \minus{} \tan\frac {(4k \plus{} 1)\pi}{16}\right| < \frac cu < \frac C{a^4}$, which has only finitely many solutions by Thue, Siegel, or Roth (but not Liouvile, since $ \tan\frac\pi{16}$ has degree 4 over $ \mathbb{Q}$).

[quote="Eigenray"]Then $ x \equal{} a^2 \plus{} b^2$ is either 1 or 13, giving $ y \equal{} 1$ or $ y \equal{} 239$.[/quote]
exactly! :)

[quote="Rust"]
If $ |y| > 1$, then $ y$ odd, therefore $ (y \minus{} 1,y \plus{} 1) \equal{} 2$, therefore may be only $ y \minus{} 1 \equal{} 2$[/quote]

WHY????
from the equation
$ 2x^4 \equal{} (y \minus{} 1)(y \plus{} 1)$

we have only $ y\minus{}1$ divides $ 2x^4$ did you prove: $ y\minus{}1$ doese'nt divide $ x^4$????

@FOURRIER:Rust's argument is wrong,read here:
[quote="Rust"][quote="BaBaK Ghalebi"]@rachid:
@Rust:
your argument is wrong,let $ x \equal{} 2^{a}r$ where $ r$ is an odd number,then the case $ y \minus{} 1 \equal{} 2r^4,y \plus{} 1 \equal{} 2^a$ is valid,which you didn't solve the problem for,so your solution is not complete at all...[/quote]
Yes. Let x and y natural (if (x,y) - solution, then ($ \pm x,\pm y$) solution too).
We get, that $ x \equal{} x_1x_2, (x_1,x_2) \equal{} 1$ and
1. $ y \minus{} 1 \equal{} 2x_1^4,y \plus{} 1 \equal{} x_2^4$ or
2. $ y \minus{} 1 \equal{} x_1^4,y \plus{} 1 \equal{} 2x_2^4$.
In first case $ x_2^4 \equal{} 2x_1^4 \plus{} 2 \equal{} 2 \ or 4\mod 32$ - contradition for four degre.
In the second case $ x_1^4 \equal{} 2x_2^4 \minus{} 2 \equal{} 30 \ or \ 31\mod 32$ -contradition.[/quote]


induction
advanced fields
advanced fields theorems

From what I understand of one of Godel's incompleteness theorems, it involves coding wffs into numbers.  Each wff corresponds to a Godel number and each Godel number gives rise to a wff.  And then by using a kind of logical isomorphism between the set of wffs and N, incompleteness of any "adequate" theory is implied by the fact that computability implies definability.

Is that basically right?

Assuming it is...  Here's my question.  We use arithmetic to encode formulas.  What if we used a nonstandard model of arithmetic or some other poset to encode wffs?  What kind of results do we get, for instance, something about infinitary logic if our model of arithmetic has unlimited numbers that could correspond to an infinite string of OR or AND?

Thanks.

We are only allowed to use only up to countable set to encode.
The reason is: Human are only allowed to count up to a countable number of times.
The consistency of Peano arithmetic without addition is well know.
See for example: "S.C.Kleene: Introduction to Meta-Mathematics, Van Nostrand" for a proof.
The problem of Peano arithmetic comes from multiplication and the consistency of Peano arithmetic including multiplication is already by Gentzen in the 1930's by "Transfinite Induction", which is "Allowing to count up to more than a countable number of times" ;)


The terms $ 140, a, \frac{45}{28}$ are the first, second and third terms,
respectively, of a geometric sequence. What is the value of $ a$?

If the three terms make a geometric sequence, the middle term is the geometric mean of the first and last. The geometric mean of $ n$ numbers $ a_0$, $ a_1$, $ a_2$, and etcetera all the way up to $ a_n$ is defined as $ \sqrt[n]{(a_0)(a_1)(a_2)....(a_n)}$. Fortunately, $ n\equal{}2$ in this case, so the formula is simplified much to just the square root of a product of the first and last terms. Thus, the middle term $ a$ is $ \sqrt{140\cdot\frac{45}{28}}\equal{}\sqrt{5\cdot45}\equal{}\sqrt{225}\equal{}\boxed{15}$.


geometry
ratio
conics
ellipse
trigonometry
complex numbers
geometry proposed

In triangle ABC, the orthocenter H divides the altitude BD in the ratio BH : HD = 3 = 1.
Prove that <AKC = 90 where K is the midpoint of BD.

$ABC$ must, I'm assuming, be acute-angled, because otherwise we would have $HD>BH$ if $\measuredangle ABC >\frac \pi 2$ and the conclusion would be wrong if we had $\measuredangle ACB >\frac \pi 2$ or $\measuredangle BAC >\frac \pi 2$. I must then assume that the problem is referring to the ratio of the oriented segments $\overline{BH},\ \overline{HD}$.

The locus of $A$ s.t. $\measuredangle AKC=\frac \pi 2$ is an ellipse passing through $B,C$ and for which the ratio of the semiaxes is $2$. On the other hand, it's easy to show by a very simple computation that if $A$ moves (assume $B,C$ are fixed) on an ellipse passing through $B,C$ with the ratio of the semiaxes being $2$ s.t. $\measuredangle BAC<\frac \pi 2$, then $\frac {\overline {BH}}{\overline {HD}}=3$. If we move $A$ up or down from this ellipse, $H$ will move in the opposite direction, so the ratio won't be $3$ anymore, so the locus of $A$ s.t. $\frac {\overline {BH}}{\overline {HD}}=3$ is the same as the locus of $A$ s.t. $\measuredangle AKC=\frac \pi 2$, so we have the conclusion.

It should be easy with complex numbers but I'm not really used to work with complex numbers. Can anyone show me such a solution? (I know it's not nice...)

i dont think complex numbers is a good idea for that problem  :D trig seems *way* easier. here is some progress (tho i cant be bothered finishing it ... )
rename vertices like this (more natural  :D): ABC is a triangle, D foot of perpendicular from A to BC, H orthocentre, K midpoint of AD ...
Let A, B, C be represented by complex numbers a,b,c respectively and let ABC be inscribed in a circle with centre the origin. its well known that H is a+b+c, and D is 1/2(a+b+c - b*c/a). Using that ratio line formula, it follows
a+b+c = 1/4*a + 3/4*(1/2(a+b+c - b*c/a)) ... (*)
we also know K, the midpoint of AD is 1/2*a + 1/2*(a+b+c - b*c/a)
so we want to prove KB |_ KC, i.e.
1/2 (a + 1/2(a+b+c - b*c/a)) - b / 1/2 (a + 1/2(a+b+c - b*c/a)) - c ... (**)
is purely imaginary  :D
i dont think this follows directly from (**)  :(  :( so we may actually need to use the fact that the circumcentre of ABC is the origin, and the formula for the circumcentre is really nasty ... so instead maybe you should use the general formula for orthocentre and point where perpendicular from A to BC meets BC ... but these are probably nasty also

[quote="Arne"][color=darkred]In triangle $ABC,$ the orthocenter $H$ divides the altitude $BD$ in the ratio $BH : HD = 3 : 1.$
Prove that $\angle AKC = 90^\circ,$ where $K$ is the midpoint of $BD.$[/color] [/quote]
[b]Proof.[/b] [color=indigo]Denote $M$ be the midpoint of $AB.$ We have:
$\triangle MKB\sim \triangle HDC\Longrightarrow\begin{cases}\frac{MK}{HB}=\frac{HD}{HC}\\ \angle BMK=\angle CHD\end{cases}\Longrightarrow\begin{cases}\frac{MK}{MA}=\frac{HK}{HC}\\ \angle KMA=\angle KHC\end{cases}$ $\Longrightarrow\triangle KMA\sim\triangle KHC\Longrightarrow\angle MKA=\angle HKC$

Hence $\angle AKC=\angle AKH+\angle HKC=\angle AKH+\angle MKA=\angle MKD=90^\circ$[/color]

Let $N$ be the midpoint of $AD$
We have $NK//AB$ but $CH\bot AB$ so $CH\bot NK$
So $H$ is the orthocenter of triangle $NKC$. Then $NH\bot KC$
Otherwise $NH//AK$ so $AK\bot KC$ $q.e.d$


Hallo !

  Which do you think that is the most difficult problem in Iranian olympiads in GEOMETRY ?
 I have seen one - I have forgotten the year - where we were askedto prove equality between two angles( between 1990- 1999).
  To this problem I was sent a solution by Yetti(in PM), but using harmonic ratios.
 [u]Babis[/u]

 I wish you [color=red]a happy new year [/color]to all members of this nice comunity !

Hello Mr.Stergiu

I don't know which problem you are talking about.Would you send a link?

and Actually this isn't my new year ,Our new year will Start 3 months later. :)

anyway I wish you [color=blue]a happy new year[/color]

Thank you AMIR 2.
Sorry for the early wishes. As you see , some things are different in different countries(and this is good!).
 I'll try to find again the problem. I had seen it unsolves in KALVA ' homepage and in IRAN's Olympiads.
 There was a triangle , , an interior point O, some cevians through O  and a perpendicular from O to BCect
 [u]Babis[/u]

aslan nemishe roo daraje sakhtie ye masale nazare ghati dad

hamishe nemishe vali bazi vaghtha  mishe :)


theorems + formulas for polynomials, write as many as possible.

Check this out : [url]http://www.artofproblemsolving.com/Wiki/index.php/Polynomial[/url]


Given nonnegative integers $ a,b,c$, prove that there is no function $ f: \mathbb{N}\rightarrow\mathbb{N}$ such that

$ f(x\plus{}y)\plus{}f(x)\plus{}f(y)\equal{}xy\plus{}ax\plus{}by\plus{}c$

holds for each $ x,y\in \mathbb{N}$

First, by switching variables, we have $ a \equal{} b$.

Put in $ x \equal{} n, y \equal{} 1$ and $ x \equal{} n\plus{}1, y \equal{} 1$. Subtract the first from the second to get $ f(n\plus{}2) \minus{} f(n) \equal{} a\plus{}1$. So $ f(2k) \equal{} f(2) \plus{} (k\minus{}1)(a\plus{}1)$. Let $ x \equal{} 2k, y \equal{} 2m$. Then $ 3f(2) \plus{} (2k\plus{}2m\minus{}3)(a\plus{}1) \equal{} 4km \plus{} (2k\plus{}2m)a \plus{} c$, or $ 3f(2) \equal{} 4km \plus{} 3a \minus{} 2k \minus{} 2m \plus{} c$. This means $ f(2)$ has to be a rather large amount of values at once. Contradiction.


algebra
difference of squares
special factorizations

Here's a problem that took me a few hours to solve..... Good luck  :P

We filled out an 8x8 grid with natural numbers from 1 to 64 by starting at the top left corner with 1 and going to the right all the way in an order. When we reached the end, we continued on the left side of the second row, and so on ending with 64 in the bottom right corner. Now, we pick 8 numbers from this grid so that we pick exactly one number from each column and each row. Prove that the sum of these 8 numbers is always the same.

This one was much easier......

Find two square numbers with the difference of 2005^2 .

1
[hide]Each column forms a different arithmetic sequence. Starting from the first they are:
$8a - 7$
$8b - 6$
$8c - 5$
$8d - 4$
$8e - 3$
$8f - 2$
$8g - 1$
$8h$
We must always choose for the variable a number 1 through 8. The sum of these will always be the same. The sum of the negative numbers 1 through 7 will always be the same which means that the entire sum will always be the same.[/hide]

[hide="Easier Problem"]
There are actually two solutions: one if 2005 is the shorter leg of a right triangle and one if it is the longer leg. I am going to solve the problem for 2005 being the shorter leg:

If the shortest leg of a right triangle is $a$, then the longer leg [b]may [/b]be $\frac {a^2-1}{2}$, and the hypotenuse [b]may[/b] be $\frac {a^2+1}{2}$. Thus, if the shorter leg is $2005$, the longer leg is $\frac {2005^2-1}{2}$ and the hypotenuse is $\frac {2005^2+1}{2}$ (NOTE: There may be other Pythagorean triples, but this is the easiest to find)

Therefore, the two numbers are $(\frac {2005^2-1}{2})^2$ and $(\frac {2005^2+1}{2})^2$ or $4040148240144$ and $4040152260169$

EDIT: This is equal to $2010012^2$ and $2010013^2$
[/hide]

2
[hide]Just represent the difference of squares $(a + b)(a - b ) = 2005^2$. Choose any combination of factors of $2005^2$ such that neither $a$ or $b$ is not equal to zero. Solve for $a$ and $b$, and square them. Since all of the factors of $2005^2$ are odd, $a$ and $b$ will always come out integers.[/hide]

Hint for easier way for #2:

[hide]$x^2 - y^2 = (x+y)(x-y)$ or if you can do $2005^2 - 0^2$ :P[/hide]

Yes, Problem 2 is correct.  Tarquin's solution for problem 1# was also correct. ( I solved it differently)  :|

[hide="Problem 2"]
So far I found three solutions:

1) $2010013^2-2010012^2=2005^2$
2) $80413^2-80388^2=2005^2$
3) $5213^2-4812^2=2005^2$

For the last two solutions I used difference of squares, plugging in different factorizations of $2005^2$ : $(410^{2})(25)$ and $(401)(401*25)$.

To see how I found the first solution, view my post above. Now that I notice, though, it is the same as using difference of squares using the factors $(2005^2)(1)$.
[/hide]


If a cylindrical water tower is 75 percent full, its volume is 15205.31, then what is the weight of the water?

$ W\equal{}mg$, but to find the mass of the water, we need the density of the water, and you haven't given us any information whatsoever that would help us find this.

Several things

1) Units for volume would be nice
2) Is that volume the volume of the tower, or the volume of the water when the tower is 75% full?
3) Your sentence is a grammatically incorrect run-on which is why it is so ambiguous. It is unclear what "it" is referring to in that interior clause.

I believe that he means when the tower is filled with water $ 75%
$ of the way up, then the volume of the water is $ 15205.31$ cubic cm.  But he also hasn't given us anyway to find the weight of the water.

The density of water is $ 1 \frac{gram}{cm^3}$ so from there it would be easy

we just need units for the volume and clarity as to what "it" is referring to.

If it is referring to what mathcrazed just said, then the answer is simply $ 15205.31$ grams.
But that's not even a problem >_>

[quote="alanchou"][quote="mathcrazed"]$ W \equal{} mg$, but to find the mass of the water, we need the density of the water, and you haven't given us any information whatsoever that would help us find this.[/quote]

I believe that pure water has a density of 1 or somewhere around that[/quote]

Units are important!!

Units would be g/cc, g=grams and cc=cubic centimeters


Prove that $ 2^{k\minus{}1}\plus{}1\not\vdots \;k, \forall \;k\in\mathbb{N}\;,k>1$

[quote="BaBaK Ghalebi"]I think you mean $ k\not| 2^{k \minus{} 1} \plus{} 1$ :maybe:[/quote]

I my country , you can rewrite it in the form :

$ 2^{k \minus{} 1} \plus{} 1\not\vdots\; k$

Now, please solving this problem. The solution is so nice and not too easy  :(

[quote="TTsphn"]Posted before in this Forum in this Forum at least 3 time.[/quote]

I don't know, please give me a link  :)


Let $ {k,m,n}$ be three odd integers such that $ {k\plus{}m}$ and $ {m\plus{}n}$ are both divisble by 4,then $ {k\plus{}n}$ cannot be divisible by 4(strickly speaking).[url]http://www.nigerianolympiads.org[/url][/quote]

Let's tackle this problem with casework.

First of all, we know $ k,m,n$ are odd. This means that they are either congruent to 1 or 3 in modulo 4.

Case 1: $ k\equiv 1\pmod{4}$. Since $ 4|k\plus{}m$ and $ 4|m\plus{}n$, $ m\equiv 3\pmod{4}$, so $ n\equiv 1\pmod{4}$. Then adding the two congruences for $ k$ and $ n$ gives $ k\plus{}n\equiv 2\pmod{4}$, so $ 4$ doesn't divide $ k\plus{}n$ in this case.

Case 2: $ k\equiv 3\pmod{4}$. We then get $ m\equiv 1\pmod{4}$, so $ n\equiv 3\pmod{4}$. Adding both congruences for $ k$ and $ n$ gives $ k\plus{}n\equal{}6$ so $ k\plus{}n\equiv 2\pmod{4}$, thus $ 4$ doesn't divide $ k\plus{}n$.

In both cases, $ k\plus{}n$ isn't divisible by $ 4$. Therefore we can conclude that $ k\plus{}n$ is not divisible by $ 4$, which completes our proof.  $ \blacksquare$

[hide="Solution"]Let us assume that $ n\plus{}k$ is divisible by 4 and lead a contradiction.

According to the assumption, we have $ k \plus{} m\equiv 0 \mod 4, m \plus{} n \equiv 0 \mod 4, n \plus{} k\equiv 0 \mod 4$.

Adding all three together, we have $ 2(k \plus{} m \plus{} n)\equiv 0 \mod 4$, which means $ k \plus{} m \plus{} n$ is even.

However, $ k \plus{} m \plus{} n$ is odd since $ k, m, n$ are all odd integers; a contradiction.[/hide]


[b]1.[/b] Three people - $ A, B$ and $ C$ - are sitting in a row in such a way that $ A$ sees $ B$ and $ C$, $ B$ sees only $ C$, and $ C$ sees nobody. They were shown $ 5$ caps - $ 3$ red and $ 2$ white. They were blindfolded, and $ 3$ caps were put on their heads. Then the blindfolds were taken away and each of them was asked if they could determine the color of their caps. After $ A$, and then $ B$, answered negatively, $ C$ replied affirmatively. How was that possible?

[b]2. A variant of problem 1.[/b]
    But this time all three can see each other and red caps are placed on each of them. $ A$, being the most intelligent of the three, observes $ B$ and $ C$, and after some time, correctly states that he is wearing a red cap. How did he find that out?

[b]3.[/b] Calvin and Hobbes were walking down the street, discussing about Calvin's worms. Hobbes is trying to find out the ages(integers) of the three worms.
Calvin says, "The product of their ages is $ 36$".
Hobbes asks for more info.
Calvin says, "The sum of their ages is equal to the number on the red house across the street".
Hobbes is still unable to find their ages.
Calvin says, "The eldest one likes chocolate"'.
Hobbes says, "Ahh... Now I've got it".
How does does Hobbes find out their ages?

[b]4.[/b] Five thieves are trying to share a loot of $ 100$ gold coins. They come up with this idea to decide how the loot should be distributed. Each thief comes up with a proposal and presents it, and the order in which they present their proposals is determined by seniority. If the proposal of a particular is accepted by atleast  half the other thieves, then it is followed. Else, the thief who proposed is killed, and the next thief presents his proposal.
Now, We know that the first priority of each thief is his own life, followed by money, and lastly to kill other thieves. Assuming that all the thieves are extremely intelligent and have distinct integer ages, What distribution should the First thief propose?

P.S. Yippee...  My 100th post !!

3.easy..ans is "2,2,9"
i hav done it earlier(perhaps posted in Aops too) so no use of me explaining :)

and i think a problem similar too 1 and 2 was also posted..

4 looks interesting .i'll try.. :D

If you have solved any of these problems before, do not post a solution. Let others have a crack at it. Logic problems are a lot of fun to solve, so don't deny others the fun.  :)
BTW, The fourth problem really requires patience.

I'm nobody  :rotfl:
I think I got 1 and 2. 3rd, I don't understand even after seeing the answer. Whats the relation b/w 13, red house and 9, chocolate?
4th, thinking....
[hide]do not open[/hide][hide]1st thief must convince 2 others, so he leaves 0 in the name of the youngest 2 thieves. The youngest thieves will tend to reject any proposal for the benifit of the 'common good' coz they'd like to kill everyone else until the whole loot comes to be distributed among 1 or two. Now question is about satisfying the next two in seniority. Obviously they're concerned about their own lives and would not want to become judges of the distribution themselves. So, the first thief must propose amounts that ensure that the next two are getting at least what they would safely propose on their turns.
(To be contd.)[/hide]

For those who want the solution to the third, [url=http://www.mathlinks.ro/viewtopic.php?p=1085538#1085538]here it is[/url]. See the posts following the question.
As for the fourth, here's a [hide="hint"] try working backwards[/hide]
The first and second are pretty easy.
[b]@nuclear_alchemist:[/b] You're on the right track but you need to go more in-detail.

[quote="keyree10"]
[b]4.[/b] Five thieves are trying to share a loot of $ 100$ gold coins. They come up with this idea to decide how the loot should be distributed. Each thief comes up with a proposal and presents it, and the order in which they present their proposals is determined by seniority. If the proposal of a particular is accepted by atleast  half the other thieves, then it is followed. Else, the thief who proposed is killed, and the next thief presents his proposal.
Now, We know that the first priority of each thief is his own life, followed by money, and lastly to kill other thieves. Assuming that all the thieves are extremely intelligent and have distinct integer ages, What ... should the First thief propose?[/quote]

It's simple. Kill the Batman.

Sorry. Couldn't resist.

oops made a small mistake

i am sure now ans has to

97,0,1,0,2           0r   97,0,1,2,0

I think (97,0,1,2,0) is the right answer. Not sure though; I did this problem way back in the eighteenth century. But one thing's for certain; last guy gets a big fat goose egg.

no not always

[quote]last guy gets a big fat goose egg.[/quote]

can u justify that ?

i can disprove it

terribly sorry [b]preetham[/b]. Out of the fourth and fifth, one guy gets nothing. I did the problem again and verified it. Your solution is perfectly right. It has to be (97,0,1,0,2) or (97,0,1,2,0).

I don't understand why the '1' and '2' wala guys accept the proposal?

[color=red]The king, to test a candidate for the position of wise man, offers him a chance to marry the young lady in the court with the largest dowry. The amounts of the dowries are written on slips of paper and mixed. A slip is drawn at random and the wise man must decide whether that is the largest dowry or not. If he decided it is, he gets the lady and her dowry if he is correct; otherwise he gets nothing. If he decides against the amount written on the first slip, he must choose or refuse the next slip, and so on until he chooses one or else the slips are exhausted. In all, 100 attractive young ladies participate, each with a different dowry. How should the wise man make his decision?    [/color]

Ummm, is this a witty chutkula type puzzle or are you serious? If you're serious, this is a pretty tough one :( You'll have to check at each turn with your sample size, its mean and standard deviation. And then use the s/rt of n formula to get population standard deviation and then calc t scores and use the charts to find its probability and the probability of getting a bigger t score in the cumulative remaining turns out of 100 blah blah blah. But I don't think anyone in this forum even regards statistics as a proper branch of maths :P So, I guess there's a simpler cheat too :P

ok leave it  :D

ans cums in n/e and all
i myself donno the proof  :rotfl:

I thought so. Its actually a lil messing around with and differentiation of this equation for minima I guess-
[img]http://upload.wikimedia.org/math/8/3/2/83204805d2e6d513c103aed8a4232f46.png[/img]
Its the probability distribution function. Its never actually used in stats though.....normally, you use the implications of the formula for statistical analysis. But if the answer has the ominous 'e' its a bad sign I'm afraid. I might be able to actually solve the prob, but its probably gonna make me look back through my whole AP statistics book which I'd never want to do again :P

Hey keerie. Could you please post the explanation of the solution for the fourth question please.( the 100 thieves question).
[hide]Sorry for troubling you to post the entire expln, but this got me thinking.[/hide] :gathering:

[quote="keyree10"][b]1.[/b] Three people - $ A, B$ and $ C$ - are sitting in a row in such a way that $ A$ sees $ B$ and $ C$, $ B$ sees only $ C$, and $ C$ sees nobody. They were shown $ 5$ caps - $ 3$ red and $ 2$ white. They were blindfolded, and $ 3$ caps were put on their heads. Then the blindfolds were taken away and each of them was asked if they could determine the color of their caps. After $ A$, and then $ B$, answered negatively, $ C$ replied affirmatively. How was that possible?

[/quote]

A replied negatively, which implies that B and C are not both wearing a white cap, otherwise he could have known that he was wearing a red cap. B replied negatively, because from A's answer s/he guessed that B and C must both be wearing a red cap or any one of them is wearing a white cap. B would have known which colour s/he was wearing if s/he found that C was wearing a white cap, but since s/he replied negatively, that was definitely not the case. so C must be wearing a red cap.

[quote="keyree10"]

[b]3.[/b] Calvin and Hobbes were walking down the street, discussing about Calvin's worms. Hobbes is trying to find out the ages(integers) of the three worms.
Calvin says, "The product of their ages is $ 36$".
Hobbes asks for more info.
Calvin says, "The sum of their ages is equal to the number on the red house across the street".
Hobbes is still unable to find their ages.
Calvin says, "The eldest one likes chocolate"'.
Hobbes says, "Ahh... Now I've got it".
How does does Hobbes find out their ages?

[/quote]

hobbes must know what the number on the red house is. so if we find out the sums of the three factors whose product equals 36, then obviously we'll find out that two or more of the sums are equal. 'the eldest one' is the clue that there is one and only one largest factor.
6+6+1=9+2+2
so the ages of the worms are 9, 2, 2.

2. i can't come up with an acceptable answer. if B and C did nothing at all but stand, i don't see how it was possible for A to find out his cap colour.

and congratulations on your 100th post keyree10.

is there any order in which the thieves accept a proposed distribution?

[quote="keyree10"]
[b]4.[/b] Five thieves are trying to share a loot of $ 100$ gold coins. They come up with this idea to decide how the loot should be distributed. Each thief comes up with a proposal and presents it, and the order in which they present their proposals is determined by seniority. If the proposal of a particular is accepted by atleast  half the other thieves, then it is followed. Else, the thief who proposed is killed, and the next thief presents his proposal.
Now, We know that the first priority of each thief is his own life, followed by money, and lastly to kill other thieves. Assuming that all the thieves are extremely intelligent and have distinct integer ages, What distribution should the First thief propose?

[/quote]

ok, i assume that the answer to my previous question is no, and here is my soln.

the 4th person will try to prevent the no. of thieves alive coming down to 2 as he will get shot then.
the third person knows the picture and he knows for sure that if he gets to propose he'll get the 4th's support no matter what, as the 5th thief would want the 4th dead and all the loot both (since getting his mate dead is on his priority list), and hence he cannot get shot.
the 4th knows if the no. of thieves alive come down to 3 he'll have to agree with whatever the 3rd proposes. so he would agree to a distribution which gives him anything.
the last knows this and hence will agree with a proposal if that gives him anything at all.
the 2nd knows that the third will always disagree with him if his turn comes, even though if he gives both the 4th and the 5th a little share of the loot(a coin each), they'll agree to his proposal for otherwise they'll get nothing. so he'll try to get the 1st off the way.
the 3rd knows this and would agree to the 1st if he gives him anything.
now the 1st knows that the 2nd would not agree to his proposal, and the 3rd will agree with him if he gives him anything, and the 4th or the 5th will agree with him if he gives them a little more than the minimum that the 2nd will offer them if his turn comes.
so he will propose a distribution (97,0,1,0,2) or (97,0,1,2,0)

@Preetham:

This looks like a variation of something I've worked on sometime back.
Assuming the wise man can remember the dowry of all the slips he has seen, till any point of time, we can increase the probability from 1/100.

Think over if this increases the probability:
The wise man rejects exactly half of the proposals, here the first 50 proposals, no matter the amount.
Then he proceeds to the second half, (at the same time remembering the dowry offered in the previous proposals).

So lets say the highest dowry bid in the first half was x

The wise man accepts the first offer in the remaining half, whose dowry bid exceeds x

If not, he continues till the end, and maybe he accepts the last offer(or is there a better way to end? )

Am not sure about this, but I think this increases the probability, I worked on a similar thingy some time back.

 :roll:

[b]@illiterate:[/b] Here's the solution to the second question.

$ A$ thinks, "Let's say I have a white cap on my head. Then $ B$ would see one red cap and one white cap, and start to think, - [i]"If I were wearing a white cap like $ A$, $ C$ should have shouted out "I'm wearing red", but he hasn't. So I must be wearing a red cap."[/i] - Thinking thus, he should have shouted out "I'm wearing red", but he hasn't. So my initial assumption that I'm wearing a white cap is wrong. So, I MUST BE WEARING A RED CAP! ".


[b]Right triangle ABC with right andle at vertas C has median AD of length 5 and median BE of lenght 2rad10. Find the length of the hypotenuse of triangle ABC. Express your answer in simplest radical form.[/b]

[i]I want to show you what I did for this problem, but it is hard to explain without a diagram. And i can't make one out of latex because strangely latex won't work on my comp. Please help me.[/i]

[b]If a prism has 12 vertices, how many edges does it have?[/b]

[i]I know there is this formula: E+2=V+F. but u can't use that because u only know the vertices. How do you do this? [/i]

[hide="1"]$ AD^2\equal{}AC^2\plus{}\frac{BC^2}{4}$
$ BE^2\equal{}BC^2\plus{}\frac{AC^2}{4}$
Add the two...

$ 25\plus{}40\equal{}\frac{5}{4}(AC^2\plus{}BC^2)$
Multiply by 4/5

$ 52\equal{}AC^2\plus{}BC^2$
$ AB^2\equal{}52$
$ AB\equal{}2\sqrt{13}$[/hide]

[hide="#2"]well 12/2=6, so its a hexagonal prism, giving 18 edges[/hide]

[quote="usahw1011"][i]I want to show you what I did for this problem, but it is hard to explain without a diagram. And i can't make one out of latex because strangely latex won't work on my comp. Please help me.[/i][/quote]

I would assume you are just using LaTeX incorrectly. There is nothing to download, all you have to do is put your code in dollar signs ($ \$$).

[code]$\sqrt{x}$[/code]

$ \sqrt{x}$

Whenever you want to see someone else's code, just run your mouse over it.[/code]

meanwhile, prisms have congrent bases, and all vertices are on one of the bases.

with 12 vertices, there are 6 vertices on each base.

this means 6 edges on each base. there are also 6 edges connecting the bases.

6 edges times 2 bases plus another 6 edges connecting them equals 18
6*2+6=18
(this is just more indepth to what unimpossible said)


1.What are $ max(a,b)$ and $ min(a,b)$ . Take an example please !?

And min(a,b)=smaller of a and b

max(5,3)=5
min(5,3)=3

Then sometimes you can have like max's and min's of function. For example,

Find $ \max(\minus{}x^2\minus{}6x\plus{}7)$.

[hide="Solution"]This is a parabola, so simply find it's vertex. So the maximum happens when $ x\equal{}\minus{}3$ so the maximum is $ \boxed{16}$.[/hide]

This is just a very simple type of problem cuz the olympiad level problems are definately not like this one.   :P

[quote="modularmarc101"]Find $ \max( \minus{} x^2 \minus{} 6x \plus{} 7)$.

[hide="Solution"]This is a parabola, so simply find it's vertex. So the maximum happens when $ x \equal{} \minus{} 3$ so the maximum is $ \boxed{16}$.[/hide]
[/quote]

And adding on, you can prove that the vertex is either the minimum or maximum of a parabola. Then using a general parabola form ($ ax^2\plus{}bx\plus{}c\equal{}0$), you can easily derive the the y value that would make the parabola extreme.

An alternate, occasionally useful way of defining min and max for the two-variable case is:

$ \text{min}(x,y) \equal{} \frac {x \plus{} y}{2} \minus{} \frac {| x \minus{} y |}{2}$

$ \text{max}(x,y) \equal{} \frac {x \plus{} y}{2} \plus{} \frac {| x \minus{} y |}{2}$

The functions min and max easily extend to higher number of variables, though the definition given above does not extend. For the case of 3 or more variables, you should note that min and max provide the limit cases of the power n-mean of the list as n tends to negative infinity and to infinity, respectively.


Show in any triangle $ ABC$ the inequality

$ \frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2} \le \frac {1}{4r^2}$

edit: is fixed , thanks  :lol:

I think you meant $ \frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2} \le \frac {1}{4r^2}.$  :D

using standard notation, the inequality becomes $ \sum \frac{1}{(x\plus{}y)^2} \le \frac{x\plus{}y\plus{}z}{4xyz}.$

by AM-GM, $ LHS\le \sum \frac{1}{4xy} \equal{}RHS.$ :ddr:


quadratics
number theory unsolved
number theory

If n is a natural number,prove that the number $ (n\plus{}1)(n\plus{}2)...(n\plus{}10)$ is not a perfect square.

Let $ (n\plus{}1)(n\plus{}2)...(n\plus{}10)\equal{}A$.If one of the numbers $ (n\plus{}k)$, $ k\in\{1,2..,10\}$ is divisible by 11, then $ A$ has just once 11 in its canonical decomposition,which implies that $ A$ can't  be a perfect square then. If none of $ (n\plus{}k)$, $ k\in\{1,2..,10\}$ is divisible by $ 11$, we obtain $ A\equiv10!\equiv\minus{}1(mod11)$ by Wilson theorem, but $ \minus{}1$ isn't a quadratic residue modulo $ 11$ so $ A$ can't be perfect square in this case too. Finally we conclude that $ A$ isn't a perfect square for any $ n\in \mathbb{N}$. QED

[quote="broniran"]Let $ (n \plus{} 1)(n \plus{} 2)...(n \plus{} 10) \equal{} A$.If one of the numbers $ (n \plus{} k)$, $ k\in\{1,2..,10\}$ is divisible by 11, then $ A$ has just once 11 in its canonical decomposition[/quote]
you are wrong,maybe one of the numbers$ n\plus{}k$ is divisible by $ 11^2$


If you are burning a CD, you have 2 bars showing the percent done- 1 for the track and 1 for the whole CD. If you are burning a 6 track CD with 600 megabytes per track, at what megabyte numbers will the percents displayed on each bar be the same?

Original problem,

by Billy

PS For a hint, look below:

[hide]Think of a clock[/hide]

[hide]let x represent the # of tracks

let y represent the # of mb of the current track

(600x+y)/3600=y/600

600x+y=6y

600x=5y

120x=y

so when no tracks are written, then 0mb of the track will be the same (i dont know if you wanted this too)

track 1 done - 120 mb of track 2

track 2 done - 240 mb of track3

track 3 done - 360 mb of track 4

track 4 done - 480 mb of track 5

track 5 done - 600 mb of track 6 (cd is done)

btw, this is ignoring CD closing time[/hide]

[hide]x = the number of tracks that are done

y = the amount of megabytes done on the current track

(600x+y)/3600 = y/600

y = (600x+y)/6

6y = 600x+y

5y = 600x

So, here are the possible answers:

0 tracks done, and 0 megabytes done on the 1st track

1 track done, and 120 megabytes done on the 2nd track

2 tracks done, and 240 megabytes done on the 3rd track

3 tracks done, and 360 megabytes done on the 4th track

4 tracks done, and 480 megabytes done on the 5th track

5 tracks done, and 600 megabytes done on the 6th track[/hide]


There are n persons in a village. Each one of them is either a knight or a knave.
The number of knights is greater than the number of knives.

A stranger in the village wants to know for each person if he is a knight or a knave. He can ask the following question to a person: Are you a knight or a knave?

Prove that the stranger can identify each person by asking 2n-3 questions.

Is something wrong with the problem?
The stranger can ask every person and that's $ n$ questions...
Have you forgotten some condition?

I'm seeing a problem the opposite way. Everyone who is asked will say they are a knight. Is there some other nuance that can be added to the question?

Well, if each question can be asking if someone else is a knight or knave, it can be done in n-1 questions...so that can't be it... (ie asking one person about each other person)

The question is obviously mis-phrased, for the reason that MellowMelon stated.  Presumably, you are allowed to ask questions of the form, "Is person $ X$ a knight or a knave?" to any person (i.e. not just to person $ X$ himself).  However, as K81o7 stated, the problem can then be solved in only $ n\minus{}1$ questions: choose a given person $ X$ and ask each other person, "Is $ X$ a knight or a knave?" and then take a majority vote.  In the case that $ n$ is even, this vote necessarily results in a majority and that majority must be correct, so everyone in the majority is a knight, everyone in the minority is a knave, and $ X$ is whatever the majority says he is.  In the case that $ n$ is odd, this vote either will result in a majority (in which case proceed as above) or a tie, in which case $ X$ must be a knight and everyone who claimed that $ X$ is a knight is a knight, while everyone else is a knave.


geometry rectangle inequalities trigonometry circumcircle

M is a point inside rectangle ABCD. prove that MA*MC+MB*MD>=area of rectangle

Translate so A maps to D. Then apply ptolemy's inequality. [I had just thought of this problem recently; I thought it was new  :( ]

ns! now try a bit similar solution without ptolemy

ok I give here a great hint. and can we prove it using analytic?

[quote="srulikbd"][color=darkred]$ M$ is a point inside rectangle $ ABCD.$ Prove that $ MA\cdot MC+MB\cdot MD\ge [ABCD]$[/color][/quote]
[color=indigo][b]Proof.[/b] Construct triangle $ ADE$ outside of the rectangle $ ABCD$ such that $ \triangle BMC=\triangle DEA$
Denote $ H\in BC,\,K\in AD: \,M\in HK,\,HK\perp AD$ $ \,I\in AE: \,MI\perp AE;\,J\in DE: \,MJ\perp DE$
We have: \[ [ABCD]=HK\cdot BC=2\cdot\left([MBC]+[MAD]\right)=2\cdot\left([EDA]+[MAD]\right)\\ =2\cdot [AMDE]=2\cdot\left([MAE]+[MDE]\right)\le AM\cdot AE+DM\cdot DE\]
Hence $ \boxed{[ABCD]\le AM\cdot CM+BM\cdot DM}$[/color]

[color=darkred][b]Lemma.[/b] Let $ M$ be an interior point of the right triangle $ ABC$, where $ AB\perp AC$.

Prove that $ \{\begin{array}{ccc}O\in (BC)\ ,\ OB=OC\\\ X\in (AB)\ ,\ MX\perp AB\\\ Y\in (AC)\ ,\ MY\perp AC\end{array}$ $ \implies$ $ MB\cdot MC\ge AX\cdot AY+BX\cdot CY$,

with equality $ \Longleftrightarrow$ $ XA\cdot XB=YA\cdot YC$ $ \Longleftrightarrow$  $ \widehat{MBA}\equiv\widehat{MCA}$ $ \Longleftrightarrow$ $ OX=OY$.

[b]Remark.[/b] For any interior point $ M$ there is and the relation $ OX^{2}+OY^{2}\ge R^{2}$.[/color]

[hide="Proof."]
[color=darkblue] I"ll use the remarkable [b]C.B.S.'s inequality[/b] $ (ax+by)^{2}\le (a^{2}+b^{2})\cdot (x^{2}+y^{2})$.

Therefore, $ MB^{2}\cdot MC^{2}=$ $ (MX^{2}+BX^{2})\cdot(MY^{2}+CY^{2})=$

$ (AY^{2}+BX^{2})\cdot(AX^{2}+CY^{2})\ge$ $ (AX\cdot AY+BX\cdot CY)^{2}$ $ \implies$

$ MB\cdot MC\ge AX\cdot AY+BX\cdot CY$.

We have equality $ \Longleftrightarrow$ $ XA\cdot XB=YA\cdot YC$ $ \Longleftrightarrow$ $ \frac{MX}{BX}=\frac{MY}{CY}$ $ \Longleftrightarrow$

$ \tan\widehat{MBA}=\tan\widehat{MCA}$ $ \Longleftrightarrow$ $ \widehat{MBA}\equiv\widehat{MCA}$. But $ XA\cdot XB=YA\cdot YC$ $ \Longleftrightarrow$

$ p_{w}(X)=p_{w}(Y)$ $ \Longleftrightarrow$ $ R^{2}-OX^{2}=R^{2}-OY^{2}$ $ \Longleftrightarrow$ $ OX=OY$, where

$ p_{w}(X)$ is the power of the point $ X$ w.r.t the circumcircle $ w=C(O,R)$ of $ \triangle ABC$.[/color][/hide]
[color=darkblue][b]Proof.[/b] Denote $ \{\begin{array}{ccc}X\in (AB)\ ,\ MX\perp AB & ; & Y\in (BC)\ ,\ MY\perp BC\\\ Z\in (CD)\ ,\ MZ\perp CD & ; & T\in (DA)\ ,\ MT\perp DA\end{array}$.

Apply the above lemma to the point $ M$ in $ \triangle ABC$ and $ \triangle BCD$:

 $ \{\begin{array}{c}MA\cdot MC\ge MX\cdot MY+MT\cdot MZ\\\\ MB\cdot MD\ge MY\cdot MZ+MX\cdot MT\end{array}$ $ \implies$

$ MA\cdot MC+MB\cdot MD\ge (MX+MZ)(MY+MT)=AD\cdot AB=S$,

where $ S$ is tha area of $ ABCD$. In conclusion, $ \boxed{MA\cdot MC+MB\cdot MD\ge S}$.[/color]

[quote][color=darkred][b]Other problem.[/b] Let $ ABCD$ be a rectangle.

For a interior point $ M$ denote $ \{\begin{array}{c}X\in (AB)\ ,\ MX\perp AB\\\ Y\in (BC)\ ,\ MY\perp BC\end{array}$.

Prove that $ \boxed{MA\cdot MC+MB\cdot MD\ge \frac{1}{2}\cdot S+XA\cdot XB+YB\cdot YC}$.[/color][/quote]


linear algebra
matrix
algebra
polynomial
function

1.
If all of eigenvalues of $ A$ is $ 1$, can we show that $ A^k$ is similiar to $ A$ for any natural number $ A$ ?
2.
If $ A$ and $ B$ are two invertible matrix, can we construct a complex number $ c$ such that $ det(A\plus{}cB)\equal{}0$ ?
Thanks in advance.

I think you mean "for any natural number $ k$" for number 1.

2. $ \det(A\plus{}zB)$ is a polynomial function of $ z.$ Since $ B$ is invertible, we can rewrite this as $ \det(B)\det(zI\plus{}AB^{\minus{}1}),$ which is clearly non-constant. Invoke the Fundamental Theorem of Algebra. (Note that it doesn't matter whether or not $ A$ is invertible.)

(1) depends on the field. In characteristic $ p$, $ A^p$ is not similar to $ A$; if this is an $ n\times n$ matrix, $ A^{p^n}\equal{}I$.
In characteristic zero, it is true- put $ A$ it in Jordan form, and note that you'll always have nonzero entries in the same places on the superdiagonal for powers of $ A$.

[quote="mathangel"]Why is it not true that $ A \equal{} I$ because all eigenvalues of $ A$ are 1?[/quote]
\[ \begin{pmatrix}1&1\\0&1\end{pmatrix}\]


[quote="randomdragoon"]A blind beggar had a brother who died. What was the beggar's relation to this person?

Note: The answer is not "brother".[/quote]
sister
the beggar is a girl. her brother has died

Keep Smiling
Malay


Below are a few questions that many people missed on the sprint round last week.  See if you can solve them and post the solution (don't just tell the answer, show how you solved it).  Be sure to hide your solutions.

3.  What is the sum of the two perfect square integers that are closest to 2005?

7. Each of the letters A, B, C and D represents a different odd integer between two and ten. What is the least possible value of $ \dfrac{A * B \minus{} C}{D}$ ? Express your answer as a common fraction.

9. The points (x, y) represented in this table lie on a straight line. When the equation of this line is written in the form y =Ax + B, what is the value of A + B?
[img]table.jpg[/img]  (see table.gif below for image--If someone can help me put this in correctly, let me know)

12. A 12”-diameter pizza and a 16”-diameter pizza are each cut into eight congruent slices. Jane ate three slices of the 12” pizza. Mark ate three slices of the 16” pizza. How many more square inches of pizza than Jane did Mark eat? Express your answer as a common fraction in terms of $ \pi$.

[hide="Hint for 3."]This question was easy,but I miss understood it..
Straight to the solution...
All you have to know is that 2005 is between $ \ 40^2$(1600) and $ \ 50^2$(2500).

Then you have to try numbers out.
And add them together.

I won't go any further because it would be pretty obvious..[/hide]

as a first time user in AoPs
im not 100% sure if i did the things right to
hide the answers
so dont get mad at me if it doesnt hide

also
my answers are probably all wrong......
im still trying......

mr. beaty,
if i get this right, this better help my mathcounts grade go up

my answer to #3
[hide]91[/hide]

my answer to #7
[hide]6/7[/hide]

my answer to #12
[hide]21/2 pi[/hide]

[size=150][b]Note from Mr. Beaty:  Please show the solution, not just the answer.[/b][/size]

[quote="spiders_hate_chestnut"]

mr. beaty,
if i get this right, this better help my mathcounts grade go up

my answer to #3
[hide]91[/hide]

my answer to #7
[hide]6/7[/hide]

my answer to #12
[hide]21/2 pi[/hide][/quote]

I'll post my answers if it really helps my MC grade go up..  :lol:

[hide="Answer and Solution for 3"]I got [u]$ \boxed{3961}$[/u] (44^2=1936)+(45^2=2025)[/hide]

[hide="Answer and Solution for 7"]We know that $ \ A,B,C,D$ has to be either $ \ 3,5,7,9$.
Then I guess you just plug them in..
Answer:$ \boxed{\frac 67}$[/hide]

[hide="Answer and Solution for 9"]Even though I didn't attempt this one, I believe I have Solutions to it...
For this one, what I did was I made an equation.
$ \ v \equal{} A(t \minus{} 2) \plus{} B$ $ \rightarrow \boxed{ \minus{} 2A \plus{} At \plus{} B \minus{} v \equal{} 0}$
$ \ v \equal{} At \plus{} B \minus{} 6$   $ \rightarrow \boxed{At \plus{} B \minus{} v \equal{} 6}$
Then you subtract those two boxed equation to get $ \ A \equal{} 3$
By using substitution,
$ \boxed{A \equal{} 3 B \equal{} 1}$
[i]Answer[/i]: $ \ 3 \plus{} 1 \equal{} \boxed{4}$[/hide]

[hide=" Answer and Solution for 12"]OMG.. I can't believe I didn't attempt this one either..
[u]Answer[/u]:$ \boxed{\frac {21}2\pi}$
Solution(without $ \pi$): (8*3) - {(36/8)*3} = 10.5 which is $ \frac {21}2$ then you multiply $ \pi$[/hide]

[size=150][u]*@Mr. Beaty, try putting the table into JPG form.. Mine worked as you can see..[/u][/size]

You need to read #3 much more carefully.

Ooops..
I edited mine and put the right answer.

[quote="Kicha"]3. What is the sum of the two perfect square integers that are closest to 2005?

     [hide]Keep guessing until you get real close to the number
   44*44=1936        45*45=2025
              44+45=89[/hide][/quote]

[size=150]READ THE QUESTION AGAIN[/size] like Mr. Beaty said...

solution to #3
[hide]i read davids hint
and tried out 44, 45, 46
because they are in the middle of 40 and 50
just square each numbers and see which one two is closest to 2005
which in this case is 45 and 46
45+46=91
[/hide]

solution to #7
[hide]again, i got the numbers 3,5,7, and 9
and i knew that you would multiply the two smallest numbers
so 3x5
after that, i tried 7 as C and 9 as D
and vice versa
which got me my answer[/hide]

solution to #12
[hide]
gosh, if i missed this one
i will bang my head on the wall for an hour
(did u kno that bangin your head on the wall
for about an hour burns like 300 calories??)

the 12 and 16 are the diameters
so 6 and 8
square 6 and 8
to get 36 and 64
so the area of the 12 inch pizza is 36 pi
and the area of the 16 inch pizza is 64 pi

it says that both pizzas are divided into 8 pieces
so 36pi/8 for the area of each of the slices in 12 inch pizza
36pi/8 = 9pi/2
and 64pi/8 for the area of each of the slices in 16 inch pizza
64pi/8 = 8pi

jane ate 3 slices of 12 inch pizza
so 3(9pi/2) = 27pi/2

mark ate 3 slices of 16 inch pizza
so 3(8pi) = 24pi

"How many more square inches of pizza than Jane did Mark eat? Express your answer as a common fraction in terms of pi"
so
24pi - 27pi/2 = 48pi/2 - 27pi/2
then do the math
21pi/2
[/hide]

[quote="spiders_hate_chestnut"]solution to #3
[hide]i read davids hint
and tried out 44, 45, 46
because they are in the middle of 40 and 50
just square each numbers and see which one two is closest to 2005
which in this case is 45 and 46
45+46=91
[/hide]
[/quote]

It's not right! Read the question again like Mr.Beaty said...

Okay..
Before any of you post the answer and solution for #3, notice the part where it says[b]:[size=150]sum of the two perfect square integers[/size][/b], not sum of the two radical of perfect square integers that are closest to 2005.

[quote="spiders_hate_chestnut"]i get it
i get it
i get it
i get it
i get it
i get it

answer and solution to #3
[hide]
44 squared is 1946
and 45 squared is 2025
add it together
1946+2025=3961
if its not that
then its 89
darn it!![/hide]

From Mr. Beaty...VERY GOOD!  You Got It!  Now you get the points![/quote]

Good job. You finally figured it out.
You seemed really confused...but still you did it. :lol:


Xanthia buys hot dogs that come in packages of six, and she buys hot dog buns that come in packages of eight. What is the smallest number of hot dog packages she can buy in order to be able to buy an equal number of hot dogs and hot dog buns?

The answer is 4. The LCM of 6 and 8 is 24, so she can buy at minimmum 4 packages.

However, the answer should be 0. If you don't buy any packages of anything, you will have an equal number of hot dogs and buns, which is 0.

[quote]Xanthia [b]buys[/b] hot dogs that come in packages of six, and she [b]buys[/b] hot dog buns that come in packages of eight.[/quote]


I thought up this on the fly, it worked with each number I tried, yet I do not know how to prove it. The conjecture is:

[i]For all positive integers [/i]N[i] and primes [/i]P[i], [/i] [b](N^P) - N [/b][i] is always divisible by[/i] P.

How would I go about proving that?

[quote="Farnak"]Could we use Fermat's Little Theorem?

EDIT : I'm silly. I added a lot of unnecessary steps.

By Fermat's Little Theorem, $n^p \equiv n (mod\ p)$
$n^p - n \equiv 0 (mod\ p)$

So $n^p - n$ is divisible by $p$.[/quote]

Sweet and short

The posted problem really is Fermat's Little Theorem in the sense that it's trivially different.  I think what the problem poster wants at this point is a proof of Fermat's Little Theorem.

Idea: (source: Euler)

For each fixed $p,$ try proving this by induction on $N.$ You'll need to know the binomial theoerm, and certain divisibility properties of binomial coefficients.


UNM-PNM Round 1 2005  I just took this today.
(1-x+2x^4-3x^9+4x^16-5x^25+6x^36+x^2005)/(x^2-1) has a remainder of...


I got disconnected 4 times within 90 minutes of my class....

Use Opera (http://www.opera.com).

Also, we'll have a new classroom in mid-late September that we expect to not have difficulties with IE and Firefox for some users.

Opera is truly amazing! I also had disconnection problems but after opera came along the classroom works like a charm! Also it takes about 5 minutes to download so that is also good!


prove that for all n>0 , 4^n + 15n - 1 is divisible by 9/multiple of 9

I tried:

 (4^n + 15n - 1)/9 = k  = 4^n + 15n - 1 = 9k

4^n+1 + 60n - 4 = 36k

stuck now

thanks!!

$ 4^n \plus{} 15n \minus{} 1$ computed in $ n\equal{}1$ yields $ 18$. Assume divisible by $ 9$ until $ n$.

Now $ 4^{n\plus{}1} \plus{} 15(n\plus{}1) \minus{} 1 \equal{} 4(4^n \plus{} 15n \minus{} 1) \minus{} 9(5n \minus{} 2)$ and all is clear.

$ 4^3 \equal{}1 mod 9$ so when 3 divides n the expression is divisible by 9. similarly when n is $ 1$or $ 2 mod 3$ it is divisible by 9

hello, since $ 4^3\equiv 1 \mod 9$ we have to consider three cases
a) $ n\equal{}3k$
b) $ n\equal{}3k\plus{}1$
c) $ n\equal{}3k\plus{}2$
from here we got
a) $ 4^{3k}\plus{}15\cdot3k\minus{}1\equiv 1\plus{}0\minus{}1\equiv 0 \mod 9$
b) $ 4^{3k\plus{}1}\plus{}15(3k\plus{}1)\minus{}1\equiv (4^3)^{k}\cdot4\plus{}45k\plus{}15\minus{}1\equiv18\equiv 0 \mod 9$
c) $ 4^{3k\plus{}2}\plus{}15(3k\plus{}2)\minus{}1\equiv (4^3)^{k}\cdot16\plus{}45k\cdot30\minus{}1\equiv45\equiv 0\mod 9$
Sonnhard.

case $ n\equal{}1$
$ 4^{1}\plus{}15(1)\minus{}1\equal{}18$

assuming true up to $ n$

$ 4^{n\plus{}1}\plus{}15(n\plus{}1)\minus{}1$
$ 3*4^{n}\plus{}15\plus{}(4^n\plus{}15n\minus{}1)$
$ 3(4^{n}\plus{}5)\plus{}(4^n\plus{}15n\minus{}1)$

and then because $ 4^n\equiv 1\mod 3$ and $ 5\equiv 2\mod 3$
so $ 4^n\plus{}5 \equiv 0\mod 3$ and we are done.


Ryan's piggybank contains 5 times as many dimes as quarters, for a total value of 2.25 if q represnets the number of quarters, what equation could be used to find the number of quarters in ryans piggy bank[/code]

[hide]Let q be the number of quarters and d be the number of dimes

$(1) d=5q$

$(2) 0.25q+0.1d=2.25$
[/hide]

Just use linear systems to solve for  values of d and q.

[quote="FrankieBlueEyes"]can equation 2 be solved?[/quote]

Yeah, just substitude equation (1) in equation 2 to solve

$\boxed{q=3}$

[quote="FrankieBlueEyes"]whats a system of linear equations[/quote]

It has the form $ax+by+c=0$

[quote="FrankieBlueEyes"]whats a system of linear equations[/quote]

Well, to pass the regents, you have to know this  ;)

It is several equations in which a solution set works in all the equations.

An example would be

$x+y=2$
$x-y=0$

If the solution set is $(1,1)$.

Hope I'm explaining this correctly...


LaTeX
modular arithmetic
number theory
relatively prime
prime factorization

I often see problems like:"What is the tens digit of $x^y$?" I can find the units digit, but how do you do the tens digit?

Usually there is a pattern, but could you give an example?

i've done that with the numer 2, and there is a repeating pattern of 20..., the one for 5 goes 2 2 2 2  2 2 2... the one for 1 goes 0 0 0 00 0 0 00 0 0 0 lol... and i think the other powers of 2 also have a repeating pattern of 20 terms...

let's see...
7^2 = 49
7^3 = 43
7^4 = 01
7^5 = 07
7^6 = 49

It turns out that the last digits repeat in cycles of four, so 7^13= ...49

Well, for the last two digits, all that matters is the last two digits, so, for something without a pattern like 3^8, we have
3
9
27
81
243, dumb the hundreds
43*3=129, again, dump the hundreds
29*3=87
87*3=261, so the tens digit is a six.
Sorry for the lack of Latex, I'm lazy right now.

If you want to find the last two digits of a number generally you find it in mod 100.
If you wanted to find the last digit of $3^{1234}$:

We all know that $3^{40} \equiv 1 \pmod {100}$ so we have remainders of 1 all the way down to we have $3^{34}$. Then you kinda take it step by step.
$3^5 \equiv 43 \pmod {100}$ so we know that $3^{10} \equiv 49 \pmod {100}$ because $43^2 = 1849$.

We then know that $3^{20} \equiv 1 \pmod {100}$ because everybody knows that $49^2 = 2401$.

We already know that $3^{10} \equiv 49 \pmod {100}$ so $3^{30} \equiv 49 \pmod {100}$ and since $3^4 \equiv 81 \pmod {100}$ we multiply $81 \cdot 49 = 3969$ to find that $3^{34} \equiv 69 \pmod {100}$.

[quote="KhL1392"][quote="chess64"]Like $7^{13}$, for example.[/quote]

well in this case, we'll have a pattern, like usual
$7^1$ gives a $0$ for the tens
$7^2$ gives a $4$ for the tens
$7^3$ gives a $4$ for the tens
$7^4$ gives a $0$ for the tens
if you really want to keep going, you see that this pattern repeats for every 4 groups of powers of 7 so, based upon the power you can find the tens:
take the power and divide by 4 this will give you a remainder, and we've established
$7^1$ gives a $0$ for the tens or a remainder of 1 when the power is divided by 4
$7^2$ gives a $4$ for the tens or a remainder of 2 when the power is divided by 4
$7^3$ gives a $4$ for the tens or a remainder of 3 when the power is divided by 4
$7^4$ gives a $0$ for the tens or a remainder of 0 when the power is divided by 4

since 13/4 gives a remainder of 1, the answer is a tens digit of 0...and it checks out, $7^{13}$ is equal to 968890104$0$7

just so you know, though it may look long, with enough practice, it's really easy[/quote]

I know how to do it with most numbers, but 3 doesn't seem to have a pattern. Tarquin's solution helped though. Thanks everyone!

[quote="tarquin"]We all know that $3^{40} \equiv 1 \pmod {100}$...[/quote]

 :blush: umm, we do? :?

EDIT: Oh, is it because $3^{20} \equiv 01 \pmod {100}$, and $1^2=1$, so $3^{40} \equiv 01 \pmod {100}$?

[quote="tarquin"]The number of numbers that are relatively prime to 100 that are less than or equal too 100 is 40 so any number so $x^{40}$ is always $1 \pmod {100}$ for any integer $x$. [/quote]

I don't quite understand what you are saying because $20^{40}$ is $0 \pmod {100}$
More examples of integers that are not 1 mod 100 are 2,4,5,6...

$x$ and $100$ have to be relatively prime for
$x^{40} \equiv 1\pmod {100}$

[quote="tarquin"]The number of numbers that are relatively prime to 100 that are less than or equal too 100 is 40 so any number so $x^{40}$ is always $1 \pmod {100}$ for any integer $x$. Another example is $x^{16}$ is always $1 \pmod {17}$ because there are 16 integers relatively prime to 17 that are less than or equal to 17, also known as totatives.[/quote]

i dont feel like thinking right now so can u explain why?

[quote="tarquin"]
We all know that $3^{40} \equiv 1 \pmod {100}$ [b]so we have remainders of 1 all the way down to we have $3^{34}$[/b].[/quote]

What do you mean?

Since $3^{1234} = 3^{{40^{30}}} \cdot 3^{34}$ and you multiply remainders we have a remadiner of $1 \cdot 1 \cdot 1 ... $ times whatever $3^{34}$ is in mod 100.

[quote="tarquin"]The number of numbers that are relatively prime to 100 that are less than or equal too 100 is 40 so any number so $x^{40}$ is always $1 \pmod {100}$ for any integer $x$. Another example is $x^{16}$ is always $1 \pmod {17}$ because there are 16 integers relatively prime to 17 that are less than or equal to 17, also known as totatives.[/quote]

I don't really understand that statement. Also, is there a way to find the number of numbers less than or equal to a certain number that are relatively prime to that number? (sorry for the abusage of number)

If we looked at the multiples of three from

1, 2, 3, 4, 5, 6, 7, 8, 9

Only 1/3 of the numbers 1-9 are multiples of three. Then, the other 2/3 are not multiples of three.

So to find the totatives of a number, say 24, we take its prime factorization:

$2^3 \cdot 3$

and "fill up" the numbers.

$\frac 12 \cdot \frac 23 = 13$

Then multiply by 24. $\frac 13 \cdot 24 = 8$

And 8 is the number of numbers relatively prime to 24 less than or equal to 24. Another example is 23.

$\frac {22}{23} \cdot 23 = 22$ There are 22 totatives of 23.

I think this is also known as Euler's totient function.

i think all tens digits cycle, not sure though

It's pretty easy to prove that they do - in fact, all last-two-digit combos recycle, even if you go through all 100 (but that might be impossible).

Sorry if this is too advanced, but yes, it is true that

$a^{40}\equiv 1\bmod 100$ as long as $a$ and $100$ have no common factors except $1$.  This is the 'worst-case' scenario

But this can be a trap!  For example, we have the better result $3^{20}\equiv 1\bmod 100$.  So it always pays off to try exponentiating a few times to see when things start cycling.

[quote="dwx314"]So the unit-digit patterns operate on cycles of 4 (or it's factors)
Just a table for reference:[code]  going across= units digit of X down= y (mod 4) in X^y:
x | 0 1 2 3 4 5 6 7 8 9
--+--------------------
1 | 0 1 2 3 4 5 6 7 8 9
2 | 0 1 4 9 6 5 6 9 4 1
3 | 0 1 8 7 4 5 6 3 2 9
0 | 0 1 6 1 6 5 6 1 6 1[/code]

And tens-digit pattern operate of cycles of 40 (or t's factors)
(I hope)[/quote]

40? I don't think so, because they repeat at least every 100. 40 doesn't divide 100.


linear algebra
matrix
algebra
polynomial

[b][u]Problem[/u]:[/b] Let $ A$ be a matrix $ n\times n$ over complex field $ C$ such that $ trace(A)\equal{}trace(A^{2})\equal{}...\equal{}trace(A^{n})\equal{}0$. Prove that the characteristic polynomial of A is $ \lambda^{n}$.

[hide="hint"]For $ k\in \mathbb{N}$ we have $ Tr(A^k)\equal{}\sum_{i\equal{}1}^n \lambda_i^k$ , ,where $ \lambda_i ,i\equal{}\overline{1,n}$ are the eigenvalues of $ A$ .Using our hypothesis we obtain  a homogenuous system .Consider that $ \lambda_1 ,\lambda_2 ,...,\lambda_p$ are distinct and $ k_1,k_2 ,...,k_p$ be their algebraic multiplicities,we obtain a homogenous Vandermonde system with the unknown values $ k_1 \lambda_1,k_2 \lambda_2 ,...,k_p \lambda_p$.And from here we must obtain that $ k_1 \lambda_1\equal{}k_2 \lambda_2\equal{} ...\equal{}k_p \lambda_p\equal{}0$[/hide]


trigonometry
calculus
derivative
inequalities proposed
inequalities

hello, differentiating $ y\equal{}f(x)$ with respect to $ x$ and solving the equation $ f'(x)\equal{}0$ we get $ x\equal{}\frac{\pi}{2}$ and with the second derivative test we get that this gives the maximum, and this is $ 3$.
Sonnhard.

thank you for your idea,Dr Sonnhard Graubner . i think it is not necessary to  to use derivative in this exercise. :)  i hope that there are some other proofs  :)

i don't think that the second derivative test is simple :blush:but we still check it  :wink: .
Who help other proofs?

[quote="Victory.US"]find the maximum value of y
$ y \equal{} cos^3 x \plus{} sin^5x \plus{} sinx \plus{} \sqrt {2 \minus{} sin^2x}$,  for $ x\in[0,\pi]$[/quote]

1. $ \cos^3 x \plus{} \sin^5x \le \cos^2 x \plus{} \sin^2x\equal{}1$

2.$ \sqrt {2 \minus{} t} \le \frac{1}{2}(3\minus{}t)$

$ \Rightarrow y(x) \le 1\plus{}\sin(x) \plus{}\frac{1}{2}(3\minus{}\sin^2(x))\equal{}3\minus{}\frac{1}{2}(1\minus{}\sin(x))^2 \le 3$ , equality holdes for $ x \equal{} \frac{\pi}{2}$


[quote="math88fun"]Would it be possible to take the AMC 12 class and AIME class at the same time, or should you have completed the AMC 12 class before you begin the AIME class?[/quote]

Some students have taken both problem series at once.  The problem series classes are less intense than the subject classes because they don't have exams, so taking two problem series is probably a 5-8 hour time commitment per week for students who want to take them both.

I would recommend this to students who (a) have that much time that they want to commit and (b) are already fairly certain that they will make it to the AIME, but want to improve both their AMC 12 and AIME scores: 100-130 on the AMC 12 and 3-10 on the AIME (on average) -- if you fit into that category, it's a very reasonable combination of classes.

However, it should be said that the subject classes cover material more deeply, so students who wish to dedicate themselves to learning material most comprehensively over two or more years are advised to take subject courses when possible.  They certainly help more for students who will one day make it to the USAMO partially because the exams require more thorough, longer thinking and more organized solution-writing.


Determine the average value of numbers N which contain $ n$ ($ n>1$) digits satisfying two conditions:
(i) N contains digits {1,2,4,5} and the difference of two consecutive digits is greater than 1.
(ii) N is divisible by 11

if $ N$ is even, the answer is 333...3. because a number n satisfies the conditions if and only if 666...6-n satisfies the conditions too (it is easy to prove).

i don't know how to find the average for odd N. for $ N\equal{}3$, the average is 242. i don't have any idea what to do here.

And for $ N \equal{} 5$ it's $ \frac {239393}{5}$ and for $ N \equal{} 7$ it's $ \frac{16060605}{4}$ ... not a lot of obvious stuff going on, there.


Let $ABCD$ be an inscribed quadrilateral and $M_{1},M_{2},M_{3}$ be the midpoints of $AB,BC,CD,$ respectively. Let lines $AM_{3}$ and $DM_{1}$ intersect at $P$. Let $BP$ and $CP$ intersect with the circumcircle of $ABCD$ again at $X$ and $Y$, respectively. Knowing, that lines $AB$, $PM_{2}$ and $CD$ intersect at one point, prove that $XY\parallel AD$.

Lets make central projection wich saves circumcircle and puting point P and 'far-far' point of line BC into infty.
$M_{2}$ will be still the midle of BC and $\frac{AM_{1}}{BM_{1}}=\frac{DM_{3}}{CM_{3}}$(*).Also $AB\parallel CD\parallel PM_{2}$ (**).
Let $Q=AD \cap PM_{2}$ .As (**) Q is a midle of AD.But $AM_{1}M_{3}D$ trapezoid.So $AM_{1}=DM_{3}$. So $M_{1}M_{3}\parallel AD$.Using (*) we get that BC||AD.By the simetry XY||AD or they intersect one 'far-far' point.But this point was 'far' even before central projection.So everything OK.

We can make the condition stronger that let $M_{1},M_{3}$ be points on $AB,CD$ and $AM_{1}/M_{1}B=DM_{3}/M_{3}C$
$AB, PM_{2}$ and $CD$ concurrent at $K$
Note that $KP$ is the symmedian of triangle $AKD$
We use this small [b]lemma[/b]

[b][u]Lemma:[/u][/b]  [i]Let $M_{1},M_{3}$ be points on $AB,CD$ and $AM_{1}/M_{1}B=DM_{3}/M_{3}C$ and $AM_{3}, BM_{1}, KP$ concurrent at P where $KP$ is the symmedian of triangle $AKD$. Prove that  $BP//CD$ and $CP//AB$[/i]

By using lemma we have $XD=BC=YA$
So $XDAY$ is a  isosceles trapezium. This means $XY//AD$


geometry AMC AIME MATHCOUNTS search function probability

I've read the MC Bible...but it only gives a few things to help someone do well at MC, or rather, to memorize. Is there anything else not in the Bible that you should know or memorize?

There are 3 things you should focus on:

[b]1)Math Theory (i.e. concepts from AoPS books)
2)Practicing and doing lots and lots of problems[/b]
3)Memorizing Formulas

While formulas are important for speed, they aren't as important as the first two things. Do those first, and then start memorizing formulas.

What exactly do you eman by Math Theory?

Like Geometry and Counting etc?

You need to memorize more than that; the national test this year had a problem on something you had to prove in WOOT, which is preparation for the AIME and USAMO. For chapter and state preparation, take the Mathcounts problem series, the AMC 12 and AIME problem series, and if you are doing reasonably well at the AIME and get to MC nats, you can take WOOT. Beyond that, if you have an extraordinarily great mathematics curriculum at your school, then you have a pretty good shot at MC.

There is no golden bullet, and you can't win or do well without having lots of intelligence, but there are some strategies to help you do better.

1. Do lots and lots of old tests. Before the school test, do all of your available school rounds in ascending order of year, with on the day before doing an easy chapter. Before the chapter test, do all of your available chapter rounds in ascending order of year, with an easy state on the day before. Before the state test, do all of the state tests in ascending order of year, with an easy nats on the day before. Before nats, do all of the tests in ascending order, with one test on the day before.
2. If you don't have a team, do the team rounds by your self also. Try to allot a maximum of 80 minutes Maximum for the round.
3. If you don't have enough tests, look on the net, you can usually find school, chapter, and state tests going back to 2001. In addition, tons of people on AOPS have old Mathcounts National Tests. Ask them.
4. There are plenty of mock tests on AOPS. While they are usually too easy or too hard, they are good practice.
5. After doing a test, figure out how to do all of the problems that you didn't get. If you still can't understand, post it here, in this forum, of use the search function to see if any1 has posted in before.
6. Practice every day, whether it be on this site, an actual mc test, amc problems, or this site: http://mathcounts.saab.org/mc.cgi
7. Get LOTS of sleep before the test.
8. At target, when checking, read every single word and read the question very slowly to make sure that you read it right.
9. At sprint, try to do it really fast, and then check it over. If you are stuck on a problem, or it is gonna take too long, skip it and come back to it later.

Do lots and lots of problems ESPECIALLY Geometry and Counting&Probability. (I think we have all learned that from this year's national competition) 19 out of the 38 problems were counting/probability problems!!!!thats crazy!!!!!!!!!!!!!!

Major hint: If a question asks for a probability and it does [b]not[/b] say to express your answer as a common fraction, you can instantly deduce that the answer is either 0 or 1. :D

Note: This does not apply to mock tests unless they were written by me.

I hate it when the answer is a decimal to lots of places  (like 8211.23114) and they ask for the nearest hundred.....

[quote="tinytim"]sometimes MC will pull a stupid trick on you by asking for the answer to the nearest tenth when the answer is an integer. Like if the answer is 5, and you don't write 5.0, they count it wrong.[/quote]

I don't think they would mark that wrong cause its to the nearest tenth and further simplified. I do remember one time when it ask for like the surface area. A it didn't say simplest radical form and the answer was in simplest radical form. But that was in the early years.

I thought the rules said you can't write 5.00 or 5.0 for 5 or something like that...
Has to be exact number of decimal places...

I don't know what's up with people still saying that memorizing random stuff is still necessary (or even helps) on MathCounts or on math competitions in general, seeing all the stuff against that idea on AoPS. As a few people above me have pointed out, it is all about your ability to think, and as long as you understand where these "formulas" come from, you're fine. This isn't to say that on every problem you should re-derive every theorem or whatever you've come across, because that's silly. But if something is likely to be useful at all, you've probably used it enough times in your practice that there's no effort needed in "memorizing the formula."

I'll pick on 1=2's comment here because it's the first one I saw, that probably with the equilangular hexagon in WOOT obviously would have helped, but if you happened to be in WOOT, you probably should have learned how to prove the relation, and quickly go through the steps instead of pulling a formula out of your behind. For all you know, you could have remembered the formula wrong.

The rules state that if the answer asks for dollars, and the answer is a whole number, then either 5.00 or 5 would be excepted BUT 5.0 would not.

1.) It is always important to [b]read the question[/b]. You have no idea how many times I don't do that in high school and my test grades always suffer because of that. Circle important facts, cross out any distractors, and make sure you understand what it's asking for. Like the previous poster said, if it asks you to round, make sure you round.

2.) During practice tests, when you come across something you don't know, don't just throw up your hands and look for the solution. Finish the rest of the test, and then go back to the problem. You're not going to have time during the competition to spend ages solving a problme you don't know, but with practice months before, you get used to solving "unknown" problems.

3.) This helps during AP tests, but will also hold well for any math competition. The day or two before the competition, don't study. You may look at some math books for an hour or two, but no more than that. Any more studying will leave you nervous and anxious for the compeition, and will most likely stoke accidental mistakes.

[quote="dynamo729"]Although formulas are necessary for MC, it is not just heavy formula/algebra based, but a lot of it is creative problem solving. That being said, here is what I consider great preparation (although you may not want t take advice from me!!!!):

1. Practicing with old tests, especially recent ones.
2. Challenging yourself to do the problems you are not comfortable with. I noticed in our school team round practices that people would just do the ones they want to do. You need to practice the stuff you find challenging and practice that more than stuff you know you can do (although you still want to practice everything).
3. Speed, which is necessary in all rounds, as to leave time to check over your answers. Don't go too fast, as you may rush yourself and not catch a mistake.
4. Reading the question correctly. This could have gotten me 10 more points at state and most likely 5 more points at nationals.

Hope this helps!!!![/quote]

Almost all the people who have posted have suggested to do probs, but have suggested no place to find them. I am asking for not only Nats but Chapter and state probs

thanks,
Dhaivat Pandya(uh huh you read write that's my name) :D

[quote="Poincare"][quote="Ihatepie"]There is no golden bullet, and you can't win or do well without having lots of intelligence, but there are some strategies to help you do better.

[/quote]
Its not fully true that you need lots of intelligence to succeed but you really need to work hard (tell me about it  :oops: didn't work to hard this for chapter and didn't even make countdown but worked hard before state and got 18th lol :D)[/quote]

Working hard is what distinguishes the best from the good. However, working hard isn't everything; one needs some degree of intelligence and some creativity to succeed.

Don't do too much the day/night before the test. It really depends on you , though, what that means. For me, I have to completely disregard that, because I'm those easily energized type of people. My recommendation is that you take a test from previous years right before you go to bed. But don't overdo the thinking. Go to bed at like, 10:00 or something. But make sure that the test you take is the same level as the one you're about to. For example, if you're gonna take the state, don't take a chapter, because that will take your mind offguard, or a nats, because then your subconscious will treat the next day's test as a nats one. Don't eat too much, or party like at least 3 days leading up to  a contest. And prepare EVERYTHING beforehand so you're not rushing the next day.

Well, maybe not cram, but if you've learned them before, a nice review isn't too bad.

[quote="BOGTRO"]I'm in 5th grade right now and am going to MC next year.

I've covered all of the Intro. AoPS books, got a 29 of 30 on a recent mock test,

went to HMMT, ARML, and am going to MathPath.

I think I have the math part down, but

can someone put "Day/Night before the test" strategies and test taking

strategies? I know that some posters before me have made some.[/quote]

I personally think that right now, you shouldn't be worrying about stuff like getting sleep and cramming and stuff. Right now, for you, I think it would be best to practice a lot of problems from places like [url]http://mathcounts.saab.org/mc.cgi[/url] or from the AMC's and AIME's. In any case, you should probably focus on practice right now.

I don't mean right now, I am practicing.

Oh, I know that. I was just saying you should think about the competitions stuff closer to the actual competition.

[quote="Urc"]Never ever cram facts down at the last night...ever

On an off topic note, I lit fireworks the day before =D[/quote]

My brother got a pack of like ~300 fireworks that we randomly light when we're bored. (on an off topic note)

[quote="georgiamathdude"]You gotta buy nationals. Go the mathcounts page, look around for a bookstore or similar, and buy 'em. If you're looking for chapters and states, follow this [url=http://www.yashar.biz/mathcounts/]link.[/url]

Edit: Never mind. They ain't there no more.[/quote]

There have been books with the chapter/state/nats questions published by MC for 2000 and later. There are also books with various problems drawn from earlier years.

My suggestion: ask people who've gone to nats before for questions. After my first trip to nats, I had a collection of nats problems from 1989 to the present that was fairly complete. I've never actually gotten my hands on chapter/state tests from before 2000, though

Sometimes, incubating can be helpful. By incubating, I mean skipping (this is for sprint) the ones you do not know how to do off the top of your head and coming back to them later. This is what my current math "tutor" said (he got a perfect score at mathcounts one year and missed just one question another). However, I never did this for mathcounts, and I kind of doubt it would've worked well for me (since i'm slow and didnt even read the last ten questions at state one year and at nats), but it might for some other people.

Of all the previous advice, the two I find most helpful are to practice with many old tests and to read the question carefully.

[quote="tinytim"]sometimes MC will pull a stupid trick on you by asking for the answer to the nearest tenth when the answer is an integer. Like if the answer is 5, and you don't write 5.0, they count it wrong.[/quote]

Dang it, because of that i didn't answer 1. in the chapter sprint correctly! :wallbash_red:  :censored:  :cursing:

[quote="BOGTRO"]I'm in 5th grade right now and am going to MC next year.

I've covered all of the Intro. AoPS books, got a 29 of 30 on a recent mock test,

went to HMMT, ARML, and am going to MathPath.

I think I have the math part down, but

can someone put "Day/Night before the test" strategies and test taking

strategies? I know that some posters before me have made some.[/quote]

Did you do the Calculus portion of the test in HMMT?

Here are some things I suggest to prepare for MATHCOUNTS:

1. Do lots of practice problems. You can find tons of problems on the MATHCOUNTS website and if you surf the forums you can find some very nice problems.

2. Study the AoPS books. I suggest this order: Basics, Algebra, Number Theory, Geometry, and then Counting and Probability. I like it that way, but it's whatever works for you. And don't forget the MATHCOUNTS problem sets!

3. You might find it useful to look at old AMC 8 and if you have time, AMC 10 tests. You can find the 10's on AoPS(go to resources) and launch a search on the web for the 8's.

4. Read every single word in every single problem! It's very useful, believe me. I learned that the hard way...and if in, like, 30 seconds you don't have the faintest idea how to solve it, skip it and return to it later, after you checked all your other answers.

5. Don't blindly memorize formulas, but understand how they're derived. And of course, learn all the basic ones, like $ \frac {n(n \plus{} 1)}{2}$ if you don't know them already.

6. The day before MATHCOUNTS, do something relaxing. For some, it's fireworks. For me, it's sitting down with a good book(like Tales of Sherlock Holmes) and eating ice cream.  :D

7. Sleep well, and don't worry! Before going to bed, read something funny, like a joke book. Do the same before leaving, and definitely bring it with you, for the break.

8. Eat a good breakfast, like eggs and bacon or something like that.

9. Whatever you do, don't be anxious, like I always am! It never comes to any good. *sigh*

10. If you have a favorite pencil, bring it! (I know this sounds kind of stupid, but it's nice to have something familiar, especially if it's your first time. Like mine was when I flunked it. I knew I should have brought my green, fuzzy pencil...)

Do you have to know how n(n+1)/2 is derived?

If you want to succeed after MC, you need to understand how most/all formulas work/were derived.

If you forget a forumula(or formulas, like I ALWAYS do  :( ), you'll be able to derive it and find it. Plus, it contributes to general knowledge, and you never know when that could save your life!!  :D

I agree with not cramming down on facts, as I did with Black Saga. Didn't work very well, and I slept it off. Whatever makes you feel most comfortable (relying on your gut and not studying won't work though), such as sleeping with an AoPS book underneath your pillow, studying formulas, relaxing before the big day, etc can help. As Isabella said, a favorite lucky pencil or charm works. I suggest taking a test before you even start studying; that way you know what you need to improve on and what needs less work. Take a test in the middle of studying and see how you've progressed. If you still need a little work on the same subject, focus on that and not anything else for the time being. At the end, take another different practice test and see how you do!

What D3m0n Shad0w said about taking the tests is a good idea! You can see where you need to focus on. I think someone may have said this earlier, but you shouldn't avoid the things you aren't very good at. If you only do things that are easy for you, you won't get better at all.

Remember, your goal isn't to place 1st at MATHCOUNTS, it's to further your knowledge of mathematics. A trophy in MC is a step to that, but it isn't what you should aim for. That being said, if you don't know something, or you are afraid of it coming up on the test, then you should view that as a challenge to better yourself.

Math competitions such as MATHCOUNTS are a measure of how good you are in math. If you do well, that means you are well on your way to a healthy, educated lifestyle.  :no:


hello, do you mean $ \frac{x^{4n\plus{}1}\plus{}x^{5n}}{x^{6n}\minus{}x^{5n\plus{}2}}$?
Sonnhard.

That's what it looks like. pisqrt314, what are you asking us to do with this expression? Simplify?

[quote="Dr Sonnhard Graubner"]hello, do you mean $ \frac {x^{4n \plus{} 1} \plus{} x^{5n}}{x^{6n} \minus{} x^{5n \plus{} 2}}$?
Sonnhard.[/quote]
yes exactly,
sorry haven't been on lately, don't need the answer anymore.


trigonometry
logarithms
calculus
integration
algebra unsolved
algebra

Solve the equation
$ \cos (\pi \log_{3}(x\plus{}6))\cos(\pi log_{3}(x\minus{}2))\equal{}1$

_____________________________
Azerbaijan Land of Fire  :ninja:

We have that $ \minus{}1\le\cos\theta\le1$, so the only way the two expressions can multiply together to be $ 1$ is if they are both $ 1$ or $ \minus{}1$. For $ \cos\theta\equal{}1$, the solutions are $ 2\pi n$, where $ n$ is an integer. Also, for $ \cos\theta\equal{}\minus{}1$, the solutions are $ \pi y$, where $ y$ is an odd integer. Thus, we must have that $ x\plus{}6$ and $ x\minus{}2$ are both powers of $ 3$ (otherwise the expression would not be of the form $ \pi c$, where $ c$ is an integer). We also have that $ x\plus{}6\minus{}(x\minus{}2)\equal{}8$, so we must find two integral powers of $ 3$ with a difference of $ 8$. The only such powers of $ 3$ are $ 9$ and $ 1$, and hence the only solution to the equation is $ \boxed{x\equal{}3}$.


geometry
parallelogram
trapezoid
projective geometry
geometry proposed

The inscribed circle of a triangle $ ABC$ touches sides $ BC,CA,AB$ at $ P,Q,R$ respectively. A line through $ A$ and

parallel to $ QR$ meets line $ PQ$ at $ X$. Prove tha the line $ BX$ passes through the midpoint $ M$ afo $ QR$.

Babis

Let $ Y\equal{}PR \cap AX$.we know that $ YQ,XR$ intersect on the line $ PM$ since $ RQ \parallel{} XY$
now we want to prove that $ RX,YQ$ intersect on $ (I)$.
Let $ PM$ intersect $ (I)$ at $ T$ and let $ RT$ intersect $ AX$ at $ X'$ .we must prove that $ X\equal{}X'$.
denote $ S$ the midpoint of $ XQ$.since $ \Delta QAX \sim \Delta PQR$  we have $ \angle TPQ\equal{}\angle MPQ\equal{}\angle SAX$.
but we have $ \angle X'RQ\equal{}\angle TRQ\equal{}\angle TPQ$ which implies $ \angle SAX\equal{}\angle X'RQ$
now let $ AS$ intersect $ QR$ at $ L$.since $ AX\parallel{}QR$ the quadrilateral $ AQLX$ is a parallelogram and also the quadrilateral $ RAX'L$ is cyclic and since $ AX' \parallel{} RL$ the quadrilateral $ RAX'L$ is an isocele trapezoid which implies $ \angle LX'A\equal{}\angle X'AR\equal{}90\plus{}\frac{\angle A}{2}$ and also we have $ \angle LXA\equal{}\angle LQA\equal{}90\plus{}\frac{\angle A}{2}$  so we got $ \angle LXA\equal{}\angle LX'A$ which implies $ X\equal{}X'$.
so we have proven that $ PM,XR,YQ$ are concurrent on $ (I)$.
now use pascal's theorem for $ RRTPPQ$ we get the collinearity of $ (PP,RR),(PT,QR),(RT,PQ)$ or the collinearity of $ B,M,X$.
so we r done!

Denote $ M \equal{} PQ\cap AB$
We have $ X(ABRP) \equal{} (ABRM) \equal{} \minus{} 1$ (1)
On the other hand : $ XA// PQ$ (2)
Since (1) and (2) then $ BX$contains $ M$
 :)


1. find the circumference of the following figure: $ 12x^2 \minus{} 300 \plus{} 9y^2 \plus{} 234y \plus{} 3288 \equal{} 0$
(I don't know how to find the circumference of an ellipse)

2. Find the volume of the figure enclosed by (4,2,-3), (2,5,6,), (-3,1,5), and (3,1,4).
I could use geometry and a lot of right triangles to solve this one, but I get the feeling there is probably a nicer analytical way to do this, especially since I know almost no analytic geo.

Thanks in advance

no ideas on either of these at all?

EDIT: perhaps this belongs in the intermediate forum or something

i think the circumference of an ellipse is $ \pi \cdot(r_1 \plus{} r_2)$
you can look it up on AoPS wiki or something...

EDIT: fixed formula

thanks, that is quite easy to guess, but I just looked it up and it seems as if there is no simple exact formula for this. I need to find one that does not involve calculus or e..........

I'm not sure that's right though...it's like $ 4aE(\epsilon)$ or something. And it doesn't exactly make sense, because there are areas that are concentrated at a disproportional ratio...also, I measured the circumferences after constructing them on GeoGebra, and they were not the same.

In the stupid formula booklet my teacher gave me that I try to never consult, there is a formula for the circumference: $ \sqrt{\frac{a^2\plus{}b^2}{2}}$.

Does this give the exact value or is it just an approximation?

yea you're right. My teacher comfirmed its an approximation...also, that should have been $ 2\pi \sqrt{\frac{a^2 \plus{} b^2}{2}}$ above.

[quote="hurdler"]
2. Find the volume of the figure enclosed by $ (4,2, \minus{} 3), (2,5,6,), ( \minus{} 3,1,5)$ and $ (3,1,4)$.
I could use geometry and a lot of right triangles to solve this one, but I get the feeling there is probably a nicer analytical way to do this, especially since I know almost no analytic geo.
[/quote]

If you use Vector,it is easy to calculate.

$ A,B,C,D$ is $ A(4,2, \minus{} 3),B(2,5,6),C( \minus{} 3,1,5)$ and $ D(3,1,4)$.

$ \vec{AB} \equal{} ( \minus{} 2,3,9)$
$ \vec{AC} \equal{} ( \minus{} 7, \minus{} 1,8)$
$ \vec{AD} \equal{} ( \minus{} 1, \minus{} 1,7)$

then the volume is $ \mid\frac {(\vec{AB}\times\vec{AC})\cdot \vec{AD}}{6}\mid \equal{} \frac {155}{6}$


geometry
trigonometry
trig identities
Law of Sines
Law of Cosines

[hide]Put it in a circle. $\angle DAC$ intercepts a $80^\circ$ arc. Since $\angle BDC$ is straight and $\angle BAD$ and $\angle DAC$ are adjacent, $\angle BAD$ must intercept a $100^\circ$ arc and therefore must be $50^\circ$, making $\angle B$ equal to $70^\circ$.[/hide]

Ah, nice solution!

I was thinking about this for a while, and I didn't even think of that. Well, atleast I'll hopefully have added this to the "checks" that I do on geometry problems.

[hide="To tarquin..."]
Why must $\angle BAD$ intercept a $100^\circ$ arc? By your logic, then, wouldn't every triangle be forced to have a right angle? ;)
[/hide]

Hm, then how would one go about solving this type of problem?

[quote="shobber"]i wonder if there is really a solution without trig.[/quote]
i am desperate......
i think it is impossible solve it without trig...

You can't solve using sides and angles separately, i don't think.  There might be some really fancy geometry solution, but i think u need law of sines of law of cosines to solve it...


P is a prime number and p = 4k + 3
i) : Prove that exist a and b are positive intergers satisfy GCD(a,p) = GCD(b,p) = 1 and a^2 + b^2 + 1 devide by p

ii) : x, y are some intergers that GCD(x,p) = GCD(y,p) = 1 . Determine the number of remainder when divide                       (x^2+y^2)^2^2007  by p

i) Posted before.
ii ) Use quadric reside we have the number of remainder is $ \frac{p \minus{} 1}{2}$


1.
When the sum of the first $k$ terms of the series

\[ 1^2 + 2^2 + 3^2 + \cdots + n^2 + \cdots \]
is subtracted from the sum of the first $k$ terms of the series

\[ 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots + n(n+1) + \cdots, \]
the result is 210.  Find the numerical value of $k$.

2.
Three coplanar circles are concentric, and the lengths of their respective radii are 10, 20, and 30.  A square is circumscribed about the smallest circle, and squares are inscribed in the other two circles.  Find the length of the shortest line segment that could contain a point on each square.

3.
Find the numerical value of $k$ for which

\[ \frac{7}{x+y} = \frac{k}{x+z} = \frac{11}{z-y}. \]

4.
Spender Sam always carried only dollars and pennies, but he never carried more than 99 pennies.  When leaving home one day, Sam noted that the number of dollars he had was 150% of the number of pennies he had.  After spending 10 dollars and 78 cents, Sam discovered that the number of dollars he had remaining was 1 less than the number of pennies he then had.  With how much money did Sam leave home that day?

5.
Two cards are randomly selected from a standard 52 card deck of playing cards, without replacement.  Find the probability that at least on king and at least one heart are selected (though not necessarily on different cards).

6.
Find the ordered pair of positive integers $(a, b)$, with $a<b$, for which
\[ \sqrt{1+\sqrt{21+12\sqrt{3}}} = \sqrt{a} + \sqrt{b}. \]

7.
The length of the base of an isosceles triangle is 20.  In the plane of the triangle, a point 4 units from this base is 10 units from each leg.  Find the area of the triangle.

8.
If Arc denotes the principle value, find all values of $x$ which satisfy

\[ \sin{(4\text{Arc}\tan{x})} = \frac{24}{25}. \]

wait, why did this end up in our forum? (i know this is achieving archive posts - sry for that)
-jorian

[quote="aahmadi"]how do u do #7?[/quote]

Six right triangles are formed.

You're both big spammers so I don't see why either of you are "talking."


modular arithmetic
combinatorics unsolved
combinatorics

[quote="Nora.91"][b]Q.1.[/b] Let A = {1, 2, 3, ... 100} and B be a subset of A having 48 elements.
Show that B has 2 distinct elements x and y whose sum is divisible by 11.

[b]Q. 2. [/b]Consider the collection of all three-element subsets drawn from the set { 1, 2, 3, ... 300 }.
Determine the number of those subsets for which the sum of the elements is a multiple of 3.

[i]Thank You ! [/i] :)[/quote]

This should be in High School Basics or Intermediate.

[b]Solution:[/b]

[b]Q1.[/b]  Let $ A_i\equal{}\{n | n\equal{}i(mod 11), n \in A\}$ for $ 0 \le i \le 10$ , then assume that there is such a subset $ B$ to disprove the statement,

we have that in $ B$ we can't have more than 1 element from $ A_0$

And if B contains element(s) from $ A_i$ ($ i \neq 0$) then $ B$ cannot contain any elements from $ A_{(11\minus{}i)}$.

So $ B$ can only contains elements from $ 5$ of the subsets $ A_1,A_2,....,A_{10}$.

Each $ A_i$ contains $ 9$ elements except $ A_1$ which contains $ 10$ elements. So the maximum number of elements in $ B$

$ \equal{}1\plus{}4 \times 9\plus{}10\equal{}47$, contradiction, done.

[b]Q2.[/b] We first list all combinations .
$ 0\plus{}0\plus{}0\equal{}0(mod 3)$
$ 1\plus{}1\plus{}1\equal{}0(mod 3)$
$ 2\plus{}2\plus{}2\equal{}0(mod 3)$
$ 0\plus{}1\plus{}2\equal{}0(mod 3)$

Now in the set there are $ 100$ elements congruent to $ i (mod 3)$ for each $ i \in \{0,1,2\}$

So the number of three-element subsets $ \equal{}3 \times \frac{100 \times 99 \times 98}{3!}\plus{}100 \times 100 \times 100\equal{}1485100$

[quote="stephencheng"]This should be in High School Basics or Intermediate.[/quote]  Your assessment of difficulty is appallingly bad -- these questions would be seriously inappropriate for HSB, whereas they fall comforatbly within the universe of "easy olympiad-type problems."

Also, to make this post more useful: use \pmod{m} to get $ \pmod{m}$; if you don't want parentheses, then \mod works well.  Also \equiv gives $ \equiv$.


abstract algebra
Ring Theory
superior algebra
superior algebra solved

Please do not provide a proof for this question, just an answer.

If $\phi: R\rightarrow S$ is a ring homomorphism, and $I=\ker(\phi)$, then is $I$ a subring of $R$?

I am asking this because I believe that I have proved that we have closure over multiplication and addition, which would prove that $I$ is indeed a subring of $R$, but I found something online which disagrees with this.

However, I believe that that is because my textbook does not require that a ring $R$ have a multiplicative identity, but it is a convention in some texts.

Anyways, does someone want to clear this up?  I should, in the interests of ethicality, tell you all that the question is to prove that $I$ is a subring of $R$, so I am essentially asking you guys as to whether this is a provable assertion.

Answer is NO.  :(
The key idea depends on you play with the identities. :)

Can you say why, and are you assuming that the ring R has multiplicative identity 1_R?

It looks to me like it's a ring, just not necessarily a ring with a multiplicative unit.

yep, grobber's absolutely right. indeed, if you do not require a ring to have multiplicative identity, then every ideal is a ring.


number theory
relatively prime
number theory unsolved

Let $n$ be a positive integer with $n \geq 3$. Show that
$n^{n^{n^{n}}}-n^{n^{n}}$
is divisible by 1989.

Are you sure about that problem? It is very possible that I'am wrong but:
In particular, we wish to prove that:
$n^{n^{n}(n^{n^{n-1}}-1)}\equiv 1 \mod{9}$
It's easy to check that $2^{d}\equiv 1 \mod{9}$ is equivalent to $6|d$.
But! If we take $n=11$ then $n^{n}(n^{n^{n-1}}-1)$ is not divisible by $3$. Indeed $2^{a}\equiv 2 \mod{3}$ when $a$ is odd and $n^{n-1}$ is odd...

[b] TomciO [/b] - I think that your solution is not quite good, because in particular, we wish to prove that:
$n^{n^{n}}( n^{n^{n-1}}-1) \equiv 0 \mod{9}$
And it's not the same, what you wrote.

1) First, we show that $3| n^{n^{n}}-n^{n}$. If $n\equiv 0 (mod\ 3)$ or $n\equiv 1 (mod\ 3)$  it's obvious, if $n\equiv 2(mod\ 3)$ then $n^{n^{n}}-n^{n}\equiv (-1)^{n^{n}}-(-1)^{n}=0 (mod\ 3)$ (because $n^{n}$ and $n$ are of the same parity).

2) Now we show that $16| n^{n^{n}}-n^{n}$. If $n$ is even, then it's obvious. If it's odd we can easily notice that $n^{4}\equiv 1 (mod\ 16)$, so it sufficies to show that $n^{n}\equiv n (mod\ 4)$, what's also true (if $n=2k+1$ then $n^{n}=(n^{2})^{k}\cdot n\equiv 1\cdot n =n (mod 4)$).

3) Now we see that $\varphi(9)=6$, $\varphi(13)=12$ and $\varphi(17)=16$ divide $n^{n^{n}}-n^{n}$.

4) We notice that $1989=3^{2}\cdot13\cdot17$, so we have to show that  $n^{n^{n^{n}}}-n^{n^{n}}$ is divisible by $q$ for $q=9,13,17$. We have \[{n^{n^{n^{n}}}-n^{n^{n}}=n^{n^{n}}\left(n^{{n^{n^{n}}-{n^{n}}}}-1\right)}\] For all those $q$ we see that if the first term is not divisible by $q$ then $n$ is relatively prime to $q$, so by the Euler's theorem and previous remark $\left(\varphi(q)|{n^{n^{n}}-{n^{n}}\right)}$ second term is divisible by $q$. So for all $n\geq3$ the number $n^{n^{n^{n}}}-n^{n^{n}}$ is divisible by 9, 13 and 17 and therefore it's divisible by 1989.


If $ a, b, c  $ are real numbers such that $ a+b+c=0,  $ prove that
 \[\displaystyle \left|a^7+b^7+c^7\right|\ge
\frac{7\sqrt{2}}{10}\left|a^5+b^5+c^5\right| \sqrt{a^4+b^4+c^4} \; .
\]
Can this inequality be improved ? (Why ?)

Very nice... but remember the forum is called [b]in[/b]equalities ;) .

In fact, a nice way to deal with numbers adding up to 0 is the following fact:

[color=blue][b]Lemma 1.[/b] Let a, b, c be three real numbers. Then, a + b + c = 0 if and only if there exist real numbers u, v, w such that a = v - w, b = w - u and c = u - v.[/color]

[i]Proof.[/i] If such numbers u, v, w exist, then a + b + c = (v - w) + (w - u) + (u - v) = 0. On the other hand, if a + b + c = 0, then just take u = 0, v = - c and w = b and get v - w = - c - b = a (since a + b + c = 0), w - u = b - 0 = b and u - v = 0 - (- c) = c. Thus, Lemma 1 is proven.

Now, using a = v - w, b = w - u and c = u - v, we have

$a^7+b^7+c^7=7\left(u-v\right)\left(v-w\right)\left(w-u\right)\left(u^2+v^2+w^2-vw-wu-uv\right)^2$;
$a^5+b^5+c^5=5\left(u-v\right)\left(v-w\right)\left(w-u\right)\left(u^2+v^2+w^2-vw-wu-uv\right)$;
$a^4+b^4+c^4=2\left(u^2+v^2+w^2-vw-wu-uv\right)^2$,

what can be seen either by direct verification or by some fumbling around with third roots of unity. The latter equation yields

$\sqrt{a^4+b^4+c^4}=\sqrt{2\left(u^2+v^2+w^2-vw-wu-uv\right)^2}$
$=\sqrt2\left|u^2+v^2+w^2-vw-wu-uv\right|=\sqrt2\left(u^2+v^2+w^2-vw-wu-uv\right)$

(since we always have $u^2+v^2+w^2-vw-wu-uv\geq 0$). Now, it becomes trivial that your inequality

$\left|a^7+b^7+c^7\right|\geq\frac{7\sqrt2}{10}\left|a^5+b^5+c^5\right|\sqrt{a^4+b^4+c^4}$

is actually an identity.

  Darij

[quote="darij grinberg"]Very nice... but remember the forum is called [b]in[/b]equalities ;) .
  Darij[/quote]
Nice solution ! However  the ,,equalities" include  ,,inequalities" . Or ?


If $ a,b,c$ are distinct reals,prove that:
$ \dfrac{a^2}{(b\minus{}c)^2}\plus{}\dfrac{b^2}{(c\minus{}a)^2}\plus{}\dfrac{c^2}{(a\minus{}b)^2}\geq2$