$ \left \lceil x\plus{}n \right \rceil \equal{} \left \lceil x \right \rceil \plus{} n,$ where $ n$ is some integer .My query is it is true for whether $ x$ belongs to $ R$ or $ R\minus{}Z$ (where $ R$ is the set of real numbers and $ Z$ is the set of integers). If the second one is right then please tell me why. Thank you for your help.

I think it's when $ x \in R$, since $ x$ can be an integer too. We would just have $ \lceil x \plus{} n \rceil \equal{} x\plus{}n \equal{} \lceil x \rceil \plus{} n$

[quote="modularmarc101"]I think it's when $ x \in R$, since $ x$ can be an integer too. We would just have $ \lceil x \plus{} n \rceil \equal{} x \plus{} n \equal{} \lceil x \rceil \plus{} n$[/quote]

ok. thank you. :-)

what is a $ \left\lceil x \plus{} n \right\rceil$

what does it mean/denote

That notation is called the ceiling function.
It represents the smallest integer greater than or equal to $ x \plus{} n$. For example, $ \lceil\pi\rceil \equal{} 4$.

The opposite is the floor function. $ \lfloor\pi\rfloor \equal{} 3$.

Yes , we have if $ x$ is some real number between two integers $ n$ and $ n\plus{}1$, we have that $ \lfloor x \rfloor \equal{} n$ and $ \lceil x \rceil \equal{} n\plus{}1$.

[quote="modularmarc101"]I think it's when $ x \in R$, since $ x$ can be an integer too. We would just have $ \lceil x \plus{} n \rceil \equal{} x \plus{} n \equal{} \lceil x \rceil \plus{} n$[/quote]

But this only proves it for the special case of real numbers, namely the integers...It doesn't actually prove it for all real numbers.

$ \left\lceil x\plus{}n\right\rceil \equal{}\left\lceil x\right\rceil\plus{}n$ if and only if $ x \in R$ and $ n \in Z$

$ \lceil x\plus{}n\rceil \equal{} x\plus{}n \equal{}\lceil x\rceil\plus{}n$ if $ x, n \in Z$

Oh, sorry, I read your post wrong...I thought you said that $ x$ has to be an integer.  :blush:


I was doing a certain problem involving sequences and I had to prove the closed form of the sequence given the recursive relation using induction. I proved that for $ n \equal{} 3$ the recursive formula matches the closed form and then I assumed that the closed form is true for some integer $ k$. But the problem is, I had two expressions to substitute (the previous term and the 2nd previous term), but only one of them included the term with $ k$ as the subscript. But aren't you only allowed to state that there is only one integer $ k$ that works? If so, then wouldn't you have to assume that all values below $ k$ work too (strong induction)?

The only reason I'm asking is that I am supposed to be able to prove that problem with the normal induction.

There exists a sequence with $ a_1 \equal{} 11$, $ a_2 \equal{} 21$, and $ a_n \equal{} 3a_{n\minus{}1}\minus{}2a_{n\minus{}2}$ for $ n \ge 3$. Prove that $ a_n \equal{} 5\cdot 2^n \plus{} 1$.

Please don't prove it for me. I just want to know whether the problem requires strong induction if you could only use some sort of induction to prove it.

The inductive hypothesis you need is that the formula given is correct for some integer $ k$ and it is [b]also[/b] correct for $ k \plus{} 1$; call this $ P(k)$.  If you can show that this implies that the formula is correct for $ k \plus{} 2$, you have proven $ P(k \plus{} 1)$.  The important thing is to distinguish the formula being correct from your inductive hypothesis; just because they [b]can[/b] be the same thing doesn't mean they have to be.  In fact, "strong induction" is just regular induction, but the inductive hypothesis $ P(k)$ is "the formula is correct for $ 1, 2, ... k$."

I'll say it again because it's worth repeating:  [b]strong induction is a special case of induction.[/b]

[quote="funtwo"]am i missing something or is it possible to use the characteristic equation to find the closed form[/quote]

Sure, you could.  Or, you could see in the problem statement that it is given as $ a_n\equal{}5\cdot 2^n\plus{}1$.  Writing up a solution using the characteristic equation is often much more difficult and time consuming than simply using it to solve the given recursion and then proving the result works by induction.

Just a quick question: my base case case needs to be $ n \equal{} 3$ right?

Actually, since your inductive hypothesis is that it holds for both $ k$ and $ k\plus{}1$, you will need to show two "base cases."  The idea in induction is that you prove your inductive hypothesis is true for the smallest case, then you show that if the hypothesis is true for any case, your inductive hypothesis is true in the next case.  A sort of domino effect results.

If you want you can use $ n\equal{}3$ and $ n\equal{}4$, but you might as well use $ n\equal{}1$ and $ n\equal{}2$.  The only reason the recursion is stated to be true for $ n \ge 3$ is that they need something to start from.  But the explicit formula should hold for all $ n \ge 1$.

In my proof I already stated that it works for $ n \equal{} 1$ and $ n \equal{} 2$ and I made $ n \equal{} 3$ my base case because that is the first value you can compare between the recursive formula and the explicit formula. So should I change it just to make $ n \equal{} 1$ and $ n \equal{} 2$ my base cases?

yes.  There really is no reason to compare.  What you want to do is show that if $ a_k\equal{}2*5^k\plus{}1$ and $ a_{k\plus{}1}\equal{}2*5^{k\plus{}1}\plus{}1$, then $ a_{k\plus{}2}\equal{}2*5^{k\plus{}2}\plus{}1$ (applying the recursion).  Then, since the explicit formula works for $ a_1$ and $ a_2$, and the truth of the formula for $ a_k$ and $ a_{k\plus{}1}$ implies the truth of the formula for $ a_{k\plus{}2}$, the formula must work for all $ a_n$ where $ n$ is a positive integer.


calculus
integration
calculus computations


[hide="solution"]
answer=$\boxed5$
[/hide]
hope nobody saw :blush:  :oops:

[hide]Since there are already $4$ yellow marbles in the urn, we want a probability to select a yellow marble as $\frac4{12}$. Since there are $4$ yellow marbles and $3$ blue marbles already in there (a total of $7$), you just need to add $\boxed{5}$ more blue marbles.[/hide]

----another solution----

[hide]

  The probability of getting a yellow ball is:

 $\frac{\text{Number of Yellow Balls}}{\text{Total number of Balls}}=\frac{1}{3}$

  Thus the ratio of yellow balls to blue and yellow balls is:

$1:3$

  And the ratio of Yellows Balls to Blue Balls is:

$1:2$

  Thus we need $(4\cdot2)-3=\boxed{5}$ more blue balls. [/hide]

I like your explanations.

Thanks a lot.

[quote="symmetry"]An urn has 4 yellow marbles and 3 blue marbles.  How many blue marbles must be added to the urn so that the probability of selecting a yellow marble is 1/3?[/quote]

[hide]$\frac{4}{n+7}=\frac{1}{3}$

$4\cdot 3 = 1\cdot (n+7)$

$12 = n+7$

$\boxed{n = 5}$
[/hide]


advanced fields
advanced fields unsolved

Hello,

How could I show the simple connectedness of $\mathbb{R}^{2}-[1,+\infty[$ ?

More generally, what are the methods to address simple connectedness (in the previous example that seems really obvious but I really do not see how to formalize it).

For that example, we can pull anything in by a straight line to the origin.

In general? There's usually something obvious you can use.


inequalities
rearrangement inequality
inequalities proposed

Let $ a$ and $ b$ are positive numbers, $ n\in\mathbb N.$ Prove that:

\[ (n\plus{}1)(a^{n\plus{}1}\plus{}b^{n\plus{}1})\geq (a\plus{}b)(a^{n}\plus{}a^{n\minus{}1}b\plus{} \dots \plus{}b^{n})\]

From Chebyshev's inequality we have $ 2(a^{n \plus{} 1} \plus{} b^{n \plus{} 1})\ge (a \plus{} b)(a^n \plus{} b^n)$. Thus we need only prove that
\[ (n \plus{} 1)(a^n \plus{} b^n)\ge 2\sum_{i \equal{} 0}^n a^{n \minus{} i}b^i
\]
But from Rearrangement inequality
\[ a^n \plus{} b^n\ge a^{n \minus{} i}b^i\plus{}a^ib^{n\minus{}i}
\]
Adding these inequalities for $ i \equal{} 0,1,\dots, n$ we get the desired result.


number theory
greatest common divisor
combinatorics unsolved
combinatorics

On the segment $ [0,N]$, the endpoints and two more points are marked so that the lengths of the three segments into which the segment $ [0,N]$ is divided are integers with gcd $ 1$. If $ A,B$ are two marked points whose distance is a multiple of $ 3$, then we can divide the segment $ AB$ into three equal parts, mark one of the division points and unmark one of the points $ A,B$. Is it true that we can mark any given integer point on the segment $ AB$ with several such operations?


I doubt if this is the appropriate forum to post that question because its mathematically trivial and involves chemistry, nevertheless,
Let x denote the desired quantity. Then,
$ 0.45\cdot x\plus{}0.6\cdot (40\minus{}x)\equal{}0.48\cdot 40\Leftrightarrow x\equal{}32$.


[quote="sri340"]In the multiplication problem below, A and B stand for different digits. Find A and B

                                    A B
                                 x  B A
                         ___________
                                1  1   4
                             3 0  4   0
                      _____________
                            3  1  5   4

            A =     B =[/quote]
hi all
A=3 , B=8   or   A= 8   B = 3
mrmohamed

B would have to be 8 and A =3 because $ 3\cdot38\equal{}114$ and $ 80\cdot38\equal{}3040$


The positive difference between two consecutive perfect squares is 59. What is the greater of the two perfect squares?

let x be variable so x^2 is smaller perfect square

then,
(x+1)^2 - x^2 = 59
2x+1 = 59
2x= 58
x= 29

the greater of the two perfect squares is 30^2 = $ \boxed{900}$


Car is 125 km directly west of car B and begins moving east at 100 km/h. At the same time, car B begins moving north at 50 km/h. At what time t does the minimum distance occur?

Refer to first time you posted this.  I provided an approach to the solution.

Ok, let our car will be car $ A$ and has coords $ (0,0)$ when $ t\equal{}0$. Then car $ B$ will have coords $ (125,0)$.
Now:
$ x_a (t) \equal{} 100t;\; y_a (t)\equal{}0$
$ x_b (t) \equal{} 125;\; y_b (t)\equal{}50t$

Square of the distance between them: $ d^2 (t)\equal{}(125\minus{}100t)^2 \plus{} 50^2 t^2$

$ \frac{d}{dt}\left(d^2(t)\right)\equal{}\minus{}200(125\minus{}100t)\plus{}5000t\equal{}\minus{}25000\plus{}25000t\equal{}0$  :D $ t\equal{}1 hour$


Let $ n \equal{} 2^{p}$ be a power of $ 2$. We consider the subsets $ A$ of the set$ E \equal{}${$ 1, 2, ..., n$} having the following
property : if $ x\in A$ then $ 2x \notin A$. How many integers at most can such a subset $ A$ contain ?

Search "double-free sets of integers" on Google; one of the hits is [url]http://mathworld.wolfram.com/Double-FreeSet.html[/url]

According to it, the maximal cardinality for $ n\equal{}2^p$ is $ \frac {2\cdot 2^p \plus{} (\minus{}1)^p} {3}$.


Prove that $\mathbb{R}$ and $\mathbb{R}^2$ are not homeomorphic. ;)

Well, that's really easy: $\mathbb{R}^2$ is connected after removal of any point, but $\mathbb{R}$ is no longer connected after removal of any point.
Here is just a kinda gereneralization http://www.mathlinks.ro/Forum/viewtopic.php?highlight=homeomorphic&t=80494#

here the material

http://www.imar.ro/%7Esergium/fisiere/tanot.pdf


How to calculate the below integral:

$ \iiint_{\frac{x^2}{a^2}\plus{}\frac{y^2}{b^2}\plus{}\frac{z^2}{c^2}\leq 1}\left|xyz\right|dxdydz$


probability
logarithms
search
number theory
relatively prime

Forty teams play a tournament in which every team plays every other team exactly once. No ties occur, and each team has a $ 50\%$ chance of winning any game it plays. The probability that no two teams win the same number of games is $ m/n$ where $ m$ and $ n$ are relatively prime positive integers. Find $ \log_{2}n$.

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1750668289&t=66428

If you want/have additional discussion, you're free to revive the topic.


limit
logarithms
calculus
integration
real analysis
real analysis unsolved

Let $\{\epsilon_n\}^\infty_{n=1}$ be a sequence of positive reals with $\lim\limits_{n\rightarrow+\infty}\epsilon_n = 0$. Find \[ \lim\limits_{n\rightarrow\infty}\dfrac{1}{n}\sum\limits^{n}_{k=1}\ln\left(\dfrac{k}{n}+\epsilon_n\right)  \]

Approach it by integrals (it looks like Riemann sum), I think. And I think it is $n\to\infty$.

$\lim\limits_{n\rightarrow\infty}\dfrac{1}{n}\sum\limits^{n}_{k=1}\ln\left(\dfrac{k}{n}+\epsilon_n\right) \geq \lim\limits_{n\to\infty}\dfrac{1}{n}\sum\limits^{n}_{k=1}\ln \dfrac{k}{n} = \int_0^1 \ln x dx = -1$

And $\epsilon_n$ become sufficiently small, say $\epsilon_n\leq \epsilon$, for all $n\geq N$. Then
$\lim\limits_n \dfrac{1}{n}\sum\limits_{k=1}^n \ln\left(\dfrac{k}{n}+\epsilon_n\right) \leq \lim\limits_n \dfrac{1}{n}\sum\limits_{k=N}^n \ln\left(\dfrac{k}{n}+\epsilon\right) = \int_0^1\ln(x+\epsilon)dx = (1+\epsilon)\ln(1+\epsilon)-\epsilon\ln \epsilon-1$.

And it tends to $-1$ when $\epsilon\to 0$.

We know that orginial limit is $-1$.


1) Select from each of the following pairs of compounds the one with the highest boiling point (explaining your answer):

(a) p-Cresol or anisole.
(b) o-dichloro benzene or p-dichlorobenzene.
(c) Methyl acetate or propanoic acid.
(d) Ethyl fluoride or ethyl iodide.
(e) Phenol or thiophenol.
(f) Ethylene glicol or n-propylic alcohol
(g) Ethyl ether ot ethyl sulfide.
(h) Ethylene glicol or ethylene diamine.
(i) Ethylene glicol or 1,2-dimethoxiethane.

2) Propose reactions that might show how iso-propyl alcohol could be prepared from:

(a) an alkene.
(b) an alkyl halide.
(c) a Grignard reagent.
(d) a ketone.
(e) an alcohol.
(f) a carboxilic acid.

3) For each of the following compounds propose two possible synthesis from readily available starting materials:

(a) 1-Phenylpropan-1-ol
(b) 1-Methylcyclohexanol
(c) Cyclohexylmethanol
(d) Triphenylmethanol
(e) 1-(p-Tolyl)-ethanol
(f) 3-Phenylpentan-3-ol
(g) 2-Phenylethanol

4) Neopentyl chloride cannot be prepared from the corresponding alcohol, due to the great tendency of neopentyl cation to rearrange. Propose then a possible synthesis of neopentyl chloride.

5) Predict the relative rates of reaction of the alcohols in each of the following groups with aqueous HBr:

(a) Benzylic alcohol, p-methylbenzylic alcohol, and p-nitrobenzylic alcohol
(b) Benzylic alcohol, alpha-phenylethyl alcohol, and beta-phenylethyl alcohol.

6) What are the products of the reaction of cyclohexanol with each of the following reagents:

(a) Cold sulphuric acid.
(b) Hot sulphuric acid.
(c) $ CrO_3/H_2SO_4$.
(d) Acetic acid, $ H_3O^\plus{}$.
(e) $ CH_3MgBr$.
(f) Benzene, H+.

7) Which of the following compounds give a positive iodoform test?

(a) Ethanol
(b) Pentan-3-ol
(c) 2-Phenylethanol
(d) Pentan-2-ol
(e) Tert-butanol
(f) Sec-butylic alcohol
(g) Iso-propylic alcohol
(h) Acetone

2) Propose reactions that might show how iso-propyl alcohol could be prepared from:
(a) an alkene.
(b) an alkyl halide.
(c) a Grignard reagent.
(d) a ketone.
(e) an alcohol.
(f) a carboxilic acid.

Answer
a. Oxymercuration Demercuration from propene
b. Using Sommelet Synthesis to convert 2 halo propane into propanone and then MPV redn
    (or)
    Using aq. NaOH to convert it into propan-2-ol and then using Oppeneaur oxidation
d. Do as in b step 1 part 2.
e. Do as in b step 2 part 2

4) Neopentyl chloride cannot be prepared from the corresponding alcohol, due to the great tendency of neopentyl cation to rearrange. Propose then a possible synthesis of neopentyl chloride.

Answer: Effect a dehydration of it and then perform allylic chlorination and then hydrogenate.

8) Propose a synthesis of:

(a) n-Butyraldehyde from butan-1-ol
(b) Butan-1,2-diol from n-butanol
(c) 5-(n-propil)-nonan-5-ol from butan-1-ol
(d) Butan-2-ol from ethanol
(e) 3-methylpentan-2-ol from butan-2-ol
(f) 1-Phenylethanol from benzene and ethanol
(g) 2-methylhexan-2-ol from alcohols with four or less carbons
(h) 3-Metyl-1-phenylbutan-2-ol from benzene and alcohols with four or less carbons

9) Propose simple chemical tests that allow one to distinguish between:

(a) n-Butanol and n-octane
(b) n-Butanol and n-pentyl bromide
(c) But-2-en-1-ol and but-3-en-1-ol
(d) Pentan-2-ol and pentan-3-ol
(e) Cyclohex-2-en-1-ol, 1-methylcyclohexanol, and cyclohexylmethanol
(f) n-Butanol and tert-pentanol
(g) Ethylene chlorohydrine, ethylene chloride, and ethylene glicol
(h) Cyclohexanol, cyclohexyl bromide, cyclohexene, and cyclohexane
(i) Ethylbenzene and benzylic alcohol
(j) Tert-butanol, tert-butyl chloride, and oct-1-ene
(k) Tert-pentanol and n-butane
(l) n-Propylic alcohol and allylic alcohol
(m) Pent-2-ene and iso-propanol
(n) 3-Hexyne and iso-propanol
(o) Propyne and ethanol

For 7
I think a d f g h answer iodoform test a. because it gets oxidised to ethanal in the process
others because they are either methyl ketones or carbinols.

[quote="Madness"]e. Do as in b step 2 part 2[/quote]

How about a different synthesis?

[quote="Madness"]Answer: Effect a dehydration of it and then perform allylic chlorination and then hydrogenate.[/quote]

A dehydration of what?

10) A primary alcohol reacts with acetic acid in the presence of concentrated sulphuric acid to yield a compound Z which contains 31.38% of oxygen. Propose a possible structure for the alcohol.

11) When 5.00 mg of a colourless liquid A (composed only by carbon, hydrogen, and oxygen) are burned in oxygen, 0.01100 g of carbon dioxide and 0.00600 g of water were obtained. The molecular weight of A is 60 g/mol. Treating 0.721 g of A with sodium metal one obtained 134.6 mL of dry hydrogen (at NTP) and a white solid B that contains 28.05% of sodium. When B is treated with propene oxide the compound C is obtained. Propose structures for A, B, and C, and explain the reactions that occured.

12) Show how each of the following ethers could be prepared:

(a) Ethyl t-butyl ether
(b) Phenyl n-propyl ether
(c) Iso-butyl sec-butyl ether
(d) Cyclohexyl methyl ether
(e) p-Nitrobenzyl phenyl ether
(f) Vinylic ether

13) Propose simple chemical tests (if possible) that allow one to distinguish an alifatic ether from an:

(a) alkane.
(b) alkene.
(c) alkyne.
(d) alkyl halide
(e) a primary alcohol
(f) a secundary alcohol
(g) a tertiary alcohol
(h) alkyl aryl ether.

14) What product (if any) might we expect from the reaction of p-methylanisole with concentrated $ KMnO_4/H^ \plus{}$?

15) By reaction with HI an ether yields two alkyl halides A and B that contain, respectively, 81.4% and 89.4% of iodine. Ascertain the structure of the ether.

4) Neopentyl chloride cannot be prepared from the corresponding alcohol, due to the great tendency of neopentyl cation to rearrange. Propose then a possible synthesis of neopentyl chloride.

[b][i]1.CH3-CHO + 3 HCHO --> (CH2OH)3C-CHO
2.(CH2OH)3C-CHO + 2CH3OH + HCl --> (CH2OH)3C-CH(OMe)2
3.(CH2OH)3C-CH(OMe)2 + Red P/I2 --> (CH3)3C-CH(OMe)2
4.CMe3-CH(OMe)2 + H3O+ --> CMe3-CHO
5.CMe3-CHO + LiAlH4 --> CMe3-CH2OH + SOCl2 --> CMe3-CH2Cl (Neopentyl chloride)[/i][/b]

15) By reaction with HI an ether yields two alkyl halides A and B that contain, respectively, 81.4% and 89.4% of iodine. Ascertain the structure of the ether.

The ether is ethyl methyl ether Me-O-Et.

The molecular mass of A = 156.01
=>The mass of the alkyl part = 29 (Corresponds to C2H5)
So, A is Et-I

The molecular mass of A = 142.06
=>The mass of the alkyl part = 15 (Corresponds to CH3)
So, A is Me-I

8) Propose a synthesis of:

(a) n-Butyraldehyde from butan-1-ol
(b) Butan-1,2-diol from n-butanol
(c) 5-(n-propil)-nonan-5-ol from butan-1-ol

8.(a) PCC(Pyridinium chloro chromate)

(b)1.Heated Al2O3(ELIMINATION WITHOUT REARRANGEMENT)
2.Baeyer's reagent (cold alk.KMnO4)

(c)Both the reagent used and the reactant can be prepared.

[u]Reagent:-[/u]
butanol + SOCl2 -->butyl chloride + Mg/ether -->butyl grignard

[u]Reactant:-[/u]
butanol + acidified K2Cr2O7 -->butanoic acid + C2H5OH/H+ -->ethyl butanoate

Treat the ester with two moles of the grignard to get the product

[quote="Chemrock"]CMe3-CH2OH + SOCl2 --> CMe3-CH2Cl (Neopentyl chloride)[/quote]

Wouldn't just that be enough? And how about a synthesis without using neopentyl alcohol? Your others answers are correct.


[hide="Solution"]Because 5 and 18 are across from each other, the numbers between 5 and 17, inclusive, make half of the cards in the deck. There are 13 integers between 5 and 17, inclusive, so the total amount of cards is 26.[/hide]

Oh, how simple.  That helps a lot!   :lol:

[quote="AdithyaGanesh"]How do you solve a problem such as this one?

The cards in a deck marked with consecutive positive integers are equally spaced around a circle.  If the card marked $ 5$ is directly opposite the card marked $ 18$, how many cards are in the deck?[/quote]

Alternatively, any numbers across from each other will have the same difference, 13 in this case, so we have {1, 14}, {2,15}, etc. Up to {13, 26}.

This assumes of course that the cards start from 1 and work their way up.

I don't know why I didn't think of that.  Thanks guys!


calculus
quadratics
trigonometry
analytic geometry
graphing lines
slope
algebra

Find the largest value for $\frac{y}{x}$ for pairs of real numbers $(x,y)$ which satisfy \[(x-3)^2 + (y-3)^2 = 6.\]

$\textbf{(A) }3 + 2 \sqrt 2\qquad
\textbf{(B) } 2 + \sqrt 3\qquad
\textbf{(C ) }3 \sqrt 3\qquad
\textbf{(D) }6\qquad
\textbf{(E) }6 + 2 \sqrt 3$

I was thinking of a nice connection between the problem and circles and tangents.... though i am a bit sketchy right now... will post a solution later (of course .. for those who know --- calculus might be a way out!)

Suppose $\frac{y}{x} = k$. Then $y = kx$. Substituting back into the equation, we get

$(x-3)^2+(kx-3)^2-6 = (1+k^2)x^2-6(1+k)x+12 = 0$.

The maximum of $k$ should occur when this quadratic has only one solution, namely when the discriminant is $0$:

$36(1+k)^2 - 48(1+k^2) = 0$
$3(1+k)^2-4(1+k^2) = 0$
$-k^2+6k-1 = 0$

which, by the quadratic formula, gives the solution $k = \frac{-6-\sqrt{36-4}}{-2} = 3+2\sqrt{2}$, which is A.

This problem can be solved by eye!

We are to remark two-typed triangles, triangle rulers.

The desired answer is $\tan 75^\circ=2+\sqrt{3}$, that is $B.$ :lol:

Suppose $\frac{y}{x}=k$
$(x-3)^2+(kx-3)^2=6$
$\Rightarrow (1+k^2)x^2-6(1+k)x+12=0$
The discriminant must be greater than or equal to 0.

Thus, $D\ge0$
$\Rightarrow (6(1+k))^2-4*12*(1+k^2)\ge0$
$\Rightarrow k^2+1\ge6k$
$\Rightarrow (k-(3+2\sqrt{2}))(k-(3-2\sqrt{2}))\ge0$
Thus $3-2\sqrt{2}\le k\le 3+2\sqrt{2}$
Thus maximum value of $\frac{y}{x}$is [b]A[/b]

[quote="Bictor717"]I drew the circle, the tangent $y=kx$, the segment connecting the origin to the center, and the perpendicular from the center to the x-axis. I found the sides of the triangles and used the $\tan(a+b)$ formula to find the slope of the tangent line, which is $k$. (The slope of a line is $\tan\theta$, $\theta$ being the angle the line makes with the x-axis.)
[hide="I got"]$3+2\sqrt 2$[/hide]

How did you get $75^\circ$, kunny?[/quote]

Sorry for that my answer is incorrect. :blush:  :P

Let $\theta$ be the angle between the line $y=x$ and $y=mx$, we have $\tan \theta =\frac{\sqrt{6}}{2\sqrt{3}}=\frac{1}{\sqrt{2}}$ Thus the desired value of $m$ is $\tan (\theta +45^\circ)=3+2\sqrt{2}$

kunny

Why are none of you hiding your answers?

[hide="Answer"]We have a circle with radius $\sqrt{6}$ and center at $(3,3)$. If $\frac{y}{x}=n$, we have $(x-3)^2+(nx-3)^2=6\Rightarrow (n^2+1)x^2-6(n+1)x+12=0$. By quadratic formula, $\frac{6(n+1)\pm \sqrt{36(n^2+2n+1)-4(12)(n^2+1)}}{2(n^2+1)}$

$36(n+1)^2-48(n^2+1)=0\Rightarrow 3(n+1)^2-4(n^2+1)\Rightarrow n^2-6n+1=0$
$n=3\pm 2\sqrt{2}\Rightarrow \boxed{A}$[/hide]

the answer i got was also A. but i had to use calculs mainly b/c i was sorta crammed for time and did not feel like thinking of another way to do it. but yeah i like the solutions give so far.

I did it with 2 right triangle and the sum of tangents.

[quote="paladin8"]Suppose $ \frac {y}{x} \equal{} k$. Then $ y \equal{} kx$. Substituting back into the equation, we get

$ (x \minus{} 3)^2 \plus{} (kx \minus{} 3)^2 \minus{} 6 \equal{} (1 \plus{} k^2)x^2 \minus{} 6(1 \plus{} k)x \plus{} 12 \equal{} 0$.

The maximum of $ k$ should occur when this quadratic has only one solution, namely when the discriminant is $ 0$:

$ 36(1 \plus{} k)^2 \minus{} 48(1 \plus{} k^2) \equal{} 0$
$ 3(1 \plus{} k)^2 \minus{} 4(1 \plus{} k^2) \equal{} 0$
$ \minus{} k^2 \plus{} 6k \minus{} 1 \equal{} 0$

which, by the quadratic formula, gives the solution $ k \equal{} \frac { \minus{} 6 \minus{} \sqrt {36 \minus{} 4}}{ \minus{} 2} \equal{} 3 \plus{} 2\sqrt {2}$, which is A.[/quote]

That was pretty dang clever.

[quote="JesusFreak197"]Why are none of you hiding your answers?

[hide="Answer"]We have a circle with radius $ \sqrt {6}$ and center at $ (3,3)$. If $ \frac {y}{x} \equal{} n$, we have $ (x \minus{} 3)^2 \plus{} (nx \minus{} 3)^2 \equal{} 6\Rightarrow (n^2 \plus{} 1)x^2 \minus{} 6(n \plus{} 1)x \plus{} 12 \equal{} 0$. By quadratic formula, $ \frac {6(n \plus{} 1)\pm \sqrt {36(n^2 \plus{} 2n \plus{} 1) \minus{} 4(12)(n^2 \plus{} 1)}}{2(n^2 \plus{} 1)}$

$ 36(n \plus{} 1)^2 \minus{} 48(n^2 \plus{} 1) \equal{} 0\Rightarrow 3(n \plus{} 1)^2 \minus{} 4(n^2 \plus{} 1)\Rightarrow n^2 \minus{} 6n \plus{} 1 \equal{} 0$
$ n \equal{} 3\pm 2\sqrt {2}\Rightarrow \boxed{A}$[/hide][/quote]

So after using quadraitic formula, you get $ 36(n \plus{} 1)^2 \minus{} 48(n^2 \plus{} 1) \equal{} 0\Rightarrow 3(n \plus{} 1)^2 \minus{} 4(n^2 \plus{} 1)\Rightarrow n^2 \minus{} 6n \plus{} 1 \equal{} 0$
How do you simplify $ \frac {6(n \plus{} 1)\pm \sqrt {36(n^2 \plus{} 2n \plus{} 1) \minus{} 4(12)(n^2 \plus{} 1)}}{2(n^2 \plus{} 1)}$ to get it?

The circle and the line y=kx have only one point of intersection for a largest value.
Therefore, you want only one root, and setting the discriminant=0 gives you the x for that root.


inequalities
logarithms
number theory
relatively prime
number theory proposed

For every positive integer $ n$, let $ \phi(n)$ denotes the number of positive integers less than $ n$ that is relatively prime to $ n$ and $ \tau(n)$ denote the sum of all positive divisors of $ n$. Let $ n$ be a positive integer such that $ \phi(n)|n\minus{}1$ and that $ n$ is not a prime number. Prove that $ \tau(n)>2009$.

[b]Problem. [/b] Show that if $ n \not \in \mathbb{P} \cup \{1\}$ and $ \varphi(n) \mid n \minus{} 1$ then $ \sigma_1(n) > 2009$.

[i]Solution.[/i] Since $ 2 \mid \varphi(n)$ for all $ n \in \mathbb{N} \setminus \{0,1,2\}$ then it is true that $ \text{lpf}(n) > 2$. Again we have that $ \mu(n)^2 \equal{} 1$ since otherwise we'll have that $ p^2 \mid n$ for some $ p \in \mathbb{P}$ and $ p \mid \varphi(n) \mid n \minus{} 1$, that is a contradiction. Now the chain of inequalities $ 2^{11} > 2009 \ge \sigma_1(n) \ge 12^{\omega(n) \minus{} 3} \cdot 8 \cdot 6 \cdot 4 \equal{} 2^{2\omega(n)}3^{\omega(n) \minus{} 2} \ge 2^{\frac {7}{2}\omega(n) \minus{} 3}$ holds. It implies that $ \omega(n) \in \{2,3\}$. If $ \omega(n) \equal{} 2$ then there exist $ (p,q) \in \mathbb{P}^2$ such that $ p > q > 2$ and $ \frac {pq \minus{} 1}{(p \minus{} 1)(q \minus{} 1)} \in \mathbb{Z}$, but $ 1 \equal{} \frac {(p \minus{} 1)(q \minus{} 1)}{(p \minus{} 1)(q \minus{} 1)} < \frac {p(q \minus{} 1)}{(p \minus{} 1)(q \minus{} 1)} <$ $ \frac {pq \minus{} 1}{(p \minus{} 1)(q \minus{} 1)} < \frac {p}{p \minus{} 1}\frac {q}{q \minus{} 1} \le \frac {3 \cdot 5}{2 \cdot 4} < 2$, that is a contradiction. It remains the last case $ \omega(n) \equal{} 3$ that is false at the same way (for example see the easy and more general one [url=http://web.archive.org/web/20031228113634/www.kalva.demon.co.uk/imo/isoln/isoln921.html]IMO1992/1[/url]).[]

[i]Additional Note.[/i] This question is related to a famous unsolved problem in number theory, the Lehmer conjecture (i.e. it does not exist a positive composite integer $ n > 1$ such that $ \varphi(n) \mid n \minus{} 1$). It was first posed in 1932, and since then some progress was made, e.g. defining $ S: \equal{} \{n \in \mathbb{N} \setminus \left(\mathbb{P} \cup \{0,1\}\right): \varphi(n) \mid n \minus{} 1\}$ then $ |S \cap [1,x]| \equal{} O\left(x^{\frac {1}{2}}(\ln(x))^{\frac {3}{4}}\right)$ ( see [url=http://www.math.dartmouth.edu/~carlp/PDF/paper15.pdf]here[/url] for a proof by Carl Pomerance ). For a more advanced lecture, you can also see [url=http://www.math.waikato.ac.nz/~kab/papers/lehmer4.pdf]here[/url] where it is proved that $ |S \cap [1,x]| \equal{} O\left(x^{\frac {1}{2}} (\ln(\ln(x)))^{\frac {3}{2} \plus{} e}(\ln(x))^{ \minus{} c}\right)$ for all $ e > 0$ and where $ c \equal{} 0,129398..$ is a fixed constant .


superior algebra
superior algebra solved

Let G be a group and let x and y be elements of G. Prove that if x is in <y> then <x> is contained in <y>.

Isn't it obvious?
$x\in\langle y\rangle$ implies that $x=y^m$ for some $m\in \mathbb{Z}$.
Thus
$\langle x\rangle = \{y^{mk} | k\in \mathbb{Z}\} \subset \langle y\rangle$

Actually, it's clear that if $x\in H\le G$, then $\langle x\rangle\le H$. Are these problems your homework, liltubbs? :)

[quote="liltubbs25"]I cant get help with homework in this forum?[/quote]
You can, of course, but it is not the primary purpose of the forum. In general the problems that are posted here are either the problems you cannot get help with from other standard sourses like your teacher, help room, friends, etc. or the problems you like so much that you want others to share the pleasure of solving them. Imagine what this forum would become if every student who doesn't know how to solve his standard homework assignment would post it here  :) .


If $\displaystyle \frac{xy}{x+y} = a$, $\displaystyle \frac{xz}{x + z} = b$ and $\displaystyle \frac{yz}{y + z} = c$, where $a$, $b$, and $c$ are other than zero, then $x$ equals:
\[ \text{(A)}\ \frac{abc}{ab + ac + bc} \qquad \text{(B)}\ \frac{2abc}{ab + bc + ac} \qquad \text{(C)}\ \frac{2abc}{ab + ac - bc} \]
\[ \text{(D)}\ \frac{2abc}{ab + bc - ac} \qquad \text{(E)}\ \frac{2abc}{ac + bc - ab} \]

Yup, I agree with shobber.

[hide="Hint"]
$\frac{1}{a}=\frac{x+y}{xy}$
$\frac{1}{b}=\frac{x+z}{xz}$
$\frac{1}{c}=\frac{y+z}{yz}$
[/hide]


Three semicircles of radius $ 1$ are constructed on diameter $ AB$ of a semicircle of radius $ 2$. The centers of the small semicircles divide $ \overline{AB}$ into four line segments of equal length, as shown. What is the area of the shaded region that lies within the large semicircle but outside the smaller semicircles?
[asy]import graph;
unitsize(14mm);
defaultpen(linewidth(.8pt)+fontsize(8pt));
dashed=linetype("4 4");
dotfactor=3;

pair A=(-2,0), B=(2,0);

fill(Arc((0,0),2,0,180)--cycle,mediumgray);
fill(Arc((-1,0),1,0,180)--cycle,white);
fill(Arc((0,0),1,0,180)--cycle,white);
fill(Arc((1,0),1,0,180)--cycle,white);
draw(Arc((-1,0),1,60,180));
draw(Arc((0,0),1,0,60),dashed);
draw(Arc((0,0),1,60,120));
draw(Arc((0,0),1,120,180),dashed);
draw(Arc((1,0),1,0,120));
draw(Arc((0,0),2,0,180)--cycle);

dot((0,0));
dot((-1,0));
dot((1,0));

draw((-2,-0.1)--(-2,-0.3),gray);
draw((-1,-0.1)--(-1,-0.3),gray);
draw((1,-0.1)--(1,-0.3),gray);
draw((2,-0.1)--(2,-0.3),gray);

label("$A$",A,W);
label("$B$",B,E);
label("1",(-1.5,-0.1),S);
label("2",(0,-0.1),S);
label("1",(1.5,-0.1),S);[/asy]$ \textbf{(A)}\ \pi\minus{}\sqrt3 \qquad
\textbf{(B)}\ \pi\minus{}\sqrt2 \qquad
\textbf{(C)}\ \frac{\pi\plus{}\sqrt2}{2} \qquad
\textbf{(D)}\ \frac{\pi\plus{}\sqrt3}{2}$
$ \textbf{(E)}\ \frac{7}{6}\pi\minus{}\frac{\sqrt3}{2}$

[hide]
When two of the smaller semi circles overlap their intersection must be in the middle because they are the same size. Therefore the overlap region $ \equal{} 2\cdot\frac {1}{2}\left(\frac {2\pi}{3} \minus{} \frac {1}{2}\sin\left(\frac {2\pi}{3}\right)\right) \equal{} \frac {2\pi}{3} \plus{} \frac {\sqrt {3}}{2}$

$ \Rightarrow A \equal{} \frac {1}{2}\pi 2^2 \minus{} 3\left(\frac {1}{2}\pi 1^2\right) \plus{} \frac {4\pi}{3} \minus{} \sqrt {3} \equal{} \frac {7\pi}{6} \minus{} \frac {\sqrt {3}}{2}$[/hide]

[hide="Pretty Picture Solution"][i]Please allow the image a few moments to load[/i]
[img]http://img.photobucket.com/albums/v194/mukmasterxps/solution.png[/img]

$ \frac {1}{2} \left( \pi 2^2 \right) \minus{} \frac {5}{6} \left( \pi 1^2 \right) \minus{} 2 \left( \frac {1}{2} \cdot 1 \cdot \frac {\sqrt {3}}{2} \right)$

$ \equal{} \frac {7}{6} \pi \minus{} \frac {\sqrt {3}}{2} \implies \boxed{\textbf{(E)}}$[/hide]

Here's how I organize my work (This has become an important part of my philosophy):
Area of
Big semicircle: $ 2 \pi$
Little semicircle: $ \frac {1}{2} \pi$
60 degrees sector of little circle: $ \frac {1}{6} \pi$
60 degree segment: $ \frac {1}{6} \pi \minus{} \frac {\sqrt {3}}{4}$
So the final answer is Big semicircle- 3 x little semicircle + 2 x sector + 2 x segment =
$ 2 \pi \minus{} 3 \cdot (\frac {1}{2} \pi) \plus{} 2 \cdot (\frac {1}{6} \pi) \plus{} 2 \cdot (\frac {1}{6} \pi \minus{} \frac {\sqrt {3}}{4}) \equal{} \frac {7}{6} \pi \minus{} \frac {\sqrt {3}}{2}$


he..:)
i have a kwestion...:) i hope u can help me.
i need experience in problem solving,  I want to start with good books for maths. I have schoolbooks but there are not as ' challenging' as i thought..

Is it a good idea to buy the AoPS books volume1/2?

Yes! It's definitely a good idea. :) (Mods, move this topic.)

If your only mathematical experience is from school textbooks, the Art of Problem Solving books are what you're looking for.  How much experience do you have?  Are you looking for help with MathCounts level problems?  Can you do most MathCounts problems easily, but have trouble with some of the AMC 10/12 problems?

AoPS vol. 1 is (mostly) for MathCounts/ easier AMC 10-12 problem level study.
AoPS vol. 2 is (mostly) for harder AMC 10-12/AIME/easy Olympiad level study.

You should still get both volumes though, even if you can qualify for the AIME and stuff.  I qualified for the AIME last year, and I'm still learning a lot of stuff from just volume 1.

if all you have used are schoolbooks, you should buy both books, but put more emphasis on volume 1 for now.

I have both books myself, and they're GREAT! They come with a seperate detailed solutions book, and each section has examples and excercises, and questions at the end of each section that come from math contests such as Mandlebrot, Olympiad, ect. (If you have any more, feel free to add)


Consider the set $S=\{1,2,...n\}$ and define $S_k=\{X \subseteq S, \ k \notin X, X\neq \phi\}$.
Determine the value of the sum :

\[ \displaystyle S_k^*=\sum_{\{i_1,i_2,...,i_r\}\in S_k} \frac{1}{i_1\cdot i_2 \cdot...\cdot i_r}. \]

[hide="Comment"]It seems that it appeared before, in the APMO.[/hide]

If I understand the question, the proof is straightforward.

$\displaystyle S_k^{*}=\frac{\displaystyle\prod_{j=1}^n (1+\frac{1}{j})}{1+\frac{1}{k}}-1$
$\displaystyle =\frac{k}{k+1}\prod_{j=1}^n (\frac{j+1}{j}) -1$
$\displaystyle =\frac{k}{k+1} (n+1) -1$.


calculus
integration
algebra
function
domain
Ring Theory
superior algebra

In an integral domain $ R$, if the maximal ideals are finite,  then $ R$ is a field ??

HINT  nonunit $ b \neq 0 \implies r\mapsto br\; \text{is 1\minus{}1 not onto} \implies |bR|$ infinite $ \implies$ max ideals infinite

oh, i am  sorry, i mean the following:
 ring $ R$ is integral domain, and $ R$ has only finite many maximal ideals, then $ R$ is a field???

No. Any local ring would do, e.g. $ k[x]_{(x)}$.


[b]Να βρεθεί η οξεία γωνία                $\widehat{\phi}$,      [/b] [b]η οποία ικανοποιεί τη σχέση:
[/b][b]$\frac{\sin{(B-\phi)}}{\sin{(C+\phi)}}$ $=$ $\frac{cosB}{cosC}$[/b],[b] συναρτήσει των [/b]$\widehat{B}, \widehat{C}$ [b]όταν αυτές  είναι μη μηδενικές και ακόμη τέτοιες ώστε $\widehat{B}+\widehat{C}< \pi$[/b]

[u][b][i]Nick Rapanos[/i][/b][/u]

Niko mipws exei kapoia parenthesi pouthena? Dioti an einai etsi opws tin edwses (xwris parenthesi) tote metasximatizetai stin $\phi(sinB+sinC)=0$ diladi $\phi\cdot 2sin\big(\frac{B+C}{2}\big)cos\big(\frac{B-C}{2}\big)=0$ kai epeidi $0<B+C<\pi$ ara $0<sin\big(\frac{B+C}{2}\big)<1$ kai epeidi $0<|B-C|<B+C<\pi$ ara $0<cos\big(\frac{B-C}{2}\big)<1$ sinepws $\phi=0$.

Apo tin alli an ennoeis na lithei i eksiswsi $\frac{\sin(B-\phi)}{\sin(C+\phi)}=\frac{sinB}{sinC}$

tote auti einai isodinami me tin $\frac{sinB cos\phi-sin\phi cosB}{sinC cos\phi+sin\phi cosC}=\frac{sinB}{sinC}$ pou einai isodinami meta tis prakseis me tin $sin\phi(sinC cosB+sinB cos C)=0$ diladi tin $sin\phi(sin(B+C))=0$ kai afou $0<B+C<\pi$ ara $0<sin(B+C)<1$ ara $sin\phi=0$ ki epeidi h gwnia $\phi$ einai oksia ara $\phi=0$ (oso oksia mporei na thewrithei auti i gwnia).

Den kserw ean ennoouses kati apo auta ta dio...

Alexandros

Νίκο το ίδιο βρήκα και εγω σήμερα το πρωί στο σχολείο .....
Για κοίτα μήπως έγινε κάπου λάθος μέχρι να προκύψει ?

Συγγνώμη ρε παιδιά!Λόγω κεκτημένης ταχύτητας μπερδεύτηκα και αντί να βάλω συνημίτονο έβαλα ημίτονο....
Sorry
Τώρα το διόρθωσα..

Βρήκα  $\tan \phi = \frac{1}{2}( \tan B-\tan C )$ .


Happy Haloween everyone. Anyone dressing up today?

[quote="Iversonfan2007"][quote="joml88"]This should go in F&G.  Please post accordingly in the future.[/quote]

TRICK OR TREAT[/quote]

trick

As I grow older, I think trick or treating is a little boring. I enjoy giving out candy to little kids.

I'm giving out candy right now, wearing a sombrero!

I found a way to save money and be safe: eat out every halloween!

um someone lock all these haloween topics cept one

it rained out during halloween so I went to a mall to trick or treat grr..

LoL and Halloween rocks! i dressed up as an old man with a beard! japanese too! even tho im not japanese :lol:

[quote="jackdillon"]i was a scary monster with a pig mask on (i made a little kid cry.....(accident))[/quote]

So was I, except it wasn't a pig mask. I made about two kids cry at the haunted house we made for little kids. It was fun.


number theory
prime factorization
prime numbers

Find  number of integers $x,y>0$ which satisfy

$\frac{1}{x}+\frac{1}{y}=\frac{1}{24}$

[hide]
$\frac{x+y}{xy}=\frac{1}{24}$

Cross multiply and simplify to get $xy-24x-24y=0$

Then rewrite it as $xy-24x-24y-576=576\Longleftrightarrow (x-24)(y-24)=576$.

The rest is basically some case work...
[/hide]

it's amazing how useful simon's factoring trick is :roll:

I understand how you get $(x-24)(y-24) = 576$; then do you have to count the divisors of $576$? Is there an easy way to do this without making a list?

Tarquin I think there are only $\ 21$ such pairs.  The problem restricts $\ x$ and $\ y$ to be positive.

To answer Pocket Sand's question, we use prime factorization for $576$. Since $576=2^6\cdot 3^2$, we have $(1+6)(1+2)=21$ positive divisors.

In general, if $N=n_1^{p_1}n_2^{p_2}...n_k^{p_k}$, where $n_1$, $n_2$, ... $n_k$ are prime numbers, then the number of positive divisors for $N$ is $(1+p_1)(1+p_2)...(1+p_k)$.

edit, is the answer just 21? I think I get it now.


Hey guys, I found this online:

http://www.isabio.com/?_lng=eng&_csq=48&_bsq=352

These are videos of the solutions to all of this year's AMC 8 problems. It was pretty helpful for me. It makes them all seem so easy though! I feel dumb...

This website looks new and it seems like they're going to be making solutions videos for all the AMC's, which is pretty sweet. I'm excited.

hmmm.. it shows up as jumbled text for me..

I didn't see it either, but I have a lot of doubt that this is not legal.

Works for me. I'm so happy, I guessed #25 correctly.
These are pretty good solutions. You should see them.

You have to use Internet Explorer to see it correctly for some reason.

DIE NON STANDARD HTML.

I e-mailed them and they said that for now it only works for Internet Explorer and that the Firefox version is actually in the works right now.

Sounds like they don't yet have enough people on the development team to get it all working at once, but they're working on it.

OMG i know the guy who made those videos..
it may be broken text because it's probably in korean..
here is another link..

the website was remade by Dr. Choi (the guy explaining everything)
and it reopened in October...

I watched the videos the day they came out, and i found out that i got at least a 20...

he has a pretty decent american accent, but that doesn't matter..
dr. choi is awesome!!

sorry if this is off topic..

if you want to know more about isabio, pm me..


number theory
greatest common divisor
algorithm
modular arithmetic
number theory proposed

Let $ A$ be a subset of the set $ \{1, 2,\ldots,2006\}$, consisting of $ 1004$ elements.
Prove that there exist $ 3$ distinct numbers $ a,b,c\in A$ such that $ gcd(a,b)$:
a) divides $ c$
b) doesn't divide $ c$

a) We can always find two coprome integers
b) assume theese two coprime integers are n,n+1. We can write them in the form $ 2^m\cdot l$. We see that one of them is odd and one is even. Hence if we find another even integer, their gcd will be also even, and theese two integers will be pair (a,b) and c - odd integer. Second case is when we choose all odd integers and one even. If even integer has odd divisor >1, then we take this odd integer and it's odd divisor as (a,b) and c=1. If this even number is $ 2^m$ for natural n, then we take for example (a,b)=(9,3) and $ c\equal{}2^m$. I hope there are no mistakes in my proof...

coprome=coprime?
If so, should then a and b be coprime?

Typo ;P coprome=corprime, we take (a,b) coprime => gcd(a,b)=1, and obviously 1|c.

[hide="Solution"]a)

I will prove that there exists at least two elements $a$ and $b$ such that $\gcd(a,b)=1$ in set $A$.  If I prove this, since there are $1002$ other elements, and $1$ divides all positive integers, we can choose any of the other $1002$ elements to be $c$, and we would have $\gcd(a,b)=1|c$.
Assume on the contrary that we have a subset with all pairs of numbers $a,b$ such that $\gcd(a,b)>1$. Assume $d$ is the minimal element in set $A$.  Then since $\gcd(d, d-1)=1$ and $\gcd(d,d+1)=1$ by Euclid’s algorithm, we have to have the next element that can have a gcd of $1$ with $d$ being at least $d+2$.  We must have a minimal last term being when we have a difference of $2$ among the terms giving us $\{d, d+2, \cdots, d+(1004-1)*2\}$ or $\{d, d+2, \cdots, d+2006\}$.  Therefore $d\le 0$, but we have $d\ge 1$ from $d\in A$ so an obvious contradiction.
Therefore it is impossible for all numbers to have a gcd greater than $1$ therefore at least two have a gcd of $1$ giving us $1|c$ and our proof is complete.
b)

 First off remark that we can’t have $A$ containing all even numbers or all odd numbers.  If we did, then $A$ would have to have at most $1003$ elements (from the set $\{1, 3, 5, \cdots, 2005\}$ and $\{2, 4, \cdots , 2006\}$ having $1003$ elements) implying that we must have at least one element that is the opposite parity of the rest.  We have three cases:
Case 1:  1003 odds, 1 even
Since there are $1003$ odd numbers, we have all the numbers $\{1, 3, 5, 7, 9, 11, \cdots 2005\}$ being a part of $A$.  Let $a=3, b=9$ and $c=5$ giving us an obvious contradiction since $3$ does not divide $5$.

Case 2:  At least two evens and odds

Let $a$ and $b$ both be even and let $c$ be odd.  Therefore letting $\gcd(a,b)=2a'$ and $c=2c'+1$, we must have $2a'|(2c'+1)$.  However, $(2c'+1)\equiv 1\pmod 2$ giving us an obvious contradiction since we can’t have $\gcd(a,b)$ dividing $c$.

Case 3:  1 odd, 1003 evens
Again, let $c$ be the odd number and $a$ and $b$ be two even numbers.  The proof of this is the same as the proof for case 2, since an even number can’t divide an odd number.
--
Since we have accounted for all cases, our proof is complete $\blacksquare$.
   [/hide]


I need help on doing these:

Create additional fields to include the goal seek scenarios:
1. How many points each particular student needs on the Final Exam to improve to the next possible higher grade, i.e., if a student current grade is B, what does it takes to get an A or if the current grade is C, then how many points that student needs to receive a B, etc.

So
Tools=>Goal Seek=>
Set Value
To Value: I try to do the IF function here (If less than 80, then "90", if less than...)
By Changing Cell

THEN pop-up: saying no formula in the To Value box
???

Tips hints on doing the followings???

2. How many points each particular student needs on the Final Exam to maintain the CURRENT grade, i.e., if a student current grade is B then how many points are needed to maintain the B, etc.

Note that the Final Exam is 25% of the grade and all quizzes are equal.

---------------------
Add additional data to show how many students current receive A, how many receive B, C, D, and F.

THANK YOU

Woops, Sorry, please delete this thread, I did not mean to spam I hit some button (don't know which yet) accidentally.

 :(


How do ticket machines detect the ticket? If the ticket is expired by even 1 hour the machine rejects it. How is this possible?

This is the ticket machine found in platforms. Maybe around the world the ticket machine is similar. It looks like this and you insert the ticket at the red part and the gates open. I dont understand how it can be so accurate:S

[url=http://img208.imageshack.us/my.php?image=41429847cp7.jpg][img]http://img208.imageshack.us/img208/7866/41429847cp7.th.jpg[/img][/url]

I'd like to return the question and ask why it couldn't be that accurate.

All the machine needs to know is the time, which anything can easily get.  And the ticket needs to store what time it is supposed to expire.


Hello
$ \binom{11}{9}\equal{}55$
Thank u.

More specifically,

$ \binom{11}{9}\equal{}\binom{11}{2}\equal{}\frac{11!}{2!9!}\equal{}\frac{11 \times 10}{2 \times 1}\equal{}\boxed{55}$.

[quote=wamofan]nice bump :coolspeak:[/quote]

he bumped a whole lot of threads

[quote=RohanQV]I'm practicing writing my solutions. :mad:[/quote]

ok

[quote=RohanQV]I'm practicing writing my solutions. :mad:[/quote]

11 choose 9 is a computation, not a solution. Similarly, (66666/22222)^4 is a computation, and not a solution. Also, notice that those threads were already answered several times. If you want to practice writing solutions, maybe find [b]actual[/b] problems.


What is the number of ordered pairs (m,n) for which mn  :ge: 0 and  m:^3: + n:^3: + 99mn = 33 :^3:

dunno, at least 34.....
it works if m+n = 33... right?
so... (0,33) (1,32) (2,31).... etc... (33,0)

what are the answers? there's probably more solutions

hm... 35... the 34 i just mentioned plus...
if m and n are both -33

dunno how to get that analytically....i just got out the 99 test and saw the answers were 2 , 3, 33, 35, 99

and i was at 34 so i just thought of different cases.... maybe its 99 tho, who knows


Gauss
number theory
number theory unsolved

I still need help with this one:
Use:
$ 2\cdot4\cdot...\cdot(p \minus{} 1)\equiv(2 \minus{} p)(4 \minus{} p)\cdot...\cdot(p \minus{} 1 \minus{} p)\equiv( \minus{} 1)^{\frac {(p \minus{} 1)}{2}}\cdot1\cdot3\cdot...\cdot(p \minus{} 2)$ mod $ p$
 and
 $ (p \minus{} 1)!\equiv \minus{} 1$ mod p [b][Wilson's Theorem][/b]
 to prove
 $ 1^2\cdot3^2\cdot5^2\cdot...\cdot(p \minus{} 2)^2\equiv( \minus{} 1)^{\frac {(p \minus{} 1)}{2}}$ mod $ p$

 [b]Relevant equations[/b]

 Gauss lemma
 wilson's theorem [$ (p \minus{} 1)!\equiv \minus{} 1$ mod$ p$]

 [b]The attempt at a solution[/b]
 need assistance

 Thanks
[color=red][b]SOLVED[/b][/color]

$ (p\minus{}1)! \equal{} (2 \cdot 4 \cdot 6 \cdot ... \cdot (p\minus{}1)) \cdot (1 \cdot 3 \cdot ... \cdot (p\minus{}2))$.  Can you figure it out from here?

So you've got these two products

$ E \equal{} 2 \cdot 4 \cdot 6 \cdot ... \cdot (p\minus{}1) \bmod p$
$ O \equal{} 1 \cdot 3 \cdot 5 \cdot ... \cdot (p\minus{}2) \bmod p$

and you know the following two things:

$ E \equiv (\minus{}1)^{ \frac{p\minus{}1}{2} } O \bmod p$
and
$ EO \equiv \minus{}1 \bmod p$ (Wilson's Theorem.)

What can you conclude about the value of $ O^2$?

[hide="If youre stuck, you need to..."] Multiply the first equation by $ O$ and apply the second equation. [/hide]


function ratio integration calculus derivative logarithms algebra

How do you find the antiderivative of a rational function or some ratio of functions using u-substitution?  I don't know what to let u=...

Here are some, if you could tell me how to do them, or help me do them, I would really appreciate it:

[tex]\displaystyle\int{\frac{x+1}{x^2+2x+19}}dx[/tex]

[tex]\displaystyle\int{\frac{e^x-e^{-x}}{e^x+e^{-x}}dx[/tex]
(I know that this is hyperbolic tangent...)

Thanks!

Notice that in both cases the numerator is some constant multiple of the derivative of the denominator. ie you have the integral of cf'(x)/f(x). Now this is looking a bit like the chain rule - what do you get when you differentiate ln(f(x))?

[tex]\frac{d}{dx}{\ln{f(x)}}=\frac{1}{f(x)}*f'(x)[/tex]

Therefore:

[tex]\displaystyle \int{\frac{1}{f(x)}*f'(x)}=\ln{f(x)}[/tex]

Okay, so I am going to guess that the first one is: [tex]\ln{x^2+2x+19}[/tex]

Checking: [tex]\frac{d}{dx}\ln{x^2+2x+19}=\frac{2x+2}{x^2+2x+19}[/tex]

This is off by a factor of two, so the correct solution is:

[tex]\frac{1}{2}\ln{x^2+2x+19}+C[/tex]

So this is sortof a guess-and-check method.  How do you do it using u-substitution?

Well, its exactly the same thing. In the first case, let u = x^2 + 2x + 19. Then du/dx = 2x+2, so we want the integral of dx = du/(2x+2), and we end up with the integral of 1/(2u) du.


rotation
geometry
geometric transformation

A regular pentagon is rotated $ d^\circ$ clockwise around its center until
it coincides with its original image. What is the smallest positive
measure of degrees in $ d$?

A pentagon has five-way rotational symmetry. The smallest possible degree measure for such a rotation would thus be $ \frac {360}{5} \equal{} \boxed{72}$.


MIT college Harvard AMC AIME USA(J)MO USAMO

This post may strike you as advertising. All, right it is advertising - but as someone with more than 3000 posts on AoPS, I think I've earned the right. The appearance of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=106561]this topic[/url] helps provide me with an opportunity, but I was thinking of doing this anyway.

The governing focus - the obsession, perhaps - of this College forum seems to be how to get into MIT (or Harvard, or Caltech). And for some of the readers, that's as it should be - but that's not true for everyone who reads this, or everyone who is a member of AoPS. What if you just don't have the kind of record to get into an extremely selective institution? What if your verbal SAT is 250 points lower than your math SAT? What if your GPA is only 3.4? You're very good at math contests (that's why you read AoPS) but that means you're one of the ~3000 California students who qualified for the AIME, not one of the 50 or 60 who made it to the USAMO - or maybe you just missed making it to the AIME. And you're not president of 14 clubs, and you don't spend all your free time on a charity you founded yourself. You know you can shine - at math, or science, or engineering - but you need someone to give you an opportunity.

I'm here to promote my own institution, California State University, Long Beach (also known as Long Beach State, also known as "The Beach"). I'm not the informational brochure, and I'm not going to deliver the official recruiting pitch - you can find those easily enough if you look for them, starting [url=http://www.outreach.csulb.edu/]here.[/url] I am primarily speaking to residents of California, and expect that you are considering one or more of the campuses of the University of California.

The key difference between us and any of the UC campuses: they are Ph.D. granting research universities, while we are an "urban comprehensive" university that grants masters degrees but not doctorates. That means that our faculty members have about twice the average teaching load of UC faculty members. This cuts down on the number of world-famous researchers we have (although we do produce research), but look at this from a student point of view. If you want to get involved in research (this is particularly important in the sciences), you, the undergraduate, may well get first claim on the professor's time since there are no doctoral students around. The heavier teaching load means that if you have a favorite professor, you may get a chance to take several classes from him or her - not impossible at a UC, but harder to arrange. Over the last four years, there's one student who took eight different classes from me - and that's not even counting the Putnam preparation course. I'll admit that that's an extreme case, but 3 or 4 classes from the same professor isn't that unusual. Every math class you would take here has an enrollment limit of no more than 34 students - including the calculus courses. Walk down the corridors of our office building - you'll see open doors and faculty you can talk to. And likely, faculty who know who you are.

Here at CSULB, we do offer the Putnam exam, and we have a 1-unit course to prepare students for the it. (I teach that myself.) Four out of the last five years, we've had a team rank between 110 and 130 (out of about 400 participating colleges) and three times in the last five years, we've placed an individual student among the top 500, once among the top 200. I'll claim that that's a better record than several of the UC campuses. (It's not Berkeley's record, of course.)

There are a lot of majors that don't require much math. Out of about 4500 CSULB freshmen, about 250 take "regular" Calculus I or higher as their first math course. About 100 of those are in the "or higher" category - mostly students with AP Calculus AB credit. So if you're on the serious math track, you're already in a group much smaller than the (rather big) campus. We wind up awarding 40-50 bachelors degrees a year in math, of whom about half are planning to be high school teachers. Something like two to four students a year go directly from our undergraduate program to Ph.D. programs, with a few more from our M.S. program. The existence of the M.S. program means that you can keep going if you run out of undergraduate math courses. (Quite a few of our M.S. students wind up as community college professors.)

What about admissions? Admission is strictly "by the numbers", so you can figure it out for yourself. We have an early deadline - for admission for Fall, 2007, the deadline to apply is the last day of November, 2006. You must have taken (and passed) the UC-CSU pattern of high school courses (your counselor probably knows that by heart.) As for the numbers: take your high school GPA and multiply by 800. Add your math SAT and your verbal SAT. That's your "index." (There's an alternate version that uses ACT scores.) Note that your GPA is extremely important in this index. Go ahead and use your "weighted" (for AP and honors courses) GPA. Assuming you're from California but not from Long Beach (or a few other nearby communities) your index needs to be at least a "floor" score that we adjust to adjust the size of our freshman class. Mostly the floor score has been about 3500 or 3600. For a few students (the stated criteria are "high school valedictorian or National Merit Semifinalist", with "valedictorian" being interpreted in such a way as to allow multiple students from the same school), we have a very attractive scholarship program called the President's Scholars Program.

I'm not directly involved in either admissions or in selecting President's Scholars, but if you're applying, I'd appreciate knowing - and in some cases I at least know whom to talk to, even though it's not my decision.

I'll be happy to answer any of your questions, either on this topic or by private message.

I never lived in CA and don't know anything about Professor Merryfield's university, but I hope I can still contribute something useful here.

My high school is a pretty good school. We won the Texas State UIL Academic Competition (overall) the past two years, we sent several people to USAMO, we have a good record at National Science Bowl, plenty of National Merit finalists. So, great. That's nice for the people who win the UIL medals, go to NSB and USAMO and get National Merit. But there are plenty of smart people who don't do that kind of stuff. I did all of that but USAMO, and I am at Princeton now. And there were like 10 people ahead of me in GPA. But not all of those people go to top universities. The admissions are very selective; if Harvard rejected 80% of valedictorians who applied (or whatever joml88 said in the other thread), what hopes does 15th in his/her class, only secretary of his/her graduating class, not president, etc. have?

You have the hope of going to a place that's not internationally renowned, not what always appears in movies, not Ivy League, but can still provide a great education. I have a friend who's at least as smart as I am but went to University of Texas instead of somewhere more "prestigious." But was that bad? From what I've heard, he has great relationships with his professors, perhaps even more opportunities than I do, friends, hobbies, whatever you want. Be aware that great opportunities may not always be bright and shiny objects begging you to grab them.

Here's another random tidbit: Andrew Wiles is a professor at Princeton.  :o  :o  :o But guess what? Kent Merryfield has helped me more with math than Andrew Wiles has, so far. In fact, Professor Merryfield helped me understand what I was doing a lot last semester when I was studying Fourier Analysis; Professor Wiles hasn't done anything for me besides stand near me a few times in the math common room. So don't get blinded by famous reputations. To cheesily quote Lord of the Rings, "All that is gold does not glitter."

Who was in top 200 at Long Beach?  I have a friend at Long Beach and I suspect it's him.  His name is Triet. He's got Outstanding baccalaureate award a few years ago and he's a graduate student at Long Beach now.

And by the way, how could you interest quite a few student into taking Putnam? SJSU scrarcely scrape up a team to compete in Putnam.

Triet got his B.S. in May, 2006, so he was ineligible for the 2006 Putnam - and it wasn't him. He's still here in our M.S. program this year. The top 200 finish was back in 2001 - someone from the Bay area who was in our "President's Scholars" program. The other two top 500 finishes, in 2004 and 2006, were by a veteran of the Southern California ARML team.

As for that interest in doing the Putnam - it's only 6-8 a year, and I get them because I actively recruit them.

Oh, and SJSU had [i]both[/i] a top 200 Putnam score and another top 500 this year - definitely the top performance there by any CSU, and better than every UC except Berkeley. I wouldn't complain too much!

[quote="Kent Merryfield"]

I'll be happy to answer any of your questions, either on this topic or by private message.[/quote]

Who proposes problems for Putnam?
Only proffesors or can the proposer be a graduate student?Must be american?

Hi,

I am currently a high school senior and a member of the SoCal ARML team which Professor Merryfield coaches. Don't take this as biased, but just knowing a professor like him could make the difference in your college career.

I have seriously considered Long Beach State in my college search and have applied there. I don't wish to sound like a hypocrite (because I'll not go to Long Beach), but I would have chosen Long Beach over all but 3 other schools which have accepted me. That means that I would have gone to Long Beach over any of the UC's, WashU in St. Louis, Cornell, and Columbia. The reason for that decision would have been the comparison of the marginal cost of the price tag to the marginal benefit in the quality of education (none of the schools have very good financial aid). My parents would obviously have had a problem with this because I think that parents care more about the name of the school than the kids, but I knew what there was at Long Beach and I liked it. You can't judge a school's math department by how the school is ranked overall.

What really drew me is the dedication of the professors in the math department to bringing out the potential in their math students, as can be evidenced from the Putnam record. I would consider Long Beach State's Putnam record even better than Berkley's considering the number of interested math majors and their average talent in math at Long Beach as opposed to Berkley. I know that a student who is truly passionate about math but who might not have great grades or accomplishments in high school would be able to use his/her talent to the fullest at Long Beach.

Also consider that you might be the best or near the best if you go to Long Beach. If you are a crappy math major at Princeton, you might end up working as a quant, but if you're near the best at Long Beach, you have good chances of ending up in grad school at UCLA or UC Berkley.

As you can guess, I strongly recommend Long Beach State as a school in which you can pursue a study of mathematics. It is important that Professor Merryfield is a very good professor, but what is more important is that he is dedicated to helping all people who are passionate about math and that he'll succeed in fully bringing out your talent in math.

In conclusion, I can understand that many of you are drawn to more famous names and if you have the money to pay for those schools, go ahead, but definitely pick Long Beach over any of the UC's (unless you have a really good scholarship at Berkley or something). At Long Beach, you'll have much more attention than you probably had in your high school, while at the UC's you'll end up struggling to get into classes of 400+ students.

Go Beach!

Djordjo

[quote="JBL"]This is totally the wrong place for that question, but hopefully a moderator will split these two posts off.  The Putnam is written by three professors on a rotating basis, so each year one professor is writing for the third time, one for the second and one for the first.[/quote]
Thank you .
I didn't know that (I'm from Europe  :) )

The suggested correct forum for specific questions about the Putnam exam is the [url=http://www.artofproblemsolving.com/Forum/index.php?f=80]Putnam Subforum[/url] in the College Playground Forum.

But thanks for visiting this forum, and feel free to come back here with any questions you have about college admission, college programs, and the like.

How many international student would be accepted?

International students have to pay the full tuition, which is quite a bit more than in-state students pay. A different body within the university handles international admissions than domestic admissions, but my impression is that generally students who are qualified and who have the means to pay do get admitted.

http://www.csulb.edu/divisions/aa/grad_undergrad/cie/international_admissions/index.html

Since it's that time of year again, I thought I'd bump this old thread.

[quote="Rewrite of fifth paragraph above"]Here at CSULB, we do offer the Putnam exam, and we have a 1-unit course to prepare students for the it. (I teach that myself.) We have frequently had a team rank between 110 and 130 (out of about 400 participating colleges), and improved that to a team rank of 78. Four times in the last six years, we've placed an individual student among the top 500, once among the top 200. I'll claim that that's a better record than several of the UC campuses. (It's not Berkeley's record, of course.) [/quote]

No, shbmsc90 did not wind up attending our institution. (He went to Princeton.) But do read his post.

Just to update that Putnam paragraph again: someone in the top 500 five years out of the last six, including twice in the top 200. And a team rank of 42 for 2007.

But 2/3 of the 2007 team is graduating. We need some new talent. So let me hear from you, 11th graders out there!


For a given integer $ n$, find the number of solutions to the Diophantine equation $ \frac {1}{a} \plus{} \frac {1}{b} \equal{} \frac {1}{n}$.

cf.
[url=http://en.wikipedia.org/wiki/Egyptian_fraction]Egyptian fraction[/url] - Wikipedia
[url=http://mathworld.wolfram.com/EgyptianFraction.html]Egyptian Fraction[/url] - MathWorld

This problem implies that $ (a\minus{}n)(b\minus{}n)\equal{}n^2$ if $ a,b\in{N}$ $ \Rightarrow$ answer is number of divisors of $ n^2$


One day there was a robery and the buglars took a car. There were four witnesses, but non of them remembered the whole number of the plate. Nevertheless, the first witness remembered that the number had four digits, the second witness remembered that the two first digits were equal, the third witness remembered that the last two digits were also equal, but different from the first two. The last witness remembered that the number was a perfect square. What was the number in the plate?

[quote="paladin8"]Hmm, the last witness must've been a mathematician :D[/quote]most likely! :D

7744 that was an easy one form you after a long time


Evaluate the limit as n goes to infinity of the following expression:
$ 1\plus{}(\minus{}1)^{n}$

Does the limit exist in the following case? I don't think this expression doesn't have a limit because it depends if n is even or odd. So do I put DNE?

Lim f(x) as x approaches a is c if for any positive (small) k you give me, I can give you an m (close to a) such that for any n between m and a, the positive difference between f(n) and c is less than the k you gave me.
In this case, suppose the limit was L, then if you give me a k of let's say 0.1, I can't give you an m because either 1 or -1 must be greater than 0.1 away from L, so for any m I give you, there will be a value of f at a point greater than m (between m and infinity) that's more than 0.1 away from m.
So the limit doesn't exist.

this sequence has 2 limit points ,so it has no limit .


group theory
abstract algebra
function
superior algebra
superior algebra solved

Let $G$ be a finite group and $H$ a subgroup of $G$ with index $(G:H)=n$.

a) Find the largest normal subgroup $K$ which lies completely in $H$.
b) Prove the the index $(G:K)$ divides $n!$.

Misha

I think it should be $K=\bigcap_{g\in G} (gHg^{-1})$. Let's show that it's the group we're looking for and that $G/K$ can be embedded in $S_n$.

First of all, it's clear that this is, indeed, a normal subgroup of $G$. That's because it's the intersection of some groups and $xKx^{-1}$ is $\bigcap_{g\in G} (xgH(xg)^{-1})=K$. Now take $K'$ a normal subgroup (normal in $G$) s.t. $K'\subseteq H$. We have $gK'g^{-1}=K'\Rightarrow K'\subseteq gHg^{-1},\ \forall g\in G$, and this finishes the proof that this is the group we're looking for.

Let's show now that $G/K$ can be embedded in $S_n$.

First consider the bijection $\alpha_g:(G/H)_{\ell}\to (G/H)_{\ell}$ defined by $\alpha_g(xH)=gxH$. We have $\alpha_g=\alpha_{g'}$ iff $gxH=g'xH,\ \forall x\in G$, which is equivalent to $g^{-1}g'\in K$ or, in other words, $g'K=gK$, so the function $gK\mapsto \alpha_g$ is an injective map from $G/K$ to $S_{|G:H|}$, Q.E.D.

Thanks grobber! The first part is definetely true. For the second part I don't know what you denote with $S_n$. Can you please explain it to me.

Misha

All right. Now I understand.  What a beautiful solution!!! I had a rather long one.

Misha

[quote="Misha123"]I don't know what you denote with $S_n$.
[/quote]

Probably the symetric group of set with n elements

I understood it already, but thanks Moubinool, it is really the symmetric group.

Misha

[quote="grobber"]
First consider the bijection $\alpha_g:(G/H)_{\ell}\to (G/H)_{\ell}$ defined by $\alpha_g(xH)=gxH$. We have $\alpha_g=\alpha_{g'}$ iff $gxH=g'xH,\ \forall x\in G$, which is equivalent to $g^{-1}g'\in K$ or, in other words, $g'K=gK$, so the function $gK\mapsto \alpha_g$ is an injective map from $G/K$ to $S_{|G:H|}$, Q.E.D.[/quote]

Please explain why is $gxH=g'xH,\ \forall x\in G$ equivalent to $g^{-1}g'\in K$. Thanks.

You could do that set-theoretically by rearranging terms. But instead, I will explain the motivation behind this argument:

A group action of a group $G$ on a set $M$ is normally defined by assigning to every $g \in G$ and every $m \in M$ an element $g(m) \in M$ such that certain rules ($e(m)=m$, $(gh)(m)=g(h(m))$) apply. This definition is easily checked to be equivalent to just give a group homomorphism $h: G \to S_M$ from $G$ to the group $S_M$ of bijections of $M$ to itself (i.e. automorphisms as a set); the relation between $h$ and the group action is given by $g(m) = (h(g))(m)$, and those rules correspond to $h$ being well-defined and a group homomorphism.

Now with the above in mind, if $H$ is a subgroup of $G$, you get a group action of $G$ on the cosets $G/H$ by $g(xH) =(gx)H$. Thus there is a corresponding homomorphism $h: G \to S_{G/H}$. Now it's kernel $K$, which especially is a normal subgroup, satisfies $G/K \cong \mathrm{im(h)}$ by the homomorphism theorem, thus $|G/K| = |\mathrm{im(h)}| \leq |S_{G/H}| = |G/H|!$.


Just today, the AAPT announced [url=http://www.aapt.org/physicsteam/quarterfinalists.cfm]the names of the 407 quarter-finalists[/url]. It's not terribly difficult and not a crowning achievement by any means, but congrats nonetheless.

I don't know what the cutoff score was, but looking at the score distribution on WebAssign (which represents a limited number of test-takers, anyway), I predict that it's probably higher than that of last year's Fnet=ma exam.

Good luck to all who will be taking the next exam!

I made it.  I know it's not much of an achievement, but I managed to not make it last year, so it is nice.

(Somehow, I got much better at physics once I stopped studying it . . .  :maybe: )

[quote="Rajiv"]Congrats!

You could try these as sample papers...just for some exercise...

Good Luck! :thumbup:[/quote]

It was nice of you to post that - those are some very interesting problems.

Now I have a question for you: how does the selection process in India work?  Is that only the first in a series of tests?

[quote="calc rulz"]I qualified. I was surprised to find that no one else from my high school qualified, when I figured that 2-5 others would. My physics teacher suggests that there's more competition this year?[/quote]
Yea dude, same here. I thought several people from my school would qualify, but it turned out to be only me and this other kid...okay actually one more kid.

[quote="genericme"]Now I have a question for you: how does the selection process in India work?  Is that only the first in a series of tests?[/quote]

The selection in India consists of:
1st stage:   NSEP about 300 students are selected from country
2nd stage:  INPhO about 30 students are selected.
3rd stage:   Orientation cum selection camps: here,the team is finally selected

INPhO is 2nd level exam(the question paper presented here is of INPhO)

For more details,please check [url=http://olympiads.hbcse.tifr.res.in/subjects/physics]Physics olympiad[/url]

i made it too- a nice surprise :)
when it says
"Administer the exam between February 23 and March 4."
does it mean that we can choose?
also, anyone know what the approximate cutoff for semifinalist was from last year?

Wait what format is the second round in? Multiple choice still?

[quote="serialk11r"]By the way, do you need calculus on it? If so I'm pretty dead :([/quote]
Yes, but nothing too advanced. Correct me if I'm wrong, but basic stuff from AP Calculus should be enough for quarterfinals. At least that's the way it appeared on last year's exam.

I made it!
And so did 3 others in my school. :o
My school is administering the test in a couple days. How about you guys?

[quote="serialk11r"]
Wait what topics are you talking about? The quarterfinal round test topics? What is it on even?[/quote]

I believe it's on mechanics & electrostatics (or at least that's what the quarterfinal round covered last year).

[quote="serialk11r"][quote="dasherm"]I'm not sure if it's required, but a little single-variable stuff does help A LOT.

And what's this about them changing the topics last minute?  It shouldn't be a problem for many people, but it certainly adds to my perception of the test as disorganized (another example: I am listed on the site as from Santa Monica HS, though I am from Michigan).[/quote]
Wait what topics are you talking about? The quarterfinal round test topics? What is it on even?[/quote]

My teacher forwarded to me part of the email he received from the AAPT:

"The quarterfinal exam will have questions drawn from mechanics, including, but not necessarily limited to, kinematics, dynamics, oscillating systems, momentum conservation, energy conservation, and gravitation; and questions drawn from electricity, including, but not necessarily limited to, Coulomb's law, electric fields, electric potential, Gauss's law, capacitors, charge conservation, and DC circuits.

The previous statements restricting the quarterfinal test to mechanics and electrostatics were unfortunately not updated in the 2009 description materials."

Hmmm...that seems pretty manageable (as in I only have to cram study about 1/2 the topics :(). Thanks!

serialk11r, someone asked if calculus was required for the physics olympiad.  What I said before applies to all three exams, but moreso to the second two (qf, and semis), as they include topisc that are more amenable to calculus.

Do you guys know how much do I have to get on the QF test to qualify for semifinal? I'm not sure about my F=ma round score,let's say it's 20.

[quote="hillarryous"]Last year, the cutoff was around 40 or so (Fnet=ma + quarterfinal score combined). This year, I have no idea. I can't tell you anything specific until the semifinalists are announced, but let's just say that this year's quarterfinal was significantly different from last year's.[/quote]
Thanks for the reply. I took it yesterday and I also think it's quite different. But isn't the QF scored out of 100?how can the combined cutoff be about 40?

I took [url=http://talk.collegeconfidential.com/high-school-life/457717-usapho-2.html]this post on CC[/url] to imply that someone who got $ 25 \pm 5\plus{}16.75 \approx 40$ made it to semis last year. I think we can safely assume that 40 to 50 was last year's cutoff.


(Me) Suppose that $ a,b,$ and $ c$ are digits, not all nine and not all zero, and the repeating decimal $ 0.\overline{abc}$ is expressed as a fraction in lowest terms.  How many different denominators are possible?

$ \text{(A)} \ 4 \qquad \text{(B)} \ 5 \qquad \text{(C)} \ 6 \qquad \text{(D)} \ 7 \qquad \text{(E)} \ 8$

[hide]Answer: $ (\text{D})$ $ 7$

$ 999\equal{}37\cdot3^3$ so $ 999$ has $ 8$ divisors. Since $ a,b,c$ are not all $ 9$, we have $ 7$ possible denominators.[/hide]

This looks familiar to some problem...

[hide="Solution"]$ 0.\overline{abc} \equal{} \frac {\overline{abc}}{999}$, where $ a,b,c$ are (nonrepeating) digits. The denominator can be $ 999$, or reduced into a divisor of $ 999$. Since $ 999 \equal{} 3^3\cdot37$, it has $ 4 \cdot2 \equal{} 8$ factors. However, one of these factors, $ 1$, cannot be a denominator, since $ a,b,c$ are not all nine. Thus, the answer is $ \boxed{\text{(D) }7}$.[/hide]
Edit: Darn, Illos beat me.  :P

Yes, it is essentially a previous AMC problem from 2002 [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1091297#1091297]here[/url]. It uses exactly same idea except I extended $ 0.\overline{ab}$ to $ 0.\overline{abc}$.


algebra
polynomial
vector
linear algebra
matrix
linear algebra unsolved

[quote="eugene"]Let $A$ be a linear transformation on an $n$-dimensional vector space over $\mathbb{C}$ with characteristic polynomial $(x-1)^n$. Prove that $A$ is similar to $A^{-1}$[/quote]Am I missing something obvious or is the problem trivial? I only get that $A = A^{-1} = I_n$

it's easy, man, but not that easy :). There are plenty of matrices different from $I_n$ with characteristic polynomial $(x-1)^n$. How did you get $A=I_n$ anyway? :)

Since we can Jordanize the matrix and work with each Jordan cell separately, it suffices to prove this in the particular case when $(x-1)^n$ is also the minimal polynomial, i.e. when $A$ has a single Jordan cell of size $n$ (and, in fact, we assume WLOG that $A$ is a Jordan matrix). Moreover, in this particular case, all we have to prove is that the minimal polynomial of $A^{-1}$ is also $(x-1)^n$.

Let $B$ be the matrix having $1$ on the diagonal just below the main one and $0$ everywhere else. We have $A=I_n+B$, and $A^{-1}=I_n-B+B^2-\ldots$ (of course, the sum stops when we reach exponent $n$, since $B^n=O_n$). We have $I_n-A^{-1}=B(I_n-B+\ldots)=BC$ ($C$ is the sum between the brackets). It's now clear that the minimal polynomial of $I_n-A^{-1}$ is $x^n$ (and thus that of $A^{-1}$ is $(x-1)^n$, which is what we wanted) since the minimal polynomial of $B$ is $x^n$ and $C$ is invertible.

I've realized the fact but i'm not sure:i'd like to hear you explanations if it is not difficult for you: why is it follows that the matrices are similar if they have the same minimal polynomials
Thank you,Grobber

It's because if both matrices have $(x-1)^n$ as minimal polynomial, there is only one possibility for the Jordan form: $1$ on the main diagonal and $1$ right below the main diagonal, and $0$ everywhere else. Since the matrices have the same Jordan form, they must be similar.

What was your solution?

Thanks,Grobber. Actually my solution was almost the same the main goal is to prove that the matrices $A$ and $A^{-1}$ have the same minimal polynomials, the only difference was in proof of this fact: I used more general assertion:
If for matrix $A$ it's characteristic and minimal polynomials coincide then the minimal polynomial of the matrix $A^{-1}$ is $p(0)^{-1}x^{n}p(x^{-1})$, where $p(x)$- characteristic polynomial of $A$.
In our case for $p(x)=(x-1)^n$ the following equality holds $p(x)=p(0)^{-1}x^{n}p(x^{-1})$.
I was just curious about you stating of matrices similarity based on the fact that they have the same minimal polynomials: it seemed to me that yo mean that there is some more general fact like: if minimal polynomials of two matrices coincide then they are similar which holds in more wide set of matrices in comparision with Jordan blocks and such a special form of their minimal polynomials
By the way, does there exist a fact like i mentioned?


Do you mean zeros at the end or also in the number?

(i.e., in het first case, the answer with $x=7!=5040$ would be one, and in the second case it would be two)

errrrr... impossible to know? i mean, for 0's at the end, it's rather trivial, but in the entire number :s
unless you ask Maple or some other program ^^

emmm , ok..
so how much zeros in the end ??

Can you explain me why is this the answer ?

Well, you want to know the number of 0's. Where can you get a 0 from? a "10", right? But  10 = 5*2. Thus, we have to look for the factors 2 and 5. But, since there are more 2's than 5's, we search for the factors 5 using [url=http://mathworld.wolfram.com/LegendresFormula.html]legendre's formula[/url]. ;)

Sorry if my explanation is not very clear, don't hesitate to ask more questions if you need a precision :)

Thanks, I understood it by myself too.
I have one more question.
The original question that I couldn't solve was:
$x=2005!$
How much zeros in $x$?
And there was 4 answers:
a)222
b)401
c)500
d) answers 1 and 3 are wrong
so , for solving this question I need to know whether $x$ have more than 500 zeros or not.

Thanks for your help, but can you explain to me why there are more than 500 zeros?
In other words why is there at least one zero that isn't in the end of the number (not in 500 zeros that we talked about before)?
 :blush:


Hello,

I'm doing a research and I'm using this TeXnicCenter but sometimes it just go mad. Like everything is OK and building is great but then an 8 appear in the table of contents or I just add a \footnote{} and then doesn't build anything and when I remove it everything return great.

Why is this and how to fix it?

Thanks in advance

What does the log say? Please give a [b]complete[/b] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157945]minimal example[/url] of where things go wrong.

Well that log is kind strange it says 5 errors and build normally then I build again the same thing and it says 8 errors and I didn't edit anything.

See what I mean by footnote
\paragraph {\mdseries{\indent TEXT }}
This works nice but when I add a footnote even though I've added a lot before already
\paragraph {\mdseries{\indent TE\footnote{note} XT }}
Not it doesn't build and errors increase to 20.

And just for your information when it builds OK there is a screen apprear and has the following text " "[DocOpen("%bm.pdf")][FileOpen("%bm.pdf")]" Cannot execute the command" and I need to open the PDF file manually.

I didn't get what the minimal thing is but I can try to understand it.

The point is to give a COMPLETE minimal document. That means a document that we can compile so it must include stuff before \begin{document} as well as text afterwards. Minimal means make it small so you take out all the irrelevant stuff that has no effect on the problem.

As the link I gave you says, merely doing this can often solve the problem. If you take something out and then everything is OK, that something must be causing the problem.

So post a document which is
1) [i]small[/i]: no irrelevant stuff
and
2) [i]complete[/i]; it has everythng in it so we can compile it immediately.

Do you mean like create a new file and delete extra stuff and try to remove things one by one until we know where is the problem?

Yes, you can do that if you want to find the source of the error yourself. Or post a small version for us to try out. If it's already quite small then there's no problem. Otherwise the relevant page will do. Replace any text you don't want the world to read with nonsense.

[quote="hasan4444"]Well that log is kind strange it says 5 errors and build normally then I build again the same thing and it says 8 errors and I didn't edit anything.[/quote]But what about telling us what the log says? How can we know what the errors are unless you tell us? Look for the [i]nameofyourfile[/i].log in Windows Explorer.

OK if you want the log file here is it I attached it.
I have no problem sending it to you but I don't want to annoy you with the whole file.

There's lots of errors. For example this is on line 94:
[quote]l.94 \paragraph
               { \mdseries{ \indent Scientists has created and built some ba...
I suspect you've forgotten a `}'[/quote]It thinks you have forgotten a } which may be possible. TexnicCenter checks this for you. If you click on a { and it is red then there's a missing }

As there are lots of other errors so best to post a minimal tex document.

OK

Thanks a lot for your posts and replies


My friend and I are two of the best chess players in our grade. We play all the time and he always does the same opening as white.  He does:

1. e4
2. Bc4
3. Either Nf3 of Qf3

Then he tries to move his knight or queen to f7.

Any suggestions on how to counter this?

lol he tries to 4 move checkmate you?
you can move one knight to H6, or you can just move the pawn at f7 to f8

if he plays Qf3

simply ignore him, and play Nf6

assuming that you have played 1..e5 of course
[img]http://images6.theimagehosting.com/opening1.th.JPG[/img]

if you played 1...c5, you could play e6
thus stopping any threat to f7

[img]http://images6.theimagehosting.com/opening3.th.JPG[/img]

however, if he plays Nf3 , there are several options for you
if you played 1...c5, you could continue with normal sicilian openings, either dragon or accelerated dragon, or paulsen, or simple najdorf with 2..d6
if you played 1..e5 you could try going in to the italian, if he has already played Bc4

[img]http://images6.theimagehosting.com/opening2.th.JPG[/img]

whoa how do u get those pictures? like do u play urself on a program and just hit Alt Print screen or is there another way?

but anyway, the best way not to fall for these kinds of trick is to make/learn a black opening (french helps the counter a lot because it blocks the bishop from assualting the f7 pawn)

my friend Bogdan Girmacea gives game analysis once a week if anyone cares to read it :

http://wcn.tentonhammer.com/modules.php?op=modload&name=News&file=index&catid=&topic=11

[quote="Lena"]my friend Bogdan Girmacea gives game analysis once a week if anyone cares to read it :

htthttp://wcn.tentonhammer.com/modules.php?op=modload&name=News&file=index&catid=&topic=11[/quote]

eddited you have htt twice at the beginning

[url]http://wcn.tentonhammer.com/modules.php?op=modload&name=News&file=index&catid=&topic=11[/url]

wait chances are that if you don't always lose in the 4 moves, you havefound a pretty good counter. :P


[i]You have $n$ items in a list.  You choose a number $p>0$, and so you cross out every $p-th$ item.  You keep going down the list (then restart from top) and keep doing the same until you have one item left.  You ignore the crossed out items when you encounter them again.  The last item left is expressed in terms of $n-d$, where $d\in\mathbb{N}\cup\{0\}$.  Find $d$ in terms of $n$ and/or $p$.[/i]

Masoud Zargar

do you know the answer to this? how did you come up with the problem? can you prove there is a relation?

Well, there is only one solution for sure, since it converges to only one.  I came up with this problem while playing a game with a friend of mine at the mall (exactly), and no, I do not have a solution to this, yet. There must be a relation, since if the number of $n$ changes, then the $d$ changes by a certain degree, falling in a different spot, thus there is a relation.:)

P (d=x) = 1/n

By saying "choose p>0" , i suppose random choosing. If it is not a random choosing process, then the method of choosing obviously affect what the last item is. If you say there is a relation between n and d ( and/or p) you are saying that if i tell you there are 10 items, then the last item MUST be some particular item in the list which is absurd(assuming random choosing).

yes, that is along the lines of what I was asking

it's funny how this is the first time I have seen an actual math problem inspired by something real :lol:

For the last two sentences, are you asking anything other than this: "The last term crossed out is $d$.  Compute $d$ in terms of $n$ and $p$." ?

I'd also like to clarify the process a little bit:  Suppose n = 5, p = 3.  So I have $1\,2\,3\,4\,5$.  Then I go 1-2-[b]3[/b] so $1\,2\not3\,4\,5$,4-5-[b]1[/b] so $\not1\,2\not3\,4\,5$, 2-4-[b]5[/b] so $\not1\,2\not3\,4\not5$, 2-4-[b]2[/b] so $\not1\not2\not3\,4\not5$ and thus $d = 4$.  Is that the process you want?

Yes, exactly.  Sorry if the problem was not clear enough. :oops:

so you essentially mean that you have 1 through n arranged in order around a circle and you count in order 'round the circle, crossing out every p-th term that hasn't been crossed out... that's what i'm understanding. and $d$ is then the number of elements larger than the last one left.

Some results, where the row is $n$, the column is $p$ and the entry is the last number to be deleted:

$\begin{tabular}{c|ccccccc} &1&2&3&4&5&6\\ \hline 1&1&&&&&&\\ 2&2&1&&&&&\\ 3&3&3&2&&&&\\ 4&4&1&1&2&&&\\ 5&5&3&4&1&2&&\\ 6&6&5&1&5&1&4&\\ 7&7&7&4&2&6&3&5\\ \end{tabular}$

I used mathematica with the following code:
[code]KillNext[L_,p_,i_] := Delete[L, 1 + Mod[p+i-1, Length[L]] ]

ListKill[n_,p_] := For[i=0;L=Table[j,{j,1,n}], Length[L]>1, i=Mod[p+i-1,Length[L]+1],  L=KillNext[L,p,i]; Print[L]][/code]

This is actually in the [url=http://www.research.att.com/~njas/sequences/A032434]Encyclopaedia of Integer Sequences (A032434)[/url], which has more information about it.

Can anyone find the relationship between the values in JBL's table?

By the way, [url=http://www.mathlinks.ro/Forum/topic-79552.html]this[/url] is the same problem, but asks for a complex thing. :)

in looking for a formula, we can use mods and stuff to count how many times you skip a crossed out number, use a recursive formula. There might be a geometric result if you arrange the terms in a circle.


The following problem is confusing me so any help would be appreciated!

Consider M_22 with the basis

{$\left( \begin{array}{cc}2 & 5 \\ 2 &-1 \end{array}\right)$, $\left( \begin{array}{cc}-2 &-2 \\ 0 & 1 \end{array}\right)$, $\left( \begin{array}{cc}-3 &-4 \\ 1 & 2 \end{array}\right)$, $\left( \begin{array}{cc}-1 &-3 \\ 0 & 1 \end{array}\right)$}

and consider R^4 with the basis ${(7, 0,-3, 1), (2,-1,-2, 2), (-2, 0, 1, 0), (5, 17, 0, 0)}$.  Find the matrix which corresponds to the linear transformation L with respect to these two bases.

It would be nice to know what $L$ is first.

Sorry, I forgot to include it.

L of
(a  b
 c  d)
=

(-10a - c + 2d, 5a + b - d, 2c + d, 9a + 17b + 2d)

Here is a "cookbook" answer.
To find the matrix of a linear mapping $L: V\to \mathbb R^{n}$ with respect to the basis $u_{1},\dots,u_{m}$ in $V$ and the basis $w_{1},\dots,w_{n}$ in $\mathbb R^{n}$, you should make the matrix $W$ whose columns are $w_{1},\dots,w_{n}$ and the matrix $X$ whose columns are $Lu_{1},\dots,Lu_{m}$ and compute the product $W^{-1}X$. In practice, it may be more convenient to make the long block matrix $[W|X]$ and then to bring it by the row transformations to the form $[I|Y]$ where $I$ is the identity matrix. Then $Y$ is what you need.
I asked for $L$ because I thought that there might be some nice trick that would allow you to see the answer at once but it seems that you'll have to undertake a routine computation involving two $4\times 4$ matrices. I only hope you are not required to do it by hand :lol:

Do you know by any chance how I could use Mathematica to find out the matrix?

It allows you to invert matrices and to multiply them, which is all you need though I do not remember the syntax right away (something like [i]Inverse[W].X[/i] :?). So, just compute $X$ by hand and make a short program consisting of two assignment operators and one operator like the above. Just remember that your vectors are columns and Mathematica requires you to define matrices row by row (like {{1,2},{3,4}} for $\left[\begin{matrix}1&2\\3&4\end{matrix}\right]$)


A car travels 40 kph for 20 kilometers, 50 kph for 25 kilometers, 60 kph for 45 minutes and 48 kph for 15 minutes. What is the average speed of the car, in kph?

[hide="Hint"]By definition, average velocity $ \bar{v}\equal{}\frac{d}{t}$, so just figure out the total distance traveled and the total time required.[/hide]

In order to solve this problem we need to know that,
$Distance = Rate * Time$
 $Average Speed = \frac{Total Distance}{Total Time}$

Knowing these two facts we can solve this problem!
$Total Distance = 20 kilometers + 25 kilometers + (60 * \frac{3}{4}) + (48 * \frac{1}{4})  = 102 kilometers$

$Total Time = \frac{1}{2} + \frac{1}{2} + \frac{3}{4} + \frac{1}{4} = 2 hours$

Plugging our numbers back into our equation we get : $\frac{102 kilometers}{2 hours} = 51 \frac{kilometers}{hours}$


Alcumus
videos
probability
expected value
Support

After some Alcumus Problems I am given, when I click to see a related video, the screen says 'You Cannot Access this Video' even though I'm being tested on those problems.  I really want to see the videos so I can learn to solve these problems but I cannot access these videos and the topics do not show up on my 'Report.'  Help!

Are you continuing to experience these problems? When I look at your report things appear fine.

There are indeed topics that do not show in your report, just like there are topics that could appear on your report with problems you haven't yet done.

It still is quizzing me on stuff that don't appear on my report or my videos...

Please report the problem on the solution page when you see such an issue.

It still isn't working, and seems to be only quizzing me on problems where I cannot see the related videos and cannot see on my report.  I have reported this many times and it still is not working.

Hello,

Your numerous reports were this morning, please allow us time to investigate.

That being said, I have looked at your report and it appears you are primarily getting questions in topics for which you are "Ready" to work on. Per design, you will also get some problems that you have already passed and some which you are not yet ready for, however, the majority of problems will be in your "ready" topics, which appears to be working properly.

The problems you have reported are from topics from which you are currently marked "not ready" for. You will get a handful of these problems. This helps Alcumus determine if maybe you are now ready to work on the problems.

Again, you are primarily getting problems from topics for which you are ready. For example, so far you have only gotten 6 Expected value problems, which you are not yet ready for and will not yet see the videos for, but 30-50 problems from topics in which you are ready. There are a couple topics in your ready section which have only a few problems available, so their count is less. You may also be encountering a problem in which you are out of problems in the "ready" category so more problems than usual from the "not ready" category are being shown.

Ok, I think I understand what you are saying.

Thanks!

I really want to see the Hockey Stick 1 video, but Alcumus continues to badger me with my non access to that video

So around how long will I have to get the correct problem until it lets me access the videos?  I'm getting tired of doing only problems which have videos I cannot access.


[quote="236factorial"]If greeting cards cost 2.50 for a box of 12, 1.25 for a packet of three, or 50 cents each, then the greatest number of cards that can be purchased for 14.75 is

a) 60
b) 61
c) 65
d) 66
e) 67[/quote]

[hide]C)65

so you use 12.50 for 60 cards, then you use 1.25 for three cards, and use 1.00 on two cards
60+3+2=65[/hide]

[hide]The best value for cards is 12 for $\$$2.50. With $\$$14.75, we can buy 5 full boxes(60 cards), with $\$$2.25 leftover. The next best value for cards is 3 for $\$$1.25, and $2.25 will buy one full packet (3 cards) with$1 leftover. This dollar can be used to purchase 2 cards at 50 cents each. 60+3+2=[b]65 cards[/b][/hide]

[quote="easyas3.14159..."][hide]The best value for cards is 12 for $2.50. With$14.75, we can buy 5 full boxes(60 cards), with $2.25 leftover. The next best value for cards is 3 for$1.25, and $2.25 will buy one full packet (3 cards) with$1 leftover. This dollar can be used to purchase 2 cards at 50 cents each. 60+3+2=[b]65 cards[/b][/hide][/quote]

careful with your dollar signs...

(by the way, the answers are correct so far).


[hide]He receives 45 cents in change.  The least amount of coins he can receive is $\boxed{\boxed{\boxed{3}}}$ becuase he can revceive a 1 quarter and 2 dimes.[/hide]

math92, check your answer.

[hide]
The change will be 45 cents! The minimum number of coins is 3 because you can have 1 quarter and 2 dimes.[/hide]

[quote="MCrawford"]Jose pays for his $\$$1.55 lunch with two $\$$1.00 bills and receives correct change.  What is the least number of coins he can receive?[/quote]

[hide=":) Answer"]

$\$1 +\$1 = \$2.$

$\$2 - \$1.55 = .45$

One quarter and two dimes =$\$0.45$

3 coins is the minimal.

[/hide]

[hide=" :D "]He will recieve $\$$0.45 in change, which is a quarter, and 2 dimes. The least amount of coins is 3.[/hide]

[hide="My Answer"]
[b]His cost is [/b]$\$1.55$. [b]He pays with two [/b]$\$1.00$ [b]bills. That means his change is [/b]$\$0.45$. [b]The smallest amount of change he could get is 3 coins (1 quarter and 2 dimes)[/b][/hide]

[quote="apusey"]Couldn't you just get a IOU so they wouldn't pay you in coins because they have none you would recive no coins and get bills later.[/quote]
i guess you could but thats not what the question is asking for. :P

Right
[hide]1 + 1 = 2
2 - 1.55 = .45
.25 + .1 + .1 = .45
the answer is [b]3[/b] because a quarter and two dimes[/hide]

[hide=" :D  "]
45=25+10+10, so the answer is 3[/hide]

 There. Happy, math92?

[quote="pascal_1623"][hide=" :mad: "]
45=25+10+10, so the answer is 3[/hide][/quote]
why are you so angry? :?


Two triangles $ ABC$and $ A'B'C'$ are positioned in the space such that the length of every side of $ \triangle ABC$ is not less than $ a$, and the length of every side of $ \triangle A'B'C'$ is not less than $ a'$. Prove that one can select a vertex of $ \triangle ABC$ and a vertex of $ \triangle A'B'C'$ so that the distance between the two selected vertices is not less than $ \sqrt {\frac {a^2 \plus{} a'^2}{3}}$.

let $ O$ be an arbitrary point in the space,and for any point $ X$ let $ x$ denote the vector from $ O$ to $ X$.let $ S = \{A,B,C\}$,$ \sigma_1 = a + b + c$ and $ \sigma_2 = a\cdot b + b\cdot c + c\cdot a$.define $ S',\sigma_1',\sigma_2'$ analogously in terms of $ A',B'$ and $ C'$.
given $ (P,P')\in S\times S'$.note that $ PP'^2 = |p|^2 - 2p\cdot p' + |p'|^2$.summing over all 9 possible pairs yields the total $ t$ with:

\begin{eqnarray*}t & = & \sum_{P\in S}3|p|^2+\sum_{P'\in S'}3|p'|^2-2\sigma_1\cdot\sigma_1' \\
& = & \sum_{P\in S}3|p|^2 + \sum_{P'\in S'}3|p'|^2 + \left(|\sigma_1 - \sigma_1'|^2 - |\sigma_1|^2 - |\sigma_1'|^2\right) \\
& \geq & \sum_{P\in S}3|p|^2 +  \sum_{P'\in S'}3|p'|^2 -  |\sigma_1|^2  -  |\sigma_1'|^2 \\
& = & \left(\sum_{P\in S}2|p|^2 - 2\sigma_2\right) +  \left(\sum_{P'\in S'}2|p'|^2 - 2\sigma_2'\right) \\
& = & |a - b|^2  +  |b-c|^2  +  |c - a|^2  +  |a' - b'|^2 + |b' - c'|^2 + |c' - a'|^2 \\
& = & AB^2  + BC^2  +  CA^2  +  A'B'^2  +  B'C'^2  +  C'A'^2 \\
& \geq & 3(a^2 + a'^2)\end{eqnarray*}

thus one of the nine distances is greater or equal to:

$ \sqrt {\frac t9}\geq\sqrt {\frac {a^2 + a'^2}3}$

as desired.the equality case is never reached.because $ A,B$ and $ C$ do not all lie on a line,the nine values $ PP'^2$ we sum to find $ t$ cannot all be equal to each other.