modular arithmetic
algebra
polynomial
number theory unsolved
number theory

prove that
$1^a+2^a+...+n^a$
is always divisible by $1+2+...+n$
for all integer $n$ and odd $a$.

one could consider the interesting recurrence relation $S_{\alpha +1}=n+\sum_{i=1}^{\alpha} S_i$

i dunno ...but i think it would be possible to solve it using this relation

[edit : corrected the relation]
[edited 2nd time : now its true]

You can prove that $n|2(1^m+2^m+\cdots+(n-1)^m)$ when $n$ is even and $n|1^m+2^m+\cdots+(n-1)^m$ when $n$ is odd $((n-1)^m+1^m\equiv 0\pmod n,$ so in the first case $RHS\equiv2\cdot(n/2)^m\equiv n\cdot(n/2)^{m-1}\equiv0\pmod n$ and in the second case $RHS\equiv 0\pmod n).$ Then $n(n+1)|2(1^m+2^m+\cdots+(n-1)^m)$.

@ spx2: Your reccurence isn't true. The correct one is \[ S_m=\frac{1}{m+1}\left((n+1)^{m+1}-1-\sum_{j=1}^{m-1}{m+1 \choose j}S_j\right).  \]

bodan,i corrected the relation:
the proof is somewhat like this :

$S_{\alpha +1}=\sum_{i=1}^{n} i^{\alpha +1}=1+\sum_{i=2}^{n} i^{\alpha +1}=$
$1+(n-1)+\sum_{i=2}^{n} (i^{\alpha +1} -1)$*
----
also $i^{\alpha +1} -1=\sum_{j=1}^{\alpha} i^j$ **

*,** => $S_{\alpha +1}= n+\sum_{i=2}^{n} \sum_{j=1}^{\alpha} i^j=$
$= n+ \sum_{j=1}^{\alpha} \sum_{i=2}^{n} i^j=n+\sum_{i=1}^{\alpha} S_i$

[this time its right]

How can \[ \left(\frac{n(n+1)}{2}\right)^2=n+\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}?  \] $S_n$ is a polynomial of $(n+1)^{\mathrm{th}}$ degree so the $LHS$ in your equation is of higher degree then the $RHS?$


The pages of a book are numbered 1 through $n$. When the page numbers of the book were added, one of the page numbers was mistakenly added twice, resulting in an incorrect sum of 1986. What was the number of the page that was added twice?

[hide]
the sum of first $n$ number is: $\frac{n(n+1)}{2}$
we want to find $n$ s.t.:
$\frac{n(n+1)}{2}<1986<\frac{(n+2)(n+1)}{2}$
which is close to square root of 2*1986.
notice:
$\frac{62*63}{2}<1986<\frac{64*63}{2}$
thus $n=62\Rightarrow 1986-\frac{62*63}{2}=\boxed{33}$
[/hide]

[hide="Answer"]The closest triangular number $<1986$ is $\frac{62(63)}{2}=1953$, so the extraneous number is $\boxed{33}$.[/hide]

Amir, the problem asks for the page that was added twice, which would be $1986 - \frac{62(63)}{2}$, not $62$  :D

heh, yeah  :)
i guess i'm too sleepy to think
edited

[hide]First, we need the largest $n$ such that $\frac 1 2 n(n+1)<1986$. The missing page would be $1986-\frac 1 2 n(n+1)$.

Some quick testing yields $n=62$ as the largest $n$, so our answer is $1986-\frac 1 2(62)(63)=\boxed{033}$. [/hide]

This is way to ez for the 6th problem of AIME. Do you think it is because there was less competition in the 1980's?

[hide=Solution]We must have $\frac{n(n+1)}{2}$ close to $1986$ but less than it. Doing some trial an error starting with $n=50$ and then using $n=60$ and $n=61, 62$ gives that $n=62$ is what we need. The answer is therefore $1986-\frac{62 \cdot 63}{2}=\boxed{33}$.[/hide]


Let a,b,c be given positive real numbers
Prove that
$(a+b+c)^9abc \geq (ab+bc+ca)^6$


How do you make a 100 mL buffer with pH = 4.5, using only oxalic acid and sodium hydroxide?

[quote="Carcul"]Oxalic acid is the simplest dicarboxilic acid, i. e., it is a diprotic acid of formula

$HO_{2}C-CO_{2}H$.

So start by finding the two pKa values of this acid and then ask yourself: what species must be present in solution in order for a reasonable buffer to be possible?[/quote]
Ok, so we have that $\text{p}k_{a_{1}}=1.23$, and $\text{p}k_{a_{2}}=4.19$.

But how do we get the oxalic acid to have $\text{p}k_{a_{2}}=4.19$ (which is closest to 4.5)?

By the way, since we aren't using water we should use the anhydrous molar mass for oxalic acid, right?

Essentially, you are going to want a mixture of oxalate and hydrogen oxalate, with more of the oxalate--if you notice, this makes sense because we are on the basic side of the pK2.  So the question becomes, how do we determine the quantitative amounts?

The answer: Henderson-Hasselbach.  You can find it easily on Wikipedia, so I won't reproduce it here.  It provides a way of doing problems of this type.  Note that the most difficult step here is deciding which species, and thus pK value, should be used.

...So we should have

$\frac{[\text{C}_{2}\text{O}_{4}]}{[\text{H}_{2}\text{C}_{2}\text{O}_{4}]}=2.042$.

But how is the sodium hydroxide used in this problem?

Two comments: 1) Not $H_{2}C_{2}O_{4}$--we need to have species separated by only one proton (why?).  Thus, you should have $HC_{2}O_{4}$.  I didn't check your calculation though.

2) The sodium hydroxide is used to take the pH to the appropriate spot--it does this by converting the oxalic acid to hydrogen oxalate, and then to oxalate.  Does this make sense?

Man, I'm bad at chemistry.  :blush:

[quote="gregx007"]Two comments: 1) Not $H_{2}C_{2}O_{4}$--we need to have species separated by only one proton (why?).  Thus, you should have $HC_{2}O_{4}$.  I didn't check your calculation though.[/quote]
Oh, right, I see.

Isn't the calculation just $10^{(4.5-4.19)}$?

[quote="gregx007"]2) The sodium hydroxide is used to take the pH to the appropriate spot--it does this by converting the oxalic acid to hydrogen oxalate, and then to oxalate.  Does this make sense?[/quote]
Are you saying that

$H_{2}C_{2}O_{4}+NaOH\longrightarrow NaHC_{2}O_{4}+H_{2}O$
$NaHC_{2}O_{4}+H_{2}O\longrightarrow Na^{+}+C_{2}O_{4}\,^{2-}+H_{3}O^{+}$?

[quote="rnwang2"]
Are you saying that

$H_{2}C_{2}O_{4}+NaOH\longrightarrow NaHC_{2}O_{4}+H_{2}O$
$NaHC_{2}O_{4}+H_{2}O\longrightarrow Na^{+}+C_{2}O_{4}\,^{2-}+H_{3}O^{+}$?[/quote]

Close.  I actually made a typo in my post, I should have put $HC_{2}O_{4}^{-}$.  Your second reaction should be $NaHC_{2}O_{4}+NaOH\longrightarrow 2Na^{+}+C_{2}O_{4}\,^{2-}+H_{2}O$ (you just have to put in enough sodium hydroxide to do this last step).

So 1 mol of sodium hydroxide will yield 1 mol of oxalate when it reacts with 1 mol of oxalic acid?

If so, since we need

$\frac{[\text{C}_{2}\text{O}_{4}]}{[\text{H}\text{C}_{2}\text{O}_{4}]}=2.042$,

we can have

$[\text{C}_{2}\text{O}_{4}]=0.170 \, M$ and
$[\text{H}\text{C}_{2}\text{O}_{4}]=0.0833\, M$.

...

How do we determine how much sodium hydroxide to add to the oxalic acid for a total volume of 100 mL?

...

Ok, I know the solution to this problem now.  :o

[quote="rnwang2"]So 1 mol of sodium hydroxide will yield 1 mol of oxalate when it reacts with 1 mol of oxalic acid?[/quote]

Hydrogen oxalate to be precise. Oxalic acid is diprotic.


trigonometry
inequalities
special factorizations

Now I feel really stupid and my brain isn't working today, but I simplified a big ugly problem to this trig inequality and I can't seem to prove it algebraically :(. The inequality is

$2>\tan\theta(1-\sin\theta)$ for $0<\theta<\frac\pi2$

(sinx)-(sinx)^2<2cosx

since 0<x<pi/2

(sinx)-(sinx)^2<2sqrt(1-(sinx)^2)
(sinx)^2(1-sinx)<4(1+(sinx))

(.5sinx)^2<1+(2sinx)/(1-sinx)
this obviosuly must be true whenever 0<sinx<1

[hide]$\tan \theta(1-\sin \theta)=\sin \theta \cdot \frac{1-\sin \theta}{\cos \theta}=\sin \theta \cdot \frac{1-\cos (\pi /2 -\theta)}{\sin (\pi /2 -\theta )}$.

Let $t=\tan \frac{\theta}{2}$. Then $0<t<1$ and the expression is $\frac{2t}{1+t^2}\cdot \frac{1-t}{1+t}=\frac{2t(1-t)}{(1+t)(1+t^2)}$.

By AM-GM or completing the square, $t(1-t)\leq \frac 14$, and clearly $(1+t)(1+t^2)\geq 1$ when $0<t<1$. Since both equalities can't happen at the same time, $\frac{2t(1-t)}{(1+t)(1+t^2)}<\frac 12 <2$.[/hide]


algebra
difference of squares
special factorizations
number theory

1. Prove for any prime p, (2p)C(p) - 2 is divisible by p^2.

2. Find all triples of positive integers (x,y,z) so
x^2 + y^2 +z^2 +2xy +2x(z-1) + 2y(z+1) is a perfect square.

3. Consider 2n distinct positive integers a_1,...,a_2n not exceeding n^2 (n>2). Prove that some three of the differences a_i - a_j are equal.

Part a is a little ambiguous, do you mean?

(2p)C(p-2)?  Or (2p)C(p)-2?

If it's the first one:[hide](2p)C(p-2)=(p-1)(p)*(2p)C(p-2) which p clearly divides.

If it's the second one, it just isn't true.  5, just as an example, doesn't work.[/hide]

argghh. sorry, i mistyped the program. it actually is (2p)C(p) - 2, but more importantly, i meant it to be divisible by p^2, NOT p. Sorry again.

For the first one :

[hide]

Lemma 1 :

C(2n,n) =  :Sigma: (k=0..n) C(n,k)C(n,n-k) = :Sigma: (k=0..n) C(n,k)

It can be proved by a counting argument (choosing n people from n men and n women).

Lemma 2 :

if p is prime the p | C(p,k) for 0<k<p

It's well known (p at the numerator can not be cancelled by a number in the denominator)

So each term in  :Sigma: (k=0..n) C(p,k)C(p,p-k) is divisible by p except for k=0 or n. C(p,0)=C(p,p)=1.

QED    [/hide]

Solution to #2:

[hide]expression=(x+y+z) :^2: +2(y-x)=N :^2:

so 2(y-x)=N :^2: -(x+y+z) :^2:

              =(N+x+y+z)(N-x-y-z)

A difference of squares muse either an odd number or divisible by 4. 2(y-x) is not odd, so it must be divisible by 4.

let y-x=k (k>0). x+y+z=2x+k+z. so x+y+z must be greater than k.

The factors a,b of 2(y-x) must both be even so

a=N+x+y+z and b=N-x-y-z a>b. suppose b>2, a<k. contradiction.

b=2, a=k, still contradiction.

so k=0. x=y. all solutions are (x,x,z) where x, z= any positive integer.[/hide]

MithsApprentice wrote:Magnara, Actually [hide]5[/hide] does work.

[hide](10C5=252) -2 = 250

250 is divisible by 5.[/hide]

Haha, funny story.  I worked it out to be 10*9*8*7*6*5, for whatever reason.

Question 3:[hide]6-3=3, 5-2=3, 4-1=3.  Every set of 2n integers will start at one, and go to at least 6 as long as n>2 (as the question stated).  QED.[/hide]  Once again, that seems a little too easy so I probably understood it wrong again.[/hide]

Magnara, I don't think you understand what a set is.  A set is an (unordered) collection of items.  In a set of integers, there is no requirement that the numbers be consecutive, or start of 1.

In Question 3, we are given 2n (different) integers between 1 and n2 (inclusive), but we don't know anything else about the numbers.

God, I can't believe I did that.  I had a long day and I was half asleep when I did that.  Sorry.

As an apology:

[hide]

I got a combinatorial number theory book, and I think it's having an effect on me.

For a set of 2n elements there are (2n)C(2) ways of choosing two elements to make a difference.  There are n :^2: -1 possible differences of the integers up to n :^2: .  So it suffices to show that 2(n :^2: -1)+1<(2n)C(2), by Pigeonhole.  This is easily shown as it reduces to 2n :^2: -1<2n :^2:-n.[/hide]Better?  Or did I still mess it up?


analytic geometry
graphing lines
slope
search
geometry
ratio
rotation

1) Find the distance between 2 parallel lines with slope 6 whose x-intercepts are 8 units apart.

2) Simplify $ \sqrt{14\plus{}8\sqrt{3}}$

For #1 just use the formula

$ \frac{|b_{1}\minus{}b_{2}|}{\sqrt{a^{2}\plus{}1}}$

Where $ b_{1}$ and $ b_{2}$ are initial values and $ a$ is the slope of either parallel line (since they're identical).

For #2 not 100% sure but try re-writing $ 14$ and $ 18\sqrt{3}$ purely as roots, then simplify.

[hide="hint for 2"]Since 3 is prime, the answer must be in the from $ a\plus{}b\sqrt{3}$. Try squaring and expanding what we just wrote and solve.[/hide]

[hide="#1"]would'nt it just be 8 because theyre parallel and their x-intercepts are 8 apart?[/hide]

[hide="#2"]you know that the answer has to be of the form $ a\plus{}b\sqrt{3}$ (do u see why?) square this expression to get $ (a\plus{}b\sqrt{3})\Longrightarrow a^{2}\plus{}2ab\sqrt{3}\plus{}3b^{2}$ now equate the appropriate expressions:
$ a^{2}\plus{}3b^{2}\equal{}14$
$ 2ab\sqrt{3}\equal{}8\sqrt{3}$
SOLVE![/hide]

I got the following algebra for #2, but then, I get stuck..

$ a^{2}\plus{}3b^{2}\equal{}14\\ ab\equal{}4\Rightarrow a\equal{}\frac{4}{b}\\ By Substitution:\\ (\frac{4}{b})^{2}\plus{}3b^{2}\equal{}14\\ \frac{16}{b^{2}}\plus{}3b^{2}\equal{}14\\ 16\plus{}3b^{4}\equal{}14b^{2}$

[hide]Let $ x\equal{}b^{2}$
Then,
$ 3x^{2}\minus{}14x\plus{}16\equal{}0$
Try that now.[/hide]

Prime Factorization -

If b1 and b2 are initial values, what are initial values?

They are initial values (or y intercepts) of your 2 respective parallel lines. Just take their absolute difference for the top part of the formula I gave.

Prime Factorization-

I don't understand where this equation came from.  How did you derive it?

The derivation requires a lot of algebra so I'll just explain the basic idea.

The shortest distance between two parallel lines is a line that intersects both parallel lines at 90 degrees. The slope of this line is the negative reciprocal of the slope of the parallel lines.

You then use a bunch of algebra to find the two intersection points between the 2 parallel lines and the new line, then you apply the shortest distance formula.

If that's way too confusing, search for my "work over the summer thread" and check the geometry section. I have a derivation there that should be a little clearer.

[hide="Number 1"]

$ y_{1}\equal{}6x$
$ y_{2}\equal{}6(x\plus{}8)$
$ y_{3}\equal{}\minus{}\frac{1}{6}x$
$ y_{\text{1 and 2}}$ are the parallel lines and $ y_{3}$ is the line that crosses the parallel lines.
Intersection of $ y_{2}$ and $ y_{3}$:
$ \minus{}\frac{1}{6}x\equal{}6x\plus{}48$
$ \minus{}x\equal{}36x\plus{}288$
$ \minus{}37x\equal{}288$
$ x\equal{}\minus{}\frac{288}{37}$
$ y\equal{}\minus{}\frac{1}{6}(\minus{}\frac{288}{37})$
$ y\equal{}\frac{48}{37}$
By the Pythagorean Theorem,
$ d\equal{}\sqrt{(\minus{}\frac{288}{37})^{2}\plus{}(\frac{48}{37})^{2}}$
$ d\equal{}\sqrt{\frac{2304}{37}}$
$ d\approx7.891$
[/hide]

I noticed that most of you assumed that $ \sqrt{14\plus{}8\sqrt3}\equal{}a\plus{}b\sqrt3$.  However, this is not always the case; we only know that $ \sqrt{14\plus{}8\sqrt3}\equal{}\sqrt{a}\plus{}\sqrt{b}$, if it can be simplified.

[hide="to simplify"]We know that $ a\plus{}b\equal{}14$ and $ 2\sqrt{ab}\equal{}8\sqrt3$, so thus, we have:

$ a\plus{}b\equal{}14$

$ ab\equal{}48\Rightarrow a\equal{}\frac{48}{b}$

Substituting, we have:

$ b\plus{}\frac{48}{b}\equal{}14$
$ b^{2}\minus{}14b\plus{}48\equal{}0$
$ (b\minus{}6)(b\minus{}8)\equal{}0$

Thus, $ b\equal{}8$ or $ b\equal{}6$ and we have that $ \sqrt{14\plus{}8\sqrt3}\equal{}\sqrt8\plus{}\sqrt6\equal{}2\sqrt2\plus{}\sqrt6$[/hide]

Here's another explaination for #1...

So there are two parallel lines with slope 6, and their x-intercepts are 8 apart. That means the distance between them is the distance between them on a line perpendicular to both of them (as I drew in the attached picture). So the triangle shown in the bottom of the picture is a right triangle, and its hypotenuse is the distance between the x-intercepts. Drawing the altitude to the hypotenuse of a right triangle makes two smaller right triangles similar to the big one. Since both of the parallel lines have slope 6, the ratio of d to e is 6 (see picture to see what segments these are). Also, by the Pythagorean Theorem, the ratio of b to d is $ \frac{\sqrt{37}}{6}$. Since the small right triangle is similar to the large one, the ratio of a to c is also $ \frac{\sqrt{37}}{6}$. a=8, so $ c\equal{}\frac{48}{\sqrt{37}}$

i have a much simpler way but i dont get the same answer. and i got this answer after a million tries so dont give me a different solution. just tell me whats wrong

[hide="my solution"]we have the slope of the line is 6. if we create a triangle such that the hypotenuese is on the line, one leg is the segment connected by the x intercepts, and the height is the end of the hypotenuese to the x intercept of the leftmost line. using the sequence thingy $ \sqrt2\sqrt5\sqrt{10}$ ect, we get the height is $ 8\sqrt{37}$. the area of that triangle is $ 32\sqrt{37}$. if we rotate it so that the hypotenuse of 48 is the base and the height and the distance between the lines is x, we get $ 48x\equal{}32\sqrt{37}$. then you solve for x. i get 8.11 instead of 7.89 of whatever you guys got[/hide]

[hide]I'm not sure that I understand what triangle you're drawing in the beginning... because it seems like the height of the triangle would be 48. Also, what are you rotating?[/hide]

As far as I know, it's a right triangle. The idea is that the shortest distance between, say, a point and a line is a perpendicular line. The reason why is because if you choose any other distance, it forms the hypotenuse of a right triangle, and the hypotenuse is always greater than the length of any of its legs. Hopefully that made some sense  :D

no i mean the triangle where the right angle is at the  x intercept of the rightmost line. ill redo my solution.

we have that is triangle ABC. we have C being the x intercept of the rightmost line, B being the x intercept of the leftmost line, and A such that $ AC\perp BC$. now we have AB=48, BC=8, and $ AC\equal{}8\sqrt{37}$. draw a perpendicular from C to AB. let that be D. the area of triangle ABC is $ 32\sqrt{37}$. we can also see ABC with base AB and height DC. DC is the distance between the 2 parallel lines. we have $ (AB)(DC)/2\equal{}32\sqrt{37}$ AB=48 so $ 24(DC)\equal{}32\sqrt{37}$ so $ DC\equal{}\frac{4}{3}\sqrt{37}$. so its 8.11. do you guys understand my solution now?

Thanks for rewriting it; I understand it now :) . And I also found the problem: in the very beginning, you switched the lengths of the hypotenuse and the vertical leg. I checked and your solution works when $ AB \equal{}8\sqrt{37}$ and $ AC\equal{}48$. It is a nice solution, btw.

oh *slaps head*. thanks for figuring that out though. :)


A)Find the number of ways of dividing mn things in n equal groups.

B)There are p points in a plane, no three of which are in the same straight line with exception of q, which are all in the same straight line; find the number (1) of straight lines, (2) of triangles which result fro joining them.

C)Find the sum of all the numbers greater than 10000 formed by using the digits 1,3,5,7,9, no digit being repeated in any number.  :?:

answer to first question is

{n/[(m-1)!]}

plz reply and tell whether it is correct or not............ :wink:

[quote="anurags92"]A)Find the number of ways of dividing mn things in n equal groups.
[/quote]
[\frac{(mn)!}{(n!) ^m}] \frac{1}{m!}
(My soln. was pretty long and needs to be condensed into this somehow. Found this at- http://cat.learnhub.com/lesson/page/4634-permutations-and-combinations-basics
[quote="anurags92"]
B)There are p points in a plane, no three of which are in the same straight line with exception of q, which are all in the same straight line; find the number (1) of straight lines, (2) of triangles which result fro joining them.
[/quote]
(1) pC2-qC2
(2) pC3-qC3
[quote="anurags92"]
C)Find the sum of all the numbers greater than 10000 formed by using the digits 1,3,5,7,9, no digit being repeated in any number.  :?:[/quote]
If you keep any one digit fixed at a particular place, you'll notice that the other digits can be arranged in 4! ways.
Thus for any place the sum is (1+3+5+7+8)X4!
= 576
now if you add up, it'll be 6 in units place, carry over 57 finally giving 639936

I hope I'm right?

B) (1) answer is pC2 - qC2 + 1.

     (2) pC3 - qC3.

What? No! I'm sure I wrote 9 and the correct answer. Someone editted my post to defame my credentials!   :bruce:

(  :rotfl: )


Let [x] denote the floor function. Find in terms of n:

[Sqrt(1)] + [Sqrt(2)] + [Sqrt(3)] + ... + [Sqrt(n^2 - 1)], where n=2, 3, ...

[quote]it will be 3(1)+5(2)+7(3)+...(2n-1)(n-1) [/quote]

How'd you get this?

you could look at the finite differences.

1, 4, 9, 16, 25, 36, ...

Maybe this some sort of wierd notation I know nothing about :) What exactly is 3(1) (for instance)?

[quote]Maybe this some sort of wierd notation I know nothing about  What exactly is 3(1) (for instance)?[/quote]

I think 3(1) just means 3*1=3. But I still don't understand the solution. Could someone be a bit more specific about the first step?

i get the same answer but here is a proof of it:

lattticccee  point proof (in a way)

here is a lemma which is quite easy to prove so i will kinda leave it out for now:

>> [sqrt x] counts the number of positive perfect squares less than or equal to x.

Using this fact we can get the sum quite nicely.

So this is how we do the sum.

[sqrt 1] + [sqrt 2] + ... + [sqrt n^2 - 1 ] counts the number of p.squares <= 1, 2, 3... n^2 - 1 and adds them all up.

However, we can just count the number of terms in which k^2 is counted for all k between 1 and +inf (but since the count is 0 for k = n we will just run it from 1 to n).

so the sum becomes  :Sigma: [k = 1 to n] (n^2 - k^2)

= n^3 - n(n+1)(2n+1)/6 = n(n-1)(4n+1)/6.

Sorry this is written so poorly counting arguments are often hard to write out... since they are more intuitive.


inequalities
function
inequalities proposed

$ a > 1,$ $ b > 1$ and $ c > 1.$ Find the minimal value of the expression:
\[ \frac {a^3}{a \plus{} b \minus{} 2} \plus{} \frac {b^3}{b \plus{} c \minus{} 2} \plus{} \frac {c^3}{c \plus{} a \minus{} 2}
\]

By Holders Inequality we get  $ \sum {\frac {a^3}{a \plus{} b \minus{} 2}}\geq\frac {(a \plus{} b \plus{} c)^3}{3[2(a \plus{} b \plus{} c) \minus{} 6]} \equal{} \frac {t^3}{3[2(t \minus{} 3)]}$ if $ t \equal{} a \plus{} b \plus{} c > 3$

 But function $ f(x) \equal{} \frac {x^3}{x \minus{} 3}$ is increasing on $ (3,\frac {9}{2}]$  and decreasing on $ [\frac {9}{2}, \plus{} \infty)$   so  $ f(t)\geq f(\frac {9}{2}) \equal{} \frac {243}{4}$

  Consequently the minimal value is $ \frac {81}{8}$ , equality occurs for $ a \equal{} b \equal{} c \equal{} \frac {3}{2}$


limit
floor function
integration
algebra
binomial theorem
real analysis
real analysis unsolved

What approach would I use to prove the limit:

\[ \lim_{x\rightarrow \infty }{\left( \frac{1}{x^{k\plus{}1}} \cdot \displaystyle\sum_{i\equal{}1}^{x}{i^{k}}\right)}\]

I asked this to myself after investigating Faulhaber's formulae when I was aked for the limit:

$ \lim_{n\rightarrow \infty }{\frac{1^{10}\plus{}2^{10}\plus{}...\plus{}n^{10}}{n^{11}}}$ which equals $ \frac{1}{11}$.

Hint: Apply Stoltz' theorem.

And binomial theorem afterward will be enough.

[hide]$ \lim \frac{x^k}{x^{k\plus{}1}\minus{}(x\minus{}1)^{k\plus{}1}}\equal{}\lim \frac{x^k}{x^{k\plus{}1}\minus{}x^{k\plus{}1}\plus{}(k\plus{}1)x^k\plus{}\ldots}\equal{}\frac{1}{k\plus{}1}$[/hide]

See the discussion [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=178381]here[/url] for an alternative to Faulhaber's formula.

I wonder why you concern about an explicit formula for the summation $ 1^k \plus{} \cdots \plus{} n^k$, if you merely want to evaluate given limit.

If $ x > 1$, then $ \frac{1}{(n\plus{}1)^k} \leq \frac{1}{x^k} \leq \frac{1}{n^k}$ where $ n \equal{} \lfloor x \rfloor$. So we have

$ \lim_{x\to\infty} \frac{1}{x^{k\plus{}1}} \sum_{1 \leq i \leq x} i^k \equal{} \lim_{n\to\infty} \frac{1}{n^{k\plus{}1}} \sum_{i\equal{}1}^{n} i^k \equal{} \int_{0}^{1} x^k \, dx \equal{} \frac{1}{k\plus{}1}.$

Hmm.  So for some reason I thought you needed this result to establish the antiderivative of $ x^k$ but I'd forgotten completely about the fundamental theorem of calculus.


Express $ 1 \frac{1}{4}\%$ as a common fraction.

[hide="Solution"] We can rewrite $ 1 \frac{1}{4}$ to $ \frac{5}{4}$. Since we are dealing with percents we can divide this by 100.

Thus we have $ \frac{5}{4 \cdot 100}\equal{}\frac{1}{80}$

$ \therefore$ Our desired answer is $ \frac{1}{80}$[/hide]


linear algebra
matrix
induction
linear algebra unsolved

Suppose $\mathbb F$ is an infinite field. Prove that there exist a $n\times \infty$ matrix which every $n$ columns of this matrix are linearly independent.

We can regard the columns as points of $P^{n-1}(\mathbb F)$, the projective $n-1$-space associated to $\mathbb F$. The condition we want them to satisfy is that no $n$ of them belong to the same projective hyperplane. We can prove that any $n\times m$ matrix satifying the required condition can be extebded to an $n\times(m+1)$ one with the same property. This follows from the fact that $P^{n-1}(\mathbb F)$ is not a finite union of hyperplanes, which we prove by induction:

Let $\mathcal H_i,\ i=\overline{1,t}$ be finitely many hyperplanes of $P^{n-1}(\mathbb F)$. Since the hyperplanes are in one-to-one correspondence with the points, and there are infinitely many of these, we can find a hyperplane $\mathcal H_0$ different from all the $\mathcal H_i$, and $\mathcal H_0$ is isomorphic to $P^{n-2}(\mathbb F)$. The $\mathcal H_i$ cut it along what we can regard as finitely many hyperplanes of $P^{n-2}(\mathbb F)$, which, by the induction hypothessi, do not cover it, and we are done with the induction step. It thus suffices to prove the assertion for $n=2$, and this is easy.


Humans are earth destroyers!!!

[url]http://www.poodwaddle.com/worldclock.swf[/url]

Brilliant!!! :) But how do they get the info???  :maybe:

i must say information cud be hypothetical orrr even incorrect---------bcause there r so many rurals in india itself that u can't say what's happening ----
unless u have some high resolution satellite---[i]that might be the answer[/i]

But even then, I doubt they have any access to the current wordwide data. I suspect it is just some model of what is going on based on the information of some official census for two consequitive years in a few countries and linear extrapolation. To bring it up as a "real time process" is a brilliant idea though. Unfortunately, I doubt it will stop anybody from giving birth to a second or third child or from buying a huge SUV :( Seems like the only two discouraging messages the average human understands are physical suffering and unaffordable price. Well, with this rate both may be much closer than we think...

Its Brilliant.. But it should be highly averaged


Given $ a, b, c \geq\ 0$ and $ ab\plus{}bc\plus{}ca\equal{}3$. Prove that: $ \frac{a}{(2a\plus{}b)^2}\plus{}\frac{b}{(2b\plus{}c)^2}\plus{}\frac{c}{(2c\plus{}a)^2} \geq\ \frac{1}{3}$  :)

Sorry, I was wrong in the way to creat it  :oops:
Another result:
Given $ a, b, c, k \geq\ 0$ and $ ab \plus{} bc \plus{} ca \equal{} 3$. Prove that:
$ \frac {1}{(a \plus{} kb)^3} \plus{} \frac {1}{(b \plus{} kc)^3} \plus{} \frac {1}{(c \plus{} ka)^3} \geq\ \frac {3}{(1 \plus{} k)^3}$

Indeed, it's weaker than the open question: (with the same condition)
$ \sum\ \frac {1}{(a \plus{} kb)^2} \geq\ \frac {3}{(1 \plus{} k)^2}$
But with this weaker, we can solve it by Am-Gm and CS  :)

[quote="nguoivn"]Given $ a, b, c, k \geq\ 0$ and $ ab \plus{} bc \plus{} ca \equal{} 3$. Prove that:
$ \frac {1}{(a \plus{} kb)^3} \plus{} \frac {1}{(b \plus{} kc)^3} \plus{} \frac {1}{(c \plus{} ka)^3} \geq\ \frac {3}{(1 \plus{} k)^3}$

we can solve it by Am-Gm and CS  :)[/quote]

Dear nguoivn,

I tried to apply AM-GM this way

$ LHS \geq \frac{3}{(a\plus{}kb)(b\plus{}kc)(c\plus{}ka)} \geq \frac {3}{(1 \plus{} k)^3}$

Can you please tell me if this is the correct way?

Thank you very much. :)

It's not trues, manlio. Example, with $ ab\plus{}bc\plus{}ca\equal{}3$, we have an old and strong result: $ (a\plus{}2b)(b\plus{}2c)(c\plus{}2a) \geq\ 27$  :wink:

[quote="nguoivn"]It's not trues, manlio. Example, with $ ab \plus{} bc \plus{} ca \equal{} 3$, we have an old and strong result: $ (a \plus{} 2b)(b \plus{} 2c)(c \plus{} 2a) \geq\ 27$  :wink:[/quote]
$ (3a\plus{}7b)(3b\plus{}7c)(3c\plus{}7a)\geq1000$ stronger ( with the same conditions ). :wink:

[quote="nguoivn"]Sorry, I was wrong in the way to creat it  :oops:
Another result:
Given $ a, b, c, k \geq\ 0$ and $ ab \plus{} bc \plus{} ca \equal{} 3$. Prove that:
$ \frac {1}{(a \plus{} kb)^3} \plus{} \frac {1}{(b \plus{} kc)^3} \plus{} \frac {1}{(c \plus{} ka)^3} \geq\ \frac {3}{(1 \plus{} k)^3}$

.. we can solve it by Am-Gm and CS  :)[/quote]

Dear nguoivn, when you have any time can you please post your solution or give a full hint to solve it. :oops:

Thank you very much.

[quote="nguoivn"]Another result:
Given $ a, b, c, k \geq\ 0$ and $ ab \plus{} bc \plus{} ca \equal{} 3$. Prove that:
$ \frac {1}{(a \plus{} kb)^3} \plus{} \frac {1}{(b \plus{} kc)^3} \plus{} \frac {1}{(c \plus{} ka)^3} \geq\ \frac {3}{(1 \plus{} k)^3}$
[/quote]
This inequality is [b]Problem 6.26[/b], page 287 in new book:

Vasile Cirtoaje, Vo Quoc Ba Can, Tran Quoc Anh, [url=http://canhang2007.wordpress.com/2009/12/08/new-book-inequalities-with-beautiful-solutions-v-cirtoaje-v-q-b-can-t-q-anh/][b][i]Inequalities with Beautiful Solutions[/i][/b][/url], GIL publishing house, 2009.


geometry
3D geometry
greatest common divisor
MATHCOUNTS

[size=75][color=darkblue]Moderator rcv's note:  Certain participants felt a need to take a simple request to an extreme.  This split / moved thread is the result.[/color][/size]

[quote="Fanatic"]There is say, a 5x3 square, made of 15 equal unit squares. A line starts on one vertex and goes to the vertex that is diagonal from the original vertex (Basically 15 unit squares in 5x3 with a diagonally line through it touching 2 vertexes). How many squares (not of any size) would the line touch?

Now, I remember that I originally learned this formula as, if the length and width are A and B, then the diagonal would pass A+B-1. However, I remember learning somewhere that if its 7x4 or 15x8 or a very tall or wide rectangle, then the formula is different (more complicated version of the first formula I stated)

What's the formula? And no this isn't like a trick question that you see on a mathcounts test, I'm just trying to re-learn a formula I SHOULD KNOW!!! Especially since I plan on nabbing the 4th spot on the Indiana team (not going to happen, I know :()  :blush:[/quote]

Yeah thats the worst.
For a cube:

$A+B+C-GCF(A,B)-GCF(A,C)-GCF(B,C)+GCF(A,B,C)$

how about a $A\text{x}B\text{x}C\text{x}D$ hyperprism?

following the pattern it would seem it's

$A+B+C+D-GCF(A,B)-GCF(B,C)-GCF(C,D)-GCF(D,A)-GCF(A,C)-GCF(B,D)+GCF(A,B,C)+GCF(B,C,D)+GCF(C,D,A)+GCF(D,A,B)-GCF(A,B,C,D)$

and i think it's right but it is very long.

The general formula is:

[b]sum of dimensions - sum of GCFs of every combination of 2 dimensions + sum of GCFs of every combination of 3 dimensions - sum of GCFs of every combination of 4 dimensions ... $\pm$ GCF of all dimensions[/b]

but that would be a pain to work out for more than 4 dimensions

or in (mostly) math terms:

for an n-dimensional rectangular hyperprism:

$\sum_{i=1}^{n} (-1)^{i+1}(\text{sum of GCFs of all combinations of }i\text{ dimensions})$

as an added bonus, it comes out really nicely as a cube with lengths x and n dimensions.

$x(\sum_{i=1}^n (-1)^{i+1}(n\text{ nCr }i))$

for any that don't know, $n\text{ nCr }i$ means the number of ways n things can be taken i at a time.

[quote="chess64"][quote="nat mc"]I hope you get the basic principle[/quote]

lol  :rotfl:  :rotfl:[/quote]

I thought the exact same thing after seeing that right after my post...

But since mathcounts DID go into higher dimensions (2005 masters round) it might be useful to extend these kinds of problems to higher dimensions as well.

lol, what I meant is that I hope that Fanatic gets the picture on how to do it


let $x_i$ s be positive integers such that
$x_1x_2x_3x_4...x_n \mid (x_1+k)(x_2+k)(x_3+k)...(x_n+k)$
for every $k \in N$,prove that
$\{x_1,x_2,x_3...,x_n \}=\{1,2,3...,n\}$

I don't think the statement is correct. Probably you want to say that $x_i$ are positive integers. In this case I've posted it some long time ago. I still remember Vess' s great idea to use the fact that there are long progressions of primes to prove a stronger result.

"the fact that there are long progressions of primes"

It has a name, you know :).


can any of you suggest some good and affordable bachelor's degree in astronomy and astrophysics or physics in distance learning and related open universities?
 :lol:
thanks

soumya post it in [url=http://www.mathlinks.ro/index.php?f=143]college[/url] section.

i can't see the college section anywhere....

or maybe i am too lazy to...

more help and spoonfeeding required from pardesi aka K.V.S.

isn't the link i gave working .if not get into mathlinks then try again


calculus
geometry
3D geometry
integration
calculus computations

1. A lobster tank in a restaurant is 4 ft long by 3 ft wide by 2 ft deep. Find the
   water force on the bottom and on each of the four sides?

2. A cone with height 12 ft and radius 4 ft, pointing down ward is filled with water to a depth of 9 ft. Find the work required to pump all the water out of the top??

Answer

1.The force on the bottom is $(P+Dgh)A$  (P is atm. pressure)

The force on the walls can be calculated using $\int^{H}_{0}(P+Dgy)l\,dy$

2.Calculate the total initial potential energy using $dU=DgyAdy,$ with $A=\pi r^{2}$ with $r=ytan\theta$

I haven't learn solving using this method, is there any other method


Ένα αρκετά γνωστό ανοικτό πρόβλημα λέει το εξής: Να βρεθεί το μικρότερο κ ώστε τα σημεία του επιπέδου να μπορούν να χρωματιστούν με κ χρώματα ώστε κάθε δυο σημεία που βρίσκονται σε απόσταση 1 να έχουν διαφορετικό χρώμα.

Άσκηση 1: Δείξτε ότι $ 4 \leqslant k \leqslant 7$. (Κατάλληλη και για μαθητές)

Άσκηση 2: Βρείτε καλύτερα φράγματα. (Ανοικτό πρόβλημα)

Πρόσφατα οι Shelah και Soifer παρατήρησαν ότι η απάντηση ίσως να επηρεάζεται από τα αξιώματα της θεωρίας συνόλων που χρησιμοποιούμε. Έδωσαν λοιπόν το ακόλουθο παράδειγμα.

Να βρεθεί το μικρότερο κ ώστε τα σημεία του επιπέδου να μπορούν να χρωματιστούν με κ χρώματα ώστε κάθε δυο σημεία $ (x,y)$ και $ (x',y')$ με $ y\minus{}y' \in \mathbb{Q}$ και $ x \minus{} x' \minus{} \sqrt{2} \in \mathbb{Q}$ να έχουν διαφορετικό χρώμα.

Άσκηση 3: Δείξτε ότι στην ZF με το αξίωμα της επιλογής αρκεί να χρησιμοποιήσουμε δύο χρώματα.

Άσκηση 4: Δείξτε ότι στην ZF + (Εξαρτημένη επιλογή) + (Κάθε σύνολο είναι Lebesque μετρήσιμο) χρειαζόμαστε υπεραριθμήσιμο αριθμό χρωμάτων.

Hint: Όλες οι απαντήσεις των ασκήσεων (πλην της άσκησης 2 που δεν έχω απάντηση  :lol:  ) είναι σχετικά σύντομες.

Στέλνω μια λύση για το 1

Το να δείξεις ότι 7 χρώματα φτάνουν για να βάψεις το επίπεδο είναι εύκολο.  Μπορείς να φτιάξεις το σχήμα που έχουν οι εξαγωνικές κηρύθρες των μελισσών ή αλλιώς οι ζώνες κάλυψης των κυψελών της κινητής τηλεφωνίας. Οι γειτονικές κυψέλες έχουν άλλο χρώμα. Οπότε πάμε σε 1 χρώμα για το κέντρο και 6 για τις γειτονικές κυψέλες και είμαστε ΟΚ.

Για το ότι 3 χρώματα δε φτάνουν επίσης γίνεται. Παρουσιάζω το σκεπτικό:

Έστω Α, Β και Γ τα 3 χρώματα. Έστω ότι βάφουμε με χρώμα Α ένα σημείο. Αυτό σημαίνει ότι ο κύκλος με κέντρο Α και ακτίνα 1 αποτελείται υποχρεωτικά μόνο από τα χρώματα Β και Γ.

Ας δούμε 2 σημεία του μοναδιαίου κύκλου που απέχουν 1. Αυτά έχουν υποχρεωτικά χρώματα Β το ένα και Γ το άλλο (ας λέμε και τα σημεία έτσι). Ας πάρουμε το συμμετρικό του Α ως προς το τμήμα ΒΓ και ας το πούμε Α’. Έχει προφανώς χρώμα Α και βρίσκεται πάνω στον κύκλο με κέντρο Α και ακτίνα $ \sqrt {3}$.
Ας φανταστούμε τώρα ένα από τα 2 σημεία του κύκλου (Α,$ \sqrt {3}$) που απέχει από το Α’ απόσταση 1 (δηλαδή μήκος χορδής και όχι τόξου) και ας το πούμε Χ.

Εδώ μπορεί να υπάρξει άτοπο αν αναλογιστούμε τι χρώμα θα έχει. Ας θεωρήσουμε τα σημεία το κύκλου (Α,1) που σχηματίζουν με το σημείο Χ ισόπλευρο τρίγωνο πλευράς 1. Τα σημεία αυτά είναι αναγκαστικά χρώματος Β και Γ αφού είναι στον εσωτερικό κύκλο. Άρα το Χ πρέπει να έχει χρώμα Α. Όμως απέχει 1 από το A’ χρώματος Α και άρα έχουμε άτοπο.

Αν θυμάμαι καλά είχα λύσει παλαιότερα αυτό το πρόβλημα γι αυτό και απέφυγα να στείλω αμέσως λύση (μην το χαλάσω στους άλλους).

Dem θα με ενδιέφερε να μάθω λίγα στοιχεία για το status του προβλήματος (τι πρόοδος έχει γίνει κλπ)

Γιώργο, πολύ ωραία. Μου αρέσει πολύ η περιγραφή σου για τα τρία χρώματα που δεν αρκούν. Στα περισσότερα βιβλία αν και δίνουν ουσιαστικά την ίδια απόδειξη δεν φαίνεται η ιδέα πίσω από την κατασκευή.

Για το πρόβλημα τώρα κανείς δεν έχει βελτιώσει τα φράγματα εκτός κάτω από ορισμένες συνθήκες. Π.χ. έχει δειχθεί ότι οποιοσδήποτε "πολυγωνικός χρωματισμός" (όπως π.χ. αυτός με τις κυρήθρες, χωρίς όμως την απαίτηση όλα τα πολύγωνα να είναι όμοια) θέλει τουλάχιστον έξι χρώματα. Έχουν επίσης μελετηθεί παρόμοια προβλήματα όπου απαγορεύεις περισσότερες αποστάσεις ή κοιτάζεις τι γίνεται στο χώρο ή γενικότερα στο $ \mathbb{R}^n$. Για το αρχικό πρόβλημα όμως το 4 και 7 παραμένουν. (Ο Ron Graham μάλιστα νομίζω πως προσφέρει και χρηματικό έπαθλο σε όποιον καταφέρει να τα βελτιώσει.)

Η ιστορία του προβλήματος καθώς και διάφορα αποτελέσματα σχετικά με αυτό ή παρόμοια προβλήματα περιγράφεται πολύ όμορφα στο δεύτερο μέρος του βιβλίου του Alexander Soifer "The Mathematical Coloring Book: Mathematics of Coloring and the Colorful Life of Its Creators". (Μπορείς να διαβάσεις αρκετά μεγάλο κομμάτι του βιβλίου από το googlebooks.)


quadratics
inequalities
algebra
polynomial
inequalities proposed

For $ x,y,z, A,B,C,a,b,c$ real numbers, the quadratic inequality

$ Ax^2 \plus{} By^2 \plus{} Cz^2 \geq ayz \plus{} bzx \plus{} cxy$

iff

1)$ A,B,C$ are positive number

2) $ 4AB > c^2$ and cyclics

3) $ 4ABC \geq Aa^2 \plus{} Bb^2 \plus{} Cc^2$

See Lemma 3 here: http://forumgeom.fau.edu/FG2008volume8/FG200823.pdf

[quote="manlio"]For $ x,y,z, A,B,C,a,b,c$ real numbers, the quadratic inequality

$ Ax^2 \plus{} By^2 \plus{} Cz^2 \geq ayz \plus{} bzx \plus{} cxy$

iff

1)$ A,B,C$ are positive number

2) $ 4AB  \ge  c^2$ and cyclics

3) $ 4ABC \geq Aa^2 \plus{} Bb^2 \plus{} Cc^2\plus{}abc$

See Lemma 3 here: http://forumgeom.fau.edu/FG2008volume8/FG200823.pdf[/quote]
The lemma is trivial but nice, my friend. Of course that 1) must be satisfied. Now, we need to have
\[ f(x)\equal{}Ax^2\minus{}(bz\plus{}cy)x\plus{}By^2\minus{}ayz\plus{}Cz^2 \ge 0\]
for all real $ x,y,z$.
Consider $ f$ as a quadratic polynomial of $ x$, since we need to have $ f(x) \ge 0$ for all $ x \in R$, therefore the discriminant of $ f$ must be $ \le 0$ for all $ y,z \in R$, which is equivalent to
\[ (bz\plus{}cy)^2 \le 4A(By^2\minus{}ayz\plus{}Cz^2)\]
\[ \Leftrightarrow (4AB\minus{}c^2)y^2 \minus{}2(2Aa\plus{}bc)yz\plus{}(4AC\minus{}b^2)z^2 \ge 0\]
From now, we need to have this inequality for all $ y,z \in R$, therefore $ 4AB\minus{}c^2 \ge 0, 4AC \minus{}b^2 \ge 0$, and the discrimanant must be $ \le 0$, that is
\[ (2Aa\plus{}bc)^2 \le (4AB\minus{}c^2)(4AC\minus{}b^2)\]
\[ \Leftrightarrow 4A^2a^2\plus{}4abcA \plus{}b^2c^2 \le 16A^2BC\minus{}4ABb^2\minus{}4ACc^2 \plus{}b^2c^2\]
\[ \Leftrightarrow Aa^2\plus{}Bb^2\plus{}Cc^2\plus{}abc \le 4ABC\]
Similarly, by consider the quadratic polynomial with $ y$, we will get $ 4BC \ge a^2$.
Our theorem is proved. :)


number theory
least common multiple
modular arithmetic

What is the smallest positive integer that is divisible by 5, but yields a remainder of 1 when divided by 2, 3 or 4?

As this is a simple problem, guess and check will do:

5, 10, 15, 20, [b]25[/b], so 25 is the answer.

whenever theres the word least you will most likely use LCM and here in this case is 24 and +1=25

 :showoff: :showoff: :showoff: :showoff:

[hide="Modular Arithmetic"]
We have $x\equiv 0\mod 5$ and $x-1\equiv 0\mod 2,3,4$.

So we have $x-1\equiv 0\mod 24$ and $x-1\equiv 4\mod 5$. We see now that $x=\boxed{25}$[/hide]


quadratics
trigonometry
geometry
angle bisector

Solve for $ x$: $ x^2 \equal{} \minus{}3 \plus{} 4i$.

$ \minus{}3\plus{}4i$ is a complex number like any other complex number.  Call it $ z$.  Any complex number can be written $ z\equal{}re^{i\theta}$, where $ r\equal{}\sqrt{a^{2}\plus{}b^{2}}$ and $ \theta\equal{}\tan^{\minus{}1}(b/a)$.  This yields $ r\equal{}5$ and $ \theta\approx 127^{\circ}$.  I would just call it $ \theta_{0}$ or something for notational convenience.

Now we have $ \sqrt{z}\equal{}\sqrt{r}e^{i\theta_{0}}\zeta^{k}$, where $ k\equal{}0,1$ and $ \zeta\equal{}\sqrt{1}$ (remember, 1 has two square roots!).

[hide="Simpler solution"]$ x\equal{}\sqrt{\minus{}3\plus{}4i}\equal{}a\plus{}bi$.

$ \minus{}3\plus{}4i\equal{}(a^2\minus{}b^2)\plus{}2abi$.

$ \minus{}3\equal{}a^2\minus{}b^2$ and $ 4i\equal{}2abi\implies ab\equal{}2\implies a\equal{}\frac{2}{b}$

$ \minus{}3\equal{}\frac{4}{b^2}\minus{}b^2\implies \minus{}3b^2\equal{}4\minus{}b^4\implies b^4\minus{}3b^2\minus{}4\equal{}0$

$ b^2\equal{}\frac{3\pm5}{2}\equal{}4,\ \minus{}1\implies b\equal{}2,\ i$

$ b\equal{}2\implies a\equal{}1$
$ b\equal{}i\implies a\equal{}\minus{}2i$

$ \boxed{x\equal{}1\plus{}2i,\ \minus{}1\minus{}2i}$.[/hide]

[hide="more complicated using geometry, but interesting"]
Use Demoivre's Theorem to find one of the roots, from which we easily get the other. Let $ O$ be the origin, and $ Z$ be one the roots. Let $ B$ be $ \minus{}3 \plus{} 4i$ and let $ C$ be $ 5 \plus{} 0i.$ Picture these on the  complex plane. $ Z$ lies on the angle bisector of $ \angle BOC$ and has magnitude $ \sqrt5$ by Demoivre's Theorem. Since $ \triangle BOC$ is isosceles, the angle bisector to $ BC$ is the same as the median to $ BC$ and thus lies on the line through $ 0 \plus{} 0i$ and $ 1 \plus{} 2i$ - the midpoint of $ BC$. Finding a point $ \sqrt5$ away from the origin on this line, we note that $ Z \equal{} 1 \plus{} 2i$ is a root, so the other root is the negative of that.
[/hide]


Find all positive integers $ x$ and $ y$ such that $ (x^2 \plus{} y)(y^2\plus{}x)$ is the fifth power of a prime.

Suppose that $ (x \plus{} y^2)(y^2 \plus{} x) \equal{} p^5$ for some prime $ p$.
Then each of $ x \plus{} y^2$ and $ y^2 \plus{} x$ must represent a power of $ p$ as well, and one of them must divide the another. Without loss of generality assume that $ x < y$ and thus $ x^2 \plus{} y \equal{} p^k$ (where $ k\equal{}1$ or $ 2$) divides $ x \plus{} y^2\equal{}p^{5\minus{}k}$. Then $ x^2 \plus{} y\equal{}p^k$ divides also $ x \plus{} y^2 \minus{} y(x^2 \plus{} y) \equal{} x \minus{} yx^2 \equal{} x(1 \minus{} xy).$ Note that $ x$ and $ 1 \minus{} xy$ are coprime and $ p^k > x^2 > x$ cannot divide $ x$. Therefore, $ p^k \equal{} x^2 \plus{} y$ divides $ 1 \minus{} xy$ and thus $ 1 \minus{} xy \plus{} x(x^2 \plus{} y) \equal{} 1 \plus{} x^3 \equal{} (1 \plus{} x)(1 \minus{} x \plus{} x^2)$ as well.

Note that $ p^k \equal{} x^2 \plus{} y$ is greater than both $ 1 \plus{} x$ and $ 1 \minus{} x \plus{} x^2$ and cannot divide either of then separately. Therefore, $ p$ divides $ (1 \plus{} x,1 \minus{} x \plus{} x^2)$ which in turn divides $ 3$. Hence, the only option we have is $ p \equal{} 3$ and $ x^2 \plus{} y \equal{} 3$ or $ x^2 \plus{} y \equal{} 3^2 \equal{} 9$ that yields the only solution: $ (x,y) \equal{} (2,5)$.

Therefore, there are two solutions $ (x,y) \equal{} (2,5)$ and $ (x,y) \equal{} (5,2)$.


[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=2091103030&t=228789]Look here for rules![/url]

Also, I have decided that it will take one "challenge day" for the pill to work, so if a person takes the pill during the first challenge, they will die during the second challenge.  More suspense. :D

Will need 12 to start.

I will play.

Also, don't you mean Poison Pill 2

Oh geez I forgot to finish mine... Never mind :P

Sign in

[quote="spaceguy524"]Also, don't you mean Poison Pill 2[/quote]

Nahh... Bugi forgot to finish the second one.

Also, please send in which pill (A or B) is the poisoned pill.

[quote="abcak"]only 10 spots remain now[/quote]

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me joins

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I guess I'll join...................

I'm back after a LONG time away from AOPS

Can anyone inform me about the conclusion of Legend of the Thirteenth moon?

LTM failed. :(
LTM 2 failed in after-sign up stage :mad:
LTM 3 has a small chance of being started and modded by me :D

I ...ummm.....will join? yeah, i think i will...

Me joins.
 :D
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12.
One spot left

[quote="ernie"]"Oh my god... my stomach is gurgling, this feels AWESOME!" jjfun1 screamed, audible from a mile away.  "Ughhh... my stomach doesn't feel too well," softly whispered algebra1337.  "Uh-oh, we must move both of you to the Facility, you both have SWINE FLU!"  jjfun1 exclaims, "YAY!"  She apparently didn't realize the dangers of swine flu.  Luckily, the pill which algebra1337 took saved his life.

Meanwhile, King! and EggyLv.999 were locked together, in an interrogation room, the size of an airplane bathroom.  "Oh wow, this stinks, being stuck in here with you, Eggy!"  EggyLv.999 coolly replied, "Well, if we wanna get out of here, we better take our pills!"  King!, disgusted, replied, "Sigh... I guess that's the only way out of here."  He swallowed his pill.  "EWWWWW!!! I PUT THE POOP ON THE WRONG PILL!" EggyLv.999 next took his pill, laughing and eventually choking on the pill.  After being brought to the emergency room, EggyLv.999 was fine.  King!, on the other hand, went straight to the bathroom, the same toilet cf249 used 2 days ago, and threw up poop in the toilet.
[/quote]

:rotfl: :rotfl:

[quote="ernie"]"Oh my god, this game is SO MUCH FUN!"  You could hear the peppiness of jjfun1 from a mile away.  However, jjfun1 had met her match, none other than the emo of our competition, algebra1337.  "Oh my god, yeah!  So like, you gotta take this pill, and then you gotta die!  Yeah!"  algebra1337 responded unemotionally, "Yeah I'll take it.  Life isn't worth living anyways."  jjfun1 went first, swallowing her pill with record intensity.  algebra1337 went next, swallowing the pill with no emotion.  "This was fun while it lasted, I wish I could do it again!" jjfun1 exclaimed with glee.  "That's what she said," Mewto55555 coolly replied.[/quote]

im not emo...am i? :|

Yes you are.  In fact, in your avatar, you're the green one. :P

Well, now, if he was the green one, he wouldn't be emo. Algebra1337 would have to be poking himself with the stick to be emo.

edit: *gun*

"Oh man, I gotta take two pills in a row!  This stinks!" EggyLv.999 whined like a 6 year old wanting that Spongebob Squarepants toy.  He was placed into the interrogation room.  "What am I here for?  I didn't do nothing!"  The whining would not stop.  And finally, Mewto55555 entered the room.  "Hmmm... if I did my math correctly, there is a 50.00000000000001% chance that I live.  But you could die before you take the second pill... right?"  EggyLv.999 felt so bored, he was stuck with a psychologist [and a really boring one too] for the rest of the night!  The next day [Day 5], EggyLv.999 opened his eyes, only to find himself still in the interrogation room.  "HEY, CAN'T I COME OUT YET?" ernie replied, "Well, you would come out, only we lost the key!"  EggyLv.999 exclaimed, "I wish I would die right now!"  His wish came true, a day later.  Finally, ernie found the key, Mewto55555 coolly took his pill, and walked away.  The limp body of EggyLv.999 still remains in the interrogation room to this day.

[b][u]The Dead:[/b][/u]
[color=red]Chompy, died Day 2, 02:00, of internal bleeding [challenged by spaceguy524][/color]
[color=olive]cf249, died Day 3, 18:30, of coughing up cow poop [challenged King!][/color]
[color=darkblue]jjfun1, died Day 4, 23:30, of hyperactivity with the swine flu [challenged algebra1337][/color]
[color=olive]EggyLv.999, died Day 6, 12:00, of boredom poisoning [challenged by King!][/color]

[b][u]The Alive:[/b][/u]
[color=red]spaceguy524[/color]
[color=olive]King![/color]
[color=orange][b]Mewto55555- Oh dear!  Now my percentage of living has gone down significantly![/b][/color]
[color=cyan]Bugi [/color]
[color=violet]cgyao15 [/color]
[color=darkblue]algebra1337[/color]

not if you just revived it.
I poison poison pill.

Well, Mewto55555 hasn't been on in a while, so I'll resume this Thursday.

You're right.  I, with my pwnage mod skills, modkill Mewto55555.  It is still the same day... only it's Bugi's turn.


Εστω Α1,Β1,C1 οι προβολες απο το σημειο Lemoine [b]K [/b] του τριγωνου ABC στις πλευρες BC,CA,AB αντιστοιχα.ΝΔΟ η διαμεσος ΑΜ του τριγωνου  ABC ειναι καθετη στην B1C1. :)

Θα δώσω ένα hint:
[hide]Δείτε πού βρίσκεται το περίκεντρο (και άρα και το ορθόκεντρο :wink: ) του τριγώνου $ AB_1C_1$[/hide]

[hide]εγω σκεφτομουν το εξης..
αφου $ KB_1C_1A$ εγγραψιμο, $ \angle(KB_1C_1) \equal{}\angle(KAC_1) \equal{} \angle(MAB_1)$
ομως $ \angle( C_1B_1A)$  συμπληρωματικη με $ \angle( KB_1C_1) \equal{} \angle(MAB_1)$  αρα $ \angle(AM,B_1C_1) \equal{} \frac {\pi}{2}$[/hide]


[hide]Square the first equation to get $\sin^2 x+\cos^2 x+2\sin x\cos x=1.44$.  Simplifying this gives $2\sin x\cos x=\sin 2x=.44$[/hide]

hmh...
I have a start, i believe
[hide]
Rewrite $sin(2x)$ as $2sin(x)cos(x)$
Then, you know that $sin(x) = 1.2-cos(x)$ and $cos(x) = 1.2 - sin(x)$
After filling in you have $sin(2x) = 2(1.2-sin(x))(1.2-cos(x))$
Which reduces to
$2(1.44-1.2sin(x)-1.2cos(x)+sin(x)cos(x)$
Then, if you go back to the first equation, you can tell that $1.2sin(x) + 1.2cos(x) = 1.2*1.2$
So, after filling that in you have
$2(1.44-1.44+sin(x)cos(x))$
so
$2(0+sin(x)cos(x))$
now this is where i'm stuck, i can figure out a way to get it from the first equation
[/hide]
if someone could complete this it would be great

oh duh, okay i know how to complete it
[hide]
Multiply the first equation by $sin(x)+cos(x)$ to get
$sin^2(x)+cos^2(x)+2sin(x)cos(x)=1.2sin(x)+1.2cos(x)$
now from earlier you know that $1.2sin(x)+1.2cos(x)=1.44$ and cos squared plus sin squared equals 1
it becomes
$1+2sin(x)cos(x)=1.44$
$sin(x)cos(x)=.22$
now fill that in and you get
$2(0+.22)$
$\boxed{.44}$
[/hide]
duh, i can't believe i didn't get that before

Very similar to the "$x+y=m$, $xy=n$, so $x^2+y^2=?$" questions.

[hide="Answer"]$(\sin x+\cos x)^2=1.44=\sin ^2x+\cos ^2x+2\sin x\cos x=1+\sin 2x$
$\sin 2x=\boxed{0.44}$[/hide]


Let a,b,c >0  such that  $a+b+c=3$.Prove that :
$\sqrt{ \frac{2a}{a+1}}+\sqrt{ \frac{2b}{b+1}}+\sqrt{ \frac{2c}{c+1}}\leq 3$

It follows directly from Jensen, not to mention the fact that it is in a wrong forum.

If $ x,y,z$ are positive numbers such that $ x \plus{} y \plus{} z \equal{} 3.$ then

$ \sum{\sqrt {\frac {2x}{x \plus{} 1}}}\leq\sum{\frac {2x}{\sqrt {(x \plus{} y)(x \plus{} z)}}}\leq\sum{\frac {8}{9 \minus{} yz}} \leq 3.$

The last one due to Sefket Arslangic be recorded as [b]Bài toán 4.90[/b] at page 177 of the following book:

Phạm Văn Thuận, Lê Vĩ, [url=http://www.mathlinks.ro/viewtopic.php?t=156796]Bất đẳng thức, Suy luận & Khám phá[/url], NHÀ XUẤT BẢN ĐẠI HỌC QUỐC GIA HÀ NỘI, 2007.

See also here : http://www.mathlinks.ro/viewtopic.php?t=32541

here is one more soln to it

cyclic (2a/a+1)^1/2 <=cyclic (a)^1/4          [am,gm]

now by power mean we have 1/4=m<1

hence

{cyclic (a)^1/4}/3<={(a+b+c)/(3)}^1/4 = 1 hence

{cyclic (a)^1/4}<=3

and we r done........ :wink:  :P


AMC
AIME
geometry
circumcircle
trigonometry
trig identities
Law of Cosines

Three parallel chords of a circle have lengths 2, 3, 4 and subtend angles x, y, x + y at the center (where x + y < 180 :^o:). Find cos x.

Have fun!

This isn't all that elegant, but its fast:

[hide]For simplicity, let p = x/2 and q = y/2. With r the radius, we know that sin(p)=1/r, sin(q)=3/(2r), and sin(p+q)=2/r. Equating each to 1/r, we have sin(p) = 2/3*sin(q) = 1/2*sin(p+q).

2sin(p) = sin(p+q) = sin(p)cos(q) + sin(q)cos(p).

Substituting, we get

2sin(p) = sin(p) :sqrt: (4-9sin^2(p))/2 + 3/2*sin(p)cos(p).

Let cos(p) = r.

4 = :sqrt:(9r^2 - 5) + 3r.

(4-3r)^2 = (9r^2 - 5).

r = cos(p) = 7/8.

We want cos(x), or cos(2p), which is 17/32.

[/hide]

[hide]Because the angles given work out nicely, we can move the three segments around in the circle to form a triangle with sidelengths 2, 3, and 4 that is inscribed in the circle. This is the key insight. From here, there are 2 ways (that I see) to proceed (call the triangle ABC with AB=2, BC=3, CA=4, and the center of the circle O).

Way 1 (the harder way) use R = abc/4[ABC] to find the circumradius of the triangle (after using heron to find [ABC]). Then, use law of cosines on triangle AOB, which has sidelengts 2, R, and R, to find the cosine of angle AOB.

Way 2 (the better way) We recognize that since AOB is a central angle and since ACB is inscribed in the circle, ACB = x/2. Use law of cosines on triangle ACB to find that cos(x/2)=(3^2+4^2-2^2)/(2*3*4) = 7/8. Then, use double angle formula to find cos(x) = 2 cos :^2: (x/2) - 1 = 17/32.[/hide]

Oops. It asked for cos x, not sin x. All fixed now :D

My bad...:-p


I have two problems that I can't do.. I'm self-studying extra for law of cosine but I'm really bad at the bearing ones.

For the first problem, I got to drawing but I'm not sure how to go further. For the second one, I'm not sure on wording neither.

1. A bridge is to be build across a small lake from B to C.  The bearing from B to C is S $41^\circ$ W. From a point A, 100 meters from B, the bearings to B and C are S $74^\circ$ E and S $28^\circ$ E, respectively. Find the distance from B to C. (Begin your drawing with point A)

[hide="um"]

I have drawing as PDF but I'm not sure if that's right.. If not, someone can post a right drawing.. I can't work that one because all I know is <CAB = 46 and AB = 100.. But to use law of cosine/sine, I need another angle.. That's where I'm struggling at.[/hide]

If my drawing is right, that's what I mean.

2. A pilot has just started on the glide path for landing at an airport where the length of the runway is 9000 feet. The angles of depression from the plane to the ends of the runway are $17.5^\circ$ and $18.8^\circ$.

a) Find the air distance the plane must travel until touching down on the near end of the runway.
b) Find the ground distance the plane must travel until touching down.
c) Find the altitude of the plane when the pilot begins the descent.

[hide="Questions"]

1. Why are there two angles of depression? What are two angles for?
2. What's the difference between the ground distance and air distance? Which one corresponds to which angles of depression?
3. I might have more questions but I have to know what the question means first.[/hide]

Thank for helping me out on these two problems. They're lablede as HARDEST problems on my book. ;) And these are the even ones so no solution/answer provided..

[quote="seungsoo"]I saved the drawing and solution for the first problem on Word. By the way, why does the problem involve the law of sines?[/quote]

i think he said law of cosine.

xxreddevilzxx, you can only solve the problems using the law of sines.

I attached the solution to the second problem, but NO ANSWERS. Do the work to get the answers.

argh!

I can't see your thing at alll!!!! :?

Can you make them PDF?


Some suggestions, because I don't know what you mean:

1) If you want to space things out nicely, you should try using [ code ] tags -- every character (including spaces) is given the same width and nothing gets compressed.
2) If you want to make it look really nice, use tex.  Use dollar signs: [code]$  TEX HERE  $[/code] where nbsp; is replaced by $.  To make the fraction $\frac{a}{b}$, use \frac{a}{b}.

Check [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php]this[/url] out for $\LaTeX$ help.

For the division sign put in a \div where you want it.

Btw,  I think your dollars signs are a little messed up in your post, Joel... to have TeX and a dollar sign in the same post you will have to use this: $ \$ for the dollar sign (the sequence of keys for that is dollar sign, space, slash, dollar sign)

Well enouhg with the LaTeX stuff.  Does anyone have an answer to the question?

Isn't this just dividing twice?  I don't see whats so special, unless I'm missing something.

[quote="e39m5"]Whats the remainder of:

$\frac{27}{4}\div9$

Thanks[/quote]I'm not sure what you're asking here. If you're dividing an integer $n$ by an integer $m$ you have $n=qm+r$ where $0\le{r}<m$. I don't see what your problem would have to do with this. $\frac{27}{4}\div9=\frac{27}{36}=\frac{3}{4}$. Dividing three by four you have $3=0\cdot4+3$, so the remainder is three. That's probably not what you're asking.

You may want to fix that up so it looks right :)

What will happen is that the thing being divided by 9 will actually be an integer (not what you had). Then remainder will make sense.

im working on fixing it, what is the correct answer when I divide that big fraction?

Theres no such thing. You only get a remainder when you divide an integer by another integer. I'm pretty sure that the index -18 should be an 18 instead.. perhaps either you are misreading the problem or it was just a typo in the question..

Haha, Nope, thats what the question said.  Pretty exact if I could get the spaces in.  I might believe there was a type if it wasn't a test to get into the school with the AAST math team :).

e39m5

I'd much rather accept a typo then to accept the question asking for something which doesn't exist :)

On the test it said with not when, I thought that was a typo, and corrected it here.  But for the actual math to have a type is too hard to believe.  Especially on an entrance test for that caliper school.

But as I said, if a test said something like 'what is the 3rd digit of the number 25', I'd be more likely to believe it was a typo, since the question doesn't make sense.

If you have a hard copy of the test.. perhaps you could scan a picture of the question in? I'd like to see it exactly as written..

nope, i dun have a hard copy, but what I said is exactly what it said except it said with instead when.  Going to have to trust me.  I just took remainder as "What's left after the division?"

[quote="e39m5"]Whats the remainder of:

$\frac{27}{4}\div9$

Thanks[/quote]
To me, this question looks something like: Find 1+(1+(1+(1+(1
That's right, no end parenthesis, something like that, or like: What is the positive root of $x^2+2x+1=0$
The question is kind of weird. :?

Well, if you're 100% you are correct in copying the problem.. then you'll have to accept theres a mistake in it.. theres nothing more to say about it.
If it was +18, it would make a much more typical and solvable problem.

[quote="AntonioMainenti"]I'm not sure what you're asking here. If you're dividing an integer $n$ by an integer $m$ you have $n=qm+r$ where $0\le{r}<m$. I don't see what your problem would have to do with this. $\frac{27}{4}\div9=\frac{27}{36}=\frac{3}{4}$. Dividing three by four you have $3=0\cdot4+3$, so the remainder is three. That's probably not what you're asking.[/quote]

I'm pretty sure AntonioMainenti is correct....
Change the problem to $\frac{27}{4}\times\frac{1}{9}$.
That is the same as $27\times\frac{1}{9}\times\frac{1}{4}$
Now the 27 and the 9 cancels out leaving $3\times\frac{1}{4}$
This is the same as $3\div4$ which leaves a remainder of 3. But then, I am not great at math so.....  it probably doesn't make sense......

Then theres also the idea:
 (No LaTeX Here)

1 / 9 has a remainder of 1.

2 / 9 has a remainder of 2.

3 / 9 has a remainder of 3.

etc etc etc.

Does that number / 9 have a remainder of itself?

Just another thought.

I wonder what the real answer is.  I thought you guys could whip one up in a jiffy.  I guess it stumped you as much as me.

e39m5

[quote="e39m5"]Well, it was a question on a test.

It asked:
${ What \hspace{.1cm} is \hspace{.1cm} the \hspace{.1cm} remainder \hspace{.1cm} when \hspace{.1cm} (8 \times 10^{-18} + 1^{18}) \hspace{.1cm} is \hspace{.1cm} divided \hspace{.1cm} by \hspace{.1cm} 9?}$

Eventually, I got it to come out to a FAIRLY simple division problem.

$\frac{125000000000000001}{125000000000000000}\div9$

So what is the remainder of that problem?  I tjust multiplied by $\frac{1}{9}$ and put that down as my answer.
Anybody have an idea?  Remainders seemed so simple until this problem.
e39m5[/quote]I'm sorry but your question has gone from odd to completely incomprehensible.

[quote="e39m5"]...I wonder what the real answer is.  I thought you guys could whip one up in a jiffy.  I guess it stumped you as much as me.

e39m5[/quote]
The real answer is: Your question is incomprehensible. Yes we did whip one up in a jiffy and you didn't. And no it did not stump us, only you got stumped because your question again is incomprehensible.

Thank you sir, this discussion is over.

Please review your basics by reading a book in elementary number theory like 'Fundamentals of Number Theory' by LeVeque, or 'An Introduction to Elementary Number Theory' by G.H.Hardy & Wright.

[quote="3X.lich"]The real answer is: Your question is incomprehensible. Yes we did whip one up in a jiffy and you didn't. And no it did not stump us, only you got stumped because your question again is incomprehensible.

Thank you sir, this discussion is over.

Please review your basics by reading a book in elementary number theory like 'Fundamentals of Number Theory' by LeVeque, or 'An Introduction to Elementary Number Theory' by G.H.Hardy & Wright.[/quote]Let's all calm down here.. no need to be condescending. And I don't think a text in number theory is what is needed here; he just needs to figure out what the problem is actually supposed to be. I suspect that you were in fact trying to show off by advising him to read number theory texts in order to review the "basics."

[quote="AntonioMainenti"] And I don't think a text in number theory is what is needed here; he just needs to figure out what the problem is actually supposed to be..[/quote]

I tried pointing out that the problem doesn't make sense a few times and what it should actually be, but it seems that e39m5 is convinced the problem must be correct and that we just can't work out the answer.. I think 3X.lich was just trying to reemphasise that fact.

Anyway, I don't think this needs any more discussion. The question doesn't make sense, lets move on.

[quote="e39m5"]So there is absolutely no chance the remainder could be what is left (remains) after you divide by nine?

The answer is there is no answer?

e39m5
Confused[/quote]The final answer is that you will only have a "remainder" when you're dividing one integer by another, which is not what you're asking.

\end of thread

Wow, you all are getting so upset and I think you are just over-analyzing.  One way we could define remainder is as follows:

For two positive real numbers, $a$ and $b$, the remainder when $a$ is divided by $b$ is the smallest non-negative real number $r$ such that we can write $a = mb + r$ for some integer $m$.  That's exactly what the remainder is for integers, only we replace "real number" with "integer" everywhere it appears.  In that case, the answer would just be $27/4$ (since this number is less than 9).  If faced with this question on an exam, that's what I would put down.  I don't think it's well phrased at all -- to me it sounds a little too much like it wants to know what the quotient is, rather than the remainder -- but that is a reasonably acceptable way to interpret the question.

I don't see anything wrong with the question. And am equally perplexed as to why it is thought by many in this forum that the question is incomprehensible. Please pardon my miniscule knowledge of mathematics :(

I do see a solution in sight and is making perfect sense to me unless proven otherwise by the experts in the forum ;) .

I  will propose with confidence that the answer is 'zero' remainder. 8 X 10^(-18) when divided by 9 gives a remainder of 8. While 1^(18) wil give us a remainder of 1. the sum of remainders in 9 i.e., divisible by 9. So, the remainder for the entire sum will be zero. I don't know if I 've unknowingly employed any of the Theorems in the above reasoning. I just used a simpler example to boost my case (5 + 4)/3 will leave a remainder of zero when you individually find remainders of 5 and 4 respectively and add them. The result is divisible by three. Also, if we look closely at the question the sum of the digits in 1.000....8 is 9.

Please correct me if I'm wrong.
Your inputs will highly be appreciated.

In order to make sense of what you said, you need to define remainder as you are using it.  What do you mean when you say that the remainder when $8\times10^{-18}$ is divided by 9 is 8?  Under what definition of remainder?


modular arithmetic
function
quadratics
number theory proposed
number theory

It is a conjecture by me. It shall be great if someone can prove/disprove it.

It is as follows:
We have a square number. We sum the digits of this, again and again until we obtain a single digit.
The digit so formed will not be $ 2,5,$ or $ 8$.

For eg. $ 7^2 \equal{} 49$
so $ 4 \plus{} 9 \equal{} 13$, and $ 1 \plus{} 3 \equal{} 4$. And so on.

PS: it shall be better if solution is purely based on elementary number theory.

$ n\equiv S(n)\pmod 9$, where $ S(n)$ is its sum of digits. That's the divisibility criterion for 9.

Now, as $ x^2\equiv 2\pmod 3$ is impossible,  $ 2,5,8$ can't be reached.
Also 3 and 6 can't be reached.

It's a well known fact that, if $ S(n)$ denotes the sum of the decimal digits of a positive integer $ n$, then $ S(n) \equiv n$ $ (mod 9)$.
Then, no matter how many times we apply the sum of digits function, the remainder upon division by $ 9$ does not change.

Since $ 2$, $ 5$ and $ 8$ are not quadratic residues mod $ 9$, we could never get those numbers at the end of the process. The same argument applies to numbers $ 3$ and $ 6$.


geometry
integration
calculus
combinatorics unsolved
combinatorics

Well,this problem is basiclly trivial but almost all of the contestants failed to write their solution in a right way.
Problem.
We have a really smart Robot(Lets name it Mathlinker :D )  standing on a chessboard.Mathlinker moves on the borders of the cells of the chessboard.
Mathlinker has two memory with the names of $A,B$.
When mathlinker moves up a cell, The value of $A$ will be added by $1$ and when it moves down a cell,The value of $A$ will be decreased by $1$.
When Mathlinker moves a cell to right,the value of $A$ will be added to $B$ and
When Mathlinker moves a cell to left ,the  value of the $A$ will be decreased from $B$.
We know that at the beggining $A=B=0$ and Mathlinker had returned to the beggining point after its journey .
Prove that the Value of $B$ ,after returning to the beggining point will be the equal to the area that mathlinkers has travelled Around it.

Area of each cell is equal to $1$.

It is not surprising that nobody wrote it correctly, because one has to consider the case when the path intersects itself, and squares that are enclosed more than once.

$(A,B) = (y, \int y dx)$, at any step during the motion, so that $B$ records the oriented area swept out relative to the starting point.   To solve this problem without calculus, one can show that for any square, $B$ records the number of times it is circled with positive orientation (anti-clockwise), minus the number of times it is circled with negative orientation (clockwise).

Exactly ,fleeting guess,thats why its hard to write it.
 :)


Expand and collect terms and sub in $abc=1$ where we can. It gives
$3+2(a+b+c)+(a^{2}+b^{2}+c^{2})+(a^{2}b+ab^{2}+a^{2}c+ac^{2}+b^{2}c+bc^{2})+(a^{2}b^{2}+a^{2}c^{2}+b^{2}c^{2}) \ge 7(ab+ac+bc)$
Now add the inequalities
$a+ab^{2}\ge 2ab$ and symmetric forms
$1+a^{2}b^{2}\ge 2ab$ and symmetric forms
$\frac{a^{2}+b^{2}}{2}\ge ab$ and symmetric forms
We get the desired inequality.

What about proving : $LHS \geq 3(a+b+c)^{2}$.

davron

[quote="Davron"]What about proving : $ LHS \geq 3(a \plus{} b \plus{} c)^{2}$.

davron[/quote]
It's wrong. :(  Try $ a\equal{}3,$ $ b\equal{}\frac{1}{2}$ and $ c\equal{}\frac{2}{3}.$ :wink:


-a posistive integer n is a yoyo if the difference between any two consecutive digits is at least 7. compute the number of 8-digit yoyos.

-there are 4 black balls and 1 whtie ball in a hat. a turn consists of picking a ball from the hat and replacing it with one of the opposite color. computer the probablility that after a sequence of turns, there are 5 black balls in the hat before there are 5 white balls.

Hmm... I've used a creative method for #2. Tell me if I'm right.

[hide]
Presently, the list is BBBBW. Let the probability of winning this be $ x$. Clearly, if we replace W with B, we win automatically, so that adds $ \frac 15$ automatically to our probability.

If we replace B with W (with probability $ \frac 45$), then the list becomes BBBWW. Let the probability of winning this be $ p$. Then $ x \equal{} \frac 15 \plus{} \frac {4p}5$.

Now let's focus on winning with this configuration first (forget the above for a moment). If we replace a W with a B (with probability $ \frac 25$), then the probability of winning this is $ x$ (because we're back at our starting position). If we replace a B with a W (with probability $ \frac 35$), the list becomes BBWWW, and the probability of winning this will be $ 1 \minus{} p$ because of symmetry. Thus $ p \equal{} \frac {2x}5 \plus{} \frac {3(1 \minus{} p)}5$. Now substituting our value of $ x$ from the previous equation yields:

$ p \equal{} \frac {2(\frac 15 \plus{} \frac {4p}5)}5 \plus{} \frac {3(1 \minus{} p)}5$

Simplifying, we get
$ p \equal{} \frac {17}{25} \minus{} \frac {7p}{25}$, and
$ \frac {32p}{25} \equal{} \frac {17}{25}$, and $ p \equal{} \frac {17}{32}$. Now substituting this value of $ p$ for the equation for $ x$ yields:

$ x \equal{} \frac 15 \plus{} \frac {17}{40} \equal{} \boxed {\frac 58}$.

PLEASEBERIGHTPLEASEBERIGHT
[/hide]

[quote="Yongyi781"]Hmm... I've used a creative method for #2. Tell me if I'm right.

[hide]
Presently, the list is BBBBW. Let the probability of winning this be $ x$. Clearly, if we replace W with B, we win automatically, so that adds $ \frac 12$ automatically to our probability.

If we replace B with W (with probability $ \frac 12$), then the list becomes BBBWW. Let the probability of winning this be $ p$. Then $ x \equal{} \frac 12 \plus{} \frac p2$.

Now let's focus on winning with this configuration first (forget the above for a moment). If we replace a W with a B (with probability $ \frac 12$), then the probability of winning this is $ x$ (because we're back at our starting position). If we replace a B with a W (with probability $ \frac 12$), the list becomes BBWWW, and the probability of winning this will be $ 1 \minus{} p$ because of symmetry. Thus $ p \equal{} \frac x2 \plus{} \frac {1 \minus{} p}2$. Now substituting our value of $ x$ from the previous equation yields:

$ p \equal{} \frac {\frac 12 \plus{} \frac p2}2 \plus{} \frac {1 \minus{} p}2$

Simplifying, we get
$ p \equal{} \frac 34 \minus{} \frac p4$, and
$ \frac {5p}4 \equal{} \frac 34$, and $ p \equal{} \frac 35$. Now substituting this value of $ p$ for the equation for $ x$ yields:

$ x \equal{} \frac 12 \plus{} \frac 3{10} \equal{} \boxed {\frac 45}$.

PLEASEBERIGHTPLEASEBERIGHT
[/hide][/quote]
But when there are 4 black balls and 1 white ball in the hat, you only pick the white ball with probability $ \frac{1}{5}$.

Oh yeah, I forgot about that. I'll fix that now.


inequalities
LaTeX
function
algebra
domain
inequalities unsolved

I am sorry but I am just a TeXacker and do not now LaTeX at all so I attached the question in a PDF file.

Please do not give your solutions just comment on the approach that you would take to prove this inequality.

Fix $ x_n^2$ as the highest, expand the square, and move everything except $ (n\minus{}1)x_n^2$ to the R.H.S. Since $ x_n^2$ is the highest square, all other $ x_i^2$ are in $ [\minus{}x_n^2,x_n^2]$ so each $ x_i$ is in $ [\minus{}|x_n|,|x_n|]$

Then consider the R.H.S.= $ \minus{}2x_1(x_2\plus{}x_3\plus{}....\plus{}x_n)$+Blah. We will maximize it, and show that the maximum is smaller than $ (n\minus{}1)x_n^2$.

Now, this is a linear function in $ x_1$ so it will take it's maximum in the domain of $ [\minus{}|x_n|,|x_n|]$ at one of the endpoints. Similarly for the rest of the $ x_i$. Therefore, we may restrict our attention to when each $ x_i\equal{}\pm x_n$ with $ i \ne n$

Finish from here :P


group theory
abstract algebra
superior algebra
superior algebra theorems

let G' a commutator subgroup periodic and abelian
let x a element of G'
let G a group soluble finitely generated
then x^G (normal closure) is an abelian subgroup of finite exponent
WHY? HOW TO DEMONSTRATE?)

By definition $ x$ has finite order and because of $ x^{G}\subseteq G'$ the closure of $ x$ in $ G$ must be abelian. Since $ x$ is contained in the abelian, normal subgroup $ G'$ each $ y$ in $ x^{G}$ can be written as $ y=\prod_{g\in I}(x^{g})^{n_{g}}$ with $ I$ a subset of the finite set of generators of $ G/G'$, in particular $ y^{ord(x)}=1$. Therefore $ Exp(x^{G})=ord(x)$ is finite.


Two circles meet at $A$ and $B$. A line through $A$ meets the circles again at $C$ and $D$. $M$, $N$ are the midpoints of the arcs $BC$, $BD$ which do not contain $A$. $K$ is the midpoint of the segment $CD$. Prove that $\angle MKN = 90^\circ$.

Here is my solution:
Let $X$ be a symetric piont to $N$ by $K$. We have $CX = DN = BN$, $CM = BM$ and $\angle MCX = \angle MCA+\angle XCK = 180\circ-\angle MBA+\angle KDN = 180\circ-\angle MBA+180\circ-\angle NBA = 360\circ-\angle MBN$
, so $\bigtriangleup MBN \equiv \bigtriangleup MCX$ and $MX = MN$ and we know that $KX = KN$, thus$\angle MKN = 90\circ$. :D

Sorry, but what do you mean with a symmetric point to $N$ by $K$?

$X$ is the point on line $NK$ such that $KX = KN$ and $X \neq N$. How should I say it(my english isn't good)?


In this example I found:

[img]http://www.wtamu.edu/academic/anns/mps/math/mathlab/int_algebra/int_alg_tut24ex8b.gif[/img]

Shouldn't that 10 become a 1 if you devide 10 into 10?

You would be correct if it was simply $ \frac{10}{10}$. However, $ \frac{10^{\minus{}15}}{10^5}\equal{}10^{\minus{}15}\cdot10^{\minus{}5}\equal{}10^{\minus{}20}$.


calculus
derivative
function
calculus computations

I came across with this article in the Mathematics Teacher (Vol 101 No 9 page 657).

There is this line which says that the local maximum of f $ f(c) \equal{} \minus{} sc^3 \plus{} s^2c^2 \minus{} 4A^2$

is

$ (\frac {2}{3}s, \frac {4}{27}s^4 \minus{} 4A^2 )$

can someone please explain how did this happen?
Sorry, I have a very weak calculus background. (I don't have calculus of several variables.)

Thanks in advance.

hello, you must compute the first derivative with respect to c
$ f^'(c) = - 3sc^2 + s^22c$. By solving the equation $ f^'(c) = 0$ we get $ c = \frac {2}{3}s$ for $ c\neq0$. Inserting this in the seconde derivative $ f^{''}(c) = - 6sc + 2s^2$ we have $ f^{''}\left(\frac {2}{3}s\right) = - 2s^2 < 0$, hence follows
that $ c = \frac {2}{3}s$ gives a local maximum, from here we have
$ f\left(\frac {2}{3}s\right) = \frac {4}{27}s^4 - 4A^2$.
Sonnhard.

hi. just have 2 more questions

1.) What happened to $ 4A^2$? Does it always imply that the if you get the derivative of a function with respect to say $ x$, all the terms without $ x$ becomes $ 0$?

2.) Why do we need to get the  second derivative? As far as I can remember, the second derivative is used to test concavity (if i remember correctly).


inequalities
triangle inequality
inequalities proposed

find the largest positive number $ {\lambda}$ for every $ { x,y,z \in R}$ which satisfying $ {x^2 \plus{} y^2 \plus{} z^2 \equal{} 1}$ , we have $ {abs(\lambda xy \plus{} yz) \le \frac {\sqrt {5}}{2}}$

by AM-GM and Triangle Inequality,

$ x^{2} \plus{} (\frac{\lambda^2}{\lambda^{2}\plus{}1}\plus{}\frac{1}{\lambda^{2}\plus{}1})y^{2} \plus{} z^{2} \equal{} 1$ $ \ge \frac{2}{\sqrt{\lambda^{2}\plus{}1}} |\lambda xy \plus{} yz|$ (equality holds when $ x^2 \plus{} z^2 \equal{} y^2$).

to maximize $ \lambda$, $ |\lambda xy \plus{} yz|$ should be maximized.

hence $ \max(\lambda)\equal{}2$ :pilot:


calculus
integration
trigonometry
calculus computations

[quote="kunny"]Calculate $ \displaystyle \left|\frac {\int_0^{\frac {\pi}{2}} (x\cos x + 1)e^{\sin x}\ dx}{\int_0^{\frac {\pi}{2}} (x\sin x - 1)e^{\cos x}\ dx}\right|$.[/quote]

offer some hints?

[quote="kunny"]That's exactly same as mine. :lol:[/quote]

Thanks.. :lol:


You fill in the missing words based on the first letters. You don't have to be right, just make it funny.

Taking the first letters ofmy phrase we get

GCWKUA

Now the next person will come up with a random phrase that starts with GCWKUA. Then they make up a new one.

THis was what GCWKUA orignally stood for (dont open until you make a phrase!)
[hide]GIant Chickens will kill us all[/hide]

GCWKUA[b]MWEF[/b]

[hide]Grace carried worried kaleidoscopes up a mountain while eating fish.[/hide]

No you don't get it!

I take yur GCWKUAMWEF

Giant cookies will kill underage aliens making weird,evil faces.

Now I make a DIFFERENT ONE

BLMWPE

[hide]Beasts like mewto55555 will pwn everyone[/hide]

BLMWPEBIAB

[hide]Banana-Lover Monkeys Will Put Each Banana In A Box[/hide]

No you make a [b]completely different[/b] phrase....

Brown lunch menues will pwn everyone's big, idiotic, annoying butt.

MEWTO

[hide]Many evil witches talk often[/hide]

My empty wall taked over.

WWWWWWW

Where was Will when we were working?

When will whippersnappers walk while whistling well?

LEVANS
[hide]
large elephants vehemently attack ninja squirrrels :ninja: [/hide]

Lol you didn't exactly make this game up. I've played it before.

LEVANS

Leave empty vans at North Street.

SIAA

[hide]Skiing is amazingly awesome (haha can you tell I ski?)

[/hide]

Sam Is An Actor

PACMAN

Peter, A Clown, Murdered Andrew Nolan.

Poachers acting cool make annoying nuisances.

[hide="LWPTD"]Llamas wearing pink tutus dance.[/hide]

Light-weight planes totally destroy.

[hide="WWYSPRTTCPAAMAOAITFF"]When will you stop posting random topics that could possibly annoy admins, mods, and other AOPSers in the Fun Factory?[/hide]

Wizened wives yell significantly, pestering rabbits, turtles, toads, cows, pandas, and apples; meanwhile, an otter aroused irritatedly to fend flies.

[hide="BAWPAA"]Black and white platypuses are awesome.[/hide]

bad autopsies will pown ariel aerobics

[hide="sqacabs"]speak quietly and carry a big stick[/hide] - Roosevelt

suckers quietly attack cabs and bad scientists.

[hide="IDMWWOR"]It doesn't matter who's wrong or right[/hide]

I Did Move While We Only Read

IEBFPLOSH
I'm Enough Bored For Posting Lots Of Spam Here

Wickedestjr should go meet Matt Carry who hates pie.

ETTISMMPMC

[hide]ernie thinks that I shoulc make my posts more creative. [/hide]

E. T. talked in Spanish (Mexican), making people more confused.

[hide="SAVVVFTHIYPBTASVVVFTH"] Smilies  :D are very very very fun to have in your post because they are so very very very fun to have  :D.[/hide]

Sam Anderson. Very very very funny. That hat is yellow Pete, but that annoying scavenger, very very very funny. That's horrible.

ITTMLPWVRWDYT

[hide]I think that my last post was very random. What do you think? [/hide]

It talks. Tommy mails Larry pencils. Wanda Vondo rates Wanda Dondo's yellow tea.

[hide="TAOPSPHSNCTSFDST"]The Art of Problem Solving people have still not changed their servers for daylight savings time.[/hide]

That art of problem solving person has shot Norbert Corky the stupid fat dairy supreme telemarketer.

DYTAWNIITTSPAB

[hide]Do you think anybody would notice if I typed the same post as before?[/hide]

Do you think all water needs ice?  I think that soap places about bars.

[hide="IJWTWDAICBT"]I just went to Washington D.C. and I came back Thursday.[/hide]

I just want to water dance and I can't be there.

[hide="IYRTOBISTSEHSESDUF"] :)  If you read this one backwards it stays the same emas ht syats ti sdrawkcab eno siht daer uoy fI :) [/hide]

If your really that obnoxious Bart, I should try shooting Edgar. Harry should encourage Susie dat Uxbridge farted.

WSISN

[hide]What should I say now? [/hide]

ooh, you were close.

I wish I could fly and wish to vault.

ITIWPC
[hide]I think I was pretty close. [/hide]

I think I was Polish-Christian.

[hide="OAWNTGMCWHP"]Once again, Wickedestjr needs to get more creative with his phrases.[/hide]

Omar annoying water nugget that guts march crazy what has pretended.

TWR

[hide]That was random. [/hide]

That water roars.

[hide="TPATHPISW"]The phrase above this hidden phrase is somewhat random.[/hide]

[quote="Wickedestjr"]If your really that obnoxious Bart, I should try shooting [b]Edgar[/b]. Harry should encourage Susie dat Uxbridge farted.

WSISN

[hide]What should I say now? [/hide][/quote]

You know my name?

Well, yes, but I wasn't talking about u when I made that post.

The pink amateur tennis/hockey player is so wierd.

IAFTMAP

[hide]I almost forgot to make a phrase. [/hide]


Problem 1: Let a;b;c in R; prove that:
$\frac{(\frac{a-b}{a+b})^3+(\frac{b-c}{b+c})^3+(\frac{c-a}{c+a})^3}{\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}}\geq -3$

 you can see Problem 2 at here:
[url]http://ineqdinhthi.wordpress.com/2008/08/13/ba-bai-toan-hay-ma-khong-kho/#comments[/url]

[quote="litbon"]Problem 1: Let a;b;c in R; prove that:
$ \frac{(\frac{a\minus{}b}{a\plus{}b})^3\plus{}(\frac{b\minus{}c}{b\plus{}c})^3\plus{}(\frac{c\minus{}a}{c\plus{}a})^3}{\frac{a\minus{}b}{a\plus{}b}\plus{}\frac{b\minus{}c}{b\plus{}c}\plus{}\frac{c\minus{}a}{c\plus{}a}}\geq \minus{}3$

 you can see Problem 2 at here:
[url]http://ineqdinhthi.wordpress.com/2008/08/13/ba-bai-toan-hay-ma-khong-kho/#comments[/url][/quote]
It's true because $ \frac{a\minus{}b}{a\plus{}b}\plus{}\frac{b\minus{}c}{b\plus{}c}\plus{}\frac{c\minus{}a}{c\plus{}a}\equal{}\minus{}\frac{a\minus{}b}{a\plus{}b}\cdot\frac{b\minus{}c}{b\plus{}c}\cdot\frac{c\minus{}a}{c\plus{}a}.$ :wink:


Τόσος καιρός και κανείς δεν ενδιαφέρθηκε για αυτήν την άσκηση; Ελπίζω η λύση που έχω υπόψη μου να είναι σωστή.

Θα τη βάλω στο τέλος του μήνα, αν μέχρι τότε συνεχίζετε να αδιαφορείτε.

Κώστας Βήττας.

[quote="vittasko"]ΠΡΟΤΑΣΗ  2. Δίδεται τρίγωνο ΑΒΓ και έστω Δ τυχόν σημείο επί της πλευράς ΒΓ. Δια σημείου Ρ επί της ΑΔ, φέρνουμε τις κάθετες ευθείες επί των ΒΡ, ΓΡ, οι οποίες τέμνουν τις πλευρές ΑΒ, ΑΓ, στα σημεία Ε, Ζ, αντιστοίχως. Προσδιορίστε τη θέση του σημείου Ρ, ώστε να είναι ΕΖ//ΒΓ.[/quote][url=http://www.mathematica.gr/forum/viewtopic.php?f=112&t=189]εδώ[/url]


What is the positive difference between the probability of a fair coin landing heads up exactly 2 times out of 3 flips and the
probability of a fair coin landing heads up 3 times out of 3 flips? Express your answer as a common fraction.

There are $ C(3,2)\equal{}3$ ways that 2 heads and 1 tail could appear.

Difference: $ 3(\frac{1}{2})^3\minus{}(\frac{1}{2})^3\equal{}\frac{3}{8}\minus{}\frac{1}{8}\equal{}\boxed{\frac{1}{4}}$


probability
analytic geometry
graphing lines
slope
algebra
polynomial

Feel free to move this if it's not in the right forum

Okay, I would like to graphically depict a norm of our business. We drill oil and gas wells and approximately one in three is successful. That means we will occasionally have a run of bad luck.

I want to show, in a bell-shaped curve or something, what the likelihood is of missing (or hitting) 2 in a row, 3 in row, 4 in a row, 5 in a row, etc.

I suspect w/ one in three success rate, 60% of the time we'll miss 2 in a row, 50% 3 in a row, 40% 4 in a row... etc.

I think that the data should be presentable in a bell shaped curve, but putting all this together just goes way beyond my math/statistics kung fu at this point in my life... I figure Excel could do it, but I'm lacking that ability as well.

Make sense? Any help out there?

1. u are definitely in the wrong forum

2. ur maths/statistics kungfu is v lacking

3. for your problem, u need to know more than the probability of hitting/missing in one drill. u could either assume the number of drills to be a fixed number or assume the number of drills is an infinite number. of cos, the first approach will be simple. you also need to better define your problem. are u looking at the probability of hitting/missing 2 in a row at least once in all the drills? or the probability of hitting/missing 2 in a row say 10 times in 100 drills?

4. i'm surprised that i even bother to answer.

1. Sorry.

2. Werd.

3. Say 12 drills

4. Well I appreciate it. Any more help?

So, is the following question the one you'd like to answer?

"Given that I will drill 12 wells, and each well has an independent probability of 1/3 of hitting oil, what is the probability that over that time I have a run of 2 (or 3, 4, ...) failed (or successful) drills?"

Another question might be,

"What is the probability that all of the next 2 (or 3, 4, ...) wells I drill fail (or succeed)?"

A third possibility is,

"Suppose I'm expecting to drill many wells, about how often do I expect to run into streaks of 2 (or 3, 4, ...) wells which fail (or succeed)?"

The second is the easiest to answer, but it's also probably the least interesting.

perhaps you should google for "probability of streaks" (or something including "streaks") for your answer.

here's [url=http://www.freakonomics.com/blog/2005/08/20/what-do-the-kansas-city-royals-and-my-ipod-have-in-common/]an article[/url] explaining something similar to your problem. though not exactly the same.

Okay

We just had a run of dry holes where I work, 4 in fact.

I want to show that is to be expected.

I'll try and draw what I'm looking to present. Order of dry/success is not important... well... maybe it is.

I want to show, if we do 12 wells, what is the likelihood that we'll have a streak of say 6 or 7 dry holes..

It may not be a bell curve... Still looking to present this in the best way....

It may be best to present it w/ X being number of dry holes in a row and Y being % likelihood of occurance. That would be a curved line w/ an average slope of -1 I think.

Does that make sense?

I agree that it probably is to be expected, some of the time (although maybe not from a total of 12) -- you might look at the article mao_zai linked, which gives a very rough calculation of "how expected" such events are (in that case, he deals with a 19-game losing streak by the Kansas City Royals).

Your intuition about the shape of the curve is good, although the exact shape will depend a great deal on what numbers we actually plug in (for probability of hitting oil, for instance).  On the other hand, your intuition about the slope of the curve is probably pretty bad.  For your simple numerical example: given a probability of 1/3 that any hole is good, over 12 holes, the likelyhood $p(n)$ of getting a maximal run of $n$ dry holes is as follows:

$p(0) = 0.0002\%$
$p(1) = 1.03\%$
$p(2) = 11.68\%$
$p(3) = 22.67\%$
$p(4) = 22.08\%$
$p(5) = 16.21\%$
$p(6) = 10.73\%$
$p(7) = 6.50\%$
$p(8) = 3.90\%$
$p(9) = 2.31\%$
$p(10) = 1.35\%$
$p(11) = 0.77\%$
$p(12) = 0.77\%$

Thus, if you really get 1 in 3 holes on average, the probability of getting 12 in a row successful is miniscule.  However, the odds that you have 4 failures in a row out of 12 tries (and not more than 4 in a row) is a solid 22%.  The odds that you have a run of 4 or more failures out of 12 holes dug is a good $66\%$.

I've written a mathematica program that can perform this calculation in any reasonablely-sized case, replacing 12, 4 and 1/3 with values of your choice.  (If you want numbers much larger than 12 I can give you the code -- it's not particularly efficient as it actually enumerates every case, so the calculation time grows at least exponentially.)

[quote="JBL"]I agree that it probably is to be expected, some of the time (although maybe not from a total of 12) -- you might look at the article mao_zai linked, which gives a very rough calculation of "how expected" such events are (in that case, he deals with a 19-game losing streak by the Kansas City Royals).

Your intuition about the shape of the curve is good, although the exact shape will depend a great deal on what numbers we actually plug in (for probability of hitting oil, for instance).  On the other hand, your intuition about the slope of the curve is probably pretty bad.  For your simple numerical example: given a probability of 1/3 that any hole is good, over 12 holes, the likelyhood $p(n)$ of getting a maximal run of $n$ dry holes is as follows:
[/quote]

That's great.  Thank you very much for the help.  Would I be able to make the following inferences:

If a "program" has 12 projects each w/ a 1 in 3 chance of success, would it be safe to say that AT LEAST 60% of the "programs" would have streaks of 4 or more dry holes?

Also, will 35% of programs have max streaks of 3 or less dry holes, i.e. can you add up the numbers or would only 22% of programs have 3 or less?  Basically, is there any value to the sum of the numbers?

Also, what formula did you use?

Thanks again for your help.

Also, I did read that article.  It seemed to suggest that streaks that are statistical oddities do happen and do so according to statistical probabilities.

[edit] you posted more after I read it--or I just missed the bottom part.  Is the program in excel?  I would like to be able to run it for 10 wells, 6 wells, etc...  Probably not more than 12.

[edit 2]  How does this look?

[img]http://www.pbase.com/investorguy/image/74286351.jpg[/img]

i dun think jbl came up with a formula. my guess is that, his program listed out all this possiblities, $2^{12}$ (which is about 4000 of them), and for each case, e program found out the maximal number of dry runs. and viola. with great amount of calculation he came out with the figures.

not sure if it's possible to come up any nice formulas though.

Mao-zai correctly describes the process I used: my program simply enumerates all possible lists of twelve holes by success and failure; for each list, I calculated the probability that it occurs and the length of its longest string, and then I added.  This is something that in principle one could do by hand -- for $n = 3$, there are only 8 lists, for example, and it would probably take 10 or 15 minutes to do all the arithmetic.  For $n = 6$, there are 64 lists, and that might take a couple of hours.

The numbers I gave are the probabilities of that being [i]exactly[/i] the longest run.  So, you are correct: the probability of getting a run of 4 or more failures in 12 tries with a 1/3 individual success rate is at least 60%, and the probability of having no such streak (longest streak of failures is 3 or less) is the sum $p(0)+p(1)+p(2)+p(3) \approx 35\%$.

Your charts look good, except for two problems in the first one:
(1) there is an error in coordinates: the left-most end of your chart is 13 and the right-most end is 1, while the numbers I give range from 12 failures to 0.  So, in fact, the 87% in that diagram corresponds to the probability of getting a run of at least 3 dry wells.
(2) the title is misleading.  It's not the probability of at least $n$ failures in 12 (that's relatively easy to calculate), it's the probability that there is a string of at least $n$ consecutive failures in 12 tries.  For example, in the value for $n = 3$ we count the possibility {F, F, F, S, F, F, F, S, F, F, F, S} with 9 failures and 3 successes, because the longest string of failures is 3.

Unfortunately, the program is in the Mathematica programming language, and I doubt it could be easily transformed into an Excel program.  On the other hand, if you are willing to just pick a couple of values for the number of runs, I can give you the formulas for "the likelyhood that the maximal string of failures is $n$, where each event has a probability $p$ of success."  Then data I gave you, for instance, were the outputs at $p = \frac{1}{3}$ for 12 runs with $n$ running over all 13 possibilities.  This won't give you the flexibility to change the number of runs, but it will give you formulas (although they're kind of nasty-looking polynomials) into which you can plug $p$, which should let you see the shape of the curve over a variety of possibilities.

maybe there's a "simple" recursive formula.

let $P(n,r)$ be the probability of getting a maximal streak of $r$ dry holes in $n$ drills. let $Q(n,r)$ be the probability of getting a maximal streak of $r$ dry holes with $n$ drills and with the maximal $r$-streak ending the $n$ drills.

my guess is, $P(n+1,r)=\frac{1}{3}P(n,r)+\frac{2}{3}Q(n,r-1)$, and $Q(n+1,r)=\frac{2}{3}Q(n,r-1)$.

jbl, care to check this with your prog?

Nice thought, mao_zai, but I think your recursion fails because of strings like $S, F, F, F, S, S, F, F, F$.  That string is counted in $P(9, 3)$.  We arrive here from $S, F, F, F, S, S, F, F$ with probability $\frac{2}{3}$, but this is counted in $P(8, 3)$ and not in $Q(8, 2)$.  I'm not immediately sure how to build a recursion to  avoid this problem.


complex numbers
algebra
binomial theorem

Find the remainder when

2007C0+2007C3+2007C6+2007C9+...+2007C2004+2007C2007

is divided by 1000.

is there a way of doing this without complex numbers?

has anybody found a possible lead for a solution?

Please do not bump topics without contributing anything new (e.g. what you did above), especially if it has been less than a day since the problem was posted. This applies to your other topics as well.

As for the problem itself, can $ \sum_{k\equal{}0}^{n}\binom{n}{n\minus{}k}\equal{}2^n$ be used somehow?

$ \sum_{i \equal{} 0}^{669}{\binom{2007}{3i}}$

Perhaps a modified form of the above sum?

[quote="i_like_pie"]As for the problem itself, can $ \sum_{k \equal{} 0}^{n}\binom{n}{n \minus{} k} \equal{} 2^n$ be used somehow?[/quote]

So would it be the last three digits of $ 2^{2007}$?

And.... are we allowed to use calculators for this?

If so, we can use it to figure out the pattern of the tens digit and hundred digit. Since we know the ones digit will be 8.

I think for powers of 2, there is a repeat for the tens digit every 20 terms starting with the 5th term (2^5), so the tens digit would be 6.

Power of 2: 02 04 08 16 32 [u]6[/u]4

I'm not sure about hundred's digit tho....

I don't think so, because it's $ 3i$ at the denominator, not $ i$.

[quote="darkmatter47"][quote="i_like_pie"]As for the problem itself, can $ \sum_{k \equal{} 0}^{n}\binom{n}{n \minus{} k} \equal{} 2^n$ be used somehow?[/quote]

So would it be the last three digits of $ 2^{2007}$?[/quote]
No, because $ k$ is increased by $ 3$ each time. I was suggesting a similar approach; the exact same thing won't work.

$ \frac{1}{3}\cdot (2^n\plus{}(1\plus{}\omega)^n\plus{}(1\plus{}\omega^2)^n)$

We can show that this satisfies the recurrence relation \[ a_n\equal{}3a_{n\minus{}1}\minus{}3a_{n\minus{}2}\plus{}2a_{n\minus{}3}\]

I feel like any solution that we would have to do some roundabout thing that is motivated by the complex numbers solution.

[quote="The Zuton Force"]This is a problem that, to almost everyone, seems contrived at first, but it's come up on AoPS Forums so much it's almost standard.

[hide="Hint"]Use the Binomial Theorem on $ (1 \plus{} r_1)^{2007}$, $ (1 \plus{} r_2 )^{2007}$, and $ (1 \plus{} r_3 )^{2007}$, and add them by collecting terms of the form $ C \binom{2007}{n}$.

Here, $ r_1$, $ r_2$, and $ r_3$ are the three roots of $ r^3 \equal{} 1$.[/hide][/quote]

[hide="hint helps a lot but..."]
So I know that

$ r_1\plus{}r_2\plus{}r_3\equal{}0$,
$ r_1^2 \plus{}r_2^2\plus{}r_3^2\equal{}0$,
$ r_1 ^3\plus{}r_2^3\plus{}r_3^3\equal{}3$,
$ r_2^4\plus{}r_2^4\plus{}r_3^4\equal{}0$
$ \cdots$

So now I know
$ 3(\binom{2007}{0}\plus{}\binom{2007}{3}\plus{}\cdots \binom{2007}{2007}) \equal{} (1 \plus{} r_1)^{2007}\plus{}(1 \plus{} r_2 )^{2007}\plus{}(1 \plus{} r_3 )^{2007}$

But I don't know what to do next.
[/hide]

use the fact that nCk = (n-1)Ck + (n-1)C(k-1)

[quote="RisingStar"]But I don't know what to do next.[/quote]

See the post by [b]Altheman[/b]. Basically, $ r_1$, $ r_2$, and $ r_3$ are the third roots of unity -- that is, they are solutions to $ x^3 \equal{} 1 \implies (x \minus{} 1) (x^2 \plus{} x \plus{} 1) \equal{} 1$. $ r_1 \equal{} 1$ and $ r_2$ and $ r_3$ are complex numbers. [hide="Continued"]If you draw a hexagon inscribed in the graph of $ |z| \equal{} 1$ on the complex plane, with one vertex at $ 1$, then you see that $ r_2 \equal{} \text{cis } 120$ and $ r_3 \equal{} \text{cis } ( \minus{} 120)$, and that if you add 1 to each, you get $ 1 \plus{} r_2 \equal{} \text{cis } 60$ and $ 1 \plus{} r_3 \equal{} \text{cis } ( \minus{} 60)$.

Thus, $ (1 \plus{} r_2)^{2007} \equal{} \text{cis } (2007 \cdot 60) \equal{} \text{cis } 180 \equal{} \minus{} 1$, and similarly $ (1 \plus{} r_3)^{2007} \equal{} \text{cis } (2007 \cdot ( \minus{} 60)) \equal{} \text{cis } 180 \equal{} \minus{} 1$.

Clearly, $ (1 \plus{} r_1)^{2007} \equal{} 2^{2007}$

So, we have $ 3S \equal{} 2^{2007} \minus{} 2 \implies S \equal{} \frac {2^{2007} \minus{} 2}{3}$

Can you find the remainder when this is divided by 1000?
[/hide]

[quote="Altheman"]I feel like any solution that we would have to do some roundabout thing that is motivated by the complex numbers solution.[/quote]

Sounds about right.  The solution by recursion is valid (and completely elementary), but I agree that it is somewhat without motivation.

To the OP:  there isn't a reason to be afraid of complex numbers.

This thread was never taken to completion. If you read my last post, you will see a numerical expression for the value of your combinatorial sum.

The remainder modulo 1000 of that same value is calculated [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=138253]in this thread[/url], so that would give you the final answer, if you still care for it  :)

What is the explanation for using complex numbers?  The method makes sense, but what is the "process" of interpreting the problem as such?


Is there a specific notation to denote factors?

For example, when showing your work for the Rational Zero Test of polynomials, how can we denote factors of $ a_0$ and $ a_n$?

Assuming there are a countable number of them (I can't imagine why there wouldn't be), just enumerate them.

Like $ x\equal{}a_{1}\cdot a_{2}\cdots a_{n}$, where each $ a_{i}$ denotes a factor of $ x$.

If you're finding the factors of $ a_{0}$, you may want to list the factors as $ b_{i}$ but that's just to avoid confusion.

I don't think there is any standard notation, so do what JRav suggested.