function
inequalities
algebra
floor function
number theory solved
number theory

Hi,
here are 2 nice russian problems:

(i)
let a,b be integer>0 so that a*b*(a+b) is a
multiple of a^2+ab+b^2.
Prove that abs(a-b)^3 >= a*b

(ii)
let n>10000. Prove that there exist integer a,b so
that n< a^2+b^2 <n+3*(n^1/4)

hey
here is my solution for problem no 2
Let us represent n in the form n = k^2 + m, where 0 <= m <=2k and
note that k = [ \sqrt n] and m <= 2 \sqrt n. Let us represent m in the same way:
m = s^2 +t, where 0 <= t <= 2s. Then
 s = [ \sqrt m]  <=  \sqrt m <=  \sqrt 2* n^(1/4). Note that
if t = 0, then there is nothing to prove as n itself is a sum of two squares.
Suppose t >= 1.
Let us consider now the number N = k^2 + (s + 1)^2. We have
N − n = 2s + 1 − t <= 2s <= 2 \sqrt 2 * n^(1/4) < 3 * n^(1/4)
QED.([x] represents the greatest integer function)
HOPE IT'S GOOD. :D
nice problem alekk(say cud u post france 2003 national olympiad)
best regards

Hey, about the first one, what if a=b=3 (for instance)? Then we must prove that 0>=9, which is false!.

(just a typing advice) try to use |x| instead of abs x

let a,b be integer>0 so that a*b*(a+b) is a
multiple of a^2+ab+b^2.
Prove that abs(a-b)^3 >= a*b

Let a,b be postive integers such that ab(a+b) is a multiple of a2+ab+b2.

Prove that |a-b|3 \geq ab.

Hey, can someone please tell me what I?ve done wrong here ? (because if I?m right the problem is too easy and the ineq is TOO weak):

First let?s assume a>b.

Let d=(a,b). Then a=dm, b=dn with (m,n)=1 and m>n. We have d^2*(m^2+mn+n^2)/d^3*mn(m+n), so m^2+mn+n^2/d*mn(m+n). Because (m,n)=1 we get the following: if p is a prime which divides mn and also divides m^2+mn+n^2, then it divides one of m or n, let,s say it divides m. Then it must also divide n^2, so it divides n, but (m,n)=1, so p=1, so (mn, m^2+mn+n^2)=1 (*). By the same way we prove that (m+n, m^2+mn+n^2)=1 (**), because (m+n)^2= m^2+mn+n^2+mn. From (*) and (**) and the initial condition we have m^2+mn+n^2/d, so d>=m^2+mn+n^2. What we have to prove translates to d^3(m-n)^3> d^2*mn iff d(m-n)^3>mn. But since m-n>1 and d>=m^2+mn+n^2 this inequality is obvious. Find anything wrong?

hi grobber,

your soln seems right. Perhaps the problem was:
**
let a,b be distinct integers so that  ab(a+b) is a multiple
of a^2+ab+b^2.
prove that |a-b|^3>=3ab
**
i don't remember exactely the problem but i'm sure it was
a russian one year 2000 or 2001.

Heya there!

I just looked at kalva daemon and found the problem it is from Russia 2001 and it is different a bit, it's without the abs and without the 3 it is like so:

 m > n are positive integers such that m^2 + mn + n^2 divides mn(m + n). Show that (m - n)^3 > mn.

cheers! :D

Thanks. I found the problem in the same place, but I still think there's sth wrong.  :D


It's very easy. This one's equal

$\sum_{sym}\frac{ab(a+b)}{c^2} \ge 2(a+b+c)$

$\Leftrightarrow \frac{(ab-c^2)(a+b)}{c^2} \ge 0$

Chebyshev.

I hope there is a solution with AM-GM only.

Yes of course. Multiply hung's $\sum_{cyc}\frac{ab(a+b)}{c^2} \ge 2(a+b+c)$ by $a^2b^2c^2$ to obtain $\sum_{sym}a^3b^4\ge\sum_{sym}a^3b^2c^2$
<=> $\sum a^3(b^2-c^2)^2\ge0.$ :)

Thank you, Spanferkel. Your solution only uses simple ideas. I like it.

[quote="pvthuan"]Let $a,b,c>0$ such that $a+b+c=1$, prove or disprove that
\[ \frac{ab}{c^2}+\frac{bc}{a^2}+\frac{ca}{b^2}\geq2+\frac{ab}{c}+\frac{bc}a+\frac{ca}{b}.  \][/quote]
$\leftrightarrow \sum a(\frac{b^2}{c^2}+\frac{c^2}{b^2})\ge\sum 2a$ :)


invariant
number theory
least common multiple
prime factorization

[quote="Arvind_sn"]$a$, $b$, and $c$ are positive, distinct integers. You are given that $abc=2001$. What is the largest possible value of $a+b+c$?[/quote]

[hide]If you're looking for largest value, then the difference between the numbers should be as large as possible..2001+1+1=2003[/hide]

[quote="Arvind_sn"]$a$, $b$, and $c$ are positive, distinct integers. You are given that $abc=2001$. What is the largest possible value of $a+b+c$?[/quote]
[hide]$1+667+3=\boxed{671}$[/hide]

oh crap. Then it's [hide]667+3+1=671[/hide]

Let's see

[hide]We have that $2001 = 3\cdot 23\cdot 29$. Now suppose $1\notin\{a,b,c\}$. That would mean that $a,b,c$ are three distinct positive divisors of 2001, none of them equal to one. This is clearly only possible if $(a,b,c)$ is a permutation of $(3,23,29)$: their sum is invariant equal to $3+23+29 = 55$. Now suppose $1\in\{a,b,c\}$ and assume WLOG $a=1$. We then have $bc=2001$. Now, 2001 has $(1+1)(1+1)(1+1) = 8$ divisors, but $b\neq 1\neq c$ (because $a,b,c$ are distinct), so we have 3 couples of divisors left: $(3,667)$, $(23,87)$ and $(29,69)$. The sum $a+b+c$ is then $667+3+1 = 671$, $87+23+1 = 111$ and $69+29+1=99$ respectively. Therefore, the largest sum is 671.[/hide]

[hide]Ah! you beat me to it Gödel. well....here's an FYI for discovering the prime factorization..

It helps to realize that the sum of the digits of 2001 is = 3 therefore making 2001 divisible by 3.
So we obtain $2001= 3\cdot667$ , next apply a prime decomposition theorem that states that any composite number has prime factors less than or equal to its square root.
Now,
$\sqrt{667}\approx$ 25 (to nearest floor integer)
Trying the next prime, we obtain $667 = 23\cdot29$
Hence the unique prime factorization of 2001 is $2001=3\cdot23\cdot29$

...applying Gödel's answer above. we get $(a,b,c)= (1,3,667)$[/hide]

[quote="Franz Joseph"]This reminds me of another problem.

The sum of three positive integers is $2005$. What is the minimal value of their Least Common Multiple?[/quote]

$802$. Produced by $(401, 802, 802)$.

Obviously the theoretical minimum occurs when the three numbers are equal.  This requires that $2005$ be divisible by $3$, which it is not.

Instead, we look for the smallest non-unity positive integer that divides $2005$, which is $5$.  Then

$2005 = 401+401+401+401+401$

Which we can group into two pairs and a remaining $401$ for an lcm of $802$.

[quote="José"]Well, we already have the result, but what about a rigorous proof?[/quote]

Given a fixed product, we wish to maximize the sum.  By Karamata's, $(667, 3, 1)$ majorizes every other distinct triple $(a, b, c)$ such that $abc = 2001$.

[quote="Franz Joseph"]This reminds me of another problem.

The sum of three positive integers is $2005$. What is the minimal value of their Least Common Multiple?[/quote]

Does this work?

[hide]Since $3$ does not divide $2005$ perfectly we deduce that the three numbers $a,b,c$ are not all equal. Assume, without lost of generalization, that $a \leq b \leq c$ $\rightarrow$ $a \leq 2LCM(a,b,c)$, $b \leq LCM(a,b,c)$ and $c \leq LCM(a,b,c)$ $\rightarrow$ $2a+2b+2c= 4010 \leq 5LCM(a,b,c)$ and thus $802 \leq LCM(a,b,c)$[/hide]

Wouldn't it be easier to just use prime factorization? You obtain the numbers 3,23,29, so then 3 numbers there then we can incorporate 1. To do that we multiply the 2 larger numbers because that gives us a larger result in sums. 1,667,3


$1.$ Select any hundred consecutive $3$-digit numbers. Raise each of them to the $100th$ power. Now Sum them up. What are the last two digits of the sum? If possible, answer with reason.

$2.$ Ten $1's$, twenty $2's$ and thirty-five $3's$ are listed on a blackboard. Iteratively, you can do any of the following.
a) Rub off a $1$ and a $2$, and write a $3$ in the place
b) Rub off a $2$ and a $3$, and write a $1$ in the place
c) Rub off a $3$ and a $1$, and write a $2$ in the place
Please, note that you can not rub off two numbers of the same value.

This process continues till there is a way to rub off two different numbers (and writing the third number in that place). What would be the value of the last number(s) left on the blackboard? Please answer with reason.

$3.$ $3^{3^3}$ = $3^{3^3}$ = $3^{27}$ = $7,625,597,484,987$
$3^{3^3}$ has got $13$ digits. How many digits are there in the value of $9^{9^9}$? If possible, provide answer with reason.

[quote="samath"]Number 2:

Take the number of 1's, 2's, and 3's mod 2 (0,0, and 1, respectively).  Whenever you rub off 2 numbers, you add a third number, meaning that you toggle each (meaning that 0's become 1's and 1's become 0's).  Then after each switch, the amounts of each number in mod 2 are (0,0,1), (1,1,0), (0,0,1), etc.  When there is only one number left, the amounts of the other two are equal to 0 mod 2, meaning that a 3 must be remaining.[/quote]

My mistake... ignore the comment.
/************* ignore this *********************/
Take the number of 1's, 2's, and 3's mod 2 (0,0, and 1, respectively)

Should it not be (1, 0, 1) ??
/************* ignore this *********************/

[quote="4everwise"][hide="Problem 3"]Note that $x=3^{9^9}= 3^{3^{18}} \cdot 3^{3^{18}}=\left(3^{3^3}\right)^{12}.$

Because $3^{3^3}\approx 10^12 \cdot 7.625,$ the answer is about $10^{216}\cdot (7.625)^{18}$ and there are 216+16=232 digits.
[/hide][/quote]

$x=3^{9^9}$,

But the question asked for $9^{9^9}$ ? Am I missing something here?

[quote="4everwise"][quote="xxxyyyy"]But the question asked for $9^{9^9}$ ? Am I missing something here?[/quote]I edited it to make it less confusing. $9^{9^9}$ was $x^2$ and $3^{9^9}$ was $x$. :)[/quote]

Solution 3
[hide]Solution 3
Note that $9^9$ = $387420489$
Now let $x = 9^{387420489}$

or we have, $log(x) = 387420489 \times log(9)$
or, $log(x) = 387420489 \times 0.954243$
or, $log(x) = 369693099.63$

so, we can say that $x = 9^{9^9}$ will have $369,693,100$ digits.[/hide]

[quote="Altheman"][hide="1"]clearly, the numbers in some order have the all the different remainders mod 100

then we want $\sum_{k=1}^{100}k^{100} \mod 100$

look mod 4...

it repeats every four, so 25 cycles and we have it is

$25(1+2^{100}+3^{100}+4^{100})\equiv (1+(-1)^{100})\equiv 2$

---

in mod 25

it repeats every 25...so 4 cycles

$4(1^{100}+2^{100}+3^{100}+4^{100}+5^{100}+...+100^{100})$

the multiples of 5 drop out, and using euler...note $\phi(25)=20$

$\equiv 4( 1^0=2^0+3^0+4^0+0+...\equiv 4(80)=\equiv 20$

then using crt, it becomes $70\mod 100$[/hide][/quote]

would anyone else like to take a shot at it... I am not sure if this is correct.

[hide]If we take any 3 digit number to the 100th power, \[ (100a+10b+c)^{100}=...+1000bc^{99}+c^{100} \] So the last two digits will only depend on the 100th power of the last digit. Starting from any three digit number, there are exactly 10 numbers who have the same last digit. Since the there are 10, we only need to find the last digit of $c^{100}$. The powers of any digit forms a pattern so we can say that the last digit of $2^{100}$ is 6 and the last digit of $3^{100}$ is 1 and so on. \[ 10(1+6+1+6+5+6+1+6+1)=330 \] The last two digits is 30. [/hide]

[hide="alternate solution for 1"]
1^100 = 1
there are ten numbers selected that end in 1. 1*1 = [b]10[/b]
units digits of powers of 2 go 2,4,8,6,2,4,8,6......
call u(x) the units digit of x^100
u(2) = 6
there are ten selected numbers ending in 2. 6*10 = [b]60[/b]
the answer to this problem should then be
the last two digits of
10*(u(1)+u(2)+u(3)+u(4)...u(9))  (zero doesnt matter)
u(1) = 1
u(2) = 6
3 = 1 (I'm omitting the u() now)
4=6
5=5
6=6
7=1
8=6
9=1
so the answer is the last two digits of (10*(1+6+1+6+5+6+1+6+1) = [b]30[/b]
hold on
I don't think that's right
somebody correct my error (if I screwed up) please.[/hide]

oh whoops i didnt read the post the guy in fron of me posted
and i think we had the exact same thing...


geometry
ARML
inequalities
USAMTS
number theory

Oh shoot.  I was in such a hurry to get my prize choices in that I forgot to take a screenshot of the list.  All right, I'll try to remember.

Here is what's listed in the Prizes Announcement:

[quote]# Mathematica for Students software, generously donated by Wolfram Research.
# An assortment of books, including the following new prizes this year:

    * The Millennium Problems by Keith Devlin
    * A Mathematical Mosaic by Ravi Vakil
    * Introduction to Number Theory and Inequalities by CJ Bradley
    * A Mathematical Olympiad Primer by Geoff Smith
    * Plane Euclidean Geometry by Gardiner and Bradley
...plus 8 other mathematics books to choose from.[/quote]

Those eight texts are:

[list]
[*] The Music of the Primes
[*] The Elegant Universe
[*] The Code Book
[*] Challenging Problems in Geometry
[*] Journey Through Genius: The Great Theorems of Mathematics
[*] Count Down: Six Kids Vie for Glory at the World's Toughest Math Competition
[*] Some ARML problem book or something
[*] Another Olympiad book, I think, containing problems from the British Olympiad[/list]

Aside from the T-Shirt, and assuming no other prizes ran out in the first few hours, the prizes are as follows:

The Code Book (Simon Singh)
The Music of the Primes (du Sautoy)
Challenging Problems in Geometry (Salkind/Posamentier)
ARML & ARML Power Contests 1995-2003
The Elegant Universe (Brian Greene)
Count Down (Steve Olson)
Mathematica for Students
The USSR Problem Book
Journey Through Genius (William Dunham)
The Millennium Problems (Keith Devlin)
A Mathematical Mosaic (Ravi Vakil)
Intro to Number Theory and Inequalities (CJ Bradley)
A Mathematical Olympiad Primer (Geoff Smith)
Plane Euclidean Geometry (Gardiner and Bradley)

I chose the music of primes, the elegant universe, and another mathematica (hopefully I'll have time to learn to how to use it in the summer).

Timwu: So now you have two licenses?

By the way, I chose Mathematica, Journey Through Genius and the Olympiad Primer.

I chose three books. I forget which ones.

This is to anyone who has gotten the british olympiad books.
Are they good compared to other books?
Thanks.

[quote="barkashdanbar"]dear gold winners.
you already have one. please save mathematica for bronze winners[/quote]

How do you know that? They may have not done USAMTS until this year :P
Also, some people have said if you wait until 12:01 AM the day the prizes come out,  it should still be availibe, so theres nothing to whine about.

[quote="abacadaea"][quote="barkashdanbar"]dear gold winners.
you already have one. please save mathematica for bronze winners[/quote]

Also, some people have said if you wait until 12:01 AM the day the prizes come out,  it should still be availibe, so theres nothing to whine about.[/quote]

That was just one year, so it is most likely not a general rule of thumb to rely on if you really want something.

Also, would it be better if I got mathematica in case I want to use it in the future? I have not felt the need to have mathematica this year, but it seems to be the most worthy prize up there, although I could pick 3 good books instead of 2.

I chose [i]"Music of the Primes"[/i] and [i]"Introduction to Number Theory and Inequalitites"[/i] although I suspect after reading "A Friendly Introduction to NT" by Silverman the NT component will be review. Haha I should solve the Riemann Hypothesis after reading MotP

[quote="abacadaea"][quote="barkashdanbar"]dear gold winners.
you already have one. please save mathematica for bronze winners[/quote]

How do you know that? They may have not done USAMTS until this year :P[/quote]

This is my first year doing USAMTS and I'm a gold winner.  I'm looking forward to getting Mathematica, which, by the way, I don't have.  Occasionally I use Maxima though (http://maxima.sourceforge.net).  You can do a lot in wxMaxima (which is free).

So now I have Maple, Mathematica, and two copies of Matlab, both outdated, but one less outdated.

Thank you math competitions. :)

Could somebody update the list of remaining prizes?

Thanks. :)

mathematica was still available a few minutes after the bronze prizes came out (when i checked)

[quote="FantasyLover"]Could somebody update the list of remaining prizes?[/quote]When you log in to the USAMTS website, it will display the remaining prizes on your "My USAMTS" page.

[quote="gauss202"][quote="Differ"]So now I have Maple, Mathematica, and two copies of Matlab...
[/quote]

Which do you like the best?[/quote]

I haven't used any of them. :P

This is my first time doing USAMTS. As an 8th grader, I got a 55 with round 2 (therefore giving me honorable mention).

I got the chance to pick my prize yesterday, and I chose [i]Challenging Problems in Geometry[/i]. I hope it'd be helpful for high school/olympiad geometry prep, which is one of the things I wanted. If any of you have this book, what did you guys think about it?

It is very helpful. While it is not an incredibly difficult book, it does cover many of the methods used in high school competitions. It should help you "ease in" to them.


Έστω $ABC$ οξυγώνιο τρίγωνο με βαρύκεντρο $G$. Να δείξετε ότι
αν $BG+AC = CG+AB$ τότε το $\triangle ABC$ είναι ισοσκελές .

 :)

ΘΕΤΩ για ευκολια στην γραφη  diamesow (b)=x kai diamesow(c)=y
Εχουμε...
c+2y/3=b+2x/3
(3/2)(c-b)=x-y
(3/2)(c-b)(x+y)=x^2-y^2=(3/4)(c^2-b^2)
2(c-b)(x+y)=(c-b)(c+b)
 :arrow: c=b,teliosame....
 :arrow: c diaforetiko b ,exoume...
2(x+y)=c+b  (1)
kai 2(x-y)=3(c-b)  (2) (apo tin arxiki sxesi)
Lynontas tis 1+2 exoume
x=c-b/2  .y=b-c/2
kai tote x^2=(c-b/2)^2
.......................(antika8istonas ton tipo tou tetragonou tis diamesou....)
2a^2=2c^2+2b^2-4bc
a^2=(b-c)^2 atopo  apo tin trigonimi anisotita......

ΡΑΛΛΗΣ :D

Σωστή φυσικά η λύση και sorry αν ήταν πολύ  απλή η άσκηση
αλλά εμπιστεύτηκα την πηγή μου .  :)


Find the equation of circle passing through the points of intersection  of direct common tangents of
$ x^{2} \plus{} y^{2} \equal{}16$ and $ x^{2} \plus{} y^{2}\minus{}12x\plus{}32\equal{}0$
yes this can be done i know but it's very tidious(my process)
is there any specific geometrical property of direct common tangents that i can use here

What do you mean by "direct common tangents"?

 :)

A pair of lines tangent to the two given circles that do not intersect "between" the circles.  The other pair of common tangents intersect somewhere between the centers of the circles.

The point of intersection, let it be A, of the direct common tangents is the point that divides the centers (0, 0) and (6, 0) in the ratio 2:1 externally.
So, A is (12, 0).
The chord of contact of A, with respect to the circle ${x^2} + {y^2} = 16$ is given by $12x + 0y = 16 \Rightarrow 3x - 4 = 0$.
Let this line cut the circle ${x^2} + {y^2} = 16$  at the points B and C.
Then any circle through B and C is given by ${{x}^{2}}+{{y}^{2}}-16+2k(3x-4)=0$.
If this passes through A, we have $144-16+2k(36-4)=0\Rightarrow 2k=-4$
Hence the equation to the required circle is ${{x}^{2}}+{{y}^{2}}-12x=0$


inequalities
quadratics
algebra
three variable inequality

Let $ a,b,c$ be nonnegative real numbers, prove that
\[ (a^2\plus{}b^2\plus{}c^2\plus{}ab\plus{}bc\plus{}ca)^2 \ge 4(a\plus{}b\plus{}c)(a^2b\plus{}b^2c\plus{}c^2a).\]
When do we have equality?
:)

Oh @@ . It holds when :
$ a=b=c$and ${ b=\frac{2}{\sqrt{5}+1}}a;c=0$
Pqr ?

<=>$ \sum (a \minus{} b)^2( a \plus{} b \minus{} c \plus{} \frac {c^2}{a \plus{} b \plus{} c} ) \ge 4(a \minus{} b)(a \minus{} c)(b \minus{} c)$
Assume that :$ c \equal{} min(a,b,c)$
Easy to prove that :
$ * a \plus{} b \minus{} c \plus{} \frac {c^2}{a \plus{} b \plus{} c} \ge a \plus{} b \minus{} 2c$
$ *a \plus{} c \minus{} b \plus{} \frac {b^2}{a \plus{} b \plus{} c} \ge (a \minus{} c) \minus{} (b \minus{} c) \plus{} \frac {(b \minus{} c)^2}{a \plus{} b \minus{} 2c}$
$ *b \plus{} c \minus{} a \plus{} \frac {a^2}{a \plus{} b \plus{} c} \ge (b \minus{} c) \minus{} (a \minus{} c) \plus{} \frac {(a \minus{} c)^2}{a \plus{} b \minus{} 2c}$
Setting :$ x \equal{} a \minus{} c;y \equal{} b \minus{} c$
So we need to prove this ineq with $ c\equal{}0$
<=>$ (\minus{}b^2\minus{}ab\plus{}a^2)^2 \ge 0$
Done ;)

[quote="mitdac123"]<=>$ \sum (a \minus{} b)^2( a \plus{} b \minus{} c \plus{} \frac {c^2}{a \plus{} b \plus{} c} ) \ge 4(a \minus{} b)(a \minus{} c)(b \minus{} c)$
[/quote]

why?  :maybe:

Thank you very much.

More general, for any real $ k$ and any nonnegative real numbers $ a,b,c$, the inequality holds:

$ [\sum a^k(a\plus{}b)]^2 \ge 4(\sum a^k)(\sum a^{k\plus{}1}b)$.

The case $ k\equal{}\minus{}1$ is also nice.

With AM-GM we have :
$ (a^2\plus{}b^2\plus{}c^2\plus{}ab\plus{}ac\plus{}bc)^2 \le 4(a^2\plus{}b^2\plus{}c^2)(ab\plus{}ac\plus{}bc),$
so we have to prove that:$ (a^2\plus{}b^2\plus{}c^2)(ab\plus{}ac\plus{}bc) \le (a\plus{}b\plus{}c)(a^2b\plus{}b^2c\plus{}c^2a)$
wich is true after expanding and use AM-GM and rearangement.

;). Your solution isn't right :) .Note that :
$ a,b,c \ge 0$
$ \sum a^3b  \ge \sum a^2b^2$ when $ a \ge b \ge c$ or $ a,b,c$ be side length of a triangle :) .But it's not always true for all $ a,b,c$

[quote="Vasc"]More general, for any real $ k$ and any nonnegative real numbers $ a,b,c$, the inequality holds:

$ [\sum a^k(a \plus{} b)]^2 \ge 4(\sum a^k)(\sum a^{k \plus{} 1}b)$.

The case $ k \equal{} \minus{} 1$ is also nice.[/quote]

How to slove it?

More general, the following statement holds:

[i]Let $ p,q,r$ be nonnegative real numbers, and let $ a,b,c$ be real numbers. Then[/i]

$ [(q \plus{} r)a \plus{} (r \plus{} p)b \plus{} (p \plus{} q)c]^2 \ge 4(p \plus{} q \plus{} r)(pbc \plus{} qca \plus{} rab)$.

[hide]
For $ (p,q,r) \equal{} (b,c,a)$ and $ p,q,r) \equal{} (a(a \minus{} b \plus{} c),b(b \minus{} c \plus{} a),c(c \minus{} a \plus{} b))$ we find the inequalities posted by can_hang2007:
$ (a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca)^2 \ge 4(a \plus{} b \plus{} c)(a^2b \plus{} b^2c \plus{} c^2a)$,
$ (a^2b \plus{} b^2c \plus{} c^2a)^2 \ge abc(a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)$ - $ a,b,c$ site-lengths of a triangle.

Also, for $ p,q,r) \equal{} (a \minus{} b \plus{} c,b \minus{} c \plus{} a,c \minus{} a \plus{} b)$ we find
$ \sum a^2(a \plus{} b)(b \minus{} c) \ge 0$ - $ a,b,c$ site-lengths of a triangle.
See also  $ (p,q,r) \equal{} (b^k,c^k,a^k)$,  $ (p,q,r) \equal{} ((b \minus{} c)^8,(c \minus{} a)^8,(a \minus{} b)^8)$,  $ (p,q,r) \equal{} (|b \minus{} c|,|c \minus{} a|,|a \minus{} b|)$ etc.
[/hide]

[quote="Vasc"][i]Let $ p,q,r$ be nonnegative real numbers, and let $ a,b,c$ be real numbers. Then[/i]

$ [(q \plus{} r)a \plus{} (r \plus{} p)b \plus{} (p \plus{} q)c]^2 \ge 4(p \plus{} q \plus{} r)(pbc \plus{} qca \plus{} rab)$.[/quote][b]Proof[/b] Without loss of generality, we assume $ a \equal{} \max\{a,b,c\}$, then

$ [(q \plus{} r)a \plus{} (r \plus{} p)b \plus{} (p \plus{} q)c]^2 \minus{} 4(p \plus{} q \plus{} r)(pbc \plus{} qca \plus{} rab)$

$ \equal{} [(q \minus{} r)a \plus{} (r \plus{} p)b \minus{} (p \plus{} q)c]^2 \plus{} 4qr(a \minus{} b)(a \minus{} c)\geq0$.

For $ a,b,c,p,q,r$ be nonnegative real numbers, this inequality was also founded by Ji Chen at 16 years ago, see :

Ji Chen, Sharpening the Gerber inequality (Chinese), [url=http://math.fjnu.edu.cn/zxsx/more.asp?typeid=5]Fujian School Mathematics[/url], 1992, No. 5, pp. 8 - 9.

There, the proof is same, and the inequality be used to solve the following problem :

[i]Let $ P$ is a point in the interior of $ \triangle ABC$, $ r _1,r_2,r_3$ are the distances from $ P$ to the sides $ a,b,c$, and $ \Delta$ is the area. Then

$ r_2r_3 \plus{} r_3r_1 \plus{} r_1r_2\leq \frac {4\Delta^2}{2bc \plus{} 2ca \plus{} 2ab \minus{} a^2 \minus{} b^2 \minus{} c^2},$

with equality if and only if $ r_1 : r_2 : r_3 \equal{} (b \plus{} c \minus{} a) : (c \plus{} a \minus{} b) : (a \plus{} b \minus{} c)$. [/i]

i.e.

[b][url=http://www.mathlinks.ro/Forum/viewtopic.php?t=25662]Mittenpunkt[/url] is the solution to the problem of

finding a point $ P$ inside $ \triangle ABC$  such that $ r_2r_3 \plus{} r_3r_1 \plus{} r_1r_2$ is the maximum possible. [/b]

It's easy to prove : the inequality

$ [(q \plus{} r)a \plus{} (r \plus{} p)b \plus{} (p \plus{} q)c]^2 \ge 4(p \plus{} q \plus{} r)(pbc \plus{} qca \plus{} rab)$

holds for all real numbers $ a,b,c$ if and only if $ pqr(p \plus{} q \plus{} r)\geq 0$.

[hide="Remark."]This inequality holds for all sides $ a,b,c$ of a triangle also if and [i]only if[/i] $ pqr(p \plus{} q \plus{} r)\geq 0$.[/hide]
By the way, the following inequalities are also true and nice :

if $ n$ be nonnegative integer, and $ a,b,c$ be real numbers, then

$ \left[(b \plus{} c)(b \minus{} c)^n \plus{} (c \plus{} a)(c \minus{} a)^n \plus{} (a \plus{} b)(a \minus{} b)^n\right]^2$

$ \geq4\left[(b \minus{} c)^n \plus{} (c \minus{} a)^n \plus{} (a \minus{} b)^n\right]\left[bc(b \minus{} c)^n \plus{} ca(c \minus{} a)^n \plus{} ab(a \minus{} b)^n\right]$;

if $ x,y,z$ be [i]positive real numbers[/i], then

$ \left[\frac {(y \plus{} z)(z \plus{} x \minus{} y)}{x^2} \plus{} \frac {(z \plus{} x)(x \plus{} y \minus{} z)}{y^2} \plus{} \frac {(x \plus{} y)(y \plus{} z \minus{} x)}{z^2}\right]^2$

$ \geq 4\left(\frac {z \plus{} x \minus{} y}{x^2} \plus{} \frac {x \plus{} y \minus{} z}{y^2} \plus{} \frac {y \plus{} z \minus{} x}{z^2}\right)\left[\frac {yz(z \plus{} x \minus{} y)}{x^2} \plus{} \frac {zx(x \plus{} y \minus{} z)}{y^2} \plus{} \frac {xy(y \plus{} z \minus{} x)}{z^2}\right].$

[quote="Ji Chen"]The inequality

$ [(q \plus{} r)a \plus{} (r \plus{} p)b \plus{} (p \plus{} q)c]^2 \ge 4(p \plus{} q \plus{} r)(pbc \plus{} qca \plus{} rab)$

holds for all real numbers $ a,b,c$ if and only if $ pqr(p \plus{} q \plus{} r)\geq 0$.[/quote]Yes. It is true.

I proved this inequality writting it as

$ [(2p\plus{}q\plus{}r)a\minus{}(p\plus{}r)b\minus{}(p\plus{}q)c]^2\plus{}4p(p\plus{}q\plus{}r)(a\minus{}b)(c\minus{}a) \ge 0$.

Setting $ (a,b,c)\equal{}(x,y,z)$ and (p,q,r)=(-a+b+c,a-b+c,a+b-c), we get the following known statement:

If $ a,b,c$ are the site-lengths of a triangle, and $ x,y,z$ are real numbers, then
$ (x\minus{}y)(x\minus{}z)a^2\plus{}(y\minus{}z)(y\minus{}x)b^2\plus{}(z\minus{}x)(z\minus{}y)c^2 \ge 0$.

[quote="can_hang2007"]Let $ x,y,z$ be nonnegative real numbers, prove that

$ (x^2 \plus{} y^2 \plus{} z^2 \plus{} yz \plus{} zx \plus{} xy)^2 \ge 4(x \plus{} y \plus{} z)(x^2y \plus{} y^2z \plus{} z^2x)$.[/quote]$ (x^2 \plus{} y^2 \plus{} z^2 \plus{} yz \plus{} zx \plus{} xy)^2 \minus{} 4(x \plus{} y \plus{} z)(x^2y \plus{} y^2z \plus{} z^2x)$

$ \equal{} (z \plus{} x)^2(x \minus{} y)(x \minus{} z) \plus{} (x \plus{} y)^2(y \minus{} z)(y \minus{} x) \plus{} (y \plus{} z)^2(z \minus{} x)(z \minus{} y)\geq0$,

see: http://www.mathlinks.ro/Forum/viewtopic.php?t=186038

It's easy to see

$ \sum_{cyc}(z^2 \plus{} x^2 \plus{} kzx)(x \minus{} y)(x \minus{} z)\geq 0$

holds for all nonnegative real numbers $ x,y,z$ if and only if $ \minus{} 2\leq k \leq 2$.

When $ k \equal{} \minus{} 2$, we have a full symmetrical inequality:

$ (z \minus{} x)^2(x \minus{} y)(x \minus{} z) \plus{} (x \minus{} y)^2(y \minus{} z)(y \minus{} x) \plus{} (y \minus{} z)^2(z \minus{} x)(z \minus{} y)$

$ \equal{} (x^2 \plus{} y^2 \plus{} z^2 \minus{} yz \minus{} zx \minus{} xy)^2\geq 0$.

Setting $ y \minus{} z \equal{} a,z \minus{} x \equal{} b,x \minus{} y \equal{} c$, the above inequality can be generalized as :

for all real numbers $ a,b,c$ such that $ a \plus{} b \plus{} c \equal{} 0\neq abc$, the inequality

$ |b|^pcb \plus{} |c|^pac \plus{} |a|^pba\leq 0$

holds if and only if $ p \geq \minus{} 1$.

Prove that if $ a,b,c\geq0$, then
\[ (a^2\plus{}b^2\plus{}c^2\plus{}ab\plus{}bc\plus{}ca)^2\geq4(a\plus{}b\plus{}c)(a^2b\plus{}b^2c\plus{}c^2a).\]
I like this problem and here is the solution.

We just need to prove the inequality when $ a\geq b\geq c$. Let $ p\equal{}a\plus{}b\plus{}c$, $ q\equal{}ab\plus{}bc\plus{}ca$, and $ r\equal{}abc$. Notice that
\[ a^2b\plus{}b^2c\plus{}c^2a\equal{}\frac{pq\minus{}3r\plus{}\sqrt{p^2q^2\minus{}4q^3\plus{}2p(9q\minus{}2p^2)r\minus{}27r^2}}2.\]
For simplicity, let $ p\equal{}1$, the desired inequality reads
\[ \frac12(1\minus{}q)^2\geq  q\minus{}3r\plus{}\sqrt{q^2\minus{}4q^3\plus{}2(9q\minus{}2)r\minus{}27r^2}.\]
That is,
\[ 1\minus{}4q\plus{}q^2\plus{}6r\geq2\sqrt{q^2\minus{}4q^3\plus{}2(9q\minus{}2)r\minus{}27r^2}.\]
Squaring both sides gives
\[ (1\minus{}4q\plus{}q^2\plus{}6r)^2\minus{}4(q^2\minus{}4q^3\plus{}18qr\minus{}4r\minus{}27r^2)\geq0.\]
Regrouping yields
\[ (q^2\plus{}4q\minus{}1)^2\plus{}144r^2\plus{}4r(7\minus{}30q\plus{}3q^2)\geq0\]
If $ 7\minus{}30q\plus{}3q^2\geq0$, then the inequality is obviously true.

If $ 7\minus{}30q\plus{}3q^2\leq0$, or $ \frac13\geq q\geq\frac{15\minus{}\sqrt{104}}3$ then view the desired inequality as a quadratic equation of $ r$, we need to show that its discriminant is non-negative. That is
\[ \Delta\equal{}\minus{}16(3q^2\plus{}54q\minus{}13)(3q\minus{}1)^2\leq0.\]
This is the end of the proof.


calculus
integration
logarithms
trigonometry
calculus computations

Evaluate integral, \[\int_{0}^{1}\frac{e^{\arctan x}+x\ln(1+x^{2})}{1+x^{2}}dx.\] It's not very hard, but I can't understand why my result is not same with the one in the book (without solution).

Yes, I don't know what makes the result with $\frac{\pi}{4}$ in that book. But I think it's good.


let $ a_{1}$, $ a_{2}$, ... $ a_{n}$ be positive integers with no 9's in their decimal representations.

Prove $ \frac{1}{a_{1}}\plus{}\frac{1}{a_{2}}\plus{}...\plus{}\frac{1}{a_{n}}\leq 30$

[quote="Rust"]Posted before, consider $ S_{k}=\sum_{i|10^{k}\le a_{i}< 10^{k+1}}$. Obviosly $ S_{k+1}<\frac{9}{10}S_{k}$, therefore
$ S < S_{0}(1+x+x^{2}+...) =\frac{S_{0}}{1-x}= 10S_{0}=\frac{2283}{84}= 27.17857...,x =\frac{9}{10}$.[/quote]

I'm very very sorry if this is a stupid question, but can you show me why $ S_{k+1}<\frac{9}{10}S_{k}$? Thank you so much!

For any member $ \frac{1}{x}$ in $ S_{k}$ we have 9 members in $ S_{k\plus{}1}$, they are $ \frac{1}{10x\plus{}i}, i \equal{} 0,1,2,3,4,5,6,7,8.$ Sum of these members less than $ \sum_{i\equal{}0}^{8}\frac{1}{10x\plus{}i}<\frac{9}{10x}$, therefore $ S_{k\plus{}1}<\frac{9}{10}S_{k}.$


number theory
prime numbers
number theory unsolved

Hello,

Isn't it frustrating when you know you are just one step away of the solution of your problem but anyway, you can't solve it. Well, I feel this is what is happening to me.

The problem is: Find all the values of $n\in\mathbb{Z}^+$ such that there exist some $k\in\mathbb{Z}$ with $n!+(n+1)!=k^2$.

I believe that the solution is unique, and is $n=4$. Using the Bertran's postulate (if $m>1$, then, there is a prime $p$ such that $m<p<2m$) I can prove that there is no odd value of $n$ for which the equation has some solution for $k$. But the problem is, what happens with the even values.

Note that $n!+(n+1)!=(n+2)n!=k^2$. Suppose that $n=2m$, then, the equation becomes $(2m+2)(2m)!=k^2$. Suppose $m>2$, the two cases that we are excluding can be analyzed separately. One is a solution, the other is not. What I want is to find a prime in the factorization of the product in the left that is just once, therefore it cannot be a square. We know that there is a prime $p$ such that $m<p<2m$. If $p\neq m+1$ we are done, but if $p=m+1$ what ca we do, we have the prime twice. Well, we also know that there is a prime $q$ such that $p=m+1<q<2m+2$. Again, if $q\neq2m+1=2p-1$ we are again done, because $q$ is the prime we want, but what happens if $2p-1=q$. The family, I believe, is infinite.

Then, if someone can solve the mess I did, perfect, or if someone can find another solution to the problem, cool! :cool:

Best regards,

All I can say is that I think Betrand's postulate can be strengthened to from n to kn, where k is a constant less than 2. How much less is a problem, but once you lower it, technically you do it then by brute force following your method.

[quote="ZetaX"]There are other versions of Bertrand's theorem, one of them saying that there is a prime $m<p<2m-2,m>3$.[/quote]

Yep, that would solve the problem. Thank you. By the way, do you have any reference to this version of Bertrand's theorem?

Best regards,

[quote="ZetaX"]I have this: http://mathworld.wolfram.com/BertrandsPostulate.html[/quote]

Again, Thank you very much. I had seen the proof of the original postulate, but not this one.

Best regards,

[quote="djimenez"]
We know that there is a prime $p$ such that $m<p<2m$. If $p\neq m+1$ we are done, but if $p=m+1$ what ca we do, we have the prime twice. Well, we also know that there is a prime $q$ such that $p=m+1<q<2m+2$. Again, if $q\neq2m+1=2p-1$ we are again done, because $q$ is the prime we want, but what happens if $2p-1=q$. The family, I believe, is infinite.
[/quote]

$p=3$ is the only prime $p$ such that the next following prime is $2p-1$.
Anyway to show this you'll need the Bertrand postulate in that powerfull form or you may use some results of the prime numbers distribution.

He meant that 2 is the only number for which there are no other prime numbers smaller than 2p-1 and greater than p.

[quote="djimenez"]A brief programation excercise shows that the following are the pairs $(p 2p-1)$ of odd primes Best regards[/quote]
Well this kind of primes if my mainds doen't fails are so called Siphie Germain Primes (see Last Fermat Theorem Partial Prrof) But I think it's easy in some sense to probe that we need to consider primes $p$ such that $2p\pm 1$ are primes, so not only S.G. Primes.
In fact the problem is equivalent to find $p$ primes such that
\[
2(p+1)(p+2)\cdots (2p-2)
\]
is a perfect square.

B.R.

[quote="RobertuX"][quote="djimenez"]A brief programation excercise shows that the following are the pairs $(p, 2p-1)$ of odd primes Best regards[/quote]
Well this kind of primes if my mainds doen't fails are so called Siphie Germain Primes (see Last Fermat Theorem Partial Prrof) But I think it's easy in some sense to probe that we need to consider primes $p$ such that $2p\pm 1$ are primes, so not only S.G. Primes.
In fact the problem is equivalent to find $p$ primes such that
\[
2(p+1)(p+2)\cdots (2p-2)
\]
is a perfect square.

B.R.[/quote]

No, the Sophie Germain Primes are the other way around, those are pairs of primes $(p,q)$ such that  $q=2p+1$, and in our case we have $q=2p-1$. Actually, if you prove that the intersection (for $p$) of those families is infinite, you will be solving (by a stronger argument) a very old problem in number theory, the existence of infinite number of twin primes.

Best regards,

[quote="djimenez]
No, the Sophie Germain Primes are the other way around, those are pairs of primes $(p,q)$ such that  $q=2p+1$, and in our case we have $q=2p-1$. Actually, if you prove that the intersection (for $p$) of those families is infinite, you will be solving (by a stronger argument) a very old problem in number theory, the existence of infinite number of twin primes.
Best regards,[/quote]
First, I never talk about TWINS primes, and Second, u r right with S.G,. Primes, but I included, I mean if $p$ and $2p+1$ both primes is a case, and another one is $p$ and $2p-1$ both primes. Sorry for the possibility of confusion!

Well,

The reference that I find online for Germain Primes include only the case of $(p,2p+1)$, and not the other. On the other hand, the Twin primes are pairs of primes whose difference is two! for example, 5 and 7, 17 and 19, 71 and 73, etc. There is a conjecture that there are infinitely many of such pairs of prime, but it is not known yet. I heard that a guy like ten years ago proved that there are infinitely many pairs of twin  pairs of a prime and a pseudo-primes. A pseudo-prime is a number that is the product of two different primes, that is $n=pq$ where both, $p$ and $q$ are primes.

Saludos hasta Iberia!

[quote="djimenez"] the Twin primes are pairs of primes whose difference is two! for example, 5 and 7, 17 and 19, 71 and 73, etc. There is a conjecture that there are infinitely many of such pairs of prime, but it is not known yet.Saludos hasta Iberia![/quote]
mmmm, Remember that I am a teacher, and I know a lot respect prime numbers so, OF COURSE, I know what a TWIN prime number is, but What is the relation with this problem? I mean, thx, to explain me what a twin number are but please, we want to solve a CONCRETE problem, so we can talk about some concepts using some private messages but don't confuse people about another concepts which has NOT relation with the problem.

Saludos desde Iberia hacia .... Mexico?
No te molestes, es que estoy algo enojado hoy por aca!


The sequence $1, 2, 1, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2, \ldots$ consists of 1’s separated by blocks of 2’s with n 2’s in the nth block. The sum of the first $1234$ terms of this sequence is

$\text{(A)}\ 1996 \qquad \text{(B)}\ 2419 \qquad \text{(C)}\ 2429 \qquad \text{(D)}\ 2439 \qquad \text{(E)}\ 2449$

[hide="I think this is right"]We approach this problem by counting the number of 1's in the sequence. (Why do we count 1's and not 2's?)

Let $n$ be the $n^{\text{th}}$ 1 in the sequence, and also the $n^{\text{th}}$ 1 in the equation $\frac 12 n(n+1)$. We know that $\frac12 (49)(50)<1234<\frac 12(50)(51)$, there are $49$ 1's in the first $1234$ terms of the sequence. The rest of the terms are all 2's.

Thus, our answer is $2(1234)-49=2419$.[/hide]

[hide]Hmm, I got 2419. In the first $\binom{n}{2} - 1$ terms there are $n-2$ 1's, so in the first $1224 = \binom{50}{2} - 1$ terms there are 48 1's and hence the 48th triangular number times 2's, and the sum is $48 + \binom{49}{2} * 2 = 2400$. In the next 10 terms there is a 1 and the rest of the terms are 2's, so the sum of the ten terms is 19. Therefore the sum of the first 1234 terms is 2419.[/hide]

[quote="jhcreinhardt"]what a coincidence.  i did this amc just yesterday in class.

each "set" of 1s and 2s is sums to a different odd number.  to find the sum, you take the n odd numbers and square it, then subtract one (because the sets add to 3, 5, 7, etc., but 1 is not included).

We then find the greatest triangular number not greater than 1234+1 (the one is added to balance the equation since there is no "1" in the sum of odds) since the "sets" have one more term in each successive set.  By trial and error using n(n+1)/2, we see that the 49th triangular number is 1225.  This leaves 10 more elements, consisting of a 1 and 9 2s, adding up to 19.  We also add the sum of the first 49 odd numbers (49^2=2401), but we subtract 1 to get 2400.

Our final sum is 2400+19=

[b]2419[/b][/quote]Excellent method! I would never have thought of that!  :D

[quote="eclectic"]Weird, I got D. First I calculate the number of whole sequences 12... - (2+(2+(n-1)))*n/2=(n+3)n/2. Obviously, 51*48<2468<52*49. Thus the number is 48. 2468-2448=20 - a sequence of 1 and 19 2's.  S=(3+(3+47*2))*24+19*2+1=2439.[/quote]Eclectic,
The number of terms in $ n$ "blocks" (or sequences, as you call them) is
$ n*(n\plus{}3)/2$, yielding $ 48$ complete blocks, as you correctly pointed out. After that, I think you got your numbers mixed up, and subtracted the wrong pair.
If all $ 1234$ terms had been $ 2$, the sum would be $ 2468$, but $ 49$ of those terms are $ 1's$, so $ 2468$ overestimates the sum by $ 49$. The sum is $ 2468\minus{}49\equal{}2419$. That's one way of calculating it.
Alternatively, forty-eight "blocks" includes $ 48*51\equal{}1224$ terms, which would add up to $ 2448$ if all the terms were $ 2$. Considering there were forty-eight $ 1's$, the forty-eight blocks should add up to $ 2448\minus{}48\equal{}2440$. That sum can also be calculated as $ n*(n\plus{}2)$ for n=48. The next $ 1234\minus{}1224\equal{}10$ terms will include a $ 1$ and nine $ 2's$, adding up to $ 19$ more.

Thanks  KMST,
I really made a stupid mistake - I substracted 2468 from 2448 not 1234 from 1224.

What is the best way to organize your information for problems like these? They are very easy to make dumb mistakes on, and quite annoying. I always leave out a number or am 1 off or something like that. (This is a general question). Although for this one I got it right because they were 10 apart, but many times I don't.

[quote="jhcreinhardt"]This leaves 10 more elements, consisting of a 1 and 9 2s, adding up to 19.  We also add the sum of the first 49 odd numbers (49^2=2401), but we subtract 1 to get 2400.

Our final sum is 2400+19=

[b]2419[/b][/quote]
I don't get this; I thought that all those 10 elements should be 2's since we can group the terms of the sequence as
1,
21,
221,
2221,...

with n elements in the nth group and with a 1 as the last element in each group, and the last complete group is when n=49 which consists of 48 two's and one 1 at the end... so if we continue with n=50 and only compute the first 10 terms, aren't all of those 10 terms 2's?

EDIT: Ah, never, we two grouped the sequence differently; he grouped it as 1,2 ..., 1,2,2 ..., 1,2,2,2 and subtracted 1 from it at the end... so same result  :P

By the way, how to construct a formula for the nth term of the sequence?

[quote="mathwizarddude"]
By the way, how to construct a formula for the nth term of the sequence?[/quote]

Every grouping starts with 1. The lengths of the groupings increase by one: 1,2,3...etc. Therefore, every term that can be written as $ ((N)(N\plus{}1))/2\plus{}1$ is a 1, and everything else is a two.

[quote="4everwise"]The sequence $ 1, 2, 1, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2, \ldots$ consists of 1’s separated by blocks of 2’s with n 2’s in the nth block. The sum of the first $ 1234$ terms of this sequence is

$ \text{(A)}\ 1996 \qquad \text{(B)}\ 2419 \qquad \text{(C)}\ 2429 \qquad \text{(D)}\ 2439 \qquad \text{(E)}\ 2449$[/quote]

Consider $ (1, 2), (1, 2, 2), (1, 2, 2, 2), (1, 2, 2, 2, 2), (1, 2, 2, 2, 2, 2),( 1, 2, \ldots)$ consists of 1’s separated by blocks of 2’s .

Suppose the $ (n\minus{}1)^{th}$ group is $ (1,\underbrace{2,\cdots ,2}_{n\minus{}1})$

Let $ 2\plus{}3\plus{}4\plus{}\cdots\plus{}n\le 1234\implies n(n\plus{}1)\le 2470$

we can get $ 49\times 50$ greatly approach $ 2470$ , i.e. $ n\equal{}49$ and up to $ 48^{th}$ group.

Since $ 1234\minus{}\frac{48\times 49}{2}\equal{}10 .\therefore 49^{th}$ group can only contain $ 1\quad 1$'s and $ 9\quad 2$'s
i.e. Totally $ 49\quad 1$'s is in $ 1234$ terms .
Thus , $ 1234\times 2\minus{}49\equal{}\boxed{2419}$

[quote=SenorIncongito]Can someone please give an easy solution to this problem?[/quote]

There's not always a slick solution to a problem. What question do you have about the given solutions?

[quote=SenorIncongito]I don't understand "Let $ 2+3+4+\cdots+n\le 1234\implies n(n+1)\le 2470$

we can get $ 49\times 50$ greatly approach $ 2470$ , i.e. $ n=49$ and up to $ 48^{th}$ group.

Since $ 1234-\frac{48\times 49}{2}=10 .\therefore 49^{th}$ group can only contain $ 1\quad 1$'s and $ 9\quad 2$'s
i.e. Totally $ 49\quad 1$'s is in $ 1234$ terms .
Thus , $ 1234\times 2-49=\boxed{2419}$" of the solution above. Why do you even have to add 2+3+4 in the first place?And then why the next steps?[/quote]

There are 2 terms in the first block, 3 in the next, and so on.

[quote=SenorIncongito][quote=HumanCalculator9][quote=SenorIncongito]Can someone please give an easy solution to this problem?[/quote]

I have a somewhat slick solution
The sum is at most 2468 (1234*2)
The 1s are at positions 1, 3, 6, 10, ..., so there are 49 1s
Answer: 2468-49 = $\boxed{2419}$[/quote]

Why is the sum at most 2648(1234^2)?[/quote]

each digit is at most 2, and there are 1234 digits, so the maximum sum is 2*1234 = 2468


[i]EDIT[/i]: [b]You can see the proofs of the theorems and problems  [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=721&t=393335]here[/url], in the PDF file (containing more problems).[/b]

 [i]Lifting The Exponent Lemma[/i] is a new topic in number theory and is a  method for solving some exponential Diophantine equations. In this article we  analyze this method and present some of its application.

We can use Lifting The Exponent Lemma (this is a long name, let's call it [b]LTE[/b]! :P) in lots of problems involving exponential equations, specially when we have some prime numbers (and actually  in some cases it "explodes" the problems). This lemma  shows how to find the greatest power of a prime $p$ -which is often $\geq 3$- that
divides $a^n \pm b^n$ for some positive integers $a$ and $b.$ The proofs of theorems and lemmas in this article have nothing difficult and all of them use elementary mathematics. Understanding the theorem’s usage and its meaning is more important  to you than remembering its detailed proof.

For two integers $a$ and $b$ we say $a$ is divisible by $b$ and write $b |a$ if and only if there exits some integer $q$ such that $a = qb.$

We define $\|  x \| _p$ to be the greatest power of a prime $p$ that divides $x$; i.e. if $\|  x \| _p=\alpha$ then $p^{\alpha} \mid x$ but $p^{\alpha+1} \nmid x.$ We also write $p^{\alpha} \|  x, $ if and only if $\|  x \| _p = \alpha.$ So obviously we have $\|xy\|_p=\|x\|_p \|y\|_p$ and $\|x+y\|_p \leq \|x\|_p+\|y\|_p.$

In lots of articles about this subject, we see the $v_p(x)$ sign. To be coherent with these subjects, we can define $v_p$ in terms of $\|  x \| _p,$ So that $\|x\|_p = p^{-v_p(x)}.$

[i]Example. [/i] The greatest power of $3$ that divides $63$ is $3^2.$ because $3^2=9 \mid 63$ but $3^3 =27 \nmid 63.$ i.e. $ 3^2 \|  63$ or $\|  63 \| _3 = 2,$ and we write $v_3(63)=2.$

[i] Example.[/i] Clearly we see that if $p$ and $q$ be two different prime numbers, then $ \|   p^ \alpha q^ \beta \| _p = \alpha,$ or $p^\alpha \|  p^ \alpha q^ \beta,$ and we write $v_p(p^ \alpha q^ \beta)=\alpha.$

[b]First Form of LTE. [/b] Let $x$ and $y$ be (not necessary positive) integers and let $n$ be a positive integer. Given an arbitrary prime $p$ (i.e. we can have $p=2$) such that $ \gcd(n,p) = 1$, $p \mid  x - y$ and neither $x$ nor $y$ is divisible by $p$ (i.e. $p \nmid x$ and $p \nmid y$).  We have
\[
v_p(  x^n - y^n ) = v_p(  x - y ).
\]

[b]Second Form of LTE. [/b] Let $ x,y$ be two integers, $n$ be an odd positive integer, and $p$ be an odd prime such that $ p\mid x + y$ and none of
$x$ and $y$ are divisible by $p.$ then:
\[ v_p( x^n + y^n ) =v_p( x + y ) + v_p( n ) \]

[b]LTE for the case $p=2.$[/b] Let $x$ and $y$ be two odd integers such that $4|x-y.$ Show that
\[ v_2( x^n - y^n ) = v_2( x-y ) + v_2( n ).\]

[b]Theorem.[/b] Let $x$ and $y$ be two odd integers and let $n$ be an even positive integer. Then
\[v_2( x^n - y^n ) = v_2( x - y )+v_2( x + y )+ v_2( n )-1.\]

[b]Some Problems to Solve With This Lemma.[/b]

$\boxed 1$ Find all positive integers $n$ such that there exist $x,y \in \mathbb Z$ with $\gcd(x,y)=1$ and integer $k>1$ for which $3^n=x^k+y^k.$

$\boxed 2$ Find all $n \in \mathbb N$ such that $3^k|2^n-1$ where $k$ is a positive integer.

$\boxed 3$ Let $p$ be a prime. Solve the following equation in positive integers:
\[a^p-1=p^k\]

$\boxed 4$ Solve the equation $(n-1)!+1=n^m$ in $\mathbb N.$

$\boxed 5$ We know that $4(a^n+1)$ is a perfect cube for every $n \in \mathbb N.$ Prove that $a=1.$

$\boxed 6$ Prove that there exist infinitely many positive integers $n$ for which
\[n|1^n+2^n+\cdots+k^n \qquad k\in \mathbb N, k>1\]

[hide="My bad, this is nonsense"]Shouldn't your original lemma 1 be $ \|x^n \minus{} y^n \|_p \equal{} \|x \minus{} y \|_p$ as well?

$ x^n \minus{} y^n \equal{} (x \minus{} y)(x^{n \minus{} 1} \minus{} y^{n \minus{} 1}) \plus{} xy(x^{n \minus{} 2} \minus{} y^{n \minus{} 2})$.
So if $ p \neq 2$, and $ p|n$, then $ (p,n \minus{} 1) \equal{} (p,n \minus{} 2) \equal{} 1$, so the previous result applies. This means that the valuation of $ x \minus{} y$, $ x^{n \minus{} 1} \minus{} y^{n \minus{} 1}$, and $ x^{n \minus{} 2} \minus{} y^{n \minus{} 2}$ are all the same. So you can take out the common factors in $ x \minus{} y$ and $ x^{n \minus{} 2} \minus{} y^{n \minus{} 2}$ and no more, since $ p|x^{n \minus{} 1} \minus{} y^{n \minus{} 1}$ but not $ xy$.[/hide]

This is known as the Lifting the Exponent lemma, see e.g. [url]http://reflections.awesomemath.org/2007_3/Lifting_the_exponent.pdf[/url]

[quote="Soarer"]Shouldn't your original lemma 1 be $ \|x^n \minus{} y^n \|_p \equal{} \|x \minus{} y \|_p$ as well?

$ x^n \minus{} y^n \equal{} (x \minus{} y)(x^{n \minus{} 1} \minus{} y^{n \minus{} 1}) \plus{} xy(x^{n \minus{} 2} \minus{} y^{n \minus{} 2})$.
So if $ p \neq 2$, and $ p|n$, then $ (p,n \minus{} 1) \equal{} (p,n \minus{} 2) \equal{} 1$, so the previous result applies. This means that the valuation of $ x \minus{} y$, $ x^{n \minus{} 1} \minus{} y^{n \minus{} 1}$, and $ x^{n \minus{} 2} \minus{} y^{n \minus{} 2}$ are all the same. So you can take out the common factors in $ x \minus{} y$ and $ x^{n \minus{} 2} \minus{} y^{n \minus{} 2}$ and no more, since $ p|x^{n \minus{} 1} \minus{} y^{n \minus{} 1}$ but not $ xy$.[/quote]

I don't know where is the problem.(Can't understand).
please explain more about it.
you can write  $ x^n \minus{} y^n \equal{} (x \minus{} y)(x^{n \minus{} 1} \minus{} y^{n \minus{} 1}) \plus{} xy(x^{n \minus{} 2} \minus{} y^{n \minus{} 2})$  like this:  $ x^n \minus{} y^n \equal{} k.(x \minus{} y)$ where $ (k,x\minus{}y)\equal{}1$.I think you are exactly saying this.

amparvardi

[quote="amparvardi"]

you can write  $ x^n \minus{} y^n \equal{} (x \minus{} y)(x^{n \minus{} 1} \minus{} y^{n \minus{} 1}) \plus{} xy(x^{n \minus{} 2} \minus{} y^{n \minus{} 2})$

[/quote]

Hi, just ignore that post of mine. I don't know what's wrong with me when I wrote $ x^n \minus{} y^n \equal{} (x \minus{} y)(x^{n \minus{} 1} \minus{} y^{n \minus{} 1}) \plus{} xy(x^{n \minus{} 2} \minus{} y^{n \minus{} 2})$ , which is clearly incorrect.

[quote="ArefS"][hide="@amparvardi"]imo09.blogfa.com ???[/hide][/quote]

Hi Aref,

Yeah, I got it from the blog http://www.imo09.blogfa.com , but it was in Persian, so I translated it to English.

For $5,$if $a^n+1$ has no odd factor,then let $a^n+1=2^{b_n}$ and then from the fact $a^m+b^m=2^n\implies a=b,$it follows that $a=1$
Now let $p$ be an odd prime factor of $a+1,p^x||a+1$ and let's choose $n$ so that $n$ odd.We must have $3|v_p(a^n+1).$then let's choose $n$ such that $p^y||n$ with $3\not| x+y,$then we get a contradiction.

[quote="broniran"]@cnyd Why if $a+b=3^n\Rightarrow k=1$? What if $\frac{a^k+b^k}{a+b}=1$ and $k>1$? How do you cope with this case?[/quote]
How it is possible?
$a^k+b^k>a+b$ for $k>1$

more generally for prime $p,gcd(a,b)=1,$if $p^m=a^n+b^n,$then $p=2,a=b=1$ or $p=3,a=2,b=1,n=3$

Does anybody know of any official reference for this Lemma, that it's not Mathematical Reflections? I mean, something on AMM, Mathematic Magazine, etc. maybe even in some more general terms and with a different name?

This is a really big bump but there seems to be a typo in the definition of $||x||_p$? In particular, [hide=this][quote]We define $\|  x \| _p$ to be the greatest power of a prime $p$ that divides $x$; i.e. if $\|  x \| _p=\alpha$ then $p^{\alpha} \mid x$ but $p^{\alpha+1} \nmid x.$ We also write $p^{\alpha} \|  x, $ if and only if $\|  x \| _p = \alpha.$[/quote][/hide] seems imply that $||x||_p$ is the same thing as $v_p,$ but that's impossible because of [hide=this][quote]In lots of articles about this subject, we see the $v_p(x)$ sign. To be coherent with these subjects, we can define $v_p$ in terms of $\|  x \| _p,$ So that $\|x\|_p = p^{-v_p(x)}.$
[/quote][/hide]

@above: That's true, but note that the notion $\|x\|_p$ does not appear anymore after we have defined $v_p$ so that it doesn't really matter as long as we have the right notion of $v_p$.

Hello everyone! Can you help me with this problem? I am sorry that AoPS doesn't allow me to use LaTeX because I'm new user .
Let p be a prime, a and n are positive integers for which it is true that \frac{p^a-1}{p-1}=2^n. Find \tau(na).
I would like to see solution, that uses [color=#f00]LTE[/color].


What is the least possible product obtained from multiplying three distinct members of the set $ \left\{\minus{}\frac{1}{2}, 9, 5, 12, \frac{1}{3}\right\}$?

We want a negative with $ 2$ positives, so we want $ \minus{} \frac {1}{2}*9*12 \equal{} 9* \minus{} 6 \equal{} \minus{} 54$.


Another method could be(maybe what chess64 was hinting about):-

 $1^3+3^3+5^3+7^3+\cdots+(2n-1)^3=x$

Add $2^3+4^3+...(2n)^3$ to both sides

$(1+2+3+4....+n)^2=x+8\sum{n^3}$

And the rest is easy.


Let $ a,\ b,\ c\ (a\geq b\geq c)$ be relatively  prime natural numbers. Find $ (a,\ b,\ c)$ such that $ (a \plus{} b \plus{} c)\left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right)$ is integer.

Bertram, $ (a\plus{}b\plus{}c)\sum_{cyc}\frac{1}{a}\geq 9$, so the set $ \{6,9,10\}$ is incorrect; it should be $ \{9,10,11\}$.

[quote="JBL"][quote="kunny"]... prime natural numbers relatively....[/quote]
"Relatively prime natural numbers."  ("Relatively prime" acts as a single adjective -- you can't break them apart and maintain the meaning.)[/quote]

Thank you for your advice.

kunny

[quote="bertram"][quote="kunny"]Let $ a,\ b,\ c\ (a\geq b\geq c)$ be prime natural numbers relatively. Find $ (a,\ b,\ c)$ such that $ (a \plus{} b \plus{} c)\left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right)$ is integer.[/quote]

[hide="An attempt"]Obviously $ (a \plus{} b \plus{} c)\left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right) \equal{} 3 \plus{} \frac {ab(a \plus{} b) \plus{} ac(a \plus{} c) \plus{} bc(b \plus{} c)}{abc}$ so we need to find $ a,b,c$ with $ abc|(ab(a \plus{} b) \plus{} ac(a \plus{} c) \plus{} bc(b \plus{} c)\iff a,b,c|(ab(a \plus{} b) \plus{} ac(a \plus{} c) \plus{} bc(b \plus{} c)\iff a|(b \plus{} c),b|(a \plus{} c),c|(a \plus{} b)$
 where we used the fact that $ a,b,c$ are relatively prime. Now if we set $ a \equal{} \max\{a,b,c\}$ we either have $ b \plus{} c \equal{} a$ or
$ b \plus{} c \equal{} 2a\implies a \equal{} b \equal{} c \implies a \equal{} b \equal{} c \equal{} 1$, which is one solution.

Now let's restrict to the case $ b \plus{} c \equal{} a$. By the formulas derived above we have $ b|2a \minus{} b\implies b|2a\implies b|2$ and $ c|2a \minus{} c\implies c|2a\implies c|2$. Then $ b$ and $ c$ cannot both be 2, so we either have $ \{a,b,c\} \equal{} \{2,1,1\}$ or $ \{a,b,c\} \equal{} \{3,2,1\}$. Thi gives a total of three solutions: $ \boxed{\{a,b,c\}\in\{\{1,1,1\},\{1,1,2\},\{1,2,3\}\}}$ with $ (a \plus{} b \plus{} c)\left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right)\in\{9,10,11\}$.[/hide][/quote]

Your solution is correct, bertram. :) but you seem to have forgotten $ a\geq b\geq c.$ Please correct it.

[quote="kunny"]Your solution is correct, bertram. :) but you seem to have forgotten $ a\geq b\geq c.$ Please correct it.[/quote]

Ah, my bad :( As I'm not able to edit my post anymore I will post it again:

[hide="Hopefully complete Solution"]

Obviously $ (a + b + c)\left(\frac {1}{a} + \frac {1}{b} + \frac {1}{c}\right) = 3 + \frac {ab(a + b) + ac(a + c) + bc(b + c)}{abc}$ so we need to find $ a,b,c$ with

\begin{align*} abc|(ab(a + b) + ac(a + c) + bc(b + c) & \iff a,b,c|(ab(a + b) + ac(a + c) + bc(b + c)) \\
& \iff a|(b + c),b|(a + c),c|(a + b) \end{align*}

where we used the fact that $ a,b,c$ are relatively prime. Now  $ a = \max\{a,b,c\}$ and so we either have $ b + c = a$ or
$ b + c = 2a\implies a = b = c \implies a = b = c = 1$, which is one solution.

Now let's restrict to the case $ b + c = a$. By the formulas derived above we have $ b|2a - b\implies b|2a\implies b|2$ and $ c|2a - c\implies c|2a\implies c|2$. Then $ b$ and $ c$ cannot both be 2, so we either have $ (a,b,c) = (2,1,1)$ or $ (a,b,c) = (3,2,1)$. This gives a total of three solutions: $ \boxed{(a,b,c)\in\{(1,1,1),(2,1,1),(3,2,1)\}}$ with $ (a + b + c)\left(\frac {1}{a} + \frac {1}{b} + \frac {1}{c}\right)\in\{9,10,11\}$.[/hide]


function
limit
algebra
domain
complex analysis
complex analysis unsolved

is there a function f, analytic on the closed unit disc such that $f(\frac{1}{n})=\frac{1}{(n+1)}$, $\forall n$ ?

No. Let us show that if $f(z)$ is analytic in some neighborhood of $0$, then
\[f(z) = \frac{z}{1+z}.\]
Consider the function $h(z) = f(z)-\frac{z}{1+z}$. Note that
\[h(1/n) = \frac{1}{n+1}-\frac{1/n}{1+1/n}= 0.\]
Consider the Taylor series for $h(z)$. If $h(z) \not\equiv 0$, then
\[h (z) = a z^{k}+O(z^{k+1}),\]
for some $a\neq 0$, and some natural $k$. Hence,
\[\lim_{n\to \infty}\frac{h(1/n)}{(1/n)^{k}}= \lim_{z\to 0}\frac{h(z)}{z^{k}}= a \neq 0.\]
But the left hand side equals 0. Therefore, $h(z)\equiv 0$.
We conclude that $f(z) = \frac{z}{1+z}$. This function has a singularity at point $-1$, so [b]it is not analytic[/b] on the closed unit disc.

[quote="kouboy"]nice proof yury. how did you think of considering $f(z)=\frac{z}{z+1}$ and showing that if the given fn is analytic in a small disk around 0, it has to be f?[/quote]
Because this is what you get when you set $z=1/n$ . I would just say the difference $f(z)-\frac{z}{1+z}$ is zero on a set which has a point of accumulation in the domain of analyticity and hence they are identically equal by any of the weak uniqueness theorems.

thanks kalle and yury. i should have done this myself  :blush:

[quote="Kalle"] I would just say the difference $f(z)-\frac{z}{1+z}$ is zero on a set which has a point of accumulation in the domain of analyticity and hence they are identically equal by any of the weak uniqueness theorems.[/quote]
I just wanted to prove this theorem.  :roll:


calculus
derivative
trigonometry
calculus computations

Find the tangent line of $ y \equal{}  4 \plus{} \cot x \minus{}2\csc x$
at $ x \equal{} \frac{\pi}{2}$

I would like an elegant solution.

Rewrite the given formula as $ f(x) \equal{} 4 \plus{} \frac {\cos x \minus{} 2}{\sin x}$

We have $ f'(x) \equal{} \minus{} \left(\frac {1 \minus{} 2\cos x}{\sin^2 x}\right)$

We have $ f(\frac {\pi}{2}) \equal{} 2$ and $ f'(\frac {\pi}{2}) \equal{} \minus{} 1$ so that the line that you want is given as follows:

$ y \minus{} 2 \equal{} \minus{} 1(x \minus{} \frac {\pi}{2}) \implies y \equal{} \minus{} x \plus{} 2 \plus{} \frac {\pi}{2}$

[hide=" Alternative solution:"]

Differentiating, $ f'(x) \equal{} \minus{}\csc^{2} x \plus{} 2 \csc x \cot x$

So, $ f'(\frac{\pi}{2})\equal{} \minus{}1$

$ f(\frac{\pi}{2}) \equal{} 4 \plus{} \cot (\frac{\pi}{2}) \minus{} 2 \csc (\frac{\pi}{2}) \equal{} 2$

So the tangent is $ y\equal{}\minus{}x \plus{} c$, where $ y\equal{} 2$ when $ x\equal{} \frac{\pi}{2}$.

Substituting, $ 2 \equal{} \minus{}\frac{\pi}{2} \plus{}c$,

so $ c\equal{} \frac{\pi}{2} \plus{} 2$ and the required equation is

$ y\equal{}\minus{}x \plus{}2 \plus{} \frac{\pi}{2}$.

[/hide]


function
algebra
polynomial
binomial theorem
real analysis
real analysis unsolved

Find all functions $f:R->R$ that verify $f(x+y^n)=f(x)+f(y)^n$.

Is n a natural number? If so for the case of n being even f is identically x. But I guess thats 2 obvious.

Indeed....And now a friend gave me a nice solution to it.

An who is this friend, my friend killer?  ;)

I do not know if killer considers me his friend, but I will post the solution that I told him. Consider k a natural number. We have $ (f(x)+kf(1))^n=( f(x+k))^n=f((x+k)^n)=\sum_{i=0}^{n}_{\frac{n!}{i!(n-i)!}}k^{n-i}f(x^i)}$. Develop the LHS of this equality using again the binomial theorem, consider the equality as a polynomial equality in k, identify coefficients and conclude.
  By the way, here is a nice question: is there an aditive real function which is not liniar and with the property that for any real x there is an integer $n_x>1$ such that $ f(x^{n_x})=(f(x))^{n_x}$?


search
function
advanced fields
advanced fields unsolved

Yes, that theorem is called "search function" after the great mathematician Doa Search :wink:  :roll:

Or to phrase it otherwise: it was posted before very often.


group theory
abstract algebra
superior algebra
superior algebra unsolved

[i]Let $G$ be a cyclic group of order $m$, and let $H$ be a subgroup of $G$. If $a$ is a generator of $G$, then there exists a unique divisor $d$ of $m$ such that $H$ is the cyclic subgroup generated by $a^{d}$, and $H$ has order $m/d$.[/i]

In the proof of this theorem, $S$ is defined as the set of integers $u$ such that $a^{u}\in H$. Then is proved that $S=dZ$, then it states "and so $H$ is the cyclic subgroup generated by $a^{d}$. Since $a^{m}=1 \in H$, we have $m \in S$, and so $d$ is a positive divisor of $m$. It follows that $H$ has order $m/d$."

I didn't understand pretty much all the quoted part of the proof.

Type again, please!  :wink:

Is it lemma 2.15 in ''J. Rotman,An Introduction to the Theory of Groups ''?  :maybe:

No, I got it from a Number Theory book by Nathanson.

[quote="Stewie"] $S$ is defined as the set of integers $u$ such that $a^{u}\in H$. Then is proved that $S=dZ$

[/quote]
Because $S$ is subgroup of $\mathbb{Z}$ and  all subgroup of $\mathbb{Z}$ is kind that.

[quote] then it states "and so $H$ is the cyclic subgroup generated by $a^{d}$.[/quote]
Routine!  :maybe:


calculus
integration
Euler
real analysis
real analysis unsolved

Existence is obvious.

Integrating by parts 'k' times gives :

$I=\int_{0}^{\infty}e^{-x}ln(x)dx = \int_{0}^{\infty}e^{-x}x^k/k! (ln(x) - 1 - 1/2 - ... -1/k)dx = \int_{0}^{\infty}e^{-x}x^k/k! ln(x/k)dx + (ln(k) - 1 - 1/2 - ... -1/k)$

$x = kt$, ... $\int_{0}^{\infty}e^{-x}x^k/k! ln(x/k)dx \leq 1/\sqrt k\rightarrow 0$

$I = -\gamma$, Euler constant


we have sequence of $n$ integers and positive integer $s$, we want to find a the longest subsequence of form $a_p,a_{p+1},..a_q$ such that $a_p+...+a_q \geq s$

Start with p=q=0. Then just keep incrementing q and adding the qth number to the current sum. Whenever you get too high, subtract the pth number and increment p. Whenever you get the right sum, check if its the best so far.

could you tell something more, I don't understand everything, what do you mean ''get too high"??

Consider, say, the sequence:
4 1 1 5 1 1 2 2 1
where you want a sum of 6.
Then you would go through the following (p,q):
(0,0), sum = 4, too small
(0,1), sum = 5, too small
(0,2), sum = 6, this is best so far
(0,3), sum = 11, too big
(1,3), sum = 7, too big
(2,3), sum = 6, worse than before
(2,4), sum = 7, too big
(3,4), sum = 6, worse than before
(3,5), sum = 7, too big
(4,5) sum = 2, too small
(4,6) sum = 4, too small
(4,7) sum = 6, this is best so far
(4,8) sum = 7, too big
(5,8) sum = 6, same as before

So the longest sequence is 1,1,2,2 (or 1,2,2,1).

sorry, I should describe the problem better
we have $s \in \mathbb{Z}_+$ and sequence of $n$ integers (they can be positive and negative), we want to find the longest connected subsequence so it looks like that $a_p,a_{p+1},..a_{q-1},a_q$ and such that $a_{p}+a_{p+1}+..+a_q \geq s$ there is inequality!

Oops, sorry, I completely misread it as equality for some reason. Ignore my above posts.

[quote="JSteinhardt"][quote="Zorro"]anybody interested??
i can tell that the solution is faster than $O(n^2)$:)[/quote]

It's O(N). I think you are mistating the problem though, because the answer should just be the sum of the array. If you have the inequality flipped though, it's still O(N). Just go through the array basically the same way as the guy said who thought that it should be equality...you can do the same thing.[/quote]

You can have negative numbers , so the solution isn't the entire array.


number theory
greatest common divisor
function
algebra
domain
number theory unsolved

For any positive integers m and n, ( f(m),f(n) )=f((m,n))  ((m,n) is the greatest common divisor of m and n). k is the least number for that there exist another two positive integers r and s (r is less than s), satisfy the equation f(k)^2=f(r)f(s), Please prove that: s is divisible by k, k is divisible by r.

I'm truely sorry for my poor English, I hope you can understand me.[/img]

what is $ f$?? you started talking about $ f$ without an intro. do we know anything else about $ f$?

It could be the fibonacci sequence, since
$ \mbox{gcd}(F_mF_n) \equal{} F_{\mbox{gcd}(m,n)}$
for all $ m,n \in \mathbb{Z}_{>0}$.

To 2F:
f is a function whose domain is positive integer.

To 3F:
Fibonacci sequence is only one of these f
For example f(x)=x is also right.

Shouldn't some family of f's be forbidden? For instance f(x)=42 satisfies
(f(m),f(n))=f((m,n))=42
f(13)^2=f(7)f(19)

he stated the problem incorrectly. here's the correct problem: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=323180


:?: How would I Prove this:
The opposite angles of any quadrangle inscribed in a circle sum to two right angles?

Let $ ABCD$ be the quadrilater.
$  m(A)+m(C)=\frac{1}{2}m(arcDCB)+\frac{1}{2}m(arcDAB)=\frac{1}{2}m(circle)=\frac{1}{2}\cdot 360=180$


From 1 to 1998 ,how many natural numbers times 72 become perfect square?

[hide]
$ 72\equal{}3^2.2^3$
So $ 72\times2\times a^2$ is a perfect square

So the number of perfect squares between $ 1$ and $ 999$
$ 31^2\equal{}961$ and $ 32^2\equal{}1024$

So there are $ 31$ such numbers[/hide]


geometry
circumcircle
geometric transformation
geometry unsolved

$ X,Y,Z$ are points on $ BC,AC,AB$ sides of $ ABC$ triangle respectively such that $ XYZ$ is similar to $ ABC$ ($ A$ corresponds to $ X$ etc.). Prove that circumcenter of $ XYZ$ is equidistant to orthocenters of $ ABC$ and $ XYZ$.

$ O$ is the circumcenter and $ H$ the orthocenter of the $ \triangle ABC.$ $ A', B', C'$ are midpoints of the sides $ BC, CA, AB.$ The medial $ \triangle A'B'C' \sim \triangle ABC$ is the pedal triangle of $ O$ with respect to the $ \triangle ABC,$ $ O$ is its orthocenter and the 9-point circle center $ N$ of the $ \triangle ABC,$ the midpoint of $ OH,$ is its circumcenter. Any other $ \triangle XYZ \sim \triangle ABC$ with $ X \in BC, Y \in CA, Z \in AB$ is obtained from the medial $ \triangle A'B'C'$ by spiral similarity with center $ O,$ hence $ O$ is the orthocenter of any $ \triangle XYZ.$ See [url]http://www.mathlinks.ro/viewtopic.php?t=175292[/url].

Suppose a figure $ \mathcal F$ moves in such a way that it remains similar to the original figure $ \mathcal F_0$ and one point $ O \in \mathcal F$ is fixed. If any point $ P \in \mathcal F$ different from $ O$ describes an arbitrary curve $ \Gamma_P,$ then any other point $ Q \in \mathcal F$ different from $ O$ describes a curve $ \Gamma_Q \sim \Gamma_P.$ (Yaglom, Geom. Transformations II. pp 68-69).

For the $ \triangle XYZ \sim \triangle A'B'C',$ the orthocenter $ O$ is fixed and $ X \in BC$ describes a line. Therefore, the circumcenter $ P$ of the $ \triangle XYZ$ also describes a line. This line goes through $ N$ (the original circumcenter position). The right $ \triangle OXA' \sim \triangle OPN$ are similar by SAS, because $ \angle PON \equal{} \angle XOA',$ $ \frac{OP}{ON} \equal{} \frac{OX}{OA'}$ $ \Longrightarrow$ $ PN \perp ON \equiv OH$ $ \Longrightarrow$ $ PO \equal{} PH.$

[quote="yetti"]$ O$ is the circumcenter and $ H$ the orthocenter of the $ \triangle ABC.$ $ A', B', C'$ are midpoints of the sides $ BC, CA, AB.$ The medial $ \triangle A'B'C' \sim \triangle ABC$ is the pedal triangle of $ O$ with respect to the $ \triangle ABC,$ $ O$ is its orthocenter and the 9-point circle center $ N$ of the $ \triangle ABC,$ the midpoint of $ OH,$ is its circumcenter. Any other $ \triangle XYZ \sim \triangle ABC$ with $ X \in BC, Y \in CA, Z \in AB$ is obtained from the medial $ \triangle A'B'C'$ by spiral similarity with center $ O,$ hence $ O$ is the orthocenter of any $ \triangle XYZ.$ See [url]http://www.mathlinks.ro/viewtopic.php?t=175292[/url].

Suppose a figure $ \mathcal F$ moves in such a way that it remains similar to the original figure $ \mathcal F_0$ and one point $ O \in \mathcal F$ is fixed. If any point $ P \in \mathcal F$ different from $ O$ describes an arbitrary curve $ \Gamma_P,$ then any other point $ Q \in \mathcal F$ different from $ O$ describes a curve $ \Gamma_Q \sim \Gamma_P.$ (Yaglom, Geom. Transformations II. pp 68-69).

For the $ \triangle XYZ \sim \triangle A'B'C',$ the orthocenter $ O$ is fixed and $ X \in BC$ describes a line. Therefore, the circumcenter $ P$ of the $ \triangle XYZ$ also describes a line. This line goes through $ N$ (the original circumcenter position). The right $ \triangle OXA' \sim \triangle OPN$ are similar by SAS, because $ \angle PON \equal{} \angle XOA',$ $ \frac {OP}{ON} \equal{} \frac {OX}{OA'}$ $ \Longrightarrow$ $ PN \perp ON \equiv OH$ $ \Longrightarrow$ $ PO \equal{} PH.$[/quote]
Thanks yetti for giving us a nice solution! :)
This problem is from Bulgaria 1999,T5


inequalities
algorithm
Gauss
vector
number theory
greatest common divisor

here is one problem i posted in the inequalities section by mistake:)
but here it is

Q)
An Imp jumps on an infinite chessboard. Moves possible for the Imp are described by two pairs of integers: (a,b) and (c,d) - from square (x,y) the Imp can move to one of the squares: (x+a,y+b), (x-a,y-b), (x+c,y+d), (x-c,y-d). We want to know for which square different from (0,0) to which the Imp can jump from (0,0) (possibly in many moves) the value |x|+|y| is the lowest.

Task
Write a program which

reads from standard input two pairs (a,b) and (c,d) of integers, different from (0,0), describing moves of the Imp,
determines a pair of integers (x,y) different from (0,0), for which the Imp can jump (possibly in many moves) from square (0,0) to square (x,y) and for which the value |x|+|y| is the lowest.
writes out to standard output the value |x|+|y|.
Input
Ten test cases. Each test consists of four numbers a,b,c,d in one line, separated by spaces.
-100000 <= a, b, c, d <= 100000

Output
For every test case your program should write a single line with a number equal the lowest possible value |x|+|y|.

Example
Input:
13 4 17 5
[and 9 test cases more]
Output:
2
[and 9 answers more]

any ideas on this one

There is a famous algorithm by Gauss for solving this (finding shortest vector in a 2d lattice). Its basically the GCD algorithm; subtracting a multiple of one vector from the other to reduce its absolute value.

could u possibly write down the thing in more detail please:).
i wud be very interested in a formal proof or method to solve this as i have been trying this problem for quite some time now

ok i guess i found a document when i searched on google
thanks a lot:)
best regards
vasu


Yes, cause we [i]obviously[/i] have the time to count all our ff posts...

This is my 101st FF post.

And I just subtracted my posts that count from my total number of posts.

So no choice for 500+?

But really, many of my FF posts were legitimate games that had already finished, but were not migrated over to the new "Games" forum.

EDIT: This is my 1024th post in FF.

probably like 50??
not that much once i think about it...  :D

Er, so, how can you count?  If you subtract post count from total posts (in profile), then you count test forum posts also...

You can still get a pretty close estimate, unless you live on the test forum.

No joke-I have 42 posts in the FF Forum.

146th... I think. I posted a lot when I joined ML, but then stopped.

HOW DOES ANYBODY HAVE MORE POSTS NOT IN THE FUN FACTORY
I actaully have the same
buy 5525 or whatever posts not in funny factory and only 174 in
thats unbelievable

This is my 581st post in the Fun Factory. You can use the search function and select the "search in forum" option (right below author option) to find posts in one forum only.

ummm...alot? i've only joined for maybe a couple months, beginning of summer  :|

One of the old timers comes and pwns you all!

I have posted 4957 times here. I remember the good times (late '06 to summer of '07). :)

A lot  :)
The message is too small. Please make the message longer before submitting.


Let $ a,b,c > 0$ and $ abc \equal{} 1$, then
$ \sum\frac{a(1\plus{}c\plus{}a\plus{}ab)\plus{}2bc}{(a\plus{}2b)^2}\plus{}\frac{a^2b\plus{}b^2c\plus{}c^2a}{ab\plus{}bc\plus{}ca}\geq3$

[quote="aadil"]its not true try a=b=c=1[/quote]

when $ a \equal{} b \equal{} c \equal{} 1$, $ LHS \equal{} 3\equal{} RHS$, ineq is true.


advanced fields
advanced fields unsolved

Give an example a covering map $\varphi: X\to \mathbb{T}^{2}$ where $X$ is simply connected.

There's really only one example, and it's worth knowing: $\mathbb{T}^{2}=\mathbb{R}^{2}/\mathbb{Z}^{2}$. The quotient map from $\mathbb{R}^{2}$ to the torus is what you're looking for.


Suppose you are asked to play a game involving a fair coin. You throw it repeatedly until either you get HHT, in which you win 1 dollar and the game ends, or in the case you get HTH, you lose 1 dollar and the game ends, and if neither HHT nor HTH appears, you continue to throw the coin. How much are you willing to pay for the game?

[hide]Either you immediately get $ \text{HHT}$, which happens with $ \frac{1}{8}$ probability, and win $ 1$ dollar. Or you get $ \text{HTH}$ with $ \frac{1}{8}$ probability, and lose $ 1$ dollar. If you get anything else, which happens with $ 1\minus{}2\cdot\frac{1}{8}\equal{}\frac{3}{4}$ probability, you try again.

That is, $ E(\text{winnings})\equal{}\frac{1}{8}\minus{}\frac{1}{8}\plus{}\frac{3}{4}\left(\frac{1}{8}\minus{}\frac{1}{8}\plus{}\cdots\right)$, which is equivalent to multiplying $ \frac{3}{4}$ by itself infinitely many times, which approaches $ 0$.

So you wouldn't want to pay anything, I guess.[/hide]

You've misunderstood the game (or understood it correctly by simplified it improperly): if the first few flips are THTH..., the game ends after the fourth flip.  So, for example, if the first two flips are HH then you're guaranteed to win.

There have been plenty of other threads on this phenomenon -- a recent one was here: http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=472401278&t=238892
(I found this by searching for the words "state diagram."  There's lots more to be found where that came from.)

[hide]Expanding on the above post, if your first flip is H, you have a $ \frac {1}{2}$ probability of another H, which is a guaranteed win (since you will eventually get a T for HHT) and a $ \frac {1}{2}$ probability of T. If you do get a T for your second flip, you have a $ \frac {1}{2}$ probability of losing by getting an H and a $ \frac {1}{2}$ probability of "resetting" the chain by getting a T. All the T's up to the first H can be disregarded because neither HTH nor HHT start with T. The expected value if you stop after three throws (not counting T's at the beginning) is

$ a \equal{} \$1*\frac {1}{2} \minus{} \$1*\left(\frac {1}{2}\right)^2 \equal{} \$0.25$

The expected amount won from the game is

$ a \plus{} \frac {1}{4}\left(a \plus{} \frac {1}{4}\left(a \plus{} \frac {1}{4}\left(a \plus{} \frac {1}{4}\left(\ldots\right)\right)\right)\right)$

$ a \plus{} \frac {1}{4}a \plus{} \frac {1}{4^2}a \plus{} \frac {1}{4^3}a \plus{} \ldots$

$ a\sum_{n \equal{} 0}^{\infty}{\left(\frac {1}{4}\right)^n}$

$ \frac {a}{1 \minus{} 1/4}$

$ \$\frac{1}{3}\equal{}\$0.33\ldots$[/hide]

We can even do this without an infinite series (and this is one of the nice things about state diagrams -- finite computation for some geometric-based infinite series):
[hide]Let $ a$ be the expected value of the game.  With probability $ \frac {1}{2}$ our first flip is T and the expected value in this case is $ a$.  Otherwise, we start with H.  With probability $ \frac {1}{4}$ our first flips are HH and the expected value in this case is $ 1$.  Otherwise, the first two flips are HT and the expected value in this case is $ \frac {a \minus{} 1}{2}$.  In total, this gives us $ a \equal{} \frac {a}{2} \plus{} \frac {1}{4}\left(1 \plus{} \frac {a \minus{} 1}{2}\right)$.  We can solve this easily to see that it also gives $ a \equal{} \frac{1}{3}$.[/hide]

An intuitive way I guess.
[hide="Solution"]For the game to end, an H is required as the first of the triplet that ends it.

If the 2nd one is H, I win one dollar.

If the 2nd one is T, I have a 50% chance of losing one dollar.

Therefore I would win $ \ 2/3$ of all games played, and the other party would win $ \ 1/3$,
hence $ \$$ $ \ 1/3$ is the amount I would be willing to pay. [/hide]


fin all positive integer(x,y,z) such that $\sqrt(2006/(x+y))+\sqrt(2006/(y+z))+\sqrt(2006/(z+x))$

Surely he meant this one.. http://www.mathlinks.ro/Forum/viewtopic.php?t=37234


a, b, and c are the roots of a 3rd degree polynomial P(x) and

1)
 $a+b+c=k$
 $a^2+b^2+c^2=2k$
 $a^3+b^3+c^3=3k$

2)
$P(0)=k-12$

find k.

[hide]Notice that $(x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+ac+bc)x-abc$.
When the P(0), only the constant term matters.
$(a+b+c)^3=a^3+b^3+c^3+3a^2b+3a^2c+3b^2c+3b^2a+3c^2a+3c^2b+6abc$ so we have $(k^3-3k)/3=a^2b+a^2c+b^2c+b^2a+c^2a+c^2b+2abc$.  Now notice that $2k(k)=(a^2+b^2+c^2)(a+b+c)$ so $2k^2=a^3+b^3+c^3+b^2a+c^2a+a^2b+c^2b+a^2c+b^2c$ or $2k^2-3k=b^2a+c^2a+a^2b+c^2b+a^2c+b^2c$
Now we subtract $2k^2-3k$ against $(k^3-3k)/3$ gives $(-k^3+6k^2-6k)/3=-2abc$ so the constant term is $(-k^3+6k^2-6k)/6=-abc$.
Looking at P(0), we have $(-k^3+6k^2-6k)/6-6k/6=-12$ or $-k^3+6k^2-12k=-72$ so we have $k^3-6k^2+12k-72=0$.  Trying the Rational Roots Theorem, we get k=6 as a root.  We factor out (x-6) to get $x^2-12$.  The roots to that are $k=2\sqrt{3}$ or $-2\sqrt{3}$.[/hide]


algebra
polynomial
quadratics
limit
function
algebra unsolved

Complete the square and let $ x$ become large.  (If the quadratic coefficient is not a square, let $ x$ become large.  Alternately, multiply by the quadratic coefficient, complete the square, and let $ x$ become large.)

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=148696]A significant generalization[/url] is well-known.

[quote="t0rajir0u"]Complete the square and let $ x$ become large.  (If the quadratic coefficient is not a square, let $ x$ become large.  Alternately, multiply by the quadratic coefficient, complete the square, and let $ x$ become large.)

[/quote]

Please show in more detail

If the leading coefficient is square, write as $ (ax \minus{} h)^2 \plus{} k$.  As $ x$ becomes large, the closest square to $ (ax \minus{} h)^2 \plus{} k$ is $ (ax \minus{} h)^2$, but the difference between $ (ax \minus{} h)^2$ and the two squares next to it increases without bound while $ k$ stays fixed.  Hence $ k \equal{} 0$.

I'm out of time; I'll post the second half later.

[quote="Carlez Tevos"]Let $ w$ be a quadratic polynomial with integer coefficients. Suppose that for each integer $ x$ the value $ w(x)$ is a perfect square. Prove that $ w$ is the square of a polynomial.[/quote]
$ f(x) \equal{} ax^2 \plus{} bx \plus{} c$ where $ a,b,c\in Z$
Exist an integer $ x_o$ such that $ f(x)$ is increasing on $ [x_o, \plus{} \infty)$
$ \lim_{n \to\infty}\sqrt {f(n \plus{} 1)} \minus{} \sqrt {f(n)} \equal{} \sqrt {a}$
But the sequence $ \{\sqrt {f(n \plus{} 1)} \minus{} \sqrt {f(n)}\}_{n\geq n_o}$ is a positive integer sequence so  for n is large enough then
$ \sqrt {f(n \plus{} 1)} \minus{} \sqrt {f(n)} \equal{} \sqrt {a}$
It follows that $ f(n) \equal{} (\sqrt {a}n \plus{} t)^2$ for n is large enough .
Because $ \deg f \equal{} 2$ and the equality $ f(n) \equal{} (\sqrt {a}n \plus{} t)^2$ hold for infinite n so $ f(x) \equal{} (\sqrt {a}x \plus{} t)^2$ .

[quote="t0rajir0u"]If the leading coefficient is square, write as $ (ax \minus{} h)^2 \plus{} k$.  As $ x$ becomes large, the closest square to $ (ax \minus{} h)^2 \plus{} k$ is $ (ax \minus{} h)^2$, [/quote]

You can't be sure you can write it like this. Maybe $ a$ doesn't divide $ b$

But then we can bound $ f(x)$ between two functions of the form $ (ax \minus{} h)^2$ to derive a contradiction.  Anyway, this is a fairly weak result so there are a lot of different ways to prove it; consider the general result.

First, you could let the quadratic equation be:

$ ax^2 \plus{} bx \plus{} c$

if you sub. x=0, you know c = some $ n^2$

now consider the discriminant

as the equation must equal a perfect square for any value of x

$ b^2 \minus{} 4ac \equal{} 0$

i.e.
$ b^2 \equal{} 4ac$

b = (4ac)^(1/2)

b is an integer and

we know that c = some n^2

therefore:

when a doesn't = c

a = some $ m^2$

therefore we have

$ m^2x^2 \plus{} 2mnx \plus{} n^2$

which is a perfect square

when $ a \equal{} c$

we have

$ n^2x^2 \plus{} 2n^2x \plus{} n^2$

i.e.

$ n^2(x^2 \plus{} 2x \plus{} 1)$

which is a perfect square


geometry
incenter
Euler
geometry proposed

Hello!

Let $ ABC$ be a triangle, $ I$ is the incenter, $ F_{e}$ is the Feuerbach point and $ K\in AF_{e}$ such that
$ F_{e}$ is the midpoint of $ AK$. Prove that $ (KI$ is the interior/exterior bisector of the angle $ \angle BKC$.

Best regards, Petrisor Neagoe :)

Dear Mathlinkers,
1.  let A'B'C' be the medial triangle of ABC; then <BKC = <C'FeB'.
2.  Let DEF be the contact triangle of ABC.
3.  Let A* be the midpoint of AFe ; it is known thet the circle which have A*D for diameter passes through Fe ;
see : http://perso.orange.fr/jl.ayme/   vol. 4 Symetrique de OI… p. 14-15.
4.  According to the Feuerbach's theorem (http://www.mathlinks.ro/Forum/viewtopic.php?t=301201  vol. 1 the Feuerbach's theorem), the Euler circle and the incircle of ABC are tangent at Fe.
5.  From this figure, FeD is the angle bissector (inner or outer) of <C'FeB'….
Sincerely
Jean-Louis


1) Outline a synthesis of 4-nitropyridine.

Yes, the purpose is to start from pyridine.

How abt chichibabin reaction, by protecting the ortho groups, followed by oxidation. :huh:

Complete your explanation with a reaction scheme.

well, you can start by reacting pyridine with an methyl halide to give a quarternary pyridinium salt. which on heating gives ortho methyl substituted pyridine.similarly protect the other ortho pos. now carry out chichibabin reaction foll by oxidation. now, decarboxylate to give req. compound. i dunno if this really works or how practical this method is, but i just thought this was a nice way to protect ortho positions.  :huh:


The postmark deadline for USAMTS Year 19 Round 1 is October 9.  Visit http://www.usamts.org for the problems and submission guidelines.

Hmm, submitted on Monday, no check mark. Should I be worried?

i submitted in late august...
I'm read for more cool problems.  :) When does the next round come out?

[quote="ghjk"]I submitted since Monday but no checkmark.Is there any problem with my submission? :([/quote]

We haven't filed all the papers yet.

What happens if your solutions don't get through?

When I e-mailed my solutions, the server said it was rejected. So I e-mailed it again (first on accident to usamts@usamts.org, then to solutions@usamts.org).

Find another way to send them.  You can fax them or mail them today.

So we can still submit the solutions tomorrow, right?

Considering the fact that it's the 8th today and the due date is the 9th, obviously you can send them in tomorrow.

[quote="ccy"]Hmm...If I added Asymptote images to your solutions with \begin{asymptote}...\end{asymptote}, would I send pdf+tex, or pdf+txt+generated asy+compiled asymptote pdfs?[/quote]

Final PDF is absolutely necessary - that's what we'll print and send to the graders.

If the Asymptote is embedded in the TeX file, just send that.  If your Asymptote is in separate files, and you then include the generated PDFs, then also send the compiled image PDFs for us to include if we publish one of your solutions.

How long after the submission deadline are solutions posted?  Is it a month because they need to grade all the submissions to find exemplary solutions to post, or are there non-student-produced solutions that get posted sooner than that?  Thanks!

The math jam on the 11th has "official" solutions, after which there will be student solutions self-posted later. There won't be official student solutions until everything's been graded.

edited because I am forgetful and ignorant

[quote="solafidefarms"]The math jam on the 11th has "official" solutions, but actually there will be student solutions self-posted on the 10th.[/quote]No:

[b]Do not discuss Round 1 until after the conclusion of the Math Jam on Thursday.[/b]

This is so that we can contact students on Wednesday and/or Thursday AM who made a bona fide attempt to submit their solutions via email or fax on time, but because of some technical glitch (on our end) the solutions didn't get processed properly.  If the fault was indeed on our end, or was something legitimately unforeseen by the student, then we'll give those students a chance to resubmit.

Note that the above paragraph does not cover "I tried to fax at 11:58 p.m. but got a busy signal" or "I tried to email at 11:59 p.m. but my internet went down."  Do not wait until the last possible moment to submit your solutions.

Sorry, but I don't know where to post this.

I am new to USAMTS, and I sent the solutions through Snail Mail.

I also found out today, that we were supposed to write it on A personalized paper, but I wrote it on blank sheets of paper.

Will I get penalized for this?

[quote="DPatrick"][quote="mathgeniuse^ln(x)"]I also found out today, that we were supposed to write it on A personalized paper, but I wrote it on blank sheets of paper.

Will I get penalized for this?[/quote]No, it's no problem.  You won't get penalized.  It helps out our graders if you write your solutions on the solution forms that are on the [url=http://www.usamts.org]USAMTS website[/url].  But it's OK not to use the forms, as long as your username and ID# are clearly marked on [b]every[/b] page, you write on only one side of the page, and each problem starts on a new page.[/quote]

Oh, we had to write our username.

I thought it said write our name on each of the pages.

My username is similar to my name itself, so I wonder if it is possible to do anything with that.

It doesn't matter -- as long as the ID# and something else identifying are on there, we'll figure it out.

Should we be worrying if we don't have a check mark yet?

Sorry if this has already been asked.

[quote="elephantmaster"]Should we be worrying if we don't have a check mark yet?[/quote]If you submitted your paper more than a week ago, you should definitely have the checkmark by now.  If you submitted your paper within the few days before the deadline, then we may not have processed it yet (since a ton of submissions come in right around the deadline).  We hope to have them all done by Monday or Tuesday.


[quote="warut_suk"]Find all positive integers $n$ such that
$x_1x_2...x_n + y_1y_2...y_n \leq \sqrt{x_1^2+y_1^2}\sqrt{x_2^2+y_2^2}...\sqrt{x_n^2+y_n^2}$
for all reals $x_i,y_i$[/quote]
For n=1 it doesn't work,but does work for $n\ge 2.$
Use induction.
It's true for  n=2.
Suppose that it works for all integers $k\leq n$
then $(x_1x_2...x_n + y_1y_2...y_n)\sqrt{x_{n+1}^2+y_{n+1}^2}\ge (x_1x_2...x_nx_{n+1} + y_1y_2...y_ny_{n+1})$ the last inequality is correct because $\sqrt{x_{n+1}^2+y_{n+1}^2}\ge max\{x_{n+1},y_{n+1}\}$
Note that inequality can hold for $y_i=0$.


calculus
integration
real analysis
real analysis unsolved

Prove that if $ x$ is a rational multiple of $ pi$, then $ cos (nx)$ for integer $ n$ takes on only finitely many values, and the values in this sequence repeat after some point..

[quote="pleurestique"]isn't this obvious?[/quote]

that's the best answer to this problem.   :rotfl:

[quote="mathwizarddude"]"Obvious" is the most dangerous term in mathematics; it doesn't help anything I'm afraid...  :roll:[/quote]

it's pretty obvious.  if $ x\equal{} p\pi/q$, then $ 2qx \equal{} 2\pi p$, which is an integral multiple of $ 2\pi$, so ....

Pardon my mathematical brethren; doing math turns you into a bit of a smartass sometimes, I'm afraid.

1) Any periodic sequence takes on only finitely many values.

2) $ \cos(nx)$ is a periodic sequence when $ x$ is a rational multiple of $ \pi$.

#2 is proved using blahblahblah's calculation, which shows that the sequence has period dividing $ 2q$.

I encourage you to prove #1 on your own.  Though it is "obvious" to those who have years of experience with formal proof, it's a good exercise for those who do not.


number theory
least common multiple
logarithms
number theory proposed

Prove that for all $n \in N$ the following equality holds \[(n+1)\left[\binom {0}{n},\binom {1}{n},...,\binom {n}{n}\right]=[1,2,..,n+1],\] where $[\cdot]$ mean the lcm

It isn't hard and I love it. May you will give me the shorter solution than mine. I hope so :)

[i]Edited by Myth[/i]

At least, I see why $RHS\mid LHS$, but I don't know why $LHS\mid RHS$ at the moment :(

Suppose $n+1=p^s\cdot n'$ and $k=[\log_p n]$. We will prove that $p^{k-s}\mid C_n^{p^k-1}$ if $k\geq s$ (it shows that $RHS\mid LHS$).
Indeed, we are to show \[k-s=\sum_{l=1}^k([\frac{n}{p^l}]-[\frac{n-(p^k-1)}{p^l}]-[\frac{p^k-1}{p^l}])=\] \[=\sum_{l=1}^k([\frac{n}{p^l}]-[\frac{n+1}{p^l}]+1).\] Divide the sum into two parts $\sum_{l\leq s}+\sum_{l>s}$ to complete the proof.

I will think about $LHS\mid RHS$.

SORRY :blush:  :blush:  :blush:  :blush:


geometry
AMC
AIME
MATHCOUNTS
AMC 12
AMC 12 B
USA(J)MO
\/closed

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We will host FREE Math Jams to discuss the classes.  The Math Jam schedule can be found [url=http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php]here[/url].  There are Math Jams on Wednesday, May 20; Wednesday, May 27; and Tuesday, June 2.

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Yes, but at different times of the year, they hold different classes.

That's correct -- we start classes at different times throughout the year, but we don't always offer all the classes.

When is the registration deadline for the Advanced Mathcounts Class?

The Advanced MATHCOUNTS/AMC 8, (Fridays, Jun. 5 - Aug. 28) is probably your best choice.

Note how he is not confident in making states.  The Advanced MATHCOUNTS/AMC 8 Class is for those confident they will make states or nationals.  Instead, the MATHCOUNTS/AMC 8 Basics would probably be a better fit, though it may clash timewise with the geometry class.

if i had a 144 amc 10a, 121.5 amc12b, and a 5 aime, would the aime class or the amc12 class be best suited for me?

[quote="saszs"]I have a geometry class taught by my dad on mondays to thursdays at 7PM to 830PM

I am not confident in making states so what other class should I take.

my dad said not to take any introduction classes(he prefers it if he teaches me only)[/quote]

You could take the Algebra 2 or Algebra 3 class.

I wish there would be an intermediate Mathcounts.

For those who can make chapter but not state.

[quote="AIME15"][quote="jonathanchou711"]

I wish there would be an intermediate Mathcounts.

For those who can make chapter but not state.[/quote]

Those should take the MATHCOUNTS/AMC 8 Basics class, unless you're in an extremely tough chapter (cough SCV cough)[/quote]

Is there any chance in the future going to be a Intermediate MATHCOUNTS/AMC8  Class?

The Basics class [i]is[/i] for those who can't make states but can make chapter. Those who can should take the Advanced. There's no need for an Intermediate.

[quote="rrusczyk"]nikeball -- I agree with gauss202; AIME class.

Others: We do not anticipate creating a third level of MATCOUNTS class at this time.[/quote]
it's nikeballa :)
but thanks, i'll consider signing up.
by the way, i'm in eighth grade.

my dad

USAMO qualifer of 1991 says that richard rusczyk's book has a perfect order for teaching but not good enough

(he's really picky)

well though would a 37 in the chapter put me in advanced.


limit
trigonometry
logarithms
Support
calculus
calculus computations

[i][u]Evaluate :[/u][/i]

   $ \lim_{x \ \to \ 0} \frac{tg(tgx))\minus{}sin(sin(x))}{tgx \minus{} sinx }$

  [i] I have a nice proof for that problem  :lol: [/i]

Solution-

$ \lim_{x \ \to \ 0} \frac {tg(tgx)) \minus{} sin(sin(x))}{tgx \minus{} sinx }$

Dividing by $ tgx$ ,we get

$ \lim_{x \ \to \ 0} \frac{\frac{tg(tgx))}{tgx} \minus{} \frac{sin(sin(x))}{\frac{sinx}{cosx}}}{1\minus{}cosx}$

$ \equal{} \lim_{x \ \to \ 0}\frac{1\minus{}cosx}{1\minus{}cosx}$

$ \equal{}1$

[quote="solenoid"]nguyenvuthanhha, what is your solution? Mine, a standard one, is as follows.

Since $ \sin x \equal{} x \minus{} \frac {x^{3}}{6} \plus{} o(x^{3})$ and $ \tan x \equal{} x \plus{} \frac {x^{3}}{3} \plus{} o(x^{3})$ (as $ x\rightarrow 0$), we have:
$ \lim_{x\rightarrow 0}\frac {\tan (\tan x) \minus{} \sin (\sin x)}{\tan x \minus{} \sin x} \equal{} \lim_{x\rightarrow 0}\frac {\tan x \plus{} \frac {\tan ^{3}x}{3} \plus{} o(\tan ^{3}x) \minus{} \sin x \plus{} \frac {\sin ^{3}x}{6} \minus{} o(\sin ^{3}x)}{\tan x \minus{} \sin x} \\
\equal{} \lim_{x\rightarrow 0}\frac {\tan x \minus{} \sin x}{\tan x \minus{} \sin x} \plus{} \lim_{x\rightarrow 0}\frac {o(\tan ^{3}x) \minus{} o(\sin ^{3}x)}{\tan x \minus{} \sin x} \plus{} \lim_{x\rightarrow 0}\frac {\frac {\tan ^{3}x}{3} \plus{} \frac {\sin ^{3}x}{6}}{\tan x \minus{} \sin x} \\
\equal{} 1 \plus{} 0 \plus{} \lim_{x\rightarrow 0}\frac {1}{\tan x(1 \minus{} \cos x)}\left( \frac {\tan ^{3}x }{3} \plus{} \frac {\sin ^{3}x}{6}\right) \\
\equal{} 1 \plus{} \lim_{x\rightarrow 0}\frac {x}{\tan x}\cdot \lim_{x\rightarrow 0}\frac { x^{2}}{1 \minus{} \cos x}\cdot \lim_{x\rightarrow 0}\frac {1}{x^{3}}\left( \frac {\tan ^{3}x}{3} \plus{} \frac {\sin ^{3}x}{6}\right) \\
\equal{} 1 \plus{} 1\cdot 2\cdot \left( \frac {1}{3} \plus{} \frac {1}{6}\right) \equal{} 2.$[/quote]

Nice solution solenoid  :lol:

[quote="Aashay THE ROKR"][quote="bogdanno"][quote="Aashay THE ROKR"]

$ \lim_{x \ \to \ 0} \frac {\frac {tg(tgx))}{tgx} \minus{} \frac {sin(sin(x))}{\frac {sinx}{cosx}}}{1 \minus{} cosx}$

$ \equal{} \lim_{x \ \to \ 0}\frac {1 \minus{} cosx}{1 \minus{} cosx}$

[/quote]

You can't do that[/quote]

But why?[/quote]

when you take the limit on the top, $ cos(x)$ is also modified, so you have to take it into account. You treat the quotient as a whole. Sometimes it works as in $ \lim f\cdot g \equal{} \lim f \cdot \lim g$ (with some conditions) but there's no rule for what you wrote there.

But isn't the following law correct?

$ \lim f\cdot g \equal{} \lim f \cdot \lim g$

I have used the same law.Even Solenoid has used it in his 3rd last step.What is wrong in it?

Composition is not the same thing as multiplication.

Aashay THE ROKR, calculate the limit: $ \lim_{x\rightarrow 0}\frac{1\minus{}\cos x}{x^{2}}$. Which solution is correct and WHY?

(1). According to the Taylor theorem we have $ \cos x \equal{} 1 \minus{} \frac{x^2}{2}\plus{}o(x^2)$. Hence

$ \lim_{x\rightarrow 0}\frac{1\minus{}\cos x}{x^{2}}\equal{}\lim_{x\rightarrow 0}\frac{1\minus{}1\plus{}
\frac{x^{2}}{2}\minus{}o(x^{2})}{x^{2}}\equal{}\lim_{x\rightarrow 0}\frac{\frac{x^{2}}{2x^{2}}\minus{}\frac{o(x^{2})}{x^{2}}}{1}\equal{}\frac{1}{2}\minus{}\lim_{x\rightarrow 0}\frac{o(x^{2})}{x^{2}}\equal{}\frac{1}{2}.$

(2). Since $ \lim_{x\rightarrow 0}\cos x\equal{}1$, we have $ \lim_{x\rightarrow 0}\frac{1\minus{}\cos x}{x^{2}}\equal{}\lim_{x\rightarrow 0}\frac{0}{x^{2}}\equal{}0.$

I'm afraid that your solution of nguyenvuthanhha's problem is similar to (2).

Your approach to 2) is wrong as the denominator is tending to zero,so we cannot substitute zero.Instead we can use L Hospital rule.This problem cannot be compared with nguyenvuthanhha's one.It is quite different.

I have solved the problem in the following way.
\begin{align*} L: & = \lim_{x\to 0}\frac {\tan (\tan x) - \sin (\sin x)}{\tan x - \sin x} = \lim_{x\to 0}\frac {\tan (\tan x) - \sin (\sin x)}{\sin x\left (\frac {1}{\cos x} - 1\right )} \\
& = \lim_{x\to 0}\frac {\tan (\tan x) - \sin (\sin x)}{x(1 - \cos x)} = 2\lim_{x\to 0}\frac {\tan (\tan x) - \sin (\sin x)}{x^3} \\
& = 2\lim_{x\to 0}\frac {\sin (\tan x) - \sin (\sin x)\cdot\cos (\tan x)}{x^3} \\
& = \frac {2}{3}\lim_{x\to 0}\frac {\cos (\tan x) - \cos (\sin x)\cos ^3x\cos (\tan x) + \sin (\sin x)\sin (\tan x)}{x^2} \\
& = \frac {2}{3} + \frac {2}{3}\lim_{x\to 0}\frac {1 - \cos (\sin x)\cos ^3x}{x^2} = \frac {2}{3} + \frac {1}{3}\lim_{x\to 0}\frac {\sin (\sin x)\cos ^2x + 3\cos (\sin x)\sin x}{x} \\
& = \frac {2}{3} + \frac {1}{3}(1 + 3) = \fbox{2}. \end{align*}

[quote="Aashay THE ROKR"]Your approach to 2) is wrong as the denominator is tending to zero,so we cannot substitute zero.[/quote]

Yes, I can't write $ 1$ instead of $ \cos x$. Similarly, in your solution of nguyenvuthanhha's problem, you can't write $ 1$ instead of $ \frac{sin(sin(x))}{\sin x}$.

I just try to explain to you where the mistake is in your solution... - I hope you understand it now.

[quote="Aashay THE ROKR"]Instead we can use L Hospital rule.[/quote]

Yes, but it is easy to give examples of problems in solutions of which it is hard to use l'Hospital rule. Try this one:

$ \lim_{x\rightarrow 0}\frac{\sqrt{1\plus{}2\tan ^{2}(x\sqrt{14})}\minus{}\cos (6x)\minus{}8\sin
^{2}(2x)}{\left( \sqrt[3]{1\plus{}6\sin x}\minus{}\sqrt[7]{1\plus{}7\tan x}\right) ^{3}\ln
\left( 1\plus{}\tan (2x)\right) }$

Okk..Now I am getting some idea on it.So due to $ cosx$ present there,we cannot use

$ \frac{sin(sinx)}{sinx}\equal{}1$

I don't understand why Akshay's solution is wrong. Why can't you split the top as

$ \lim f\cdot g \equal{} \lim f\cdot\lim g$

Since all limits in question exist here, what's the problem? I've read that you can always split like that as long as the limits exist for either term.

For

$ \lim_{x\rightarrow 0}\frac{1\minus{}\cos x}{x^{2}}\equal{}\lim_{x\rightarrow 0}\frac{0}{x^{2}}\equal{}0.$

You cannot split top and bottom as the bottom limit does not exist. I do not see such problems here.

No, it's exactly the same problem as your example- the limits exist, but they lead to a $ \frac{0}{0}$ form. Those manipulations are simply not precise enough.

[quote="Aashay THE ROKR"]Can anyone tell me what is wrong with my solution? :([/quote]

It may seem surprising to you but whenever you apply any standard result you can do it only once in the entire question and it is  in this step you have to apply limits in the entire problem not like you you have done.Once a result is applied the next step must be the answer and if you don't have the answer and suppose you apply some standard result again in the next step or the step after that your answer is bound to be erroneous.I think lot of counter examples have already been given in support of what I have said though people have not been able to pin-point what I have said.