Find $a$ if $a$ and $b$ are integers such that $x^2 - x - 1$ is a factor of $ax^{17} + bx^{16} + 1$.

[hide="Woah!"]Let's work backwards! Let $F(x)=ax^{17}+bx^{16}+1$ and let $P(x)$ be the polynomial such that $P(x)(x^2-x-1)=F(x)$.

First, it's kinda obvious that the constant term of $P(x)$ must be $-1$.  :P Now, we have $(x^2-x-1)(c_1x^{15}+c_2x^{14}+\cdots+c_{15}x-1)$, where $c_{15}$ is some random coefficient. However, since $F(x)$ has no $x$ term, it must be true that $c_{15}=-1$.

Let's find $c_{14}$ now. Notice that all we care about in finding $c_{14}$ is that $(x^2-x-1)(\cdots+c_{14}x^2+x-1)=\text{something}+0x^2+\text{something}$. Therefore, $c_{14}=-2$. Undergoing a similar process, $c_{13}=3$, $c_{12}=-5$, $c_{11}=8$, and we see a nice pattern. :D The coefficients of $P(x)$ are just the Fibonacci Sequence with alternating signs! Therefore, $a=c_1=F_{16}$, where $F_{16}$ denotes the 16th Fibonnaci number and $a=\boxed{987}$.[/hide]

say $r_i$ are the roots of $x^2-x+1$, use newton's sums to find $s_{17}$ and $s_{16}$, by factor theorem we get $a*r_i^{17}+b*r_i^{16}+1=0$ or $a*s_{17}+b*s_{16}+2=0$, then solve in integers

[quote="4everwise"][quote="chess64"][hide="hint"]
One of the many many possible solutions is to just keep substituting $x^2=x+1$ into the equation $ax^{17}+bx^{16}+1$.
[/hide][/quote]Does that turn out nicely? I haven't tried it yet.[/quote]

Yes, and it requires absolutely no creative thinking, except a little bit at the end. :D

[hide="Here is a shorter solution."]
[color=darkred]Denote $F_0(x)=1,\ F_1(x)=x,\ F_{n+2}(x)=F_{n+1}(x)+F_n(x),\ n\in N$

and $F_n=F_n(1),\ n\in N\ .$ ([u]the Fibonacci's sequence[/u]). Therefore,

$x^2-x-1=0\Longrightarrow x^n=F_n(x),\ n\in N\Longrightarrow x^{n+2}=F_{n+1}\cdot x+F_n,\ n\in N\ .$

$0=ax^{17}+bx^{16}+1=a(F_{16}\cdot x+F_{15})+b(F_{15}\cdot x+F_{14})+1\Longrightarrow$

$(aF_{16}+bF_{15})\cdot x+(aF_{15}+bF_{14}+1)=0,\ x\not\in Q\Longrightarrow$

$aF_{16}+bF_{15}=0$ and $aF_{15}+bF_{14}+1=0\Longrightarrow$

$a=F_{15},\ b=-F_{16}\Longrightarrow \boxed {a=987,\ b=-1597}\ .$

I hope didn't made some mistakes ![/color][/hide]

Just start with $x^2-x-1$ and keep strategically adding multiples of $x^2-x-1$ in the form $ax^n(x^2-x-1)$ to cancel out other terms until you have a degree 17 polynomial. It comes out in 2 minutes.

During the process, the fibonacci number pattern becomes obvious, but rather than trusting it completely, I just used it to assist in my addition.

[quote="PenguinIntegral"][quote="chess64"][hide="hint"]
One of the many many possible solutions is to just keep substituting $x^2=x+1$ into the equation $ax^{17}+bx^{16}+1$.
[/hide][/quote]
You, sir, are a genius.[/quote]

Wait are you being sarcastic? Because if you are it actually comes out nicely. $ax(x+1)^8+b(x+1)^8+1$ and then square the thingies $ax(x^2+x+1)^4+b(x^2+x+1)^4+1$ and then substitute again and then just keep doing that.

This problem set me thinking: since the characteristic polynomial of the recursion giving the Fibonacci sequence is $x^2-x-1=0$ and the coefficients of the quotient polynomial are simply the Fibonacci numbers....so can this be generalised to any homogeneous recursion with a quadratic characteristic polynomial?

To be rigorous:
Let a recursion be as follows: $k_1a_n=k_2a_{n-1} + k_3a_{n-2}$ Then the characteristic polynomial is $p(x)=k_1x^2-k_2x-k_3$. Then... following the above mentioned pattern, let q(x) be such that $p(x)q(x)=ax^n+bx^{n-1}+1$ Then q(x) will have its coefficients as terms (with alternating signs) of the aforementioned recursive sequence.

[hide ="Further generalisation?"] If we have a recursion where each term linearly depends on k preceding terms, then the characteristic polynomial would have degree k and if this is multiplied with the polynomial having coefficients as sign alternating terms of the recursion, we would have a polynomial of the form $a_1x^n+a_2x^{n-1}...a_kx^{n-k+1}+1$.[/hide]

For simplicity, let $f(x) =ax^{17} + bx^{16} + 1$ and $g(x) = x^2-x-1$. Notice that the roots of $g(x)$ are also roots of $f(x)$. Let these roots be $u,v$. We get the system\begin{align*} au^{17} + bu^{16} + 1 &= 0, \\ av^{17} + bv^{16} + 1 &= 0. \end{align*}If we multiply the first equation by $v^{16}$ and the second by $u^{16}$ we get\begin{align*} u^{17} v^{16} a + u^{16} v^{16} b + v^{16} &= 0, \\ u^{16} v^{17} a + u^{16} v^{16} b + u^{16} &= 0. \end{align*}Now subtracting, we get\[a(u^{17}v^{16} -u^{16} v^{17}) = u^{16}-v^{16} \implies a = \frac{u^{16} - v^{16}}{u^{17}v^{16} -u^{16} v^{17}}.\]By Vieta's, $uv=-1$ so the denominator becomes $u-v$. By difference of squares and dividing out $u-v$ we get\[a= (u^8+v^8)(u^4+v^4)(u^2+v^2)(u+v).\]A simple exercise of Vieta's gets us $a= \boxed{987}.$


Every one finds an awesome question while practising and gets inspired by that question and finds new ways to solve questions thnx to the method that question involves....What are ur favourite questions?(they can be fun as as well as complicating ;)  :lol: )

One of my favourites is the census-taker queston,the one in which the product of the ages of threee children is 36 and u have to find their individual ages...THat question was nice. :lol:

I enjoy the occaisional extremely difficult problem that prompts one to ask, on reading the solution, "how the heck did they think to do [i]that[/i]?!"  I think this one's a good example:

ARML Individual 1997, #8:
If $\sqrt[3]{\sqrt[3]{2}-1}$ is written as $\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}$, with a, b, and c rational, compute $a+b+c$.

Solution:
[hide]
Let $x=\sqrt[3]{2}$.  Then
\begin{eqnarray*}
(x+1)^3 &=& x^3+3x^2+3x+1 \\
&=& 3x^2+3x+3 \\
&=& 3(x^2+x+1) \\
&=& 3*\frac{x^3-1}{x-1} \\
&=& \frac{3}{x-1}
\end{eqnarray*}

So $x-1=\frac{3}{(x+1)^3}$, and $\sqrt[3]{x-1}=\frac{\sqrt[3]{3}}{x+1}$.  The LHS is exactly what we're looking for, and rationalizing denominators on the RHS gives $a+b+c=\frac{1}{3}$.
[/hide]

My all-time favourite is IMO 2001 problem 2:

\[\frac{a}{\sqrt{a^2 + 8bc}} + \frac{b}{\sqrt{b^2 + 8ca}} + \frac{c}{\sqrt{c^2 + 8ab}} \geq 1.\]

this is one of my favorites!:

given a $n\times n$ board we have $m \leq n$  on each row and on each column. prove there are $n$ pieces one on each row and on each column.

From the 2001 putnam (the best year problem ever).

$a^{b+1}-(a+1)^b=2001$. Find all possible values of $(a,b)$.

I ran out of time, and didn't finish my proof of uniqueness.

this is a good and easy one....posted in getting started....
 What are all integers k for which x^2 + kx + k + 17 = 0?

Why do you like that question that much, shadysaysurspammed?
It doesn't look particularly interesting to me :huh:

my favorite problem is:
find all $n$  s.t. : $2^n+1\equiv 0$ $(\text{mod}$ $n^2)$

Which is IMO 1990 problem 3. :) I think it looks nicer in this form:

Find all $n$ such that \[\frac{2^n + 1}{n^2}\] is an integer.

[quote="shadysaysurspammed"]How'd you do that?are there finite answers?[/quote]

you can prove that the only answer is n=3.

and yes it was an IMO problem

[quote="boxedexe"][quote="Drunken_Math"]My favorite problem is this:

What is the last two digits of

$1+11+111+1111+\cdots+1111\cdots1111$

Where the last number has 100 1's.

I don't know why I like it...  :?[/quote]

Hmmm....this is a problem that fuels my interest.  Could you please tell me how I would solve this problem, or problems of this sort? Thanks in advance.:)

Masoud Zargar[/quote]

multiply the whole thing by $9$. you will have 9+99+999+...+999..99
every number is in form of $10^n-1$~very easy!

How about calculating mod 100? The answer is equivalent to 99*11+1=100*11-10.


Every combination of v=(1,-2,1) and w=(0,1,-1) has components that add to______________. Find c and d so the c[b]v[/b] + d[b]w[/b] = (4,2,-6).

I am reading over the section that seems to be tell me how to but I still am stuck! I have the answers right in front of me though...
[hide]a. components of c[b]v[/b] add to 0
b. components of ever c[b]v[/b] + d[b]w[/b] add to 0
c. choose c=4 and d=10[/hide]

Could someone help me?

This is so easy I suspect you are stuck in the [i]definitions[/i] of vectors, etc ... is that a good book you are working with ?

Well it has lectures to accompany it on OCCW but I was just wondering if someone could help with this because I have a second book which I am looking at too...
I think my REAL problem is the definitions in here:
Every combination of v=(1,-2,1) and w=(0,1,-1) has components that add to______________.
I wouldn't get what they were asking if the vectors were (5,9,0) and (3,4,-4) either. ah th3 alarms going off

A linear combination of two vectors $(a,b,c)$ and $(x,y,z)$ is $k(a,b,c)+l(x,y,z)=(ka+lx,kb+ly,kc+lz)$. You are asked to show that the components of this combination add to $0$, i.e that $ka+lx+kb+ly+kc+lz=0$.

Thanks! Err...sorry about the alarm thing it was a false alarm...
I hope this is all I don't get... :blush:


Suppose all the pairs of a positive integers from a finite collection \[A=\{a_{1}, a_{2}, \cdots \}\] are added together to form a new collection \[A^{*}=\{a_{i}+a_{j}\;\; \vert \; 1 \le i < j \le n \}.\] For example, $A=\{ 2, 3, 4, 7 \}$ would yield $A^{*}=\{ 5, 6, 7, 9, 10, 11 \}$ and $B=\{ 1, 4, 5, 6 \}$ would give $B^{*}=\{ 5, 6, 7, 9, 10, 11 \}$. These examples show that it's possible for different collections $A$ and $B$ to generate the same collections $A^{*}$ and $B^{*}$. Show that if $A^{*}=B^{*}$ for different sets $A$ and $B$, then $|A|=|B|$ and $|A|=|B|$ must be a power of $2$.

Note_to_self_for_bugfix: it should be pairs of different integers.

it's a very nice problem
have you a solution for it ?

I agree that it is a beautiful problem, but I have no solution myself. I also think it is more combinatorics than number theory.

i'll try to find a solution.
my first idea is to use induction in $ k\equal{}MaxA\cup B$any way it was just a first idea,  :blush:

Either [i]collection[/i] means multiset, or the problem is incorrectly stated as for $ A \equal{} \{3,6,7,8,9\}$ and $ B \equal{} \{4,5,6,7,8,9\}$ we have [i]as sets[/i] $ A^* \equal{} B^* \equal{} \{9,10,11,12,13,14,15,16,17\}$.
Assuming collection means multiset, the problem is true. $ A^*$ and $ B^*$ clearly have $ \binom n2$ and $ \binom m2$, respectively, elements, from where $ m \equal{} n$ follows immediately.
I've been trying to recall the solution for this restated problem for a while and here it is:
Let $ A \equal{} \{a_1,\ldots,a_n\}$, $ B \equal{} \{b_1,\ldots,b_n\}$.
Let $ f(x) \equal{} x^{a_1} \plus{} x^{a_2} \plus{} \ldots \plus{} x^{a_n}$ and $ g(x) \equal{} x^{b_1} \plus{} x^{b_2} \plus{} \ldots \plus{} x^{b_n}$. Then $ A^* \equal{} B^*$ means $ f(x)^2 \minus{} f(x^2) \equal{} g(x)^2 \minus{} g(x^2)$.
Note that $ f(1) \equal{} g(1) \equal{} n$.
Denote $ h(x) \equal{} f(x) \minus{} g(x)$ and $ q(x) \equal{} f(x) \plus{} g(x)$. Then $ h(1) \equal{} 0$ and $ q(1) \equal{} 2n$. Then we have $ \boxed{h(x)q(x) \equal{} h(x^2)}$. We have $ h(1) \equal{} 0$. Let $ h^{(l)}(x)$ the $ l$th derivative of $ h$. Since $ h\neq0$, there exists a $ k$ such that $ h^{(k)}(1)\neq0$. Take the smallest such $ k$. Differentiate the 'boxed' relation $ k$ times and take $ x \equal{} 1$. Because $ h^{(i)}(1) \equal{} 0$ for $ 0\le i\le k \minus{} 1$, ignoring the terms involving these derivatives (see the [i]note[/i] below) we get $ h^{(k)}(1)q(1) \equal{} 2^kh^{(k)}(1)$. Because $ h^{(k)}(1)\neq0$, we have $ f(1) \plus{} g(1) \equal{} 2n \equal{} 2^k$, from where $ n$ is a power of $ 2$.

[i]Note:[/i] By an easy induction we have $ (hq)^{(k)} \equal{} \displaystyle\sum_{i \equal{} 0}^k\binom{k}{i}h^{(i)}q^{(k \minus{} i)}$, where $ h,q$ are $ k$ times differentiable functions (in our case, polynomials).

The above solution 'hides' the following idea: $ n \equal{} 2^r$, where $ r$ is the greatest positive integer such that for any $ 0\le i\le r$ holds: $ \displaystyle\sum_{j \equal{} 1}^na_j^i \equal{} \sum_{j \equal{} 1}^nb_j^i$. I'm thinking about this approach.

[u]Edit[/u]: Under these sad circumstances, I dedicate this solution to Mr. Vornicu's [i]Mother[/i], my [i]Mother[/i] and all the [i]Mothers[/i] in the world.
[u]Edit 2[/u]: Edited thanks to [i]othman[/i]'s observation below.

this is a very good solution thanks   :P but $ q(x)\equal{}f(x)\plus{}g(x)$  :lol:

this is a slightly different version of [b]freemind[/b]'s solution:

after it has been established that $ f(x)^2 \minus{} f(x^2) \equal{} g(x)^2 \minus{} g(x^2)$ and $ f(1) \minus{} g(1) \equal{} n \minus{} n \equal{} 0$ we could have concluded that:
\[ f(x) \minus{} g(x) \equal{} (x \minus{} 1)^{k}h(x)
\]
for some $ k\geq 1$ and $ h(x)$, so that $ h(1)\neq 0$.
So
$ f(x) \plus{} g(x) \equal{} \frac {f(x)^2 \minus{} g(x)^2}{f(x) \minus{} g(x)} \equal{} \frac {f(x^2) \minus{} g(x^2)}{f(x) \minus{} g(x)} \equal{} \frac {(x^2 \minus{} 1)^kh(x^2)}{(x \minus{} 1)^kh(x)}$,or
\[ \boxed{f(x) \plus{} g(x) \equal{} (x \plus{} 1)^k\frac {h(x^2)}{h(x)}}
\]
Setting $ n \equal{} 1$ into the boxed equation, we get $ n \equal{} 2^{k \minus{} 1}$

P.S:

To [b]Peter:[/b]

I am more inclined to think that this one belongs to algebra.

Can't there be multiplicity such that $a_j+a_k=a_i+a_l$ for some numbers??


[quote="hunter_cvp_ic"]Let $a,b,c\geq 0$.Prove that:
           $\sum_{cyc}\frac{ab}{3a+4b+5c}\leq \frac{a+b+c}{12}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=89774


[quote]Seeking methods to memorize the first 100 natural numbers[/quote]
I can memorize the first $ 1000000$ natural numbers. Want me to recite them?  :D

Seriously, what is the point of memorizing squares beyond single digits? I can find the value of $ 87^{2}$ if I want to; why clutter my memory with it?

Man, you are a genius.
I am trying to help some average students.
Besides, what is your method of memorizing so many numbers?
Is that the amazing memory method?
The last question you posted really makes sense. :!:
Let me tell you how to use the squares later....
A rich man doesn't know how to use his treasure is such a waste of resources.   :)
******************************************************************************
Let me apologize for my mistake and my rudeness.  :blush:
Too sleepy was me when I typed the second line....
It should be " Seeking methods to memorize the squares of the first 100 natural numbers."

[quote="luimichael"]I am trying to help some average students.[/quote]
And how exactly the memorization of squares is supposed to help them? Just teach them how to multiply integers.

If it was really the idea behind, I would then teach them how to use the calculator for simplicity!  :roll:
Okay, I am talking about some tricks in rapid calculation.
My opinion is that, teaching students some tricks in calculation can arouse their interests in Mathematics and also improve their efficiency and accuracy in calculations.
Of course, some weak students are not able to grasp these methods. Then we don't need to force them to learn these methods. Ask them to follow the traditional ways first...... :)
I am talking about how to do rapid calculations base on the knowledge of the squares of some numbers.

Well, computing the squares of 2-digit numbers without paper can, indeed, help to develop counting skills but why to memorize anything? What is wrong with old good formula $ (10a\plus{}b)^2\equal{}100a^2\plus{}20ab\plus{}b^2$?
If I want the square of 67, say, I would close my eyes and say: "60^2=3600, keep it, 420 times 2 is 840, add to 3600, get 8440; keep it, add 7^2=49, get 8489". On the other hand, adding and subtracting fractions with 1-digit numerators and denominators is a bit easier and also gives you a good practice, so you may want to do it instead :)

You can do a lot with squaring binomials, but perhaps in different ways.  For example, fedja suggested that

$ 67^2 \equal{} (60 \plus{} 7)^2 \equal{} 3600 \plus{} 840 \plus{} 49 \equal{} 4489$

(I assume that is what you meant, anyway :P).  We could also perform this calculation in two other related ways:

$ 67^2 \equal{} (70 \minus{} 3)^2 \equal{} 4900 \minus{} 420 \plus{} 9 \equal{} 4489$
$ 67^2 \equal{} (65 \plus{} 2)^2 \equal{} 4225 \plus{} 260 \plus{} 4 \equal{} 4489$

Perhaps someone might prefer these, especially the second for those who have mastered the squares of numbers ending in $ 5$.

Another cute trick:  If you know the square of a number, finding the square of the next number is not so bad.

$ 68^2 \equal{} 67^2 \plus{} 67 \plus{} 68 \equal{} 4624$

Finally, a general-purpose method for testing one's arithmetic is to test it $ \bmod 10, 3, 8, 11$.  For example, suppose we calculated that

$ 68^2 \equal{} 4634$

How would we quickly determine if we were right?  Well, first of all, $ \bmod 10$ we have

$ 8^2 \equiv 4 \bmod 10$

so the calculation checks out there.  However,

$ 2^2 \equiv 1 \bmod 3$
$ 4634 \equiv 2 \bmod 3$

so we know we have made a mistake.  :)

Let's say that a student has memorized the squares up to $ 25^2$.  Then to square the numbers from 26 to 75, I use the formula
\[ (50 \plus{} n)^2 \equal{} (25 \plus{} n)100 \plus{} n^2.
\]
For example, to square 67, I add 17 to 25 to get 42, then I add 4200 and $ 17^2$ to get 4489, and that's the answer.  To square 43, I subtract 7 from 25 to get 18, then I add 1800 and $ 7^2$ to get 1849, and that's the answer.

To square the numbers from 76 to 100, I use the formula
\[ (100 \minus{} n)^2 \equal{} (100 \minus{} 2n)100 \plus{} n^2.
\]
For example, to square 86, I subtract 14 from 86 to get 72, then I add 7200 and $ 14^2$ to get 7396, and that's the answer.

To square the numbers from 100 to 125, we can use that last formula in the form
\[ (100 \plus{} n)^2 \equal{} (100 \plus{} 2n)100 \plus{} n^2.
\]
For example, to square 112, I add 12 to 112 to get 124, then I add 12400 and $ 12^2$ to get 12544, and that's the answer.

[quote="luimichael"]Suppose you are very familiar with the squares with integers ending in 5:
$ (a \plus{} b)^2 \equal{} a^2 \plus{} 2ab \plus{} b^2$.
In particular, we may think of
$ 26^2 \equal{} 625 \plus{} 50 \plus{} 1 \equal{} 676$;
$ 36^2 \equal{} 1225 \plus{} 70 \plus{} 1 \equal{} 1296$;
$ 46^2 \equal{} 2025 \plus{} 90 \plus{} 1 \equal{} 2116$;
$ 56^2 \equal{} 3025 \plus{} 110 \plus{} 1 \equal{} 4136$;
$ 66^2 \equal{} 4225 \plus{} 130 \plus{} 1 \equal{} 4356$; (something like saying out 3,4,5,6 in a row.)
$ 76^2 \equal{} 5625 \plus{} 150 \plus{} 1 \equal{} 5776$;
$ 86^2 \equal{} 7225 \plus{} 170 \plus{} 1 \equal{} 7396$;
$ 96^2 \equal{} 9025 \plus{} 190 \plus{} 1 \equal{} 9216$.
As a last remark, don't forget that each of these squares end in 6!
******************************************************************************
Squares of 91, 92, 93, ......, 99:
Take a look at the correspondence first:

[size=150]9[color=violet]1[/color]^2 = 8[color=violet]2[/color]81     and $ 81 \equal{} 9^2$ (9 is the complement of 10)
9[color=violet]2[/color]^2 = 8[color=violet]4[/color]64     and $ 64 \equal{} 8^2$ (8 is the complement of 10)
9[color=violet]3[/color]^2 = 8[color=violet]6[/color]49    ...
9[color=violet]4[/color]^2 = 8[color=violet]8[/color]36    ...
95^2 = 9025 (a watershed)
9[color=violet]6[/color]^2 = 9[color=violet]2[/color]14    ...
9[color=violet]7[/color]^2 = 9[color=violet]4[/color]09    ...
9[color=violet]8[/color]^2 = 9[color=violet]6[/color]04    ...
9[color=violet]9[/color]^2 = 9[color=violet]8[/color]01    ...[/size]
******************************************************************************[/quote]
What did you mean by the word "complement?"
I thought $ 9$ was the complement of $ 1$ and $ 8$ was the complement of $ 2$... ($ 9 \plus{} 1 \equal{} 8 \plus{} 2 \equal{} 10$)

Thanks for pointing out my mistake.
You are right. As you mentioned:
1 is the complement of 9 out of 10.
2 is the complement of 8 out of 10.
a and b are said to be complements of each other if a + b = 10, provided that a and b are both natural numbers.
In a more general setting, we may say that a and b are complements if $ a \plus{} b \equal{} 10^n$, provided that a,b and n are natural numbers.
There are many applications of the complements in rapid calculation.
*****************************************************************************
Some other examples about the squares:
Squares of numbers of the form 5*10+a:
$ (5*10\plus{}a)^2 \equal{} 2500\plus{}100a\plus{}a^2\equal{}(25\plus{}a)*100\plus{}a^2$
Therefore,
$ 51^2\equal{}(25\plus{}1)*100\plus{}1 \equal{} 2601$;
$ 52^2\equal{}(25\plus{}2)*100\plus{}4 \equal{} 2704$;
$ 53^2\equal{}2809$ where $ 28 \equal{} 25 \plus{}3$ and $ 9\equal{}3^2$,
......
You may try out other cases and check it with a calculator.
Enjoy!
*****************************************************************************
In fact, we have
$ 36*76\equal{}(3*7\plus{}6)*100\plus{}6^2\equal{}2700\plus{}36\equal{}2736$;
$ 48*68\equal{}(4*6\plus{}8)*100\plus{}8^2\equal{}3200\plus{}64\equal{}3264$;
$ 89*29\equal{}(16\plus{}9)*100\plus{}9^2\equal{}2581$;
....... and so on.
Can you make a generalization?
*****************************************************************************
About the squares of the numbers of the form 10a+8:
$ a^2\equal{}(a\plus{}b)(a\minus{}b)\plus{}b^2$
We may use the identity in this way:
$ 28^2\equal{}30*26\plus{}4\equal{}784$; (a =28 and b = 2)
$ 38^2\equal{}40*36\plus{}4\equal{}1444$;
$ 48^2\equal{}50*46\plus{}4\equal{}2304$;
$ 58^2\equal{}60*56\plus{}4\equal{}3364$;
$ 68^2\equal{}70*66\plus{}4\equal{}4624$;
$ 78^2\equal{}80*76\plus{}4\equal{}6084$;
$ 88^2\equal{}90*86\plus{}4\equal{}7744$;
$ 98^2\equal{}100*96\plus{}4\equal{}9604$.

luimichael, there is a website for these things: [url]http://mathforum.org/k12/mathtips/beatcalc.html[/url]

Thanks a lot!
The information on the above-mentioned webpage is very useful to me.  :)  :)  :)


I need this fore someone....can you help me?:)

$ 1\minus{}\frac{1}{2}\plus{}\frac{1}{3}\minus{}...\plus{}\frac{1}{2n\minus{}1}\minus{}\frac{1}{2n}\equal{}\frac{1}{n\plus{}1}\plus{}\frac{1}{n\plus{}2}\plus{}...\plus{}\frac{1}{2n}$

thx a lot!

Let $ S\equal{}\sum_{k\equal{}1}^{2n}\frac{(\minus{}1)^{k\minus{}1}}{k}, H_n\equal{}\sum_{k\equal{}1}^n\frac{1}{n}$.
Then $ S\plus{}H_n\equal{}H_{2n}$, therefore $ S\equal{}H_{2n}\minus{}H_n\equal{}\sum_{k\equal{}n\plus{}1}^{2n}\frac{1}{k}.$


Let $AA_{1}, BB_{1}, CC_{1}$ be bisectors in the triangle $ABC$ , let $G_{1}, G_{2}, G_{3}$ be the intersection points of medians in the triangles $AB_{1}C_{1}, BC_{1}A_{1}$ and $CA_{1}B_{1}$ respectively. Prove that the straight lines $AG_{1}, BG_{1}, CG_{1}$ intersect in a common point.

$AG_1$ passes through midpoint of $B_1C_1$ and, if $X\in AG_1\cap BC$, we need to find the ratio $\frac{BX}{CX}$.
Let $b, c$ be the intersections of parallels to $BC$ through $B_1, C_1$ respectively; obviously, $B_1b=C_1c\ (\ 1\ )$.
But $\frac{C_1c}{BX}=\frac{AC_1}{AB}$ and $\frac{CX}{B_1b}=\frac{AC}{AB_1}$. Multiply the last relations side by side, keep in mind the relation $(1)$ and get the required ratio.
Find other two relations on $\frac{CY}{AY}, \frac{AZ}{BZ}$ respectively, multiply the three of them and get the result $1$, meaning $AX, BY, CY$ are concurrent.

But the problem is just a special case of the [b]cevian nest theorem[/b].

Best regards,
sunken rock

Dear Mathlinkers,
for the cevain nest theorem, you can see
http://jl.ayme.pagesperso-orange.fr/  vol. 3,  the cevian nest theorem

Sincerely
Jean-Louis


Ten letter "words" are to be formed using at most 5 a's, 4 b's, 4 c's, and 3 d's (and only these four letters). How many different possible "words" can be formed with no a's and at least 3 b's, or with 5 a's and the same number of c's and d's?

how can there be 5 a's and the same number of c's and d's if the number of c + d must be 5?

[quote="t0rajir0u"]The second case also allows some number of $b$s.

[hide="Atlantean hint"] The number of ways to make a word out of $x_{1}$ copies of some letter, $x_{2}$ copies of some other letter, and so forth, is

$\frac{(x_{1}+x_{2}+...+x_{n})!}{x_{1}! x_{2}! x_{3}! ... x_{n}!}$ [/hide][/quote]
Ah, thanks, lemme try that.

Edit: Got it! Thanks =)

Atlantean?

Also, just to be overly pedantic in the service of absolute clarity: that formula gives the number of words using [i]exactly[/i] $x_{1}$ of the first letter and [i]exactly[/i] $x_{2}$ of the second letter, and so on.


calculus
algebra
partial fractions
calculus computations

Not sure if you can post calculus here, but i saw no warning, so apologies if your not. This problem is nagging me, i just cant seem to get it.
Integrate: ${\displaystyle \frac {(x-1)}{(x+1)(x^2+1)}} dx$

Do you know the algebraic technique called partial fractions?

Hmmm, yes i tried that already but couldn't seem to make it work.....

The number of unknown coefficients in a partial fractions setup must equal the degree of the denominator. In this case, that's three. You want to find $A,B,$ and $C$ such that

$\frac {(x-1)}{(x+1)(x^2+1)} = \frac A{x+1}+\frac{Bx+C}{x^2+1}.$

Now why didnt i try that!  :blush: Thanks!


AMC USA(J)MO USAMO quadratics vector calculus geometry

[quote="symmetry"]The National Council of Teachers of Mathematics has established its own definition of highly qualified teachers, which in some ways exceeds the federal mandates. Math teachers, in the NCTMs view, should have: * Completed math coursework equivalent to that required for a major, if they had to teach high school. * Acquired math proficiency equal to a minor in college, if they teach middle school. * Finished at least three college-level courses that emphasize math essential to younger students, if they teach in elementary school. * Expertise in working with students from diverse backgrounds. * An ability to conduct effective lessons and gauge their impact. * A commitment to lifelong professional development and improvement.

What is your view?[/quote]
This should be the criterion for a [b]regular[/b] teacher.

They should have qualified for USAMO (or other national olympiad) ... :D jk, then there would barely be any teachers

[quote="AntonioMainenti"]To be good pre-university math teachers, they should ignore the horrible cirricula they have to teach from.[/quote]

Hear, hear.

And they should keep calculators away from elementary school.

This sounds like Careers in Mathematics. So off it goes.

[quote="white_horse_king88"]They should also delete calculators from Middle School. In fact, I find it horrific when teachers give you problems that need a calculator...

If that was the standard for all subjects, my school would have quite a few less teachers.[/quote]I wouldn't get rid of calculators, just use them less.  I mean, calculators are very useful, and kids need to know how to use them...

They can learn how to use them later - they don't need to know how when they are in 4th grade.  If I had my druthers, I'd wait to hand out the calculators until the student can prove the quadratic formula on her own - it takes a real understanding of algebra to get there.  If you give out the calculators earlier, the students never understand fractions.  And if you can't understand fractions, you'll never understand math.

At a recent board meeting someone suggested integrating Mathematica with teaching.  I think the same thing would happen - put Mathematica in a student's hands too early and his understanding of algebra would never develop.

Calculators and mathematica can be fantastic tools, but if used improperly (as the former definitely is), they are devastating.  It's impossible to fix the calculator problem at this point, I think.  They are far too entrenched and many teachers love them because teaching kids how to push buttons is way easier than teaching them fractions.  Of course, that button pushing is probably perfect training for the jobs they'll have if they don't learn how to think. . . .

Only 2 people in my precalculus class knew how to derive the quadratic formula two days ago ... that's a sad story. Both are in AOPS... :)

Unfortunealty, I dont recall exactly how the quadratic formula is recalled. I know that in the former Soviet Union, they did not use calculators in school, only sometimes they got to use logarithmetic tables, thats what my parents told me, even though that was a while ago, when calculators were not to advanced.

to derive the quadratic formula, you just complete the square of the quadratic equation ax^2 + bx + c = 0.

(1) NCTM guidelines for highly qualified teachers---My view?  I like them.  They also don't differ much from the certification criteria used in some (many?) states.  The real problem is that many teachers are assigned to each outside of their certification area(s).

(2) Use of calculators vs. use of Mathematica in teaching---I can't speak to the use of technology in teaching elementary mathematics.  But at least for precalculus and up, there's a significant difference between using calculators and using Mathematica, as long as no extra front-ends are used with Mma.  My experience is that students view calculators as simple and intuitive, and are then confused when changes in the input syntax produce changes in the output.  In contrast, students seem to realize that Mma does only what you tell it to do; it's interpreting the input syntax. It's much more difficult to use Mma without having some conceptual understanding.  So given the choice (which is not often, because of monetary and space constraints), I prefer to teach using Mathematica.

hem hem hem.....

I kind of agree with Mr. Rusczyk about calculators.

I was in upper class math thing last year, more than half of the guys there were older than me. Anyway, they don't even know how to do distributing properly..

It's kind of sad like white_horse_king88, that I had to suffer reteaching everything my twin sister every basic concept of distributing and solving radicals this summer.

And, even worse, in middle schools they hand out calculators to teach kids how toplug" quote " plug in , basically everything in to come up with solutions....

my point = mr.rusczyk

I teach and advise at a university where probably 50% to 60% of the math majors intend to make a career of teaching at the high school or middle school level, and some of the others who weren't planning to make that their career wind up doing that anyway. I also frequently encounter career-changers and people with college degrees in other subjects who want to become math teachers.

A depressingly large number of the latter group consists of people who have had very little formal instruction in mathematics since high school - the minimum college general education requirements, or perhaps the courses necessary for a business major. Ask one of them, "Why math? Why is that the subject you want to teach?" You'll hear that he or she "always enjoyed" math in school; that he or she did well in it in high school. The cynic might note that one who always enjoyed math might have done more to seek it out during his college years; the cynic might also note that it's well known that there's a shortage of math teachers.

My message that it's going to take three or four years to work through all of the courses and all of the stages of development needed to finish that coursework "equivalent to that required for a major" is not well received. At this point in California, there is an exam that allows candidates to bypass the courses; the exam is probably too easy.

The shortage is real. The people I'm describing do get hired. Those that are completing the math major are being snatched up without going through the controlled and mentored environment of student teaching and being handed "internships" which are essentially just being dumped into a classroom and told to cope with it.

Every year, we get calls, a week before classes start - maybe even a week [i]after[/i] classes start - from schools asking if we know a prospective teacher who is ready to go and able to start [i]right now[/i]. And of course we don't know that; we know people who aren't ready to begin, and people who have already been hired.

There's an institutionalized ritual hazing that goes on in which the newest teachers tend to get the hardest and least desirable jobs. First-year math teachers often teach schedules consisting entirely of 9th grade pre-algebra, or the two-year version of Algebra I, or "consumer math" - classes filled with students whose attitudes are already hardened against mathematics and who expect to do poorly.

Math majors are required to take a considerable variety of courses - vector calculus, linear algebra, differential equations, abstract algebra, and several others. None of those courses are taught in high schools. In fact, most high school teachers don't teach precalculus or calculus - the structure is a pyramid, and the largest number of students are at the lowest level of course. So why is it important to study material that you'll never teach, that is far above the level of any of the courses you're teaching?

Part of it is perspective. Part of it is confidence. Perspective is about seeing the items you're teaching as not being isolated and arbritrary obstacles to be climbed one at a time, but as embedded within the whole structure of mathematics. And confidence? Part of this topic has been about pedagogical methods, particularly about calculators. It takes a [i]confident[/i] teacher to say, "I'm smarter that this calculator, and you can be smarter than that calculator, too. Let's see what we can get by thinking!"

In summary? Teaching is a hard life, a hard job. It drains you, physically, mentally, emotionally. Don't let anyone tell you that it's easy. In particular, it's not easy to do it well. And if you want to teach math, don't do it just becuase you've heard that's where the jobs are. Do it becuase you have are deeply, irrevocably attracted to the subject itself and you want to pass some of that on to another generation.

There's a particular shortage of teachers with any real problem-solving background - the kind of background that brings the members of AoPS together. It would be nice to have more of you take up the teaching profession.


A box is one-third full of marbles. If one-fourth of the
marbles are removed, what percent of the box is filled with
marbles?

Since $ \frac14$ are removed, we have $ \frac13(1\minus{}\frac14)\equal{}\frac13\times{\frac34}\equal{}\boxed{\frac14}$.

$ \frac13\cdot(1 \minus{} \frac14)\equal{}\frac14$
it wants percent so
25

Thanks for finishing it off (I missed the part about it having to be a percent).


Euler
projective geometry
geometry proposed
geometry

Let $ ABC$ be a triangle.$ X;Y$ are two points on $ AC;AB$,respectively.$ CY$ cuts $ BX$ at $ Z$ and $ AZ$ cut $ XY$ at $ H$$ (AZ \perp XY)$
$ BHXC$ is a quadrilateral inscribed in a circle.
Prove that $ XB\equal{}XC$

Problem is very easy.Here is my solution.
Proof:
Denote $ T\in XY\cap BC$,$ I\in CH\cap AB$ ,hence division $ (T,Y,H,X)$ is harmonic.Now consider pencil $ C(T,X,H,Z)\cap AB\equal{}(B,Y,I,A)\equal{}\minus{}1$,now using $ \angle AHY\equal{}90^o$,we obtain that
$ \angle IHY\equal{}\angle YHB\Rightarrow \angle XBC\equal{}\angle XHC\equal{}\angle IHY\equal{}\angle YHB\equal{}\angle XCB$,thus $ XB\equal{}XC$.

[quote="Erken"] [color=darkred]Problem is very easy. Here is my solution. Denote $ \{\begin{array}{c} T\in XY\cap BC \\
\ I\in CH\cap AB\end{array}$ . Hence the division $ (T,Y,H,X)$ is harmonically. Now consider the pencil $ C(T,X,H,Z)\cap AB =$ $ (B,Y,I,A) = - 1$ . Now, using $ m(\widehat {AHY} = 90^{\circ}$ we obtain that $ \widehat {IHY}\equiv\widehat {YHB}$ . If the quadrilateral $ BHXC$ is cyclically, then $ \widehat {XBC}\equiv\widehat {XHC}\equiv\widehat {IHY}\equiv\widehat {YHB}\equiv\widehat {XCB}$ . In conclusion, $ XB = XC$ . [/color][/quote]

[color=darkblue]Very nice the your proof, Erken !
This problem is no hardly, but is no ... [b]very[/b] easily !
[b]This problem is only nicely ![/b] [/color]

lemma:Let $ ABC$ be a triangle and $ D,E,F$ are points on $ BC,CA,AB$ such  $ AX,BY,CZ$ meet on point. $ X$ is projection of $ D$ on $ EF$.then $ \angle BXD\equal{}\angle DXC$
by lemma, $ \angle BHZ\equal{}\angle ZHC$$ \equal{}>$ $ \angle YHB\equal{}\angle CHX$. Since $ \angle YHB\equal{}\angle XCB,\angle CHX\equal{}\angle CBX$, we are done

To Leonhard Euler:
Please correct your proof and by the way your lemma follows from the property of harmonic division and our solutions are very similar. :wink:
To Virgil Nicula:
Thank you!Yes,the problem is nice,but i can't tell that it is not easy.

AH cut BC =E,XY cut BC =F
(E,F,B,C)=-1 such that <EHF=90
HE is besector ange BHC
<XBC=XCB(prove)


A man born in the first half of the nineteenth century was $x$ years old in the year $x^2$. In what year was he born?

Since we are looking for a man orn in the 1800s, we need the first two digits of the square to be $18$. The only one is $43^2=1849$

If the man is $43$ in $1849$, he is was born in $1849-43= 1806$

Yeah, I just checked and it was from the 1954 ASHME.

Yes same answer,
Only $x^2=1849$ fits, so
the answer is $1849-43$,
which is,
$\boxed{1806}$

P.S: Cool avatar 4everwise! :D

[quote="Silverfalcon"]Wasn't this one of AHSME problems?

I think I remember seeing this in Parting Shots in AoPS Volume I.[/quote]
Yeah that's where i got it from.


Let $y_{1}, y_{2}, y_{3},\ldots$ be a sequence such that $y_{1}=1$ and, for $k>0,$ is defined by the relationship:
\[y_{2k}=\begin{cases}2y_{k}& \text{if}~k~ \text{is even}\\ 2y_{k}+1 & \text{if}~k~ \text{is odd}\end{cases}\]\[y_{2k+1}=\begin{cases}2y_{k}& \text{if}~k~ \text{is odd}\\ 2y_{k}+1 & \text{if}~k~ \text{is even}\end{cases}\]Show that the sequence takes on every positive integer value exactly once.

Who's up for some binary!?!

[hide="Dont click"]Not me[/hide]

Nice Problem.

Solution
[hide]
Consider this situation, where:
$y_{1}=1$
$y_{2k}=2y_{k}\text{ if k is even \& }2y_{k}\text{ if k is odd}$
$y_{2k+1}=2y_{k}+1 \text{ if k is even \& }2y_{k}+1\text{ if k is odd}$

The sequence of integers would look like this:
$y_{1}=1,y_{2}=2,y_{3}=3,....y_{k}=k$
including every integer in order.

Now consider this one:
$y_{1}=1$
$y_{2k}=2y_{k}\text{ if k is even \& }2y_{k}+1\text{ if k is odd}$
$y_{2k+1}=2y_{k}+1 \text{ if k is even \& }2y_{k}\text{ if k is odd}$

All what happened was the cases were switched, but all the numbers are included. It could be further explained but quite apparent and i don't feel like it  :) [/hide]


Some topics may have already been created to look at likely AIME scores.  This poll has more options, and since people shouldn't post until they get their official score, this should be a more accurate representation of the AIME I.

[quote="Zmastr"]I think they're coming out this week.  The AMC website said results would be mailed ~10 days after the test.[/quote]

Ill go ask my teacher tomorrow...
I think we shouldve already gotten

I asked him today, and he hadn't gotten it. Eh.

hmm.. seems that the aime results have been distributed at different times?
anyways i haven't gotten mine...
cound amc director clarify?

Forum will be locked today, tomorrow, and part of Thursday morning anyway.

See, I heard that official scores wouldn't come until after AIME II?
We're on spring break anyway, so it doesn't make much of a difference.

EDIT: hey i was actually right.

AIME results for AIME I and AIME II will be mailed out starting approximately April 11.  This is because we include a list of USAMO qualifiers along with the results, the answer pamphlet and the certificates of participation.  See the AIME Teachers' Manual, Section V, page 5.

Steve Dunbar
MAA Director, American Mathematics Competitions

aww dang, gotta wait another 2 weeks...(anxiety building..)

oh well, im  going to be on spring break so i probalby wont even think about it that much..


I've taken a course in world history (but it didn't cover Europe), and I am planning to take the SAT II World History test this December. Does anyone know which (Barrons, Princeton Review, etc) test prep book is the most helpful, or have any other tips to help me? Thank you!

Since you have a good six months until the test, I'd recommend getting your old textbook and reading it over. Nothing beats the good ol' textbook. I don't know much about test prep books for World History, so I can't give much more advice.  :(  Good luck though!  ;)

Thank you for replying! :P

I've already turned in my textbook, but I have outlines for all the chapters (except Europe and Soviet Union). So I think I'll wait until June to take it, because Europe is a big part of the test, supposedly.

Wait a second. What course in World History doesn't cover Europe? Europe is basically the embodiment of the world powers up till the 19th and 20th century! The Soviet Union is important, too! Especially in the 20th century, the Soviets were considered the source of evil in the world, which indirectly started the Cold War! Wow, that's a lot of cirriculum your course didn't cover. I'd say your notes aren't sufficient, so you should try to borrow a textbook from the library to supplement those notes.

I wouldn't wait until June to take it. Although that gives you a year to prepare, history (in my opinion) is one of the easiest tests, since it is basically fact-retention.

My course has 3 parts, The 3rd part (next year) covers Europe and the Soviet Union. :/

Oh I see. So this course would take you 3 years to complete?  :huh:

Well, actually 2 years. The first two parts (everything except SU and Europe) are in year one, then the second (Soviet Union, Europe) can be stretched out over a year, or done in one semester. I'm doing one year.

That's cool. With a year of emphasis on the Soviet Union and Europe, you'll have no trouble. Probably all you have to worry about is your test date. Around halfway into the course, try to pick up a few test prep books and try the sample tests. If you can do well, sign up as soon as possible. If there's quite a lot of material that you haven't learned yet, aim for the June test. Other than that, I think you're set.  :D

I dont know why anyone would take that for their SAT2 there are tons of better tests to take

[i]edited to tone down the language--moderator[/i]

[quote="⌡accidental_freak"]I dont know why the f**k anyone would take that for their SAT2 there are tons of better tests to take[/quote]
1. I dont know why the f**k youre using the word f**k.
2. top of the top schools want you to take SATIIs. I cannot think of "tons of better tests" other than SAT I (and/or ACT) for college prep.


geometry
geometric transformation
reflection
3D geometry

A point source of light is placed on the axis of a cone with semi-vertical angle $ \alpha$, height $ h$ and radius $ r$. The inside surface of he cone is silvered.
Determine the minimum value of $ \alpha$ which will ensure all light rays emitted by a point source $ S$ on the axis of the cone be reflected from the conical surface not more than once.

[hide="Hint:"]Use property of reflection, $ i \equal{} r$.
[/hide]

I got the solution after a great struggle!

[hide="Solution:"]
The exterme case of reflection will be at the vertex of cone. Let semi-vertical angle be $ \theta$. The incident ray falls along the axis with incidence angle $ i$ and is reflected with same angle.

The reflected ray must go along the second surface after reflection from the extreme end of one surface.

$ \color{red} \therefore 2i \equal{} \theta \\
 \\
\theta \equal{} \frac {\pi}{2} \minus{} i \\
 \\
\color{black} \therefore 3i \equal{} \frac {\pi}{2} \\
 \\
\theta \equal{} \frac {\pi}{6} \equal{} 90^o$

Anyone with a different idea?[/hide]


Sam the Sham is an unfair shoe salesman. He has a pair of shoes that was priced at $ \$60$ last week. To entice customers, he wants to put a ``$ 25\%$ off'' tag on the shoes but he still wants the customer to pay $ \$60$ for the shoes. He raises the price of the shoes so that after the $ 25\%$ discount the shoes will cost $ \$60$. In dollars, what must the new price of the shoes be before the discount?


We say x E positive Integers (Z+) is a cube, if y E Z+such that x = y^3

a, b E Z+ such that gcd(a,b) = 1.  Prove the following assertion:

if c^3 = ab, for some c E Z+, then a and b are cubes

Just write:

$c = \sqrt[3]{ab}$

And use the fact that $a, b, c \in \mathbb{Z}^+$.

[quote="shobber"]should this be in the Round Table? :?[/quote]

No...

[hide]

Let $a=p_1^{\alpha_1}p_2^{\alpha_2}...p_n^{\alpha_n}$ and $b=q_1^{\beta_1}q_2^{\beta_2}...q_m^{\beta_m}$ where all the $p$'s and $q$'s are distinct because $(a,b)=1$.

Suppose one of $a,b$ is not a cube. WLOG, let it be $a$. Then for some $1 \le i \le n$, $\alpha_i$ is not divisible by $3$.

Now consider the product $ab = p_1^{\alpha_1}p_2^{\alpha_2}...p_n^{\alpha_n} q_1^{\beta_1}q_2^{\beta_2}...q_m^{\beta_m} = c^3$.

But since $c^3$ is, well, a cube, all of the $\alpha$'s and $\beta$'s have to be divisible by $3$, contradicting the assumption.[/hide]


The sum of the digits of a two-digit number is subtracted from the number. The units digit of the result is $ 6$. How many two-digit numbers have this property?

$ \textbf{(A)}\ 5\qquad
\textbf{(B)}\ 7\qquad
\textbf{(C)}\ 9\qquad
\textbf{(D)}\ 10\qquad
\textbf{(E)}\ 19$

[hide="Solution"]$ N\equal{}\overline{ab}\equal{}10a\plus{}b$
$ 10a\plus{}b\minus{}a\minus{}b\equal{}9a\equiv 6\pmod{10}$.
$ 0<a\leq 9$ and $ 9\equiv \minus{}1\pmod{10}$.
$ \minus{}a\equiv 6\implies a\equiv \minus{}6\equiv 4\pmod{10}$.
$ a\equal{}4$.
$ b\in[0,9]$.
So there are $ \boxed{\textbf{(D)}\ 10}$ solutions.[/hide]

Another way to look at it:

the number is $ AB$. The number minus the sum would be: $ 10A\plus{}B\minus{}(A\plus{}B)\equal{}9A$. Now 9A just has to be a two digit number. So we have that if $ a\equal{}2$, its two digits... all the way to $ a\equal{}11$. you then have $ 11\minus{}2\plus{}1\equal{}10$

[quote="Nandubetax"]Another way to look at it:

the number is $ AB$. The number minus the sum would be: $ 10A \plus{} B \minus{} (A \plus{} B) \equal{} 9A$. Now 9A just has to be a two digit number. So we have that if $ a \equal{} 2$, its two digits... all the way to $ a \equal{} 11$. you then have $ 11 \minus{} 2 \plus{} 1 \equal{} 10$[/quote]

This doesn't work.  AB is a two-digit number.  9A doesn't necessarily have to be, the problem just says the units digit of 9a has to be 6.  The only way to make the units digit of 9a equal 6 is if a = 4. Also A is a digit so A can't go up to 11..

[hide="Solution"]Let the number be $ N.$  Let the digits be $ a$ and $ b$ respectively.

$ N\equal{}10a\plus{}b$
$ N\minus{}a\minus{}b\equal{}10a\plus{}b\minus{}a\minus{}b \Rightarrow 9a$

9a has to have a units digit of 6 where $ a \in [0, 9].$
$ 9a \equiv 6 \pmod{10}$
The only $ a \in [0, 9]$ that works is 4, so $ a\equal{}4.$

Now, $ b$ can be any digit $ 0 \leq b \leq 9$, and $ 0, 1, 2 \dots 9$ is 10 numbers, so correspondingly there are $ \boxed{\textbf{(D)}\ 10}$ solutions.[/hide]

[quote="AdithyaGanesh"][hide="Solution"]Let the number be $ N.$  Let the digits be $ a$ and $ b$ respectively.

$ N \equal{} 10a \plus{} b$
$ N \minus{} a \minus{} b \equal{} 10a \plus{} b \minus{} a \minus{} b \Rightarrow 9a$

9a has to have a units digit of 6 where $ a \in [0, 9].$
$ 9a \equiv 6 \pmod{10}$
The only $ a \in [0, 9]$ that works is 4, so $ a \equal{} 4.$

Now, $ b$ can be any digit $ 0 \leq b \leq 9$, and $ 0, 1, 2 \dots 9$ is 10 numbers, so correspondingly there are $ \boxed{\textbf{(D)}\ 10}$ solutions.[/hide][/quote]

Isn't that pretty much exactly the same as the solution I posted about a year ago?

Sorry, but can someone please show me how the division you did in mods is justified? It's obvious that it works but I'm a little bit nervous to do this kind of division.

They didn't do any division in mods..  they just went through and checked which values of A yielded 6 as the units digit of 9a.


vector
real analysis
real analysis unsolved

Suppose that $B(z) = b_{0}+b_{1}z+b_{2}z^{2}+\cdots$ is a square summable power series, i.e.
$\sum_{n=0}^\infty |b_{n}|^{2}< \infty.$

Suppose further that multiplication by $B(z)$ is an isometry on the space of square summable power series, i.e., for any $A(z)$ which is also a square summable power series, $C(z) = A(z)B(z)$ is also a square summable power series and $\|C\| = \|A\|$. (The norm is $\|A\|^{2}= \langle A(z), A(z)\rangle$; this inner product is defined below.)

Prove that $B(z), zB(z), z^{2}B(z), \ldots$ form an orthonormal sequence in the space of square summable power series, i.e.
$\sum_{n=0}^\infty |b_{n}|^{2}= 1$
and
$\langle z^{j}B(z), z^{k}B(z)\rangle = 0$ iff $j \ne k$

(The inner product is what you'd expect: $\langle A(z), B(z) \rangle = \sum_{n=0}^\infty a_{n}\overline{b_{n}}$)

There is nothing special about the space of square summable power series. If $e_{1}, e_{2}, e_{3}, \dots$ is an orthonormal sequence in an inner product space, and $B$ is an isometry, then the sequence $Be_{1}, Be_{2}, Be_{3}, \dots$ is also an orthonormal sequence.

In this problem, $e_{k}= x^{k}$ (clearly, vectors $x^{k}$ are orthogonal unit vectors).


quadratics
MATHCOUNTS
Pythagorean Theorem
geometry
algebra
system of equations

Suppose there is a right triangle with leg lengths a and b and a hypotenuse of 122.  We know that the product of the 2 legs is 2640.  Find the sum of the 2 legs.

[hide]Hint: Messing with the Pythagorean Theorem might help.[/hide]

[hide]given that ab=2640, we can easily see that a^2+b^2=122^2.  If we add 2ab to this equation we get (a+b)^2=122^2+2640.
So a+b=sqrt(17524)... don't wanna simplify[/hide]

Oh ya, thx Moldy, wasn't thinking straight  :P  :roll:

one could further extend this problem and try to find the length of the legs, using basically the same properties used to find the sum, as well as a system of equations

Extending on that, we have to solve the quadratic:
$x^{2}-142x+2640=0$

[hide]Using that equation, a=120 and b=22[/hide]

wow!  I doubt anyone would ever need to know that quadratic  :lol:

[quote="alanchou"]Suppose there is a right triangle with leg lengths a and b and a hypotenuse of 122.  We know that the product of the 2 legs is 2640.  Find the sum of the 2 legs.[/quote]
Since everything they gave you is an integer, you might also consider Pythagorean Triplets.
[hide]The final answer ended up being a 22-120-122 triangle. Divide by 2 and we have 11-60-61. This turns out to be a member of the Pythagorean Triplet sequence:
$\begin{array}{ccccc}3&-&4&-&5 \\ 5&-&12&-&13 \\ 7&-&24&-&25 \\ 9&-&40&-&41 \\ 11&-&60&-&61 \\ &&\cdots&& \end{array}$
Pythagorean Triplets come up all the time in MATHCOUNTS (and other contest math).  If you are doing MATHCOUNTS, this is why it's important to memorize some Pythagorean Triplets, and to think about when they might be applicable.[/hide]


What is the probability that a fair coin will land on heads [u]exactly[/u] $51$ times out of $100$? $N+1$ times out of $2N$?

[hide]
$C_{n+1}^{2n}$ distinct ways to accomplish this
$2^{2n}=4^{n}$ distinct outcomes of $2n$ coin flips

$P(\text{n+1 heads})=\frac{C_{n+1}^{2n}}{4^{n}}$[/hide]


Two mathematicians were each born in the last 500 years. Each lived to be 100 years old, then died. Each mathematician is equally likely to be born at any point during those 500 years. What is the probability that they were contemporaries for any length of time?

We can illustrate the probability on a Cartesian plane.  Let $ x$ equal the first mathematician and let $ y$ equal the second.

[asy]draw((0,0)--(0,6),EndArrow);
draw((0,0)--(6,0),EndArrow);
label("$500$",(0,5),W);
label("$500$",(5,0),S);
label("$100$",(1,0),S);
label("$100$",(0,1),W);
draw((0,0)--(1,0)--(5,4)--(5,5)--cycle, red);[/asy]

This best illustrates the probability.  Now, we must find the area of that trapezoid.

To find the area, we must find the bases and the height.

The long base is $ \sqrt{500^2 \plus{} 500^2}$, or $ 500\sqrt2$, while the short base is $ \sqrt{400^2 \plus{} 400^2}$, or $ 400\sqrt2$.

Thus, we have a triangle to solve with the Pythagorean theorem.

[asy]draw((0,0)--(sqrt(5000),0)--(0,sqrt(5000))--cycle);
label("$50\sqrt2$",(sqrt(1250),0),S);
label("$100$",(sqrt(1250),sqrt(1250)),NE);[/asy]

The height is $ 50\sqrt2$.

The area of the trapezoid is $ \frac12 \times b_1b_2 \times h$.

Thus, the area is $ 450\sqrt2 \times 50\sqrt2$, or $ 45000$.

The area of the whole square is $ 250000$.  Thus, the answer is $ \frac{45000}{250000}$, or $ \boxed{\frac{9}{50}}$.

Can anyone clarify?

@dipenm It means they lived at the same time

[hide=the answer is]$\frac{9}{25}$[/hide]

[hide=hint]make a graph[/hide]

This is basically the same problem as [url=https://artofproblemsolving.com/community/c63h292724p1582202]this[/url].
My answer to it is [url=https://artofproblemsolving.com/community/c63h292724p21886441]this[/url].

[quote=vectorfunction][quote=Warbunny]Ernie, you're answer was incorrect but nice try :(
Here's the work![hide]Ernie had the right idea but his graph was incorrect. I made a 5x5 graph where each interval was a factor of 100 years. My graph was the same as Ernie's but he forgot that y would be 100(just dying) while x would be 0 (just born). The same thing applies for the rest of Ernie's graph so the area of success would be the red trapezoid and another on of the same shaped flipped over to the top left. The finding of the probability in Ernie's solution was valid except heres a short cut. Since my graph is 5x5 the total area(place of success would be 25. Now to find the probability we will use 1-(area of possible failure)/(total area). Now both triangles are 4x4 so we solve 1-((4*4/2)+(4*4/2)/5x5. Which is equal to 1-16/25 so the final solution is...[/hide]
Solution[hide]9/25![/hide] :winner_first:
PS: how do you make graphs on the forum? (thanks!) :|[/quote]

Asympote, don't ask me how to use it though, I don't know[/quote]

Essentially, you could open the wiki and search up all of the functions.[quote=brianyuan13]Hey go away[/quote]

Is not needed

[quote=vectorfunction]Morbid...

Anyways, You can always make a formula to make your life easier,

Call the lifespan (Or the questions version of the lifespan) $L$ and the Range in where they can be born $R$. The Formula would be the area in which they could not be born, and then that number would be subtracted by 1. Because of this, the formula would be $1-$[b]$ \left(\frac{L}{R} \right)^2$[/b]

Bolded: Triangle area is $2 \cdot \frac{1}{2} \cdot \frac{L}{R} \cdot \frac{L}{R} = \left(\frac{L}{R} \right)^2$

P.S> $\LaTeX$ is annoying[/quote]

P.P.S. You should probably put hide marks over it, and yes, $\LaTeX$ is annoying.


algebra polynomial geometry circumcircle

Let $a,b,c$ be positive reals such that $a\geq b$, $a\geq c$ and $b+c>a$, Solve for $x$:
\[ b\sqrt{x^2-c^2} + c\sqrt{x^2-b^2} =ax.  \]

[hide="hint (maybe)"]
Not sure if this helps but the restrictions look like  a triangle with greatest side $a$. I dont know what to do after realizing this so i dont turn it into a 4th degree polynomial though.[/hide]

Rationalization on LHS ...

Answer
[hide]
$x = \frac{2abc}{\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}}$
[/hide]

[hide]Consider triangle ABC with circumcircle O and side lengths a, b, and c. Extend AO past O to form diameter AD.

Then, if R is the circumradius, we have:

$AC=b$
$AB=c$
$BD = \sqrt{4R^2 - c^2}$ ($\angle DBA = 90$)
$CD = \sqrt{4R^2 - b^2}$ ($\angle DCA = 90$)
$BC = a$
$AD = 2R$

Then by Ptolemey's Theorem, as quad ABDC is cyclic, \[ AC * BD + AB * CD = BC * AD  \] or \[ b\sqrt{4R^2 - c^2} + c \sqrt{4R^2 - b^2} = 2aR  \]

Thus $x = 2R = \frac{2abc}{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}$.[/hide]


[quote="lowest_reynolds"]Alright fellow Canadians!  Who else is taking the Canadian Open Mathematics Challenge on Wednesday, November 22?
Let's try and make an effort towards Winter Camp/CMO qualification!!![/quote]
I am, I would have taken it last year too but i found out about it when it was too late. :(
On another note, I'm hoping on getting a good standing. :)

Hellus yeah, in between two 3-hour hugeass exams on Monday and Friday.

[quote="rem"][quote="Leao"]I hope I will do well enough to qualify for CMO this year  :)[/quote]
I am sure you can make it. Last year you got 63-65 which is at most 4 pts off the cut-off. If you just study a bit and don't make too many mistakes you will easily qualify for CMO. Good luck. :)[/quote]
Thanks  :lol: I will certainly try my best.

[quote="blootzboy"]how do you get invited to winter camp[/quote]

You either own COMC (ie 75+ -- about top 10 in Canada), or have gone to IMO or owned CMO the year before.
It's tough because only 15 people are invited there.

someone please delete this message, it ended up in the wrong forum

ty
-jorian

His last post was like a week ago though.

I'll be free after this Saturday guys!

Well, under the school results, our school had your total score and your scores from part A and part B.

The contest seems great, yet didn't have time to pwn the 4b (iii). anyone has answers for partA (full list)?

I ran out of time- didn't even get a chance to read question 4 in part B. Can someone please PM me answer to part A #8.

[quote="blootzboy"]i messed up really badly on the COMC cuz i found out i made a ton of mistakes so i'm most likely not writing CMO but if i get like honourable mention or better on USAMO do you think it's still possible to make the IMO team?[/quote]
Maybe you didn't make as many mistakes as you think. :)

i did, i found out after i checked it over in science class

no..... I got part A question 8 wrong. That means i will probably get like 60/80. That stinks.

[quote="rem"]I think the cutoff will be a bit lower, like 66.[/quote]
I hope its lower, just in case. :rotfl:

lol, the lower the better. My math teacher is saying the cutoff will be 70 because the first 8 of part A and the first 3 of part B are really easy, and any good math student can get all of them. That really scared me (because I messed up on A6)  :ninja: Btw, do any of you have experience with how generous the markers are with bs?  :rotfl: I think I got part B number 4bi but for the rest of that question, I just totally bsed.

Some one PM me the answer to 3b) i didnt get it on the actual contest  :( but i think i might have solved it.[/code]

[quote="rem"]I think the cutoff will be a bit lower, like 66.[/quote]

how can u know the cutoff will be lower??
btw, do u know the cutoff for a provincial medal?

Thank you rem leao and vincent five for the answer to 3b). After leao told me the answer i knew the answer i had was wrong cuz he was 99% sure, after redoing the question i got the answer and i know how to solve it now

screw Part B 3b.....
I used 20 mins to prove it not an isoceles right triangle, and....had no clue after that!

I took the test on wednesday.  i got all the part A questions, except for question 8 i stupidly put the odds of 2 successive heads not coming up, as opposed to their odds of coming up... will they penalise too badly for that if i showed my work?

B1 and B2 were fairly easy, as was B3 i, but B3 ii was too difficult for me to solve in that time frame.. also question B4i was quite easy i think.  Usually the question 4's are more approachable than they were this year, so i suspect the cut-off points will be lower.


I came up with a proof to the following with tons of contradictions but it seemed a bit much.  I was wondering if there is an easier way:

[b]Note[/b]: $ f$ is an increasing function on $ A \subseteq \mathbb{R}$, if $ f: A \to \mathbb{R}$ and $ \forall ~x,y \in A, ~x<y \implies f(x)<f(y)$.

[b]Prove[/b]: If $ f: A \to \mathbb{R}$ is an increasing function, then $ f^{\minus{}1}: \text{Rng}(f) \to A$ is an increasing function.

$ \text{Rng}(f)$ is the range of a function $ f$.  For example, the range of $ f(x)\equal{}x$ is $ \text{Rng}(f)\equal{}\mathbb{R}$.

[hide="Hint"] What you want to prove is that the condition that $ f$ is an increasing function can be written $ x < y \Leftrightarrow f(x) < f(y)$.  Then let $ f(x) \equal{} X, f(y) \equal{} Y, x \equal{} f^{ \minus{} 1}(X), y \equal{} f^{ \minus{} 1}(Y)$. [/hide]

How would you go about proving that it is a function?  Its obvious that $ \text{Dom}(f^{\minus{}1})\equal{}\text{Rng}(f)$ but essentially I got stuck trying to prove that it was a function by the second part using the definition of a function.

That is, $ (x,y) \in f^{\minus{}1} and (x,z) \in f^{\minus{}1} \implies y\equal{}z$.

[quote="skimnc"]$ \text{Rng}(f)$ is the range of a function $ f$.  For example, the range of $ f(x) \equal{} x$ is $ \text{Rng}(f) \equal{} \mathbb{R}$.[/quote]
If so, here is solution
[hide="solution"]
Assume that $ u<v$ and $ u,v\in \text{Rng}(f)$ then $ u\equal{}f(x),v\equal{}f(y)$ for some $ x,y\in A$. We have $ x<y$ because $ f(x)<f(y)$ and $ f$ is increasing, but $ x\equal{}f^{\minus{}1}(u),y\equal{}f^{\minus{}1}(v)$ we're done.[/hide]


Given that $ 6x \plus{} y \equal{} 15$, the value of $ 3x$ can be written in terms of $ y$
as $ ay \plus{} b$ for some numbers $ a$ and $ b$. What is the simplified value
of $ a \plus{} b$?

$ 6x\equal{}15\minus{}y\\
3x\equal{}7.5\minus{}.5y$
So $ a\equal{}\minus{}.5, b\equal{}7.5 \Rightarrow a\plus{}b \equal{} \boxed{7}$

We can get 3x on one side alone by moving y over and dividing by 2. We have 3x=-0.5y+7.5, so a=-0.5 and b=7.5. a+b=7

Somebody already posted the exact same solution. Please do not keep repeating these. It's spam.

Already told 'em, izzy.

I am very, truly sorry for this :spam:.


number theory
greatest common divisor
algorithm
LaTeX
least common multiple
prime factorization

just in case any of you didn't already know this...

take any two numbers, say 801 and 356,

divide the first number by the second, then take the remainder. 801/356 R=89

then divide the second number by the remainder you just found, and take the remainder again. 356/89 R=0

once you reach a remainder of 0, take the last divisor you used, and it is your GCD! in this case the GCD is 89.

This is called the [b]Euclidean Algorithm[/b].

Here is an example with latex, and an explanation of the extended euclidean alg:

$ (1116,972) \\
(972,144) \\
(828,144) \\
(684,144) \\
(540,144) \\
(396,144) \\
(252,144) \\
(144,108) \\
(108,36) \\
(72,36) \\
(36,36) \\
(36,0)$
Thus the GCD of $ 972$ and $ 1116$ is $ \boxed{36}$.

[hide="Click for faster method"]
Subtracting $ 144$ $ 6$ times was boring work.  If we don't want to do this, we find the largest multiple of $ 144$ that is less than $ 972$, which in this case was $ 6$, and then subtract that.  To see why this works, we have two numbers $ m$ and $ n$, where $ m$ is less than $ q \cdot n$, but greater than $ (q \minus{} 1) \cdot n$.  If we use the normal algorithm, we get
$ (m,n) \\
(m \minus{} n,n) \\
(m \minus{} 2n,n) \\
\ldots \\
(m \minus{} qn,n)$.  Because $ m$ is greater than $ (q \minus{} 1) \cdot n$, we can subtract $ n$ from $ m$ $ q$ times so that $ n$ is the greater number.  This extended algorithm is called the Extended Euclidean Algorithm.
[/hide]

erm...i think mine is slightly different from that subtracting method

Not the extended part; the click for faster way.  click it.

[quote="mewto55555"]Also it is the product divided by the LCM. And Quincy, with regards to your signature:

I herd you likez Mudkipz.[/quote]

WHAT is the product divided by the LCM?
my friend says that mudkip thing, where is that from?

First off I have no idea where it comes from. I just herd [size=9]you like mudkipz[/size] it. 2nd off lets take an example. Product of 8 and 20 is 160. LCM is 40. 160/40=4 which is also the GCF of 8 and 20. If you just think about it for a minute it makes sense.

yes, the gcd(a,b)*lcm(a,b)=a*b. in words, The greatest common divisor of two numbers times their least common multiple is equal to the product of the two.

notice: this only holds for nonnegative integers: gcd(-1,1)*lcm(-1,1)=1*1 which does not eqaul -1*1.

wouldn't that method take a lot longer though?

[quote="QuincyCello"]WHAT is the product divided by the LCM?[/quote]

Well, what did I just prove? :wink:

[quote="QuincyCello"][quote="AIME15"]LCM is the least common multiple.

Mudkip is a POKEMON![/quote]

not exactly what i meant...i meant what do you get when you divide the product by the LCM? and of course i know mudkip is a pokemon, i'm obsessed with it. (pokemon AND mudkip) i meant my friend also says "i heard you likez Mudkipz", and i was wondering where they both got it from...but more importantly i'd like to know what dividing the product by the LCM gives you[/quote]

so I herd sum things about u lieking mudkipz

it's a meme. It's stupid. Slap people that say it.

[quote="math154"][hide="Easy Proof"]Let $ a \equal{} p_1^{k_1}p_2^{k_2}p_3^{k_3}\cdots p_n^{k_n}$ and

$ b \equal{} p_1^{j_1}p_2^{j_2}p_3^{j_3}\cdots p_n^{j_n}$,

where $ p_1,p_2,...,p_n$ are all the prime factors of $ a,b$ and

$ j_1,j_2,...,j_n,k_1,k_2,...,k_n$ are all nonnegative integers.

The product of $ a,b$ is obviously equal to

$ p_1^{j_1 \plus{} k_1}p_2^{j_2 \plus{} k_2}p_3^{j_3 \plus{} k_3}\cdots p_n^{j_n \plus{} k_n}$.

and

$ \gcd(a,b) \equal{} p_1^{\min(j_1,k_1)}p_2^{\min(j_2,k_2)}p_3^{\min(j_3,k_3)}\cdots p_n^{\min(j_n,k_n)}$,

$ \operatorname{lcm}(a,b) \equal{} p_1^{\max(j_1,k_1)}p_2^{\max(j_2,k_2)}p_3^{\max(j_3,k_3)}\cdots p_n^{\max(j_n,k_n)}$

which means that

$ \gcd(a,b)\operatorname{lcm}(a,b) \equal{} p_1^{j_1 \plus{} k_1}p_2^{j_2 \plus{} k_2}p_3^{j_3 \plus{} k_3}\cdots p_n^{j_n \plus{} k_n} \equal{} ab$,

as desired.
[/hide][/quote]

This is really really helpful if you know the gcd or lcm though.  And you have the product.

Yes, I do remember.

Everybody here should sign up for that class!!!

A lot of questions that people are asking here relate to number thoery.

[quote="QuincyCello"]errrrrrr...i can't tell really....you just proved that it's extremely complex and most likely much slower than the Euclidian Algorithm[/quote]

What I just proved is that $ ab\equal{}\gcd(a,b)\operatorname{lcm}(a,b)$. Hm...this is very complicated.

GASP

OMG

I just found a revelation!!! When you divide the equation by $ \operatorname{lcm}(a,b)$, you find that the product divided by the LCM is the GCD!!!!!!! OMG!!!!

Hint: What did you ask a few posts ago? :wink:

[quote="math154"][quote="QuincyCello"]errrrrrr...i can't tell really....you just proved that it's extremely complex and most likely much slower than the Euclidian Algorithm[/quote]

What I just proved is that $ ab \equal{} \gcd(a,b)\operatorname{lcm}(a,b)$. Hm...this is very complicated.

GASP

OMG

I just found a revelation!!! When you divide the equation by $ \operatorname{lcm}(a,b)$, you find that the product divided by the LCM is the GCD!!!!!!! OMG!!!!

Hint: What did you ask a few posts ago? :wink:[/quote]

That's just using some elementary property.....

your saying if ab=c, then a=c/b, which is quite obvious :D

sorry to burst ur buubble

Hopefully you were being sarcastic there. My post was obviously sarcastic. :wink: I would feel sorry for you if you weren't being sarcastic.

What if you have uber big numbers?

Like: find gcd of (283479283792874,1827468273648726534857634876)

hehehe prime factorize THAT!

Hm? The answer is 2.

@Everyone: I merely posted the proof as a response to a question that somebody apparently didn't know they posted. I never said the proof would help you find the GCF or anything.

[quote="QuincyCello"][quote="hongsquare"]there is a MUCH easier way to find gcd
primefactor all the numbers
then
you can take the least power of EACH COMMON prime factor[/quote]

errr...actually this way is way easier

and faster[/quote]

GIve an example where its better.....

Yes, Quincy...if you want people to actually listen to you, back up your statements.

guys, this discussion is really sorta pointless
pplz have given ways of finding GCD, which are all true ways
some ways are more useful than others in certain situations

to take the last few posts for an example, finding the GCD of 6 and 15 might be easier using prime factorization than multiplying and dividing by the LCM


[b]Nα λυθεί στο σύνολο των θετικών ακεραίων η εξίσωση

                                           α^4 + 4^α = p ,  όπου p πρώτος .[/b]


How can i open a ps file (post script) with what program?

can you just give me a link to start downloading, i don`t know which to download

I asked something  :bomb: ..  :(  please

[quote="Beat"]I asked something  :bomb: ..  :(  please[/quote]
Woah, don't get mad.
The reason he linked to that page is the download depends on what kind of computer you use.
If you tell me your operating system I will link to a download.

Im not mad, im sad and i want somebody to help me. well i really don`t know what do you need: operating system?

Your flag says you are from Belarus. Do you speak russian?

I speak rusian but not very well, like english.
My windows is xp.. something else?
I`ll be very thankful if you help me

You will want to go to [url=http://www.cs.wisc.edu/~ghost/doc/AFPL/get853.htm]this page[/url] and scroll down to the part that says "Windows 95, 98, ME, NT, 2000, or XP." From that, you choose a Ghostscript file to download, depending on whether you need a 32-bit version or a 64-bit version.

Also, it's probably a good idea to install GSview, which is right below that.

Adobe Acrobat (at least the Standard edition; not sure about the free Reader) can change .ps files into .pdf, which it can in turn read.

<snipped>

[url=http://view.samurajdata.se/]This website[/url] lets you view a .ps (or .pdf or .doc) files without requiring you to download anything.

[quote="Valentin Vornicu"]Adobe Acrobat is far from being a freeware, and, indeed using the Distiller you can create a .pdf from a .ps file.[/quote]

My mistake - as I have the standard version, I am not particularly familiar with the free Reader.


integration
algebra
polynomial
calculus
calculus computations

A hemispherical tank of radius 6 feet is positioned so that its base is circular. How much work is required to fill the tank with water through a hole in the base if the water source is at the base?

V of hemisphere: 2/3 (pi) r^3.
Weight of Water: 62.4 pounds per ft^3

W= integrate [(weight)(Volume of slice)] from 0 to 6.

Please help, I've tried and tried, but I just can't get the right answer.

Thanks in advanced:  :ninja:

I'll write $ \rho$ for that density of $ 62.4\text{ lb}/\text{ft}^3.$

Let $ y$ be distance (in feet) above the flat base of the tank. Take a slice, $ dy$ in thickness of the water in the tank at that height. This is a disk whose radius is $ \sqrt{36\minus{}y^2}.$ Hence the volume is

$ dV\equal{}\pi\left(\sqrt{36\minus{}y^2}\right)^2\,dy\equal{}\pi(36\minus{}y^2)\,dy.$

The weight of that much water is simply the volume time the density:

$ dW\equal{}\rho\pi(36\minus{}y^2)\,dy.$

Finally, the work required to live that much water from the base to that height is the weight time the height lifted, or

$ d(\text{work})\equal{}y\,dW\equal{}\rho\pi y(36\minus{}y^2)\,dy$

To calculate the total work, we simply add up all of these slices.

$ \text{work}\equal{}\int_0^6\rho\pi y(36\minus{}y^2)\,dy$

Since that's a polynomial integral, I'll leave the remaining details to you.

(Note that the formula $ V\equal{}\frac{2\pi r^3}3$ was useless in this instance.)


[quote="SquirrelGirl"]So, I'm not sure if this is the right place to put this problem, but it's from my trig/precalc class. Here goes...

Put the following in standard form:

3x^2 - 2(squareroot3)xy + y^2 + 2x + 2(squareroot3)y = 0

I know this is a parabola because b^2 - 4ac = 0

and I get that
x = .866x prime - .5y prime
y = .5x prime + .866y prime

but I'm getting stuck when I go through and distribute and multiply it out. The xy term doesn't cancel for me. I don't know if I'm making a simple mistake or what, but it's confusing me. Please help me![/quote]

LATEXFIED
Put the following in standard form:
$ 3x^2 \minus{} 2\sqrt{3}xy \plus{} y^2 \plus{} 2x \plus{} 2\sqrt{3}y \equal{} 0$
THE WORK HE/SHE has so FAR
$ x \equal{} x\frac{\sqrt{3}}{2} \minus{} \frac{1}{2} y'$
$ y \equal{} \frac{1}{2}x' \plus{} \frac{\sqrt{3}}{2}y'$


Prove that for all positive integers $n$, \[\lfloor \sqrt{n}+\sqrt{n+1}+\sqrt{n+2}\rfloor =\lfloor \sqrt{9n+8}\rfloor.\]

The proposition is proven, when
$0\leq | \sqrt{9n+8}-\sqrt{n}-\sqrt{n+1}-\sqrt{n+2} | \leq 1$, i.e.
$-1\leq \sqrt{9n+8}-\sqrt{n}-\sqrt{n+1}-\sqrt{n+2} \leq 1$.
First, i will show that
$-1 \leq \sqrt{9n+8}-\sqrt{n}-\sqrt{n+1}-\sqrt{n+2}$:
Therefore, you need estimate 1:
$3\sqrt{n+3/2}\geq\sqrt{n}+\sqrt{n+1}+\sqrt{n+2}$ (proof by estimate 2: $\sqrt{n+1}\geq\sqrt{n(n+1)}$ and calculation)
So, you have to show that
 $-1 \leq \sqrt{9n+8}-3\sqrt{n+3/2} \Leftrightarrow 3\sqrt{n+3/2}\leq \sqrt{9n+8}+1 \\
9n +27/2\leq 1+9n+8+2\sqrt{9n+8} \\
11/4\leq\sqrt{9n+8} \\
8 \leq 9n+8 \\
0 \leq 9n$
, what is true for all $n>0$.
Now, you have to show that  $\sqrt{9n+8}-\sqrt{n}-\sqrt{n+1}-\sqrt{n+2}  \leq 1$.
Therefore, you need estimate 3: $3\sqrt{n+1}\leq \sqrt{n}+\sqrt{n+1}+\sqrt{n+2}$ (proof by calculation).
So, you have to show that
$\sqrt{9n+8}-3\sqrt{n+1}\leq 1 \\
12n+11-6\sqrt{(9n+8)(n+1)}\leq 1$
Now you need estimate 4: $(2n+2)^2\leq (9n+8)(n+1)$ (again proof by calculation) and you have to show that
$12n+11-6(2n+2)\leq 1 \\
0\leq 5n^2+9n+4$
, what is obviously true for alle $n>0$
Because of that, $-1\leq \sqrt{9n+8}-\sqrt{n}-\sqrt{n+1}-\sqrt{n+2} \leq 1$ is true and hence, $\left\lfloor \sqrt{9n+8} \right\rfloor = \left\lfloor \sqrt{n}+\sqrt{n+1}+\sqrt{n+2} \right\rfloor$ is true.

I think it is not enough to prove that     $ 0\leq | \sqrt{9n\plus{}8}\minus{}\sqrt{n}\minus{}\sqrt{n\plus{}1}\minus{}\sqrt{n\plus{}2} | \leq 1$. For example, $ 4,5\minus{}3,6<1$ but $ \lfloor 4,5\rfloor\ne\lfloor 3,6\rfloor$. Here is my proof:

Since $ 9n\plus{}8$ is never a square, there is no integer in the interval $ (9n\plus{}8,9n\plus{}9)$. I will proof that $ \sqrt{9n\plus{}8}<\sqrt{n}\plus{}\sqrt{n\plus{}1}\plus{}\sqrt{n\plus{}2}<\sqrt{9n\plus{}9}$.

The first inequality can be proven by squaring and rearranging terms a few times. It's very boring, so I won't write it here. The second inequality is equivalent to $ \sqrt{n}\plus{}\sqrt{n\plus{}2}<\sqrt{9n\plus{}9}\minus{}\sqrt{n\plus{}1}\equal{}3\sqrt{n\plus{}1}\minus{}\sqrt{n\plus{}1}\equal{}2\sqrt{n\plus{}1}$, which is obvious after squaring. Hence, proved.

I am unsure about the above proof; the lower bound $\sqrt{9n+8} < \sqrt{n} + \sqrt{n+1} + \sqrt{n+2}$ is not trivial, and actually false for real $n$ in the interval $[0, 0.6]$, so I don't think it's easily proven 'by squaring and rearranging terms a few times'. I agree with the upper bound $\sqrt{n} + \sqrt{n+1} + \sqrt{n+2} < \sqrt{9n+9}$, which follows from the concavity of the square root function, and from this the inequality $\lfloor \sqrt{n} + \sqrt{n + 1} + \sqrt{n + 2} \rfloor \le \lfloor \sqrt{9n+8} \rfloor$ can be shown. To show that we actually have the reverse inequality $\lfloor \sqrt{n} + \sqrt{n + 1} + \sqrt{n + 2} \rfloor \ge \lfloor \sqrt{9n+8} \rfloor$ as well, I want to make use of the fact that $9n + 8$ is never a square, from which we get that $ \lfloor \sqrt{9n+8} \rfloor =  \lfloor \sqrt{9n+7} \rfloor$ for all $n \in \mathbb{N}$. It therefore suffices to show  $\sqrt{n} + \sqrt{n + 1} + \sqrt{n + 2} > \sqrt{9n+7}$.

To prove this inequality, I claim that $\sqrt{n} + \sqrt{n+2} > 2\sqrt{n + 0.8}$ for all $n \ge 1$. Taking this latter claim for granted at the moment, we deduce $\sqrt{n} + \sqrt{n + 1} + \sqrt{n + 2} > 3\sqrt{n + 0.8} = \sqrt{9n + 7.2} > \sqrt{9n+7}$, and we are done.

So it comes down to proving $\sqrt{n} + \sqrt{n+2} > 2\sqrt{n + 0.8}$, to finish off the proof. We will show this by starting with the inequality $n > 0.45$ and then repeatedly add or multiply by something positive, factorize both sides, or take square roots.

\begin{align*}
n &> 0.45 &\text{By assumption} \\
3.2n & > 1.44 &\text{Multiply both sides by } 3.2\\
8n &> 4.8n + 1.44 &\text{Add } 4.8n \text{ to both sides} \\
4n^2 + 8n &> 4n^2 + 4.8n + 1.44 &\text{Add } 4n^2 \\
4n(n+2) &> (2n + 1.2)^2 &\text{Factorize both sides } \\
2\sqrt{n}\sqrt{n+2} &> 2n + 1.2 &\text{Take the square root}\\
2n + 2 + 2\sqrt{n}\sqrt{n+2} &> 4n + 3.2 &\text{Add } 2n + 2\\
(\sqrt{n} + \sqrt{n+2})^2 &> 4(n + 0.8) &\text{Factorize}\\
\sqrt{n} + \sqrt{n+2} &> 2\sqrt{n + 0.8} &\text{Take the square root}
\end{align*}


It's happened twice already:

I am reminded that there is a new private message in my inbox and my Private Messages button is light yellow but I when I click on it I find that I have no new private messages. What's going on? Are there messages that aren't getting delivered, or is my internet connection not functioning properly?

The people who sent you the messages decided to delete them before you actually got to see them so that's why it will say 1 message but when you refresh the page you will have no messages.

Thanks for the information.

Wait...so a user can actually delete a message they sent after they sent it, as long as it hasn't been seen my the person who is supposed to receive the message yet?
So, to delete, do you click the Outbox, select and delete, I guess?

Is it possible for us to at least know who sent the PM after they delete it?

The standard email I get about having a PM (I have that part of my preferences in what must be the default setting) does contain the user name of the person who sent the message.

Another question about private messages: are pm's completely "private" or can the administrators view the messages of other users?

I can change your password. Therefore I can log in as you, and I could read your PMs. [b]But[/b] that does not happen. We respect people's privacy. Actually we do more about that than most sites out there ;)


Ba Salam

Sale no ra be tamame Irani haye aziz be khosos mathlinkihaye gerami pishapish tabrik migoyam. :lol:

Farz konid baraye adade hagigye $ x,y,z$ darim $ x\plus{}2y\plus{}3z\equal{}3$ .maximome ebarate zir ra biyabid?

\[ xy\plus{}yz\plus{}zx\]

Farz konid baraye adade hagigi darim $ x\plus{}2y\plus{}3z\equal{}3$ .sabet konid?

a)\[ xy\plus{}yz\plus{}zx\leq \frac{9}{8}\]
b)\[ x^2\plus{}y^2\plus{}z^2\plus{}xy\plus{}yz\plus{}zx\geq \frac{9}{10}\]

Do namosavie gashang va marhale dovomy .

man injori halesh kardam vali be nazaram rahe ghashangtari ham dashte bashe:
tu namosavi be jaye $ x$ , $ 3 \minus{} 2y \minus{} 3z$ gozashtam bad be dast omad:
$ \minus{} 16y^2 \minus{} 32yz \plus{} 24y \minus{} 16z^2 \plus{} 24z \minus{} 9 \leq 0$
 ye moadele daraje 2 barhasbe $ y$ ke deltash 0 va zaribe $ y^2$ manfie pas kochektar ya mosavie  $ 0$   :)  :!:
(roye b fek nakardam vali fekonam ba hamin ravesh hal beshe)

Rahe hale asliye man baraye in soal Moshtag giri ast vali ye rahe kotah ham hast

a)\[ (x\plus{}2y\plus{}3z)^2\minus{}8(xy\plus{}yz\plus{}zx)\equal{}(2y\plus{}z\minus{}x)^2\plus{}8z^2\]

b)\[ 10(x^2\plus{}y^2\plus{}z^2\plus{}xy\plus{}yz\plus{}zx)\minus{}(x\plus{}2y\plus{}3z)^2\equal{}\frac{3}{4}(z\plus{}3x)^2\plus{}\frac{1}{4}(4y\plus{}x\minus{}z)^2\plus{}2(x\plus{}y)^2\]


A triangle has two equal sides.  The third side has a length of 13 feet, 2 inches.  If the perimeter is 40 feet, what is the length of one of the equal sides?

[hide="Solution"]
$40-13-\frac{1}{6}=2*equal side$
$\frac{161}{12}=equal side$
[/hide]
Laugh at me if this is wrong!

no need to laugh, i got the same thing  :D

I also got 161/12 = one side but how do I convert this fraction to feet and inches?


hello, to 1) we will give a proof by induction. For $ n\equal{}2$ we get
$ 1\plus{}1/9>7/6\minus{}1/14$ which is true because $ 1\plus{}1/9\minus{}7/6\plus{}1/14\equal{}1/63>0$.
Now we assume that
$ 1\plus{}\frac{1}{3^{2}}\plus{}\frac{1}{5^{2}}\plus{}\cdots\plus{}\frac{1}{(2n\minus{}1)^{2}}>\frac{7}{6}\minus{}\frac{1}{2(2n\plus{}1)}$
is true, adding $ \frac{1}{(2n\plus{}1)^2}$ on both sides we get
$ 1\plus{}\frac{1}{3^{2}}\plus{}\frac{1}{5^{2}}\plus{}\cdots\plus{}\frac{1}{(2n\minus{}1)^{2}}\plus{}\frac{1}{(2n\plus{}1)^2}>\frac{7}{6}\minus{}\frac{1}{2(2n\plus{}1)}\plus{}\frac{1}{(2n\plus{}1)^2}$ and we have to show that
$ \frac{7}{6}\minus{}\frac{1}{2(2n\plus{}1)}\plus{}\frac{1}{(2n\plus{}1)^2}>\frac{7}{6}\minus{}\frac{1}{2(2n\plus{}3)}$, this is true because$ \frac{1}{2(2n\plus{}3)}\plus{}\frac{1}{(2n\plus{}1)^2}\minus{}\frac{1}{2(2n\plus{}1)}\equal{}\frac{2}{(2n\plus{}3)(2n\plus{}1)^2}>0$
for all $ n$ with $ n \in \mathbb{N}$ and $ n>1$.
Sonnhard.

2.) You have

$ \frac{n}{4}<\frac{1^{2}}{1\times 3}\plus{}\frac{2^{2}}{3\times 5}\plus{}\cdots\plus{}\frac{n^{2}}{(2n\minus{}1)(2n\plus{}1)}\equal{}\frac{n}{4}\left (1\plus{}\frac{1}{2n\plus{}1} \right )<\frac{n}{2}$ Which is obvious


inequalities
number theory unsolved
number theory

Any one ?
It 's really a beautiful pro :)


For some reaon, the time that this site shows at the top of the page seems to be 8 minutes behind...

Yeah, our system clock runs a bit slow for some reason.  Every once in a while we have to give it a nudge, as I just did, so it should be more or less accurate now.

And just to forstall further comments -- yes, we know that there are programs that we can install that will automatically update our clock.  When we do our planned big server upgrade this summer, we probably will.  For now, it's not a big enough deal for us to worry about it.

But it drove Dave batty for about a year.  :rotfl:


Find the coefficient of x2  in the polynomial (1-x)(1+2x)(1-3x)...(1+14x)(1-15x).

I tried brute force and looking for patterns and anything possible that I could think of.. but I still can't get it! I even missed it on the AIME last year. how sad. any suggestions?

I bet there are lots of ways to do this one.  One way (and not necessarily the best) involves [hide]expressing coefficients of a general polynomial in terms of the roots[/hide].

[quote]I understand how to express a polynomial in terms of its roots, but wouldn't that mean taking the sum of the roots 13 at a time??[/quote]

13 at a time is very similar to 15-13=2 at a time.

[quote="JBL"]Another thing you can try is [hide]multiply it out!  But this isn't too bad, because you get to drop all terms greater than $x^2$.  So, we get $1 - x$, then $1 +x - 2x^2$, then $1 -2x -5x^2$, and just keep going.[/hide]  It shouldn't take [i]too[/i] long -- certainly not a full 12 minutes.  What number was this on the AIME?[/quote]

i did that on the real AIME, but in doing that, you run the risk of small arithmetic errors, which is what happened to me. so if you mess up on one small step, your whole answer messes up. would you happen to have any other suggestions?

It seems like there is a better way, maybe with Fibonacci, than just brute forcing it by summing all possible multiplications of 2 of those numbers.  That'd be 105 numbers, too much and too prone to one little mistake.

You can always view the math jam that discusses that problem [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=26#7]here[/url].

They use symmetric sums like Joel suggested although they don't go into too much detail.

Summing up the 105 numbers isn't actually too bad, if you use a little math rather than just doing it on a calculator.
[hide="How?"]
Well, we have -1 times (2, -3, 4, -5, ..., up to -15). Then theres 2 times (-3, 4, -5, .. up to -15). Etc. Its quite easy to sum the bits in brackets.
[/hide]

And this may help you come up with a slightly better solution:
[hide="An improvement"]
The above may look something like (-1+2-3..-15)*(-1+2-3..-15). How do we get from that to the answer?
[/hide]

[hide][quote="TripleM"]
Well, we have -1 times (2, -3, 4, -5, ..., up to -15). Then theres 2 times (-3, 4, -5, .. up to -15). Etc. Its quite easy to sum the bits in brackets.
[/quote]
That's pretty good.
[/hide]

[hide]
[quote="TripleM"]
And this may help you come up with a slightly better solution:
The above may look something like (-1+2-3..-15)*(-1+2-3..-15). How do we get from that to the answer?
[/quote]

This, too, is a great idea (it can follow from the earlier one if you made the stuff in parentheses the same).  It multiplied out will give you $2*[what you're looking for]+1^2+2^2+3^2+...$  Then you can get it from there without much difficulity.  To facilitate it even more, the equation for the sum of consecutive squares is (n)(n+1)(2n+1)/6[/hide]

an easy way on the actual contest:

to get x^2, you have to get exactly 2 of the x terms when you choose a term from each factor.  Because the other terms besides x terms are 1's, we dont have to worry about the coeff. changing.

Notice 2 and 3, 4 and 5, etc. can be paired to make the summing even easier.

Now you are summing:

-1 (2-3+4-5 ... +14-15)  = -1(-7)
+2 (-3+4-5+... +14-15) = 2 (-9)
...
-13(+14-15) = -13(-1)
+14(-15) = 14(-15)

The result is (note the pattern):

(-1)(-7) + (2)(-9) + (-3)(-6) + (4)(-10) + (-5)(-5) + 6(-11) + (-7)(-4) + 8(-12) + (-9)(-3) + (10)(-13)  + (-11)(-2) + (12)(-14)  + (-13)(-1) + (14)(-15)

which is easily summed.


A woman owns an indestructable car. Also, the lock is unpickable.

She drives the car to the mall. She rolls up the windows, gets out of the car, and locks the doors. When she is in the middle of shopping, she realizes that she left her purse on the passenger seat of the car. But when she goes back to the car, the purse is stolen. How?

[hide="haha this is pretty easy :lol: "]she didn't close the door[/hide]

[hide="another answer"] the thief was her passenger riding in the backseat.[/hide]

[hide="pretty simple"]well, it is given that she locks the doors, but she does not close them...so is that it?[/hide]

[hide]Her keys are pickpocketed, or her husband's secret lover takes his keys.[/hide]

[hide="another bad solution but works"]
it was a convertible. So it had no top. The thief just jumped in and stole it.[/hide]


abstract algebra
superior algebra
superior algebra unsolved

Let R be a ring with unity.
I've heard that every R-module M admits a projective resolution,
that is, an exact infinite sequence of the form:

$ \ldots \longrightarrow P_2 \longrightarrow P_1 \longrightarrow P_0 \longrightarrow M \longrightarrow 0$

where each $ P_i$ is a projective module.

How do I prove it?

Are you aware of the fact that a free module is projective?

Yes. I tried to construct a 'free resolution', but
it didn't work either.

Can you prove this lemma?  The full statement follows by induction.

For every homomorphism $ \phi: A\to B$, there exists a free module $ F$ and homomorphism $ \psi: F\to A$ such that $ \rm{image}\ \psi \equal{} \ker{\phi}$. (hint: this is equivalent to showing that every module is the image of a free module!)

Hmmm... why not? Getting a surjection $ F \to M \to 0$ from a free module $ F$ to a given module $ M$ is easy - just take $ F$ to be the free module generated by all the elements of $ M$. Now take the kernel, and do the same for the kernel, and keep going... where do you think this fails?

EDIT: Sorry for the duplication, LydianRain.

Ok, here's my trial:

Let $ M$ be an R-Module. Take $ L_0$ a free module and $ \Phi_{0} : L_0 \rightarrow M$ a
surjective homomorphism (using the lemma LydianRain said). Then

$ L_0 \longrightarrow M \longrightarrow 0$ is exact, and $ L_0$ is projective.

Next step (as suggested by randomgraph) is taking $ L_1$ a free module and
$ \Phi_{1} : L_1 \rightarrow ker(\Phi_0) \subseteq L_0$ another surjective homomorphism.
Now

$ L_1 \longrightarrow L_0 \longrightarrow M \longrightarrow 0$ is exact, and so on..

I think that's it. I tried something similar earlier, but the sequence was:

$ ker(\Phi_0) \longrightarrow L_0 \longrightarrow M \longrightarrow 0$

Then I wasn't sure that $ ker(\Phi_0)$ is projective, so it didn't work.

Thank you both. :)

For not being sure: indeed, submodules (and thus general kernels) of projective modules aren't always projective (e.g. the $ \mathbb Z / 4\mathbb Z$ module $ 2\mathbb Z / 4\mathbb Z \subset \mathbb Z / 4\mathbb Z$ isn't).


geometry
combinatorics proposed
combinatorics

[quote="nayel"]In a given convex pentagon, two arbitary points are chosen. prove that there exists four vertices of the pentagon such that the triangle they form contain both points.

i have an easy solution. so i'm confused if mine is correct or not. i'll post it only if someone else posts one first. :D[/quote]

FOUR vertices? Do you mean triangles that they form? In that case it would be obvious and hardly a Russian Olympiad problem.

sorry :blush:   the term triangle should be replaced with quadrilateral.

Now I understand why u made this introduction....
Look the problem is obvious!Just case chasing...So yr solution must be the same with my solution and of course it is true!
The point is that such a problem can't be a Russian Problem or actually it can!At the $Primary School$ :rotfl:  :P

I am joking-I think that your source has given the problem incorrectly!

[quote="Nick Rapanos"]I think that your source has given the problem incorrectly![/quote]

i found this problem in the book "Mathematical Olympiads 2000-01" by T. Andeescu and one more writer. :wink:


function
calculus
trigonometry
combinatorics proposed
combinatorics

Prove that there exists no function $f: [0,1] \to [0,1]$ which is:
- surjective
- continuous
- and for every $x \in [0,1]$ there exist $y,z \in [0,1]$ such that $y \ne z$ and $f(y)=f(z)=x$.

Edit: I'm sorry, is this supposed to be in the Calculus forum? Because it seems more of a discrete math question.

We're talking about continuous functions -- there's nothing discrete about it.  The statement given is wrong, however: $f(x) = |\sin 2\pi x|$ meets all three conditions.  The statement you probably want is to replace your third condition with, "For each $x \in [0, 1]$, there are exactly two members of the set $f^{-1}(x)$."  This problem has appeared on the forum before, so you might try searching for it.

Yes, this condition is what I meant. As for searching in the forum, honestly I wouldn't know where to start.

I would begin by considering the places at which the function achieves the value 1.  Then you're going to be using the Intermediate Value Theorem.  Is that enough help?


integration
algebra
functional equation
real analysis
real analysis solved

Show that

pi^2/sin^2(piz) = \sum_{-\infty}^{\infty} 1/(z-n)^2.

(Note: One can make it a three-liner without much effort)

[quote="Mindspa"]Show that
pi^2/sin^2(piz) = \sum_{-\infty}^{\infty} 1/(z-n)^2[/quote]
Let
f(x)=\sum_{-\infty}^{\infty} 1/(x-n)^2 - pi^2/sin^2(pix) for x\in R\Z
f(x)= 0 for x \in R
f is 1-periodic continous on R\Z,let see the continuity at x=0
f(x)=1/x - pi/sin(pi.x) + \sum_{n=1 to +oo}(1/(n-x)+1/(n+x))
when x->0, f(x) ~  -pi/3 + 2\sum_{n=1 to +oo} 1/n = 0
f is continuous at x=0, by 1-periodicity  f is continuous on R and verify
the functional equation f(x/2)+f((1+x)/2)=4f(x)
Lemma: f:R-->R continous 1-periodic such that f(x/2)+f((1+x)/2)=4f(x)
for any x then f=0
Proof:  E={f:R->R, continous, 1-periodic} equiped with the
norm ||f||=sup_{x\in R}|f(x)|
let u:E->E , f \in E , u(f)(x)=f(x/2)+f((1+x)/2) is R-linear map
||u(f)|| \leq 2||f||
4||f|| \leq 2||f|| => ||f||=0 =>f=0

for any x in R , f(x)=0
pi^2/sin^2(pix)=\sum_{-\infty}^{\infty} 1/(x-n)^2
Indeed it is nice result Laurent series
http://www-history.mcs.st-andrews.ac.uk/history/Mathematicians/Laurent_Pierre.html
 \cap  \cap  \emptyset  \cup \beta  \int  \cap  \emptyset  \emptyset  \lambda


[b]I had some questions about the section for work experience included in the common application for undergraduates college admissions. Won't you please tell me your thoughts and advice?  :( [/b]
............................................................................................................................

Are you allowed to put down jobs that you had after high school?
Would you recommend spending a few years right after high school to look for a job to have something to put down in this area? If you would recommend doing so, how long would you work after school?

If you were a high school freshmen who had never worked before, what would jobs would you look into doing?
If you were looking for a more intellectual job, where would you start looking? Supposing you had not learned many skills outside of school, where would you look?

[hide="Other Questions"]...................................................................................................
In your experience, what have been the most common mistakes with students who apply to college?

What are some common traits of people you know or have heard about who succeed in getting into the colleges that are more difficult to get into?

Where or who else would be good to ask these questions to?
[/hide]

You're allowed to put any jobs you've had up to the point that you're writing the application.  If you're applying after you've had a job after high school, then you can clearly put that job down.

I do not recommend foregoing college to get a job [i]just[/i] to be able to fill in a box on an application.  Colleges are usually looking at what you've done IN high school.  Taking time off after high school before going to college is fairly rare.  I know that MIT lets you petition to take a year off before college but this only applies once you're accepted to MIT.  In general, the common route is to go straight to college after high school.  It's rather unlikely that you have a legitimate reason to do otherwise.

Had I been looking for a job my freshman year I probably would've looked into working at one of the on-campus places such as the convenience store or the library.  My other major option would have been to do research for money.  It all depends on where you're going to college though.

[quote]
If you were looking for a more intellectual job, where would you start looking? Supposing you had not learned many skills outside of school, where would you look?[/quote]

Once again, the places I'd look pretty much only apply to MIT students.  If you go to any top university, I imagine it'd be pretty much the same.  They should have their own channels for you to get jobs (job fairs, mailings lists, etc.).  I mean, companies actively seek out individuals at these schools and thus finding out how to go about getting these jobs is easy.  You just need to make it happen.

[quote]In your experience, what have been the most common mistakes with students who apply to college? [/quote]

I'd say limiting your options too early.  One major cause of this is the issue of financing your education.  Financial aid is often better than you'd expect and, overall, it'll probably be a better financial investment.

[quote]What are some common traits of people you know or have heard about who succeed in getting into the colleges that are more difficult to get into? [/quote]

Hard working, motivated, self-disciplined,...

Thanks joml88.

May I ask you a few questions about what you said?

[b]I do not recommend foregoing college to get a job just to be able to fill in a box on an application. Colleges are usually looking at what you've done IN high school.[/b]
How would they know that you didn't do it in high school? Are saying that people older than high school seniors are at a disadvantage when they apply to undergraduate colleges if they put down things that they've done after high school?

[b]It's rather unlikely that you have a legitimate reason to do otherwise[/b]
Is it frowned upon by colleges to wait? Or is waiting just a bad idea for some other reasons?

[b]My other major option would have been to do research for money.[/b]
As a high school student, where can you go to do research or more intellectual jobs than working at a convenience store?

Lastly, what are some other ways that students limit their options too early?
Thanks for your time.

First, let me re-emphasize how rare I find it for someone to take a year off before applying to college (I'm assuming we're talking about top universities here).  I've never heard of it happening.  However, taking a 'gap year' is much more common which is where you apply, get accepted, and then take a year off.  So basically, I don't think you should even be seriously considering this option.

[quote][b]My other major option would have been to do research for money.[/b]
As a high school student, where can you go to do research or more intellectual jobs than working at a convenience store?[/quote]

My apologies.  I didn't read your question correctly.  My answer referred to if I were a [b]college[/b] freshman.  While you're in high school, I would put getting a job at a pretty low priority.  There are far better things for you to do with your time.

[quote="aufha"]Is it frowned upon by colleges to wait? Or is waiting just a bad idea for some other reasons?[/quote]
I :love: working with students who come (or return) to college some years after finishing high school (and who are sometimes older than me). Their reasons vary: some want to start a new career, some did military service after high school, some flamed out of college the first time and eventually got tired of low-level jobs. The best thing about such students is that they actually know why they are in college.

But if you don't have a [b]serious[/b] reason to postpone college (much more serious than the desire to fill in every box on an application), then you should not. Unused knowledge and skills do deteriorate over time.

[quote="mlok"]I :love: working with students who come (or return) to college some years after finishing high school (and who are sometimes older than me).[/quote]
One subcategory of this - the following is a composite of several different students I have known. Female. Mid or late 30's in age. Old academic records show that she was always a good student, someone with high grades, but in her first round of college, showed little apparent interest in math/science/engineering. But now that's what she's come back for. And the first semester back is very, very hard, and her grades that first semester are lower than those on her old record. But it gets better - call that first semester the "kicking the rust out of the pipes" experience. As mlok said, "Unused knowledge and skills do deteriorate over time." The only thing that saves her is that she always had an aptitude for this; she just didn't previously know she had the interest.