I am reviewing free-response questions from past AP Chemistry Exams en route to taking this year's AP Chemistry Exam.

Can anyone help me with part (b) (i) of #8 of the 2003 Free-Response section located here:

[url]http://www.collegeboard.com/prod_downloads/ap/students/chemistry/chemistry_frq_03.pdf[/url]

It asks to explain why propane and propanone do not have the same value for their standard heats of vaporization.

How exactly can I word it to explain the difference in the types of intermolecular forces the different species have?

Thanks!

Propane is $\text{CH}_{3}\text{CH}_{2}\text{CH}_{3}$.  Propanone is $\text{CH}_{3}\text{COCH}_{3}$.

Because a propane molecule is symmetric and its own atoms are covalently bonded, its only intermolecular forces are London dispersion forces, caused by temporary dipoles resulting from electrons moving in the molecule and which are relatively weak.  However, a propanone molecule is asymmetric in addition to containing atoms covalently bonded to each other; thus, its intermolecular forces are dipole-dipole forces, caused by permanent dipoles resulting from electrons being drawn from one part of the molecule to another and are relatively stronger.  Because its intermolecular forces are stronger, propanone has a higher $\Delta H_{vap}^{\circ}$ than propane.

Can anyone help me in cutting down the length of this explanation?  I tend to be too wordy... which could get me points taken off.  Is there ANY way ANY of you notice so that I could be briefer and more succinct, concise, and to the point?

Maybe I don't have to explain why each compound has that particular type of bonding?  Something like, "This molecule has these forces; propanone's are stronger, so it has a higher standard heat of vaporization."

Propane is a gas at room temperature. You can buy it readily, but only in pressurized containers. Propanone (more usually called acetone), is a liquid at room temperature. You can also buy it, at drugstores or hardware stores, in a screw-top plastic bottle.

Now, consider another commonly available product: isopropanol, $\text{CH}_{3}(\text{CHOH})\text{CH}_{3}.$ This is also a liquid at room temperature, but has a higher heat of vaporization and a higher boiling point than propanone. Like propanone, isopropanol is polar, but its dipole moment is probably a little smaller. What other force or forces explain its higher boiling point?

Thanks for the info on their phases and where they are commonly available, etc.

Isopropanol definitely has hydrogen-bonding forces, which are quite strong, because it involves a $\text{H}$ bonded to an $\text{O}$.

You state that a propane molecule is symmetric while an acetone molecule is not.  However, acetone is certainly symmetric: just write it's formula backwards!  I think I see what you are getting at, but that's not actually useful in this problem.  If you draw out the structure, you'll see why.  Also keep in mind that propane does not actually exist in a symmetric form, it makes a zig-zag chain.

The important explanation here is London dispersion forces vs. dipole-dipole (and possibly some dipole-induced dipole) interactions.  That's really all you have to mention.

As a related question that I've started wondering about, can a covalently-bonded molecule that is symmetric and hence has a net zero dipole moment interact through dipole-dipole forces?  I would think so, but such compounds are by definition non-polar... anyway, it seems to be an interesting question.

For example, what about something like $C(C=C=C=O)_{4}$?

Ok, my explanation I should replace "symmetric" with "nonpolar" and "asymmetric" with "polar"?


quadratics
number theory
prime factorization

$xy$$=$ $82896$

$x$+$y$ $=$ $578$

I keep on simplifying an equation to this and just have to brute-force the answer. Is there an easier way to do this?

[hide="answer to equation"]
$x$ $or$ $y$ $are$ $264$ $or$ $314$
[/hide]

Find the Prime Factorization of $82896$, then play around with it to get 2 factors that have a sum of $578$.

I don't see an easier way than that... :|

easy... think more dude!
hint: 2 equations + 2 variables = solvable

im gonna laugh if no one gets the right way within 3 hours

wrong wrong! u don't use prime factorization... it would just be like brute forcing in a smaller scale

i know how to do it if you are given a calculator.

i think it ends up being quadratic if u use substitution...i hate those, but it is solvable with a calculator...

[hide]
for $xy = 82896$ and $x+y = 578$ the units digits of the two numbers have to be $4$.

and since you can see that the prime factorization of $82896$ results in $2^{4}* 3 * 11 * 157$ and (a little vague here) the only way to make the units digit of $157$ equal $4$ is to multiply it by $2$, the answers are $157 * 2 = 314$ and $2^{3}* 3 * 11 = 264$.
[/hide]


This problem has, I know, at least two quite different solutions.

Does there exist a strictly increasing function $f(x)$ from the positive rationals to themselves satisfying $f(f(x))=3x$ for all $x$, but $f(x)\not\equiv x$?

There is a unique increasing function $f: \mathbb{N} \rightarrow \mathbb{N}$ such that \[ f(f(n)) = 3n \] for all $n$ - that's not difficult to prove.

I don't immediately see how to continue.


Problems on Induction from the INMOTC . The second was was for The Pre RMO Camp Selection.It contains some Easy but good for Practice Problems.

I would some more Questions as soon I Get It.

Nice Aman.
But if you notice, all the induction problems have been taken from Engel directly.

Sorry Aravind !! I haven't Done Induction from Engel.
Also almost all(3/4th) of the Problems in the Polynomials PDF are from Engel

Hey can anyone post the answers for Qs 10,11,12 of RMO selection camp problems posted by aman


function
real analysis
real analysis unsolved

1. Prove theorem: any infinite integer set is countable (don't use the fact that every infite subset of a countable set A is countable cuz that's wat I'm trying to prove in the end.)

2. Definition: A is at most countable if A is finite or countable. What does "at most countable" mean? how to apply this in solving problems?  :huh:

3. general question: how to prove set A ~ J when you cannot find a function s.t f(n) = a? (n-->all natural numbers, a is an element of set A. )

thx a lot  :lol:

At most countable means finite or countably infinite.  If you prove the set is countable, then it is at most countable.  For the first one, what are you trying to prove in the end, since the point you mentioned is pretty useful.

Bernstein's theorem allows you to construct a one-one map $ f: A\rightarrow J$ and a one-one map $ g: J\rightarrow A$.  If you can construct these two, you don't need to worry about making them onto, and $ A\sim J$.  Otherwise, it depends on how $ A$ and $ J$ are defined.

thx for the reply, it is very helpful. =)

For the 1st question, I'm actually trying to prove that every infite subset of a countable set is also countable. So Let A = a countable set, E =an infite subset of A. Then, store the positions of E in A in a new set nk. For example:A ={1.41, 2, -3, 5, 7,...}, E ={ 1.41, -3, 7, ..} then nk= {1, 3,5,...}. Since nk is a set of positive integers, if I can show N ~ nk (which is question 1), then f(nk)=E so I can N~E. So again, how do u prove question 1? ==T

number 3's kinda related to number 1,so if u show me how to do #1 I can probably generalize the solution.  :)


Let $ ABC$ be an acute triangle.Prove that:
$ (cos\frac{A}{2})^3+(cos\frac{B}{2})^3+(cos\frac{C}{2})^3 \le \frac{9\sqrt{3}}{8}$
And, we also have the inequality than stronger:
$ (cos\frac{A}{2})^3+(cos\frac{B}{2})^3+(cos\frac{C}{2})^3 \le \frac{9\sqrt{3}}{8}-\frac{3}{10}(\frac{3\sqrt{3}}{8}-cos\frac{A}{2}cos{\frac{B}{2}cos\frac{C}{2})}$

[quote="quykhtn-qa1"]Let $ ABC$ be an acute triangle.Prove that:
$ (cos\frac {A}{2})^3 + (cos\frac {B}{2})^3 + (cos\frac {C}{2})^3 \le \frac {9\sqrt {3}}{8}$
And, we also have the inequality than stronger:
$ (cos\frac {A}{2})^3 + (cos\frac {B}{2})^3 + (cos\frac {C}{2})^3 \le \frac {9\sqrt {3}}{8} - \frac {3}{10}(\frac {3\sqrt {3}}{8} - cos\frac {A}{2}cos{\frac {B}{2}cos\frac {C}{2})}$[/quote]
What about the folowing problem?
let $ ABC$ be an acute triangle and $ a \ge b \ge c$.Prove that:
$ \frac{h_a^2}{m_b^2}+\frac{h_b^2}{m_c^2}+\frac{h_c^2}{m_a^2} \ge 3$  :)


It's a nationwide competition with a bunch of schools in it. Basically there are 6 tests a year and 7 questions per test. They generally get harder as the year progresses. I'm from Flour Bluff, you'll find me as Raphael-Joel Lim on the newsletter :)

I've never heard of it either . . . then again, my district is relatively new to math competitions.

MAybe you could tell us what it stands for. Ive never heard of the acronym, but maybe ive heard of it in ia full title. :D

You can read all about it at:

http://www.asan.com/asa/asma1.htm

I also take the AtPac, and the Sigma, as well as the AMC 8  and 10.

I know a lot of people take the AMC's, but does anyone take either of the other ones? :)


trigonometry
geometry
trig identities
Law of Cosines
inequalities proposed
inequalities

you will have to believe me that after squaring it reduces to $(ab+bc-ac\sqrt{2})^{2}\geq  0$  :D

:rotfl:  god ... I didn't tried this  , I didn't think this is such a humdrum .  :blush:  I was searching for a nice solution :)

for a nice solution i think you must try to find a geometrical solution (i find it) :)

[hide]
Create points $O,A,B,C$ in the plane such that $OA=a$, $OB=b$, $OC=c$, $\angle AOB=\angle OBC=45^\circ$.

(oh, and i guess i have to add that none of the lines $\overline{AB},\ \overline{BC},\ \overline{AC}$ pass through $O$, to prevent stupid drawings from appearing...)

Since $AB+BC\geq AC$, and the Law of Cosines gives $AB=\sqrt{a^{2}+ab\sqrt{2}+b^{2}}$, $BC=\sqrt{b^{2}-bc\sqrt{2}+c^{2}}$, and $AC=\sqrt{a^{2}+c^{2}}$, we're done.

If you want to know the equality case, it occurs when $C$ lies on $\overline{AB}$, so the areas of $\triangle OAB$ and $\triangle OBC$ add up to the area of $\triangle OAC$. In order for this to happen,
\[\frac{ab}{2}\cdot \frac{1}{\sqrt{2}}+\frac{bc}{2}\cdot \frac{1}{\sqrt{2}}=\frac{ac}{2}\]
\[b=\frac{ac\sqrt{2}}{a+c}\]
[/hide]


induction
combinatorics unsolved
combinatorics


modular arithmetic
Miscellaneous Problems

Four consecutive even numbers are removed from the set \[A=\{ 1, 2, 3, \cdots, n \}.\] If the arithmetic mean of the remaining numbers is $51.5625$, which four numbers were removed?

Let the sum of the remaining numbers be $ A$. Then, because there are
$ n \minus{} 4$ numbers remaining after $ 4$ numbers are removed, have
$ \frac {A}{n \minus{} 4} \equal{} 51.5625 \equal{} \frac {825}{16}$
Now, the sum of the $ 4$ even numbers removed is greater than
$ 10 \equal{} 1 \plus{} 2 \plus{} 3 \plus{} 4$ but less than $ n \plus{} (n \minus{} 1) \plus{} (n \minus{} 2) \plus{} (n \minus{} 3)$ because the numbers removed are all even and belong to the set $ \{1,2,...,n\}$.
The sum of all the numbers in set $ \{1,2,...,n\}$ is $ \frac {n(n \plus{} 1)}{2}$.
Then, $ \frac {n(n \plus{} 1)}{2} \minus{} (n \plus{} (n \minus{} 1) \plus{} (n \minus{} 2) \plus{} (n \minus{} 3)) < A < \frac {n(n \plus{} 1)}{2} \minus{} 10$
Simplifying, get:
$ \frac {(n \minus{} 3)(n \minus{} 4)}{2} < A < \frac {(n \minus{} 4)(n \plus{} 5)}{2}$
Then:
$ \frac {n \minus{} 3}{2} < \frac {A}{n \minus{} 4} < \frac {n \plus{} 5}{2}$
But $ \frac {A}{n \minus{} 4} \equal{} \frac {825}{16}$ so:
$ \frac {n \minus{} 3}{2} < \frac {825}{16} < \frac {n \plus{} 5}{2}$
Clearing denominators, we get $ 98.125 < n < 106.125$, and as $ n$ is a
positive integer, $ 99 < \equal{} n < \equal{} 106$.
On the other hand, if the four consecutive numbers are
$ 2(k \minus{} 1),2(k),2(k \plus{} 1),2(k \plus{} 2)$($ k$ is an integer) their sum is $ 2(4k \plus{} 2)$ the sum of all the numbers from $ 1$ to $ n$ is:
$ A \plus{} 2(4k \plus{} 2) \equal{} \frac {n(n \plus{} 1)}{2}$
But $ A \equal{} \frac {825}{16}(n \minus{} 4)$, so:
$ \frac {825}{16}(n \minus{} 4) \plus{} 2(4k \plus{} 2) \equal{} \frac {n(n \plus{} 1)}{2}$    (*)
Because $ 2(4k \plus{} 2)$, $ \frac {n(n \plus{} 1)}{2}$ are integers, $ \frac {825}{16}(n \minus{} 4)$ is an integer, so $ 16|(n \minus{} 4)$ so $ n\equiv 4\pmod {16}$. Then, the only such value of $ n$ such that $ 99 \le n  \le 106$ is $ n \equal{} 100$.
Then, substituting $ n \equal{} 100$ into (*) have:
$ 2(4k \plus{} 2) \equal{} 100$ then $ k \equal{} 12$ and so the four consecutive even integers removed are $ 22,24,26,28$.


Let $ P(x) \equal{} (x \minus{} 1)(x \minus{} 2)(x \minus{} 3)$.  For how many polynomials $ Q(x)$ does there exist a polynomial $ R(x)$ of degree $ 3$ such that $ P(Q(x)) \equal{} P(x)\cdot R(x)$?

[hide]$ 22$ polynomials[/hide]
Thanks in advance!

[size=75][EDIT = Added the correct answer.][/size]

let $ Q(x) \equal{} u \implies (u \minus{} 1)(u \minus{} 2)(u \minus{} 3) \equal{} (x \minus{} 1)(x \minus{} 2)(x \minus{} 3) * R(x) \therefore$ we have a few  cases, either $ u \equal{} 1,2,3$ at $ x \equal{} 1,2,3$ or $ u \equal{} (1,2,3)$ at the zeros of $ R(x)$. $ R(x)$ has $ 3$ zero's so there are
$ 3 *3*3 \equal{} 27$. We subtract 5 to discount the case where the selected roots are $ 1,2,3$ because in this cause u is linear when it is clearly a quadratic $ \implies \boxed{22}$ cases.

EDIT: fixed

[quote="mathemonster"]let $ Q(x) \equal{} u \implies (u \minus{} 1)(u \minus{} 2)(u \minus{} 3) \equal{} (x \minus{} 1)(x \minus{} 2)(x \minus{} 3) * R(x) \therefore$ we have a few  cases, either $ u \equal{} 1,2,3$ at $ x \equal{} 1,2,3$ or $ u \equal{} (1,2,3)$ at the zeros of $ R(x)$. $ R(x)$ has $ 3$ zero's so there are
$ (6C3) \equal{} 18$ ways to choose $ 3$ roots that correspond to u, $ 3! \equal{} 6$ ways to assign them to values of u for a total of $ 108$ polynomials. We subtract 1 to discount the case where the selected roots are $ 1,2,3$ because in this cause u is linear when it is clearly a quadratic $ \implies \boxed{107}$ cases.[/quote]

I don't know if you checked the answer, but that's off...by a lot. I fail to see your flow of ideas here, and I know for sure this is not correct.
In addition, this is an old AHSME/AMC 12 problem so please refer to the contest section of the site, to get a solution, or look it up in the Wiki.

[quote="mihail911"][quote="mathemonster"]let $ Q(x) = u \implies (u - 1)(u - 2)(u - 3) = (x - 1)(x - 2)(x - 3) * R(x) \therefore$ we have a few  cases, either $ u = 1,2,3$ at $ x = 1,2,3$ or $ u = (1,2,3)$ at the zeros of $ R(x)$. $ R(x)$ has $ 3$ zero's so there are
$ (6C3) = 18$ ways to choose $ 3$ roots that correspond to u, $ 3! = 6$ ways to assign them to values of u for a total of $ 108$ polynomials. We subtract 1 to discount the case where the selected roots are $ 1,2,3$ because in this cause u is linear when it is clearly a quadratic $ \implies \boxed{107}$ cases.[/quote]

I don't know if you checked the answer, but that's off...by a lot. I fail to see your flow of ideas here, and I know for sure this is not correct.
In addition, this is an old AHSME/AMC 12 problem so please refer to the contest section of the site, to get a solution, or look it up in the Wiki.[/quote]

I got this problem from the Intermediate Algebra book, so I don't the problem number.  Searching didn't work since most of the problem would probably have been typed in $ \text{\LaTeX}$.  Could someone please tell me the question number or solution?

Thanks a lot!

[hide="So Close"]
Maybe someone can point out where I go wrong, but I think I have the basic idea.

We have that $ \deg (P(x) \cdot R(x)) \equal{} 6 \Rightarrow \deg Q(x) \equal{} 2$, or $ Q(x)$ is quadratic.

Now, we replace $ x$ in $ P(x)$ with $ Q(x)$ to obtain the following equality:
$ (Q(x)\minus{}1)(Q(x)\minus{}2)(Q(x)\minus{}3) \equal{} (x\minus{}1)(x\minus{}2)(x\minus{}3)(x\minus{}a)(x\minus{}b)(x\minus{}c)$
Here $ a,b,c$ are the roots of $ R(x)$.

Now, in order for the two sides to be equal, we must have each $ (Q(x)\minus{}i), \ i \in \{1,2,3\}$ factor into either one root of $ P(x)$, two roots, or no roots.
If each $ (Q(x)\minus{}i)$ factors into one root, then there are $ 6$ possible choices for each root to be paired with each $ (Q(x)\minus{}i)$. If one $ (Q(x)\minus{}i)$ factors into two of such roots, then there are $ 3$ choices for which two roots it factors into, $ 2$ choices for which other $ (Q(x)\minus{}i)$ factors into another root, and $ 3$ initial choices for which $ (Q(x)\minus{}i)$ factors into 2 such roots for a total number of polynomials as

$ 6 \plus{} 2 \cdot 3 \cdot 3 \equal{} 24$

So I either over counted or did it wrong. Are $ Q(x)$ and $ R(x)$ necessarily monic polynomials, that is, is the coefficient of their highest degree necessarily 1? Are any of the above mentioned cases impossible?
[/hide]

[[2005 AMC 12A Problems/Problem 24]]

stuck, your first case included the instances where $ (x \minus{} i)|Q(x \minus{} i)$ and $ (x \minus{} 4 \plus{} i) | Q(x \minus{} i)$ for $ i \equal{} 1,2,3$, which causes $ Q(x)$ to be a linear polynomial ($ Q(x) \equal{} x$ and $ Q(x) \equal{} 4\minus{}x$, respectively).


Given the numbers $ 1,2,2^2,...,2^{n\minus{}1}$. For a specific permutation $ \sigma \equal{} X_1, X_2, ... X_n$ of these numbers we define $ S_1 (\sigma) \equal{} X_1, S_2 (\sigma) \equal{} X_1 \plus{} X_2, S_3 (\sigma) \equal{} X_1 \plus{} X_2 \plus{} X_3, ...$ and $ Q(\sigma) \equal{} S_1 (\sigma) S_2 (\sigma) \cdots S_n (\sigma)$. Evaluate $ \sum 1/Q(\sigma)$ where the sum is taken over all possible permutations.

Only thing I know about this problem is that there are $ n$ terms and sum of all subsets of $ n$ terms is expressed as $ 2^n$. But this problem has $ 2^0,2^1,...2^n$ so I'm somewhat confused. And I don't see any nice way to manipulate $ Q$s. Any hint please?  :)

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=204839774&t=125603
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=472621896&t=168641

I'm adding the words
Canada
Canadian
permutation
reciprocal
1989
to this post to make future searches easier.

Thanks. I'm going through the years (what I feel like) and doing them so thanks for bringing them up. Mods can lock this.  :)


[quote="pavel kozlov"]
What is the same way?
[/quote]
[hide="I think this one"]
Taking $m=3, a=x_{1}, b=x_{2}, c=x_{3}$ in the following we obtain point c).
[u]Slight generalization[/u]. Let $m,n \ge 1$ be integers and $x_{i}> 0$, $\forall i \in \overline{1,m}$ such that $\prod x_{i}\ge 1$. Then the following inequality holds:
$n^{m}\prod_{i=1}^{m}\sum_{j=0}^{n}x_{i}^{j}\ge (n+1)^{m}\prod_{i=1}^{m}\sum_{j=0}^{n-1}x_{i}^{j}$.
Notice that it suffices to prove that $n\sum_{i=0}^{n}a^{i}\ge \sqrt{a}\cdot(n+1)\sum_{i=0}^{n-1}a^{i}$, for any $a \in \{x_{1}, ..., x_{m}\}$. Multiplying this relations for all $x_{i}$ yields the result, since $\prod \sqrt{x_{i}}\ge 1$ is obvious. The last inequality can be proven for any $a > 0$. First substitute $x = \sqrt{a}$. After some computations, we see that $n\sum_{i=0}^{n}x^{2i}-$ $(n+1)\sum_{i=1}^{n}x^{2i-1}\ge 0$ is $(n-1)\sum_{i=0}^{n-1}((i+1) \cdot x^{2i+1}-$ $(n-i) \cdot x^{2(i-1)}) \ge 0$, or equivalently $(n-1)^{2}$ $(\sum_{i=1}^{n}(ni-i^{2}+i) \cdot x^{2(n-i)}+$ $\sum_{i=1}^{n-1}(ni-i^{2})\cdot x^{2(n-i)-1}) \ge 0$. Since all the coefficients are positive, the relation is proven.
[/hide]

[quote="rogue"]Let $a,b,c>0$ and $abc\ge1.$ Prove that
a) $\left(a+\frac{1}{a+1}\right)\left(b+\frac{1}{b+1}\right) \left(c+\frac{1}{c+1}\right)\ge\frac{27}{8}.$[/quote]
By AM--GM ,we have

$\left(a+\frac{1}{a+1}\right)\left(b+\frac{1}{b+1}\right) \left(c+\frac{1}{c+1}\right)$

$\ge\left(1+\frac{3a-1}{4}\right)\left(1+\frac{3b-1}{4}\right) \left(1+\frac{3c-1}{4}\right)$

$=\frac{27}{64}\left(a+1\right)\left(b+1\right)\left(c+1\right)$

$\ge\frac{27}{8}\sqrt[3]{abc}\ge\frac{27}{8}.$

perhaps you meant:
$ \frac{27}{64}(a+1)(b+1)(c+1)\geq \frac{27}{8}\sqrt{abc}\geq \frac{27}{8} $ Am I right?

[quote=sqing]Let $a,b,c>0$ and $abc\ge1.$ Prove that $$\left(a+\frac{1}{a^4+2}\right)\left(b+\frac{1}{b^4+2}\right) \left(c+\frac{1}{c^4+2}\right)\ge\frac{64}{27}.$$[/quote]
It's enough to prove this inequality for $abc=1$, which is very ugly, but it's true.

Let $a,b,c>0$ and $abc\ge1.$ Prove that
$$\left(2a+\frac{1}{b+1}\right)\left(b+\frac{1}{c+1}\right) \left(c+\frac{1}{a+1}\right)\ge\frac{16}{3}$$
$$\left(a+\frac{1}{b+1}\right)\left(b+\frac{1}{c+1}\right) \left(c+\frac{1}{a+1}\right)\ge\frac{27}{8}$$


linear algebra
matrix
superior algebra
superior algebra theorems

If $A\in M_n(\mathbb{R})$ s.t $detA=1$, prove that there exist $X,Y$ invertible matrices in $M_n(\mathbb{R})$
$A=X\cdot Y\cdot X^{-1}\cdot Y^{-1}$

Moubi, could you please post your proof? I read in a french book ( Problemes choisis...) a proof, but it was a whole subject about it. Do you have a short proof for this problem?

Apply the

Theorem: $A$ real nxn matrix suppose $rank(A)=n$ assume A is not scalar matrix
Let $b_1,b_2,...,b_n ; c_1,c_2,...,c_n$ given non zero  scalars
Assume $det(A)= b_1.b_2...b_n.c_1.c_2...c_n$ then there exist $B,C$ matrices
with eigenvalues $b_k, c_k$ s.t. $A=BC$

" Problmes choisis " who is the author ? Editor ?

This problem is equivalent to $G/[G,G]\simeq (\mathbb{R}^*,\cdot)$ where $G=GL_n(\mathbb{R})$.
Or better said $GL_n(\mathbb{R})^{ab}=\mathbb{R}$.
I know no proof for this but I'm eager to see one.


You are shopping for a new pair of shoes. As you scan some advertisements, you
find five different offers for the type of shoe you are interested in.

Store A: Your shoe is normally 50 dollars, but is now on sale for 30% off.
Store B: Your shoe is normally 60 dollars, but is now on sale for 40% off.
Store C: Your shoe is normally 80 dollars, but is now on sale for 55% off.
Store D: Your shoe is normally 90 dollars, but is now on sale for 60% off.
Store E: Your shoe is normally 100 dollars, but is now on sale for 64% off.

Based on these sales, which store is selling the shoe you’re interested in for the cheapest price?
A. Store A
B. Store B
C. Store C
D. Store D
E. Store E

[hide]store A is 1 dollar cheaper haha[/hide]


[quote="manlio"]Let a_0 and a_1 be positive integers and for n \geq 2 define

a_n=(a_(n-1)) 2 /a_(n-2)

For what choices of a_0 and a_1 will all the a_n be integers?[/quote]

a_n >0 for any n \in N={1;2;...}

t(n)=lna(n), t(n)-t(n-1)=t(n-1)-t(n-2), t(n)=bn+c with initial conditions

a(n)=a(0).(a(1)/a(0))^n is positive integer for any n, if a(0)|a(1)

It seems STT math problem
Mr.Omarjee

In my opinion your solution, Moubinol, is not complete. You must prove that it is both necessary and sufficient that a_0 divide a_1.

I think that Moubinol solution is not complete. He has only proved that the condition "a_0 divides a_1" is sufficient but not that it is necessary.

Am I right?

This is my reasoning for the secon part. I hope it is correct.

Suppose that a_0 doesn't divide a_1. Then there is a prime p and  natural numbers n and k such that k<n and p^n divides a_0, p^k divides a_1 but p^(k+1) does not divide a_1. Then we count the number of factor of p in a_(n+1) which must be positive for a_(n+1) to be an integer.
By our formula for a_n we see there are n+1 factors of a_1 in the numerator of a_(n+1) and n factors of a_0 in the denominator.
Thus there are (n+1)k -n^2 \leq (n+1)(n-1) -n^2 =-1 factors of p in
 so a_(n+1) is a fraction that in reduced terms has a factor of p in the denominator.

Hence it is both necessary and sufficient that a_0 divides a_1.

Have you a more simple proof for this second part ?

Moubinol did not state that his result was also necessary, but it is an easy step from his final deduction to find that indeed a_0 | a_1 is necessary.

[quote]By our formula for a_n we see there are n+1 factors of a_1 [/quote]

Hmm I'm not sure which formula for a_n, to which you refer, yields that there are n+1 factors of a_1. (I'm probably not seeing a step)

The following may be simpler...

First assume that all the a_n's are integers.

Assume, for the sake of contradiction, that a_0 does not divide a_1.
There exist x, y such that a_1 / a_0 = x / y and (x, y) = 1, y > 1.

Define a_n / a_(n-1) = b_n for all positive n.
Clearly, b_n = b_(n-1) = ... = b_1 = x / y for all positive n.
Moreover, b_n * b_(n-1) * ... * b_1 = a_n / a_0 for all positive n.

Thus, x^n / y^n = a_n / a_0.
-> a_n = a_0 * x^n / y^n.

The LHS is an integer, so RHS is also an integer.
So y^n | (a_0 * x^n). But (y^n, x^n) = 1, so y^n | a_0.

Since this must be true for any positive integer n, we must have y = 1. Contradiction.

Thus, a_0 | a_1.

I think your solution is better than mine because more simple and neat.
Have you checked my solution? Is it right?

P.S. I said that
                       "by our formula for a_n we see that there are n+1
                        factor of a_1 in a_(n+1)"

                        because   a_n=a_0(a_1/a_0)^n

                        and then  a_(n+1)=a_0(a_1/a_0)^(n+1)

Hello.

[quote]Then there is a prime p and natural numbers n and k such that k<n and p^n divides a_0, p^k divides a_1 but p^(k+1) does not divide a_1. Then we count the number of factor of p in a_(n+1) which must be positive for a_(n+1) to be an integer. [/quote]

As opposed to what you have said, what if there exists such n, k such that p^n divides [b]a_1[/b], p^k divides [b]a_0[/b] but p^(k+1) does not divide [b]a_0[/b]?

Then it seems that the number of p's in a_n, according to the formula, is not negative as you showed.[/b]

You are right. My solution is not good.
Luckily your solution is good and also very nice.
Thank you very much.

I don't know why, but I find this problem extremely simple. So, let's suppose that (a_0)^n-1|(a_1)^n for all n. Let p a prime such that p|a_0 and let v and w the exponent of p in a_0 and a_1. Then (n-1)v<=nw for all n and v<=(1-1/n)w for all n. It follows that v<=w and thus a_0|a_1.

Harazi's solution to this problem is genial by its simplicity.

Congratulation to Harazi.


Find the value of $\frac{1\times2\times4+2\times4\times8+...}{1\times4\times7+2\times8\times14+...}$.

[hide="Not sure if you mean this"]$\frac{1\times2\times4+2\times4\times8+\dots }{1\times4\times7+2\times8\times14+\dots }= \frac{(1\times2\times4)(1+2\times2\times2+\dots )}{(1\times4\times7)(1+2\times2\times2+\dots )}=\frac{1\times2\times4}{1\times4\times7}=\frac{2}{7}$[/hide]

Why was it moved to the high school forum?

[quote="nutz_for2.718281828"]Most likely because it was high-school level.[/quote]
But it is my homework.......... I am a junior 1 student.

SHAME not doing your homework!!! :P

Well, easy highschool level problems may be mosted in the Mathcounts section, but all medium and hard ones must be posted here or the pre-olympiad section.


Six people are going to sit in a row. Grant does not want to sit next to Lee. Homer wants to sit next to Marge. Bill and Ted can sit anywhere. How many ways can these six people be seated?

This probably isn't the best way to do it, but it can be done indirectly.
Call the elements $G, L, H, M, B, T$. Since Homer wants to sit next to Marge, we can group $H$ and $M$ together, and count them as $HM$ and $MH$.

How many ways we can arrange them with Grant and Lee next to each other?
$(4\times3\times2\times1)\times4=96$ (We multiply by four to account for $GL$, $LG$, $MH$ and $HM$ (as one element).

However, we do NOT want these cases. So let's look at all possible cases.
$(5\times4\times3\times2\times1)\times2=240$ (We multiply by two to account for $MH$ and $HM$ as one element)

However, since those 96 case are the ones we DON'T want (because we counted $LG$ and $GL$), we subtract them so we get $240-96=\boxed{144}$

Is this right?


A total of $119$ residents live in a building with $120$ flats. We call a flat $overpopulated$ if there are at least $15$ people living there. Every day the inhabitants of an overpopulated flat have a quarrel and each goes off to different flat in the building. Will the process necessarily stop?

[hide="Hint"]Try finding some monovariant that describes crowdedness[/hide]

The monovariant is the number of handshakes of all people with everyone in their rooms.


what is the only correct answer?

A. Let S be a non-empty, bounded-below subset of $ \mathbb{R}$, and let $ \beta \equal{} infS.$
        then it is sure that:
        i) it does exist $ x\in S$ such that $ x<\beta\plus{}\epsilon \ \forall \epsilon >0$
        ii)  the set $ [\beta,\beta\plus{}1[ \ \cap S$ is not empty
        iii) for all $ \epsilon >0$ there exists $ x\in S$ such that $ \beta \minus{}\epsilon <x\leqslant\beta$

B. The set $ \displaystyle\bigcup_{n\equal{}1}^{\plus{}\infty} [1/n,1[$:
i) is closed
ii)is open
iii)is neither closed nor open (I mean is not closed and is not open)

A. Consider a set that consists of a single point.

B. Try to write the union as one interval.

-for A: if I consider a set of one element, this element must be $ \beta$; then both i) and ii) are true: the first because there exists $ \beta\in S$ such that $ \beta<\beta+\epsilon$, for all $ \epsilon >0$; the second because $ [\beta,\beta+1)\cap\{\beta\}=\{\beta\}$ is not empty.

please, tell me what's wrong :(

Sorry, I misread the problem. (i) and (ii) are actually true, and (iii) fails when $ \beta\notin S$.


find sum $ {{C_n}^1}sinx\plus{}{{C_n}^2}sin2x\plus{}{{C_n}^n}sinnx$

Maybe your problem would be $ C_n^1 \sin x\plus{}C_n^2 \sin 2x\plus{}\cdots \plus{}C_n^n \sin nx$.

$ \binom{n}{1}\sin x+\binom{n}{2}\sin 2x+\cdots+\binom{n}{n}\sin nx  \& =\sum_{k=0}^{n}\binom{n}{k}{e^{kxi}-e^{-kxi}\over 2i}\\
\hspace*{2.57in}= {(1+e^{xi})^n-(1+e^{-xi})^n\over 2i}$
with
$ 1+e^{xi}=1+\cos x+i\sin x=2\cos^2 {x\over 2}+2i\sin {x\over 2}\cos {x\over 2}=2\cos {x\over 2}e^{xi/ 2}$
and analog
$ 1+e^{-xi}=2\cos {x\over 2}e^{-xi/ 2}$.

So
$ \binom{n}{1}\sin x+\binom{n}{2}\sin 2x+\cdots+\binom{n}{n}\sin nx=2^n\cos^n{x\over 2}\cdot {e^{nxi/ 2}-e^{-nxi/ 2}\over 2i}\\
\hspace*{2.57in}=2^n\cos^n{x\over 2}\sin {nx\over 2}$


let a,b,c>0
prove that
 \sum a ^3 + 6(\sum (cyclic)a 2 b) \geq 3( \sum (cycle)ab 2 )+12abc.
try it  :D  :D (may be it's easy)

Yes, the inequality is interesting. But what is more interesting is the question why nobody solved it since it was posted four months before.

In fact, you ask to prove the inequality

$a^3+b^3+c^3 + 6\left(a^2b+b^2c+c^2a\right) \geq 3\left(ab^2+bc^2+ca^2\right) +12abc$

for every three positive real numbers a, b, c.

My proof simply consists of rewriting the inequality in the form

$b\left(b+c-2a\right)^2 + c\left(c+a-2b\right)^2 + a\left(a+b-2c\right)^2 \geq 0$,

what is trivial.

BTW, I didn't use Fanfan's killer program to find that  :D

  Darij


can someone please send me the 03 and/or the 04 national sprint and target rounds?? if you don't wish to post it on the site can you send it to me via email? url=mailto:nikhilnidhi@hotmail.com]nikhilnidhi@hotmail.com[/url]

can someone send me targets? smarteryaman@gmail.com

thanks! that would be great. Is it possible someone could also send me solutions or even answers? Also if anyone has 2004 that would be greatly appreciated

can someone send me nats sprint and target for 03 and 04 (w/ answers or solutions if possible) at [hide]kadeel@gmail.com[/hide] Thanks  :)

can someone send me 1998/1999 state+chapter target and sprint? Send to mcpreet@gmail.com

Someone give me any nationals test...i dont really care which one, but plz have solutions.

[hide]ameem91@gmail.com[/hide]

does anyone have the answer keys for 1991-1998 nats?


[quote="minsoens"]Don't you need to know the number of available toppings?[/quote]
[quote="DiscreetFourierTransform"]How many ways are there to order two pizzas with [b]up to 5 toppings[/b] on each?[/quote]

Let's say that there are 11 different toppings, that way you can use 10 distinct toppings.

maybe im not understanding the question correctly :( ... but
assume you have $n$ different toppings available and you can choose the number of toppings for each pizza.
EDITed:
$\binom{n}{5}$ different ways for one pizza.
if the topping combinations are same for both pizzas, then: $\binom{\binom{n}{5}}{2}$
if the topping combinations are different, then: $\binom{n}{5}$

total: $\boxed{\frac{\binom{n}{5}\left(\binom{n}{5}+1\right)}{2}}$

No, I don't think that's right, because you can have UP TO 5, not exactly 5 combinations of toppings.

[quote="DiscreetFourierTransform"]How many ways are there to order two pizzas with up to 5 toppings on each?[/quote]

[hide]I just assumed that there are $x$ possible toppings. Then, for one pizza, it would be $\dbinom{x}{5}$ toppings possible.
[hide="If the pizzas have to be distinct..."]Then there would be $\dbinom{x}{5}-1$ possbile toppings for the pizzas, making the total $\left(\dbinom{x}{5}\times\left(\dbinom{x}{5}-1\right)\right)$

Then, for 4, you have that, and for three and so on, to get:

$\\ \left(\dbinom{x}{5}\times\left(\dbinom{x}{5}-1\right)\right)\times\left(\dbinom{x}{4}\times\left(\dbinom{x}{4}-1\right)\right)\times\left(\dbinom{x}{3}\times\left(\dbinom{x}{3}-1\right)\right)\times\left(\dbinom{x}{2}\times\left(\dbinom{x}{2}-1\right)\right)$

[/hide]

I probably misunderstood somewhere. [/hide]

ok lets suppose there are $n$ toppings...

there are $S=\sum_{k=1}^{5}\binom{n}{k}$ different pizzas

clearly, the order you get the pizzas does not matter, so there are $\binom{S}{2}+S$ ways to get two pizzas

(EDITED $S$, not $S^{2}$...)

[quote="rishabh"]i didnt get the s^2 part. shouldnt it just be s(s-1)/2?

ps- sorry, i dont know latex :blush:[/quote]

woops... that should be $\binom{S}{2}+S$, remember, that order does not matter, so if we get AB, that is the same as BA, but if we get AA, that has to be counted differently


Hello, my first post here so....

Well, I thought you brains here could help me a little

Can you solve this problem for me or at least tell me something about it?

2x=1 (mod m)

x^2=-1 (mod m)

I know it works if  m=5 and x=3

This gives $ 2^2 \equiv \minus{} 1\mod m$, giving $ m \equal{} 1,5$.

[moved to Intermediate]

Hope it is correct we have x^2+1=0 ( mod m)   (x+1)^2-2x=0(mod m)  but 2x=1 so  (x+1)^2=1( mod m) , we know that are at most two solution of x ( it has degree 2) one is x=0 mod m ( replacing in the first equation we get that it is false except m=1) or x=m-2(mod m)  m-2=-2   (-2+1)^2=1   ( mod m) so x=m-2 ( mod m) then replacing we get 2(m-2)=1( mod m)   0=5 mod m so we finally get m=5 and x=3 and also m=1 x is every natural  ...Hope you understand and it's ok
Daniel

[quote="lambruscokid"]or x=m-2(mod m)[/quote]

Can you explain ? :)

Multiply the first equation by $ x$ and replace $ x^2$ with $ \minus{}1$.

Square the first mod, multiply the second by 4:

$ 4x^2 \equiv 1 \mbox{ mod } m$
$ 4x^2 \equiv \minus{}4 \mbox{ mod } m$

Hence,

$ \minus{}4 \equiv 1 \mbox{ mod } m$
$ \minus{}5 \equiv 0 \mbox{ mod } m$

Hence, $ m | \minus{}5$. Since $ m$ must be positive, either $ m \equal{} 1$ or $ m \equal{} 5$.


If all angles are measured in degrees, the ratio of three times the
measure of $ \angle A$ to four times the measure of the complement of $ \angle A$ to half the measure of the supplement of $ \angle A$ is $ 3: 14: 4$.
What is the number of degrees in the measure of the
complement of $ \angle A$?

Let $ x\equal{}\angle A$.  Three times the measure of $ \angle A$ is $ 3x$.  Four times the measure of the complement of $ \angle A$ is $ 4(80\minus{}x)$.  Half the measure of the supplement of $ \angle A$ is $ \frac{1}{2}(180\minus{}x)$.  Thus, $ \frac{3x}{4(90\minus{}x)}\equal{}\frac{3}{14} \implies 42x\equal{}12(90\minus{}x) \implies 54x\equal{}1080 \implies x\equal{}20$.  Thus, the answer is $ 90\minus{}20\equal{}\boxed{70}$.  To check our work, we compare another pair:  $ \frac{4(90\minus{}x)}{\frac{1}{2}(180\minus{}x)}\equal{}\frac{14}{4}\equal{}\frac{7}{2} \implies 8(90\minus{}x)\equal{}\frac{7}{2}(180\minus{}x) \implies 180\equal{}9x \implies x\equal{}20$ again, so we know our answer is correct.


ehhh...i know... that's not nice..
after full expansion and JUST A LITTLE computations(doing nothing else!)
we get :
$ \sum{\frac{a^{3}}{b\plus{}c}}\leq\frac{a^{4}\plus{}b^{4}\plus{}c^{4}}{a^{2}\plus{}b^{2}\plus{}c^{2}}{\sum{\frac{a}{b\plus{}c}}}  \Longleftrightarrow \sum a^2b^2 \le \sum a^4$
we r done!

[quote="2424"]With $ a,b,c \geq 0$. Prove that
   $ \sum{\frac {a^3}{b \plus{} c}} \leq \frac {a^4 \plus{} b^4 \plus{} c^4}{a^2 \plus{} b^2 \plus{} c^2}{\sum{\frac {a}{b \plus{} c}}}$[/quote]
The inequality is equivalent to
$ \sum \frac{a[a^4\plus{}b^4\plus{}c^4\minus{}a^2(a^2\plus{}b^2\plus{}c^2)]}{b\plus{}c} \ge 0.$
Since $ a^4\plus{}b^4\plus{}c^4\minus{}a^2(a^2\plus{}b^2\plus{}c^2)\equal{}c^2(c^2\minus{}a^2)\minus{}b^2(a^2\minus{}b^2),$ it can be written as
$ \sum \frac{c^2a(c^2\minus{}a^2)}{b\plus{}c}\minus{}\sum \frac{ab^2(a^2\minus{}b^2)}{b\plus{}c} \ge 0,$
or
$ \sum \frac{a^2b(a^2\minus{}b^2)}{a\plus{}c}\minus{}\sum \frac{ab^2(a^2\minus{}b^2)}{b\plus{}c} \ge 0.$
This rewrites as
$ abc\sum \frac{(a\minus{}b)^2(a\plus{}b)}{(a\plus{}c)(b\plus{}c)} \ge 0,$
which is obviously true. Equality holds if and only if $ abc\equal{}0,$ or $ a\equal{}b\equal{}c.$


let a,b,c > 0
and $ a^2\plus{}b^2\plus{}c^2\plus{}2abc\equal{}1$
Find MAX of $ a\plus{}b\plus{}c$

hello, we have $ a\plus{}b\plus{}c\le \frac{3}{2}$ and th equal sign holds for $ a\equal{}b\equal{}c\equal{}\frac{1}{2}$.
Sonnhard.

you can use substitution a=cosx, b=cosy, c=cosz, x+y+z=pi


I have a very difficult inequality :(  but I don't know it's proof. Hope you can solve it for me.
   Let ABC be an acute triangle. Prove that
      ( 1/prod cos[(B-C)/2]) +1 >= 3/( \sum cosA).
   In fact, I think it's right but it may be wrong in some triangles but I don't check. Please prove or disprove it. :D

This is really a difficult one.I changed into three sides of a triangle and have a terrible inequality.But it is interesting when I found that your inequality may be right for all triangles from that one.:maybe: However,I have not proved completely yet.But with some condition,your inequality is always right,for example:
      i)4b>a+c>3b/2 where a \geq b \geq c
  or ii)9(ab+ac+bc) \geq 8(a 2 +b 2 +c 2 ).
     where a,b,c are sides of a triangle.
I will replied if i find anything else interesting.  :P


Let x,y,z be real positive numbers such that x^3+y^3+z^3=1.

Prove that x^2+y^2+z^2  \geq x^5+y^5+z^5 +2(x+y+z)(xyz)^2

and  1/x^2  + 1/y^2  + 1/z^2  \geq x+y+z +2(x^4+y^4+z^4)/(xyz)

I try to do something with the first:

x^2+y^2+z^2 = (x^2+y^2+z^2)(x^3+y^3+z^3) = x^5+y^5+z^5 + x^2.y^3+x^2.z^3+y^2.x^3+y^2.z^3+z^2.x^3+z^2.y^3.

Now, we have  x^2.y^3+x^2.z^3 >= 2(xyz)^3/2.(x)^1/2 and the two similar. From that
x^2.y^3+x^2.z^3+y^2.x^3+y^2.z^3+z^2.x^3+z^2.y^3 >=
2(\sqrt x + \sqrt y + \sqrt z)(xyz)^3/2.

Finally, from x, y, z <= 1 we have \sqrt x >= x, \sqrt y >= y, \sqrt z >= z, (xyz)^3/2 >= (xyz)^2 and from that we get the result.

Namdung

$ \sum \frac{1}{x^2} = \left( \sum \frac{1}{x^2} \right) \cdot \left( \sum x^3 \right) = \sum x + \sum_{sym} \frac{x^3}{y^2} = \sum x + \frac{\sum_{sym} x^5y^2}{x^2y^2z^2} \ge \sum x + \frac{\sum_{sym} x^5yz}{x^2y^2z^2} = \sum x + 2 \frac{\sum x^4}{xyz}$


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i was reading this book (i forgot its name) but it had this cool mental math trick.

How do you do 92*95 in your head

[hide="solution"]1.  subtract those numbers from 100 and wrte them below the corresponding numbers:
92 * 95
8.......5

2. now now subtract either of the diagonals (it doesnt matter which) so:
92-5=87 or 95-8=87

3.  multiply the two bottom numbers:
8*5=40

so the answer is 8740[/hide]

That is pretty cool :D . What do you do if its a three digit number by a two digit one or multiplication of two three digit ones.

I saw the same trick in "Speed Mathematics", to which I forgot the author's name.

Yeah, i've read about that in speed mathematics, the authors Australian as far as my memory serves.

Link: [url=http://www.amazon.com/gp/product/0471467316/002-8030360-1823246?v=glance&n=283155]Book![/url]

Fast mental math is really about memorization.  You have to know your perfect squares, cubes, powers,...
Me, I use the distributive property ad infinitum to do mental math and extract factors to go from fractions to decimals.
For example, to find the decimal representation for$\frac{1}{11}$, you could say
$\frac{1}{11}=\frac{9}{99}=.\bar{09}\cdots$

I think I know what you're talking about, but unfortunately I don't remember it either...

I would rather multiply each number by $2$ and then divide the product by $4$ at the end, though.
(For example, $35\cdot 45=\frac{1}{4}\cdot 70\cdot 90 = \frac{6300}{4} = 1575$)

well, ya know, you could always just say:
92 * 95 = (2*95)*10^0 + (9*95)*10^1 = 190 + 8550 = 8740
pretty much just as quick

The math on this works out perfectly:

You're suggesting
X * Y can be solved by subtracting both X and Y from 100, multiplying both results, then adding this to the result from one number subtracted from the other, juxtaposing the latter in front of the former to form a 4-digit number

This is represented algebraically as:
(10a + b) * (10c + d) = 100(10c + d - (100 - 10a - b)) + (100 - 10a - b) * (100 - 10c - d)
simplifying:
100ac + 10ad + 10bc + bd = 1000c + 100d - 10000 + 1000a + 100b + 100(100 - 10c - d) - 10a(100 - 10c - d) - b(100 - 10c - d)
continuing, using only the right side of the equation:
=1000c + 100d - 10000 + 1000a + 100b + 10000 - 1000c - 100d - 1000a + 100ac + 10ad - 100b + 10bc + bd
(this can all be done in one step, but I'll show all that's being eliminated step by step using different typesets)
=[b]1000c[/b] + [u]100d[/u] [i]- 10000[/i] + 1000a + 100b + [i]10000[/i] [b]- 1000c[/b] [u]- 100d[/u] - 1000a + 100ac + 10ad - 100b + 10bc + bd
=[b]1000a[/b] + [i]100b[/i] [b]- 1000a[/b] [i]- 100b[/i] + 100ac + 10ad + 10bc + bd
=100ac +10ad + 10bc + bd

QED

sorry about lack of more math-ly text... unsure how to implement

[quote="robinhe"][quote="tz"]I remeber there was a trick for multiplying numbers that end in 5. But I keep forgeting it. Does anyone know it?[/quote]

If you are talking about $45\times 45$

It ends in 25 automatically.
Take the 4 and multiply it by 4+1 which is 5

So 45×45=2025

This goes for all numbers expressed as $10x+5$ where $x$ is an integer.

Sorry I have no easy way to put this.[/quote]

That's because $(10x+5)^2=100x^2+100x+25$, which factors into $100x(x+1)+25$. So the first part of the number is x(x+1) and the second part of the number is 25... Hope that counts as an "easy way to put this". ;)

Yeah I guess

But I was wondering if there was a more concise way to express that trick (as in non-algebraic) :)

I guess you learn new tricks everyday in Math, right?

Wow! I wonder if that will work for any numbers. I guess I'll try. :10:

All these tricks come from Vedic Mathematics. that's right, from India. they were discovered a long time and were recorded in mantras.

I'll research Vedic Mathematics! Thanks. :yinyang:

Fantastic mental methods...
Would you share more?

[quote="fireemblem13"]let's see. you probably know this one. When multiplying two number, $a$ and $b$, where $a=10x-n$and $b=10x+n$ the product is the square of the multiple of ten minus the square of the difference, $(10x)^2-n^2$.
Ex: $38*42=40^2-2^2=1596$

if you simplify $(10x)^2-n^2$ you get  $100x^2-n^2$. If you multiply the binomials, $10x-n$ and $10x+n$, you get that polynomial, $100x^2-n^2$.

also if you're multiplying two numbers, $10a+x$ and $10a+y$ where $x+y=10$, the product is equal to $100a(a+1)+xy$
Ex: $14*16=[1*2] [24]=224$
[this denotes a secton of the product][/quote]

for the second trick, you can use the fist trick

so 14*16 =15^2-1=224

here's another one. when you're trying to find the equation between two points, $(a,b), (c,d)$ arrange the coordinates in addition type format.
``   a   b
__c_d_
$(a-c)y=(b-d)x+$[b]/[/b]$ad-bc$[b]/[/b]
does that make sense. i don't know how do to absolute values though
you can find a lot more stuff, ranging from really easy stuff like addition and subtraction all the way to trig and calculus.

[quote="numbergeek"]All these is given in a book called Vedic Maths by Swami Ram Krishna Tirath Ji

Search for Nikhilam Sutra.
You shall find many more tricks like how to find value of 1/19
or say 3/18 correct to 10 decimal places in 6 secs[/quote]
wow, i will have to look into reading that book. :o

I shall domonstrate one of the case -
Take for example 2/19 and you want to find its value to 15 decimal places -

Now 2/19 can be written as 0.2/1.9.

Now 1.9 is approximately 2. Remember since we have made an approximation here so on dividing by 2 we shall have an approximate result not correct but the method is such that it corrects the approximation. Just see how the method goes -

0.2/2 :-

0.1 as quotient and 0 as remainder
Make two rows. In first row write the quotient and in the second row write the remainder -

Quotients - 0.1
Remainder   0

Now for next digit write remainder as ten'digit and quotient as unit's digit. So it becomes 01 to be divided by the divisor ie 2 again.

So remainder is 1 and quotient is 0.
So it becomes  -
Quotient -    0.10
Remainder -   01

For next division remainder ie 1 as ten's digit and quotient ie 0 as unit's digit. So 10 is to be divided by 2, quotient is 5 and remainder is 0

So it becomes  -
Quotient -    0.105
Remainder -   010

Similarly next 05 is to be divided by 2. 2 as quotient and 1 as remainder.

So it becomes  -
Quotient -    0.1052
Remainder -   0101

Now in this mental one line answer continue the process for as long as you want.....

So it becomes  -
Quotient -    0.10526315789.........................
Remainder -   01010011110..........................

and so on and so forth......
The quotient line is the answer. Try for 3/19, 4/19 etc.
If your fraction is improper say 23/19
Then make it 1 + 4/19 and then find 4/19 and add it to 1.0
There are modifications if Denominator is 18,17 etc.

isnt it cool ????


function
calculus
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real analysis theorems

Ok, so I've been doing some basic stuff with hyperbolic functions the last couple of days, and I was just wondering what they're actually used for.  The only thing that I've seen so far that seems like it really has much use for other things is using the inverse hyperbolic functions to evaluate certain integrals, but it seems like they should have more applications than that...

They're some pretty important integrals.

Anyway, some of this might be interesting to you: http://en.wikipedia.org/wiki/Hyperbolic_function


Suppose you have 9 camels, and they're trudging through the desert. At some time they have a water break at an oasis, and all the camels stop to get water, obviously. They're thirsty. After an hour, they have to leave and jumble into a random order, randomly. Obviously. What is the probability that no camel is immediately behind the one it was behind when it came?

Since no one has replied I will give a hint: Figure out the probability that a camel is behind the camel which it was previously behind and then subtract this from 1.

There was a similarly disguised problem previously posted in the Intermediate forums.

Let $S_i$ be the set of all arrangements in which camel $i$ is behind the same camel as before.
Then we want:
\[{\displaystyle 9!\\
-\sum_{1\le i\le9} \#(S_i)\\
+\sum_{1\le i<j\le9} \#(S_i\cap S_j)\\
-\ldots\\
+\ldots\\
\vdots\\
-\#(S_1\cap S_2\cap S_3\cap S_4\cap S_5\cap S_6\cap S_7\cap S_8\cap S_9})\]
This is principle of inclusion-exclusion on the sets of arrangements of the camels, where $\#(S_i)$ represents the number of "elements" in set $S_i$.

Divide the number above by the total number of rearrangements, $9!$.

Is there a way besides Inclusion Exclusion?

...

I think I've done a similar problem by Counting in Two Ways before, but I forget what exactly I counted.  :blush:


function
limit
real analysis
real analysis unsolved

Let $ f,g: R\to{R}$ . If $ g$ is continous, $ \lim_{x\to\infty}{g(x)} \equal{} \infty$, $ \lim_{x\to\infty}{f(g(x))} \equal{} l,$ prove that $ \lim_{x\to\infty}{f(x)} \equal{} l$.

Let $ M > 0$ and $ m \equal{} \min_{x\geq M}f(x)$ , and the function $ h: [M,\infty)\to [m,\infty)$ ,$ h(x) \equal{} g(x)$ ,$ h$ is surjective.Thus $ \forall y \in [m,\infty)$  $ \exists x \in [M,\infty)$ such that $ h(x) \equal{} y$
Let $ (y_n)_n$ be an arbitrary sequence such that $ y_n \rightarrow \infty$.We may assume that $ y_n \in [m,\infty)$ .Because $ h$ is surjective $ \Rightarrow$ there is a sequence $ (x_n)_n \in [M,\infty)$  , $ x_n \rightarrow \infty$ such that $ h(x_n) \equal{} y_n$.Therefore $ \lim_{n\to\infty}f(y_n) \equal{} \lim_{n\to\infty}f(h(x_n)) \equal{} l$ for any sequence $ y_n \rightarrow \infty$.Therefore $ \lim_{x\to\infty}f(x) \equal{} l$


number theory
prime numbers
irrational number

1If d ia integer number , un which d = p1 · · · pk, where k >= 1 and pi's are positive prime numbers, consider Z[pd] = {a + b*d^1/2}. Mostre que:
a) d^1/2 is a irrational  number

Reworded hard to read problem:

If $d$ is an integer, in which $d=p_{1}p_{2}\cdots p_{k}$, where $k\ge1$ and $p_{i}$'s are positive prime numbers, consider $\mathbb{Z}[\sqrt{d}]=\{a+b\sqrt{d}\}$. Show that $\sqrt{d}$ is irrational.

PS: Doesn't this belong to the Olympiad Section under Algebra forum?

I don't think that it belongs to the Olymiad section, it was an ewxercise from my number theory book

[quote="ZetaX"]As some of the others of you, this makes no sense:
Three possibilities what the problem is I think:
- show that $\sqrt{d}$ is irrational
- show some properties about the ring $\mathbb{Z}[\sqrt{d}]$
- both

When it is just the first or an easy property for the second, then it's good here.
But before anything can be said, we need the exact statement.

PS: I don't know how that much people think such stuff is Algebra... Number Theory concerns with the properties of integers as ring, which is definetely the case here.
In general, some people post everything into Algebra, and this must be stopped![/quote]Under your PS part, I only asked a simple question and you can say "No, this is not Algebra, its NT;" you don't have to be dramatic about how much people think such stuff is Algebra. Also, if you think that wrong posts in Algebra can be stopped, we wouldn't need moderators :). There are many ways to [b]reduce[/b] such wrong posts like PM to the original post creator and advice him/her. There will always be new members that post in the wrong forum, no way you can stop it.

I didn't want to be dramatic  :blush:
I just wanted to make clear for all were it belongs. and where not :)

And you are right, that way works, but there are hard cases (and Algebra sometimes lookes like a big amount of problems of all sections ;) ).
A guide for new users could help, but that requires some general definition of Algebra, Number Theory and so on...

(but lets stop the offtopic now or continue another way)

But something more to parrot (could be PMed, but it gives this post at least some sense): is your problem caused by bad english, by problems in writting the things down here or by something else¿ Because the questions itself and the forums look appropriated, but sometimes it's hard to understand what the question really is...


That's not so hard.
Step 1: For two vertices $U,V$ denote $p(U,V)$ the shortest path joining them(we easily see it's unique). Then $|p(U,V)\leq n+1$.
One easily sees that $p(U,V)\bigcap p(U,W)=\{U\}$ or one of this paths is contained in another (else we get a cycle with at most $2n$ vertices).
Step 2: Pick up a random vertex $U$. A vertex $V$ is called far if for any $W$ joined to $U$ we have $|P(U,W)|\leq | P(U,V)|$.
The we easily see that if $V_1,\cdots V_m$ are all far from $U$ then
$P(U,V_i)/\{U\}$ partition the vertices of the graph except $U$
Step 3: From Step 2 we deduce each vertex is joined to at least two vertices. Now take $U$ $V$ with $V$ far from $U$ and let $W$ be joined to $U$ s.t. we don't have $W\in P(U,V)$. Then $P(U,V),P(U,W)$ and the edge $VW$ form a cycle. That's possible only for $|P(U,V)|=|P(U,W|=n+1$.
Hence we established $|P(U,V)|=n$
Step 4: Finally, we see that if $t$ is the total number of vertices, using Steps 2,3 we see that $m=\frac{t-1}n$ hence independent on $U$.
But one easily proves that the degree of $U$ is exactly $m$
QED

[quote="pbornsztein"]In a graph, it is possible to go from any vertex to any other passing through at most $n-1$ vertices in between. The shortest cycle in this graph has length $2n+1$. Prove that all the vertices have the same degree.

Pierre.[/quote]

Pierre, which source are you using for the LMO problems ? I have the Russian LMO 1961-1993 by Fomin. And I have the materials from Crux which featured the town round from time to time during the 80s.  :)

In 2002, Crux published a selection of LMO problems from 1965 to 1984.
A friend of mine has a book with problems from these years but those given by Crux appear in the book as a list of problems with no solution (in the book). I don't know why...
You may see that the October and November issues of Crux 2004 contain solutions of some of these problems (the ones I solved  :D ).

Pierre.


induction
number theory proposed
number theory

i adore this problem :P

Let n be a positive integer. ¿Does n^2 have more divisors =1(mod 4) or =3(mod 4)

Actually, if I am not mistaken, all numbers have at least as many divisors of the form $4k+1$ as divisors of the form $4k+3$. I guess this should be even more adorable :).

I'm sure you can try to prove it yourself. Just look at the prime factors of the form $4k+3$ and those of the form $4k+1$. It should be a simple counting argument.

i will definitly try. it doesnt look that simple. but are you positive about it???' i would hate to end up trying to prove something thats not true :blush:

[quote="marko avila"]i think the proof to it is not any diferent than the original problem. what do you think??? you can find the solution to this problem in the kalva page. :lol:[/quote]

Can you show?

Grobber's result follows immediately from the repressentation of a number n as sum of two perfect squares. It is 4(D-E) where, D is no. of divisors p of n =1 mod 4 and E is that =3 mod 4. Since D-E has to be non-negative, result follows.

Bomb


In right triangle $ ABC$, $ BC \equal{} 5$, $ AC \equal{} 12$, and $ AM \equal{} x$; $ \overline{MN} \perp \overline{AC}$, $ \overline{NP} \perp \overline{BC}$; $ N$ is on $ AB$. If $ y \equal{} MN \plus{} NP$, one-half the perimeter of rectangle $ MCPN$, then:
[asy]defaultpen(linewidth(.8pt));
unitsize(2cm);
pair A = origin;
pair M = (1,0);
pair C = (2,0);
pair P = (2,0.5);
pair B = (2,1);
pair Q = (1,0.5);

draw(A--B--C--cycle);
draw(M--Q--P);

label("$A$",A,SW);
label("$M$",M,S);
label("$C$",C,SE);
label("$P$",P,E);
label("$B$",B,NE);
label("$N$",Q,NW);[/asy]$ \textbf{(A)}\ y \equal{} \frac {1}{2}(5 \plus{} 12) \qquad \textbf{(B)}\ y \equal{} \frac {5x}{12} \plus{} \frac {12}{5}\qquad \textbf{(C)}\ y \equal{} \frac {144 \minus{} 7x}{12}\qquad$
$ \textbf{(D)}\ y \equal{} 12\qquad \qquad\quad\,\, \textbf{(E)}\ y \equal{} \frac {5x}{12} \plus{} 6$

[hide="Solution"]
Clearly $ \triangle AMN\sim\triangle NPB\sim\triangle ACB$, so $ \frac{x}{12}\equal{}\frac{NM}{5}\implies NM\equal{}\frac{5x}{12}$.
$ AM\plus{}MC\equal{}12$, so $ MC\equal{}12\minus{}x$.

$ \frac{5x}{12}\plus{}12\minus{}x\equal{}\boxed{\textbf{(C)}\ \frac{144\minus{}7x}{12}}$.
[/hide]

hello, we have $ \overline{NP}\equal{}12\minus{}x$ and the realtion $ \frac{\overline{MN}}{x}\equal{}\frac{5}{12}$, from here we get
$ y\equal{}\frac{5}{12}x\plus{}12\minus{}x\equal{}\frac{144\minus{}7x}{12}$.
So the answer is $ \boxed{C}$
Sonnhard.


number theory
least common multiple
algebra unsolved
algebra

Does there exist a sequence of positive integers $ a_{2003}>a_{2002}>..>a_2>a_1$ with $ a_{1}\equal{}2003$ such that the folowing two conditions are satisfied:
1)All integers in the interval $ (2003;a_{2003})$ are either a member of this sequence or a non-prome
2)$ A\equal{}\frac{2004}{a_1}\plus{}\frac{2004}{a_2}\plus{}...\plus{}\frac{2004}{a_{2003}}$ is an integer?

[Moderator edit: Locked! Discuss the same problem at http://www.artofproblemsolving.com/Forum/viewtopic.php?f=38&t=252173 .]

The sequence must because of the Bertrand postulate and (1) contain primes greater than 2004. There is actually one such prime which only occurs once in the sequence. It is possible to rewrite A as a fraction using the least common multiple of the terms in the sequence.
$ \frac{\sum_{k\equal{}1}^{2003}(2004\cdot  lcm({a}_{1},    ,{a}_{2003})/{a}_{k}}{lcm({a}_{1},    ,{a}_{2003})}$
The unique prime is a primefactor in the lcm and in every term except for one in the numerator. This means that the numerator isnt divisible by this prime and therfore cannot be an integer.

[quote="negative"]Does there exist a sequence of positive integers $ a_{2003} > a_{2002} > .. > a_2 > a_1$ with $ a_{1} \equal{} 2003$ such that the folowing two conditions are satisfied:
1)All integers in the interval $ (2003;a_{2003})$ are either a member of this sequence or a non-prome
2)$ A \equal{} \frac {2004}{a_1} \plus{} \frac {2004}{a_2} \plus{} ... \plus{} \frac {2004}{a_{2003}}$ is an integer?[/quote]
Please delete your help,here is the newest problem in this issue of MYM in our country,ofcourse the test is running
So ,I think mods should warm him

IMHO, There should be some censoring on the posts of negative before they become open for other ppl to reply into.


Brazilian Math Olympiad 2004 round 1 problem 3:

The measures of the altitudes of a triangle are 12, 15 and 20. So the greatest internal angle of the triangle is:

a)72º
b)75º
c)90º
d)108º
e)120º

[hide]Let the area $S$ of the triagle be $S=30d,$ for some real number $d$

Then $2S = 60d = a\cdot 12 = b\cdot 15 = c\cdot 20$

So
$a=5d$
$b=4d$
$c=3d$

So $a: b: c = 5: 4: 3$
Hence it's a right triangle with $\angle A=90^\circ$[/hide]

[hide]The area of a triangle is equal to $\frac{bh}{2}$.  Therefore we have
$\frac{12a}{2}=\frac{15b}{2}=\frac{20c}{2}$, where a, b, and c are the side lengths of the triangle.  Therefore, we see that the sides are in the ratio $3: 4: 5$.    Therefore the triangle is a right triangle and the greatest internal angle is 90 degrees.[/hide]


[hide]

It can be simplify to

$(\sqrt{x-1}+\sqrt 9)-(\sqrt{x-1}+\sqrt 4)=1$

$\implies 3-2=1$ which is true for all real $x\geq 1$

[/hide]

hint

[hide]Write x as x-1 so as to get $(a+b)^2$ and thus remove the big roots[/hide]

sorry guys , i forgot to give the x>=1. Thx for pointing it out

I get $x\geq1$ as the answer.
There are lots of solutions? :?


In a convex pentagon $ ABCDE$ in which $ BC\equal{}DE$ following equalities hold:

\[ \angle ABE \equal{}\angle CAB \equal{}\angle AED\minus{}90^{\circ},\qquad \angle ACB\equal{}\angle ADE\]

Show that $ BCDE$ is a parallelogram.

Let $ EB \cup DA \equal{} G$ and $ EB \cup CA \equal{} F$, and $ \angle CAB \equal{} \angle EBA \equal{} \phi, \angle ADE \equal{} \angle ACB \equal{} \varphi$. Then, by sine law to triangles $ DEA$, $ EAB$ and $ ABC$ with $ DE \equal{} CB$ in mind, we get that $ \angle AEB \equal{} \angle GAE$ ($ \frac {DE}{EA} \frac{EA}{AB} \frac{AB}{CB}\equal{}\frac{\sin \angle EAG}{\sin \varphi} \frac{\sin \phi}{\sin \angle EAB} \frac {\sin \varphi} {\sin \phi}$).
Let $ X$ and $ Y$ be intersectio point of perpendicular line from $ E$ to $ DE$ with $ DA$ and $ CA$ respectively. Then, $ \angle XEA \equal{} \phi$, so $ \angle EXD \equal{} \phi \plus{} \angle EAG$. $ \angle BFY \equal{} 2\phi$ so $ \angle FYE \equal{} \phi \plus{} \angle GEA \equal{} \phi \plus{} \angle EAG \equal{} \angle EXD$ so $ EX\bot CB$, hence $ CB\parallel{} DE$.

Nice problem


Let $(a_{n})$ be a sequence of real numbers,$a_{n}>0$ for every $n\in N$.
Prove that for every for every $n\in N$,the ecuation $a_{1}x+a_{2}x^2+a_{3}x^3+...+a_{n}x^n=1$ have just one solution,$b_{n}$,in $(0,\infty)$.
Prove that the sequence $(b_{n})$ is convergent,and compute $\lim_{n \rightarrow \infty}b_{n}$ in case when $a_{1}=2$ and $a_{n}=1$ for every $n \geq 2$.

the first part is just descartes' rule of signs,

the next is

$2x+x^3+...+x^n=1$
$2x+x^3*\frac{x^{n-2}-1}{x-1}=1$

for x positive, it is obvious that $x<1$

so the sequence is bounded, and should be strictly decreasing because each greater n, there is another term added, forcing x to be smaller, so it converges

i cannot solve the final equation right now...it is just geometric


(>^^)>
<(^^<)

Those are the dancing kirbies

[hide="VERY NASTY, do not look. please"]
(.Y.)

Don't feel like explaining what that it  :roll: [/hide]

lol, funny. {^\[u]v[/u]/^}
Doesn't that sorta look like a squirel?

(^.^)==|;;;;;;;;;;;;;;;>

Oversized sword or something

[quote="modeler"][quote="sbanerjee"]I made a cool smiley: (9^-^)=o(*_*)

Anyone know any other cool TEXT simleys??[/quote]

This is cooler : (9^-^)==o)>.<)[/quote]

Oh, yeah, great.

I presume you can call this ASCII art.
Yeah......

Way smaller than yours.  This is from AoPS.  Its a hidden smiley
[img]http://www.artofproblemsolving.com/Forum/images/smiles/mrgreen.gif[/img]

\(^O^)/                          it looks weird in this font tho...
(:::::[X]:::::)                   bandaid!
(*-*)
 )__)                     and
(__(                           this
 )__)                               is
(__(                                    a
 \_/                                        snake!
:3 yay!

[quote="kimmystar94"]where did u find this??
[img]http://www.artofproblemsolving.com/Forum/images/smiles/mrgreen.gif[/img]
ps. i just copied this from ur quote.. :D[/quote]
Where-AoPS
How-A long story.
:)

umm....

i wanted to know HOW..... :D
i hope you know that..

[quote="minicon"]Well, I was wondering if the forums got their smileys off of phpBB.com or if they downloaded them to their own site.  So I randomly chose Mr. Green (Probably because its the the first one.).  I control clicked, and clicked copy image location, opened a new tab and pasted.  It showed
http://www.artofproblemsolving.com/Forum/images/smiles/icon_mrgreen.gif
and I wondered why it had the "icon_" part, so I deleted that and tried and got this:
[img]http://www.artofproblemsolving.com/Forum/images/smiles/mrgreen.gif[/img][/quote]

That will be my avatar from now on. :)

I just made a (laughing) owl by accident.
(^"\/"^)
BTW It doesn't work well in this font.  :(
(~\/O) winking penguin?
...
{-]====[ <]======>
watch?
...
♣
clubs?
 :lol:

o~("|~o
o~|")~o
o~("|~o
o~|")~o

open to interpretation

/\
  /  \
  /__\
    ||

Tree, but it doesn't really work on AoPS

(^-^)--@

Modified Boxer Kirby

I saw my friend made a really cool cat, but I don't know how to make it.


Hi,

I just have a general question. I am working through introduction to Counting and Probability and i came across the "Counting with symmetries" section. The "example" that I have a question about was the one with the keys. It says "How many distinct ways can 4 keys be placed on a keychain? Two arrangements are not considered different if the keys are in the same order (or can be made to be in the same order without taking the keys off the chain.)"

I can see how the rotational symmetry, but i dont exactly get how there is just only 1 flip arrangement, i though you could flip it 4 ways. So how is this so? And also, how am i supposed to figure out how many symmetries are in a question (especially in competitions when there isnt much time)?

for each way the keys can be arranged on the keyring, we can make 8 arrangements. We can move the keys around the keyring so eg. 1234 becomes 4123, 3412, 2341. We can also turn the keyring itself round 180 degrees to reverse the order of the keys eg. 1234 becomes 4321 etc. This means that for all $ 4!\equal{}24$ arrangements of the keys, groups of $ 8$ are in fact the same arrangement, so there are $ 24/8 \equal{} 3$ distinct arrangements of 4 keys on a keyring.

[quote="KrazyFK"]for each way the keys can be arranged on the keyring, we can make 8 arrangements.[/quote]

How did you figure this out?  :maybe:

[quote="KrazyFK"]I explain what the arrangements are in the next sentence. There are 4 arrangements that can be made from rotating the keys, and each of these can be made into another by flipping the whole keyring 180 degrees, and therefore reversing the order of the keys.[/quote]

Ah, okay. Thankyou =)

I don't think there is any "trick" to finding all the ways something can be flipped, just thinking about it logically and maybe trying to visualise it.


[hide]
$\frac{1}{2}+\frac{7}{8}=\frac{11}{8}$
the mean means average so dived it by two
$\frac{\frac{11}{8}}{2}=\boxed{\frac{11}{16}}$[/hide]

[quote="hwenterprise"]What is the mean of $\frac{1}{2} \text{ and } \frac{7}{8}$?

Express your answer as a common fraction.[/quote]

[quote="anirudh"]11/16[/quote]


algebra
polynomial
real analysis
real analysis unsolved

Is there  nice form or good expression   fo this product ?
\[ \prod\limits_{1 \leq \ell < k \leq d}\left( \exp\left(\frac {2\pi(\ell \minus{} 1)i}{d}\right) \minus{} \exp\left(\frac {2\pi(k \minus{} 1)i}{d}\right) \right)\]
$ d \in \mathbb{N}$

Yes, there is a nice formula.Hint: just consider the polynomial $ P(x) = x^d - 1$ and obeserve that if $ x_1,x_2....x_d$ are  roots of this polynomial then $ |P^'(x_1)\cdot P^'(x_2)...P^'(x_d)| = (\prod\limits_{i\ne j}(x_i - x_j))^2$ :wink:


algebra
polynomial
function
Diophantine Equations

Find all pairs $(a,b)$ of different positive integers that satisfy the equation $W(a)=W(b)$, where $W(x)=x^{4}-3x^{3}+5x^{2}-9x$.

[quote="Peter"]This seems not from Poland 2003...[/quote]

It's correct. Check it at "http://www.mimuw.edu.pl/~chel/Olimp/2etang03.html"

I think we need change the problem statement as the official version :

The polynomial $ W(x) \equal{} x^{4}\minus{}3x^{3}\plus{}5x^{2}\minus{}9x$ is given. Determine all pairs of different positive integers a,b satifying the equation $ W(a) \equal{} W(b)$.

[quote="Peter"]Strange, that's a different paper from the one on Kalva. :?:[/quote]

What Kalva offers is the Final Round problems of PMO. The above problem is from Second Round.

suppose that $ W(a) \equal{} W(b)$ then since $ a$ and $ b$ are different we have $ (a^2 \plus{} b^2)(a \plus{} b) \minus{} 3(a^2 \plus{} ab \plus{} b^2) \plus{} 5(a \plus{} b) \minus{} 9 \equal{} 0.$

so  $ (a^2 \plus{} b^2 \plus{} 5)(a \plus{} b \minus{} 3) \equal{} 3(ab \minus{} 2)$.

if $ ab \equal{} 2$ then $ a \equal{} 1,b \equal{} 2$ or $ (a \equal{} 2,b \equal{} 1)$  which are solutions.

if $ ab \equal{} 1$ then $ a \equal{} b \equal{} 1$ ,there are no solution in this case .

now suppose $ ab > 2$. and let $ ab \equal{} x^2$.and $ a \plus{} b > 2$.($ a \plus{} b \equal{} 2$ gives $ a \equal{} b \equal{} 1$ and $ a \plus{} b \equal{} 1$ doesn't have

solution)

$ 3 \equal{} \frac {(a^2 \plus{} b^2 \plus{} 5)(a \plus{} b \minus{} 3)}{(ab \minus{} 2)} \ge \frac {(2ab \plus{} 5)(2\sqrt {ab} \minus{} 3)}{(ab \minus{} 2)} \equal{} \frac {(2x^2 \plus{} 5)(2x \minus{} 3)}{(x^2 \minus{} 2)}$.

so after expanding (for  $ x > 3/2$ ) we get $ 4x^3 \minus{} 9x^2 \plus{} 10x \minus{} 9 \le 0$. and consider the function $ f(x) \equal{} 4x^3 \minus{} 9x^2 \plus{} 10x \minus{} 9$, we

have $ f'(x) \equal{} 12x^2 \minus{} 18x \plus{} 10 > 0$ so $ f$ is stricly increasing therefore $ x < 2$ (because $ f(2) > 0$).then $ ab \le 3$ so

$ ab \equal{} 3$ then $ a \equal{} 1$ and $ b \equal{} 3$ (or $ a \equal{} 3$ and $ b \equal{} 1$) which not verify the equation.

and if $ x \le 3/2$ then $ ab \le 9/4\equal{}2,...$ so $ ab \le 2$.

without loss of generality asume a>b. from the above post we see that $ 3 =\frac{(a^{2}+b^{2}+5)(a+b-3)}{(ab-2)} \rightarrow 3 > a+b-3 \rightarrow a+b\le 5$. Now just plug in all a,b that satisfy this and find that only 1,2 works.


Eight football teams participate in a tournament of one round (each teams plays each other team once).  There are no draws.  Prove that it is possible at the conclusion of the tournament to be able to find 4 teams, say $A$, $B$, $C$ and $D$ so that $A$ defeated $B$, $C$ and $D$, $B$ defeated $C$ and $D$, and $C$ defeated $D$.

There are $28$ matchs in the tournament, then there is a total of $28$ victories. It follows that the average number of victories is $\frac {28} 8 = 3,5$. Thus, at least one team, say $A_1 = A$, has won at least $4$ matchs, say against $A_2,A_3,A_4,A_5$.

Now, there are $6$ matchs between these $4$ teams. As above, we deduce that the average number of victories in these matchs is $1,5$, so that at least one team say $A_2 = B$, at won at least two of its matchs, say against $A_3$, and $A_4$.

Wlog, we may assume that $A_3=C$ has defeated $A_4 = D$, and we are done.

Pierre.


OK I'm interested in doing some research this summer in mathematics (or science). I've done a lot of work on smooth dynamical systems and I think that this would be what I could research (considering there is still a lot to discover) in math. However, I don't know where to start or how to conduct serious math research. What should I do? Also, I know a lot of science from independent study, and I am a very excellent experimenter, could this be used to do research? Obviously I would need to have someone give me an idea that I could follow, but where do I find an idea that is original, worthwhile to pursue, and that the "mentor" lets say wouldn't already try and research? I assume that most of these people in the Intel Talent Search didn't do all of their research on their own, or at least had some help starting or workign out the intricacies of their project. OK, now I'm rambling. Help, anyone?

OK then, could someone help the both of us out? I'll give you candy...

I'm trying to motivate myself to write the detailed answer that this question deserves.  That will probably happen by Sunday.  In the meantime, do a search of the forums for "research" or related words and see what pops up.

Here is my question from another thread that nobody answered. If anyone could help me out that would be great:

I was wondering, how do people manage to get project titles like "Isolation and Expression of an Epinephrine-synthesizing Enzyme (PNMT) from Entamoeba Parasites" (actual title of project for ISEF)? Where do they learn that kind of stuff? Is this something taught in a standard biology class? I think the only words that I actually know is "parasites" and "enzyme."

Wow we are developing quite a thread here. I await your response JBL :)

[quote="mathnovice"]how do people manage to get project titles like "Isolation and Expression of an Epinephrine-synthesizing Enzyme (PNMT) from Entamoeba Parasites" (actual title of project for ISEF)? [/quote]

They get an opportunity to work in some PhD's lab who has spent the past x years and y hundreds of thousands of dollars in grant money working on "Isolation and Expression of a Norepinephrine blah, blah, blah", and that person suggests to the young ISEF wannabe that they use the lab techniques and equipment already in place to look at the same thing, only for Epinephrine instead of Norepinephrine.

And that's considered "research?" I would expect more from a ISEF finalist than merely applying known techniques to a different problem. I mean that might deserve some merit for, say, a mathematics or technology project. However, if "research" means just carrying out an experiment and summarizing the data, how much credit does that person deserve? Anyone can carry out a process, not everyone can come up with some new technique or discover something new.

Where do find a PhD willing to let you do that anyway? Do you just call someone up and ask? Why would they even want to spend their valuable time teaching you how to do something when you might damage the equipment or something? After all, they have no stake in your success.

I think usually professors get graduate students and maybe undergraduates as well to do ``gruntwork'' that they would rather not do. Then that professor may have another vague idea and might suggest that the graduate student come up with a project and perform the experiment. That's basically what I have heard and read, but it might be wrong anyway. Does it sound plausible or true?

[quote]They get an opportunity to work in some PhD's lab who has spent the past x years and y hundreds of thousands of dollars in grant money working on "Isolation and Expression of a Norepinephrine blah, blah, blah", and that person suggests to the young ISEF wannabe that they use the lab techniques and equipment already in place to look at the same thing, only for Epinephrine instead of Norepinephrine.[/quote]

Sadly, this seems to be the case all too often; and more often than not, it is nearly impossible to tell whether the student actually did original work worthy of ISEF (or another big science-fair-like competition) if one isn't an expert in the field involved.  This sort of thing really spoils it for those of us who don't have access to that kind of assistance and those who actually do come up with something novel.  (Either the judges can't tell and assume that the student came up with everything on his/her own - all too easy sometimes because many students tend to overstate their cases - or they assume that the student couldn't possibly have done anything novel, because so many people do get that kind of help.)  It really bothers me that some people get access to such facilities and "mentoring" from professors and win, while others who are perhaps better original thinkers do not.  I could give some specific examples, but I think I'd better stop ranting before I get in too deep here.

On a more positive note, I'd like to give the encouraging news that it [i]is[/i] possible to get to ISEF even if you don't have that kind of help.  If you can find a good math project and prove something original, you don't need fancy lab equipment.  On the other hand, it can be pretty difficult to find a good, open/new yet still solvable math problem.  I might suggest looking at some of the threads in the "Advanced Problems" forum here at AoPS for inspiration.  The basic idea for my own project actually was inspired by a problem on this forum (although it naturally changed a lot as I did research and found out what was already known about the topic).

[quote]The basic idea for my own project actually was inspired by a problem on this forum (although it naturally changed a lot as I did research and found out what was already known about the topic).[/quote]

Oh, so you essentially created your own unsolved problem and then solved it? (checking around to make sure that the problem was truly unsolved, of course)

JBL was one of the Intel Science Talent Search Finalists. He did some Game Theory project.

Okay.  The basic solution to doing advanced research in high school (or even as an undergraduate in college) is to find a mentor who can help you along.  This is not necessary, but it is very, very helpful.  One can do this in one of two ways: either (1) one has either a school research coordinator or a parent/relative/acquaintance who knows professors/other researchers and can arrange a meeting between one and them or (2) one goes around university websites (preferably those near one's home) and finds the names of professors who might work in a related area and e-mails them and politely asks them if they might be able to help.  I personally got to take route (1), since my school has a fabulous, long-established research program with links to a variety of universities and professors in New York City.  However, it is certainly possible to go the second route, and it is probably what you will have to do.  It's more work, but it can bring you equally good results.

Now, there is a [i]lot[/i] of science (and math) to be done.  If you talk to a scientist, there is certainly some direction that they've thought of pursuing in their research but haven't pursued.  Often, they would be happy to hand that over to an eager student to work on, provided that student knew what he or she was doing and presented him or herself appropriately.  You are highly unlikely to come up with a question so irresistable that your mentor would steal your idea for research, so that's not much of a worry.  (This next bit will be largely addressed to the comments of texas and setofallsets.)  Now, whether you think this is "gruntwork" or whatever texas137's description amounts to or actual science is your own business.  I am inclined to think that the relatively derogatory way in which texas137 and setofallsets referred to research suggested by a professor related to his own research is in fact valid science.  It is rather unreasonable to expect high school students, having taken all of high school biology and chemistry, to come up independantly with a project on the cutting edge of science, write grant proposals to fund it themselves, etc., etc.  There is an obvious knowledge gap between research scientists and high school students, and the research scientists are best prepared to fill that knowledge gap in their own area of specialty, which would best prepare the high school students to work in a related area.  I worked on a problem very closely related to several that my mentor has worked on in the past.  Does that at all devalue the work I did?  I don't think so.  Biochemistry projects are even more constrained by this same problem, compounded by the difficulty of carrying out biochemical research in one's living room.

Anyhow, mathnovice, what do you imagine science is if not "carrying out experiments and summarizing the data."  That's what science is all about!!  Coming up with a new scientific techniqe is a rather unlikely thing for a student to do, and making a new discovery is premised on carrying out an experiment and seeing what the resulting data is.

As for setofallsets' "positive note," it is true that one can do research without a mentor.  This is more true in mathematics than in other fields.  However, it's very hard.  New problems tend to require a vast amount of background knowledge while old unsolved problems tend to be very very hard.  (I know someone who wrote his Intel paper about the Goldbach conjecture ... it did not go too well.)  This is not to say that it's impossible: I have a good friend who was a semifinalist this year with a project he did without a formal mentoring relationship.  But it's much harder.  I truly do recommend trying to find a mentor.

I rambled quite a bit, there, I'm afraid.  I hope Setofallsets and Texas137 don't take my comments as personal in any way, because they aren't meant as such.  I just don't think disparaging someone from seeking out a mentor is really in that person's best interest.  If anyone has any questions that I didn't address, I'd be happy to answer them.

By the way, those of you interested in doing research -- how old are you (what grade are you in)?

By the way, mathnovice, that title isn't really that complicated.  Remember: in science, everything gets a big name :).
[b]
Isolation[/b]: they only messed around with this one thing, or in some other way isolated it, so that they weren't having more than one variable at a time.  They may have removed it from the cell, for example.
[b] and [/b]
[b]Expression[/b]: how something is expressed.  For example, I have genes for brown and blue eyes and curly and straight hair.  The brown and the curly are expressed, while the others aren't.  For an enzyme, this might have to do with whether or not it was present, or something.
[b] of an
Epinephrine-synthesizing[/b]: something which synthesizes (creates) epinephrine, which is a large name.
[b]Enzyme[/b]: enzyme you say you know.  This is the thing that is doing the synthesizing.
[b](PNMT)[/b]: the name of the enzyme.
[b] from
Entamoeba[/b]: we've got another name here.
[b]Parasites[/b]: this is the thing whose name is entamoeba.

I mean, I could have given my project a big name like that.  I stuck with the easy-to-pronounce but uninformative "Analysis of a Vector Game."

I didn't mean to imply that mentoring is a bad thing, nor that all (or even a large portion of) projects done in that context are unworthy.  But it does happen, often enough to be disturbing, that someone will do a project in that context that is not nearly so good as it [i]seems[/i], for one reason or another.  This is not bad in a purely scientific context - data is data and is worthwhile, no matter who comes up with the ideas for how to gather it or what to gather it on - but it is unbalancing in the context of a science fair.  No, I don't expect that students will be able by themselves to come up with projects on the cutting edge of science, but I do think that a good amount of creativity should be displayed by the student involved.  Maybe I'm just a bit more cynical than JBL about this sort of thing.  I suppose one should take my earlier entry with a grain of salt.

To answer Fierytycoon's question:  Yes and no.  I took a known item - a particular combinatorial game (Wow - combinatorial game theory seems to be hugely popular, probably because novel things are accessible to advanced high school students there.) and the known results about it (although when I began I didn't know what all of the known results were, because I hadn't yet done any searching, so I ended up initially duplicating many of them before I began working on original research) - and I asked (and worked on answering) new questions about it, questions that (I feel, at least) were relevant and very interesting.  I'm not sure if this is the normal route for high school students doing math research, though.

PS:  JBL - what on earth did the guy write about the Goldbach conjecture??

I did not mean to sound derogatory! I think the people who do well on things like ISEF know a tremendous amount about their topics and have sometimes made very real contributions to scientific knowledge. But I am trying to bring it down to real life scale. It's easy to look at project topics like the sample one and think that the students who do them are some sort of rarified research geniuses who have been obsessed with things like epinephrine-synthesis from an early age. I don't believe that is the case. These are high school students, who may have only had a few weeks or months to work on their projects. Sometimes they accomplish much because they have "stood on the shoulders of giants" so to speak. If you add an inch of progress on top of someone else's research project that has already come a mile, it sounds more impressive and specific than making an inch of progress starting from ground zero. But either way, it's still an inch of progress.

I also think JBL raised an excellent point that sometimes an impressive, high-falutin' title really translates down to something much easier to understand.

Another consideration when looking for opportunities to work in research labs - you may run into profs who would be delighted to have you, but have to say no because of liability concerns. Unless you are a student, employee, or have some other official affiliation with the college where the prof is working, there may be problems with insurance. I had a student on the US physics team in the past who needed more lab experience. I contacted the local colleges to try to set something up, and was blocked by this issue. I also work in a hospital and have been approached by young people who would like to follow me around and see what I do. Same issue there. I have to say no. Students can only do this as part of an official course with specific local schools that have a formal arrangement with the hospital, or if they go through a whole process to be cleared as volunteers.

OK, here is my dilemna. I would really like to do some research, but I am not as proficient at math as many of you on this forum (I only recently started doing this type of math). However, I am very good at analyzing data, picking up math concepts, and I know a fair amount on various harder math concepts (smooth dynamical systems being the main one). I have only had high school physics, chemistry, and biology, but I have read some other stuff as well. Is there any hope for me to do any experimentation and research? It wouldn't have to be in math.

[quote="scm007"]OK, here is my dilemna. I would really like to do some research, but I am not as proficient at math as many of you on this forum (I only recently started doing this type of math).[/quote]

Contest math and research math are distinct, so you indeed have the possibility of doing an interesting research project. I'm still learning how to find local research mentors (which my son will need DESPARATELY in view of his liberal-arts-major parents), so all I can advise is to ask around among any and every possible interested adult to see if anyone knows anyone who might help you. Ask a lot more local people than you think you need to ask, and you should eventually turn up something. Part of what is rewarded in science fair projects is the learner's dedication in either getting connections to sources of help (e.g., by getting into a school with an established mentor program) or doing the difficult by self-effort (e.g., as Setofallsets did) or scrounging up a new source of help (as the guy who tried to make a project on the Goldbach conjecture would have been better advised to do). Know what you can do without help, and know what you need help for, and look for the help all the while.

Good luck.

scm -- if you want to do research in science, I really recommend that you just start e-mailing professors at universities near where you live and telling them you'd be interested in working with them and seeing if you can get any positive responses.  Of course, you should probably target professors working in an area you are interested in/know something about.

If you want to do math, you could do the same thing, or you could try and research it on your own.  If you know what you want to do, start researching to see what problems you think you can tackle in that area and get working :).

What grade are you in?

[quote="Setofallsets"]What on earth did the guy write about the Goldbach conjecture??[/quote]

Nothing worth mentioning :)


EBDF is a square
Angle B=90 degrees
BC=12cm
AB=6cm
What is BF?

Refer to picture below.

EDIT: Click the picture to make it more clear.

Note that D is on the side of the triangle.

[hide="hint"]Look for similar triangles, and set the length DF to be x. [/hide]


MATHCOUNTS
logarithms
floor function
modular arithmetic

Look at the number 2^9876543. Determine (1) the number of digits it has; (2) its last three (rightmost) decimal digits; and (3) its first three (leftmost) decimal digits.

After you solve that...

Is there a way to do this for any other powers of 2?

Good Luck!

[quote="USAMO13"][quote="pythag011"]Take log 10.[/quote]

That doesn't explain anything...[/quote]

the laws of logarithms state that $ log_x$ $ a^b=b * log_x$ $ a$

so, similarly:

$ log{}_1{}_0$ $ 2{}^9{}^8{}^7{}^6{}^5{}^4{}^3=9876543 * log{}_1{}_0$ $ 2$

@ USAMO13
Think about the definition of a logarithm. Why does this apply to (1)? How many powers of 10 are there in the number?

What happens after $ 2{}^{9}{}^{8}{}^{7}{}^{6}{}^{5}{}^{4}{}^{3}=9876543 * log{}_{1}{}_{0}$ ?
(I know what  happens I just want full solutions for practice for nationals :wink: )

[hide="1"] The number of digits in a number x is $ (\lfloor\log{x}\rfloor) \plus{} 1$ (or at least I think it is; haven't done this type of problem in forever?).

Now, $ \log{2^{9876543}} \equal{} (9876543)(\log{2})$. Unfortunately, as Victor (or Vincent if you're Temperal) just told me, you'd definitely need to use a calc to get the exact value.[/hide]

For #2, $ 2^x\pmod{1000}$ repeats every 100 terms, except for 001, 002, and 004.If you want to know how you get 100, use the Chinese Remainder Theorem on (mod 125) and (mod 8).

Why are you even pretending these are nats problems when it's obvious that none of these three would ever appear at nats?

@xpmath - don't they repeat every 400 terms (not 100) from Euler's modular theorem?

$ \gcd(2,100)\equal{}2\ne1$...for $ x\ge3$, $ 2^x\equiv0\pmod8$, and $ \gcd(2,125)\equal{}1$ so we can use Euler's to get $ \varphi(125)\equal{}100$, and thus the period is $ 100$ excluding $ 2^0,2^1,2^2$.

Forgot that only 1/8 of the residues are actually reachable for x>=3  :oops:

[quote="abacadaea"][quote="USAMO13"]GO AWAY algebra2000...nike admitted it..[/quote]

I hope you realize that arguing like this gets you absolutely nowhere given that most likely do not know each other in real life.
@mods- please lock this topic.[/quote]

i'm pretty sure i do know him in real life. also, this proves that he was lying; he said that he was the state winner for MC and he made nats from Maine, and that he made USAMO. too bad no middle schoolers from MAINE made USAMO...

yeah. lock this.


[quote="hzbrl"]Let p be an odd prime number. Find the positive integers x and y with xless than or equal to y
for which
sqrt(2p) - sqrt(x) - sqrt(y)
is the smallest possible.[/quote]

... and nonnegative, otherwise the problem is a bit nonsense.

http://www.mathlinks.ro/viewtopic.php?t=48300
http://www.mathlinks.ro/viewtopic.php?t=150984

These two topics contain exactly the same proof by myself. I may delete one of them in some time (they come from the time when MathLinks and PEN were two different forums).

  darij


How many different arrangements are there of all the letters of the word STREET?

There are 6 letters, so the answer is $ \frac{6!}{2!2!} \equal{} \boxed{180}$.

The reason for the 2!2! as the denominator is because the 2 pairs of the same letter. By dividing the factorial of the number of letters in the word by the factorial of how many letters are repeated, you  make up for the extra letters


Does anybody know the formula for the number of ways that N distinguishable items can be placed in K indistinguishable boxes?

I think you mean the number of ways to rearrange $ n$ objects into $ k$ boxes, where order matters?

It should be $ \binom{n}{k}k!$, or $ \frac{n!}{(n\minus{}k)!}$

No, that's not what I mean. Here's a link to the number sequence, but it's a little confusing. http://www.research.att.com/~njas/sequences/A019538

"Number of ways n labeled objects can be distributed into k nonempty parcels."

They include a recursive definition ( $ T(n, k) \equal{} k*(T(n\minus{}1, k\minus{}1)\plus{}T(n\minus{}1, k))$ ) for how to calculate it, but I am interested in knowing if there is a non-recursive method.


Let $x,y,z \in ( \frac{1}{3}, \infty)$. Show that the biggest number between: $3x-y^{2}+z^{3}$, $3y-z^{2}+x^{3}$, $3z-x^{2}+y^{3}$ cannot be smaller than $\frac{1}{27}$.

Maybe the smallest, in which case it is trivial.

[quote="harazi"]Maybe the smallest.[/quote]

Well, in the book it says "the biggest".

It may say that, but just take y and z fixed and x very large. I really doubt the conclusion will hold!

Yes harazi, you are right!  :lol:

Prove that if $x,y,z>\frac{1}{3}$ then $max(3x-y^{2}+z^{3},3y-z^{2}+x^{3}, 3z-x^{2}+y^{3}) \geq \frac{1}{27}$

(by Razvan Satnoianu)

I have also edited the first post!


I'm weak at combinatorics, and I have a NSML regional math meet coming up soon that covers combinatorics (and the scary thought is I'm trying to compete for the senior division and I'm only a (fre)shman/sophomore who skipped ninth grade (but in the meets, I'm considered as a ninth grader), so I thought it may be wise to start a marathon here. (Intermediate difficulty=maximum, also please do not post problems that are lower than High School Basics level)

Just like any other marathon, please answer the problem of the most recent poster (with full solutions!) and post a new combinatorics problem. I hope this will help me get better at combo. Thanks.

Anyways, I'll start with a problem: (which is from one of our math practices)

One fourth of the people last generation are bad drivers. 30% of the people in this generation are bad drivers. If someone is a bad driver, it's 80% likely that their children will be, also. I am a good driver. What are the chances that my parents are good drivers?

[hide]The probability is $ 1\minus{}0.25*0.2/0.7 \equal{} 13/14.$[/hide]
Next Problem:
There are 27 candidates in elections and $ n$ citizens that vote for them.
If a candidate gets $ m$ votes, then $ 100m/n\le m \minus{} 1$. What is the smallest possible value of $ n$?
PLEASE solve this fast as I can't wait to solve the next problem! :D


inequalities
integration
calculus
Euler
real analysis
real analysis unsolved

Let $ \gamma (t) \equal{} x(t)\bold{i}\plus{}y(t)\bold{j}$ be a simple smooth closed curve for $ a\leq t\leq b$. Let $ L > 0$ then all curves so that $ \int_{a}^{b}\sqrt{[x'(t)]^{2}\plus{}[y'(t)]^{2}}dx \equal{} L$ we have $ \oint_{\gamma}(xdy\minus{}ydx)\leq\frac{L^{2}}{2\pi}$.

How many different ways can you derive this? I have seen the Fourier Analysis approach to it. Anybody knows how it can be derived with Calculus of Variations?

Use the Lagrange Multiplier to handle the constraint.
$ I\equal{}\int_{a}^{b}\mathfrak{~L}\,\,dt$, where $ \mathfrak{~L}\equal{}x(t)y'(t)\minus{}x'(t)y(t)\plus{}\lambda\left(\sqrt{x^{\prime}(t)^{2}\plus{}y^{\prime}(t)^{2}}\minus{}\frac{L}{b\minus{}a}\right)$.
Apply Euler-Lagrange equations and find the quadrature $ x^{2}(t)\plus{}y^{2}(t)\plus{}C_{1}x(t)\plus{}C_{2}y(t)\plus{}C_{3}\equal{}0$

i remember reading a proof of this in docarmo's book.


$non-isosceles\ \triangle ABC,\ M\in (BA,\ N\in (CA,\ BM=CN=BC\Longrightarrow$ $MN\perp OI,\ MN=\frac {a}{R}\cdot OI.$

See http://www.mathlinks.ro/Forum/viewtopic.php?t=52563

APMO 2004, question 5, if I'm not mistaken.

This was posted by Gil: http://www.mathlinks.ro/Forum/viewtopic.php?t=20406 . Cf. http://www.mathlinks.ro/Forum/viewtopic.php?t=30955 as well.

  darij


:( I need a sequence formula for the following numbers i.e. I need the formula that'll allow me to map the first number to the second number i.e the multiples of 5:

2 - 5
3 - 10
4 - 15
5 - 20
6 - 25
7 - 30
8 - 35
9 - 40
10 - 45
11 - 50
etc etc

Thanks
:oops: