AMC
USA(J)MO
USAMO
AMC 12
AMC 10
AIME
AMC 12 A

There are some posts in the forum predicting the USAMO index for 2009. Now the AMC state stats are out, and I did some preliminary analysis on a few “heavy” states, and we have some surprises and predictions….

We picked five states, CA, NY, IL, VA (due to TJ) and MA, and below are the number of papers with scores >= 120 in AMC12 and AMC10 respectively:

-----------12A-----12B------10A------10B
CA-------62-------69--------268------188
NY-------23-------21--------102-------46
IL--------13-------11--------40--------16
VA-------16--------6--------54--------14
MA-------26-------20--------86--------54
TOTAL---140------127-------550-------318

GRAND TOTAL:
AMC12----267
AMC10----868

From this table, you can clearly see the AMC10 takers scored much higher than AMC12 takers this year.  This will most likely present some irony situations:

1) When calculating the USAMO index for the remaining 160 qualifiers, the USAMO index will likely be higher for those AMC10 takers than the first 330 qualifiers whose USAMO index will be based on their AMC12 scores.
2) It is very unlikely for any 10th graders and below who took AMC12 to be selected for the 160 qualifiers when their USAMO index is to be calculated based on AMC12 scores to compete against those based on AMC10 scores.

The real victims are the 10th graders and below who took AMC12 this year. They took a harder test on AMC12, and they had to compete against their peers who took a relatively easier test on AMC 10, therefore their USAMO index would be lower.

The goal of the AIME and USAMO process is to ensure the most qualified 500 students to be selected based on their knowledge, skills and merit, not the choice of contest they took. In many cases, they do not have choice as their schools or teachers already decide which one to take. To make it a fair process, as some people suggested in the past, we sincerely hope and pray the following suggestions be seriously considered, especially for this year::

1) Purely use the AIME scores to select the remaining 160 qualifiers. If there are more than 160 qualifiers, then raise the AIME score by one point to decide the 160 qualifiers.
2) Increase the pool of the first cut from 330 to 360(?) so those who took AMC12 will have a fairer chance.
3) Give some sort of booster to the 10th graders and below who took AMC12.

It might make some sense for a moderator to combine this with kfeng40's [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=265400]other topic[/url].

Of course, I basically agree with what he's saying. The only problem with doing anything about it this year is that many students made the particular choices they made because of the rules that are in place. To change those rules now could be seen as a breach of faith. But the larger point that the system is broken and has misplaced incentives - that much I agree with.

It would be more fair if the index was higher for the 10, because then the AMC 12 score and the AMC 10 score would kinda balance out (since the 10 is easier than the 12)

I think this is a very good idea. Under the current system as I see it, several flaws are apparent, many of which have been addressed before. One of the biggest of these flaws is actually the system for accepting 10th graders and below into the USAMO.
Specifically, the AMC10 score and the AMC12 score should not be counted the same. Especially this year, when the AMC10 was (many people say) relatively easier in comparison to the AMC12, many people who took the AMC12 and are in the grades 10 and below will get cut from USAMO, even though they have higher mathematical potential (arguably) than the AMC10 takers who make it into the USAMO.

For some substantial proof, take a look at the [url=http://en.wikipedia.org/wiki/USAMO#Participant_selection_process]Wikipedia site for USAMO[/url]. This site states the unfairness of the situation with numbers:

[quote="The article"]
7. In advising young students (in grade 10 or below) who desire to be selected for the USAMO whether to take the AMC 12 contest or the AMC 10 contest, please be aware of the following facts:
a. In 2007, among 506 students invited to take the USAMO, 229 were in 10th grade and below. Those students had scored 6 or greater on the AIME.
b. Among those 229 students, 87 had their AIME qualifying high score based on the AMC 12 and 142 had their AIME qualifying high score based on the AMC 10.
c. In 2007, among 8,312 students who took the AIME, 2,696 were in grades 10 and below. Of those, 998 qualified for the AIME from the AMC 12 and 1,698 qualified from the AMC 10.
[/quote]

This shows the fact that it is easier to make it into the USAMO via the AMC10 for sophomores and below. This should [i]not[/i] be the case. In a perfect system, the number of AMC10 takers who make the USAMO should be equal to, if not slightly less than, the number of AMC12 takers who make it to the USAMO.

Also, another very important point to take into consideration is that [i]many schools do not purchase the AMC10 tests[/i]. My school is one of the many who only allow its students to take the AMC12. This gives the students [i]no choice[/i] but to fall into the unfair situation described above. I, for one, scored a 126 on the AMC12 and a 7 on the AIME (why is another story), which puts me very borderline for the cutoff. I just feel that it is not right for someone who scored a 132 on the AMC10 and a 7 on the AIME to make it to USAMO when I don't. (I know that last statement sounds really selfish, but keep in mind that I'm only giving an example, this situation occurs to [i]many[/i] other students.) I scored the highest as a sophomore or below on the AMC12 this year in my state.

So all in all, I think the system should be changed. It could be argued that a perfect system doesn't exist, but I believe that it is the AMC's responsibility to make the system [i]as fair as possible[/i].

Which reminds me of something. On the AIME answer sheet, they ask for your score and whether you took the AMC10 or 12. I got a 124.5 on 12 and 126 on 10. So what was I supposed to bubble?

kfeng, please keep similar posts confined to one topic in the future - I can't merge the first post of a thread, so I can't fix it now.

It is interesting to see some posts on this very topic and I tend to agree with most of you that the current USAMO selection rule equating the AMC12 scores to AMC10 scores is not fair. Question is if there is anything that can be done for this year so the most qualified 160 students can and will be selected for the 2nd cut.

I am not sure if this is the right channel to voice this concern. Does anyone know if the MAA officials even read some of the posts here and how quickly they can implement the change?

Bottom line: there may never ever be perfect rules to all, but [b]as long as everyone is measured on the common basis[/b], no matter what that is, it will be relatively fair to all. In this case, it seems it is not, so a change is highly warranted. And by working together we can, and yes we can  :lol:  :lol:  :lol:

 :lol:

The AMCDirector reads the forum but he probably knows about this already.

MAA officials *do* read posts here, but if you have direct suggestions they should go through the direct channels. Nothing will be done this year; I would be very, very surprised if something did. In my opinion, which has changed over the years, it is the student's responsibility to understand the qualification process. The only potential issue is that of schools restricting the test you take based on your math level.

1) Definitions:
> USAMO Index I = AMC12 score + 10 * AIME score
> USAMO Index II = AMC10 score + 10 * AIME score

2) Prework:
There is no general stats yet for the entire USA and Canada (perhaps purposely) in 2009, so we have to base on the state stats available from the MAA site: http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2009-12a/09-1-1012-StateStats.shtml

We exported all state stats into Excel, and tallied them. To compare with the AMC12 stats, we used the 2008 High School Summary book from the MAA site: http://www.unl.edu/amc/e-exams/e6-amc12/archive12.shtml

We used AMC12 score 121.5 as benchmark because we feel this maybe the threshold in order to have a chance to be selected in 2009, otherwise students will need to score higher in AIME in order to catch up.

3) Some Data:
Below is the # of AMC12 papers scores >= 121.5 for 2008 and 2009 respectively:

---------------USA-----------Canada-------TOTAL
2008---------541--------------46------------587
2009---------520--------------50------------570

NOTE: there are about 100 students duplicate in both AMC12A and AMC12B each year, therefore the estimated # of individuals who scored >= 121.5 in AMC12 is about 450 in 2009 and 467 in 2008.

4) Analysis and Predictions:
> The 2009 AMC12 is about the same level of difficulty as 2008 AMC12, perhaps a little harder, but it is within the margin of error.
> Almost everyone agreed the AIME 2008 was a very easy one. Most people seem to agree the AIME 2009 is not easier than AIME 2008.
> The 2009 AMC10 was way too easier than 2008 AMC10.

Based on the analysis above, we can draw a few predictions:
> The USAMI Index I will likely be a few points lower in 2009 than in 2008. It was 204.5 for USAMO Index I in 2008, thus it could be between 201.5 and 204.5 for USAMO Index I in 2009.
> The USAMO Index II will be higher in 2009 than in 2008 due to the overwhelming larger # of students who scored higher in 2009. It was 202.5 for USAMO Index II in 2008, thus it will be a lot higher. [b]And USAMO Index II will likely be higher than the USAMO Index I in 2009 [/b](very irony situation).

5) Conclusion:
The true victims of the current USAMO selection process are those 10th graders and below, who scored USAMO Index I between 195 and 201.5, just barely missed the first cut. They had to compete against their peers with their USAMO Index II. It is simply not fair. For most of those students, if they took AMC10 instead, it is very likely they would score at least one extra problem (+6 points), and this will put most of them qualify for USAMO.

AMC, AIME and USAMO are once a year thing. Serious students have been working extremely hard year round for this moment. Because the test they took (many of them did not have a choice to choose which one to take), and they were not given the opportunity to be measured by the common standards, it is simply not fair nor right. We sincerely hope and pray the MAA officials will hear the pledge of those students and make the right decision, because it is simply the right thing to do.

Thanks in advance!!!!

[quote="buzzer11"]For some substantial proof, take a look at the [url=http://en.wikipedia.org/wiki/USAMO#Participant_selection_process]Wikipedia site for USAMO[/url]. This site states the unfairness of the situation with numbers:

[quote="The article"]
7. In advising young students (in grade 10 or below) who desire to be selected for the USAMO whether to take the AMC 12 contest or the AMC 10 contest, please be aware of the following facts:
a. In 2007, among 506 students invited to take the USAMO, 229 were in 10th grade and below. Those students had scored 6 or greater on the AIME.
b. Among those 229 students, 87 had their AIME qualifying high score based on the AMC 12 and 142 had their AIME qualifying high score based on the AMC 10.
c. In 2007, among 8,312 students who took the AIME, 2,696 were in grades 10 and below. Of those, 998 qualified for the AIME from the AMC 12 and 1,698 qualified from the AMC 10.
[/quote]

This shows the fact that it is easier to make it into the USAMO via the AMC10 for sophomores and below. This should [i]not[/i] be the case. In a perfect system, the number of AMC10 takers who make the USAMO should be equal to, if not slightly less than, the number of AMC12 takers who make it to the USAMO.

[/i].[/quote]

You  can not just simple compare the number.  Did you think why is  that "among those 229 students, 87 had their AIME qualifying high score based on the AMC 12 and 142 had their AIME qualifying high score based on the AMC 10"? Simple fact is majority of 10th grade and below took AMC 10.

[quote="kfeng40"]1) Definitions:
5) Conclusion:
The true victims of the current USAMO selection process are those 10th graders and below, who scored USAMO Index I between 195 and 201.5, just barely missed the first cut. They had to compete against their peers with their USAMO Index II. It is simply not fair. For most of those students, if they took AMC10 instead, it is very likely they would score at least one extra problem (+6 points), and this will put most of them qualify for USAMO.

AMC, AIME and USAMO are once a year thing. Serious students have been working extremely hard year round for this moment. Because the test they took (many of them did not have a choice to choose which one to take), and they were not given the opportunity to be measured by the common standards, it is simply not fair nor right. We sincerely hope and pray the MAA officials will hear the pledge of those students and make the right decision, because it is simply the right thing to do.

Thanks in advance!!!![/quote]

I think you are right that this year the index for 10th and below will be higher than 11th and 12th.   But I do not agree with what you said.
From AMC web sites:
Selection for the USAMO will be made according to the following rules:
“The selection process is designed to favor students who take the more mathematically comprehensive AMC 12A and AMC 12B contests. “
Remember that Only about 1/3 or less was chosen from 10th and below.

Also don’t forget that only few students among 10th and below were only offered AMC 12.  You need talk to your school teachers and ask them to organize AMC 10 test.

[quote="johnk4521"]
Remember that Only about 1/3 or less was chosen from 10th and below.

Also don’t forget that only few students among 10th and below were only offered AMC 12.  You need talk to your school teachers and ask them to organize AMC 10 test.[/quote]

My point pertains to sophomores. your "1/3 or less" statement doesn't work for underclassmen.

And also, its more than only a few students who are offered only the AMC12. Some school teachers will not organize the AMC10 test when all of their students are already taking the AMC12. But no matter what the details are, the fact stands that the test is unfair for some people, and this should be fixed.

"Selection for the USAMO will be made according to the following rules:
The selection process is designed to favor students who take the more mathematically comprehensive AMC 12A and AMC 12B contests. “

I am very surprised some people still refer to this quote.  This is simply not true based on the posts and the threads discussed here. I try to locate the quote on the AMC web site, but can not. Can someone put a link there? I highly double AMC would mislead anyone with such a quote on their web site.

The fact of matter is: compare AMC10 scores with AMC12 scores point by point is not fair, this is especially true for this year given the AMC12 test is relatively harder while AMC10 is a lot easier than usual. Some people raise the concerns that need to be looked at by the AMC so a relatively fair process be established and followed. I agree that everyone's index, whatever is called, should be based on the same test, whether it is AIME, AMC12 or AMC10.

http://www.unl.edu/amc/e-exams/e8-usamo/usamo.shtml

See #7 at the bottom.

It [b]was[/b] designed to favor those students until the expansion - the tiebreaker index had rarely come into play before then. They haven't finished removing all the instances of that, it seems.

[quote="buzzer11"]But shouldn't the tests still favor the AMC12 takers? Why should the test favor AMC10 takers because of the expansion?[/quote]
I think the AMC 10 was designed so that it is easier for 10th graders and lower to advance. The AMC 10 is available to people for a smaller duration of time (after 10th grade, they can't take the AMC 10 anymore, but can still take the AMC 12), and thus should be taken advantage of.

Yes, but that's not equivalent to easier for AMC10 takers to advance.

What can be said the interaction of the floor score with tiebreakers system with the expansion of the USAMO?

The system only ran for a few years before the expansion, so there wasn't really a lot of experience with it. There wasn't a need for index-as-tiebreaker in those years, but that was partly just luck. They would have run into trouble in some years sooner or later. Because the expansion pushed the minimum AIME score down to a level were there more students at each AIME score, it greatly increased the probability of needing the tiebreaker. In the first year, they handled a large tie by expanding further than originally intended. Each year since then, the index tiebreaker has been needed. kfeng40 started this topic by predicting that the second-round index would be higher than the first-round index. That's not all that bold a prediction, considering what happened last year.

The biggest problem I have with the current system is the perverse incentive it creates for very strong students to take the AMC 10. I was looking at the state stats for California. There were 24 students in California alone who made 150's on the AMC 10. How many 150's does that make for the country as a whole? 60? 100? More than 100? A lot in any case. I'm guessing that well over half of those students are at a level in their mathematical education where they really have no business taking the AMC 10 rather than "outgrowing" it and moving on to the AMC 12. There are students in there who have performed very well on the AMC 10 in past years and have little left to prove; there are students in there who are taking advanced courses; it wouldn't surprise me if there are students in there who have been to Red MOP.

But even as I say that such students have "no business" taking the AMC 10, it's still true that they made an entirely rational decision to do so, as they have correctly deduced that it maximizes their chances of qualifying for the USAMO. I'm not blaming them; rather, I'm not pleased with the system that makes that choice rational.

My "unified index" proposal would have equated a 150 on the AMC 10 to a 127.5 on the AMC 12A or a 130 on the AMC 12B. I said there were 24 California students with 150 on the AMC 10; there were 48 students reaching those levels on the AMC 12. (As opposed to just 2 perfect scores.) Or to pick another level: 117 in CA with AMC 10 of 138 or better; 160 with either AMC 12A of 115.5 or AMC 12 B of 118. All of this seems within reason. Adjusting between the exams to create a unified index would, in my opinion, neither particularly advantage or particularly disadvantage those who took either exam; opportunities would still be present on either side.

I encouraged the very strong freshmen and sophomores in my school to take the 12A and 10B to solidify their chances of getting into the USAMO. Indeed, they had no business taking the AMC-10. They generally scored in the 120s on the 12A and basically approached the 10B with the kind of attitude you'd expect toward a standardized test--"I guess I have to do this. Wow, if I don't get a perfect I'll be really disappointed." Oh, by the way, one of the students who did this already made USAMO last year on the strength of his AMC-12 score, but he didn't want to risk it again.

So basically, I think Kent Merryfield's characterization of the AMC-10 150s and 144s is pretty accurate. Regardless of whether the current USAMO selection algorithm is statistically fair, it certainly has an unhealthy effect on the AMC contest experience of a strong but cautious underclassman.

Weighing in as a parent of a past AMC participant, I must first make it clear that in my opinion AMC is one of the very, very best things that exist for the students of this country. I shall continue to point any student in the direction of AMC if he/she will listen -- and even he/she refuses to listen :lol:.

[quote="Kent Merryfield"]... I'm guessing that well over half of those students are at a level in their mathematical education where they really have no business taking the AMC 10 rather than "outgrowing" it and moving on to the AMC 12. There are students in there who have performed very well on the AMC 10 in past years and have little left to prove; there are students in there who are taking advanced courses; it wouldn't surprise me if there are students in there who have been to Red MOP.

But even as I say that such students have "no business" taking the AMC 10, it's still true that they made an entirely rational decision to do so, as they have correctly deduced that it maximizes their chances of qualifying for the USAMO. I'm not blaming them; rather, I'm not pleased with the system that makes that choice rational...[/quote]
I am in total agreement with the above statements. My son, who had success at every level of the AMC tests, did not even consider taking AMC10 during his 9th and 10th grade a few years ago. It was AMC12 all the way. If he were in 9th or 10th grade right now, he’ll be foolish, though un-mathematical, to not consider taking AMC10A, and probably also AMC10B if he did not think he did well enough in A.

The reason for that strategy is simple. AIME is an unforgiving step towards USAMO. Small slips in AIME will lead an able math student to a 10, 20, 30, or even 40 point fall in the USAMO Index. With the current USAMO qualification rule, by taking AMC10 the student primarily only has to worry about making the floor AIME value.

As an AMC advocate, I have a hard time advising students 10th or below who have a legitimate chance to qualify for USAMO to not take AMC10, even twice if necessary. That just does not seem to be consistent with AMC’s goals.

The statement below on the MAA web site (http://www.unl.edu/amc/e-exams/e8-usamo/usamo.shtml) is totally untrue and misleading especially for this year. This year, the 10th graders and below who took AMC12 are indeed being put into a huge disadvantage in comparison with those who took AMC10 in the 2nd selection process. This is simply the plain fact. Their USAMO index based on AMC12 scorse can not and should not be equated to the one based on AMC10 scores.

This is the issue that everyone sees now and it is AMC’s responsibility to try to adjust it so it becomes as fair as possible. The following statement should be taken out from the MAA web site so no more students and teachers will be misled again.

“Selection for the USAMO will be made according to the following rules:
1.	…..
2.	…..
3.	…..
4.	…..
5.	…..
6.	…….
7.	[b]The selection process is designed to favor students who take the more mathematically comprehensive AMC 12A and AMC 12B contests.[/b]"

how likely is it that the 2nd index will be used? because if the floor is 7 or 8, it may not even be, so only those who did well on aime go on.


The straight line AB is devided at C so that AC = 3CB. Circles are drawn with AC and CB as diameters and a common tangent to these meets AB extended at D. Show that BD equals the radius of the smaller circle.

[i]*I know this wording might be complicated but this is what the problem says[/i]

Isn't the tangent of the circles at point C according to your explanation?

$A-----------------------------C-------B$  :huh: So where is D?

This problem is too difficult for the Classroom Math middle school community, so I'm moving it.  In general, try to pay attention to the general difficulty levels of each forum -- many have forum guidelines posted to make this clear.

robinhe - Try to draw your tangent line such that it touches both circles in different places.
[hide="Answer"]
Let the center of the circle with diameter AC be E and the center of the circle with diameter CB be F. Also let the point of tangency to the larger circle be G and to the smaller be H. Let us call BF r. CE is thus 3r. By AA similarity (angles FDH and EDG and angles DHF and DGE), $\triangle DFH \sim \triangle DEG$. If we let $BC = x$, we have

$\frac{r}{r + x} = \frac{3r}{5r + x}$

Cross multiplying yields

$3r^2 + 3rx = 5r^2 + rx$
$\Rightarrow 2r^2 = 2rx$
$\Rightarrow r = x$

Thus BD is equal to the length of the radius of the smaller circle.[/hide]

Here is a picture of a common tangent (although the diagram is not the same as this problem). To see how the diagram for this problem would be, just imagine sliding the circles closer so they touch, draw in the line connecting the centers, and relabel the points.

http://www.geom.uiuc.edu/education/math5337/monge/two_circs.gif

[hide]Name the center of the circle with diameter $\overline{AC}$ E, and the center of the circle with diameter $\overline{BC}$ F.  Name the tangent point of $\odot E$ G, and the tangent point of $\odot F$ H. Let $AC=6x$ and $BC=2x$.  Then the radius of $\odot E$ is $3x$, and the radius of $\odot F$ is $x$.  $EF=3x+x=4x$, and because of similar triangles, $DF=2x$, and $DB=FG=x$.[/hide]


1) Five points $ O$, $ A$, $ B$, $ C$, $ D$ are taken in order on a straight line with distances $ OA\equal{}a$, $ OB\equal{}b$, $ OC\equal{}c$, and $ OD\equal{}d$.  $ P$ is a point on the line between $ B$ and $ C$ and such that $ AP: PD\equal{}BP: PC$.  Find $ OP$ in terms of $ a,b,c,d$.

2) Show that if $ r_1$ and $ r_2$ are the distinct real roots of $ x^2 \plus{} px \plus{} 8\equal{}0$, then l $ r_z \plus{} r_2$ l $ > 4\sqrt{2}$.

3) We are given $ 4$ bags of coins such that (a) all coins in a given bag weigh the same, and (b) the coins of a given bag weigh either $ 1,2,$ or $ 3$ ounces.  Take $ 1$ coin from bag $ 1$, $ 3$ coins from bag $ 2$, $ 9$ coins from bag $ 3$, and $ 27$ coins from bag $ 4$.  Weighing these $ 40$ coins together on a scale yields a weight of $ 95$ ounces.  Determine the weight of a coin from each of the $ 4$ bags.

Thanks in advance!

[hide="2."]$ p \equal{} r_{1} \plus{} r_{2}$
We need the discriminant to be $ 0$ or positive so
$ p^{2} \minus{} 32 \ge 0 \rightarrow |p| \ge 4\sqrt {2}$ [/hide]

#2 alternate:
by Vieta's formulas, we have the product of the roots is 8 and therefore the GM is $ 2\sqrt2$
and the AM is at least the GM (with equality only is the numbers are the same which we do not have since we have distinct roots)
so the AM is more than the GM which is $ 2\sqrt2$
and $ p + q > 4\sqrt2$

#1
EDIT: i know there was a mistake
let line segment OP equal x
by the proportion we have $ \frac {x - b}{x - c} = \frac {x - a}{x - d}$
cross multiplying gives us $ x^2 - ax - cx + ac = x^2 - bx - dx + bd$
$ - ax - cx + ac = - bx - dx + bd$
$ bx + dx - ax - cx = bd - ac$
$ (b + d - a - c)x = bd - ac$
$ x = \frac {bd - ac}{b + d - a - c}$
that is our answer

#3
the coins in bag 4 weigh 3 b/c if it weighed 2 or less, the 13 coins from the first 3 bags could only weigh 39 at most and the coins in bag 4 would only weigh 54 and 27 and both of these are far less than 95

with 27 coins weighing 3 each, we have to make the other coins weigh 14 ounces

the coins from bag 3 and 2 must weigh 1 b/c if it was more, the weight would be too much.
this leave the coins in bag 1 weighing 2 oz.

so $ \boxed{Bag\,1 - 2,Bag\,2 - 1,Bag\,3 - 1,Bag\,4 - 3}$


If a line segment makes equal acute angles with the x,y, and z axes, what is the measure of that acute angle?

[hide="Solution"]Let $ \alpha$, $ \beta$, and $ \gamma$ be the angles made with the $ x$, $ y$, and $ z$ axes respectively. It is well-known that $ \cos^2 \alpha \plus{} \cos^2 \beta \plus{} \cos^2 \gamma \equal{} 1$. Therefore, since the three angles are equal, $ \cos \alpha \equal{} \displaystyle\frac{\sqrt{3}}{3}$.

So, the angle is $ \cos^{\minus{}1} \frac{\sqrt{3}}{3}$.[/hide]


modular arithmetic
number theory proposed
number theory

[quote="rogue"]Find the $ 2008$-th rightmost digit in decimal representation of the number $ 7^{\small\overbrace{99\ldots9}^{2009}}.$[/quote]
Quite nice.

[hide="My solution"]
Let $ U \equal{} 7^{10^{2009} \minus{} 1}$ and so $ 7U \equal{} 7^{10^{2009}}$

$ \varphi(10^{2008}) \equal{} \varphi(2^{2008})\varphi(5^{2008})$ $ \equal{} 4\cdot 10^{2007}$ $ \implies$ $ 7^{4\cdot 10^{2007}}\equiv 1\pmod{10^{2008}}$

So, raising to power $ 25$ : $ 7U \equal{} 7^{10^{2009}}\equiv 1\pmod{10^{2008}}$ and ends with at least $ 2007$ zeroes and a one.

Then $ U$ ends with a $ "3"$ preceded with periodics sequence $ 285714$ at least as much as necessary to cover the $ 2008$ lasts digits.
(this sequence is easy to find making the division in a reverse manner and since $ \gcd(7,10) \equal{} 1$).

Since $ 2008\equiv 4\pmod 6$ (the period length), the $ 2008$ rightmost digits of $ U$ are $ 714,(334$ times $ 285714),3$

And the required digit is $ "7"$

[/hide]

[b][u]Edit [/u][/b]:  :censored: ***, too late!. Congrats, Stephencheng :) :)


31) What is the number of degrees in the acute angle $ \theta$ satisfying $ \cos(\theta) \equal{} \frac {1}{2}\sqrt {2 \plus{} \sqrt {2}}$?

Also another problem from the same test

16) A roll of toilet paper has a diameter of 6 in. The tissue is rolled around a 2 in diameter cardboard tube. There are 1000 sheets, each 4in by 4 in. Estimate to the nearest thousandth of an inch, the thickness of each sheet of paper.

[hide="#31"] Uh...desimplify to get $ \cos\theta = \sqrt {\frac {1 + \frac {\sqrt {2}}{2}}{2}}$. Now, this looks like the half angle formula, so we find it's 22.5 or 67.5.[/hide]

EDIT:Fixed for the other value.

[hide="Longer but more direct and more complete"]
$ \cos^2 \theta \equal{} \frac{2 \plus{} \sqrt{2}}{4}$
$ 4 \sin ^2 \theta \equal{} 4\left( 1 \minus{} \frac{2 \plus{} \sqrt{2}}{4} \right) \equal{} 2 \minus{} \sqrt{2}$

Multiplying, we get $ 4 \sin ^2 \theta \plus{} \cos ^2 \theta \equal{} \frac{1}{2}$

$ \sin ^2 (2 \theta) \equal{} \frac{1}{2}$

$ \sin 2 \theta \equal{} \frac{\sqrt{2}}{2}$

$ 2\theta \equal{} 45$ or $ 2 \theta \equal{} 135$, since $ 0 < 2 \theta < 180$

Then, $ \theta \equal{} 22.5$ or $ \theta \equal{} 67.5$[/hide]

[quote="The Zuton Force"][hide="Longer but more direct and more complete"]

Multiplying, we get $ 4 \sin ^2 \theta \plus{} \cos ^2 \theta \equal{} \frac {1}{2}$

[/hide][/quote]

Nice. Just one thing. I think you meant $ (4\sin^{2}\theta)(\cos^{2}\theta)\equal{}\frac{1}{2}$.

There aren't two different acute values of $ \theta$ which give the same value of $ \cos \theta$ - ever. Sure, if two equations hold, then their product does: but the reverse isn't always true, so you'd have to check for extraneous solutions. (There's another way of using double-angle formula that gives no extraneous solutions for acute $ \theta$.)


A set of five numbers has a median of 15, a range of 10, a mode of 12, and a mean of 16. What is the sum of the two greatest numbers in the set?

If there r 5 numbers and the mode is 12 then the first 3 numbers are 12,12,15
the mean is 16 so 16*5 is 80
80-39=41
so the sum is 41


Let $ n,k$ be positive integers and $ a_{1},a_{2},\cdots,a_{k}$ be postive reals. If $ S=\sum^{k}_{i=1}a_{i}$ Prove that:

$ \sum^{k}_{i=1}\frac{a^{n}_{i}}{S-a_{i}}\geq\frac{1}{n-1}\sum^{k}_{i=1}a^{k-1}_{k}$

Please correct all the typos. Here's a proof of the one I assume you wanted to post.

WLOG let $ a_{1}\ge a_{2}\ge \dots\ge a_{k}$. So $ S-a_{1}\le S-a_{2}\le \dots\le S-a_{k}$.

From Chebyshev's inequality,
\[ \sum_{i=1}^{k}\frac{a_{i}^{n}}{S-a_{i}}\ge \frac{k\sum_{i=1}^{k}a_{i}^{n}}{\sum_{i=1}^{k}(S-a_{i})}=\frac{k\sum_{i=1}^{k}a_{i}^{n}}{(k-1)S}\]
but we also get from Chebyshev's, $ k\sum_{i=1}^{k}a_{i}^{n}\ge S(\sum_{i=1}^{k}a_{i}^{n-1})$. Thus
\[ \sum_{i=1}^{k}\frac{a_{i}^{n}}{S-a_{i}}\ge \frac{1}{k-1}\sum_{i=1}^{k}a_{i}^{n-1}\]

well nice solution, my solution was also similar...sorry for the typos :(  i think i was completely asleep when posting the problem :P


function
integration
calculus
calculus computations

What rules do you have when you have a function with a square containing another function if you want to find the original function? Here is my problem:

$f(x)=^4\sqrt{x^3}$

-----------------------------------------------------------------

EDIT: Yes, i know i can rewrite it...

$f(x)=(x^3)^{0.25}$

-----------------------------------------------------------------

$F(x)=???$

What rules do you have that can apply here?

For rational numbers $p,\ \boxed{\int x^p\ dx=\frac{x^{p+1}}{p+1}+Const.\ (p\neq -1)}.$

I know that rule, but how can i apply this rule to this particular problem? If it was $^4\sqrt x$, but now that it is $(x^3)^{0.25}$... :?

Oh, i see that's true. Didn't realise that, but thx a lot for your answers, and for the very fast replies...


Asa ca iertati-ma daca, din cand in cand ,mai postez forumul gresit  :blush: .

Bun.Pe scurt, sunt elev in  clasa a VII-a la o scoala din Bucuresti si ma intereseaza si pe mine chestia asta, matematica cum ii spuneti voi :roll: .
1)Anul asta am posibilitatea de a ajunge la Onm, dar trebuie sa lucrez la greu!!!! :!: ...Ma intrebam ce culegeri/carti sunt recomandate?
2)Cate locuri sunt pentru fiecare judet la Onm sau cum se trece mai departe?
3)Intampin dificultati in downloadarea LATEXULui.....miktextul ala nu vrea sa mearga, la fel si textnicul  :?:  :?:  ...Am downloadat ceva numit Crimson editor...pot sa scriu si cu el?Daca nu, atunci de unde pot sa mai downloadez ceva?
4)Unde sunt instructiunile pentru Latex?

Imi cer scuze daca am postat intr-un loc gresit!
~~~Multumesc foarte mult!~~~

Edit:gata m-am prins cu Latex, mersi! :!:

salut,
sper sa te pot lamuri cu privire la cateva chestii :)
1) iti recomand sa lucrezi olimpiadele din anii trecuti
2) bucurestiul are ceva locuri pt nationala, mi se pare ca vreo 5-6 (depinde de clasa, si de ....alte chestii :) ) nu stiu sigur insa daca asta e numaru.
3) pt latex, vezi ca exista in meniul de sus al site`ului asta butonul "latex help", citeste cu atentie pe acolo . banuiesc ca deja ai facut asta, mie mi`a mers din prima cand am incercat. ia`o "step by step"
4) vezi 3)
succes


The number of real values of $x$ that satisfy the equation
\[ (2^{6x+3})(4^{3x+6}) = 8^{4x+5} \]
is:
\[ \text{(A) zero}\ \ \ \ \ \text{(B) one}\ \ \ \ \ \text{(C) two}\ \ \ \ \ \text{(D) three}\ \ \ \ \ \text{(E) greater than 3} \]

[hide]So we put all the exponents into a common base (2)
$\((2^{6x+3})(2^{6x+12})=2^{(12x+15)}$

$\\2^{(12x+15)}=2^{(12x+15)}$

Since the equations are equal on both sides, there are infinite many solutions, so the answer is E.[/hide]

[hide][
Simplifying the exponents:
$2^{6x+3}(2^{6x+12}) = 2^{12x+15}$
$2^{12x+15} = 2^{12x+15}$
Indicating that there will be an infinite number of solutions since both sides are equal.
E[/hide]

symplifiying yields

[hide](2)^(6x+3)*(2)^(2(3x+6))=(2)^(3(4x+5))
(2)^(12x+15)=(2)^(12x+15)

Becuase both sides are equal, any value satifies the two equalities, E I believe is hte correct answer.[/hide]

Do we need to prove the monotonicity of $f(x)=a^x$?

I think that we don't need too, because it's a subject that's kinda hard for this forum.


You want to color the integers from $1$ to $100$ so that no number is divisible by a different number of the same color. What is the smallest possible number of colors you must have ?

7 colors is clearly necessary because 1, 2, 4, 8, 16, 32 and 64 must all be different colors.  It is also sufficient: if $n$ has $m$ prime factors with multiplicity, color $n$ with color $m+1$.  The smallest color we use is 1 (for 1) and the largest color we use is 7 (for 64 and 96).


Find all pairs of functions $ f,g$ that takes the non-negative integers and maps into non-negative integers.
$ f(n)\plus{}f(n\plus{}g(n))\equal{}f(n\plus{}1)$

[hide="Solution"]$ f(n\plus{}1) \minus{} f(n) \equal{} f(n\plus{}g(n))$. Since the RHS is nonnegative, $ f$ is nondecreasing.

$ f(n) \equal{} 0$ for all $ n$ is a solution (with $ g$ being anything). Now eliminate that. Take the minimum nonnegative integer $ m$ such that $ f(m) > 0$. We know $ f(m) \plus{} f(m\plus{}g(m)) \equal{} f(m\plus{}1)$. Since $ f(m\plus{}g(m)) \geq f(m) > 0$, $ f(m\plus{}1) > f(m)$, and similarly $ f(m\plus{}1\plus{}n) > f(m\plus{}n)$ for all nonnegative integers $ n$. So if $ g(n) \geq 1$ for some $ n \geq m$, then $ f(n) \plus{} f(n\plus{}g(n)) > f(n\plus{}1)$, contradiction. Thus $ g(n) \equal{} 0$ for all $ n \geq m$. For such $ n$, the original equation becomes $ 2f(n) \equal{} f(n\plus{}1)$.

Let $ f(m)$ be any positive integer, and consider the solution $ f(n) \equal{} 0$ for $ n < m$, $ f(n) \equal{} 2^{n\minus{}m}f(m)$ for $ n \geq m$. For $ n < m \minus{} 1$, $ g(n)$ can be anything less than $ m \minus{} n$ without contradiction. $ g(m\minus{}1)$ must satisfy $ f(m\minus{}1) \plus{} f(m\minus{}1\plus{}g(m\minus{}1)) \equal{} f(m)$. We have $ f(m\minus{}1) \equal{} 0$. $ g(m\minus{}1) \equal{} 0$ gives $ 0 \equal{} f(m)$, contradicting the assumption that $ f(m) > 0$. $ g(m\minus{}1) > 1$ gives $ m\minus{}1\plus{}g(m\minus{}1) > m$, and then $ f(m\minus{}1\plus{}g(m\minus{}1)) \equal{} f(m)$ contradicts $ f(n)$ increasing for $ n > m$. So $ g(m\minus{}1) \equal{} 1$. We have already shown that $ g(n) \equal{} 0$ for $ n \geq m$.

So, the answers.

1. $ f(n) \equal{} 0$ and $ g(n)$ any function on nonnegative integers.

2. Given a nonnegative integer $ m$, a positive integer $ x$, and a sequence $ a_0, a_1, \ldots a_{m\minus{}2}$ where $ a_i < m \minus{} i$,
- for $ n < m \minus{} 1$ (if $ m > 1$), $ f(n) \equal{} 0$, $ g(n) \equal{} a_n$
- for $ n \equal{} m \minus{} 1$ (if $ m \geq 1$), $ f(n) \equal{} 0$, $ g(n) \equal{} 1$
- for $ n \geq m$, $ f(n) \equal{} 2^{n\minus{}m}x$, $ g(n) \equal{} 0$.[/hide]


logarithms
limit
function
induction
geometric series
real analysis
real analysis unsolved

Let $ \alpha\in\Bbb{C}$ and $ m\in\Bbb{Z}^{>0}$ be fixed numbers.

Define $ f(z): \equal{}1\minus{}\frac{1}{z^{\alpha}}\left(\sum_{k\equal{}0}^{m}\frac{\alpha^{k}}{k!}\left(\frac{\ln z}{z}\right)^{k}\right)^{z}$ and $ g(z): \equal{}z\,\frac{\alpha^{m\plus{}1}}{(m\plus{}1)!}\left(\frac{\ln z}{z}\right)^{m\plus{}1}$

Do $ f$ and $ g$ have the same asymptotic behaviour for $ |z|\to\infty$ ?
That means $ f(z)\sim g(z)$ resp. $ \frac{f(z)}{g(z)}\to 1$ for $ |z|\to\infty$ .

Gosh,  :o  this problem is a bit ridiculous/messy.  Hope I did not make a mistake:
Roughly speaking,
\[ z^{m}\cdot f \equal{} z^{m}\bigg(1\minus{}\frac{1}{z^{\alpha}}\bigg(1\plus{}\alpha\frac{\ln z}{z}\plus{}\cdots(whatever)\bigg)^{z}\bigg)\sim z^{m}\mbox{~as~}\alpha>0\] and \[ z^{m}\cdot g \equal{}\frac{\alpha^{m\plus{}1}}{(m\plus{}1)!}(\ln z)^{m\plus{}1}.\]
Altogether, $ \frac{f}{g}\sim C(\alpha,m)\frac{z^{m}}{(\ln z)^{m\plus{}1}}$ which tends to infinity.  Hence, I say no to your question.

Ah, my problem is a bit ridiculous and messy. Thank you very much for the flowers.

I don't see the reason for $ z^{m}\cdot f(z)\sim z^{m}$ as $ \alpha>0$ and I think that's not true.

[quote="Dr. Doe"]Ah, my problem is a bit ridiculous and messy. Thank you very much for the flowers. [/quote] flowers?  :huh:
[quote="Dr. Doe"]I don't see the reason for $ z^{m}\cdot f(z)\sim z^{m}$ as $ \alpha > 0$ and I think that's not true.[/quote]
How about for $ \alpha>1$:
\[ z^{m}\cdot f \equal{} z^{m}\bigg(1\minus{}\frac{1}{z^{\alpha}}\bigg(\sum_{k\equal{}0}^{m}\frac{\alpha^{k}}{k!}\left(\frac{\ln z}{z}\right)^{k}\bigg)^{z}\bigg)\]
We will let $ \ell\equal{}1\minus{}\lim_{z\to\infty}\frac{1}{z^{\alpha}}\left(\sum_{k\equal{}0}^{m}\frac{\alpha^{k}}{k!}\left(\frac{\ln z}{z}\right)^{k}\right)^{z}$, and show $ \ell\equal{}1$. Now, for large $ z$ we have (geometric series)
\[ \frac{1}{2(1\minus{}1/z)}\le\sum_{k\equal{}0}^{m}\frac{1}{z^{k}}\le\sum_{k\equal{}0}^{m}\bigg(\frac{\ln z}{z}\bigg)^{k}\le\frac{1}{1\minus{}z/\ln z}\] which gives us that \[ c(\alpha,m)\bigg(\frac{z}{z\minus{}1}\bigg)^{z}\le\left(\sum_{k\equal{}0}^{m}\frac{\alpha^{k}}{k!}\left(\frac{\ln z}{z}\right)^{k}\right)^{z}\le C(\alpha,m)\bigg(\frac{z/\ln z}{z/\ln z\minus{}1}\bigg)^{z}\]
Notice that $ z/(z\minus{}1) \equal{}\big( 1\plus{}(1/z)\big)^{z}\to e$ and $ \bigg(\frac{z/\ln z}{z/\ln z\minus{}1}\bigg)^{z}\to z$.  Next, multiplying through by $ 1/z^{\alpha}$ and taking limits yields my claim for $ \ell$.  So, $ z^{m}\cdot f(z)\sim z^{m}$.

$ \ell\equal{}0$ and not $ 1$ !

You meant $ \sum_{k\equal{}0}^{m}\left(\frac{\ln z}{z}\right)^{k}\le\frac{1}{1\minus{}\ln z\,/z}$ but this doesn't give us the next estimate.

And you meant $ (z/(z\minus{}1))^{z}\sim (1\plus{}(1/z))^{z}\to e$ and $ \left(\frac{z/\ln z}{z/\ln z\minus{}1}\right)^{z}\sim z$.

As a matter of fact $ \frac{1}{z^{\alpha}}\left(1\plus{}\alpha\frac{\ln z}{z}\plus{}o(z)\right)^{z}\to 1$, hence $ f(z)\to 0$ for $ |z|\to\infty$ .

[quote="Dr. Doe"]You meant $ \sum_{k \equal{} 0}^{m}\left(\frac{\ln z}{z}\right)^{k}\le\frac{1}{1\minus{}\ln z\,/z}$
And you meant $ (z/(z\minus{}1))^{z}\sim (1\plus{}(1/z))^{z}\to e$ and $ \left(\frac{z/\ln z}{z/\ln z\minus{}1}\right)^{z}\sim z$.[/quote]
Yes, sorry for the typo.

[quote="Dr. Doe"]As a matter of fact $ \frac{1}{z^{\alpha}}\left(1\plus{}\alpha\frac{\ln z}{z}\plus{}o(z)\right)^{z}\to 1$, hence $ f(z)\to 0$ for $ |z|\to\infty$ .[/quote]
Oh, you are right.  I forgot that the constants in my argument would be to the $ z$ power  :oops: which messes this messy idea up.  Sorry.

asymptotic for $ f(z)$ will be
$ f(z) = 1 - \frac {1}{z^{a}}a^{z}( \frac {\ln z}{z} )^{z}( \sum \frac {a^{k - 1}}{k!} (\frac {\ln z}{z})^{k - 1})^{z}$
=$ 1 - \frac {1}{z^{a}}a^{z}(\frac {\ln z}{z})^{z}\exp (z \ln ( \sum \frac {a^{k - 1}}{k!} (\frac {\ln z}{z})^{k - 1}))$
~${ 1 - a^{z} z^{\frac {az}{2} - z - a}}\ln ^{z}z$
ofcourse $ f$ and $ g$ has different asymptotic, I note that asymptotic of function $ f(z)$ is not depend  $ m$

[quote="baz"][quote="Dr. Doe"]Ah, my problem is a bit ridiculous and messy. Thank you very much for the flowers. [/quote] flowers?  :huh: [/quote]

 :rotfl:  :rotfl:

oh man ! my stomach is hurting from laughing so hard.....

When $ x \to 0$, $ \ln \sum_{k\equal{}0}^m \frac{x^k}{k!}\equal{}\ln(e^x\minus{}\sum_{k\equal{}m\plus{}1}^{\infty}\frac{x^k}{k!})\sim x\minus{}\frac{x^{m\plus{}1}}{(m\plus{}1)!}\plus{}o(x^{m\plus{}2})$,
let $ x\equal{}\alpha\frac{\ln z}{z}$
$ f(z)\equal{}1\minus{}\exp\left(z*\ln\sum_{k\equal{}0}^m \frac{x^k}{k!}\minus{}\alpha*\ln z\right)\sim z*\frac{\alpha^{m\plus{}1}}{(m\plus{}1)!}\left(\frac{\ln z}{z}\right)^{m\plus{}1}\sim g(z)$

[quote="Buffalo"]When $ x \to 0$, $ \ln \sum_{k \equal{} 0}^m \frac {x^k}{k!} \equal{} \ln(e^x \minus{} \sum_{k \equal{} m \plus{} 1}^{\infty}\frac {x^k}{k!})\sim x \minus{} \frac {x^{m \plus{} 1}}{(m \plus{} 1)!} \plus{} o(x^{m \plus{} 2})$,
let $ x \equal{} \alpha\frac {\ln z}{z}$
$ f(z) \equal{} 1 \minus{} \exp\left(z*\ln\sum_{k \equal{} 0}^m \frac {x^k}{k!} \minus{} \alpha*\ln z\right)\sim z*\frac {\alpha^{m \plus{} 1}}{(m \plus{} 1)!}\left(\frac {\ln z}{z}\right)^{m \plus{} 1}\sim g(z)$[/quote]
I believe you are right.
now I'm sure that this moment is true :
$ \ln(e^x \minus{} \sum_{k \equal{} m \plus{} 1}^{\infty}\frac {x^k}{k!})\sim x \minus{} \frac {x^{m \plus{} 1}}{(m \plus{} 1)!} \plus{} o(x^{m \plus{} 2})$,
I can rpove it only for small $ m$ and then by induction , but I think there must be more simple way :
how do you prove this :

$ \ln(e^x \minus{} \sum_{k \equal{} m \plus{} 1}^{\infty}\frac {x^k}{k!})\sim x \minus{} \frac {x^{m \plus{} 1}}{(m \plus{} 1)!} \plus{} o(x^{m \plus{} 2})$
maybe I understand because :
$ \ln e^{x} \equal{} x$ when $ x \minus{} > 0$
in other hand
$ \ln e^{x} \equal{} \ln (1 \plus{} \sum _{i \equal{} 1}^{ \plus{} oo}\frac {x^{i}}{i!}) \equal{} \sum_{j \equal{} 1}^{ \plus{} oo} ( \minus{} 1)^{j \plus{} 1}\frac {(\sum _{i \equal{} 1}^{ \plus{} oo}\frac {x^{i}}{i!})^{j}}{j}$ which must be equal to $ x$
therefore if we put
$ \ln (e^{x} \plus{} A)$ where $ A \equal{} O(x^{k})$ then I will have :

$ \ln (e^{x} \plus{} A) \equal{} \sum_{j \equal{} 1}^{ \plus{} oo} ( \minus{} 1)^{j \plus{} 1}\frac {(A \plus{} \sum _{i \equal{} 1}^{ \plus{} oo}\frac {x^{i}}{i!})^{j}}{j} \equal{} A \plus{} ...$ where $ ... \equal{} O(x^{k \plus{} 1})$
I understand  :)

[quote="Extremal"][quote="Buffalo"]When $ x \to 0$, $ \ln \sum_{k \equal{} 0}^m \frac {x^k}{k!} \equal{} \ln(e^x \minus{} \sum_{k \equal{} m \plus{} 1}^{\infty}\frac {x^k}{k!})\sim x \minus{} \frac {x^{m \plus{} 1}}{(m \plus{} 1)!} \plus{} o(x^{m \plus{} 2})$,
let $ x \equal{} \alpha\frac {\ln z}{z}$
$ f(z) \equal{} 1 \minus{} \exp\left(z*\ln\sum_{k \equal{} 0}^m \frac {x^k}{k!} \minus{} \alpha*\ln z\right)\sim z*\frac {\alpha^{m \plus{} 1}}{(m \plus{} 1)!}\left(\frac {\ln z}{z}\right)^{m \plus{} 1}\sim g(z)$[/quote]
I believe you are right.
now I'm sure that this moment is true :
$ \ln(e^x \minus{} \sum_{k \equal{} m \plus{} 1}^{\infty}\frac {x^k}{k!})\sim x \minus{} \frac {x^{m \plus{} 1}}{(m \plus{} 1)!} \plus{} o(x^{m \plus{} 2})$,
I can rpove it only for small $ m$ and then by induction , but I think there must be more simple way :
how do you prove this :

$ \ln(e^x \minus{} \sum_{k \equal{} m \plus{} 1}^{\infty}\frac {x^k}{k!})\sim x \minus{} \frac {x^{m \plus{} 1}}{(m \plus{} 1)!} \plus{} o(x^{m \plus{} 2})$[/quote]

Isn't it as clear as ...
$ \ln(e^x \minus{} \epsilon)\equal{}x\plus{}\ln(1\minus{}\epsilon*e^{\minus{}x})\sim x\minus{}\epsilon*e^{\minus{}x}\plus{}o(\epsilon)\sim x\minus{}\epsilon\plus{}o(\epsilon)$

[quote="Buffalo"]

Isn't it as clear as ...
$ \ln(e^x \minus{} \epsilon) \equal{} x \plus{} \ln(1 \minus{} \epsilon*e^{ \minus{} x})\sim x \minus{} \epsilon*e^{ \minus{} x} \plus{} o(\epsilon)\sim x \minus{} \epsilon \plus{} o(\epsilon)$[/quote]
Nice proof   :what?: .
This really work  :blush:  When I post my proof I forget something easy..


$ 13$ black and $ 12$ white pawns are placed in a $ 5\times 5$ board in a usual fashion. The first player removes any black pawn and moves any white pawn into the empty square. The second player then moves any black pawn into the empty square, if he can. The first player then moves any white pawn into the empty square, if he can.  The second player then moves any black pawn into the empty square, if he can. Play continues like this. The player who cannot move loses.

Who has a winning strategy?

How do you move your peons? Like kings? I don't understand the way they move.

I'm sorry. Well, it can move one square in either the vertical or the horizontal direction.

Anybody ? Blacks seem to be the winners , how to prove it ?

Domi

Sorry for my bad english  :blush:

Solution .

The white choose a square and blacks imagine the field without this square covered by dominoes 1X2 . The blacks only choose the second square of the domino that the white choose .

Domi

Sorry for my bad english


Factor by grouping.

a.  x^3 - 6x^2 - 4x + 24
b.  x^2 - z^2 + 8z -16

forgot how to do this

[quote="DiscreetFourierTransform"]I'll do number one as an example.
Hopefully you get the gist and can do number two by yourself.

[hide="Solution 1"]
Find terms with coefficients that are constant multiples of each other:
$1: 4$ and $6: 24$ work nicely.
Now look at the exponents, and see how the best way to GROUP the terms is (hence the name factoring by grouping)
I see
$(x^{3}-4x)-(6x^{2}-24)$
Now factor each one of these to get a nice constant multiple:
$x(x^{2}-4)-6(x^{2}-4)$
Notice how each of these has a $(x^{2}-4)$ term, so we can combine them with distributive property.
Answer =
$(x-6)(x^{2}-4)$
This isn't done, though, because the right factor can be simplified as a difference of squares.
Solution:
$(x-6)(x+2)(x-2)$
[/hide]
Number two is similar, except you have to notice that you have two variables, and need to group accordingly.[/quote]

would the solution to number 2 be -z(x+4)(x-4)

@linzk3:
No it wouldn't. If you open brackets, you would get:
$-z(x+4)(x-4)=-z(x^{2}-16)=-zx^{2}+16z$
which is not equal to the original form

Solution to b:
$x^{2}-z^{2}+8z-16 = x^{2}+(-z^{2}+8z-16) = x^{2}+(z-4)(4-z)$

@kohjhsd:
oh ok, I did it a bit differently


trigonometry
geometry
3D geometry
pyramid
geometry solved

Hi,

I have just posted this on the computer science and informatics section (http://www.mathlinks.ro/Forum/viewtopic.php?t=40157) forgive the cross-posting - but think you guys might be able to help me!

I have been trying to create an equation to plot points (perspective projection style) for wireframe shapes (for Flash). I need to start with 4 points and go up to an unlimited number creating geodesic forms and need to do this dynamically. I have been trying various things to do this for me and am now at the stage of dreaming in sine and cosine wavelengths! :)

I have found co-ordinates for two forms and need an equation that would generate this kind of data.

For example a 3sided pyramid co-ordinate points are - (4 points)
-0.212331, -58.3687, -82.3175,
82.3334, 58.3686, 0.228231,
-82.7581, 58.3686, 0.228231,
0.212331, -58.3686, 82.774

and a square's are: (8 points)
-150, -150, -150,
-150, 150, -150,
150, 150, -150,
150, -150, -150,
150, -150, 150,
-150, -150, 150,
-150, 150, 150,
150, 150, 150

Is anyone familiar with perspective projection and plotting xyz co-ordinates on an xy plane - and may be able to shed some light on this equation? ;)

Thanks all!

Maybe I don't understand the question, but this doesn't sound very difficult. Assume that the x, y, and z axes intersect in a common point and that this point is the origin. Then the x axis is y=ax, the y axis is y=bx, and the z axis is y=cx. For each of these lines (I use a as in the x axis), we want $y^2+x^2=1$, $x^2(a^2+1)=1$, and $x=\sqrt{\frac 1{a^2+1}}$. Let this value of x be (for the x, y, and z axes respectively) $x_x$, $x_y$, and $x_z$, and let $y_x=ax_x$, $y_y=bx_y$, and $y_z=cx_z$. Now we plot the point $(x,y,z)$ as $(xx_x+yx_y+zx_z,xy_x+yy_y+zy_z)$. Assuming I understood the question, no trigonometry is involved.

Good luck with the program; you have inspired me to make something myself :) . (I have thought of such a thing before, but always from the perspective of rotating, not of starting with certain angles.)

Afterthought: also, if you want it to, for instance, look up at something, multiply or divide the z values by a scaling factor.


As everyone knows, division by zero is undefined. Also, multiplying any number by zero will equal zero.

But...

1/0 * 0 = 1 because the zeroes cancel each other out.

That would imply that 0/0 = 1

But division by zero is undefined...

I'm confused now. Probably should've laid off the caffiene :/

[quote="Talon"]
1/0 * 0 = 1 because the zeroes cancel each other out.
[/quote]

You can't actually cancel out zeroes like that :)

1/0 * 0 is just undefined since 1/0 is undefined.

They don't cancel each other out. 1 / 0 * 0 = 1 / (0 / 0), so you'd still be attempting to divide by 0.

We usually call 0/0 an indeterminate form since ``different forms of zero'' so to speak can give us different answers. 1/0, on the other hand, is ``undefined'' since it always goes to something unbounded.

You can't divide by zero, because dividing by a number is to multiply by the reciprical. A reciprocal is what you multiply the number by to get one. Since anything times zero is zero, zero has no reciprocal. So, since it has no reciprocal, you cannot divide by zero.

[quote="tucsonlarry"]I think it depends on the individual situation.  For example, sin(x)/x = 1 if x is 0.  But I think sin^2(x) / x = 0 when x = 0.  They are both examples of 0/0.[/quote]

No -- the limit as x goes to zero of sin(x)/x is 1, and the limit as x goes to zero of sin2(x)/x is 0.  The expressions themselves are indeterminant.

This reminds me of something very profound. Mathcamp rule #4: "Never divide by 0 without staff supervision."

Quote from MC '04: "Whoa... you need [b]way[/b] more staff supervision if you want to divide by [b]that[/b] many zeroes."


inequalities
search
number theory theorems
number theory

Bertrand-Chebyshev theorem states that for any $n > 1$, there exists a prime number $p$ such that $n < p < 2n$.

Any proof? (the easiest possible, please...)

[url=http://planetmath.org/encyclopedia/ProofOfBertrandsConjecture.html]proof of Bertrand's conjecture[/url] by PlanetMath
[url=http://en.wikipedia.org/wiki/Bertrand%27s_postulate]Bertrand's postulate[/url] by Wikipedia
[url=http://en.wikipedia.org/wiki/Proof_of_Bertrand%27s_postulate]Proof of Bertrand's postulate[/url] by Wikipedia

I think that a lot of sites proving that I search it elsewhere come out.


This game/thing to do when you are bored is very simple.

try to be the person to reach 100.

if you do, i will give you a  :D  and a  :wink:  and also a  :rotfl: .

state 5 consecutive numbers, starting off the last person.

i will start...

1,2,3,4,5

ummmm Dragon96 i didn't know that 49 came after 36 and 64 came after 49 and that 81 came after 64 and that 100 came after 81...

 :rotfl:  :rotfl:  :ninja:

How about we just let this spammy, pointless thread die off?

Yes i agree

A Admin plz lock this Thread.. Its getting really Spammy!!

Wait a minute just let me finish to 100,000 before we close this awfully spammy thread (sorry I think I probably made it spammy hehe).

.06, 25000.045, 50000.03, 75000.015, 100000 K I'm Done.

I thought that more people would have selected "Don't select this option."

Let's count to 1000 by 4's.

1, 2, 3, 4

5,   L
6,   O
7,   C
8,   K

this, please.

This has gone far enough.  It has more spam than the original vending machine game.  Moderator, please lock.

ugh...i agree with maybach...please lock.

its sort of sad the stuff people did to this topic...i mean, if you think its boring, then don't post in it!!! seriously.

spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam spam 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[quote="kathy"]When two indistinguishable dice are rolled n times.
(1) In how many ways can a pair of each of the six numbers show at least once?
(2) In how many ways can each of the pairs of 3's ,4's ,5's and 6's show at least
once and either a pair of  aces or a pair of deuces show at least once?

I really have trouble with combinatorial math. Please help me .
Thank you very very much!![/quote]

[hide]
(1) we can do 1 - probabiloity of those 6 pairs never showing up. So $1 - \frac{5}{6}^n$
(2) so either 1,3,4,5,6 shows up at least once or 1,2,4,5,6 shows up once, but it could be that both show up so we subtract that..

\[ 1 - \frac{31}{36}^n + 1 - \frac{31}{36}^n - ( 1 - \frac{5}{6})^n \]
I am not too sure about this actually, uh, don't trust what i put for 2.

[/hide]

the first one, should it be (1-35/36)^n. or shoule it be (1-5/6)^N?

Thank you!!

I'm sorta guessing this... so don't trust my answers
[hide]
(1)
um... $n\ge 6$

6 rolls are already taken by each pair showing up (and the number of possible combinations is 6!), so we have $n-6$ spaces left for whatever combination we want (18 combinations). The $n-6$ spaces can be arranged any way  within the $n$, so we have $\binom{n}{n-6}$ combinations.
$6!\cdot 18\cdot \binom{n}{n-6}$

(2)
similarly,
$2(5!\cdot 17\cdot \binom{n}{n-5})$
[/hide]

How about I ask it like this:

what is the probability of getting the pair of 1's when we rolled two indistinguishable
dice n times
and what is the probability of getting the pair of 1's and 2's ?....
then we can use the principle of inclusion and exclusion to do it.

i meant $(\frac{5}{6})^n$

except, now I see it's indistinguishable so uh, it should actually be $1 - \frac{2}{3}^n$ for the first

and well, just switch ever instance of $36$ with $18$ in my 2nd answer.

Solving the first one:

Total no. of ways in which the dice can show up = $15$
Now no. of ways in which the dice donot show the same no.= $15-5 = 10$
Thus the probability of never obtaining a pair of six numbers = $(10/15)^n = (2/3)^n$
Thus, probability of obtaining pair of numbers atleast once = $1-(2/3)^n$

eh... it looks like my interpretation of the problem is different from others'
kathy,
is it the number of ways to get pairs of 1's AND 2's AND 3's AND 4's AND 5's AND 6's? (this is the one I solved for)
or the number of ways to get pairs of 1's OR 2's OR 3's OR 4's OR 5's OR 6's?
(this is the one I believe agolsme solved for)


algebra
polynomial
search
AMC
AIME
algebra unsolved

find all polynomials P(x) such that
P(x)+ P($\frac{1}{x})=x+\frac{1}{x}$
for all x non zezo

For any P(x) with degree > 1, $P(x)+P(\frac{1}{x})$ will give terms of the form $\frac{1}{x^{p}}, \ p > 1$, so $\deg P = 1$, and P(x)=x.

[quote="falconwing64"]mathisfun, your proof doesn't sound rigorous...you're assuming that P(x) is a nice orderly polynomial...you have to disprove floor, ceiling, absolute, etc.  ...besides that, your logic is good for that one section...

btw, $P(x)=x^{-1}$ works.[/quote]
$P(x)=x^{-1}$ is not polinomial.

[quote]mathisfun, your proof doesn't sound rigorous...you're assuming that P(x) is a nice orderly polynomial...you have to disprove floor, ceiling, absolute, etc. ...besides that, your logic is good for that one section...

btw, $P(x)=x^{-1}$ works.[/quote]

Hmm... I thought the problem tells us to restrict our search to polynomials.

I'm sorry...I tend to skip over those details...one of my main faults  :blush:
(which tends to me losing points on AIME)  :mad:


What is the positive difference between $ 3^5$ and $ 5^3$?

hello, it is $ 3\cdot3\cdot3\cdot3\cdot3\minus{}5\cdot5\cdot5\equal{}9\cdot9\cdot3\minus{}25\cdot5\equal{}81\cdot3\minus{}125\equal{}243\minus{}125\equal{}118$.
Sonnhard.

That is a shortcut. He just made it look long.

A shortcut could be $ 243\minus{}125\equal{}(250\minus{}7)\minus{}125\equal{}(250\minus{}125)\minus{}7\equal{}125\minus{}7\equal{}\boxed{118}$ but its basically the same thing.


10. The medians $AD, BE, CF$ of triangle $ABC$ intersect at $G$. The quadrilaterals $AFGE$ and $BDGF$ are cyclic. Prove that the triangle is equilateral.

2. The lines containing the sides of triangle $ABC$, inscribed in a circle of center $O$, intersect at $H$. It is known that $AO = AH$. Find the measure of $\angle CAB$.

5. Given is an integer $n \geq 1$. Solve the equation:
$\left|\tan^n x - \cot^n x\right| = 2n\left|\cot 2x\right|$.

In fact I don't understand the second problem. I think the lines haven't a common intersection. :(

[b](1)[/b] Denote a = BC, b = CA, c = AB the triangles sides and $m_a = AD, m_b = BE, m_c = CF$ its medians. Let X, Y be the diagonal intersections of the quadrilaterals AFGE and BDGF. Then $XF = XE = \frac a 4$, $XA = \frac{m_a}{2}$, $XG = \frac{m_a}{6}$. Similarly, $YD = YF = \frac a 4$, $YB = \frac{m_b}{2}$, $YG = \frac{m_b}{6}$. Since these 2 quadrilaterals are cyclic,

$XF \cdot XE = XA \cdot XG,\ \ YD \cdot YF = YB \cdot YG$

$\frac{a^2}{16} = \frac{m_a^2}{12},\ \ \frac{b^2}{16} = \frac{m_b^2}{12}$

$3a^2 = 4m_a^2,\ \ 3b^2 = 4m_b^2$

Substituting $m_a^2 = \frac 1 4 (2b^2 + 2c^2 - a^2),\ m_b^2 = \frac 1 4 (2c^2 + 2a^2 - b^2)$,

$3a^2 = 2b^2 + 2c^2 - a^2,\ \ 3b^2 = 2c^2 + 2a^2 - b^2$

$2a^2 = b^2 + c^2,\ \ 2b^2 = c^2 + a^2$

Eliminating c from these 2 equations,

$c^2 = 2a^2 - b^2 = 2b^2 - a^2,\ \ 3a^2 = 3b^2, a = b$

Substituting a = b to either equation yields $b^2 = c^2,\ b = c$.

[b](2)[/b] I do not understand the wording of the 2nd problem. However, assuming that H is the triangle ortocenter, then $AH = 2R \cos \widehat A$, where R is the triangle circumradius. If $\widehat A > 90^\circ$ is obtuse, $\cos \widehat A < 0$ and negative AH means that the orthocenter H is on the ray opposite to the ray AD, where D is the A-altitude foot. Since AO = R, |AO| = |AH| means $|2R \cos \widehat A| = R,\ |\cos \widehat A| = \frac 1 2,\ \widehat A = 60^\circ$ or $120^\circ$.


I weep for those who have valiantly tried to excite the minds of young problem-solvers. :(

However, there are MOEMS and MATHCOUNTS marathons that moderators won't lock simply because a problem was too hard! :D

Um well see here...
The people actually posting were people that obviously knew how to do the problems, and have known for a while.


please could somebody do at least some of these problems? thx

[quote="xxxxx"]1)Dan can read 1 page every 2 mins or listen to 1 page every 5 mins. If he has an hour and a half to finish 30 pages, what is the maximum number of pages to which he can listen?

2)Find the remainder when $x^{11}-x^{10}+x^{3}+x-4$ is divided by $x-2$

3)Find the units digit of $7^{21}-3^{17}+2^{51}$

4)Carlson starts building a glass elevator at the rate of 10 goops per hour. After 5 hours, he is maimed by a glass splinter. Now, one-armed Carlson can only work at 2 goops per hour, but finishes the elevator in a total of 20 hours. What is his average rate in goops per hour?

5)evaluate 5-3+7+5/12-1+1+5/4-1/3+1/7...[/quote]
[hide="1"]$2x+5y=90$ and $x+y=30$

$y=\boxed{10\text{ pages}}$[/hide]
[hide="2"]$x^{10}+x^{9}+2x^{8}-4x^{7}-8x^{6}-16x^{5}-32x^{3}-63x^{2}-126x-251-\frac{506}{x-2}$

The remainder is $\boxed{506}.$[/hide]
[hide="3"]The units digit of powers of numbers repeats every 4 powers. This means the units digit of $7^{21}-3^{17}+2^{51}$ is the same as the units digit of $7^{1}-3^{1}+2^{3},$ which is $\boxed{2}.$[/hide]
[hide="4"]$\text{Average}=\frac{\text{total goops}}{\text{total time}}=\frac{50+30}{20}=\boxed{4\text{ goops per hour}}$[/hide]
[hide="5"]$5-3+7+\frac5{12}-1+1+\frac54-\frac13+\frac17=\boxed{10}$[/hide]

[quote="perfect628"]On 5 it was an infinite series.[/quote]
Can you express it as a function or sum? I don't see the pattern.

[quote="Farenhajt"]I would bet that ${5\over 12}$ should read ${5\over 2}$.

[hide="Then"]we have

\begin{eqnarray*}&& \left(5+{5\over 2}+{5\over 4}+\dots\right)+\left(-3-1-{1\over 3}-\dots\right)+\left(7+1+{1\over 7}+\dots\right)\\ &=& \frac{5}{1-{1\over 2}}-\frac{3}{1-{1\over 3}}+\frac{7}{1-{1\over 7}}\\ &=&{41\over 3}\end{eqnarray*}[/hide][/quote]

Ooops..sry. Ure right, it's 5/2


Functional Analysis
real analysis
real analysis unsolved

is the $l^{2}$$(IN)$ a banach space?and
find the topologique dual of it

$\ell^{2}(\mathbb{N})$ is a Hilbert space. Two key facts:

1. Any Hilbert space is a Banach space.

2. The dual of any Hilbert space is itself.

(I'm assuming that by "topologique dual" you mean the Banach space of all bounded ([i]i.e.[/i], continuous) linear functionals, with the operator norm as the metric.)


For those of you who are ranked in school according to academics, what is your rank? Also how many people are in your class?

Well, I say number 1 because I'm I'm the only person in the gifted program that had all A's that year.

There should be an option for people not in they top fifth of their class.  Some people might go to a private or magnet school, where there are many smart people.

we are unranked. it is impossible to rank our class, because of the nature of our school.

I was 8th out of 130 something...for 2 years I had the same GPA as the person who was first, but I screwed up my freshman year really badly.  It really does not bother me too much though.

If you can call what my school does with GPA's rank then yes I go to a ranked school.  This past year I ranked #1.

If you mean grade wise - I have had straight A's for who knows how long + extra classes. So, I think that's first, though you never know.

Ethically, I'd have to vote under being in the Top 10 in the freshman class at my school (based on GPA).  There's a six-way tie for #1, and I'm in it.

Well in grade nine i was ranked 3rd or 250 ish with 92.9% because two of my teachers dispised me (English and Science and they gave me 87 and 88), but this year I'm pretty sure I'm far enough ahead to make up the 2% lead the girl on top had (I had 95.9%, while the other two were hovering around 93%)

My school isn't ranked, but I don't think doing homework due that day on the bus, or 10 minutes before class would help my cause.

i'm 1 of 1.

class rankings don't mean anything, really...

I don't know...top ten but my school doesn't tell you your rank until senior year.

[quote="MagnusRex"]Well, at our school I'm 2 out of 130 or so.

Our GPA is screwed. Here's why.

For an AP or Honors, you get a 5.4 for a 98-100, a 5.2 for a 93-97, and a 5.0 for a 90-92. That's ridiculous.

Even more ridiculous is that they weight AP and Honors the same. I would be murdering the #1 girl because I take much more difficult courses than she does, but I think she beat me by one point in English once. This makes me very frustrated.[/quote]

My school is much more screwed up.  You get a multiplier of 1.0x for your GPA, where x is the number of honors classes you've taken so far, not exceeding three per semester.  So, for example, the kid who takes 5 honors classes freshman year and gets straight A's has a GPA of 1.06*4=4.24, the 6 coming from 3 honors classes per semester.  So in every class there are kids who take 4 honors classes, then if they find that they won't get an A in one of them, drop down in one of them, and become Validictorians.  I'm not making this stuff up.  I took 3 AP's and 3 honors classes last semester, and got the same weighting as someone taking just 3 honors classes.

Eum I'm too lazy to work a lot for school so I finished 16th out of 21 and that's only high school  :cool:


Find all $ (x,y)\in\mathbb Z^2$ of the equation

$ 9x^2 \plus{} 2xy \plus{} y^2 \minus{} 92x \minus{} 20y \plus{} 244 \equal{} 0.$


AMC
AIME
USA(J)MO
USAMO
modular arithmetic
LaTeX
binomial coefficients

Prove that for all $a\in\{0,1,2,\ldots,9\}$ the following sum is divisible by 10:
\[ S_a = \overline{a}^{2005} + \overline{1a}^{2005} + \overline{2a}^{2005} + \cdots + \overline{9a}^{2005}. \]

For those unfamiliar with the notation, I'm assuming $\overline{9a}$ means the 2-digit number with tens digit 9 and units digit $a$ (please correct me if I'm wrong about that, Valentin).

Yes you are correct Richard, I should have mentioned that too, but I forgot  :blush:

Does a with a little bar over it mean that it represents a two digit number with first digit a and second digit 0?

It represents simply the digit $a$. So in general we use the notation $\overline{a_na_{n-1}a_{n-2}\ldots a_0} = 10^na_n + 10^{n-1}a_{n-1} + 10^{n-2}a_{n-2} + \cdots + 10a_1 + a_0$, provided we know that each of $a_i$-s are digits.

correct me if i am wrong...

[hide="solution"]

for a^2005, it ends in the number itself, because the ending of a number in exponential form repeats itself in every four powers. (for example, powers of two ends in the sequence of 2, 4, 8, 6, which repeats over itself)

so a^2005+2a^2005+3a^2005+ ... +9a^2005 = (a+2a+3a+...+9a) = 10a = 0mod10.

there it ends.
[/hide]

[quote="xxreddevilzxx"]correct me if i am wrong...
[/quote]You are wrong several times actually.

[hide="First of all"]
[quote="xxreddevilzxx"]
for a^2005, it ends in the number itself, because the ending of a number in exponential form [b]repeats itself in every four powers[/b]. (for example, powers of two ends in the sequence of 2, 4, 8, 6, which repeats over itself)
[/quote]
Which is obviously not true, as you can see with simple examples: 5, 6, 1, 0 only end in 5, 6, 1, 0, if we raise 9 to exponential power, then we only obtain 9 or 1 as last digit, etc. :)
[/hide]

[hide="Secondly"]
[quote="xxreddevilzxx"]
so a^2005+2a^2005+3a^2005+ ... +9a^2005 = (a+2a+3a+...+9a) = 10a = 0mod10.
[/quote]
You did not understood the definition of $\overline{ab} = 10a+b$. A simple decimal representation. Thus your computations above are not correct.
[/hide]

wat i tried to point out was the fact that if any number with unit's digit of a (a being one of {0,1,2,3,...,9}) is raised to nth power, then the resulting number ends in a if n = 1mod4, whether the sequence of unit's digit in powers of a has 4 repeating sequence, 2, only 1.

so, if you add all the numbers, which would consequently all end in a, we have 0 at the end. so it is divisible by 10.

if that is still not very understandable to u, valentin, then i need to hone my skill of proof. :(

[quote="xxreddevilzxx"]if that is still not very understandable to u, valentin, then i need to hone my skill of proof. :([/quote]Now that is a different argument. It is important what you write on an examination, than what you think ;)

For this problem would an acceptable proof be just to show that for all elements of that set to show that their units digits will add to a 0.  Because any number that ends in 0 is divisble by 10.

Or, you can just use the fact that the binomial coefficients are integers (a fact of which I gave a proof recently in this forum) and then showing that ten divides the whole thing is trivial.

Yea, I just realized that if we treat the ones digit as a, and sum we get 10a^2005, which must always end in 0, which means it is divisible by 10.  Or you could go through the binomials expansion like goodyfresh said.

[quote="mna851"]Hmm...  I have a question about the origin of these questions.  I'm assuming RMO means Romanian Math Olympiad.  So, does Romania hold math olympiads for all grades, not just the high school level like the U.S. does?[/quote]You are right with both ideas. You can see this year's questions (for district - similar to AIME - after this people go to the National RMO which is like USAMO, only that all the kids gather together in week's camp and take the tests there) in this topic: http://www.mathlinks.ro/Forum/viewtopic.php?p=179586#p179586

I think this problem is quite trivial, as
[hide]
$\overline{a}^{2005} \equiv \overline{1a}^{2005} \equiv \overline{2a}^{2005} \equiv \cdots \equiv \overline{9a}^{2005} (mod 10). $

So we obtain:
$S_a \equiv 10\overline{a}^{2005}\equiv 0 (mod 10)$[/hide]

[quote="warut_suk"]I think it's quite difficult for a 7th-grade student![/quote]It was quite easy in fact. The easiest problem in the competition :)

Hmm... the standard of mathematics there must be really good

You just need to show that $x$ has the same units digit as $x^5$ and you're done.

[hide]
b is any integer.
$a \equiv 10b+a \pmod {10}$
$a^{2005} \equiv (10b+a)^{2005} \mod {10}$
$10a^{2005} \equiv 0 \equiv a^{2005} + (10+a)^{2005}  + ... + (90+a)^{2005} \pmod {10}$
Thus S_a is divisible by 10.
[/hide]

P.S: I was in 7th grade this year (it's summer vacation for me right now)! :D

[quote="Gyan"]V -  Do I need a new pair of glasses <again>  or missing something obvious? what's so special about 2005,  Isn't it  true for ANY positive integer power?  [hide](  who cares, whatever  it is, unit digit of   2^n or 12^n or 22^n  or 32^n  ... is same for any given  n, and if you add 10 numbers which have  the same unit digits, the sum  will be divisible by 10)  :?[/hide][/quote]

 It's true for any positive integer, not just 2005.  Throwing the year into the problem makes it easy to find what year the problem was given in :lol:

[quote="LynnelleYe"]nr1337, your LaTeX is messed up.[/quote]

Fixed!


[hide]
and angle must equal 90 degrees in a square so
$5x+10=90\\\\5x=80\\\\x=16$[/hide]

[quote="236factorial"]An interior angle of a square measures 5x+10 degrees. What is x?[/quote]

[hide]Every angle of a square must be  90 degrees or it is not a square. So $5x+10=90$ $ 5x=80$ therefore $x=16$[/hide]

[quote="236factorial"]An interior angle of a square measures 5x+10 degrees. What is x?[/quote]

[hide="Click here to reveal hidden answer"]Every square has four 90 degree interior angles.

That means the equation: 5x + 10 is equal to 90.

5x + 10 = 90

After subtracting 10 from both sides, the equation turns into:

5x = 80

Divide 5 from both sides:

x = 16

The answer is 16.[/hide]

[quote="risinglegend"]my answer is 16[/quote]
well, then, how did you get 16?

please remember to hide your answers and solutions

[hide]The interior angle of a square is always 90, 90-10=80=5x, so the answer is x=16[/hide]

This is classroom math. That's why. 236 posts in the high school forums, too.

[quote="ashegh"]X is 16.

excuse me but why is this problems so easy.easier than i thaught :( .[/quote]
i think easy problems can help us to think and then we can solve any problems.(there is something else,i think ashegh you are so cool so may be this problem is not easy but it is easy for you.)


This problem is from a workout a few years ago.

A stone is dropped into a well, and the splash is heard 8.9 seconds after it is dropped. The stone falls at a rate of 16t^2 feet in t seconds, and the speed of sound is 1120 feet per second. What is the number of feet from the well to the top of the water? Express your answer to the nearest whole number.

Time down + Time up = 8.9

Time up is obviously h/1120.

Time down, however, is not so easy.

It falls 16h:^2: feet in h seconds. So, if h is height, it takes +/- sqrt(h)/4 seconds to go down. Obviously it can't take -sqrt(h)/4. So...

sqrt(h)/4 + h/1120 = 8.9
*dies*
sqrt(h)/4 = 8.9 - h/1120
sqrt(h)/4 = -(h/1120 - 8.9)
h/16 = h:^2:/1254400 - 89h/5600 + 79.21
78400h = h:^2: - 19936h + 99361024
h:^2: - 98366h + 99361024 = 0

The discriminant is negative.

What.
The.
HeIl.

What did I do wrong >_< ?

Yo calculated the discriminant wrong -- it's going to be a 10-digit number minus an 8 digit number -- definitely >0

To write two equations :

Down -   D = 16t^2

Up -        D=1160(8.9-t)

By substituting for D, you'd get a much more manageable quadratic

On another forum, you mention that the solution isn't in the handbook.  However, I found the solution on page 100 of my old handbook.  [Perhaps you have a different version of the handbook.]

In any case, the solution in the handbook uses the formulas $d=16t^2$ and $d=1120(8.9-t)$ and a guess-and-check strategy.

Looking in the Handbook, I see this was the final Workout in that handbook.  This problem also assumed the use of a calculator.  Realistically, this method would probably require at least 25 iterations to yield the necessary precision.

The handbook also mentions the quadratic formula, almost as a footnote.  Knowing some algebra certainly makes things easier.
I think a problem offering such a huge advantage to the algebra student would probably only appear on the National Competition.

Since it only asks for the nearest whole number, one way to do it if you were running out of time would be just to ignore the time. That might cause the answer to be a little off, though.


Yann and Camille go to a restaurant.  If there are 10 items on the menu, and each orders one dish, how many different combinations of meals can Yann and Camille order if they refuse to order the same dish?

Yann has $ 10$ choices for her meal, and Camille has $ 9$ choices for hers, because she won't order what Yann ordered.  Then we have $ 10 \cdot 9 \equal{} \boxed{90}$ different combinations of meals.

EDIT: Beaten.

A rather roundabout way of doing this is to use complimentary counting. There are 100 ways they could choose their meals. There are 10 ways they could choose the same dish. 100-10 = 90 ways for them to order.

[hide](9*10)1)+0+1-1+0-1+1+0-1+1=90[/hide]
yup
it all makes sence
dont worry