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Thread: Contest questions

  1. #1 Contest questions 
    Moderator Moderator AlexP's Avatar
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    There was a math contest at my school and I thought I'd post the questions from it for your solving pleasure. I wanted this to be separate from the forum game thread so I can just post another of these questions after one has been solved. Here's the first, in the same way it was written... (I didn't understand it personally)

    If ##, then find the value of ##


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  3. #2  
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    the answer is , or at least that is what I think.


    Haha, I made a dumb error, its actually 28.


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  4. #3  
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    Not what the answer key has... Certainly doing better on it than I did though. I didn't even begin.
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    check my new answer, if my assumption is correct you just plug in the values a b and c into the equation. a = 4, b = -2, c = -1/2 so just go from there right? I might be missing something. I don't really understand what the # symbol is suppose to do, I don't recognize that as anything in mathmatics, unless you include programming languages.
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  6. #5  
    Forum Radioactive Isotope MagiMaster's Avatar
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    Let's see, so , and . Then,
    =
    =
    =


    The # is just made up for the problem according to the defintion given, unless I misread.

    (Edit: Oops. Messed my math up. Fixed, now I think.)
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  7. #6  
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    28 is correct. That is how you do it GenerationE...I don't know why I didn't get it before. ok, next one...

    How many base 10 four-digit numbers, , satisfy all three of the following conditions?

    i.
    ii. is a multiple of
    iii.

    (how do I make my 'less than' on the same level as the other one?)
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  8. #7  
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    That's just a simple counting argument.

    If or equivalently then there are two choices for a.

    If N is divisible by 5, then, being base 10, that restricts d to be either 0 or 5.

    Then iii there says there are only so many choices for b and c.

    Since these are all independent, you can just multiply the totals for each part together to get the overall total.
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    The reason you got, the wrong answer is because you substituted the second term incorrectly.

    If I'm reading the question right. Then I believe that the answer is 0. Do you mean by

    that if

    that

    If so the answer is obviously 0.
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  10. #9  
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    No, by , I'm pretty sure it means that a, b, c and d are the base-10 digits of N. So technically, .

    Edit: Oh. I think I read your post wrong. It looks like some commas went missing. Still, the answer shouldn't be 0, since there are numbers than can satisfy all three parts (3230 being the smallest, I think), well, assuming that I interpreted part i right (it looks like a typo).
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  11. #10  
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    Quote Originally Posted by MagiMaster
    No, by , I'm pretty sure it means that a, b, c and d are the base-10 digits of N. So technically, .

    Edit: Oh. I think I read your post wrong. It looks like some commas went missing. Still, the answer shouldn't be 0, since there are numbers than can satisfy all three parts.
    You're correct in the pre-edit part. The answer is not 0.
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    That's pretty easy like Magi said.

    the answer is just the multiplication of all the choices for each variable




    Wow, I'm full of mistakes

    if

    then it is actually

    2 * 3 * 3 * 2

    if however you meant



    then my answer is correct.
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    Quote Originally Posted by Chemboy
    Quote Originally Posted by MagiMaster
    No, by , I'm pretty sure it means that a, b, c and d are the base-10 digits of N. So technically, .

    Edit: Oh. I think I read your post wrong. It looks like some commas went missing. Still, the answer shouldn't be 0, since there are numbers than can satisfy all three parts.
    You're correct in the pre-edit part. The answer is not 0.
    The reason that I assumed 0 was because I believed, if was equal to the first number in the term, that it was impossible for a one digit number to be between 3000 and 5000.

    It was a comprehension mistake on my part. I didn't understand the wording without explanation, Magi helped me out though.
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  14. #13  
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    Quote Originally Posted by GenerationE
    if

    then it is actually

    2 * 3 * 3 * 2
    Not what I have.
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    Since you haven't posted a solution I will just ask a question.

    It's not hard, I just feel like asking, it follows along the same lines as the previous question.

    cities A, B, C, D are connected by various numbers running in between them.

    You cannot get from city A to city C without passing through B, and likewise.

    There are 147 road in between A and B

    There are 93 roads in between B and C

    There are 561 roads in between C and D

    How many unique paths are possible between (EDIT: A and D)?
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    Quote Originally Posted by Chemboy
    Quote Originally Posted by GenerationE
    if

    then it is actually

    2 * 3 * 3 * 2
    Not what I have.
    Really? neither 36 or 64? well I don't know then. Are you sure all of the less than signs are posted right in the question?
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  17. #16  
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    I checked, everything's correct. Is your problem simply a matter of multiplying the number of roads between each city, giving you ?
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  18. #17  
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    Quote Originally Posted by Chemboy
    I checked, everything's correct. Is your problem simply a matter of multiplying the number of roads between each city, giving you ?
    yes, which is why the number of possibilities in your problem multiplied together should be the right answer as well. I just can't figure out where I went wrong with the logic. A can only be 2 possible values; 3 and 4. B can only be three possible values; 2, 3, and 4. C as well can only be 3 possible values; 3, 4 and 5. and obviously D can only be 2 values; 0 and 5.

    Multiply all that out and you get 36.

    By the way my problem is just a display of the "fundamental principle for counting." Many people are tempted to add when counting things like paths between cities or number of outcomes after rolling a die 7 times.
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    I just figured out my error. but now I'm not sure how to calculate the final answer

    B and C are not independent, In fact they are mutual dependent. That is why I messed up.

    the only possible values for bc are 23, 24, 25, 34, 35, and 45.

    So maybe, 2 x (3+3) x 2 = 24

    Is this correct? I am killing myself over here.
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  20. #19  
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    Quote Originally Posted by GenerationE
    By the way my problem is just a display of the "fundamental principle for counting." Many people are tempted to add when counting things like paths between cities or number of outcomes after rolling a die 7 times.
    I never would have thought to add them so I guess I won't be tricked that way. haha.

    Well, I just remembered that I heard there was a controversy with one of the problem when it was being graded...perhaps this is that problem. The answer they give is 24, however it's multiple choice (I just wasn't giving the choices) and another choice is 36. And I agree with your logic, and that's how I did it. I say we conclude that 36 is correct and that the answer key is screwed up.

    Let f be a function whose graph is a line with properties such that , and . Which of the following statements is true?

    A.
    B.
    C.
    D.

    I realize this are most likely very easy for some of you but I think they're fun anyway and we can all learn a little by your processes for solving them.

    EDIT: ok so maybe with what you've now discovered 24 really is the right answer. I have no idea personally. Let's see if one of those superior mathematicians in this forum can solve this for us and move on with this new problem in the mean time.
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    Yeah, I figured out that variables b and c were dependent. It was really simple after that. Notice I got the right answer before you gave it out though, so the odds of me guessing the right answer are very slim. Its interesting how it is a factor of (2/3)'s less than what I had predicted of 36. just like 6 is (2/3)'s of 9 which is the product of 3. 6 is the sum of 3's.

    Numbers are crazy like that. I'm almost 100% positive 24 is the right answer it has to be.

    I know I can do this next one so I'm going to go draw it out right now.
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    I can't justify my reasoning completely but I believe the answer is D.

    f(0) must be 11 if you want your problem to maintain symmetry about x = 2.5

    It seems to me that the function is probably a parabola with a vertex that has a height f(2.5).

    This is clearly a guess, I don't have enough information to formulate anything. It didn't specify anything about concavity or second derivatives did it? If it gave you some inflection points or something then I think we could learn about the problem a little more.

    Edit: haha, I did it again. I think if it was a parabola then c would be true as well. So I think we can actually throw out the parabola shape altogether.

    If this is a correct assumption then A and B must be somehow related in a form that we can rule one of them out.
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  23. #22  
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    The answer is D. Come up with a graph that would be true for all of the conditions.
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  24. #23  
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    Oh. The answer is D. If it's a line, then the only way to satisfy all of the givens is for it to be... I won't spoil the answer yet. (Except that I think GenerationE read it while I was editing the answer out. :P)

    As for the other one, the answer is 24. There's 2 choices for a, 2 choices for b, and 1+2+3 choices for b and c together. b and c aren't independent of each other, but they are independent of a and d, so you can multiply a * (b and c) * d for 2*6*2 = 24.
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    Well darn, I should have stuck with my initial answer then. I'm kind of 3 for 3 right now. Only after you told me my second answer was wrong. So I guess I'm really like 2 for 3.

    So any more Questions?
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    Quote Originally Posted by MagiMaster
    Oh. The answer is D. If it's a line, then the only way to satisfy all of the givens is for it to be... I won't spoil the answer yet. (Except that I think GenerationE read it while I was editing the answer out. :P)

    As for the other one, the answer is 24. There's 2 choices for a, 2 choices for b, and 1+2+3 choices for b and c together. b and c aren't independent of each other, but they are independent of a and d, so you can multiply a * (b and c) * d for 2*6*2 = 24.
    Ha, I didn't even consider the definition of a line. I assumed the function could take any shape.

    It took me about 25 minutes of frustration before I noticed that variables B and C were dependent. I figured it out though.

    These are getting increasingly trickier. The first one was straight forward. The second one was if you noticed that the variables were dependent. And you had to read the third problem slow to catch the key word "line." I'm pretty enthused for another one. Or the solution to the part B on the forum game. I have a pretty long expression for the probability but I can't get n to come out when setting the expression equal to .5
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  27. #26  
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    Quote Originally Posted by GenerationE
    These are getting increasingly trickier.
    That they are.

    There exist positive integers , , and , with no common factor greater than ,
    such that . What is ?
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    Other than looking at the answer choices and ruling out obvious things like, having two even numbers in the set (A, B, C) or any two that have the same factors, I don't know how to do it.

    I reduced the expression to



    I don't know if that really helped any, I think the best approach would be to somehow write two formulas based on the given that A, B and C share no factors other than one. Obviously we need a way to formulate that statement.

    Like



    or something like this.
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  29. #28  
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    Well, your guess is as good as mine. I have 4 choices...if I must I can give them to you so you can work from them.
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  30. #29  
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    Not a complete solution, but progress, I think.

    Since

    we easily establish that:.....

    and therefore:.....

    Which implies that:.....

    When, by inspection, I get A = 2, B = 3, C = 1
    but I can't actually show that this is the only solution.
    Everything the laws of the universe do not prohibit must finally happen.
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  31. #30  
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    Well, you're correct. . And given the "with no common factor greater than 1," condition, it seems to me that what you got is the only solution. Next one...

    A list of five positive integers has mean 12 and range 18. The mode and the median are both 8. How many different values are there for the second largest element of the set?
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    I thought that 1 2 3 might be a possible solution since they are the smallest numbers that satisfy the condition of no common factor other than one.

    My guess for this second one is going to be 1

    The largest value has to be 21 and the mode and median are smaller than the mean. this tells me all the values right of 12, the mean, have to be rather close to the max.

    I don't know how to prove this one but you have at least 2 values that are 8 and one that is 21 and one that is 3.

    It has to be one. and its value is easily found to be 20.
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    simplifies to



    which equals

    this is the only possible value.
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  34. #33  
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    Not the answer I have.
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  35. #34  
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    Quote Originally Posted by Chemboy
    Not the answer I have.
    The only other posiblities is that there are 6 values. But then that would be that the range is not symmetric about the mean.
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  36. #35  
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    The answer is 6. Is there some reason the range can't be asymmetric about the mean? I'll go on though... I think this one is pretty cool...not that I got it right...

    Four distinct points, A, B, C, and D, are selected from 2008 points equally spaced around the circumference of a circle. All quadruples are equally likely to be chosen. What is the probability that chord AB interesects chord CD?
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  37. #36  
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    I define the distance between the points A, and B as being the number of points between them, counting clockwise round the circle, including both A, and B. So when A and B are next to each other the distance (A,B) = 2

    The probability P that any particular cord (C,D) does not intersect the cord (A,B) is given as the ratio (number of non-intersecting cords) / (number of possible cords).

    Therefore, when ....

    and when: ....

    So that in general, when: ....

    The overall probability P that no cord (C, D) intersects a cord (A, B) is then the product of these individual probabilities.

    So the answer is: ....

    Or at least, I think it is...
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  38. #37  
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    Quote Originally Posted by Chemboy
    Four distinct points, A, B, C, and D, are selected from 2008 points equally spaced around the circumference of a circle. All quadruples are equally likely to be chosen. What is the probability that chord AB interesects chord CD?
    I think itís
    but I might be wrong. Probability is one of my weaker areas of mathematics.
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    Quote Originally Posted by JaneBennet
    Quote Originally Posted by Chemboy
    Four distinct points, A, B, C, and D, are selected from 2008 points equally spaced around the circumference of a circle. All quadruples are equally likely to be chosen. What is the probability that chord AB interesects chord CD?
    I think itís
    but I might be wrong. Probability is one of my weaker areas of mathematics.
    Mine too. Lets get the answer for this and then move onto the next one. These questions were pretty fun to me. Besides this one, I can't think of a neat way to do it, It looks like Jane used a combination or permutation formula. I guess we don't have a factorial symbol.
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    Well, I can explain how I got the answer I gave, if you like.

    The denominator is, of course, the number of ways of choosing the quadruple ABCD. Point A can be any of the 2008 points Ė it doesnít matter which. Then there are 2007 choices for B, and after that 2006 choices for C, and after that 2005 choices for D.

    For the chords to intersect, we must have point C or D (but not both) between A and B in the clockwise sense from A to B. Suppose there are n points between A and B (not including A and B) in the clockwise sense from A to B. Then C can be any of those n points, whereupon D can be any of the 2006−n points from A to B in the anticlockwise sense. So there are ways of choosing chord CD given this set up.

    Since n can be any integer from 1 to 2005 inclusive, the number of ways of choosing chord CD with point C between A and B clockwise from A to B is But then we could have point D instead of C between A and B clockwise from A to B. This will give the same answer as before (you just swap C and D).

    Hence the total number of ways of choosing ABCD such that the chords intersect is At any rate, this would be if I havenít missed anything.
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  41. #40  
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    The answer I have to the problem is a simple fraction. What's your sum work out to? Here's the next problem though... (hint: it's kind of a two-part answer)

    If sixteen teams are in a double elimination tournament (that is, a team has to lose twice before it is eliminated), what is the possible number of games needed to determine a winner?
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  42. #41  
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    Quote Originally Posted by Chemboy
    The answer I have to the problem is a simple fraction.
    Very well.











    And



    Hence



    Happy?
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    in fact, the answer is no matter how many equally spaced points there are on the circle.
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    Yep, I am happy.
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  45. #44  
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    Quote Originally Posted by Chemboy
    Four distinct points, A, B, C, and D, are selected from 2008 points equally spaced around the circumference of a circle. All quadruples are equally likely to be chosen. What is the probability that chord AB interesects chord CD?
    I believe I have discovered a much more intuitive way of looking at the problem. Consider the point D. It must lie in one of three places: (i) on the arc AB, (ii) on the arc BC, (iii) on the arc CA. By symmetry, the probability that it lies in any of these three places must be the same. Hence the probability that D lies on the arc AB is

    I wonder why I didnít see that before.
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  46. #45  
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    That's kind of how I did it, but not as simply... I looked at all the possible orders A, B, C, and D could be in and looked at how many included C or D between A and B. Basically the same except I worked out all the combinations. The next question has been posted...
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  47. #46  
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    The current question is one of the easiest, no one wants to try? It's in the Nov 11 post.
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    Quote Originally Posted by Chemboy
    The current question is one of the easiest, no one wants to try? It's in the Nov 11 post.
    I have 31 games for the max. I never did any calculation for the minimum. I'm not sure how to really do this with out charting it out. That almost seems like cheating since this is a mathematics forum.

    I just said 8 + 8 that is of the 16 teams in the first round 8 games then 8 more. It is possible to go through those two rounds with out having a single team lose twice. then one more round of 8 now we have 8 teams left who play 4 games. now 2 teams left which play 1 game. That decides a winner in the longest manner I believe.

    So

    I don't know which method seems the best for determining the min.

    I think it is

    Just shots in the dark really. I'm not interested in this problem enough to care. That is why I stopped posting on this thread. I just don't really care for this particular problem, all the others were very interesting though.
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  49. #48  
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    The answer's "30 or 31" so I'll give it to you.

    Eight steel balls, each with a radius of one inch, are melted down and reshaped to form one large ball. What is the radius of the large steel ball?
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  50. #49  
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    Quote Originally Posted by Chemboy
    The answer's "30 or 31" so I'll give it to you.

    Eight steel balls, each with a radius of one inch, are melted down and reshaped to form one large ball. What is the radius of the large steel ball?
    Um so I'm just going to take the indirect approach. Find the total volume and then resolve for r. I know there is another way of doing it but I can't remember it.


    That should be right. I didn't write it out on paper first though so there could be a error.


    Oh, and yeah it is 31 or 30 I needed an extra +1 on my 29 because those teams would have to play twice to determine a winner. That wasn't really that bad after all.
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  51. #50  
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    Nope, the answer is 2 inches. :wink:

    By what percent does the length of time for a trip of a given length change if the average speed is increased by 25%?
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  52. #51  
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    Decreases by 20 percent.
    I demand that my name may or may not be vroomfondel!
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    Quote Originally Posted by Vroomfondel
    Decreases by 20 percent.
    I'm pretty sure that is correct but, I like to see work. I did it using real speeds and velocities and then divided my trip distance by my increased speed. I would like to see a more symbolic approach though.

    travel 60 miles in 60 minutes you go 60 miles per hour. Travel at a velocity 25% greater 75 miles. the trip length is given by 60 miles/ 75 miles per hour. Or 4/5 the amount of time. or 20%.
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  54. #53  
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    20 percent is correct.

    For this one there's a picture. It's simple a circle with another smaller circle centered inside it, with the part between the smaller circle and larger circle filled in.

    Let r be the radius of the smaller circle and let R be the radius of the larger circle. Find an expression for R in terms of r so that the areas of the washer (shaded region) and the hole are equal.
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  55. #54  
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    Quote Originally Posted by Chemboy
    20 percent is correct.

    For this one there's a picture. It's simple a circle with another smaller circle centered inside it, with the part between the smaller circle and larger circle filled in.

    Let r be the radius of the smaller circle and let R be the radius of the larger circle. Find an expression for R in terms of r so that the areas of the washer (shaded region) and the hole are equal.


    I just found it doing the obvious and subracting the middle area to find the outer area setting them both equal to area A and solving the equation for R.

    Note if it wasn't so hard to type on here I could have shown the same thing using integrals in polar coordinates. It would have to be done with double integrals and it really isn't easier but it always looks cooler when you use integrals.
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  56. #55  
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    Correct. Yes, double integrals are a lovely way to find the area of a circle. May be a little more involved but it's certainly a more interesting way to do it.

    Many three digit numerals can be formed with the digits 2, 3, 4, 5, 6, and 7 used only once in each numeral. If these numerals are arranged in order from the largest to the smallest, then what is the 75th numeral?
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  57. #56  
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    Quote Originally Posted by Chemboy
    Correct. Yes, double integrals are a lovely way to find the area of a circle. May be a little more involved but it's certainly a more interesting way to do it.

    Many three digit numerals can be formed with the digits 2, 3, 4, 5, 6, and 7 used only once in each numeral. If these numerals are arranged in order from the largest to the smallest, then what is the 75th numeral?
    First the amount of permutations is 6!/3! or 120

    so this tells us each starting number has 20 different permutations for example 765...764 and so forth. you go through the second cycle with four different choices. that is if 7 is first then its 65...64...63...62.

    This tells us the starting number is 4

    The second number then has to be 3

    and so I'm thinking the last number is 5 just by going through the counting.


    So the 75th largest number is 435.

    Took me a second at first I thought that might be a tough one.
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  58. #57  
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    Correct.

    Find the digit represented by if for some digit .
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    Quote Originally Posted by Chemboy
    Correct.

    Find the digit represented by if for some digit .
    Wow, I'm really confused on how to set something like that up with out just testing different values of the root in a calculator. That is definitely cheating so I'm going to let someone with more experience actually do the math. I'll be waiting till someone does this one so we can move on.
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    The number must be divisible by 9, so the sum of its digits must be divisible by 9. Hence .
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  61. #60  
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    Quote Originally Posted by JaneBennet
    The number must be divisible by 9, so the sum of its digits must be divisible by 9. Hence .
    Why is that?

    Two hundred stones are placed on the ground 3 feet apart, the first being 3 feet from a basket. If the basket and all the stones are in a straight line, how far does a person travel who starts from the basket and brings the stones to it one by one?
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  62. #61  
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    For the stone, you travel feet ( there and back). Then the total is just feet. Unless there is some clever way of moving the stones that does better than this, the straight line is a red herring.
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  63. #62  
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    Quote Originally Posted by MagiMaster
    For the stone, you travel feet ( there and back). Then the total is just feet. Unless there is some clever way of moving the stones that does better than this, the straight line is a red herring.
    That is exactly how I would of done it although I wouldn't of been able to evaluate the summation like you did. Why does that work . Might be a noob question, sorry.

    Its pretty neat though. Is it always going to be? Or is the equation for it something else?
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  64. #63  
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    Those are called the triangular numbers, being the number of dots in an equilateral triangle made of evenly spaced dots. See wikipedia link for more info, but yeah, it'll always work out to .
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    Correct.

    A number a is increased by 25% and then the resulting number is decreased by 20%. Another number c is decreased by 40% and that result is increased by 20%. If these two final results are equal, then a is what percent of c?
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    Quote Originally Posted by Chemboy
    Correct.

    A number a is increased by 25% and then the resulting number is decreased by 20%. Another number c is decreased by 40% and that result is increased by 20%. If these two final results are equal, then a is what percent of c?
    A is 90% of C
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  67. #66  
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    You didn't work out the fraction on the "" side correctly, it's which equals percent. I'll move on though since at least you had the right fraction.

    The graphs of and intersect in how many points?
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  68. #67  
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    Quote Originally Posted by Chemboy
    You didn't work out the fraction on the "" side correctly, it's which equals percent. I'll move on though since at least you had the right fraction.

    The graphs of and intersect in how many points?
    I didn't do any work I'm just guessing that it is infinite intersections.

    the slope of the second equation is just the graph of x with a slope slightly smaller than 1. The first equation is the graph of cosine but the oscillation's height are getting increasingly larger. I think it is always a little bit bigger than the slope of the second at the points where is equal to one, that is and so on.

    So that was just a guess if it's not right don't tell me the answer I will actually graph it and see if I get some intersections where I think they are.

    I don't know how I got lol, that is a classic example of what is known as a brain fart.
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  69. #68  
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    It's not infinite. Try again.
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  70. #69  
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    Quote Originally Posted by Chemboy
    It's not infinite. Try again.
    1

    The slope is larger than one not less than one. And Cosine 0 starts at 1 but you have to multiply it by 0 so they are both at zero when at the origin. I can't believe I keep messing this problem up. Final answer is 1.
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  71. #70  
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    Where the graphs intersect,



    or

    The latter has no solutions because

    Graphically:
    • http://www.mathsisfun.com/graph/function-grapher.php?func1=x*cos(x)&func2=x&xmin=-120&xmax=120&ymin=-80&ymax=80

    The line has a slope just slightly greater than 1 Ė so itís like the line rotated anticlockwise by a teeny weeny bit. Hence it will only intersect the curve just once.
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  72. #71  
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    Yep.

    This one has a picture accompanying it but I think you can manage without it.

    Shown are two semicircles with radii and , respectively. The triangles are inscribed in the semicircles, with one side of each triangle being a diameter of the semicircle in which it is inscribed. In terms of , what is the exact smallest possible area of the shaded region?

    The figure has, well, two semicircles with triangles inscribed in them. They're not right triangles. The shaded region is the region inside the semicircles not including the triangles.
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  73. #72  
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    Actually 2 questions.

    Why do we need exact values if all we are given is r and 2r?

    Why are there 2 semicircles? How do they interact in the picture?
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  74. #73  
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    It will be in terms of r, but it has to be exact in terms of any irrationals in it. The two semicircles are right next to each other, touching at one point. There's a continuous line formed by their straight edges. The smaller semicircle is on the left, the larger on the right.
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  75. #74  
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  76. #75  
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    Yep.

    Only three more.

    How many of the following numbers have at least four different positive even factors?

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  77. #76  
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    Well, as a half-answer, I can say that the 2008! is, more or less, a red herring. If (a + b) has a factor c, and c is a factor of a, then c is a factor of b. Any number between 2 and 2008 is definitely a factor of 2008!, so only the added number is important.
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  78. #77  
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    2008! ends in a zero. I don't know the value but I know that it ends in zero. So just adding numbers up between 4 and 2006 on top of that value, like magi said its only going to be added number that matters. But all of the even added numbers will count toward the total so the answer is greater than 1001 that is for sure.
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  79. #78  
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    Yes, it's greater than 1001...
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  80. #79  
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    I have more similar questions if anyone's interested. Forget about the previous unanswered question since I'm not sure of where the answer is anymore.

    has an inverse function, and . Find .
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  81. #80  
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    Let be a function of three independent variables , , and , such that the value of for any ordered triple is the sum of the three coordinates divided by the
    difference of the last and first coordinate. For this function find the value of

    (These start easy and get harder, by the way.)
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  82. #81  
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    So, , then and if , then and , and if , then and


    okay, which, with the substitution of the point, is which becomes or next please!
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  83. #82  
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    Quote Originally Posted by Arcane_Mathamatition
    No, this is false. Didn't read any further. Review your understanding of inverses of functions; they are not reciprocals!
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  84. #83  
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    Quote Originally Posted by Arcane_Mathamatition
    okay, which, with the substitution of the point, is which becomes or next please!
    Nope. Why do you have a ?
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  85. #84  
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    , let the inverse of be so


    and so
    so
    and

    is that better?
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