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A few problems for math league

  1. Feb 15, 2004 #1
    Let's see... I couldn't find the answers to these. They look fun- maybe you can find the solutions.

    1.) If x and y are positive integers, and 2003x=2004y, what is the least possible value of x?

    2.) An integral triangle is a triangle with positive side-lengths and a positive area. Such a triangle can have a perimeter as small as 3. What is the next smallest possible perimeter of an integral triange?
     
  2. jcsd
  3. Feb 15, 2004 #2
    Okay, here's a third one. I have to put it in seperate pieces because it keeps sticking together in one line.

    3.) [tex] a^2 - b^2 \neq 0 [/tex]
     
  4. Feb 15, 2004 #3
    [itex](a^2 - b^2)x^2 - (a^2 + b^2)x + ab = 0[/itex]
     
  5. Feb 15, 2004 #4
    What are all real solutions of a for which the sum of the squares of the roots of [tex] x^2 - 8ax + 14a^2 = 0 [/tex] is 25?

    (This is the last piece.)
     
  6. Feb 15, 2004 #5

    matt grime

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    for the first one, as 2003 and 2004 differ by one, they are coprime. Here meaning, any prime factor of 2003 must divide y. you can put the rest together from there.

    second, one find an equation for area using only side length (there is incidentally a misprint in your question, the word integer is missing from some sentence).

    are all those posts to do with the third alone?

    hint. x^2+px+q = (x-r)(x-s)

    so find p and w in terms of r and s.

    note u^2+v^2 = (u+v)^2 - 2uv.

    that should let you solve the last one.
     
  7. Feb 15, 2004 #6
    Well, I'm not sure about the first and second, but I think I understand the third one. It looks like a trinomial squared.
     
  8. Feb 15, 2004 #7

    uart

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    1. The greatest common divisor of two integers that differ by one is always one. This means that there can be no possible cancelling of factors between 2003 and 2004, which should make the only solution pretty obvious.


    2. You haven't even described what is the integer in the integral triangle? Is it the perimeter or the area or something else. ???????????????


    3. What has "b" got to do with anything in this problem? Using only the last equation by itself you can show that a equals plus or minus 5/6
     
  9. Feb 16, 2004 #8

    matt grime

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    nothing to do with trinomials.

    let's derive uart's answer:

    let r and s be the roots of

    x^x-8ax+14a^2 =0

    mutliplying out (x-r)(x-s) we see that

    -r-s = -8a

    rs = 14a^2

    we want r^2+s^2 = 25


    using the identity I gave before

    25 = 64a^2 - 28a^2= 36a^2


    hence a = +/- 5/6

    so, as uart mentioned, where's b gone?
     
  10. Feb 16, 2004 #9
    1. 2003

    2. 5
     
  11. Feb 20, 2004 #10
    1.) If x and y are positive integers, and 2003x=2004y, what is the least possible value of x?

    Wait-

    2003x=2004y
    x=2004y/2003
    x/y=2004/2003

    so x=2004 and y=2003?
    That makes sense...
     
  12. Feb 21, 2004 #11

    matt grime

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    that alone doesn't tell you that 2003,2004 are the least possible. the answer has been given in a form that does tell you that, both by me and by at least one other person.

    example find the least positive x and y such that 12x=18y.

    12/18=y/x does not say y=12 x=18, rememebr you need the minimal solution.

    it's rather obvious there how to find the minimal, so say what's going on in the larger example.
     
  13. Feb 21, 2004 #12
    12x=18y
    y/x=12/18=2/3
    y=2, x=3
    12*3=36=18*2

    2003x=2004y
    y/x=2003/2004=2003/2004 (no common factors)
    y = 2003, x = 2004
    2003*2004=2003*2004

    Seems to amount to the same thing as the "better" answers to me.

    cookiemonster
     
  14. Feb 21, 2004 #13

    matt grime

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    at what point in the answer given by Blue does he cite that 2003 and 2004 have no common factor? That is exactly what the 'better' suggestions were stating, and they were 'better' in the sense that they were correct.
     
  15. Feb 21, 2004 #14
    The last time I left a fraction in unreduced form was grade school.

    The nature of reducing the fraction is to compare the factors of two numbers, which is also the nature of the problem.

    All I'm saying is that you can say that 2003 and 2004 are coprime because they differ by 1 and therefore don't have any common factors, or you can factor both 2003 and 2004 and demonstrate that 2003/2004 is reduced. It's the same thing and so both solutions are perfectly valid, so it's unfair to call BluE's solution inferior.

    Personally, I rather prefer his solution.

    cookiemonster
     
  16. Feb 22, 2004 #15

    matt grime

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    if you prefer reduced fractions, then I'll restate the question: where does blue state that the terms are in reduced form?

    It is the standard 'write more words' thing that you get fed up with saying.

    If Blue had mentioned *any* of those things about lowest terms reduced or coprime then it's fine, but as it stands there is one last step. Just because you've never left a fraction in unreduced form since grade school (which is when?) doesn't mean we assume everyone does this. In particular the answer doesn't demonstrate that Blue now knows why this is the correct answer.
     
  17. Feb 26, 2004 #16
    Alright, I see now that it is in the fraction is reduced to the simplest form.
     
  18. Feb 27, 2004 #17
    BluE -- your math teacher should be giving you the answers plus the explanations along with your test sheet in the days after each Math League (or at least that's what happens at my school)
     
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