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Integers: 1,2,3,4,5,6,7,8,9

  1. Oct 19, 2009 #1
    I saw this somewhere, it looks like fun but i cant seem to answer it

    integers: 1,2,3,4,5,6,7,8,9

    [tex]\frac{a}{bc}+\frac{d}{ef}+\frac{g}{hi}=1[/tex]

    what is a,b,c,d,e,f,g,h,i ?
    pick from the above integers. (ONLY USE EACH OF THE ABOVE INTEGERS ONCE)
    :)

    bc means for example 35 (b=3 and c=5), not multiply.
     
  2. jcsd
  3. Oct 20, 2009 #2
    Re: fractions

    Interesting.























    We can always reorder terms such that b < e < h. Once we do that, it's easy to see that b=1 (because if b>1, the whole sum is necessarily less than 1).

    We can restrict the number of possibilities for a & c further, because, since b=1, d/ef+g/hi is at most 9/24+8/35 ~ 0.603 and therefore a/bc > 0.397. Furthermore, if c is 2, d/ef+g/hi <= 9/35+8/46 ~ 0.431 and a/bc > 0.569. If either a or c is 9, d/ef+g/hi <= 8/24+7/35 ~ 0.533 and a/bc > 0.467. If a and c are 8 and 9, d/ef+g/hi <= 7/24+6/35=0.463 and a/bc > 0.537 (so, a and c can't be 8 and 9, because both 8/19 and 9/18 are less than 0.537).

    c, f, and i can't be 5.

    Any two-digit prime number that occurs in decompositions of denominators must occur in at least two denominators. For example, we can rewrite the statement as (d*hi+g*ef)-ef*hi = -a*ef*hi/bc. Right hand side is integer. If bc is a multiple of 13, either ef or hi (or both) must also be multiples of 13.

    These principles limit the set of possibilities for the first fraction to 16 possibilities:

    7/12 8/12 9/12
    6/13 7/13 8/13 9/13
    6/14 7/14 8/14 9/14
    7/16 8/16 9/16
    8/17 9/17

    At this point, I don't see any clear ways to reduce the set of possibilities further, but we've cut the possibility space enough to start brute-force search for a solution ...
     
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