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Another simple problem

  1. Mar 24, 2003 #1
    Prove this: Among any 6 natural numbers in a row (e.g. 20,21,22,23,24,25) there's at least 2 of them which have no common divisor larger than 1.
  2. jcsd
  3. Mar 24, 2003 #2
    All you need too do in this case is prove that the LCD (Least Common Denominator) of two consecutive integers is 1. It then follows that m consecutive integers have at least m - 1 pairs of integers that have no common divisors greater than 1. But, since I am a Sadistic Mathematician, I shall leave the proof as an exercise for the reader.
  4. Mar 24, 2003 #3
    Good, Ben-CS. Except the L in LCD means largest not least. - Anyone?
  5. Mar 24, 2003 #4
    We're both wrong: It should be GCF, for Greatest Common Factor. (Those *...* fractions are messing with my mind!)
  6. Mar 24, 2003 #5


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    call the numbers x, x+1, x+2...x+5

    Now, either x or x+1 is even, along with two other numbers. Regardless, we can redefine the set that is not divisible by 2 as y,y+2,y+4 with y defined as odd. The other numbers are multiples of 2.

    Only one of these can be divisble by 3. Any multiple of 3 +/- 2 or 4 is not a multiple of 3.

    We might think that if y is a multiple of 4, that y+4 would be as well, but none of them is even, so none of them is a multiple of 4.

    For any higher multiples, if y=nz, where z>4, (n+1)z>y+4, ruling out higher multiples.

  7. Mar 24, 2003 #6
    Correct IMO, Njorl.
  8. Mar 25, 2003 #7
    Another problem:

    consider a set of 11 numbers

    Define the that sets can you do the following:
    Remove 1 arbritary number from the set then divide the set in to 2 subsets of 5 numbers each and make sure that the sum of the numbers of the first subset equals the sum of the numbers of the second subset.
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