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Posers and Puzzles

Posers and Puzzles

  1. 12 Jan '10 22:50
    Prove:

    If two numbers greater than 1 appear in the same row in Pascal's triangle, then they cannot be coprime.
  2. Standard member ua41
    Sharp Edge
    12 Jan '10 23:53
    I'm not as fluent in mathematics, but could one use the Euclidean algorithm to do such a thing?
  3. 13 Jan '10 13:47
    Originally posted by David113
    Prove:

    If two numbers greater than 1 appear in the same row in Pascal's triangle, then they cannot be coprime.
    i think i've seen this before and it was a relatively cute problem... don't remember the path towards solution though. maybe using the definition of the "choose" numbers, and seeing that some factor not cancelled in n!/k!(n-k)! must also necessarily be present in n!/(k+i)!(n-[k+i])! for any number i<(n-k)? something to that effect?
  4. Standard member TheMaster37
    Kupikupopo!
    17 Jan '10 17:12
    Here's my go at it. Might be wrong though :/

    Let A = k!(n-k-1)! and C = GCD(k+1, n-k)

    (n over k) = n! / k!(n-k)! = n! / A(n-k)

    (n over k+1) = n! / (k+1)!(n-k-1)! = n! / (k+1)A


    So the GCD of the two is Cn! / (k+1)(n-k)A = Cn! / (k+1)!(n-k)! = C(n over k) / (k+1)

    Since both of the combination-numbers are not equal to 1 we know that k is not 0, n-1 or n.

    From that we have that k+1 divides N but also is smaller than n. And thus the GCD > 1.