Pair Share III

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Remember Pair Share? It carries over seamlessly to triples. Let T be the set of all triples (m,n,p) such that mnp = m+n+p. (Digressing: notice that these triples are not sets, since if two numbers are the same, we can't let them fall together; also, they are not ordered triples or 3-vectors, since we don't distinguish between (m,n,p) and (m,p,n), etc. I believe that the technical term for this intermediate kind of entity is "bag.")

(0,0,0) is in there, of course, but so is (1,2,3). Cool! Most pairs (m,n) allow one (m,n,p) in T, but not all. Can you find (m,n) that doesn't allow any (m,n,p)?

Can you generate all and only the integer triples in T? Are there finitely or infinitely many of them?

Comments

  1. shonk9:56 PM

    I believe that the technical term for this intermediate kind of entity is "bag."

    Never heard "bag" used in this context; the term I've usually heard is multiset.

    ReplyDelete
  2. gcallah10:41 PM

    "Bag" is a synonym for "multi-set"; it was the popular term for this entity in programming circles.

    ReplyDelete
  3. Anonymous10:55 AM

    Are m, n, and p defined to be integers here?

    ReplyDelete
  4. Wabulon6:40 PM

    The question of whether the set T is finite or infinite is uninteresting: trivially, there are infinitely many triples having the T-property. Thus you can show a formula for p in terms of m and n that works for almost all pairs (m,n). The pairs for which it doesn't work supply answers to the first question I posed. The question of whether the subset of T comprising triples of integers is finite or infinite is more interesting, and is the subject of my second question.

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