Are there any consecutive integers which don't share letters when spelled out in English (e.g.One, twO)? If not, can you construct a proof?





  • Firstly consider the early numbers with "irregular" words:

    one, Two, Three, Four, Five, Six, Seven, eight, nine, ten, eleven, twelve.

  • Then the "teens" all share letters, ending with

    ninetEen, twEnty.

  • Then all the twenty-somethings share letters with each other, all the thirty-somethings with each other, etc. The only possible place to get two consecutive numbers not sharing letters is

    a multiple of ten and the one just before it. And even that's not possible because they all contain "ty" in the number. Going from Ninety-nine to a huNdred is also fine, and then to a thousaNd, a millioN, a billioN, trillioN, etc.

Even infiNity will be OK :-P

| improve this answer | |
  • $\begingroup$ You can extend it through zero to the negative integers. $\endgroup$ – Jaap Scherphuis Feb 20 at 16:31
  • $\begingroup$ @Jaap Negative numbers are trivial: they all share the letters of MINUS :-D $\endgroup$ – Rand al'Thor Feb 20 at 16:32
  • 1
    $\begingroup$ Good answer. Your bold letters are missing for the alternate pairs though $\endgroup$ – simonalexander2005 Feb 20 at 18:35
  • $\begingroup$ @simonalexander2005 I didn't want to do everything in bold because then it wouldn't be immediately clear what was paired with what. So I alternated between bold and capital letters to mark out the matching pairs. $\endgroup$ – Rand al'Thor Feb 20 at 18:46
  • $\begingroup$ There's at least one point where the N portion of the proof fails. It's at 100 9's and a googol - but the O's cover for you $\endgroup$ – TCooper Feb 20 at 18:46

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