Some of Harry Nelson's Puzzles, and Their Answers

Editor's note, added June 9, 2006

One of the regular columns in the Tentacle (see page four for some samples of Tentacle issues) was a set of puzzles supplied by Harry Nelson, who it turns out is one of the world's most prolific puzzle creators. Annually, he invents and sells a charming collection of puzzles often based on geometry, or algebra or logic, and it seems that many of the readers of these interviews would enjoy examining a small selection of his creations.

A word of caution: the three selected for inclusion here may seem easy, but don't be fooled. When you think you have an answer, you can look on Page 2 for the correct answers.

A Sample of Nelson puzzles:
  1. How should 36 be partitioned into positive summands so that their product is maximized? [Example: 36 = 24 + 12; 24 * 12 = 288.]

  2. 100, which equals ten to the power two, can be factored into two factors, 4 and 25, neither of which has any zeroes. What is the SMALLEST power of ten that CANNOT be factored into (exactly) two factors, neither of which has any zeroes?

  3. What is the LARGEST power of ten that CAN be factored into (exactly) two factors, neither of which has any zeroes?

  4. By referring to any standard dictionary, one finds that any given letter, with the exceptions of j,k and z appears in the "spelled out in English" form of one or more of the positive integers. (Zero is not a positive integer.) What is the smallest positive integer containing all 23 possible letters?


Editor's note added June9, 2006 Here in Harry's own words are the solutions to the above. Surprised?
  1. The correct answer is to partition 36 into 13 parts, each of which is equal to 36/13.

  2. Eight

  3. 1033 = 8589934592 * 116415321826934814453125.

  4. One octillion one septillion one quadrillion one billion one million two thousand five hundred sixty eight.


Solutions:
  1. First: All parts should be equal. Proof: Suppose they were not equal, and consider two of them, y and z, with y > z. Then these could be replaced by .5(y+z), which would not change their sum but would increase their product; e.g., .25(y+z)(y+z) = yz+.25(y-z)(y-z) > yz.
    Thus, all components will have the same value, x, and the sum, S, will equal nx for an integer n. Hence the product will be P =xn = x(S/x).

    Taking logs, differentiating with respect to x, and setting the result equal to zero, yields a maximum at x=e. But, since S/e is not an integer, this means that the maximum product cannot be obtained for an integer number of summands. Thus, the best partition will occur for one of the integers nearest to S/e=13.24366. Checking both 13 and 14, we find that 13 gives a larger product, roughly 563208.

  2.                10=2*5,
              100=4*25,
             1000=8*125,
            10000=16*625,
           100000=32*3125,
          1000000=64*15625,
         10000000=128*78125,
    but 58=390625, and if any factor of 108 were to have both a 2 and a 5 among its factors, it would end in 0.

  3. No proof is known for this, but statistical, numerical evidence suggests that it is true. Proof solicited.

  4. No integer smaller than "one octillion" has a spelling that contains a "c"; no integer smaller than one septillion contains a "p"; and similarly quadrillion for "q", billion for "b", and million for "m".

    Checking, by computer program, all legitimate spellings of integers up to 1000000, shows none smaller than 2568 whose spelling contains all the other 18 letters.