Solution 85: Maximizing the Difference of "Alternating Sums"

  • Published on Mar 28, 2019
  • Without a doubt, D(n) is unbounded. Does that mean that D(n)-D(n+111) is unbounded as well? If it is not, how can its maximum be achieved?
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Comments • 9

  • Sarthak Bansal
    Sarthak Bansal 5 months ago +1

    I noticed that there was no maximum for D(n) - D(n+111) because you could have n ending in a bunch of 9s and thus n+111 would end in a bunch of 0s

  • adandap
    adandap 5 months ago +2

    Looking at the solution here I'm glad I resorted to a program!

    • adandap
      adandap 5 months ago

      Sure thing Matthew. It's in Mathematica language.
      mmax= -1;
      For[n = 1, n mmax, mmax = ans] ];

    • Matthew Hoge
      Matthew Hoge 5 months ago

      adandap I'd love to see the code :D

  • Benjamin Wang
    Benjamin Wang 5 months ago +14

    The proof that only final 5 digits matter is kinda hand-wavy :/

    • LetsSolveMathProblems
      LetsSolveMathProblems  5 months ago +3

      I did gloss over some parts of the argument that only the final five digits matter. In retrospect, I probably appealed too much to intuition instead of providing a clear, rigorous proof. I sincerely apologize if my explanation led to confusion or dissatisfaction. I include below what is perhaps a much better proof.

      Assertion: Given any positive integer n, D(n)-D(n+111) = D(p)-D(p+111), where p consists of the final k digits of n, where k is at most 5.

      Proof: Given any n that has three or more digits (certainly, we are done if n has one or two digits), we will perform the following casework. If the 100's digit of n does not change upon the addition of n by 111 (that is, no rounding occurs in the 100's digit), only the final three digits of n matter, since every digit preceding the 100's digit will be identical in n and n+111.

      Therefore, we focus only on n that has a rounding-up in its 100's digits. We will have a consecutive series of rounding-ups starting with the 100's digit (and proceeding leftward), which must eventually cease at kth digit from the right, where k is 4 or greater. Certainly, all digits to the left of kth digit will be identical in n and n+111 and can be ignored. If k = 4 or 5, we are done. Let k > 5. Now note that 4th, 5th, ... , (k-1)st digits from the right of n must be all 9's to allow for the consecutive rounding-ups. So, the 4th, 5th, ... , (k-1)st digits from the right of (n+111) will all be 0's. As in the video, we can safely disregard an even number of pairs of 9's in n and, correspondingly, 0's in (n+111) and retain the same value of D(n)-D(n+111). This completes the proof.

  • vokuheila
    vokuheila 5 months ago +24

    I am not a fan of this problem at all. The solution is far from beautiful and extremely grindy.

    • adandap
      adandap 5 months ago

      @LetsSolveMathProblems I think the problem might have been (at least for me) that I didn't have enough intuition about the problem to be able to be surprised by the answer. The fact that two unbounded numbers might have a finite difference doesn't seem that surprising. After all, (n+1) - n = 1...

    • LetsSolveMathProblems
      LetsSolveMathProblems  5 months ago +1

      I admit that this problem is very "bashy" and, as far as I am aware, does not yield to any elegant approaches. I had hoped that the surprising final answer of 24 (and the corresponding value of n of 908) would compensate for the long-winded proof.