Solution 57: Remainders and Summations

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  • Published on Sep 12, 2018
  • Let's five-cycle through the double summation.
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Comments • 3

  • Zoom ChipIn
    Zoom ChipIn 9 months ago

    It looks like so complexity

  • adandap
    adandap 9 months ago +1

    I missed the initial posting of the problem, but got it solved ok before I looked at your answer. I had to play around for a while and calculate some numbers explicitly, but finally cottoned on. Perhaps it's just a matter of taste, but I preferred to think of the sum as being over a square array. Then it's not to hard to see that you get (k+1) rows consisting entirely of 4, each of which contribute 4*(5k+3) and k single ones in the first column. So S = 4 (k+1) (5 k + 3) + k
    I do like the problems that aren't calculus (which I think I'm pretty good at) - they remind me how little I actually know. (Last week's combinatorial problem with Pascal's identity completely stumped me.) Please keep it up!

  • Andy Arteaga
    Andy Arteaga 9 months ago +5

    Really nice problem!