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Going Maximum With Harmonic (Posted on 2013-12-28) Difficulty: 3 of 5
Determine the maximum value of a (base ten) positive integer N (with non leading zeroes) such that each of the digits of N, with the exception of the first digit and the last digit, is less than the harmonic mean of the two neighboring digits.

*** For an extra challenge, solve this puzzle without the aid of a computer program.

No Solution Yet Submitted by K Sengupta    
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Solution By hand Comment 2 of 2 |
It seems the number should be a palindrome with either an even or odd number of digits.  The center digit or digits will be the smallest with the largest (probably 9's) at the beginning and end.

Assuming odd number of digits:
center 1:
212, end
313, end
etc...  [it turns out a 1 doesn't get you far because the h.m. of 1 and a number is always less than 2]

center 2:
323, 73237 or 83238 or 93239
424, end

center 3:
434, 74347 or 84348 or 94349
535, end

center 4:
545, 75457 or 85458 or 95459
646, end

center 5:
656, 86568 or 96569
757, end

center 6:
767, 97679
868, end

center 7: 878 or 979
center 8: 989

It turns out we cannot exceed 5 digits assuming an odd number.

For an even number of digits you can just add an extra copy of the center digit since x is less than the HM of x and x+1.

So extend 97679 to 976679 for the solution.

  Posted by Jer on 2013-12-28 23:54:34
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