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 Going Maximum with Geometric (Posted on 2011-10-21)
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 geometric mean of the two neighboring digits.

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

 See The Solution Submitted by K Sengupta No Rating

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 Analysis and solution | Comment 3 of 4 |
There are only 22 trios of strictly descending digits abc such that b < √ac as follows:
976
965
954
953
943
942
932
921
865
854
843
832
821
754
743
732
721
643
632
621
532
521

In addition for any a>b, abb also works (for instance 9766 works) but you cannot keep going down if you use this pattern.  You can extend back up.  So any of the above can be used to make a six digit number.  The largest is 976679.

However any of the above trios could be extended to a fourth number if it can be overlapped with another of the.   The only pair that can be so combined are 953 and 532 which overlap to 9532.

Hence the solution 95322359

 Posted by Jer on 2011-10-21 16:20:27

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