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 Armstrong plus one : Armstwo (Posted on 2021-12-27)
An Armstrong number is a positive integer that equals the sum of M-th powers of their digits when the number is M-digit long.
153 is an Armstrong number, since: 13+53+33=153.
Sloane's A005188 has an article on this, in which inter-alia it is mentioned that the sequence of Armstrong numbers terminates at the 88th term.

An Armstwo number is a base ten, M-digit long positive integer which is equal to the sum of M-th powers of one greater than each of the digits.
For example, if we check for 153, we find that:
23+63+43= 288, which is NOT equal to 153.

Determine the smallest Armstwo number.

**** Heartfelt thanks to Larry for inspiring this puzzle.

 See The Solution Submitted by K Sengupta Rating: 5.0000 (1 votes)

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 Proof of finite number of solutions Comment 3 of 3 |

The smallest value of an M-digit number is 10^(M-1). If the sum of the (incremented and exponentiated) digits is to equal the original, then at least one term in the sum must be greater than or equal to the average (or else the total would be too small). The average term is at least 10^(M-1) / M

The terms themselves are of the form (d+1)^M where d is each digit in turn. So there must be at least one term where

(d+1)^M >= 10^(M-1)/M.

Solving for d, at least one d must obey:

d >= (10^(M-1) / M) ^ (1/M) - 1

Simplifying:

d >= 10 * (10M) ^ (-1/M) - 1

Now, there can never be an actual 9 in a solution because if there were, that digit would contribute 10^M which has M+1 digits — one too many.

Consider M = 61. At this point, the expression for the minimum d is just a hair over 8.0 (8.00199…) So there must be a digit that’s greater than 8, but that would require using the forbidden 9, so we’re squeezed on both sides and there is no solution.

Note that the expression for d is strictly increasing in terms of M, since its derivative is always positive. Any M > 61, then, will have an even larger d requirement, so there can’t be any solutions when M >= 61 and the total set of solutions is finite.

That being said, I do believe the three 3-digit solutions mentioned by others are the complete list, so there’s a pretty big gulf between 10^3 and 10^60 to try to bridge.

 Posted by Paul on 2021-12-27 20:26:53

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