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Summing Digits (Posted on 2007-02-26) Difficulty: 3 of 5
Prove that there is a finite number of values of n that satisfy

n = 4s(n) + 3s(s(n)) + 2s(s(s(n))) + 1

where n is a positive integer and s(n) denotes the sum of the digits of n. Also, determine analytically, all values of n that satisfy the equation.

See The Solution Submitted by Dennis    
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Some Thoughts Castaways (spoiler) | Comment 2 of 5 |
By the way, s(n) is the "casting out 9's" function.

n mod 9 = s(n) mod 9 = s(s(n)) mod 9 etc.

So, (4s(n) + 3s(s(n)) + 2s(s(s(n)))) mod 9 = 9n mod 9 = 0

So, n mod 9 must equal 1. 

There are only 21 possible n's to check: 1, 10, 19, ... 181. (Given Charlie's limit).
For all of these 21, s(n) is either 1 or 10, s(s(n)) = 1, and s(s(s(n))) = 1, so n can only be 10 or 46, and both of these work.

  Posted by Steve Herman on 2007-02-26 17:33:40
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