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Going Cyclic With Arithmetic (Posted on 2009-06-08) Difficulty: 2 of 5
Three 3-digit positive base N integers P, Q and R, each with no leading zeroes and having the restriction P < Q < R, are such that:

  • Q is the arithmetic mean of P and R, and:
  • P, Q and R can be derived from one another by cyclic permutation of digits.
Determine all possible positive integer values of N < 30 for which this is possible.

No Solution Yet Submitted by K Sengupta    
Rating: 3.0000 (1 votes)

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Solution Analytical solution (spoiler) | Comment 2 of 4 |
1) Let p (base N) = abc
       q (base N) = bca
       r (base N) = cab

   where a < b < c < N

2) p + r - 2q = 0,

so  (c + a - 2b)*N*N + (a + b - 2c)*N + (b + c -2a) = 0

3) Since the first two terms are multiples of N, then (b + c - 2a) must be also.
   But a < b < N and a < c < N, so 0 < (b + c - 2a) < 2N
   So (b + c - 2a) = N
   
4) Substituting gives

   (c + a - 2b)*N*N + (a + b - 2c)*N + N = 0
  
   Dividing by N, gives
  
   (c + a - 2b)*N + (a + b - 2c + 1) = 0
  
   Since the first term is a multiple of N, (a + b - 2c + 1) must be also.
   But a < c < N and b < c < N, so 0 > (a + b - 2c + 1) > -2N
   So (a + b - 2c + 1) = -N
  
5) Taking the two equations
     (b + c - 2a) = N
     (a + b - 2c + 1) = -N
  
   and solving for b and c in terms of a and N
  
   Gives:
     b = a + ( N - 1)/3
     c = a + (2N + 1)/3
    
 6) c < N if and only if
   
    a + (2N + 1)/3 < N
    3a + (2N + 1) < 3N
    a < (N - 1)/3
   
 
7) In order for b and c to be integers, N = 1 mod 3

   And in order for a > 0, N must >= 7
  
   So, N can be 4 + 3k where k = 1,2,3, etc.
       and a can take any value between 1 and k 

  Posted by Steve Herman on 2009-06-08 22:33:03
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