 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Going Cyclic With Arithmetic (Posted on 2009-06-08) 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) Comments: ( Back to comment list | You must be logged in to post comments.) 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 Please log in:

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