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.
I notice that the two solutions so far have used abc (base N) to denote P, and cycled alphabetically to get Q as bca and R as cab. If the cycling is done in the opposite sense, so that P is abc, Q is cab and R is bca (base N) then by using the same basic methods, further solutions can be found that are closely related to those already submitted.
Using Steve's notation, N = 4 + 3k (k = 1, 2, 3..) with a = 1,..k in each case, the other base N digits are now given by
b = a + (2N + 1)/3 and c = a + (N + 2)/3
Using Charlie's method of listing, these further solutions look like this...
N a b c P Q R <o:p></o:p>
<o:p> </o:p>
7 1 6 4 95 209 323<o:p></o:p>
10 1 8 5 185 518 851
10 2 9 6 296 629 962<o:p></o:p>
13 1 10 6 305 1037 1769
13 2 11 7 488 1220 1952
13 3 12 8 671 1403 2135<o:p></o:p>
16 1 12 7 455 1820 3185
16 2 13 8 728 2093 3458
16 3 14 9 1001 2366 3731
16 4 15 10 1274 2639 4004<o:p></o:p>
19 1 14 8 635 2921 5207
19 2 15 9 1016 3302 5588
19 3 16 10 1397 3683 5969
19 4 17 11 1778 4064 6350
19 5 18 12 2159 4445 6731<o:p></o:p>
22 1 16 9 845 4394 7943
22 2 17 10 1352 4901 8450
22 3 18 11 1859 5408 8957
22 4 19 12 2366 5915 9464
22 5 20 13 2873 6422 9971
22 6 21 14 3380 6929 10478<o:p></o:p>
25 1 18 10 1085 6293 11501
25 2 19 11 1736 6944 12152
25 3 20 12 2387 7595 12803
25 4 21 13 3038 8246 13454
25 5 22 14 3689 8897 14105
25 6 23 15 4340 9548 14756
25 7 24 16 4991 10199 15407<o:p></o:p>
28 1 20 11 1355 8672 15989
28 2 21 12 2168 9485 16802
28 3 22 13 2981 10298 17615
28 4 23 14 3794 11111 18428
28 5 24 15 4607 11924 19241
28 6 25 16 5420 12737 20054
28 7 26 17 6233 13550 20867
28 8 27 18 7046 14363 21680
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Posted by Harry
on 2009-06-10 00:49:08 |