# Enigma 1700

Spoiler for New Scientist Enigma 1700:

From a box of counters numbered 1 to 12, Joe asked Penny to select a set of six and place them on the corners of a regular hexagon and then place the remaining six counters on the mid-points of each side, so that the number on each of these counters was the sum of the numbers on the counters on the two adjacent corners. Joe then produced 16 counters and asked Penny to repeat the challenge, but this time selecting a set of eight and placing them, and the remaining eight, similarly on an octagon.

In the first case, Penny found her choice of the set of six counters was rather limited. How many choices did she have?

And how many choices did she have in the case of the octagon?

Edited 31 May 2012 for simplification.

If each side is the sum of the two adjacent corners, then the sum of all sides must be twice the sum of all corners. So the sum of all sides and corners is 3 times the sum of all corners. The sum of 1 through 12 is 78, so the sum of corners is 26.

Neither 1 nor 2 can be a sum of two smaller whole numbers so they must be corners. Neither 11 nor 12 can be added to either of two smaller whole numbers to get two whole numbers less than 13, so they must be sides.

So knowing all that, you can work out pretty quickly that there are only five possible sets of markers for the corners:

Case 1: 1 2 3 4 6 10
Case 2: 1 2 3 4 7 9
Case 3: 1 2 3 5 6 9
Case 4: 1 2 3 5 7 8
Case 5: 1 2 4 5 6 8

Case 1: 10 must be on a corner adjacent to both 1 and 2, so we have 1-11-10-12-2. (Red numbers are corners.) 3 can’t be on a corner adjacent to 1 and must be on a corner adjacent to 6, and 4 can’t be on a corner adjacent to 2 or 6, so 1-11-10-12-2-8-6-9-3-7-4-5.

Case 2: To make 12, 3 and 9 must be on adjacent corners and the next corner after 9 must be 1 or 2. 1-10-9-12-3 or 2-11-9-12-3. In the first case 4 and 7 must be on adjacent corners to make 11, and 3 and 4, 3 and 7, and 1 and 2 cannot be on adjacent corners, leaving us with 1-10-9-12-3-5-2-6-4-11-7-8. In the second, 1 and 2, 1 and 3, 2 and 7, 3 and 4 and 4 and 7 can’t be on adjacent corners, and that forces 1-5-4-6-2-11-9-12-3-10-7-8.

Case 3: Again 3 and 9 must be adjacent corners and the next corner after 9 must be 1 or 2. 1-10-9-12-3 or 2-11-9-12-3. In the first case the other corner adjacent to 1 has to be 6, then the next must be 5 to make 11, and that leaves 2 for the final corner, but 2 cannot be on an adjacent corner to 3, so this doesn’t work. In the second case the remaining corners are 1, 5, 6. But 5 and 6 cannot be adjacent corners, nor can 1 be adjacent to 5, so this also doesn’t work.

Case 4: To make 12, 5 and 7 must be adjacent corners. The other corner adjacent to 5 can only be 1. 1-6-5-12-7. Remaining corners are 2, 3, 8, and 1 and 2, 2 and 3, and 7 and 8 cannot be adjacent. This leaves only 1-6-5-12-7-9-2-10-8-11-3-4.

Case 5: To make 3, 1 and 2 must be adjacent corners. To make 12, 4 and 8 must be adjacent corners. To make 11, 5 and 6 must be adjacent. 1 and 4, 1 and 5, 2 and 4, and 2 and 6 cannot be adjacent corners. This gives two possibilities, 1-3-2-10-8-12-4-9-5-11-6-7 and 1-3-2-7-5-11-6-10-4-12-8-9.

So there are 6 different ways of doing it, not counting reflections or rotations as distinct. I was pretty sure I didn’t miss any… but I checked with a Perl script. I didn’t.

The octagon is easier. As above, the sum of all counters must be 3 times the sum of the corners. But the sum of numbers 1 through 16 is 136, which isn’t divisible by 3. So there are no solutions.