**From New Scientist #1291, 4th February 1982** [link]

I recently amused my son with a little numerical pastime. I drew a circle and marked 12 equally spaced points around the circumference. By each of the points I wrote a digit: each of the digits from 0 to 9 occurred somewhere (and, of course, there were some repeats).

The exercise consisted of starting at the centre of the circle, drawing a straight line to one of the 12 points on the circumference, drawing another straight line from there to another of the 12 points, a line from there to another of the 12 points, and finally a line from there to another of the 12 points. We would then look at the four-figure number “spelt out” (we never started with a zero). We’d then start again at the centre.

With one such sheet we did three similar exercises. The four-figure numbers spelt out were, respectively:

a product of five consecutive integers;

a number divisible by 7, 11 and 13;

an odd perfect cube.

By coincidence each of the straight lines drawn in obtaining these three numbers was a whole number of inches long.

What was the perfect cube?

[enigma146]

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I found this puzzle more convoluted to solve than I was expecting. A bit of analysis leads to an approach that gives a program that runs in a reasonable time. This Python 3 program runs in 74ms.

Solution:The perfect cube is 4913.You can remove the code from my program that constructs the orbits and you find that there are four candidate triples and each of them use the same perfect cube. But for two of these candidates it is not possible to construct a clock face corresponding to the problem statement. So in order to fully solve the problem it is necessary to show that a clock face can be constructed for at least one of the candidate triples. In fact, two of them are possible, and there are many different ways to construct the clock face for each solution.

As you point out in the comment lines of the program, the digits fall into not just “two sets of six” as the solution published in the magazine said, but two overlapping hexagons. Each 4-digit number can use digits from one set only.

If each of 0 – 9 occurs at least once, at most two are repeated. 7×11×13 = 1001 and all its 4-digit multiples have 00 in the middle, so that is one of the repeated digits. The other falls in the same set.

The product of five consecutive integers can only be 3×4×5×6×7 = 2520 (in which case 2 is also repeated) or 4×5×6×7×8 = 6720. The odd perfect cube cannot be 1331 for that would require a third repeated digit. It must be the cube of an odd number from 13 to 21.

The solution I found, clockwise from the 12 o’clock position, is:

4 6 9 7 1 2 3 0 8 0 5 2.

4913 reads clockwise from 12 o’clock, 6720 clockwise from 1 o’clock, 2002 clockwise from 5 o’clock or anticlockwise from 11 o’clock. No doubt there are other arrangements, not just rotations and reflexions of that.