# Listen With Others

## Listener No 4503: Property Management by Smudge

Posted by Steve Tregidgo on 8 Jun 2018

I don’t write blog posts for every Listener; I tend to wait until I have a tip or a solving technique to share. These tips may well be obvious to many, but maybe some solvers will have not thought of them, and posts like this will help them next time.

This ingenious numeric was most easily solved on two grids: the one provided, and my own table mapping grid numbers to rules. I started by marking out a 40×19 grid: one row for each grid number (with extra rows for 10a/10d and 23a/23d), and one column for each rule (I wrote the rule letter and the count at the top of each column). Much of the solve was spent eliminating possibilities in this grid.

To start with, and to serve as an example on a smaller scale, there are only six squares in the range 1-38 (1, 4, 9, 16, 25, 36), three cubes (1, 8, 27), two fourth powers (1, 16) and two fifth powers (1, 32). The only one of those categories where the required count matches the possibilities is the fourth powers, so we have 1 and 16. With the 1 and 16 spoken for, the squares must be 4, 9, 25 and 36, and the fifth power must be 32 (and there’s only one three-digit fifth power, so that gives us our first entry too!). That just leaves the cubes to decide between, and the very next rule takes care of that: the only possibilities are 1^1=1, 2^2=4 and 3^3=27. 1 and 4 are accounted for, 27 is the only remaining option, and so the cube must be 8.

I therefore spent half of my time crossing off boxes in my table where I knew the grid number didn’t match the rule. By the end of that process there would often be the correct number of viable choices in the rule column, or just one choice in the grid-number row, so another grid-number could be matched with a rule (I drew a circle to confirm the match) and more possibilities eliminated. This was a fairly simple process. The other half of the time was spent trying to eliminate rules based on the grid entries. For example, primorials of 2 or more digits must always end in 0 (they have factors of 2 and 5); primes of 2 or more digits are always odd (they cannot have a factor of 2). So any grid entry with a final digit can immediately be eliminated from one of those rules.

No discussion of a property-based numeric is complete without mentioning the single most useful resource for solving them: the Online Encyclopedia of Integer Sequences. Here you will find all squares, cubes, fourth and fifth powers, primes (Mersenne, Fermat or otherwise), triangular and tetrahedral numbers, factorials, primorials, palintiples (rule ‘s’, where the multiplier is not 1)… the list goes on. You can find a sequence by name, or by typing in a few of its members. And from any page, the “links” section usually has a table of several hundred or thousand entries to pore over.

For example, there are fourteen fourth powers with 6 digits, giving a relatively short list of possibilities for 16d. Six of those have zeroes that would become the first digit of a crossing entry in 1a, so that leaves just eight possibilities there. When I have a list that short, I write them all out for further elimination later. My final tip is that squared paper is helpful here: it becomes a lot easier to read down the columns, especially to discover that the Nth column only ever contains particular digits (which can feed into a crossing entry).