# Listen With Others

## Listener 4295: Codebreaker by Zag — Jaguar Solution and Comments

Posted by Jaguar on 13 June 2014

This is about as tough as numerical puzzles become, ironically precisely because of how short the clues are. There seemed for a while to be very little to go on. And then, after that, I followed a few dead ends as I misunderstood some of the instructions. Still, it is doable and after all the logic isn’t all that obtuse.

One dead end, that I followed for a while and which seems worth mentioning, is that for the first time ever the “no entry or clue answer starts with zero” really seems to matter. The clues at 1ac and 4ac tell us that if there is a zero in the entries at 19ac or 20ac then it will also belong in the answers to 1ac and 4ac and for a while it looked like this would mean one of 1,2,3,5 down starting with a zero, rendering the puzzle unsolvable! But after all there is no “clue answer” at 5 down, so the zero can safely appear there in the pre-encoded clue answer to 4 across. Phew — faith in Listener editors restored. Anyway, on with the worked solution.

I hope that this solution is clear and indeed correct! I may have missed some shortcuts along the way, of course. Feel free to point out any mistakes you might see.

* * * * * * *

We do after all start with 1 across and 20 across. Clues elsewhere, at 3 and 16 down, tell us that the grid entries are both square numbers — and the preamble informs us that they also have six different digits. This turns out to be possible in 40 separate ways (for example, 289 and 361 or 841 and 729 might work). Let us see if we can reduce this.

Indeed we can. The clue for 1 down also informs us that it’s a square root of some three-digit number (at 12dn), and so the clue answer to 1 across (which is also the grid entry to 20 across) must start with 1, 2, or 3 (because 31^2=961 is the largest possible 3-digit square number). This reduces our 40 possible pairs earlier down to 16. Now we can look at the clue to 3 down, though, and see that this is yet another square root — indeed, the square root of 20ac. Having established that 20ac is at most 361, then 3dn must start with 1 (as 19^2=361), and so 1ac’s answer (20ac’s grid entry) must end with a 1. So 20ac is 361 and we have that 1ac is either 289, 529, 729 or 784. This means that the digit 1 will encode to either 9 or 4. Also the clue answer at 1dn must be either 30 or 31, making the grid entry at 12dn start with 9 (900 or 961).

18dn enables us to sort through these, though. The clue is bizarre. “18dn: reverse of 18dn”. This means that supposing the entry to 18dn were 23, then the answer is 32 which must then encode back to 23. So in fact clue 18dn tells us that its digits are a pair of numbers that encode to each other. We’ve established that 1 encodes to 4 or 9, so therefore either 4 or 9 must encode to 1. If the pair is 91 though, we run into trouble with 3dn. This is the square root of 20ac=361 so the answer is also 19, which would also encode to 91, and we have two identical grid entries which isn’t allowed. So 18dn is 41, 1 encodes to 4 and vice versa, and 1ac is 784. A rather colourful working table for the above logic. Possible grid entries for 19ac run along the top, and 1ac on the left. Colour coding as follows: red cells indicate pairs with duplicate digits; orange cells don’t fit with the clue at 3dn and yellow cells don’t fit with 18dn. Only the green cell remains, indicating 1ac = 784 and 19ac = 361.

Encoding so far:

1 -> 4 ; 3 -> 7 ; 4-> 1, 6-> 8.

The only place that a zero can appear in the top or bottom rows is as the last digit of 19ac, so we can also slot this in. The remaining digits to be filled into the top and bottom rows are 2, 5 and 9. Two of these appear in 4ac so that the sum of the digits there is either 7, 11 or 14. In the meantime the answer to 16dn is 28 so we have that 8 encodes to one of 2, 5 or 9. Indeed, so does 0 (giving the second digit in 4ac’s answer) and finally either 2, 5 or 9 will encode to the first digit in 4ac (which is 5, 9 or 2).

We will now look at 17dn for various cases.

i) 4ac = 59 or 95, so that 19ac is 20 and we have the encoding 8->2 and {2,0}->{5,9} in some order. Now 17dn is some two-digit multiple of 14, i.e. one of 14, 28, 42, 56, 70, 84 or 98. But the entry to 17dn ends in 3, and yet all of 2, 4, 6, 8 and 0 have had their encodings fixed. So 14ac cannot be 59 or 95.

ii) 4ac = 25 or 52, so that 8->9 and {9,0}->{5,2}. This time 17dn is a multiple of 7, i.e. one of 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98. We can rule out most of these using the same trick as above to leave that 17dn can only be one of 35, 42, 77.

iia) 17dn answer is 35. Then we have that 5 encodes to 3 and 17dn encodes to 73, but also that the answer to 20ac is a multiple of 73 between 500 and 600 ending in 4 73×8=584 works but requires 8 encoding to 6 as well as to 9. Contradiction.

iib) 17dn answer is 42. Then 2 encodes to 3 and we are looking for a multiple of 13 between 200 and 299 ending in 4. The only possibility is 234 (13×18), but that has 3 encoding to 6 as well as 7. Contradiction.

iic) 17dn answer is 77. Now 7 encodes to 3, twice, and we are seeking a multiple of 33 between 700 and 800 ending in 4. None exists. Contradiction.

iii) This leaves 4ac = 29 or 92, so that 8->5 and {5,0}->{2,9}. In turn 17dn is a multiple of 11, and can be either 22, 77 or 99. We’ve actually already ruled out 77 from the previous step (the same logic applies), leaving 17dn = 22 or 99. In both cases the entry at 17dn is 33.

The relevant multiples of 33 that end in 4 are 33×8=264, 33×18=594, 33×28=924. In the first of these, 6 encodes to 6 to give the 20ac entry which isn’t allowed under the preamble rules. In the second, this would make the answer for 18dn end in 9 which is not an even number as required. The third does work nicely, so we have that 20ac answer is 924, encoding to 361, and we now have the encoding rules:

{0,5}->{9,2}, 1->4, 2->6, 3->7, 4->1, 6->8, 8->5, 9->3. Perforce this means that 7 encodes to 0.

This allows us to fill in the grid entry to 16dn as 65, and also to see that 16ac has the grid entry 603 or 613 and so has the answer 249 or 279. The first is 3×83 (83 is prime) so cannot be a multiple of 5dn (which is 2_ or 9_), but 279 = 93×3. This means that we can set 4ac = 29, 12dn = 900, 5dn = 93 and have the full encoding already of, in order:

0123456789 -> 9467128053

The grid at this stage looks as follows: We can do a bit more still. 6ac is 900 + a square ending in 9, so is 909 or 949. But 909 would make the clue at 7dn begin with 0 so this isn’t right, so 6ac is 949 entered as 313. Also 9dn = 12ac-7dn =9??-1?? must have an answer starting in 8 or 7. But if it starts 7 then the grid entry would be 0??. So 9dn starts with 8, entered as 5.

2dn is 8ac+8dn and the clue answer begins in 6, so that the grid entries must both start 3. Also we have that 13ac is 8ac-8dn = 3?5-3?? = ?7. Only 5-8 “equals” 7 (really 15-8, but this sort of “clock arithmetic paying attention to only the last digit is often very useful), so 13dn is entered as 80 and decodes to 67. Possibilities are 308 and 375, 318 and 385… but this second pair sums to 703 which is too big for the answer we have to 2dn. So 8ac is 375, 8dn is 308, and their sum at 2dn is 683, to be entered as 857. The grid now is: The last steps are probably the trickiest since we have nearly exhausted all clues. However we can return to clue 9dn and note that its clue includes two numbers sharing the same final digit, so that 9dn ends in a 0, entered as 9. Also there is a strange crossing we can exploit. Supposing for example that 12ac were 990, then 9dn becomes 599 which decodes to 800, making 7dn equal to 190. We can run through the various options to see if they work, where for now “working” means just that the answer to 9dn is at least 100 less than the entry at 12ac with matching middle digit.

We find that e.g. if 12ac is 950 then this makes 9dn 880, which would require 7dn to be less than 100 — so 12ac can’t be 950 (or rather 95_). Continuing down this short chain of logic excludes all possibilities for the grid entry at 9dn other than 549, 569, 579, 589, 599. In turn these make 7dn equal (again we don’t yet have the information needed to know the last digit, so for now call it 0) 130, 140, 140, 120, 190.

Note that 7dn is a permutation of 14ac, so that in particular the answer to 7dn contains a 9 that will be encoded as 3. So we can, after all, find that 7d is 123, 143, 193 or 13?. In the first three cases we can look at the consequences for 14ac: it must start 9 and then have a 4 and either 5, 9, 1 in some order, i.e. 14ac is 914 or 941 or 945 or 954 or 901 or 910.

To help sort through all these possibilities we need the help of 11dn. By now we can sum up our current set of numbers in the grid and find that the answer to 11dn (assuming all remaining entries were 9) is not going to exceed 200, so that the first digit of 11dn is 1, entered as 4. This makes 10ac either 24, 34, 44, or 94 according to our work above. An even multiple of this will give the answer to 15dn which we know ends in a 2 (encoding to 6). Thus we seek multiples of 8, 18, 28… only by 24, 34, 44, 94, that will fit with all the rest of the grid entries:

a) 10ac is 94: then 94×8 = 752 is the only possible answer to 15dn. This encodes to 326.But 14ac’s grid entry for 10ac = 94 can only be 910, so these two clash. Contradiction.

b) 10ac is 44: then 44×8 = 352, 44×18 = 792 could both fit at 15dn. These encode to 726 and 036, so the second is obviously no good, while the first clashes with 14ac (which for 10ac = 44 can only be 914 or 941). Contradiction.

c) 10ac is 24: then 24×8 = 192, 24×18 = 432, 24×28 = 672 and 24×38 = 912 all fit. These encode respectively to 436, 176, 806, 342. But 14ac for 10ac = 24 can only be 945 or 954. So we do have one possible fit which is 14ac = 945 and 7dn = 123. However, on summing the digits in the resulting grid we find a total of 149 plus the final missing digit in 11dn, while 11d is 4?5 giving a presumed total of 158. But then we want the missing digit to be 9, but 5 encodes to 2 rather than 9. So 10ac isn’t 24 either.

The only remaining possibility is that 10ac = 34, for the following grid: Multiples of 34 that end in 2 include 272, 612, 592 encoding on entry to 606, 846 and 236 respectively. In all cases this forces 14ac to be 9?4. We can make deductions about the final digit in 7dn and the resulting sum of the digits as follows:

i) 15dn entry is 236. Then 14ac is 924 and so 7dn is 136. The sum of all digits less the missing one is 146. The presumed sum is 1?1, so that we want the missing digit to be 5 but this would require the 5->5 encoding that isn’t allowed.
ii) 15dn entry is 846. Then 14ac is 984 and so 7dn is 135. The sum of all digits less the missing one is now 152, and the presumed sum is 1?1 still or 161, for a missing digit of 9. But 6 does not encode to 9.
iii) 15dn entry is 606. Then 14ac is 964 and so 7dn is 138. The sum of all digits this time is 149, and the presumed sum is 1?1 or 151 for a missing digit of 2. And 2 does decode to 5! So we have our answer and a self-consistent grid. Given the apparent scarcity of information, it’s incredible if not amazing that there is after all enough to get to the end of such a complex logic puzzle. Thanks, Zag, for a tour de force of elegant logic.

### 5 Responses to “Listener 4295: Codebreaker by Zag — Jaguar Solution and Comments”

1. ### Alastair Cuthbertsonsaid

I have a totally different solution which I think works

78452
28521
35834
19067
10250
60734
90361

0123456789 code to 2467138095

I’m confused, but that’s not difficult nowadays!

2. ### Jaguarsaid

I think that’s going to break down at the 7d entry: if 0 is encoding to 2 then the clue answer for 7d begins in 0.

3. Bugger!

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