Listener 3944 Dedication by Oyler
Once the initial hurdles were cleared, this was a fairly straightforward numerical puzzle, unlike some we have seen recently, with no base conversions, no powers to speak of and no coding schemes to be figured out. The message to be uncovered at the end was apposite and of homophonic significance to Oyler, but there was no mystery in decoding it. Indeed, for Gregson at least, figuring out the quote helped to shortcut the final steps in solving.
The path to the solution begins at 23A and 24D, which are reverses of each other. Since 29A is a prime, 24D must be odd. Since it is also a 2-digit cube it can only be 27. Then 23A is 72, 37D = 4.24D is 108, and 21D = 23A – 24D = 45.
Since 40A ends in 0, so also must 38D = 40A/3.
29A is a palindrome so it starts with 7. 22D is also a palindrome and must now be 575. Also 18D is a palindrome and so must be 77. Then 7A = 3.18D = 231.
Since 1A > 7D + 27D it must be greater than 231 + 100, so must start at least 3. Therefore 1D, a square, must be one of (36, 49, 64, 81).
Now 38D must be at least 40 since 3*38D = 40A > 100.
Considering possibilities for 1D and 38D we have the following possibilities for 39D (= 1D + 38D) and 5D (=38D * 39D):
1D |
38D |
39D |
5D |
36 |
40 |
76 |
3040 |
49 |
40 |
89 |
3560 |
36 |
50 |
86 |
4300 |
49 |
50 |
99 |
4950 |
36 |
60 |
96 |
5760 |
Now 37A is prime and starts with 1, so must be one of (11, 13, 17, 19). Therefore 12A = 3.37A must be one of (33, 39, 51, 57). In particular it must start 3 or 5.
Therefore 5D can only be one of 3560 or 4300. But the third digit cannot be 0 since 14A is half of 8D, an anagram of 231. Therefore we have 5D = 3560, 1D = 49, 38D = 40, 39D = 89 and 40A = 120.
8D then must be 312 since the last digit of 14A is 6. 14A thus is 156. And 15D = 7A + 8D – 5 = 538.
Now that 12A must be in (51, 57), 37A must be in (17,19). But 18A = 70? and is a multiple of 37A. Therefore 37A = 19, 18A = 703 and 12A = 57.
4A = 38D + 41D > 50, and is prime. Therefore it must be 53, 73, or 83. And 44A is a multiple of 4A, with the second digit 8 and third digit 3 (since 4A ends in 3 and 38D ends in 0). Therefore we must have 4A = 53, 41D = 13 and 44A = 583.
Now many entries fall out by simple substitution:
20A = reverse of 4A = 35;
26A = reverse of 41D = 31;
27D = a multiple of 53 with form 1?9, i.e. 159;
6D = 12A + 26D and ends with 7, so 26D = 30 and 6D = 87;
7D = 40A + 6D = 120 + 87 = 207;
15A = 572? with digits summing to 15, i.e. 5721;
11D = 2*26D = 60;
16D = 23A + 41D = 72 + 13 = 85;
34D = 41D * 41D = 169;
43A = 37A + 11D = 19 + 60 = 79;
13A = 4A + 26D = 53 + 30 = 83;
13D = multiple of 49 with form 8?3, i.e. 833;
16A = 4D + 13D > 833 + 50 = 883. So 17D = 842 (the other possibility was 441);
28A = anagram of 231 but cannot be 312 (already used), so 28A = 132;
10A = 88? + a square, second digit 6, so must be 96?;
1A = 7A + 4D + 27D = 231 + 159 + 4D = 390 + 50 + some x < 10; i.e. 440 + x,
so 2D = 493 (confirmed as 17 * 29);
3D is of form y(x+1)1(x) for some x, y < 10, and ends 3, 7 or 9 because prime and 16A >= 833 + 50. Only x=3 satisfies the digits summing to 13, so 3D = 5413;
4D = 883 – 833 = 50;
1A = 390 + 50 = 440;
Since 36D is even and divides 19D, 19D must be even. But its last digit must be 0 or 5 since 31A is a multiple of 50. Therefore its last digit is 0 and 31A is a multiple of 100.
Then 33D = 31A + 8D + 11D + 38D = 412 plus some multiple of 100. So its last two digits are 12, giving 42 the form ?29. Factoring all possible matches, we find that only 429 has three prime factors. So 42A = 429.
35A = 42A – 31A/2 = 429 – some multiple of 50, so 35A ends 9.
Since 31D is a multiple of 4A, 53 divides 31D which is of form ?91?. 9911 and 2915 are the only matching multiples of 53. But 9911 would give 31A = 900, which would give a negative value for 35A. Therefore 31D = 2915, 35A = 329, and 33D = 612.
Now we have 32A = ?615 with digits summing to 21, so 32A = 9615. Then we must have 28D = 1904 and 9D = 1904 - 87 = 1817.
At this point, with just 6 cells left unfilled, Gregson resorted to the mechanical process of decoding the hidden instruction, which yielded:
READEULERREADEULER??IST?EMA?TEROF?SAL?LAPLACE
Of the four possibilities for 19A = 493 – square (477, 457, 297, 237) only 457 fills in “likely” letters for the first two unknowns, giving:
READ EULER READ EULER HE IST?EMA?TEROF?SAL?LAPLACE
So 19A = 457.
25A must be 82 or 86, giving D or H for the next missing unknown, of which H is likelier, giving 25A = 86 and 38A = 4. Now we also have 19D = 460 and so 36D must be 92. The quote now is:
READ EULER READ EULER HE IS THE MA?TEROF US ALL LAPLACE
29A, as a prime, is one of (727, 757, 787, 797), giving A, E, I or S for the next unknown. Assuming S gives 29A = 797, confirmed by 30D = 9337 being prime.
Filling in the final blank gives the quote:
READ EULER, READ EULER. HE IS THE MASTER OF US ALL - LAPLACE