Listener Crossword 4177: Solution Notes

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Number Plates by Xanthippe

Puzzle explanation

Each two-letter entry was an abbreviation for a different state of the USA. The other 12 states remained unconverted in the rows as 21 = NC & NM, 07 = WI, 92 = KY, 36 = OH, 72 = ID, 09 = AK & WV, 10 = CA & MA, 35 = OR, 00 = AL. The first clue in row 7 resolved to (2159 − √2500), giving 2109 or 2209 (since square roots can be positive or negative). To complete the list of states, the solver had to choose 22 (NY) in that row instead of 21 (NC or NM).

Puzzle solution process

This is one possible logical path for solving the puzzle. We’ll use lower-case Roman numerals to distinguish the numbers in each row, eg, 3(ii) for the second number in row 3.

2(i) is a 3-digit number equal to 30a, so a < 34. Of the primes in that range, only a = 13 and 2(i) = 390 fit in the grid, as they share a cell. That makes x = 26u and, since r = x / u, r = 26.

7(i) is now 127n − √(2(26u − u)) = 127n − √(50u); for 50u to be a perfect square, u must be 2N2 for some integer N. Since x = 26u has 4 digits, u > 38, so u is one of {50 = 2 × 52, 72 = 2 × 62, 98 = 2 × 72}.

8(i) is 6_ and prime, so it’s one of {61, 67}, which makes t start with 1 or 7. Since t > u, it can only start with 7, so 8(i) = 67 and u is now restricted to {50, 72}.

5(i) = 13nv / 2 + 10, starting with the same digit as n, which is prime. But v = n − 5, so 5(i) = 13n(n − 5) / 2 + 10. Since 13 × 13 × (13 − 5) / 2 + 10 = 686 (not enough digits) and 13 × 29 × (29 − 5) / 2 + 10 = 4534 (too big to start with the same digit as 29), n must be one of {17, 19, 23}. The corresponding values for 5(i) are {1336, 1739, 2701}. Since h is prime, 4(i) starts with one of {1, 3, 7, 9}. 4(i) = N2 / 2, so N must be even and 4(i) is one of {18 = 62 / 2, 32 = 82 / 2, 72 = 122 / 2, 98 = 142 / 2}. That means i (a prime) starts with 2 or 8, so it’s one of {23, 29, 83, 89}. The only value for 5(i) that fits with one of those is 1336, so 5(i) = 1336, n = 17 and v = 12. Then 9(ii) = 22 (palindrome) and 3(i) = 169, which completes e = 96 in the grid.

Since q = t, it starts with 7. Since 7(ii) = 10q, 7(ii) ends with 0, so q = 70, t = 70 and 7(ii) = 700. Since t > u, u = 50, which makes s = 900, x = 1300 and 12(i) = 1700. 8(ii) is now _05_1, so to be palindromic, 8(ii) = 10501. 9(i) is now _0100; to make a digit sum of 10, 9(i) = 90100. Then w is a prime starting with 9, so w = 97. 7(i) is either 2109 = 2159 − 50 or 2209 = 2159 − (−50).

y is a prime starting with 2, so it’s one of {23, 29}. 10(ii) = 10(z + 1), with a middle digit of 2 or 3, and z is a palindrome, so z is one of {22, 88} and 10(ii) is one of {230, 890}. The corresponding values for 11(ii) = 900 − 4z − 10 are {802, 538}, but B is even, so 11(ii) = 802, z = 22 and 10(ii) = 230, making y = 23.

p is _7, so in 6(ii) = (p − 7)2 + 7 the square term must end in 00; thus 6(ii) ends in 07 and it first 2 digits are the square of the first digit of p (which is the same as the second digit of 6(ii)). That restricts p to 57 or 67 and 6(ii) to 2507 or 3607. But j is even, so 6(ii) = 2507 and p = 57. j is now _62, so c = j + 10 is _72. But 1(i) is a palindrome, so it’s 1__1, which makes c = 172 and j = 162. 3(ii) is now b − 83, where b is _09_ and 3(ii) is 2___, so b is in the range 2091 to 2099 and 3(ii) is in the range 2008 to 2016. 1(i) is 1_21 and palindromic, so 1(i) = 1221. f = h + i where f is _0 and i is _3, so h ends in 7, giving h = 17. That makes 4(i) = 72, i = 23, f = 40 and d = 40.

2(ii) is 740_ and is equal to 2o, where o is 3_01. Thus, 2(ii) = 7402 and o = 3701. Since g is a prime starting with 2, it’s either 23 or 29, so 3(ii) ends with 3 or 9, so b ends with 6 or 2. The digit sum of 2096 is 17, which isn’t less than h = 17, so b = 2092, 3(ii) = 2009 and g = 29.

6(i) is 7_7 and is equal to 3d + n + A − 10 = A + 127, so A must start with 5 or 6. For 10(i) = 723_ to have a prime digit sum, it can’t be 7236, so 10(i) = 7235, A is one of {580, 590} and 6(i) is one of {707, 717}.

The sum of the 2 digits of 4(ii) is k, so k starts with 1, making 4(ii) = 211. m = 100 − k, so m starts with 8. The digit sum of m is the same as the digit sum of e, namely 15, so m = 87 and k = 13. 5(ii) is now 3_7 but its digit sum is 10, so 5(ii) = 307. To have a digit sum of 13, 4(iii) = 58. 4(ii) = 211 = ((87 − 1) / 2 + A) / 3, so A = 590, which makes 6(i) = 717.

11(i) is _209; to have a prime digit sum, it must be one of {2209, 6209, 8209}. C starts with the same digit and is a prime ending in 1, so C = 61, which makes 1(ii) = 642, and 11(i) = 6209. Finally, 12(ii) = 233 and the grid is complete apart from the ambiguous second digit of 7(i).

Using the definitions supplied, the 3-letter encodings can be deduced as SPY = eye, FAN = cool, LAV = john, CRU = vintage, OWT = anything, RAI = music (African), ILL = sick, GAM = leg, ADD = join, NEW = fresh, JAY = bird, DZO = cross. The resulting 2-letter entries can be seen to be distinct US state abbreviations, and the digits in the first 2 columns can be converted so that all of their 2-letter entries are also distinct states. In row 7, the state could be NC or NM (from 21) or NY (from 22). The two occurrences of 21 in rows 1 and 4 account for both NC and NM, so row 7 must contain NY.

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