## REJOB by Botox

### Puzzle explanation

The letters in each clue spelt out communications code words (ALPHA, BRAVO, …, ZULU), each word appearing exactly twice. In the second grid, the digits that are converted to letters correspond to the last digit of the alphabetic position of the letter, ie, 1 = A/K/U, 2 = B/L/V and so on to 0 = J/T. In this cipher, both HOTEL and the title REJOB correspond to 85052. The resulting grid words are parts of a hotel in appropriate positions: RECEPTION, LIFT, STAIRS (twice), PENTHOUSE. The remaining digits represent room numbers on four floors, numbered sequentially on each floor, with room 13 missing for superstitious reasons.

### Puzzle solution process

This is one possible solution path.

From 21dn we can get T/E + L = TE − L, so T must be a multiple of E, and 2L = TE − T/E = (E2 − 1)T/E. The least T/E can be is 2, so E2 − 1 ≤ L ≤ 26, which makes E ≤ 5 (and E ≥ 2, otherwise E2 − 1 would be 0).

From 11ac we have D(E + L + T) = D + (E + L)T, so D = T(E + L)/(T + E + L − 1). If E is 5 then T is one of {10, 15, 20, 25} but 15 would give L = (52 − 1)15/(2×5) = 36, too big, so T could only be 10, giving L = (52 − 1)10/(2×5) = 24. That would make D = 10(5 + 24)/(10 + 5 + 24 − 1) ≈ 7.6, not an integer; so we can rule out 5 for E. If E is 4 then T is one of {8, 12, 16, 20, 24} but 12 would give L = (42 − 1)12/(2×4) = 22.5 (not an integer) and 16 would give L = (42 − 1)16/(2×4) = 30 (too big), so T could only be 8, giving L = (42 − 1)8/(2×4) = 15. That would make D = 8(4 + 15)/(8 + 4 + 15 − 1) ≈ 5.8, not an integer; so we can rule out 4. For 3, the valid combinations for (E, T, L) are (3, 6, 8), (3, 9, 12), (3, 12, 16), (3, 15, 20), (3, 18, 24), but none of those gives an integer for D. For 2, the valid combinations for (E, T, L) are (2, 4, 3), (2, 8, 6), (2, 12, 9), (2, 16, 12), (2, 20, 15), (2, 24, 18), of which the only combination giving an integer for D is E = 2, T = 12, L = 9, D = 6.

Now 21dn is HO − 15 with three digits so HO is in the range [115, 1014] and the least H or O can be is 115/26 ≈ 4.4, ie they’re both in the range [5, 26]. 13ac is (2 − C + H)O with three digits; we can’t have 2 − C + H = 2 because then H and C would have the same value, so 2 − C + H ≥ 3. O can’t be 6 (already taken); if O is 7 then 37 = 2187 is too big; therefore O = 5, with 2 − C + H = 3 (as 45 = 1024 is too big) and 13ac = 243. H ≥ 115/5 = 23, so H is one of {23, 24, 25, 26}, C is one of {22, 23, 24, 25} and 21dn is one of {100, 105, 110, 115} respectively. Now we know 4ac and 25dn are multiples of 5, so both end in 5 (not 0 because 6dn and 33ac can’t start with 0).

In the grid we have 6dn = 54 = Z + 10U, so Z ends in 4. It can’t be 4 because that would make U = (54 − 4)/10 = 5 (already taken). For 1dn we have (Z + U)(9 − U) with two digits; if Z is 24 then U = (54 − 24)/10 = 3 and 1dn would be (24 + 3)(9 − 3) = 162, too big; therefore Z = 14, U = 4 and 1dn = 90.

For 11ac we have 138A with three digits, so A ≤ 999/138 ≈ 7.2 and the only values available are {1, 3, 7}. It can’t be 1 because then 27ac would be 2P and couldn’t have three digits. If A is 7 then 1ac would be ≥ (63 + P)23 + 7 = 1456 + 23P, too big; therefore A = 3, making 11ac = 414. 27ac is then 12P with three digits and can’t be 12×9 = 108 because 28dn can’t start with 0, so P ≥ 10. 29dn is (P − 3)2 with two digits, so P ≤ 12, but 12 is already taken, so P is one of {10, 11}.

For 4dn we have M + I + K/2, so K is even. For 1ac we have K + 45I and 9__ in the grid, so I is between (911 − 26)/45 ≈ 19.7 and (999 − 2)/45 ≈ 22.2, ie one of {20, 21, 22}. For 20dn we have IN − 6 − I/3, so I is a multiple of 3, ie I = 21. Then for 4dn we have K + 24 and _4 in the grid, so K ends in 0 and is one of {10, 20}. If K is 10 then for 1ac we’d have 10 + 945 = 955, P would have to be 11, and we’d have 955 = 38H + 3, but H = (955 − 3)/38 ≈ 25.1 isn’t an integer. So K = 20, 1ac = 965 and H = 962/(27 + P); 962/(27 + 11) ≈ 25.3 isn’t an integer, so P = 10 and H = 26, which completes 25ac = 249, 27ac = 120, 21dn = 115 and 29dn = 49, and C = 25.

We have 4dn = 44 (from K − I + LO), so also 44 = M + I + K/E = M + 31, therefore M = 13, which gives 16ac = 93, 22ac = 26 and 23dn = 606. From the grid, 12dn = 21N − 9 ends in 2, so N ends in 1; it can’t be 1 because 21×1 − 9 = 12 is too short for the three-digit entry, so N = 11, 12dn = 222 and 20dn = 218.

For 31ac = 65R + 10 we have _85 in the grid, so R is odd and bounded by (185 − 10)/65 ≈ 2.7 and (985 − 10)/65 = 15. The only available values are {7, 15} but 65×7 + 10 = 465, so R = 15, which gives 31ac = 985 and 15ac = 32 and also completes 8ac = 959 and 18ac = 3341.

For 8dn = 41S we have 9_3 in the grid, so S ends in 3; the only available value is S = 23, giving 8dn = 943, 32ac = 5609 and 25dn = 215. 24ac is (9/11)G + 5, so G is a multiple of 11, the only available value being G = 22, which gives 24ac = 23 and 33ac = 549. The grid has 25 for 28dn = F + 8, so F = 17, making 7dn = 1333.

For 30dn = 17X − 47 we have _9 in the grid, so X ends in 8. But 17×18 − 47 = 259 is too big, so X = 8, giving 30dn = 89 and 14dn = 1989.

For 16ac = 117 − Y we have 93 in the grid, so Y = 24, giving 5dn = 829. For 3dn = 26W + 485 we have 51_ in the grid; 26×2 + 485 = 537 is too big, so W = 1, giving 3dn = 511 and 19ac = 9212.

For 26dn = 4J + 390 we have 4_4 in the grid, so J ends in 1 or 6. The only available value is J = 16, giving 26dn = 454 and 10dn = 9019. For 2dn = 21V + 265 we have 64_ in the grid, so V is between (640 − 265)/21 ≈ 17.9 and (649 − 265)/21 ≈ 18.3, ie V = 18, giving 2dn = 643 and 9dn = 544.

For 14ac = 15B + 26 we have 131 in the grid, so B = 7 and the only remaining value is Q = 19. We can complete the grid with 4ac = 4085, 17dn = 321 and 15dn = 3213.

We’re told that Grid B contains no zeros, so its five zeros must all be converted to letters as part of the five entries that are words. As the words are symmetrically placed and there are an odd number of them, one word must include the two middle cells in the central column, and very likely the 0 below them and the cell above (for symmetry). The other four zeros are all in the perimeter, and as the word entries occupy 66 − 32 = 34 cells in total, it’s a good bet that they take up the whole perimeter and the middle column, leaving two 4×4 blocks to be kept as digits.

The digits therein show some kind of pattern, with four columns consisting of 4321 and the others having 1111, 2222, 3334 and 4445. There are 21 − 5 = 16 numeric entries, so unless some of them occupy only single cells, they must all be two-digit numbers. Dividing the blocks into horizontal pairs, we can see four rows of (41, 42, 43, 44), (31, 32, 33, 34), (21, 22, 23, 24), and (11, 12, 14, 15), from which we might recognise a pattern of room numbers in a four-storey HOTEL — one of the thematic code words used in the clues. If HOTEL and the title REJOB match the same sequence of digits, then the cipher has equivalences H ≡ R, O ≡ E, T ≡ J and L ≡ B; comparing their alphabetic positions, 8 ≡ 18, 15 ≡ 5, 20 ≡ 10 and 12 ≡ 2, it seems clear that the cipher matches each digit to the letters whose positions end in that digit.

On that basis, 901989 in the first column (and the last) would match [IS][JT][AKU][IS][HR][IS], which yields only STAIRS. (If the entry excludes one or more of the corner cells, the possibilities are TASH, TUSH, STAIR and STASH, which are less likely as thematically-placed items.) The top row, excluding the corners used by STAIRS at either end, should match [FPZ][EOY][DNX][JT][HR][EOY][AKU][IS][EOY], which gives only PENTHOUSE; similarly, the bottom row has [HR][EOY][CMW][EOY][FPZ][JT][IS][EOY][DNX] → RECEPTION. Finally, the middle column has [BLV][IS][FPZ][JT] → LIFT, completing the hotel diagram.