Movements by Mango

Puzzle explanation

The grid contains the alphametic SPRING + SUMMER + AUTUMN + WINTER = PRIEST in base 12, using GSWETAUMIRNP for the digits 0 to 11 in order. The Four Seasons, consisting of 12 movements in all, are part of a set of concerti by Antonio Vivaldi, born on 4 March 1678. He was nicknamed il Prete Rosso (the Red Priest) because of his red hair.

Clue explanations

Conventions: * = anagram, < = reversal

No Moved letter Answer Explanation
Across
1 Ro(P)e → P-eel PEA P(erfect) E(n) A(spic)
8 learner(S) → S-loth AIL AI + L
12   WILTON (WILT + N) around O
14   STRUMAE (MATURE + S) *
15   OARED O + A + RED
16   CRISP CR + PSI <
17   RING NG after RI
18   HANDCUFF HAND + CU + FF
21   HAN H + N/A <
22   ALAS ALASKA−KA
24   MERINO MINER * + O
25   GOSSAMER GOER around MASS <
29   ABLAUT TUBA * around LA
32   NEAR EAR after N
33   SUI SUI(d)
34   MISERERE MISER + ERE
36   WIN’T W + IN + T
38   ERATO ERA + OT <
39   OPTIC OPT + I + C
40   REVISAL VIS in REAL
41   NEARER (A + R) in NE’ER
42 (G)ratin → ratin-G PER P(i)E + R(ating)
43 (M)art → a-M-id EAR (insid)E AR(t)
Down
1   PASCHAL (A + L) after (C in PASH)
2   ENTRALL (c)ENTRALL(y)
3   AGRIN (AG(e) + RN) around I
4 page(R) → Sp-R-ite IMP PIMP−P
5 fo(U)r → Po-U-nd LARUMS L + ARUMS
6 (A)new → w-A-ives TE TEN−N
7 we(T) → bas-T-e OON OO + N
9   AYRSHIRE ((d)AIRY’S HER(d)) *
10   ICE MAN I + C + NAME <
11   LADINO LAID * + NO
13   NAGOR ORANG(e) *
19   CUSUM US in CUM
20   FEMME FÉE around MM
23   AGLITTER AG + LITTER
26 unt(I)ed → un-I-ted AUSTER TEAR * around (U + S)
27   TARTANA TARTAN + A
28   AREOLAR ARE + ORAL *
29   ASTOOP AS TO + OP
30   BURPEE RUB < + PEE
31   ALWIN A + L + WIN
35   EASLE (m)EASLE(d)
37 Fro(W)stiness → W-rap ICE 2 meanings
38 see(N) → Easter-N EVE E(ggs) + V + E
40 (E)radiate → h-E-ad RA R(adiate) + (be)A(ms)

Solving the sum

This is one logical path for solving the alphametic. There are 12 clues with distinct moved letters (in 26dn either I or T could be moved, but T is the letter for 7dn), ie: 1ac P, 8ac S, 42ac G, 43ac M, 4dn R, 5dn U, 6dn A, 7dn T, 26dn I, 37dn W, 38dn N, 40dn E. Thus, the thematic number d is 12, the addition is in base 12, and the sum of moved letter + inserted letter for each entry is 11 (the letter for 0 being paired with the one for 11, 1 with 10, etc).

For the gap in 5dn, _LARUMS, the only letter that makes a new word is A, so the clue’s moved U gives us U + A = 11. Then for 6dn with moved A the inserted letter must be U, making UTE. Similarly, 26dn’s AUSTER_ can only become AUSTERE, so I + E = 11 and 40dn has I inserted to make RAI.

For the 6-letter word in the bottom row we have __IE__. Adding the letters that would make new words in the affected entries, the options are [MNPT][RS]IE[NRST][GNPRSTW], for which the only word with distinct letters is PRIEST (making PERP, ICER, EVES, TEAR). This establishes the remaining letter pairs as P+G, R+W, S+N and T+M and the top-row gaps can be completed as GWAUMN (making PEAG, WIMP, MOON, NAIL), so we know G < W < A < U < M < N. From pairings, we also know S < T < A < U < R < P.

A has at least four values below it and five above it, so it’s in the range [4, 6]. Similarly, U is in the range [5, 7]. Since A + U = 11 and A < U, A is one of {4, 5} and U is the corresponding element in {7, 6}. Since U < R < P, P must be at least 8. In the second column of digits, P + U + U + I must be at least 8 + 6 + 6 + I = 20 + I and there must be a carry of at least 1 to the leftmost column.

In the leftmost column we have S + S + A + W + (carry) = P = 11 − G, so S + S + G + W + A < 11 (where S, W and A are all non-zero, being initial digits). If G is 1 then the minimum sum is 2 + 2 + 1 + 3 + 4 = 12, too big; if G is 2 or more, the minimum sum is 1 + 1 + 2 + 3 + 4 = 11, also too big; so G = 0 and P = 11. As S < T < A, S is in the range [1, 3]. But if S is 3, the minimum sum is 3 + 3 + 0 + 1 + 5 = 12, too big; so S is one of {1, 2}. As W < A, W is in the range [1, 4]. But if W is 4, the minimum sum is 1 + 1 + 0 + 4 + 5 = 11, too big; so W is one of {1, 2, 3}. If W is 1, S must be 2; if W is 2, S must be 1; if W is 3, S must be 1 (because 2 + 2 + 0 + 3 + 4 = 11, too big).

If (S, W) = (2, 1) then the pairings give (N, R) = (9, 10). In the rightmost column we have G + R + N + R = 0 + 10 + 9 + 10 = 29 = 2×12 + 5, giving T = 5 (so M = 6) and carrying (2). The sum in the fifth column is then 9 + E + 6 + E + (2), which is odd, so it can’t give S = 2 (modulo 12).

If (S, W) = (1, 3) then the pairings give (N, R) = (10, 8). In the rightmost column we have 0 + 8 + 10 + 8 = 26 = 2×12 + 2, giving T = 2 (so M = 9) and carrying (2). The sum in the fifth column is then 10 + E + 9 + E + (2) = 2E + 21; that needs to equal 12n + 1 (for some integer n ≥ 0), so 2E + 20 = 12n, or E = 6n − 10. If n = 2 then E = 2, but that value is already assigned to T; if n = 3 then E = 8, but that value is already assigned to R; other values of n give E < 0 or E > 11.

The only remaining possibility is S = 1, W = 2, and their pairs N = 10, R = 9. The rightmost column gives 0 + 9 + 10 + 9 = 28 = 2×12 + 4, so T = 4 (and M = 7) and we carry (2). That forces A = 5 and U = 6, leaving only 3 and 8 for E and I, in some order. In the fifth column, 10 + E + 7 + E + (2) = 12n + 1, so E = 6n − 9, which is clearly a multiple of 3, so E = 3, I = 8 and the alphametic is complete.

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