Spoiler for New Scientist Enigma 1729: “Xmas Gifts” (Follow the link to see the puzzle.)
This one took me ages to figure out, partly due to a stupid mistake that had me thinking the puzzle had no solution, but it turns out not to be too bad.
If XMAS is 34 and GIFTS is 34 then XMASGIFTS is 68.
But if ARE = ORIENT then A = OINT, and if OF = ORIENT then F = RIENT. That means XMOINTSGIRIENTTS = 68. The smallest value for XMOINTSGIRIENTTS results when the letters that occur 3 times (I and T) are 1 and 2, the letters that occur 2 times (N and S) are 3 and 4, and the remaining 6 letters are 5 through 10… and that smallest value is 68. Any other possibility result in a larger sum. So I and T are 1 and 2; N and S are 3 and 4; E, G, M, O, R, and X are 5 to 10; and any letter other than these must be 11 or larger.
Now A = OINT and the largest value this can have is 10+2+4+1 = 17. But also ARE = THREE which means A = THE, and the smallest value this can have is 1+11+5 = 17. Therefore A = 17, and T, H and E must have their smallest possible values (1, 11 and 5) while O and N have their largest possible values (10 and 4). I then must be 2 and S must be 3.
XMAS = 34 then means XM = 14. Each of X and M must be 6, 7, 8, or 9, but the only way to get XM = 14 is X = 6, M = 8 or vice versa. Since X and M occur nowhere else in the puzzle we can’t resolve this ambiguity, but we don’t need to to get the ANSWER.
That leaves 7 and 9 for R and G. Suppose R = 7 and G = 9. Then THREE = ORIENT = ARE = 29, and we must also have KINGS = K+2+4+9+3 = 29 so K = 11. But we already have H = 11.
So R and G are 9 and 7; THREE = ORIENT = ARE = 31. K = 15 to make KINGS = 31, F = 21 to make OF = 31, and W = 26 to make WE = 31.
Then A, N, S, W, E, R = 17, 4, 3, 26, 5, 9.