One of the most challenging of codewords, this puzzle ran for many years in 'Tough Puzzles'. To solve an alphacipher it is necessary to discover the numbers allocated to 25 letters, when the 26th letter (the one you are asked for) will correspond to the one number unallocated.
A common solving device is the equation.
If OLIVE = 54 and EVIL = 45, you see that O = 9. But, faced with:
HARROW = 60, RHINO = 65, PINE = 42 and WARP = 25, you may not observe that E can be resolved by putting the words in an equation and cancelling out letters from both sides:
PINE 42 + HARROW 60 = 102
WARP 25 + RHINO 65 = 90
E = 12
There does not have to be the same number of words on each side of an equation. For example, if:
MELEE 77 + BEER 64 + ICE 44 = 185
CLIMBER = 115
5 x E = 70, so E = 14
Nor are complete words necessary: where you know the combined value of a number of letters, these can be used:
CKL 47 + FINITE 52 = 99
LIFE 37 + NICK 56 = 93
T = 6
Substitution is another device. If you are faced with, say:
(1) OPAL = 53,
(2) LEAP = 51,
(3) POLE = 44
(again, these can be parts of words instead of complete words), and you have values (or a joint value) for P and L, you can break out A, E and O by substitution:
Say PL = 18, then (1) OA = 35, (2) EA = 33 and (3) OE = 26
From (1) and (2) O = E + 2
Substitute in (3): 2E + 2 = 26; thus E = 12, O = 14 and A = 21
Always make a list of the numbers from 1-26, and cross them off as you allocate them to letters. This will let you see at a glance the numbers that are still available. Look for relationships between unresolved letters. For example, if you notice that Q + 9 = H and H + 5 = B, a glance at the unallocated numbers will reveal the combinations possible – if there is only one, you have resolved all three letters at a stroke. Even to know whether a letter is odd or even can be helpful. For example, if EEL = 41, L is clearly odd.
This should be enough to get you started. Other tricks you will discover, by yourself.