Each letter of the alphabet has been assigned a different value from 1 to 26. The number following each word is the sum of all the letters in it. Your task is to discover the values of all the letters and then decide the value of J, which is not used.
ABLE 38 ACID 23 ACRID 24 ADMIT 45
ADO 16 AGREE 37 AHEAD 43 AISLE 45
ALE 30 ANIL 48 ANKLE 66 ANTLER 52
ANVIL 66 APPLE 72 AQUA 36 AREA 12
ARENA 24 ARROW 27 AURA 27 AWAY 41
AWE 24 AWFUL 78 AXLE 44 AZALEA 49
The first few I found were R=1, N=12, B=8, Q=10, X=14, V=18, P=21, K=24, T=9.
Then, because 2A+E = 11 (from AREA and R=1), and other similar relationships, these were the possibilities:
A E Z Y U
2 7 15 22 22
3 5 13 19 20
4 3 11 16 18
But the first line has two 22's and the last line assigns 18 to U, but 18 is already assigned to V. That makes A=3, E=5, Z=13, Y=19, U=20.
Then further algebra among the words leads to W=16, L=22, I=11, G=23.
Then, for D, O and C, using only unused possibilities for D:
D O C
6 7 3
7 6 2
But 3 is already used by A, so D=7, O=6 and C=2.
From there it was straight algebra again, with H=25, F=17, M=15 and S=4. That leaves 26 for J.

Posted by Charlie
on 20060804 14:49:24 