Source: rec.puzzles. Used as Ken's Puzzle of the Day 5/16/95.
>From Philippe Fondanaiche -B is obviously equal to 1 which is the figure carried over from the equation A + L + r = 10 + A where r ( = 0 or 1) is carried over from P + E + r' = 10*r + N -the equation A + L + r = 10 + A triggers L = 9 and r = 1 -as P<9 and E<9 and P + E +r' = 10 + N, then N < E and N < P -E + N = A. If it is not the case, E + N = 10 + A, then L + O + 1 = O + 10 = N + 10 ==> O = N, that is impossible. -as A<9 and E + N = A, then N<4. So, the 2 possible values of N are N=2 and N=3 ¤ if N=2,then O=3 ¤¤ E=4 ==> A=6. We find the solution: APPLE=67794 LEMON=94832 BANANA=162626 ¤¤ E=5 ==> A=7. Easy to check that there is no solution for P and M ¤¤ E=6 ==> A=8. As above,no solution. ¤ if N=3,then O=4 and E=5. No solution for P and M
From Larry Baum: APPLE LEMON ------ BANANA It is immediately clear that B=1 and L=9. Since L=9, there is no carry to the tens column, so 1) E+N=A 2) N+1=O and there must be a carry to the hundreds column. There is definitely a carry to the ten-thousands column. So there are only two cases: A) no carry to thousands column; then: 3) 1+P+M=A 4) P+E=N+10 1) & 3) => P+M=E+N-1 Subtracting 4) => M-E=E-11 or 2E=M+11 Since B=1,L=9 1)=> N is not 0 and E<=6 => M+11 <= 12 => M<=1. Since B=1, M must be 0 => 2E=11 oops! So case B MUST be true: b) carry to thousands column; then: 3) 1+P+M=A+10 P+M=A+9 4) 1+P+E=N+10 P+E=N+9 1) & 3) => P+M=E+N+9 Subtracting 4) => M-E=E or M=2E So (E,M) can be: (2,4) (3,6) (4,8) E=2: P=N+7. Since P<=8, N=0 or 1. But N=0 => A=E & N=1 => N=B; either way: oops! E=3: P=N+6. Since P<=8, N=0, 1 or 2. N can't be 0 or 1 for the same reasons just mentioned. N=2 => O=3=E oops! So E=4, M=8: P=N+5. N=0, 1, 2, or 3. N=0 or 1 are out for the usual reasons. N=3 => O=4=E (and P=M=8). So N must be 2 => A=6, O=3, P=7 67794 94832 ------ 162626 is the unique solution.