Substitute digits for letters (different letters, different digits; same letters, same digits) so ALPHABET + LETTERS = SCRABBLE. No number begins with zero.

Can you manage without a computer program?

Surprisingly, this does not require much trial and error or plugging.

A  L  P  H  A  B  E  T
+  L  E  T  T  E  R  S
__________________
S  C R  A  B  B  L  E

I will label the columns 1-8 with column 1 being A+0-->S and column 8 being A+S-->E. Also, the carry over's can at most be 1 since we're only dealing with 2 addends.

Column 6 implies that either E=9 (with 1 to carry over to the column 5) or E=0 (with nothing to carry over).

Case: E=9
T+S=9     column 8
S=A+1     column 1
So A+T=8
Because there is 1 to carry over, A+T+1=B=9
Since E=9, B cannot be 9. So E is not 9. Hence E=0.

E=0
T+S=10     column 8
S=A+1     column 1
So A+T=9

Because there is nothing to carry over, A+T=B=9. So B=9.

R=P+1     column 3
L=R+1     column 7      --> L=P+2

Since A+T=9, there is nothing to carry over to column 4. Also, column 3 shows that something was carried over from column 4.

H+T=10+A     column 4
H+9-A=10+A
H=2A+1

Column 1 shows that there is something carried over from column 2 so:

2L=10+C     column 3
2(P+2)=10+C
C=2P-6

Writing all equations in terms of A and P

E=0
B=9

A
S=A+1
H=2A+1
T=9-A

P
R=P+1
L=P+2
C=2P-6

H=2A+1 means that A=1, 2, or 3

C=2P-6 along with P, P+1, P+2 means that P=4, 5, 6
P=6 means that C=6 so that's eliminated.

A=1,2,3
P=4,5

These lead to two sets of numbers. The solution is the one that such that there is no intersection between the two sets.

(A,S,T,H) = (1,2,8,3) (2,3,7,5) (3,4,6,7)
(P,R,L,C) = (4,5,6,2) (5,6,7,4)

Both 5 and 6 show up in both cases for (P,R,L,C). Two of the choices for (A,S,T,H) can be eliminated so that

(A,S,T,H) = (1,2,8,3)

By eliminating the first choice for (P,R,L,C) because C cannot be 2, we have (P,R,L,C) = (5,6,7,4).

Hence

E=0
A=1
S=2
H=3
C=4
P=5
R=6
L=7
T=8
B=9

Posted by np_rt on 2005-06-22 20:19:22