The same substitution (each letter corresponds to a particular digit) has been applied to all four summations. What are the substitutions?

A. LFOH           B. LTEL
 + EMAO            + LAHF
   MOST              HOST

C. ELRO           D. OTTH
 + OLRF            + LETH 
   MORE              FORE

I put +1's in the parentheses for the 1's that get carried over.

2H=E from D

2L(+1)=H from B
4L(+2)=2H

L+E(+1)=M from A
L+2H(+1)=M
L+4L(+1 +2)=M
So either 5L, 5L+1, 5L+2, 5L+3 = M

In order for M to be a single digit, L=1. This means that H=2,3. E=4,6. M=5,6,7,8

And from C, R+R(+1)=R. This means that R=0, or 9. From D, because H=2 or 3, there is nothing carried over when we add H+H. So 2T=R which means that R must be even. So R=0. Hence T=5.

From B, L+F=T or 1+F=5 so F=4. This means that E=6, meaning H=3.

From D, T=5 so there's a 1 carried over when we add T+T. So T+E+1=O. 5+6+1=12 so O=2.

From C, L=1, so nothing is carried over when L+L. So E+O=M. 6+2=8 so M=8.

From B, L+F=5 so nothing's carried over. E+H=S so 6+3=9, so S=9. By elimination A=7.

R=0
L=1
O=2
H=3
F=4
T=5
E=6
A=7
M=8
S=9

A formal check does confirm that everything works:

1423+6872=8295
1561+1734=3295
6102+2104=8206
2553+1653=4206

Edited on December 18, 2004, 4:27 pm

Posted by np_rt on 2004-12-18 16:26:55