Answer to December 2009 Problem
Holiday Cryptarithm

Proposed by David L. Pagni
Mathematics Teacher January 1988
Each letter of the alphabet stands for a different digit from 0 to 9 throughout this addition problem.
Can you break the code?

    M E R R Y
  +   X M A S
-- -- -- -- -- -- --
    T O A L L


You may send in your answer in the following form:
A =
E =
L =
M =
O =
R =
S =
T =
X =
Y =

 

Solution to the Problem:

There are four possible solutions: 

 85772    A=6 E=5 L=3 M=8 O=0 R=7 S=1 T=9 X=4 Y=2 
+ 4861
------
 90633

 85771    A=6 E=5 L=3 M=8 O=0 R=7 S=2 T=9 X=4 Y=1 
+ 4862
------
 90633

 84772    A=6 E=4 L=3 M=8 O=0 R=7 S=1 T=9 X=5 Y=2 
+ 5861
------
 90633

 84771    A=6 E=4 L=3 M=8 O=0 R=7 S=2 T=9 X=5 Y=1 
+ 5862
------
 90633

Correctly solved by:

1. Becca G. Mountain View High School,
Mountain View, Wyoming
2. Courtney Peterson Mountain View High School,
Mountain View, Wyoming
3. Taiya Cheney Mountain View High School,
Mountain View, Wyoming
4. Mark Kampman Mountain View High School,
Mountain View, Wyoming
5. Clint Lamb Mountain View High School,
Mountain View, Wyoming