December 2009
Problem of the Month

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  
+ 4861    R=7 S=1 T=9 X=4 Y=2
------
 90633

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

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

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

Send any comments or questions to: David Pleacher