Problem:
Let A,M, and C be digits with
(100A+10M+C)(A+M+C)=2005
What is A?
Answer Choices:
A. 1
B. 2
C. 3
D. 4
E. 5
Solution:
Since A,M, and C are digits we have
0≤A+M+C≤9+9+9=27
The prime factorization of 2005 is 2005=5â‹…401, so
100A+10M+C=401 and A+M+C=5
Hence A=4​,M=0, and C=1.
The problems on this page are the property of the MAA's American Mathematics Competitions