Problem:
How many three-digit positive integers a​ b​ c​ are there whose nonzero digits a,b, and c satisfy
0.a​ b​ c​ˉ​=31​(0.aˉ+0.bˉ+0.cˉ)?
(The bar indicates repetition, thus 0.a​ b​ c​ˉ​ is the infinite repeating decimal 0.a​ b​ c​ a​ b​ c​ ⋯)
Answer Choices:
A. 9
B. 10
C. 11
D. 13
E. 14
Solution:
The given equation means
999100a+10b+c​=31​(9a​+9b​+9c​)
which simplifies to 7a=3b+4c. Therefore 3b≡−4c≡3c(mod7), so b≡c(mod7). Given that the variables are nonzero digits, the possibilities for (b,c) are (1,1),(2,2),(3,3),…,(9,9),(1,8), (8,1),(2,9), and (9,2). In each case the value of a is uniquely determined, and the (D)13​ positive integers are 111, 222, 333, …,999,518,481,629, and 592.
The problems on this page are the property of the MAA's American Mathematics Competitions