Problem:
How many three-digit positive integers aβ bβ cβ are there whose nonzero digits a,b, and c satisfy
0.aβ bβ cβΛβ=31β(0.aΛ+0.bΛ+0.cΛ)?
(The bar indicates repetition, thus 0.aβ bβ cβΛβ is the infinite repeating decimal 0.aβ bβ cβ aβ bβ cβ β―)
Answer Choices:
A. 9
B. 10
C. 11
D. 13
E. 14
Solution:
The given equation means
999100a+10b+cβ=31β(9aβ+9bβ+9cβ)
which simplifies to 7a=3b+4c. Therefore 3bβ‘β4cβ‘3c(mod7), so bβ‘c(mod7). Given that the variables are nonzero digits, the possibilities for (b,c) are (1,1),(2,2),(3,3),β¦,(9,9),(1,8), (8,1),(2,9), and (9,2). In each case the value of a is uniquely determined, and the (D)13β positive integers are 111, 222, 333, β¦,999,518,481,629, and 592.
The problems on this page are the property of the MAA's American Mathematics Competitions