2022 AMC 10 A

Complete problem set with solutions and individual problem pages

Problem 17 Easy

How many three-digit positive integers \underline{a} \underline{b} \underline{c} are there whose nonzero digits a, b, and c satisfy

0.\overline{\underline{a}\ \underline{b}\ \underline{c}}=\frac{1}{3}(0.\overline{a}+0.\overline{b}+0.\overline{c})?

(The bar indicates repetition, thus  0.\overline{\underline{a}\ \underline{b}\ \underline{c}} is the infinite repeating decimal 0.\underline{a} \underline{b} \underline{c} \underline{a} \underline{b} \underline{c} \cdot \cdot \cdot)

  • A.

    9

  • B.

    10

  • C.

    11

  • D.

    13

  • E.

    14

Answer:D

0.\overline{\underline{a}\ \underline{b}\ \underline{c}} is a recuesive decimal, so as 0.\overline{a}, 0.\overline{b} and 0.\overline{c}.

\frac{100a+10b+c}{999}=\frac{1}{3}\cdot \left(\frac a9+\frac b9+\frac c9 \right)

\frac{100a+10b+c}{999}=\frac{a+b+c}{27}

100a+10b+c=37a+37b+37c

63a=27b+36c

3b+4c=7a

\therefore b\equiv c( \text{mod} 7).

Thus, either b=c or (b,c)=(1,8),(2,9),(9,2). Each of these will have an unique solution, yielding the 13 intergers: 111,222,333,444,555,666,777,888,999,518,629,481,592.

Therefore, the answer is 13.

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