AMC 10 Daily Practice Round 3

Complete problem set with solutions and individual problem pages

Problem 8 Easy

Call a positive integer \textit{good} if all its digits are nonzero and it is divisible by each of its digits. How many two-digit positive integers are \textit{good}?

  • A.

    9

  • B.

    14

  • C.

    15

  • D.

    16

  • E.

    18

Answer:B

Let \overline{ab} = 10a + b be a two-digit good integer, where a and b are nonzero digits. Then, a and b must both divide 10a + b. This implies a divides b and b divides 10a. Consider cases on the value of a.

Case 1: a = 1. Then, the divisibility becomes b divides 10. There are 3 such digits b, namely 1, 2, and 5.

Case 2: a = 2. Then, the divisibility becomes 2 divides b and b divides 20. There are 2 such digits b, namely 2 and 4.

Case 3: a = 3. Then, the divisibility becomes 3 divides b and b divides 30. There are 2 such digits b, namely 3 and 6.

Case 4: a = 4. Then, the divisibility becomes 4 divides b and b divides 40. There are 2 such digits b, namely 4 and 8.

Case 5: 5 ≤ a ≤ 9. Then, the only possible nonzero digit that is a multiple of a is a itself. Thus, b = a. Then, b is guaranteed to divide 10a. Thus, there are 5 good integers here. The total number of two-digit good integers is 3 + 2 + 2 + 2 + 5 =14.