2018 AMC 12B Problems/Problem 15

Problem

How many odd positive 3-digit integers are divisible by 3 but do not contain the digit 3?

$\textbf{(A) } 96 \qquad \textbf{(B) } 97 \qquad \textbf{(C) } 98 \qquad \textbf{(D) } 102 \qquad \textbf{(E) } 120$

Solution 1 (For Dummies)

Analyze that the three-digit integers divisible by $3$ start from $102$ . In the $200$ 's, it starts from $201$ . In the $300$ 's, it starts from $300$ . We see that the units digits is $0, 1,$ and $2.$

Write out the 1- and 2-digit multiples of $3$ starting from $0, 1,$ and $2.$ Count up the ones that meet the conditions. Then, add up and multiply by $3$ , since there are three sets of three from $1$ to $9.$ Then, subtract the amount that started from $0$ , since the $300$ 's all contain the digit $3$ .

We get: $3(12+12+12)-12.$

This gives us: $\boxed{\textbf{(A) } 96}.$

Solution 2

There are $4$ choices for the last digit ( $1, 5, 7, 9$ ), and $8$ choices for the first digit (exclude $0$ ). We know what the second digit mod $3$ is, so there are $3$ choices for it (pick from one of the sets $\{0, 6, 9\},\{1, 4, 7\}, \{2, 5, 8\}$ ). The answer is $4\cdot 8 \cdot 3 = \boxed{96}$ (Plasma_Vortex)

2018 AMC 12B (Problems • Answer Key • Resources)
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2018 AMC 12B Problems/Problem 15

Contents

Problem

Solution 1 (For Dummies)

Solution 2

See Also