2018 AMC 12B Problems/Problem 15

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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)

See Also

2018 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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