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

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 ll 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)

Solution 3

Consider the number of $2$ -digit numbers that do not contain the digit $3$ , which is $90-18=72$ . For any of these $2$ -digit numbers, we can append $1,5,7,$ or $9$ to reach a desirable $3$ -digit number. However, $1 \equiv 7 \equiv 1$ $(mod$ $3)$ , and thus we need to count any $2$ -digit number $\equiv 2$ $(mod$ $3)$ twice. There are $(98-11)/3+1=30$ total such numbers that have remainder $2$ , but $6$ of them $(23,32,35,38,53,83)$ contain $3$ , so the number we want is $30-6=24$ . Therefore, the final answer is $72+24= \boxed{96}$ .

Solution 4 (easy)

We need to take care of all restrictions. Ranging from $101$ to $999$ , there are $450$ odd 3-digit numbers. Exactly $\frac{1}{3}$ of these numbers are divisible by 3, which is $450\times\frac{1}{3}=150$ . Of these 150 numbers, $\frac{4}{5}$ $\textbf{do not}$ have 3 in their ones (units) digit, $\frac{9}{10}$ $\textbf{do not}$ have 3 in their tens digit, and $\frac{8}{9}$ $\textbf{do not}$ have 3 in their hundreds digit. Thus, the total number of 3 digit integers are $900\times\frac{1}{2}\times\frac{1}{3}\times\frac{4}{5}\times\frac{9}{10}\times\frac{8}{9}=96$ , or $\boxed{\text{A}}$

~mathpro12345

2018 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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