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

There are $4$ choices for the last digit ( $1, 5, 7, 9$ ), and $8$ choices for the first digit (excluding $0$ and $3$ ). We know what the second digit modulo $3$ 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 $4\cdot 8 \cdot 3 = \boxed{\textbf{(A) } 96}.$

~Plasma_Vortex

Solution 2

Solution 3

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 $3(12+12+12)-12=\boxed{\textbf{(A) } 96}.$

Solution 4

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, we have $7 \equiv 1\pmod{3},$ and thus we need to count any $2$ -digit number $\equiv 2\pmod{3}$ 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 $72+24= \boxed{\textbf{(A) } 96}.$

Solution 5

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 is $900\cdot\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{4}{5}\cdot\frac{9}{10}\cdot\frac{8}{9}=\boxed{\textbf{(A) } 96}.$

~mathpro12345

Video Solution

https://youtu.be/mgEZOXgIZXs?t=448

~ pi_is_3.14

2018 AMC 12B (Problems • Answer Key • Resources)
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