# 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 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

Note that this isn't a great solution, but a more practical one to achieve the answer.

Notice that there are 300 numbers that are 3 digits and are divisible by 3 (from 102 to 999). Now one by one apply the restrictions. The restriction for only odd numbers would mean that half the numbers are taken out. 300*1/2 = 150. Next, apply the restriction of no 3s. For the units digit, that would mean multiplying by 4/5 (remember that now you only have odd numbers to choose from), for the tens that would mean multiplying by 9/10, and for the hundreds that would mean multiplying by 8/9 (because you cant have 0 here). Thus, we get 150*4/5=120*9/10=108*8/9=96, which is A.

Sol by IronicNinja~