2018 AMC 12B Problems/Problem 15

Problem

For how many digits $C$ is the positive three-digit number $1C3$ a multiple of 3?

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 have $3$ 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 $\Rightarrow 300*\frac{1}{2} = 150$.

Next, apply the restriction of no $3$s. For the units digit, that would mean multiplying by $\frac{4}{5}$ (remember that now you only have odd numbers to choose from).

For the tens that would mean multiplying by $\frac{9}{10}$, and for the hundreds that would mean multiplying by $\frac{8}{9}$ (because you cant have 0 here).

Thus, we get $150*\frac{4}{5}*\frac{9}{10}*\frac{8}{9}=96$, which is $\boxed{A}$.

Sol by IronicNinja~

 2018 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 14 Followed byProblem 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