Difference between revisions of "2018 AMC 12B Problems/Problem 15"

Revision as of 20:07, 8 October 2021

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

Let $\underline{ABC}$ be one such odd positive $3$ -digit integer with hundreds digit $A,$ tens digit $B,$ and ones digit $C.$ Since $\underline{ABC}\equiv0\pmod3,$ we need $A+B+C\equiv0\pmod3$ by the divisibility rule for $3.$

As $A\in\{1,2,4,5,6,7,8,9\}$ and $C\in\{1,5,7,9\},$ there are $8$ possibilities for $A$ and $4$ possibilities for $C.$ Note that each ordered pair $(A,C)$ determines the value of $B$ modulo $3,$ so $B$ can be any element in one of the sets $\{0,6,9\},\{1,4,7\},$ or $\{2,5,8\}.$ We conclude that there are always $3$ possibilities for $B.$

By the Multiplication Principle, the answer is $8\cdot4\cdot3=\boxed{\textbf{(A) } 96}.$

~Plasma_Vortex ~MRENTHUSIASM

Solution 2

Let $\underline{ABC}$ be one such odd positive $3$ -digit integer with hundreds digit $A,$ tens digit $B,$ and ones digit $C.$ Since $\underline{ABC}\equiv0\pmod3,$ we need $A+B+C\equiv0\pmod3$ by the divisibility rule for $3.$

As $A\in\{1,2,4,5,6,7,8,9\},B\in\{0,1,2,4,5,6,7,8,9\},$ and $C\in\{1,5,7,9\},$ note that:

There are $2$ possibilities for $A\equiv0\pmod3,$ namely $A=6,9.$
There are $3$ possibilities for $A\equiv1\pmod3,$ namely $A=1,4,7.$
There are $3$ possibilities for $A\equiv2\pmod3,$ namely $A=2,5,8.$
There are $3$ possibilities for $B\equiv0\pmod3,$ namely $B=0,6,9.$
There are $3$ possibilities for $B\equiv1\pmod3,$ namely $B=1,4,7.$
There are $3$ possibilities for $B\equiv2\pmod3,$ namely $B=2,5,8.$
There are $1$ possibility for $C\equiv0\pmod3,$ namely $C=9.$
There are $2$ possibilities for $C\equiv1\pmod3,$ namely $C=1,7.$
There are $1$ possibility for $C\equiv2\pmod3,$ namely $C=5.$

We apply casework to $A+B+C\equiv0\pmod3:$ $\begin{array}{c|c|c||l} & & & \\ [-2.5ex] \boldsymbol{A\operatorname{mod}3} & \boldsymbol{B\operatorname{mod}3} & \boldsymbol{C\operatorname{mod}3} & \multicolumn{1}{c}{\textbf{Count}} \\ [0.5ex] \hline & & & \\ [-2ex] 0 & 0 & 0 & 2\cdot3\cdot1=6 \\ 0 & 1 & 2 & 2\cdot3\cdot1=6 \\ 0 & 2 & 1 & 2\cdot3\cdot2=12 \\ 1 & 0 & 2 & 3\cdot3\cdot1=9 \\ 1 & 1 & 1 & 3\cdot3\cdot2=18 \\ 1 & 2 & 0 & 3\cdot3\cdot1=9 \\ 2 & 0 & 1 & 3\cdot3\cdot2=18 \\ 2 & 1 & 0 & 3\cdot3\cdot1=9 \\ 2 & 2 & 2 & 3\cdot3\cdot1=9 \end{array}$ Together, the answer is $6+6+12+9+18+9+18+9+9=\boxed{\textbf{(A) } 96}.$

~MRENTHUSIASM

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 $3.$

Together, the answer is $3(12+12+12)-12=\boxed{\textbf{(A) } 96}.$

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 $450\cdot\frac{1}{3}=150.$ 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

Solution 6

We will start with the numbers that could work. This numbers include _ _ $1$ , _ _ $5$ , _ _ $7$ , _ _ $9$ . Let's work case by case.

Case $1$ : _ _ $1$ : The two blanks could be any number that is $2$ mod $3$ that does not include $3$ . We have $24$ cases for this case (we could count every case).

Case $2$ : _ _ $5$ : The $2$ blanks could be any number that is $1$ mod $3$ that does not include $3$ . But we could see that this case has exactly the same solutions to case $1$ because we have a $1-1$ correspondence. We can do the exact same for case $3$ .

Cases $4$ : _ _ $9$ : We need the blanks to be a multiple of $3$ , but does not contain 3. We have $(12, 15, 18, 21, 24, 27, 42, 45, 48, 51, 54, 57, 60, 66, 69, 72, 75, 78, 81, 84, 87, 90, 96, 99)$ which also contains $24$ numbers. Therefore, we have $24 \cdot 4$ which is equal to $\boxed{\textbf{(A) } 96}.$

~Arcticturn

Video Solution

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

~ pi_is_3.14

@@ Line 76: / Line 76: @@
 Cases <math>4</math>: _ _ <math>9</math>: We need the blanks to be a multiple of <math>3</math>, but does not contain 3. We have <math>(12, 15, 18, 21, 24, 27, 42, 45, 48, 51, 54, 57, 60, 66, 69, 72, 75, 78, 81, 84, 87, 90, 96, 99)</math> which also contains <math>24</math> numbers. Therefore, we have <math>24 \cdot 4</math> which is equal to <math>\boxed{\textbf{(A) } 96}.</math>
+~Arcticturn
 == Video Solution ==

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