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

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~ pi_is_3.14
 
~ pi_is_3.14
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==Video Solution by WhyMath==
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https://youtu.be/vdJFrAq0NDY
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~savannahsolver
  
 
==See Also==
 
==See Also==

Revision as of 07:19, 30 May 2023

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:

  1. 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 $A=2,5,8.$

  2. 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 $B=2,5,8.$

  3. 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

Solution 7

This problem is solvable by inclusion exclusion principle. There are $\frac{999-105}{6} + 1 = 150$ odd $3$-digit numbers divisible by $3$. We consider the number of $3$-digit numbers divisible by $3$ that contain either $1, 2$ or $3$ digits of $3$.

For $\underline{AB3}$, $AB$ is any $2$-digit number divisible by $3$, which gives us $\frac{99-12}{3} + 1 = 30$. For $\underline{A3B}$, for each odd $B$, we have $3$ values of $A$ that give a valid case, thus we have $5(3) = 15$ cases. For $\underline{3AB}$, we also have $15$ cases, but when $B=3, 9$, $A$ can equal $0$, so we have $17$ cases.

For $\underline{A33}$, we have $3$ cases. For $\underline{3A3}$, we have $4$ cases. For $\underline{33A}$, we have $2$ cases. Finally, there is just one case for $\underline{333}$.

By inclusion exclusion principle, we get $150 - 30 - 15 - 17 + 3 + 4 + 2 - 1 = \boxed{\textbf{(A) } 96}$ numbers.

~Zeric

Video Solution by Omega Learn

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

~ pi_is_3.14

Video Solution by WhyMath

https://youtu.be/vdJFrAq0NDY

~savannahsolver

See Also

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

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