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Difference between revisions of "2018 AMC 12B Problems/Problem 15"
(Prioritized solutions based on elegance and clarity. Let me know if you disagree.)
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<math>\textbf{(A) } 96 \qquad \textbf{(B) } 97 \qquad \textbf{(C) } 98 \qquad \textbf{(D) } 102 \qquad \textbf{(E) } 120 </math>
 
<math>\textbf{(A) } 96 \qquad \textbf{(B) } 97 \qquad \textbf{(C) } 98 \qquad \textbf{(D) } 102 \qquad \textbf{(E) } 120 </math>
  
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== Solution 1==
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== Solution 1 ==
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Analyze that the three-digit integers divisible by <math>3</math> start from <math>102.</math> In the <math>200</math>'s, it starts from <math>201.</math> In the <math>300</math>'s, it starts from <math>300.</math> We see that the units digits is <math>0, 1, </math> and <math>2.</math>
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There are <math>4</math> choices for the last digit (<math>1, 5, 7, 9</math>), and <math>8</math> choices for the first digit (excluding <math>0</math> and <math>3</math>). We know what the second digit modulo <math>3</math> is, so there are <math>3</math> choices for it (pick from one of the sets <math>\{0, 6, 9\},\{1, 4, 7\}, \{2, 5, 8\}</math>). The answer is <math>4\cdot 8 \cdot 3 = \boxed{\textbf{(A) } 96}.</math>  
  
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Write out the 1- and 2-digit multiples of <math>3</math> starting from <math>0, 1,</math> and <math>2.</math> Count up the ones that meet the conditions. Then, add up and multiply by <math>3,</math> since there are three sets of three from <math>1</math> to <math>9.</math> Then, subtract the amount that started from <math>0,</math> since the <math>300</math>'s ll contain the digit <math>3.</math>
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~Plasma_Vortex
  
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We get: <cmath>3(12+12+12)-12.</cmath>
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== Solution 2 ==
  
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This gives us: <cmath>\boxed{\textbf{(A) } 96}.</cmath>
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== Solution 3 ==
 +
Analyze that the three-digit integers divisible by <math>3</math> start from <math>102.</math> In the <math>200</math>'s, it starts from <math>201.</math> In the <math>300</math>'s, it starts from <math>300.</math> We see that the units digits is <math>0, 1, </math> and <math>2.</math>
  
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== Solution 2==
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Write out the <math>1</math>- and <math>2</math>-digit multiples of <math>3</math> starting from <math>0, 1,</math> and <math>2.</math> Count up the ones that meet the conditions. Then, add up and multiply by <math>3,</math> since there are three sets of three from <math>1</math> to <math>9.</math> Then, subtract the amount that started from <math>0,</math> since the <math>300</math>'s ll contain the digit <math>3.</math>
  
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There are <math>4</math> choices for the last digit (<math>1, 5, 7, 9</math>), and <math>8</math> choices for the first digit (exclude <math>0</math>). We know what the second digit mod  <math>3</math> is, so there are <math>3</math> choices for it (pick from one of the sets <math>\{0, 6, 9\},\{1, 4, 7\}, \{2, 5, 8\}</math>). The answer is <math>4\cdot 8 \cdot 3 = \boxed{96}</math> (Plasma_Vortex)
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We get <cmath>3(12+12+12)-12=\boxed{\textbf{(A) } 96}.</cmath>
  
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== Solution 3==
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== Solution 4 ==
  
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Consider the number of <math>2</math>-digit numbers that do not contain the digit <math>3,</math> which is <math>90-18=72.</math> For any of these <math>2</math>-digit numbers, we can append <math>1,5,7,</math> or <math>9</math> to reach a desirable <math>3</math>-digit number. However, <math>1 \equiv 7 \equiv 1</math> <math>(mod</math> <math>3)</math>, and thus we need to count any <math>2</math>-digit number <math>\equiv 2</math> <math>(mod</math> <math>3)</math> twice. There are <math>(98-11)/3+1=30</math> total such numbers that have remainder <math>2,</math> but <math>6</math> of them <math>(23,32,35,38,53,83)</math> contain <math>3,</math> so the number we want is <math>30-6=24.</math> Therefore, the final answer is <math>72+24= \boxed{96}.</math>
+
Consider the number of <math>2</math>-digit numbers that do not contain the digit <math>3,</math> which is <math>90-18=72.</math> For any of these <math>2</math>-digit numbers, we can append <math>1,5,7,</math> or <math>9</math> to reach a desirable <math>3</math>-digit number. However, we have <math>7 \equiv 1\pmod{3},</math> and thus we need to count any <math>2</math>-digit number <math>\equiv 2\pmod{3}</math> twice. There are <math>(98-11)/3+1=30</math> total such numbers that have remainder <math>2,</math> but <math>6</math> of them <math>(23,32,35,38,53,83)</math> contain <math>3,</math> so the number we want is <math>30-6=24.</math> Therefore, the final answer is <math>72+24= \boxed{\textbf{(A) } 96}.</math>
  
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==Solution 4 (easy)==
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== Solution 5 ==
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We need to take care of all restrictions. Ranging from <math>101</math> to <math>999,</math> there are <math>450</math> odd 3-digit numbers. Exactly <math>\frac{1}{3}</math> of these numbers are divisible by 3, which is <math>450\times\frac{1}{3}=150.</math> Of these 150 numbers, <math>\frac{4}{5}</math> <math>\textbf{do not}</math> have 3 in their ones (units) digit, <math>\frac{9}{10}</math> <math>\textbf{do not}</math> have 3 in their tens digit, and <math>\frac{8}{9}</math> <math>\textbf{do not}</math> have 3 in their hundreds digit. Thus, the total number of 3 digit integers are <math>900\times\frac{1}{2}\times\frac{1}{3}\times\frac{4}{5}\times\frac{9}{10}\times\frac{8}{9}=96,</math> or <math>\boxed{\text{A}}</math>
+
We need to take care of all restrictions. Ranging from <math>101</math> to <math>999,</math> there are <math>450</math> odd <math>3</math>-digit numbers. Exactly <math>\frac{1}{3}</math> of these numbers are divisible by <math>3,</math> which is <math>450\times\frac{1}{3}=150.</math> Of these <math>150</math> numbers, <math>\frac{4}{5}</math> <math>\textbf{do not}</math> have <math>3</math> in their ones (units) digit, <math>\frac{9}{10}</math> <math>\textbf{do not}</math> have <math>3</math> in their tens digit, and <math>\frac{8}{9}</math> <math>\textbf{do not}</math> have <math>3</math> in their hundreds digit. Thus, the total number of <math>3</math>-digit integers is <cmath>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}.</cmath>
  
 
~mathpro12345
 
~mathpro12345

Revision as of 17:26, 27 September 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

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

See Also

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

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