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

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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, <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>.
  
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==Solution 4==
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==Solution 4 (easy)==
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Note that this isn't a great solution, but a more practical one to achieve the answer.
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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>
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Notice that there are <math>300</math> numbers that have <math>3</math> digits and are divisible by <math>3</math> (from <math>102</math> to <math>999</math>). Now one by one apply the restrictions.
 
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The restriction for only odd numbers would mean that half the numbers are taken out <math>\Rightarrow 300*\frac{1}{2} = 150</math>.
 
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Next, apply the restriction of no <math>3</math>s. For the units digit, that would mean multiplying by <math>\frac{4}{5}</math> (remember that now you only have odd numbers to choose from).
 
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For the tens that would mean multiplying by <math>\frac{9}{10}</math>, and for the hundreds that would mean multiplying by <math>\frac{8}{9}</math> (because you cant have 0 here).  
 
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Thus, we get <math>150*\frac{4}{5}*\frac{9}{10}*\frac{8}{9}=96</math>, which is <math>\boxed{A}</math>.
 
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Sol by IronicNinja~
 
  
 
==See Also==
 
==See Also==

Revision as of 20:23, 18 December 2020

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

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 (easy)

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 are $900\times\frac{1}{2}\times\frac{1}{3}\times\frac{4}{5}\times\frac{9}{10}\times\frac{8}{9}=96$, or $\boxed{\text{A}}$

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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