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

Revision as of 23:47, 28 July 2018

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 (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 all 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 $72+24= \boxed{96}$ .

@@ Line 16: / Line 16: @@
 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)
+== Solution 3==
+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>.
 ==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
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