Difference between revisions of "2018 AMC 12B Problems/Problem 15"
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== Problem == | == Problem == | ||
− | How many odd positive 3-digit integers are divisible by 3 but do not contain the digit 3? | + | How many odd positive <math>3</math>-digit integers are divisible by <math>3</math> but do not contain the digit <math>3</math>? |
<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> | ||
− | == Solution 1 | + | == Solution 1 == |
− | + | Let <math>\underline{ABC}</math> be one such odd positive <math>3</math>-digit integer with hundreds digit <math>A,</math> tens digit <math>B,</math> and ones digit <math>C.</math> Since <math>\underline{ABC}\equiv0\pmod3,</math> we need <math>A+B+C\equiv0\pmod3</math> by the divisibility rule for <math>3.</math> | |
− | + | As <math>A\in\{1,2,4,5,6,7,8,9\}</math> and <math>C\in\{1,5,7,9\},</math> there are <math>8</math> possibilities for <math>A</math> and <math>4</math> possibilities for <math>C.</math> Note that each ordered pair <math>(A,C)</math> determines the value of <math>B</math> modulo <math>3,</math> so <math>B</math> can be any element in one of the sets <math>\{0,6,9\},\{1,4,7\},</math> or <math>\{2,5,8\}.</math> We conclude that there are always <math>3</math> possibilities for <math>B.</math> | |
− | + | By the Multiplication Principle, the answer is <math>8\cdot4\cdot3=\boxed{\textbf{(A) } 96}.</math> | |
− | + | ~Plasma_Vortex ~MRENTHUSIASM | |
− | == Solution 2== | + | == Solution 2 == |
+ | Let <math>\underline{ABC}</math> be one such odd positive <math>3</math>-digit integer with hundreds digit <math>A,</math> tens digit <math>B,</math> and ones digit <math>C.</math> Since <math>\underline{ABC}\equiv0\pmod3,</math> we need <math>A+B+C\equiv0\pmod3</math> by the divisibility rule for <math>3.</math> | ||
− | + | As <math>A\in\{1,2,4,5,6,7,8,9\},B\in\{0,1,2,4,5,6,7,8,9\},</math> and <math>C\in\{1,5,7,9\},</math> note that: | |
+ | <ol style="margin-left: 1.5em;"> | ||
+ | <li>There are <math>2</math> possibilities for <math>A\equiv0\pmod3,</math> namely <math>A=6,9.</math> <p> | ||
+ | There are <math>3</math> possibilities for <math>A\equiv1\pmod3,</math> namely <math>A=1,4,7.</math> <p> | ||
+ | There are <math>3</math> possibilities for <math>A\equiv2\pmod3,</math> namely <math>A=2,5,8.</math> <p> | ||
+ | </li> | ||
+ | <li>There are <math>3</math> possibilities for <math>B\equiv0\pmod3,</math> namely <math>B=0,6,9.</math> <p> | ||
+ | There are <math>3</math> possibilities for <math>B\equiv1\pmod3,</math> namely <math>B=1,4,7.</math> <p> | ||
+ | There are <math>3</math> possibilities for <math>B\equiv2\pmod3,</math> namely <math>B=2,5,8.</math> <p> | ||
+ | </li> | ||
+ | <li>There are <math>1</math> possibility for <math>C\equiv0\pmod3,</math> namely <math>C=9.</math> <p> | ||
+ | There are <math>2</math> possibilities for <math>C\equiv1\pmod3,</math> namely <math>C=1,7.</math> <p> | ||
+ | There are <math>1</math> possibility for <math>C\equiv2\pmod3,</math> namely <math>C=5.</math> <p> | ||
+ | </li> | ||
+ | </ol> | ||
+ | We apply casework to <math>A+B+C\equiv0\pmod3:</math> | ||
+ | <cmath>\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}</cmath> | ||
+ | Together, the answer is <math>6+6+12+9+18+9+18+9+9=\boxed{\textbf{(A) } 96}.</math> | ||
− | + | ~MRENTHUSIASM | |
− | + | == 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> | ||
− | + | 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> | |
− | |||
− | + | Together, the answer is <math>3(12+12+12)-12=\boxed{\textbf{(A) } 96}.</math> | |
− | + | == Solution 4 == | |
− | + | 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> | |
− | + | == Solution 5 == | |
+ | 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\cdot\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 | |
− | + | ==Solution 6== | |
+ | |||
+ | We will start with the numbers that could work. This numbers include _ _ <math>1</math>, _ _ <math>5</math>, _ _ <math>7</math>, _ _ <math>9</math>. Let's work case by case. | ||
+ | |||
+ | Case <math>1</math>: _ _ <math>1</math>: The two blanks could be any number that is <math>2</math> mod <math>3</math> that does not include <math>3</math>. We have <math>24</math> cases for this case (we could count every case). | ||
+ | |||
+ | Case <math>2</math>: _ _ <math>5</math>: The <math>2</math> blanks could be any number that is <math>1</math> mod <math>3</math> that does not include <math>3</math>. But we could see that this case has exactly the same solutions to case <math>1</math> because we have a <math>1-1</math> correspondence. We can do the exact same for case <math>3</math>. | ||
+ | |||
+ | 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 | ||
+ | |||
+ | ==Solution 7== | ||
+ | |||
+ | This problem is solvable by inclusion exclusion principle. There are <math>\frac{999-105}{6} + 1 = 150</math> odd <math>3</math>-digit numbers divisible by <math>3</math>. We consider the number of <math>3</math>-digit numbers divisible by <math>3</math> that contain either <math>1, 2</math> or <math>3</math> digits of <math>3</math>. | ||
+ | |||
+ | For <math>\underline{AB3}</math>, <math>AB</math> is any <math>2</math>-digit number divisible by <math>3</math>, which gives us <math>\frac{99-12}{3} + 1 = 30</math>. For <math>\underline{A3B}</math>, for each odd <math>B</math>, we have <math>3</math> values of <math>A</math> that give a valid case, thus we have <math>5(3) = 15</math> cases. For <math>\underline{3AB}</math>, we also have <math>15</math> cases, but when <math>B=3, 9</math>, <math>A</math> can equal <math>0</math>, so we have <math>17</math> cases. | ||
+ | |||
+ | For <math>\underline{A33}</math>, we have <math>3</math> cases. For <math>\underline{3A3}</math>, we have <math>4</math> cases. For <math>\underline{33A}</math>, we have <math>2</math> cases. Finally, there is just one case for <math>\underline{333}</math>. | ||
+ | |||
+ | By inclusion exclusion principle, we get <math>150 - 30 - 15 - 17 + 3 + 4 + 2 - 1 = \boxed{\textbf{(A) } 96}</math> numbers. | ||
+ | |||
+ | ~Zeric | ||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/mgEZOXgIZXs?t=448 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
==See Also== | ==See Also== |
Latest revision as of 22:26, 4 November 2021
Contents
Problem
How many odd positive -digit integers are divisible by but do not contain the digit ?
Solution 1
Let be one such odd positive -digit integer with hundreds digit tens digit and ones digit Since we need by the divisibility rule for
As and there are possibilities for and possibilities for Note that each ordered pair determines the value of modulo so can be any element in one of the sets or We conclude that there are always possibilities for
By the Multiplication Principle, the answer is
~Plasma_Vortex ~MRENTHUSIASM
Solution 2
Let be one such odd positive -digit integer with hundreds digit tens digit and ones digit Since we need by the divisibility rule for
As and note that:
- There are possibilities for namely
There are possibilities for namely
There are possibilities for namely
- There are possibilities for namely
There are possibilities for namely
There are possibilities for namely
- There are possibility for namely
There are possibilities for namely
There are possibility for namely
We apply casework to Together, the answer is
~MRENTHUSIASM
Solution 3
Analyze that the three-digit integers divisible by start from In the 's, it starts from In the 's, it starts from We see that the units digits is and
Write out the - and -digit multiples of starting from and Count up the ones that meet the conditions. Then, add up and multiply by since there are three sets of three from to Then, subtract the amount that started from since the 's ll contain the digit
Together, the answer is
Solution 4
Consider the number of -digit numbers that do not contain the digit which is For any of these -digit numbers, we can append or to reach a desirable -digit number. However, we have and thus we need to count any -digit number twice. There are total such numbers that have remainder but of them contain so the number we want is Therefore, the final answer is
Solution 5
We need to take care of all restrictions. Ranging from to there are odd -digit numbers. Exactly of these numbers are divisible by which is Of these numbers, have in their ones (units) digit, have in their tens digit, and have in their hundreds digit. Thus, the total number of -digit integers is
~mathpro12345
Solution 6
We will start with the numbers that could work. This numbers include _ _ , _ _ , _ _ , _ _ . Let's work case by case.
Case : _ _ : The two blanks could be any number that is mod that does not include . We have cases for this case (we could count every case).
Case : _ _ : The blanks could be any number that is mod that does not include . But we could see that this case has exactly the same solutions to case because we have a correspondence. We can do the exact same for case .
Cases : _ _ : We need the blanks to be a multiple of , but does not contain 3. We have which also contains numbers. Therefore, we have which is equal to
~Arcticturn
Solution 7
This problem is solvable by inclusion exclusion principle. There are odd -digit numbers divisible by . We consider the number of -digit numbers divisible by that contain either or digits of .
For , is any -digit number divisible by , which gives us . For , for each odd , we have values of that give a valid case, thus we have cases. For , we also have cases, but when , can equal , so we have cases.
For , we have cases. For , we have cases. For , we have cases. Finally, there is just one case for .
By inclusion exclusion principle, we get numbers.
~Zeric
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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