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 | |
− | + | == Video Solution == | |
+ | https://youtu.be/mgEZOXgIZXs?t=448 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
==See Also== | ==See Also== |
Latest revision as of 07:57, 28 September 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
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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