Difference between revisions of "2018 AMC 12B Problems/Problem 15"
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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> | ||
| − | == Solution == | + | == Solution 1 (For Dummies) == |
| + | 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 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 all contain the digit <math>3</math>. | ||
| + | |||
| + | We get: <cmath>3(12+12+12)-12.</cmath> | ||
| + | |||
| + | This gives us: <cmath>\boxed{\textbf{(A) } 96}.</cmath> | ||
| + | |||
| + | == Solution 2== | ||
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) | 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) | ||
Revision as of 18:19, 16 February 2018
Problem
How many odd positive 3-digit integers are divisible by 3 but do not contain the digit 3?
Solution 1 (For Dummies)
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 1- and 2-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 all contain the digit .
We get: ![\[3(12+12+12)-12.\]](http://latex.artofproblemsolving.com/b/5/7/b5779733be23b32019f497c660e0f784ee21eb1d.png)
This gives us: ![\[\boxed{\textbf{(A) } 96}.\]](http://latex.artofproblemsolving.com/8/6/c/86c5c1ac998d61ef958efebe45446d2c5765f935.png)
Solution 2
There are 
choices for the last digit (
), and 
choices for the first digit (exclude ). We know what the second digit mod is, so there are choices for it (pick from one of the sets 
). The answer is 
(Plasma_Vortex)
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
