Difference between revisions of "2018 AMC 12B Problems/Problem 15"
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Note that this isn't a great solution, but a more practical one to achieve the answer. | Note that this isn't a great solution, but a more practical one to achieve the answer. | ||
| − | Notice that there are 300 numbers that | + | 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. |
| + | |||
| + | The restriction for only odd numbers would mean that half the numbers are taken out <math>\Rightarrow 300*\frac{1}{2} = 150</math>. | ||
| + | |||
| + | 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}=120*\frac{9}{10}=108*\frac{8}{9}=96</math>, which is <math>\boxed{A}</math>. | ||
Sol by IronicNinja~ | Sol by IronicNinja~ | ||
Revision as of 19:32, 8 April 2019
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 ll 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)
Solution 3
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, 
, 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 4
Note that this isn't a great solution, but a more practical one to achieve the answer.
Notice that there are numbers that have digits and are divisible by (from to 
). Now one by one apply the restrictions.
The restriction for only odd numbers would mean that half the numbers are taken out 
.
Next, apply the restriction of no s. For the units digit, that would mean multiplying by 
(remember that now you only have odd numbers to choose from).
For the tens that would mean multiplying by 
, and for the hundreds that would mean multiplying by 
(because you cant have 0 here).
Thus, we get 
, which is 
.
Sol by IronicNinja~
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. 
