Difference between revisions of "2018 AMC 12B Problems/Problem 15"
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==Solution 4 (easy)== | ==Solution 4 (easy)== | ||
We need to take care of all restrictions. Ranging from <math>101</math> to <math>999</math>, there are <math>450</math> odd 3-digit numbers. Exactly <math>\frac{1}{3}</math> of these numbers are divisible by 3, which is <math>450\times\frac{1}{3}=150</math>. Of these 150 numbers, <math>\frac{4}{5}</math> <math>\textbf{do not}</math> have 3 in their ones (units) digit, <math>\frac{9}{10}</math> <math>\textbf{do not}</math> have 3 in their tens digit, and <math>\frac{8}{9}</math> <math>\textbf{do not}</math> have 3 in their hundreds digit. Thus, the total number of 3 digit integers are <math>900\times\frac{1}{2}\times\frac{1}{3}\times\frac{4}{5}\times\frac{9}{10}\times\frac{8}{9}=96</math>, or <math>\boxed{\text{A}}</math> | We need to take care of all restrictions. Ranging from <math>101</math> to <math>999</math>, there are <math>450</math> odd 3-digit numbers. Exactly <math>\frac{1}{3}</math> of these numbers are divisible by 3, which is <math>450\times\frac{1}{3}=150</math>. Of these 150 numbers, <math>\frac{4}{5}</math> <math>\textbf{do not}</math> have 3 in their ones (units) digit, <math>\frac{9}{10}</math> <math>\textbf{do not}</math> have 3 in their tens digit, and <math>\frac{8}{9}</math> <math>\textbf{do not}</math> have 3 in their hundreds digit. Thus, the total number of 3 digit integers are <math>900\times\frac{1}{2}\times\frac{1}{3}\times\frac{4}{5}\times\frac{9}{10}\times\frac{8}{9}=96</math>, or <math>\boxed{\text{A}}</math> | ||
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| + | ~mathpro12345 | ||
==See Also== | ==See Also== | ||
Revision as of 20:24, 18 December 2020
Problem
How many odd positive 3-digit integers are divisible by 3 but do not contain the digit 3?
Solution 1
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 (easy)
We need to take care of all restrictions. Ranging from 
to 
, there are 
odd 3-digit numbers. Exactly 
of these numbers are divisible by 3, which is 
. Of these 150 numbers, 
have 3 in their ones (units) digit, 
have 3 in their tens digit, and 
have 3 in their hundreds digit. Thus, the total number of 3 digit integers are 
, or 
~mathpro12345
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. 
