2018 AMC 12B Problems/Problem 15
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 
namely 
namely 
namely 
namely 
namely 
namely 
namely 
namely 
namely 
We apply casework to 
![\[\begin{array}{c|c|c||l} & & & \\ [-2.5ex] \boldsymbol{A\operatorname{mod}3} & \boldsymbol{B\operatorname{mod}3} & \boldsymbol{C\operatorname{mod}3} & \multicolumn{1}{c}{\textbf{Count}} \\ [0.5ex] \hline & & & \\ [-2ex] 0 & 0 & 0 & 2\cdot3\cdot1=6 \\ 0 & 1 & 2 & 2\cdot3\cdot1=6 \\ 0 & 2 & 1 & 2\cdot3\cdot2=12 \\ 1 & 0 & 2 & 3\cdot3\cdot1=9 \\ 1 & 1 & 1 & 3\cdot3\cdot2=18 \\ 1 & 2 & 0 & 3\cdot3\cdot1=9 \\ 2 & 0 & 1 & 3\cdot3\cdot2=18 \\ 2 & 1 & 0 & 3\cdot3\cdot1=9 \\ 2 & 2 & 2 & 3\cdot3\cdot1=9 \end{array}\]](http://latex.artofproblemsolving.com/0/8/7/0877f3cd5e3292ae748bebe8d31d84899ed20ca8.png)
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 ![\[900\cdot\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{4}{5}\cdot\frac{9}{10}\cdot\frac{8}{9}=\boxed{\textbf{(A) } 96}.\]](http://latex.artofproblemsolving.com/f/3/8/f38b8b5912f805d8ea8f12a0b6c4a4618d8c2738.png)
~mathpro12345
Solution 6
We will start with the numbers that could work.
PLEASE DO NOT EDIT THIS SOLUTION AS I AM STILL WORKING ON IT.
~Arcticturn
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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