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> | ||
− | ==Solution 4 ( | + | 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> |
− | We need to take care of all restrictions. Ranging from <math>101</math> to <math>999</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 | ||
+ | |||
+ | ==Solution 6== | ||
+ | |||
+ | We will start with the numbers that could work. This numbers include _ _ <math>1</math>, _ _ <math>5</math>, _ _ <math>7</math>, _ _ <math>9</math>. Let's work case by case. | ||
+ | |||
+ | Case <math>1</math>: _ _ <math>1</math>: The two blanks could be any number that is <math>2</math> mod <math>3</math> that does not include <math>3</math>. We have <math>24</math> cases for this case (we could count every case). | ||
+ | |||
+ | Case <math>2</math>: _ _ <math>5</math>: The <math>2</math> blanks could be any number that is <math>1</math> mod <math>3</math> that does not include <math>3</math>. But we could see that this case has exactly the same solutions to case <math>1</math> because we have a <math>1-1</math> correspondence. We can do the exact same for case <math>3</math>. | ||
+ | |||
+ | Cases <math>4</math>: _ _ <math>9</math>: We need the blanks to be a multiple of <math>3</math>, but does not contain 3. We have <math>(12, 15, 18, 21, 24, 27, 42, 45, 48, 51, 54, 57, 60, 66, 69, 72, 75, 78, 81, 84, 87, 90, 96, 99)</math> which also contains <math>24</math> numbers. Therefore, we have <math>24 \cdot 4</math> which is equal to <math>\boxed{\textbf{(A) } 96}.</math> | ||
+ | |||
+ | ~Arcticturn | ||
+ | |||
+ | ==Solution 7== | ||
+ | |||
+ | This problem is solvable by inclusion exclusion principle. There are <math>\frac{999-105}{6} + 1 = 150</math> odd <math>3</math>-digit numbers divisible by <math>3</math>. We consider the number of <math>3</math>-digit numbers divisible by <math>3</math> that contain either <math>1, 2</math> or <math>3</math> digits of <math>3</math>. | ||
+ | |||
+ | For <math>\underline{AB3}</math>, <math>AB</math> is any <math>2</math>-digit number divisible by <math>3</math>, which gives us <math>\frac{99-12}{3} + 1 = 30</math>. For <math>\underline{A3B}</math>, for each odd <math>B</math>, we have <math>3</math> values of <math>A</math> that give a valid case, thus we have <math>5(3) = 15</math> cases. For <math>\underline{3AB}</math>, we also have <math>15</math> cases, but when <math>B=3, 9</math>, <math>A</math> can equal <math>0</math>, so we have <math>17</math> cases. | ||
+ | |||
+ | For <math>\underline{A33}</math>, we have <math>3</math> cases. For <math>\underline{3A3}</math>, we have <math>4</math> cases. For <math>\underline{33A}</math>, we have <math>2</math> cases. Finally, there is just one case for <math>\underline{333}</math>. | ||
+ | |||
+ | By inclusion exclusion principle, we get <math>150 - 30 - 15 - 17 + 3 + 4 + 2 - 1 = \boxed{\textbf{(A) } 96}</math> numbers. | ||
+ | |||
+ | ~Zeric | ||
+ | |||
+ | == Video Solution by Omega Learn == | ||
+ | https://youtu.be/mgEZOXgIZXs?t=448 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | ==Video Solution by WhyMath== | ||
+ | https://youtu.be/vdJFrAq0NDY | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== |
Latest revision as of 08:19, 30 May 2023
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
Solution 6
We will start with the numbers that could work. This numbers include _ _ , _ _
, _ _
, _ _
. Let's work case by case.
Case : _ _
: The two blanks could be any number that is
mod
that does not include
. We have
cases for this case (we could count every case).
Case : _ _
: The
blanks could be any number that is
mod
that does not include
. But we could see that this case has exactly the same solutions to case
because we have a
correspondence. We can do the exact same for case
.
Cases : _ _
: We need the blanks to be a multiple of
, but does not contain 3. We have
which also contains
numbers. Therefore, we have
which is equal to
~Arcticturn
Solution 7
This problem is solvable by inclusion exclusion principle. There are odd
-digit numbers divisible by
. We consider the number of
-digit numbers divisible by
that contain either
or
digits of
.
For ,
is any
-digit number divisible by
, which gives us
. For
, for each odd
, we have
values of
that give a valid case, thus we have
cases. For
, we also have
cases, but when
,
can equal
, so we have
cases.
For , we have
cases. For
, we have
cases. For
, we have
cases. Finally, there is just one case for
.
By inclusion exclusion principle, we get numbers.
~Zeric
Video Solution by Omega Learn
https://youtu.be/mgEZOXgIZXs?t=448
~ pi_is_3.14
Video Solution by WhyMath
~savannahsolver
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.