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 1== | + | == Solution 1 == |
− | + | ||
+ | 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 (excluding <math>0</math> and <math>3</math>). We know what the second digit modulo <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{\textbf{(A) } 96}.</math> | ||
− | + | ~Plasma_Vortex | |
− | + | == Solution 2 == | |
− | + | == 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> | ||
− | + | 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 get <cmath>3(12+12+12)-12=\boxed{\textbf{(A) } 96}.</cmath> | |
− | == Solution | + | == 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, <math> | + | 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 | + | == 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 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 | + | 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\times\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 | ~mathpro12345 |
Revision as of 18:26, 27 September 2021
Contents
Problem
How many odd positive -digit integers are divisible by
but do not contain the digit
?
Solution 1
There are choices for the last digit (
), and
choices for the first digit (excluding
and
). We know what the second digit modulo
is, so there are
choices for it (pick from one of the sets
). The answer is
~Plasma_Vortex
Solution 2
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
We get
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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.