Difference between revisions of "2018 AMC 12B Problems/Problem 15"
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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 (exclude <math>0</math>). We know what the second digit mod <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{96}</math> (Plasma_Vortex) | 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 (exclude <math>0</math>). We know what the second digit mod <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{96}</math> (Plasma_Vortex) | ||
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+ | == Solution 3== | ||
+ | |||
+ | 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>1 \equiv 7 \equiv 1</math> <math>(mod</math> <math>3)</math>, and thus we need to count any <math>2</math>-digit number <math>\equiv 2</math> <math>(mod</math> <math>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{96}</math>. | ||
==See Also== | ==See Also== |
Revision as of 00:47, 29 July 2018
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 all contain the digit
.
We get:
This gives us:
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
.
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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