Difference between revisions of "2018 AMC 12B Problems/Problem 15"
MRENTHUSIASM (talk | contribs) (Prioritized solutions based on elegance and clarity. Let me know if you disagree.) |
MRENTHUSIASM (talk | contribs) (Rewrote Sol 1 a little so the setup is more clear. Will use the same template for Sol 2.) |
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== 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> | ||
− | + | 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> | |
− | ~Plasma_Vortex | + | 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> | ||
== Solution 3 == | == Solution 3 == |
Revision as of 05:15, 28 September 2021
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
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
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.