Difference between revisions of "2005 AMC 10B Problems/Problem 13"
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== Problem == | == Problem == | ||
+ | How many numbers between <math>1</math> and <math>2005</math> are integer multiples of <math>3</math> or <math>4</math> but not <math>12</math>? | ||
+ | |||
+ | <math>\mathrm{(A)} 501 \qquad \mathrm{(B)} 668 \qquad \mathrm{(C)} 835 \qquad \mathrm{(D)} 1002 \qquad \mathrm{(E)} 1169 </math> | ||
== Solution == | == Solution == | ||
+ | We can use the [[Principle of Inclusion-Exclusion]] to solve the problem as follows: We can count the number of multiples of <math>3</math> that are less than <math>2005</math>, add the number of multiples of <math>4</math> that are less than <math>2005</math>, and subtract the number of multiples of <math>12</math> that are less than <math>2005</math> (since those are counted twice in each of the <math>3</math> and <math>4</math> cases). Calculating, we get <math>\left\lfloor\dfrac{2005}{3}\right\rfloor+\left\lfloor\dfrac{2005}{4}\right\rfloor-\left\lfloor\dfrac{2005}{12}\right\rfloor=668+501-167=\boxed{\mathrm{(D)}\ 1002}</math> (where <math>\lfloor x \rfloor</math> denotes the [[floor function]]). | ||
== See Also == | == See Also == | ||
− | + | {{AMC10 box|year=2005|ab=B|num-b=12|num-a=14}} |
Revision as of 17:12, 10 July 2011
Problem
How many numbers between and
are integer multiples of
or
but not
?
Solution
We can use the Principle of Inclusion-Exclusion to solve the problem as follows: We can count the number of multiples of that are less than
, add the number of multiples of
that are less than
, and subtract the number of multiples of
that are less than
(since those are counted twice in each of the
and
cases). Calculating, we get
(where
denotes the floor function).
See Also
2005 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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 10 Problems and Solutions |