Difference between revisions of "2005 AMC 10B Problems/Problem 13"

(Solution)
(Solution)
Line 4: Line 4:
 
<math>\mathrm{(A)} 501 \qquad \mathrm{(B)} 668 \qquad \mathrm{(C)} 835 \qquad \mathrm{(D)} 1002 \qquad \mathrm{(E)} 1169 </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> twice 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-2*\left\lfloor\dfrac{2005}{12}\right\rfloor=668+501-167=\boxed{\mathrm{(c)}\ 835}</math> (where <math>\lfloor x \rfloor</math> denotes the [[floor function]]).
+
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> twice 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-2*\left\lfloor\dfrac{2005}{12}\right\rfloor=668+501-334=\boxed{\mathrm{(c)}\ 835}</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}}
 
{{AMC10 box|year=2005|ab=B|num-b=12|num-a=14}}

Revision as of 14:49, 19 October 2011

Problem

How many numbers between $1$ and $2005$ are integer multiples of $3$ or $4$ but not $12$?

$\mathrm{(A)} 501 \qquad \mathrm{(B)} 668 \qquad \mathrm{(C)} 835 \qquad \mathrm{(D)} 1002 \qquad \mathrm{(E)} 1169$

Solution

We can use the Principle of Inclusion-Exclusion to solve the problem as follows: We can count the number of multiples of $3$ that are less than $2005$, add the number of multiples of $4$ that are less than $2005$, and subtract the number of multiples of $12$ twice that are less than $2005$ (since those are counted twice in each of the $3$ and $4$ cases). Calculating, we get $\left\lfloor\dfrac{2005}{3}\right\rfloor+\left\lfloor\dfrac{2005}{4}\right\rfloor-2*\left\lfloor\dfrac{2005}{12}\right\rfloor=668+501-334=\boxed{\mathrm{(c)}\ 835}$ (where $\lfloor x \rfloor$ denotes the floor function).

See Also

2005 AMC 10B (ProblemsAnswer KeyResources)
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