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

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<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 ==
 
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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 23:03, 17 October 2017

Problem

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

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

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

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