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 == | == 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- | + | 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 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 twice 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 |