2005 AMC 10B Problems/Problem 13

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).

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2005 AMC 10B Problems/Problem 13

Problem

Solution

See Also