2002 AMC 12B Problems/Problem 21

Problem

For all positive integers $n$ less than $2002$, let

\begin{eqnarray*} a_n =\left\{ \begin{array}{lr} 11, & \text{if\ }n\ \text{is\ divisible\ by\ }13\ \text{and\ }14;\\ 13, & \text{if\ }n\ \text{is\ divisible\ by\ }14\ \text{and\ }11;\\ 14, & \text{if\ }n\ \text{is\ divisible\ by\ }11\ \text{and\ }13;\\ 0, & \text{otherwise}. \end{array} \right. \end{eqnarray*}

Calculate $\sum_{n=1}^{2001} a_n$.

$\mathrm{(A)}\ 448 \qquad\mathrm{(B)}\ 486 \qquad\mathrm{(C)}\ 1560 \qquad\mathrm{(D)}\ 2001 \qquad\mathrm{(E)}\ 2002$

Solution

Since $2002 = 11 \cdot 13 \cdot 14$, it follows that \begin{eqnarray*} a_n =\left\{ \begin{array}{lr} 11, & \text{if\ }n=13 \cdot 14 \cdot k, \quad k = 1,2,\cdots 10;\\ 13, & \text{if\ }n=14 \cdot 11 \cdot k, \quad k = 1,2,\cdots 12;\\ 14, & \text{if\ }n=11 \cdot 13 \cdot k, \quad k = 1,2,\cdots 13;\\ \end{array} \right. \end{eqnarray*}

Thus $\sum_{n=1}^{2001} a_n = 11 \cdot 10 + 13 \cdot 12 + 14 \cdot 13 = 448 \Rightarrow \mathrm{(A)}$.

See also

2002 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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 12 Problems and Solutions

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

Invalid username
Login to AoPS