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

2002 AMC 12B (Problems • Answer Key • Resources)
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

Page

Toolbox

Search

2002 AMC 12B Problems/Problem 21

Problem

Solution

See also