Difference between revisions of "2002 AMC 12B Problems/Problem 21"

Revision as of 09:22, 4 July 2013

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
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All AMC 12 Problems and Solutions

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Difference between revisions of "2002 AMC 12B Problems/Problem 21"

Revision as of 09:22, 4 July 2013

Problem

Solution

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