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2002 AMC 12B Problems/Problem 21

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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, &amp...

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, &...

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 (Problems • 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
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