For all positive integers n less than 2002, let a_n=\begincases 11, & if n is divisible by 13 and 14;\ 13, & if n is divisible by 14 and 11;\ 14, & if n is divisible by 11 and 13;\ 0, & otherwise. \endcases Calculate \sum_n=1^2001 a_n.

2002 AMC 12B Problem 21 — Solution

2002 AMC 12B Problem 21

Below is the professionally curated solution for Problem 21 of the 2002 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2002 AMC 12B solutions, or check the answer key.

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Concepts:divisibility least common multiple multiple

Difficulty rating: 1650

21.

For all positive integers $n$ less than $2002,$ let $a_n=\begin{cases} 11, & \text{if } n \text{ is divisible by } 13 \text{ and } 14;\\ 13, & \text{if } n \text{ is divisible by } 14 \text{ and } 11;\\ 14, & \text{if } n \text{ is divisible by } 11 \text{ and } 13;\\ 0, & \text{otherwise.} \end{cases}$ Calculate $\displaystyle\sum_{n=1}^{2001} a_n.$

$448$

$486$

$1560$

$2001$

$2002$

Solution:

Since $2002=11\cdot13\cdot14$ with $11,$ $13,$ $14$ pairwise coprime, $a_n=11$ when $182\mid n,$ $a_n=13$ when $154\mid n,$ and $a_n=14$ when $143\mid n$ (and no $n\lt2002$ is divisible by all three).

For $n\le2001$ there are $10$ multiples of $182,$ $12$ of $154,$ and $13$ of $143.$ So the sum is $11\cdot10+13\cdot12+14\cdot13=110+156+182=448.$

Thus, the correct answer is A.

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