Question

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

$448$

$486$

$1560$

$2001$

$2002$

Answer: A

Concepts:divisibility least common multiple multiple

Difficulty rating: 1650

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 $\begin{gathered} 11\cdot10+13\cdot12+14\cdot13 \\ {}=110+156+182 \\ {}=448. \end{gathered}$

Thus, the correct answer is A.

2002 AMC 12B Problem 21

Answer: A

Difficulty rating: 1650

Solution:

Problem 21 in Other Years