2002 AMC 12B Problem 21

Question

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

Because

11,

13,

14

are pairwise coprime, “divisible by two of them” means divisible by their product

Count multiples of

182,

154,

143

below

2002

; no number under

2002

is divisible by all three

Since

2002=11\cdot13\cdot14

with

11,

13,

14

pairwise coprime,

a_n=11

when

n

is divisible by

182,

a_n=13

when

n

is divisible by

154,

and

a_n=14

when

n

is divisible by

143

(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.

Problem 21 in other years