AMC12 2002 B

AMC12 2002 B · Q21

AMC12 2002 B · Q21. It mainly tests GCD & LCM, Remainders & modular arithmetic.

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 $\sum_{n=1}^{2001} a_n$.

对于所有小于 $2002$ 的正整数 $n$，令 \begin{eqnarray*} a_n =\left\{ \begin{array}{lr} 11, & \text{如果 }n\ \text{能被 }13\ \text{和 }14\ \text{整除};\\ 13, & \text{如果 }n\ \text{能被 }14\ \text{和 }11\ \text{整除};\\ 14, & \text{如果 }n\ \text{能被 }11\ \text{和 }13\ \text{整除};\\ 0, & \text{否则}. \end{array} \right. \end{eqnarray*} 计算 $\sum_{n=1}^{2001} a_n$。

(A) 448 448

(B) 486 486

(D) 2001 2001

(E) 2002 2002

Answer

Correct choice: (A)

正确答案：（A）

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 $\sum_{n=1}^{2001} a_n = 11 \cdot 10 + 13 \cdot 12 + 14 \cdot 13 = 448 \Rightarrow \mathrm{(A)}$. $\begin{array}{lr} 11, & \text{if\ }n=13 \cdot 14 \cdot k, \quad k = 1,2,\cdots 10;\\ 13, & \text{if\ }n=14 \cdot 11 \cdot k, \quad k = 1,2,\cdots 12;\\ 14, & \text{if\ }n=11 \cdot 13 \cdot k, \quad k = 1,2,\cdots 13;\\ \end{array}$.

由于 $2002 = 11 \cdot 13 \cdot 14$，可得 \begin{eqnarray*} a_n =\left\{ \begin{array}{lr} 11, & \text{如果 }n=13 \cdot 14 \cdot k, \quad k = 1,2,\cdots 10;\\ 13, & \text{如果 }n=14 \cdot 11 \cdot k, \quad k = 1,2,\cdots 12;\\ 14, & \text{如果 }n=11 \cdot 13 \cdot k, \quad k = 1,2,\cdots 13;\\ \end{array} \right. \end{eqnarray*} 因此 $\sum_{n=1}^{2001} a_n = 11 \cdot 10 + 13 \cdot 12 + 14 \cdot 13 = 448 \Rightarrow \mathrm{(A)}$。 $\begin{array}{lr} 11, & \text{如果 }n=13 \cdot 14 \cdot k, \quad k = 1,2,\cdots 10;\\ 13, & \text{如果 }n=14 \cdot 11 \cdot k, \quad k = 1,2,\cdots 12;\\ 14, & \text{如果 }n=11 \cdot 13 \cdot k, \quad k = 1,2,\cdots 13;\\ \end{array}$。

Topics