AMC12 2003 A

AMC12 2003 A · Q18

AMC12 2003 A · Q18. It mainly tests Divisibility & factors, Remainders & modular arithmetic.

Let $n$ be a $5$-digit number, and let $q$ and $r$ be the quotient and the remainder, respectively, when $n$ is divided by $100$. For how many values of $n$ is $q+r$ divisible by $11$?

设 $n$ 是一个 $5$ 位数，当 $n$ 除以 $100$ 时，商为 $q$，余数为 $r$。有多少个 $n$ 使得 $q+r$ 能被 $11$ 整除？

(A) 8180 8180

(B) 8181 8181

(D) 9000 9000

(E) 9090 9090

Answer

Correct choice: (B)

正确答案：（B）

Solution

When a $5$-digit number is divided by $100$, the first $3$ digits become the quotient, $q$, and the last $2$ digits become the remainder, $r$. Therefore, $q$ can be any integer from $100$ to $999$ inclusive, and $r$ can be any integer from $0$ to $99$ inclusive. For each of the $9\cdot10\cdot10=900$ possible values of $q$, there are at least $\left\lfloor \frac{100}{11} \right\rfloor = 9$ possible values of $r$ such that $q+r \equiv 0\pmod{11}$. Since there is $1$ "extra" possible value of $r$ that is congruent to $0\pmod{11}$, each of the $\left\lfloor \frac{900}{11} \right\rfloor = 81$ values of $q$ that are congruent to $0\pmod{11}$ have $1$ more possible value of $r$ such that $q+r \equiv 0\pmod{11}$. Therefore, the number of possible values of $n$ such that $q+r \equiv 0\pmod{11}$ is $900\cdot9+81\cdot1=8181 \Rightarrow\boxed{(B)}$.

当一个 $5$ 位数除以 $100$ 时，前三位数字成为商 $q$，后两位数字成为余数 $r$。因此，$q$ 可以是从 $100$ 到 $999$（含端点）的任意整数，而 $r$ 可以是从 $0$ 到 $99$（含端点）的任意整数。对于 $q$ 的 $9\cdot10\cdot10=900$ 种可能取值中的每一个，至少有 $\left\lfloor \frac{100}{11} \right\rfloor = 9$ 个 $r$ 的取值使得 $q+r \equiv 0\pmod{11}$。由于还存在 $1$ 个“额外”的 $r$ 取值也满足 $\equiv 0\pmod{11}$，所以对于 $\left\lfloor \frac{900}{11} \right\rfloor = 81$ 个满足 $q \equiv 0\pmod{11}$ 的 $q$ 值，各自会多出 $1$ 个 $r$ 取值使得 $q+r \equiv 0\pmod{11}$。因此满足 $q+r \equiv 0\pmod{11}$ 的 $n$ 的个数为 $900\cdot9+81\cdot1=8181 \Rightarrow\boxed{(B)}$。

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