AMC10 2003 A

AMC10 2003 A · Q25

AMC10 2003 A · Q25. It mainly tests Divisibility & factors, Remainders & modular arithmetic.

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

设 $n$ 是一个五位数，当 $n$ 除以 100 得到商 $q$ 和余数 $r$。有几个 $n$ 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

(B) Note that $n = 100q + r = q + r + 99q$. Hence $q + r$ is divisible by $11$ if and only if $n$ is divisible by $11$. Since $10,000 \le n \le 99,999$, there are $$ \left\lfloor \frac{99999}{11} \right\rfloor - \left\lfloor \frac{9999}{11} \right\rfloor = 9090 - 909 = 8181 $$ such numbers.

（B）注意 $n = 100q + r = q + r + 99q$。因此，当且仅当 $n$ 能被 $11$ 整除时，$q + r$ 才能被 $11$ 整除。由于 $10,000 \le n \le 99,999$，这样的数有 $$ \left\lfloor \frac{99999}{11} \right\rfloor - \left\lfloor \frac{9999}{11} \right\rfloor = 9090 - 909 = 8181 $$ 个。

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