AMC8 1999

AMC8 1999 · Q24

AMC8 1999 · Q24. It mainly tests Remainders & modular arithmetic, Digit properties (sum of digits, divisibility tests).

When $1999^{2000}$ is divided by 5, the remainder is

$1999^{2000}$ 除以 5 的余数是

(A) 4 4

(B) 3 3

(D) 1 1

(E) 0 0

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Since any positive integer (expressed in base ten) is some multiple of 5 plus its last digit, its remainder when divided by 5 can be obtained by knowing its last digit. Note that $1999^1$ ends in 9, $1999^2$ ends in 1, $1999^3$ ends in 9, $1999^4$ ends in 1, and this alternation of 9 and 1 endings continues with all even powers ending in 1. Therefore, the remainder when $1999^{2000}$ is divided by 5 is 1.

答案（D）：由于任何正整数（用十进制表示）都可以写成某个 5 的倍数加上它的末位数字，因此它除以 5 的余数可以通过末位数字来确定。注意到 $1999^1$ 的末位是 9，$1999^2$ 的末位是 1，$1999^3$ 的末位是 9，$1999^4$ 的末位是 1，并且这种末位在 9 和 1 之间交替的现象会一直持续，所有偶次幂的末位都为 1。因此，$1999^{2000}$ 除以 5 的余数是 1。

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