AMC12 2020 A

AMC12 2020 A · Q4

AMC12 2020 A · Q4. It mainly tests Basic counting (rules of product/sum), Divisibility & factors.

How many 4-digit positive integers (that is, integers between 1000 and 9999, inclusive) having only even digits are divisible by 5?

有多少个 4 位正整数（即 1000 到 9999 之间，包括两端）仅由偶数数字组成且能被 5 整除？

(A) 80 80

(B) 100 100

(D) 200 200

(E) 500 500

Answer

Correct choice: (B)

正确答案：（B）

Solution

For an integer to satisfy the given conditions, the thousands digit must lie in $\{2, 4, 6, 8\}$, the hundreds and tens digits must lie in $\{0, 2, 4, 6, 8\}$, and the units digit must be 0. By the Multiplication Principle for counting, the number of choices of digits is $4 \cdot 5 \cdot 5 \cdot 1 = 100$.

要满足条件，千位数字必须在 $\{2, 4, 6, 8\}$ 中，百位和十位数字必须在 $\{0, 2, 4, 6, 8\}$ 中，个位数字必须是 0。根据计数乘法原理，数字选择数为 $4 \cdot 5 \cdot 5 \cdot 1 = 100$。

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