AMC10 2020 A

AMC10 2020 A · Q6

AMC10 2020 A · Q6. It mainly tests Basic counting (rules of product/sum), Digit properties (sum of digits, divisibility tests).

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

有多少个仅由偶数位组成的4位正整数（即1000到9999之间的整数，包含两端）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 $\lbrace 2, 4, 6, 8\rbrace$, the hundreds and tens digits must lie in $\lbrace 0, 2, 4, 6, 8\rbrace$, and the units digit must be 0. By the Multiplication Principle for counting, the number of choices of digits is $4 \cdot 5 \cdot 5 \cdot 1 = 100$.

要满足条件的整数，其千位数字必须在$\lbrace 2, 4, 6, 8\rbrace$中，百位和十位数字必须在$\lbrace 0, 2, 4, 6, 8\rbrace$中，个位数字必须是0。根据计数中的乘法原理，数字的选择数量是$4 \cdot 5 \cdot 5 \cdot 1 = 100$。

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