AMC 8 2026 Problem 16
AMC8 2026 年第 16 题
整除与因数★★★☆☆
Consider all positive four-digit integers consisting of only even digits. What fraction of these integers are divisible by 4?
考虑所有由偶数数字组成的四位正整数。其中有多少比例的数能被 4 整除？
- A.
- B.
- C.
- D.
- E.
正确答案 · Correct Answer
D
解析 · Solution
解法1
被 4 整除只需看末两位。因为所有数字均为偶数，个位只能是 0、2、4、6、8。其中只有末位为 0、4、8 的数（在十位也为偶数的前提下）能被 4 整除。5 个可能个位中有 3 个可行，比例为 。
枚举所有满足条件的数。末两位必须为偶数且构成 4 的倍数，共 15 种：00,04,08,20,24,28,40,44,48,60,64,68,80,84,88。首位有 4 种选择（2,4,6,8），第二位 5 种。有效总数：4×5×15=300。全部四位偶数：4×5×5×5=500。比例： 。
Solution 1
A number is divisible by 4 if its last two digits form a multiple of 4. Since all digits are even, the ones digit must be 0, 2, 4, 6, or 8. But only numbers ending in 0, 4, or 8 can be divisible by 4 regardless of the tens digit (which is also even). So 3 out of 5 possible last digits work. The first two digits can be anything even (with the first digit nonzero), so the fraction is .
Count all valid numbers. Last two digits must be even AND form a multiple of 4. The 15 valid pairs are: 00,04,08,20,24,28,40,44,48,60,64,68,80,84,88. First digit: 4 choices (2,4,6,8; not 0). Second digit: 5 choices. Total valid: 4×5×15 = 300. Total possible: 4×5×5×5 = 500. Fraction: .
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