AMC 8 1990 Problem 1
AMC8 1990 年第 1 题
逻辑推理（跨域）★★☆☆☆
What is the smallest sum of two 3 -digit numbers that can be obtained by placing each of the six digits 4,5,6,7,8,9 in one of the six boxes in this addition problem?
将六个数字 4,5,6,7,8,9 放入此加法算式的六个方框中各一次，能得到的两个三位数之和最小是多少？
- A.947
- B.1037
- C.1047
- D.1056
- E.1245
正确答案 · Correct Answer
C
解析 · Solution
Solution 1
The minimum sum is found by choosing smaller digits to have higher place values. In this case, 4 and 5 should be hundreds digits, 6 and 7 should be tens digits, and 8 and 9 should be units digits. The minimum sum is therefore 468 + 579 = 1047.
Let the two 3 -digit numbers be and . Their sum is 100(a+d)+10(b+e)+(c+f).
To minimize this sum, a+d should be minimized, which happens if . By similar logic using the remaining digits, the sum is minimized if while . The minimum sum is therefore 100(9)+10(13)+(17)=1047
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