1996 AMC 8 第 14 题

先试着解答 1996 AMC 8 第 14 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 1996 AMC 8 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

14.

从集合 {1,2,3,4,5,6,7,8,9}\{1, 2, 3, 4, 5, 6, 7, 8, 9\} 中选出六个不同数字,填入下图的方格中,使竖直列中各数之和为 2323,水平行中各数之和为 1212。所用六个数字之和是
Six different digits from the set {1,2,3,4,5,6,7,8,9}\{1, 2, 3, 4, 5, 6, 7, 8, 9\} are placed in the squares in the figure shown so that the sum of the entries in the vertical column is 2323 and the sum of the entries in the horizontal row is 12.12. The sum of the six digits used is
2727
2929
3131
3333
3535
答案:B
知识点:数字逻辑推理
小提示:
三个不同数字和为 2323,只能是 66、88、99。
Three distinct digits summing to 2323 must be 6,6, 8,8, and 99
大提示:
共享方格在两次求和中都被计算了,所以六个数字总和等于 23+1223 + 12 减去共享数字。
The shared square is counted in both sums, so the total of the six digits is 23+1223 + 12 minus the shared digit
解答:
从 11 到 99 中选三个不同数字和为 2323,只能是 66、88、99。水平行另外三个数字最小为 1+2+3=61 + 2 + 3 = 6,所以共享方格至多是 12−6=612 - 6 = 6。因此共享数字是 66。
六个数字是 66、88、99、11、22、33,总和为 2929。等价地,23+12−6=2923 + 12 - 6 = 29。
所以正确答案是 B。
Three distinct digits from 11 through 99 summing to 2323 must be 6,6, 8,8, and 9.9. The row’s other three digits are at least 1+2+3=6,1 + 2 + 3 = 6, so the shared square (belonging to both the column and the row) is at most 12−6=6.12 - 6 = 6. Hence the shared digit is 6.6.
The six digits are then 6,6, 8,8, 9,9, 1,1, 2,2, and 3,3, whose sum is 29.29. Equivalently, 23+12−6=29.23 + 12 - 6 = 29.
Thus, the correct answer is B .
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