AMC 8 2015 Problem 23
AMC8 2015 年第 23 题
逻辑推理（跨域）★★★★☆
Tom has twelve slips of paper which he wants to put into five cups labeled , , , , . He wants the sum of the numbers on the slips in each cup to be an integer. Furthermore, he wants the five integers to be consecutive and increasing from to . The numbers on the papers are 2, 2, 2, 2.5, 2.5, 3, 3, 3, 3, 3.5, 4, and 4.5 . If a slip with 2 goes into cup and a slip with 3 goes into cup , then the slip with 3.5 must go into what cup?
Tom 想把 12 张纸条放进 5 个标有 A，B，C，D，E 的杯子中。他希望每个杯子中的纸条上的数字之和均为整数。并且，他希望这 5 个整数是连续的整数，从 A 到 E 递增。这 12 张纸条上的数字分别是 2，2，2，2.5，2.5，3，3，3，3，3.5，4，4.5．如果写有 2 的那张纸条放进了杯子 E，写有 3 的那张纸条放进了杯子 B，那么写有 3.5 的那张纸一定放进了哪个杯子？
- A.A
- B.B
- C.C
- D.D
- E.E
正确答案 · Correct Answer
D
解析 · Solution
解法1
常规解法 · 定各杯和再安置
总和 =2×3+2.5×2+3×4+3.5+4+4.5=35，五个连续整数和 35 ⇒ 中间 7 ⇒ 各杯和 $A5,B6,C7,D8,E9$。带 $.5$ 的四张（2.5,2.5,3.5,4.5）必须在杯内偶数张地凑整：两张 2.5 配成 5 落 A；剩 3.5,4.5 只能互配 =8 落 D。故 。
巧解 · 只看小数部分配对
要每杯整数，带 $.5$ 的纸条必须成对同杯。2.5,2.5 自然一对（和 5）；剩下 3.5,4.5 没有别的 $.5$ 可搭，只能彼此成对 ⇒ 同杯、和 8=D。一步定 ，无需排满全部。
Solution 1
Elementary · fix each cup sum, then place
The slips sum to . Five consecutive integers summing to 35 have middle value 7, so the cup sums are A=5,B=6,C=7,D=8,E=9. The four half-slips (2.5,2.5,3.5,4.5) must pair up within a cup to make integers: the two 2.5's pair to 5 (cup ), leaving 3.5 and 4.5 to pair with each other for 8 (cup ). So the 3.5 slip goes in cup .
Slick · pair by fractional part
For every cup total to be an integer, the half-slips must come in pairs sharing a cup. The two 2.5's form one pair (sum 5); that leaves 3.5 and 4.5, which have no other half-slip to join, so they must pair together — sum 8, i.e. cup . This pins at once, without filling in the rest.
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