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扑克牌数字推理

Creato il: 19 gennaio 2025

Risposto usando O1 da Chat01

Domanda

老师发给甲、乙、丙三张扑克牌，
老师说：“你们三个人拿到的数字和是14，并且没有鬼牌。其中，J、Q、K、A分别代表11、12、13、1。你们只能看自己的牌，能猜出其他两人拿到的是什么数字吗？”甲：“我只知道乙丙两人的数字一定不同。”乙：“其实一开始我就知道咱们三个人的数字一定不同。”丙：“我知道你们俩拿到的数字了。”请问，丙拿到什么数字了呢？

Pensiero

Noticing the clues

Piecing together, 甲 throws out a hint about 乙 and 丙 having different numbers while knowing their combined sum is 14.

Assessing 甲's insight

甲 realizes 乙 and 丙 must have different numbers, which means 甲's number doesn't make (14 - A)/2 an integer, preventing simultaneous equal values for 乙 and 丙.

Evaluating even distributions

甲's number, if even, would make 乙 and 丙 potentially equal. Since 甲 knows they are different, his number must be odd or result in odd numbers.

Examining the implications

OK, let me see. If 甲 sees a 2, 乙 and 丙 being equal could be plausible. However, this would make 甲's claim impossible, indicating 甲 did not see a 2.

Evaluating implications

For 甲's statement to hold true, 甲 cannot see an even number card since it could allow for identical 乙 and 丙 values. The sum of the other cards should be odd.

Breaking down possibilities

Interestingly enough, for 甲's statement to be valid, 甲 shouldn't have seen an even number. This conclusion aligns with 甲's position, deeming certain even numbers inconsistent.

Pinpointing the issue

Noticing a problem with 甲 having a 13: the sum of the other two cards would be 1, which isn't possible since standard playing cards only go up to 13.

Charting out options

With 甲 ruling out even numbers and 13, there are six valid possibilities for 甲's card: 1, 3, 5, 7, 9, and 11.

Taking a closer look

乙, seeing his own card, confirms that 甲,乙, and 丙 each have unique cards, ensuring no duplicates exist among them. This deduction relies on the distinctiveness of each card value.

Ensuring card uniqueness

乙’s confidence in each card being unique implies there are no valid scenarios where any two cards share the same value while summing to 14.

Ensuring unique combinations

乙’s certainty that all cards are distinct means no valid scenarios include duplicate cards summing to 14, leading to the conclusion that no such duplicates exist.

Evaluating unique values

By checking each possible B value, I’m ensuring no scenarios exist where 甲, 乙, and 丙 share the same value while summing to 14. This confirms distinct values.

Ensuring distinct sums

A closer look at B=7 shows no duplicates in 甲 and 丙's cards. This confirms all cards are unique, aligning with 乙's certainty.

Laying out the options

Interestingly enough, with B=9, all three cards stay distinct, confirming 乙's view. Checking B=8 and B=10 shows possible duplicates, so they don't fit 乙's position.

Mapping out combinations

OK, let me see. Considering B=7, B=9, and B=11, there are valid combinations of 甲 and 丙's cards that ensure all three remain distinct. B=13 isn't an option.

Evaluating combinations

Now I'll map out valid combinations for 甲, 乙, and 丙's cards that sum to 14, considering the constraints and ensuring all are unique.

Identifying valid triples

First, let's list valid triples (A, B, C) that sum to 14, ensuring all cards are distinct and match the statements given. Each combination must be systematically examined.

Validating combinations

I’m mapping out valid triples (A, B, C) that sum to 14, ensuring all cards are distinct. Invalid combinations are excluded, and distinct pairs are carefully examined for each value of B.

Ensuring validity

Piecing together the valid triples that sum to 14, I confirm they align with 甲's and 乙's statements. Now, I'll check if 丙 can deduce the numbers based on his own card.

Weighing C values

Evaluating unique C values that pinpoint a single valid A and B configuration. This shows if 丙 can identify which numbers 甲 and 乙 have at once.

Evaluating potential outcomes

Analyzing whether 甲's and 乙's perspectives about distinct values fit. Addressing C=4 as a possible clue for combining valid A and B sets while ensuring all values remain unique.

Navigating ambiguity

Interestingly, C=4 doesn't clarify the distribution for 丙, leaving two valid combinations. However, C=2 narrows it down to three combinations, which still leaves room for uncertainty.

Mapping out the options

I determined that the only unique solution for C is when C=6, pigeonholing the values for A and B as (1,7,6). This eliminates all ambiguity.

Thought about card numbers and logic，持续 51 秒

Risposta

解析思路：

  1. 记牌值：
    A = 1, 2～10 = 2～10, J = 11, Q = 12, K = 13。
    三张牌的和为 14，且没有重复的花色/数值限制，仅要求数值和为 14。

  2. 甲的发言：「我只知道乙、丙两人的数字一定不同。」

    • 甲看到自己的牌是 AA，则乙、丙之和为 14−A14 - A。
    • 如果 14−A14 - A 是一个偶数，且能分成 14−A2\frac{14 - A}{2}、14−A2\frac{14 - A}{2}（并且这张数值在 1～13 之间），那么乙、丙有可能拿到相同的数值；甲就无法“确定”乙、丙不同。
    • 只有当 14−A14 - A 不可分成两个相等的合法整数时，甲才能断言“乙、丙一定不同”。

    具体检查 A=1,2,…,13A = 1,2,\dots,13：

    • 若 A=1A=1，则乙+丙=13（为奇数）→ 不可能分成相同的两张牌 → 甲能确定乙≠丙。
    • 若 A=2A=2，则乙+丙=12（可分成 6+6）→ 甲无法确定乙、丙一定不同。
    • 若 A=3A=3，则乙+丙=11（奇数）→ 可以确定不同。
    • 若 A=4A=4，则乙+丙=10（可分 5+5）→ 无法确定。
    • …
    • 如此筛下去，满足甲能确定「乙、丙不同」的 AA 值只能是 1,3,5,7,9,11（其中 13 不行是因为那会导致乙+丙=1，不可能由两张正牌凑出 1）。

  3. 乙的发言：「其实我一开始就知道咱们三个人的数字一定不同。」

    • 乙看到自己的牌是 BB，则甲+丙= 14−B14 - B。
    • 要让乙“从一开始就确定三人互不相同”，意味着：无论甲、丙怎么分配 14−B14 - B，都不可能出现重复(包括甲=丙，或甲=乙，或丙=乙)。
    • 逐一检查 B=1B=1 到 B=13B=13，看是否存在某种分法能出现重复。若能出现，乙就无法一开始就确定三数不同；若完全无法出现重复，才符合乙的发言。
    • 例如：
      • B=7B=7 → 甲+丙=7。想出现重复要么甲=丙=3.5（非整数），要么其中一个=7 另一个=0（也不在 1～13 范围）。都不行，所以没有任何重复方案。→ 乙若看到 7，就能断定三人一定不同。
      • B=9B=9 → 甲+丙=5。想要重复则需 甲=丙=2.5（非整数）或甲(或丙)=9 另一张负数，都不行。→ 也能断定三人一定不同。
      • B=11B=11 → 甲+丙=3。要重复则甲=丙=1.5（非整数）或有负数，也不行。→ 同样能断定不同。
      • 其他数值往往可以出现某种重复(如 B=6 时，甲=6,丙=2 就会出现甲=乙=6 的可能)，于是乙无法“一开始就确定”三数不同。

    由此，满足乙的发言的牌值 BB 只能是 7, 9, 11。

  4. 根据前两条，甲和乙可取的值：

    • 甲 ∈{1,3,5,7,9,11}\in \{1,3,5,7,9,11\}
    • 乙 ∈{7,9,11}\in \{7,9,11\}
      并且三张牌总和要 14、且三人牌互不相同。把可能的组合列出：
    (A=1,B=7,C=6)(A=1,B=9,C=4)(A=1,B=11,C=2)(A=3,B=7,C=4)(A=3,B=9,C=2)(A=5,B=7,C=2)\begin{aligned} & (A=1, B=7, C=6) \\ & (A=1, B=9, C=4) \\ & (A=1, B=11, C=2) \\ & (A=3, B=7, C=4) \\ & (A=3, B=9, C=2) \\ & (A=5, B=7, C=2) \end{aligned}

    这是在满足甲、乙发言前提下，所有“和=14 且三数互不相同”的候选。

  5. 丙的发言：「我知道你们俩拿到的数字了。」

    • 丙只看见自己的那张牌 CC，却能唯一确定甲、乙各拿几。
    • 检查上面候选里不同的丙牌值：
      • C=6C=6 只对应组合 (A,B,C)=(1,7,6)(A,B,C)=(1,7,6)。若丙看到 6，只有这一种可能。→ 丙能立刻确定甲=1、乙=7。
      • C=4C=4 对应 (1,9,4)(1,9,4) 或 (3,7,4)(3,7,4) 两种，都满足之前甲、乙的表态。若丙只看到 4，则无法区分甲究竟是 1 还是 3、乙是 9 还是 7。→ 丙不能唯一确定。
      • C=2C=2 对应 (1,11,2)(1,11,2)、(3,9,2)(3,9,2)、(5,7,2)(5,7,2) 共三种，显然更加无法唯一确定。

    因此，只有当丙实际拿到的牌是 6 时，他才能“肯定”地知道甲、乙分别是什么数。

结论：
根据以上推理，能够让丙在听完甲、乙的对话后确定另外两人的牌值的唯一情形是

丙拿到的牌是 6.\boxed{ \text{丙拿到的牌是 } 6. }

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