A Confusing Logic Puzzle

Logic Level 5

Jane, Emily, and Mike are perfect logicians. One day, Jane said, "I'm thinking of four non-negative integers a , b , x , a, b, x, and y y that obey the following conditions: a x 1 a x min ( b , y ) a x 1 + min ( b , y ) b y 1. \begin{aligned} |a - x| &\geq 1\\ |a - x| &\geq \min(b, y)\\ |a - x| &\leq 1 + \min(b, y)\\ |b - y| &\leq 1. \end{aligned} Then Jane said, "I'm going to tell a a and b b to Emily and x x and y y to Mike."

Emily said, "I don't know x x , and I wouldn't know it even if I knew whether y y was the same as b b ."
Mike said, "I don't know a a , and I wouldn't know it even if I knew whether b b was the same as y y ."
Emily said, "I don't know x x , and I wouldn't know it even if I knew whether y y was the same as b b ."
Mike said, "I don't know a a , and I wouldn't know it even if I knew whether b b was the same as y y ."
Emily said, "I don't know x x , and I wouldn't know it even if I knew whether y y was the same as b b ."
Mike said, "I don't know a a , and I wouldn't know it even if I knew whether b b was the same as y y ."

Jane then interrupted, "Stop! You two could go on forever like that!"

Emily said, "I didn't know that."
Mike said, "I didn't know Emily didn't know that. If Emily had said she knew that, I wouldn't know whether Emily knew whether x x is greater than 15 15 . But now, I do."
Emily said, "Before Mike said that, I didn't know whether Mike knew which of a a and x x is bigger."

What is the maximum value of a × b + x × y ? a \times b+x \times y?


The answer is 25.

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1 solution

Rocket Singh
Dec 24, 2017

Let us go step by step. My solution is a mixture of logic and hit & trial. Therefore, if someone else can come up with an elegant solution, please do so.

  1. Given a, b, x and y are non-negative integers, we need to ascertain whether one or more of them are 0.

  2. When Jane interrupts Emily and Mike, we get to know that b is not equal to y, which is corroborated by further statements.

  3. Given the modulus condition of |b - y|, we realize that |b - y| = 1.

  4. If we assume either b or y to be 0, then we would be violating the revelation Emily and Mike had after Jain interrupted.

  5. Emily did not know they could go on forever, so she must have thought the sequence could stop. This is only possible if b = 1.

  6. For the first case, let b = 1 making y = 2, thus ensuring |b - y| = 1, where b < y. This means min(b,y) = b = 1.

Hence, |a - x| >= 1 and |a - x| =< 2. Given point 6., we can conclude that |a - x| = 2. Given all 4 numbers are distinct, we are now sure that neither a nor x is 0 or 1 or 2.

  1. Now (a,x) can be (3,5) or (4,6) or (7,9). Reversing the digits will also render the condition feasible.

  2. Though I wish I could explain further, I simply put the above sequence and got 2 wrongs answers with (3,5) and (4,6). The 3rd attempt was pure serendipity with the solution being (7,9).

*I am not able to gauge why the solution cannot be (9,7) or for that matter any other number. All I am certain is that b = 1 and y = 2. *

Can you elaborate on line 2? Why must that imply that b is not equal to y?

In particular, what is the line of reasoning for "You two could go forever like that"? Note that you're using some of the constraints, but not all of them, which makes it hard to explain why the perfect logicians cannot figure out additional stuff with the additional information.

Calvin Lin Staff - 3 years, 5 months ago

25, like december 25th :O i got 40 something looking for a maximum, i think that the statement of "...wheter x is greater than 15..." needs some clarification

Eliud Alejandro Maldonado Sanchez - 3 years, 5 months ago

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