There is a small town with a strange custom. Residents each own a single pebble, painted blue or red, which they always carry in their pockets. Those with blue pebbles always tell the truth, and those with red pebbles always lie. Each resident also knows the color of every other resident's pebble.

The town has an eccentric mayor, who is the sole resident not abiding by the custom. The mayor also carries a pebble (blue or red), but can say both true statements and false statements.

One day the town's bank gets robbed. You are a detective, sent to investigate the crime. You know for sure the getaway car was either blue or red, but you would like to know which. The witnesses at the scene make the following statements:

  • Alex: I didn't get a good look at the getaway car.

  • Beth: The getaway car was the same color as Alex's pebble.

  • Carl: Everyone whose name starts with a vowel has the same color pebble.

  • Doug: Either Beth or Grace (or possibly both) have a pebble that is a different color than my own.

  • Emily: Irina's pebble is red.

  • Floyd: Between me, Grace, and Alex, we have an odd number of red pebbles.

  • Grace: The mayor is not Alex, Doug, Floyd, or me.

  • Hal: My pebble is blue and so is Carl's.

  • Irina: At least one of Emily, Hal, or Carl has a pebble the same color as mine.

Unfortunately, you do not know the mayor's name, nor can you see the color of any witness' pebble. Can you determine the color of the getaway car?

  • $\begingroup$ Do we know that the mayor is in this group? $\endgroup$
    – DylanSp
    Commented Feb 9, 2016 at 19:00
  • $\begingroup$ @DylanSp It is possible the mayor is not one of the witnesses. $\endgroup$ Commented Feb 9, 2016 at 19:04
  • 3
    $\begingroup$ Just a side note, imagine the real-life consequences if people were to actually respond to detectives this way... :-D $\endgroup$
    – Michael
    Commented Feb 9, 2016 at 22:28

2 Answers 2


The car is...


Reasoning (using A through I to indicate the people):

Let's look first at the cycle of C-E-H-I and start by assuming none of them are the mayor.

Then, if E's pebble is blue, she's telling the truth and I's is clearly red. But if E's pebble is red, then she's lying and I's is clearly blue.

Given that, C must be lying since E and I have two different pebbles. So C's pebble is red.

Given that, H's pebble must also be red. And I can't have a red pebble because then the statement would be true, so we've got I with a blue pebble. So that gives us C-Red, E-Red, H-Red, I-Blue. But I's statement is not actually true, which is an impossible situation. That means one of C, E, H or I is the mayor. (I'm not sure it's possible to figure out which one, but it doesn't matter).

Musical interlude (for organization).

Given that, We now know that G is telling the truth and has a blue pebble. That means that D can't have a red pebble or their statement would be true, so D has a blue pebble. Since D's statement must be true and B and G have blue pebbles, then B must have red and be lying. Therefore the getaway car is a different color than A's pebble.

F says that A+F+G yields an odd number of red pebbles. Since G is blue, we either have F blue and A red or F red and A red. Either way, A is red.

Stunning conclusion:

Therefore, since A is red and the car is different than A, the getaway car must be blue.

Nice puzzle!


The car is



C has a blue pebble, but is the mayor and is lying. Suppose C is telling the truth: E cannot be blue, else I would be blue and that contradicts E's statement. If E is red, then I is blue, and that contradicts C's statement. Therefore, C is lying.

Now, suppose C is not the mayor. Then C and H have red pebbles. If I has a blue pebble, then E has a blue pebble, but that contradicts E's statement. If I has a red pebble, then it has the same pebble as C and H, which is also a contradiction. Ergo, C is the mayor.

EDIT: As pointed out in the comments, this doesn't show that the mayor is not H, I or E.

Now that we know that:

G is telling the truth. If D is red, then D's statement is true, so D's pebble must be blue. This means that B != D, so B is red, which means the car is not the color of A.

So what color is A?

We know G is blue. If F is blue, then A must be red to satisfy the number of odd pebbles. If F is red, the number of red pebbles is even, so A must also be red. Thus, A is red, and the car is blue.

  • 1
    $\begingroup$ How do you know that the mayor isn't E, I, or H? $\endgroup$
    – StephenTG
    Commented Feb 9, 2016 at 20:56
  • $\begingroup$ @StephenTG You are right. I thought I had ruled out the others, but something got flipped along the way. Any of them can be mayor. $\endgroup$
    – Matt
    Commented Feb 9, 2016 at 21:38

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