perplexus.info :: Logic : Hustler Duo Determination

Hustler Duo Determination (Posted on 2014-01-13)

Two hustlers are working a scam together. Five suspects Andy, Brad, Cal, Dave and Ethan have been identified. Each of the suspects makes a statement. The two guilty ones make false statements. The other three make true statements.

The statements are as follows:

Andy: Ethan is not one of the guilty ones.
Brad: Cal is not guilty.
Cal: Dave is innocent.
Dave: If Brad is not guilty, then Andy is guilty.
Ethan: Cal is guilty.

Which two out of the five suspects are the hustlers?

No Solution Yet Submitted by K Sengupta

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Comments: ( Back to comment list | You must be logged in to post comments.)

Too much info

Comment 2 of 2 |

B and E are contradicting, so one of them is lying and one is telling the truth.

If B is lying, then C is guilty, but then C is lying so D is guilty also.

This is too many hustlers, so B is not lying so C is not lying and E is.

Then A is lying about E, so our two hustlers are A and E.

Note that we have not used D's statement.

We expect that D is telling the truth, and check confirms this.

Posted by Steve Herman on 2017-04-03 10:30:40

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