A Math Riddle
Reposting this riddle from Tanya Khovanova's Math Blog, so that I can put the solution behind a link:
Four wizards, A, B, C, and D, were given three cards each. They were told that the cards had numbers from 1 to 12 written without repeats. The wizards only knew their own three numbers and had the following exchange.
A: "I have number 8 on one of my cards."
B: "All my numbers are prime."
C: "All my numbers are composite. Moreover, they all have a common prime factor."
D: "Then I know the cards of each of you."
Given that every wizard told the truth, what cards does A have?
Solution: There are five primes {2, 3, 5, 7, 11} in the set, so D must have two of them to deduce which remaining three B has.
C could have some subset of {4, 6, 10, 12} (composites with a factor of 2, excluding 8) or {6, 9, 12} (composites with a factor of 3). ({5, 10} doesn't have enough cards, and there aren't composites with a higher prime factor.)
One card can't eliminate "factors of two" from consideration, so it needs to be a card that eliminates both one of the factors of three (ruling out that C has those) and one of the factors of two (making it clear exactly which ones C has). That could be 6 or 12.
Either way, the last three players have the primes and the factors of two (except for 8) between the three of them, leaving A with 1, 8, 9. This entry was originally posted on Dreamwith.
comments are there.
Four wizards, A, B, C, and D, were given three cards each. They were told that the cards had numbers from 1 to 12 written without repeats. The wizards only knew their own three numbers and had the following exchange.
A: "I have number 8 on one of my cards."
B: "All my numbers are prime."
C: "All my numbers are composite. Moreover, they all have a common prime factor."
D: "Then I know the cards of each of you."
Given that every wizard told the truth, what cards does A have?
Solution: There are five primes {2, 3, 5, 7, 11} in the set, so D must have two of them to deduce which remaining three B has.
C could have some subset of {4, 6, 10, 12} (composites with a factor of 2, excluding 8) or {6, 9, 12} (composites with a factor of 3). ({5, 10} doesn't have enough cards, and there aren't composites with a higher prime factor.)
One card can't eliminate "factors of two" from consideration, so it needs to be a card that eliminates both one of the factors of three (ruling out that C has those) and one of the factors of two (making it clear exactly which ones C has). That could be 6 or 12.
Either way, the last three players have the primes and the factors of two (except for 8) between the three of them, leaving A with 1, 8, 9. This entry was originally posted on Dreamwith.
comments are there.