T O P I C R E V I E W 
TchrWill 
Posted  04/30/2013 : 18:04:08 Three students are told the following.
1Three positive integers, x, y and z, are written on three separate cards. 2The three cards are placed face down on the table in the order of x, y and z. 3The three numbers sum to 13. 4The numbers on the cards are in increasing order on the table, i.e., x < y < z. 5The three students are asked to determine the numbers on the cards. 6Student #1 is allowed to take a look at card "x" without letting anyone else see the number on the card. 7After some soul searching, the student states that he is unable to determine the numbers on all three cards. 8Student #2 is asked to take a look at card "z" without letting anyone else see the number on the card. 9Shortly, he also states that he is unable to determine the numbers on all three cards. 10Student #3 is asked to take a look at card "y" without letting anyone else see the number on the card. 11He also states that he is unable to determine the numbers on all three cards. 12You, as an observer to these events, and not having seen the numbers on each card, are asked "What is the number on the middle card?"

2 L A T E S T R E P L I E S (Newest First) 
TchrWill 
Posted  05/01/2013 : 12:35:59 Congratulations the hill1962. You are right on the money. 
the_hill1962 
Posted  05/01/2013 : 12:04:50 The number on the middle card is 4. I arrived at this by listing out the possible combinations: 1. 1 2 10 2. 1 3 9 3. 1 4 8 4. 1 5 7 5. 2 3 8 6. 2 4 7 7. 2 5 6 8. 3 4 6 Then, eliminate #8 because if it was 3 4 6, student 1 would know the cards since that is the only possibility when he sees the 3. Now, student 2 has the following possible combinations: 1. 1 2 10 2. 1 3 9 3. 1 4 8 4. 1 5 7 5. 2 3 8 6. 2 4 7 7. 2 5 6 Eliminate #1, #2 and #7 because each of these is unique given the card he/she looks at (i.e. an 8 or a 7 are the only numbers that have more than one possibility). Now, student 3 has the following possible combinations: 3. 1 4 8 4. 1 5 7 5. 2 3 8 6. 2 4 7 Eliminate #4 and #5 since those are unique (i.e. he/she would know it if the middle card is a 5 and also if the middle card is a 3. So, the only two possibilities left are 3. 1 4 8 6. 2 4 7 Each having the middle card be "4"


