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 Rec. Math - Numbers on Cards
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TchrWill
Advanced Member

USA
80 Posts

Posted - 04/30/2013 :  18:04:08  Show Profile  Reply with Quote
Three students are told the following.

1--Three positive integers, x, y and z, are written on three separate cards.
2--The three cards are placed face down on the table in the order of x, y and z.
3--The three numbers sum to 13.
4--The numbers on the cards are in increasing order on the table, i.e., x < y < z.
5--The three students are asked to determine the numbers on the cards.
6--Student #1 is allowed to take a look at card "x" without letting anyone else see the number on the card.
7--After some soul searching, the student states that he is unable to determine the numbers on all three cards.
8--Student #2 is asked to take a look at card "z" without letting anyone else see the number on the card.
9--Shortly, he also states that he is unable to determine the numbers on all three cards.
10--Student #3 is asked to take a look at card "y" without letting anyone else see the number on the card.
11--He also states that he is unable to determine the numbers on all three cards.
12--You, 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?"
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the_hill1962
Advanced Member

USA
1468 Posts

Posted - 05/01/2013 :  12:04:50  Show Profile  Reply with Quote
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"

Edited by - the_hill1962 on 05/01/2013 12:23:27
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TchrWill
Advanced Member

USA
80 Posts

Posted - 05/01/2013 :  12:35:59  Show Profile  Reply with Quote
Congratulations the hill1962. You are right on the money.
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