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[1800GAMBLER]

1. Three cards are placed face down in a row on a table. Each has a natural number

(that is, an integer 1, 2, 3, . . .) written on it. The left-hand card has the number x,
the centre card has the number y, and the right-hand card has the number z.

Three students each know all of the following facts:
(a) The numbers are all different.
(b) The numbers add to 14.
(c) x < y < z.

Student A begins by looking at the number x written on the left-hand card. She
announces that she is unable to say what the three numbers are.

Student B now looks at the number z written on the right-hand card. She announces
that she is unable to say what the three numbers are.

Student C now looks at the number y written on the middle card. She announces
that she is unable to say what the three numbers are.

Student A again says that she does not know what the three numbers are.

Student B also repeats that she is unable to say what the three numbers are.

Assuming that all students use perfect logic, show that student C (and also student
A) can now say what the three numbers are.

Answers in white.

[tpir90036]

<font color="white">
OK, well there are only 10 possibile combinations of x,y,z:

[1, 2, 11], [1, 3, 10], [1, 4, 9], [1, 5, 8], [1, 6, 7],
[2, 3, 9], [2, 4, 8], [2, 5, 7], [3, 4, 7], [3, 5, 6]

as soon as student B says she can not tell we can rule out the entries where Z = 11, 10 and 6 since she would instantly know.

[1, 4, 9], [1, 5, 8], [1, 6, 7], [2, 3, 9], [2, 4, 8], [2, 5, 7], [3, 4, 7]

now when student C says she does not know we can rule out Y = 3 &amp; 6.

[1, 4, 9], [1, 5, 8], [2, 4, 8], [2, 5, 7], [3, 4, 7]

since A still does not know we can get rid of X = 3. following that, since B still does not know we can get rid of Z = 9 and the remaining Z = 7.

[1, 5, 8], [2, 4, 8]

so while student B will still be in the dark, students A and C should now know based on their card what the other two are.
</font>

[tolbiny]

<font color="white"> Each of x y and z must be lower than 12.
If z were &gt;=10 then that person would immediatly know what the other two numbers were.
therefor x &lt;=3, y &lt;=6, z &lt;=9, any combination with x&gt;=4 adds up to more than 14, as does y=&gt;7.
y must also be &gt;=3 or else z would have to be &gt;9 (Which we know is false)
The most that x and y can add up to is 7 there fore z must be &gt;=7.
so x&lt;=3, 3&lt;=y&lt;=6, 7&lt;=z&lt;=9. If x =3 then y =4 and z=7, therefore x cannot = 3.
so x&lt;=2,
y cannot be =6 or else z would equal 7 and x=1
so x&lt;=2,3&lt;=y&lt;=5,7&lt;=z&lt;=9.
Y cannot =3 or else z would have to be 9 and x 2, so y is either 4 or 5 and x 1 or 2. </font>


i never was any good at these, i am stuck