Master Number is a game in which one person comes up with a four-digit number (called “the master number”) and another person tries to guess it. Repeated digits in the number are not allowed. Each time the second player guesses a number, the first person grades how good the guess is, writing one X for each correct digit in the correct place, and one O for each correct digit in the wrong place. For instance, if the master number is “2468” and your opponent guesses “1248”, you would score it “XOO”. Note that the location of X’s and O’s in the grade may not correspond with the location of digits in the number they are referring to.

A recent game of Master Number began as follows (the first number in parentheses shows the order of guesses):

(1)   4321   XO
(2)   5678   O
(3)   7140   XO
(4)   6914   X

What is the value of the master number?
(Prove that this is a unique solution.)

8310.

To state the obvious, the first two guesses show us three things:
Exactly two digits from 1-4 are present.
Exactly one digit from 5-8 is present.
Exactly one of 9 or 0 is present.

Assume that 9 is in the number, and 0 is not.
The fourth guess tells us that 1, 4, and 6 are not in the number.
However, that leaves only one correct digit in guess 3 (0, 1, and 4 have been rules out), but we are told that there are two.

Therefore, 0 is in the number, and 9 is not.
For guess three, then, exactly one of 1, 4, and 7 is present in the correct number.

If it is one, then 4 is not, and either 2 or 3 is in the number (from #1). Also, 7 is not, nor is 6 (#4), and one of either 5 or 8 is in the number.
Further, the X in #4 tells us that 1 would be the third digit, and the X in #3 tells us that 0 is the last digit.
There is an X and an O in #1; the O must correspond to the 1 (since we already know that it is the third digit), and the 2 or 3 will be in the correct position. It can't be the 2 in the third spot, so 3 must be the second digit.
By elimination, the 5 or the 8 must be the first digit.
If it were the 5, the response to #2 would be an X, but it's not, so it must be the incorrectly placed 8.
Therefore, 8310 is a solution that works.

We are told that this is unique, but since the solution is based on an assumption, we will rule out the other possibilities.

If we assume that a 4 is in the number, and 1 and 7 are not, then from clue 4, the digit 4 must be in position 4 (kinda nice how that works). However, since the 4 in guess three is in a different spot, the O corresponds to that, and the X represents the 0 that we already know is in the number. However, that means that both 4 and 0 occupy the fourth spot; this is not possible, so the assumption is false.

Finally, if we assume that 7 is in the number, and 1 and 4 are not, then a 6 must be in the number from guess #4 (1, 4, and 9 are eliminated already). However, that leaves us with 6 and 7 in the number, which by clue two we know cannot be the case. Again, we have a contradiction and know that the assumption is false.

Therefore, 8310 is the only number that works.

Posted by DJ on 2003-09-20 16:11:23