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Master Number (Posted on 2003-09-20) Difficulty: 4 of 5
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.)

See The Solution Submitted by Bryan    
Rating: 3.8889 (9 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution solution | Comment 1 of 9
By the transition from (3) to (4) either 7 or 0 is present while 6 or 9 is absent.

From (1) to (4), 3 or 2 is present.

From (1) and (2), either 0 or 9 is present, but not both (to make the fourth present digit, from the 3 accounted for in these two lines). But from the transition from (3) to (4) it must be the 0 that's present, not the 9.

If 0 and 7 were both present then neither 1 nor 4 would be (by (3)), and so the absence of 9 would mean that 6 would have to be present to satisfy (4). But then with 6 and 7 both present, (2) would not be correct. Therefore, since 0 is present, 7 is absent.

Then, by (3) either 1 or 4 is present.

By (4), then, 6 is absent.

Then, by (2), either 5 or 8 is present but not both.

Summary so far:
0 is present
2 or 3 is present
5 or 8 is present
1 or 4 is present

As this accounts for all four, all the "or"s are exclusive.

Either the 1 occupies the 3rd position or the 4 occupies the fourth, by the X in (4).

So in (3) neither of these can be the one in the correct position, so it must be the 0 in the 4th position.

That means the 4 can't be in the 4th position, so it must be the 1 in the 3rd position that's correct in (4).

Then, by (1), the 1 is the one that's in the incorrect position, and since the 2 cannot be in the 3rd position, already assigned to 1, it must be the 3 in the 2nd position that's correct.

The only remaining position to be filled is the 1st. But (2) has 5 in the first position and the only correct number there is in the wrong position, so it must be the 8 that belongs in the 1st position.

So in sum the answer becomes:

8310


  Posted by Charlie on 2003-09-20 10:53:44
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