All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Games
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 2 of 10 |
This is a messy proof, but it works in the end.

Since guess 3 has no correct numbers in the right places and guess 4 has no correct numbers in the wrong places, the folowing digits can't have the following numbers:
1: 9,1,4,5
2:6,1,4
3:6,9,4,7
4:6,9,1,8
For guess 1 to have a number in the correct digit, digit 2=3 or digit 3=2. From guess 3, we can similarly deduce that digit 1=7 or digit 4=0

If digit 1=7, 9,1,or 4 must exist, according to guess 4. If 1 or 4 exist, the digits 3 or 2, 7, and 1 or 4 satisfy all guesses with only 3 digits, but the master number has 4 digits. Therefore, if digit 1=7, digit 2=9, digit 3=2, and both 3 and 0 would exist so as to satisfy guess 1 and 3 without including digits 1 and 4. There would be 5 digits, which is not possible. Therefore, digit 1 does not equal 7, and digit 4=0 instead.
Since digit 4=0, the number 4 has no places to be and therefore does not exist.

If digit 3=2, 3 or 1 must exist too, according to guess 1. For 1 to exist, it must be digit 3, which is already equal to number 2. For 3 to exist, the number 1 doesn't, and so the only number left to satisfy guess 3 is number 7, which must be digit 2. Therefore, to satisfy guess 4, it must be 6720, which doesn't work with guess 2. Therefore, digit 3 doesn't equal 2, and instead, digit 2=3.
So far the number is ?3?0

According to guess 4, 6 or 1 exist (only one), and for guess 1, 1 or 2 exist. For 6 and 2 exist, both of them must be in digit 1. Therefore, only number 1 exists, in digit 3
So far, ?310

The last digit must be from guess 2, the last guess to be satisfied. Only 8 works.

The number is 8310!
  Posted by Tristan on 2003-09-20 12:04:59
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (12)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information