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

Home > Just Math
Mystery Numbers (Posted on 2003-03-29) Difficulty: 4 of 5
I have chosen 3 different whole numbers less than 10, and have found several simple combinations that lead to perfect squares. Calling the numbers x,y, and z, the following combinations all yield a perfect square as the answer. (A perfect square is a number that has a whole number square root).

(x^2)y + (y^2)z + (z^2)x

x+y+z

z-y-x

xyz

(x^2)(z-1)

There are also several more complicated arrangements that lead to perfect squares, such as

x((z^2)-1)+z((y^2)-3)-x(yz-xy)

2xz+x+z

x((z^2)+x)+z(y^2)-(x^2)(z-y)

Given that these perfect squares are all different, and range between 0 and 100 (inclusive), can you determine x,y, and z?

  Submitted by Cory Taylor    
Rating: 3.5000 (6 votes)
Solution: (Hide)
The expressions, when evaluated (in order) are 100,9,1,0,64,81,49,36. The values for x,y, and z are 4,0, and 5 respectively.

The best starting point to determine the numbers lies in the following 2 equations;

x+y+z -> because you know the square will be less than 25, and more than 5 (possibilities 16,9).

z-y-x -> tells you that z >= (x+y). Also, the only possible squares are 4,1, or 0.

Now consider that the first equation has a maximum value of 100. and you can conclude that the x+y+z equation is equal to 9, as any combination for 16 would cause the first equation to exceed 100, unless one of the numbers is zero, which would then force the other 2 numbers to be 7,9 - but these do not satisfy the z-y-x equation.

Next, knowing that x+y+z is 9, you can conclude that z-y-x must be odd (the sum of the two equations is 2z, which must be even), leaving only 1 as a valid solution.

Again using the sum of these two equations (2z), we get 2z=9+1, or z=5

Looking back at the first equation with the knowledge that z=5, and knowing that (x+y) must be 4, the only combination of (x,y) that leads to a square (100) is x=4 and y=0, completing the problem. There are of course, many other succesful methods for determining the variables' values.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
answerK Sengupta2007-11-23 04:04:14
negativesJack Squat2004-01-13 14:21:26
re(2): Zero?Cory Taylor2003-03-31 17:18:41
re: Zero?Fernando2003-03-31 16:09:41
QuestionZero?Bryan2003-03-31 05:49:08
re: SolutionFernando2003-03-29 20:53:49
SolutionRajat2003-03-29 08:01:09
Solution part 2Fernando2003-03-29 05:52:27
SolutionSolution part 1Fernando2003-03-29 05:49:56
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 (14)
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