Jump to content
BrainDen.com - Brain Teasers
  • 0

Perfect Pairs/Trios


BMAD
 Share

Question

Consider the following definitions of perfect pairs/trios. If there exist such numbers that fit the definition show how many exist, if no number exists, provide a proof: (Each number is assumed to be distinct)

1. When you add two numbers you get a certain answer. Using the same two numbers, subtract the larger from the smaller and get the same answer in the first sentence.

2. Using three numbers, add the first two numbers together then divide the sum by the third number. The result will be one of the three numbers.

3. Two numbers whose sum is equal to their quotient.

Link to comment
Share on other sites

8 answers to this question

Recommended Posts

  • 0

besides we are assuming that all numbers are distinct

1. I think Phil found the only solution to this one. (0,0) Still thinking about this one.

2. There are three two sets of solutions for this one.

A. x=0, y=0 and z!=0

B. x!=0, x+y!=0 and z=(x+y)/x or (x+y)/y x!=y

C. sqrt(x+y)!=0 and z=+/-sqrt(x+y) x!=y

3.Any point on the curve y2+xy-x=0 except (0,0) and (.5,.5)

Link to comment
Share on other sites

  • 0

1) if you allow zero this is possible. 0+0, 0-0.

2) this is possible if the third number is 2. (x+x)/2=x. or again if both numbers are zero. (0+0)/x = 0. or... (0+1)/1

3) no idea how this is possible.

Remember if it is possible, to what extent is it possible, describe the total range of numbers

Link to comment
Share on other sites

  • 0

1. I think Phil found the only solution to this one. (0,0)

2. There are three sets of solutions for this one.

A. x=0, y=0 and z!=0

B. x!=0, x+y!=0 and z=(x+y)/x or (x+y)/y

C. sqrt(x+y)!=0 and z=+/-sqrt(x+y)

3.Any point on the curve y2+xy-x=0

0,0 is not the only possibility for #1

Link to comment
Share on other sites

  • 0
On 4/22/2014 at 0:09 PM, Rob_G said:

1. I think Phil found the only solution to this one. (0,0) Still thinking about this one.

2. There are three two sets of solutions for this one.

A. x=0, y=0 and z!=0

B. x!=0, x+y!=0 and z=(x+y)/x or (x+y)/y x!=y

C. sqrt(x+y)!=0 and z=+/-sqrt(x+y) x!=y

3.Any point on the curve y2+xy-x=0 except (0,0) and (.5,.5)

(0.5,0.5) works for number 3

Edited by IAMc24
Link to comment
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Answer this question...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

Loading...
 Share

  • Recently Browsing   0 members

    • No registered users viewing this page.
×
×
  • Create New...