Perfect Pairs/Trios

[BMAD]

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.

[phil1882]

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.

[BMAD]

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

[Rob_G]

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

[BMAD]

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

[BMAD]

besides we are assuming that all numbers are distinct

[Rob_G]

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)

[Rob_G]

Oh...

x=0

y<0

[BMAD]

counting posts #7 & #8 together that is

[IAMc24]

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 December 6, 2016 by IAMc24