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# Perfect Pairs/Trios

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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.

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Posted · Report post

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.

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Posted · Report post

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

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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

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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

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Posted · Report post

besides we are assuming that all numbers are distinct

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Posted · Report post

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)

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Posted · Report post

Oh...

x=0

y<0

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Posted · Report post

counting posts #7 & #8 together that is

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