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Ramsey_KJ
Member since Sep2304

Nov0604, 07:29 PM (EST) 

1. "RE: Back to back pairs"
In response to message #0

If the pairs are denoted A(1)A(2) and B(1)B(2) then A(1)*B(1) must equal A(2)*B(2). This implies that the if the product A(1)*B(1) = P(1)^X * P(2)^Y * P(3)^Z, then the sum of X,Y and Z must be greater than 2 but the exponent of 1 can only be 1 since higher exponents give only 1. Because of the limitation that A(1), A(2), B(1), and B(2) must all be less than 10, there are only a few possible combinations, e.g. for P(n) > 3 there are no possible solutions and P(n) must equal 1, 2 or 3. It is then easy to determine the few possible combinations of A(1)(A2) and B(1)B(2) {of course you ignore the reversed orders}. I get a total of 14 including the 4 that you gave. Thanks for your problem. Have a Good Day KJ Ramsey 

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Owen
guest

Dec2304, 12:30 PM (EST) 

2. "RE: Back to back pairs"
In response to message #0

According to the previous reply if the pairs are denoted A(1)A(2) and B(1)B(2) then A(1)*B(1) must equal A(2)*B(2) where A(1), A(2), B(1), and B(2) can only be positive integer less than 10 (1 to 9). Those integers can produce the ratio 1/2 1/3 2/3 1/4 and 3/4 where A(1)A(2) and B(1)B(2) can be created ratio 1/2 : (1,2) (2,4) (3,6) (4,8) ratio 1/3 : (1,3) (2,6) (3,9) ratio 1/4 : (1,4) (2,8) ratio 2/3 : (2,3) (4,6) (6,9) ratio 3/4 : (3,4) (6,8) Then, we can built a pair of number by selecting 2 members within each ratio group ratio 1/2 : select 2 out of 4 = 4C2 = 6 ways ratio 1/3 : select 2 out of 3 = 3C2 = 3 ways ratio 1/4 : select 2 out of 2 = 2C2 = 1 way ratio 2/3 : select 2 out of 3 = 3C2 = 3 ways ratio 3/4 : select 2 out of 2 = 2C2 = 1 way 14 ways in total and they are ... 12 42 12 63 12 84 13 62 13 93 14 82 23 64 23 96 24 63 24 84 26 93 34 86 36 84 46 96 :) 

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