7Â years, 10Â months ago
7Â years ago
Let us try to enumerate all the cases and count the number of triplets.Â
Without loss of generality, we can assume that a <= b <= c.
Therefore, the maximum value 'a' can take is 4.
In the case where a = 2, 1/b + 1/c = 1/4 or c = 4*b/(4-b)
The only possible natural number solutions to this are b = 5, c = 20; b = 6, c= 12 and b = c = 8
In the case where a = 3, 1/b + 1/c = 5/12 or c = 12*b/(5b-12)
The only possible natural number solutions to this are b = 3, c = 12 and b = 4, c= 6
In the case where a = 4, 1/b + 1/c = 1/2 and the only possible case is a=b=c=4
So, let us write all the possible triplets again (2,5,20); (2,6,12); (2,8,8); (3,3,12); (3,4,6) and (4,4,4)
Hence, the number of triplets possible is 6 + 6 + 3 + 3 + 6 + 1 = 25
7Â years, 10Â months ago
Hi,
Can you please elaborate the question? Is it 1 X b or 1/b and 1 X c or 1/c. Are their any conditions on a,b,c such as they are intergers, natural numbers etc?