Math question : 1/A + 1/B +1/C = 3/7?

1/A + 1/B +1/C = 3/7


A, B, C are all different numbers.


I know the answers. A=4, B=7,C=28, but I don't know the steps.


Is there a formula or a logical way to solve this problem?

Math question : 1/A + 1/B +1/C = 3/7?
there is a logical way to indicate solutions





find smallest A such that 1/ A %26lt; 3/7


A= 4


Now if A = 4


1/B+1/C = 3/7 -1/4 = 5/28





5/28 = (1+4)/28 = 1/28+1/7


(2+3)/28 no


if we take A = 5 we may get more solutions





1/B+1/C= 3/7-1/5= 8/35


= (1+7)/35 = 1/35+1/5 allow if duplicates permitted else no


no other





take A =6 and we get





1/B+1/C = 3/7-1/6 = 11/42


no other solution as 11/42 is not sum of reciprocal
Reply:Normally you 'clear the denominator' by multipling by the product of the denoms (ABC)





-%26gt; BC+AC+AB=3ABC/7 , but I don't see how that helps here.l





How about reciprocating the whole thing:





1/[1/A+1/B+1/C] = 7/3





[A+B+C]/ABC = 7/3


[A+B+C]/(ABC/3) = 7 , which says that whatever ABC is it must have


Possible choices are


A,B,C, ABC,


1,3,7, 21 , no sol


1,4,7, 28, -%26lt; bingo....





So, the trick is to notice sum must have 7 as a factor, and the product, ABC, must have 3 as a factor.
Reply:the LCD will be 7 or multiple/s of 7


it cannot be 7 since the A,B,C are different


so it can be 6/14 or 9/21 or 12/28


but the denominator has got to be LCD of 4,7 and 28


so it has to be 12/28


so now it has to be 4/28,7/28 and 1/28


so A=4,B=7 and C=28