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1/a + 1/b + 1/c = 3/7 ; using LCD:irene12 said:Find solution to 3/7=1/a+1/b+1/c where a, b, and c are positive intergers and a<b<c .

(bc + ac + ab) / abc = 3/7

Simplifying, in terms of c:

c = 7ab / (3ab - 7a - 7b)

At this point, you can tell that 3ab > 7a + 7b

Hope that "gets you going"...

For now: http://www.youtube.com/watch?v=nGd4jkaoHRg

EDIT: in case you're worried there might be no solutions, don't: there are 6 different solutions :idea:

\(\displaystyle \text{Find solutions to: }\;\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\:=\:\frac{3}{7}\; \text{ where }a, b, c\text{ are positive intergers and }a<b<c.\)

I found a number of solutions:

. . \(\displaystyle \frac{1}{3} + \frac{1}{12} + \frac{1}{84} \;=\;\frac{3}{7}\)

. . \(\displaystyle \frac{1}{3} + \frac{1}{11} + \frac{1}{231} \;=\;\frac{3}{7}\)

. . \(\displaystyle \frac{1}{3} + \frac{1}{14} + \frac{1}{42} \;=\;\frac{3}{7}\)

. . \(\displaystyle \frac{1}{3} + \frac{1}{15} + \frac{1}{35} \;=\;\frac{3}{7}\)

. . \(\displaystyle \frac{1}{4} + \frac{1}{6} + \frac{1}{84} \;=\;\frac{3}{7}\)

. . \(\displaystyle \frac{1}{4} + \frac{1}{7} + \frac{1}{28} \;=\;\frac{3}{7}\)

Do a Google search for "Egyptian fractions".

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Denis said:Any other opinions?

My opinion is that soroban has already provided the regulars here with sufficient evidence to show that he doesn't care what we think.

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That was clever - one "get out of corner now" pass for you ......Denis said: