Question

If a, b, c \in N , and 1\leq a\leq b\leq c , find all a,b,c numbers that hold the equation always true,

\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{3}{4}

Steven Zheng Steven Zheng posted 1 month ago


Answer

Case 1, if a\geq 5

5\leq a\leq b\leq c

\therefore \dfrac{1}{5}\geq \dfrac{1}{a} \geq \dfrac{1}{b} \geq \dfrac{1}{c}

\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{ 1}{c}\leq \dfrac{3}{5}< \dfrac{3}{4}

So no solution if a\geq 5

Case 2, if a=1

\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}>\dfrac{3}{4} , no solution.

Case 3, if a=4

\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{3}{4}-\dfrac{1}{4}=\dfrac{1}{2}

Case 3-1, if b\geq 5

5\leq b\leq c

\dfrac{1}{5}\geq \dfrac{1}{b} \geq \dfrac{1}{c}

\therefore \dfrac{1}{b}+\dfrac{1}{c} <\dfrac{2}{5} <\dfrac{1}{2} , no solution

Case 3-2, if b=1

\dfrac{1}{c}=-\dfrac{1}{4} , no solution

Case 3-3, if b=4

\dfrac{1}{c}=\dfrac{1}{2}-\dfrac{1}{4}

\therefore c=4

Now we get our first solution a=b=c=4

Case 3-4, if b=3

\dfrac{1}{c}=\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{1}{6}

\therefore c=6

So a=4 , b=3 ,c=6 is the 2nd solution we find.

Case 3-5, if b=2

\dfrac{1}{c}=\dfrac{1}{2}-\dfrac{1}{2}=0 , no solution

Case 4, if a=3

\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{3}{4}-\dfrac{1}{3}=\dfrac{5}{12}

Case 4-1, if b\geq 6

6\leq b\leq c

\therefore \dfrac{1}{b}+\dfrac{1}{c} <\dfrac{2}{6} <\dfrac{5}{12} , no solution

Case 4-2, if b=1

\dfrac{1}{c} =\dfrac{5}{12} -1=-\dfrac{7}{12} , no solution

Case 4-3, if b=5

\dfrac{1}{c} =\dfrac{5}{12}-\dfrac{1}{5}

=\dfrac{25-12}{60} , c is not an integer, no solution

Case 4-4, if b=4

\dfrac{1}{c} =\dfrac{5}{12}-\dfrac{1}{4} =\dfrac{1}{6}

\therefore c=6

So a=3 , b=4 , c=6 is another solution

Case 4-5, if b=3

\dfrac{1}{c} =\dfrac{5}{12} -\dfrac{1}{3} =\dfrac{1}{12}

\therefore c=12

So a=3 , b=3 , c=12 is another solution

Case 4-6, b=2

\dfrac{1}{c} =\dfrac{5}{12}-\dfrac{1}{2}=-\dfrac{1}{12} , no solution

Case 5, a=2

\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{3}{4}-\dfrac{1}{2}=\dfrac{1}{4}

Case 5-1, if b\geq 9

9\leq b\leq c

\dfrac{1}{9}\geq \dfrac{1}{b}\geq \dfrac{1}{c}

\dfrac{1}{b}+\dfrac{1}{c} \leq \dfrac{2}{9}< \dfrac{1}{4}

\therefore b\geq9 , no solution

Case 5-2, b\leq4

\dfrac{1}{c}=\dfrac{1}{4}-\dfrac{1}{b}\leq 0 , no solution

Case 5-3, b=8

\dfrac{1}{c}=\dfrac{1}{4}-\dfrac{1}{8} =\dfrac{1}{8}

\therefore c=8

So a=2 , b=8 , c=8 is another solution

Case 5-4, b=7

\dfrac{1}{c}=\dfrac{1}{4}-\dfrac{1}{7} =\dfrac{3}{28}

c is not an integer, no solution

Case 5-5, b=6

\dfrac{1}{c}=\dfrac{1}{4}-\dfrac{1}{6} =\dfrac{1}{12}

\therefore c=12

So a=2 , b=6 , c=12 is another solution

Case 5-6, b=5

\dfrac{1}{c}=\dfrac{1}{4}-\dfrac{1}{5} =\dfrac{1}{20}

\therefore c=20

So a=2 , b=5 , c=20 is another solution

Steven Zheng Steven Zheng posted 1 month ago

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