# Number of triples $(a,b,c)$ of positive integers such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{3}{4}$?

What is the number of triples $(a,b,c)$ of positive integers such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{3}{4}$ is:

A) $16$

B) $25$

C) $31$

D) $19$

E) $34$

Note: Through trial and error I've concluded that it's greater than $19$, but I wonder if there is a better way to solve it. I believe this question was on a contest, but I don't know which.

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im guessing its 1 plus a multiple of 6...it might also be easier to do if you pretend as if a>=b>=c – mathguy Aug 29 '12 at 21:36
Fix an ordering as in Sidd's comment: $a \ge b \ge c > 1$ since $1$ can't be a solution. Then use the equality constraint $1/a + 1/b + 1/c = 1/4$ to find an upper bound on $c$. That's $1 < c \le 4$. Now enumerate all the cases for $c = 2, 3, 4$. – user2468 Aug 29 '12 at 21:47
Re contest: Google shows a relevant question in Y!Answers. – user2468 Aug 29 '12 at 21:48

Assume that $a\le b\le c$. The average of the fractions $\frac1a,\frac1b$, and $\frac1c$ is $\frac14$, so $\frac1a\ge\frac14$, and $a\le 4$. Clearly $a>1$, so $a$ must be $2,3$, or $4$.
• If $a=4$, the only possibility is $a=b=c=4$.
• Suppose that $a=3$. If $b=3$, $\frac1c=\frac34-\frac23=\frac1{12}$, so $c=12$. If $b=4$, $\frac1c=\frac34-\frac13-\frac14=\frac16$, so $c=6$. If $b\ge 5$, then $c\ge 5$ as well, so $\frac1a+\frac1b+\frac1c\le\frac13+\frac25=\frac{11}{15}<\frac34$, so there are no more solutions with $a=3$.
• Suppose that $a=2$; clearly $\frac1b+\frac1c=\frac14$, so $b>4$. If $b=5$, $\frac1c=\frac14-\frac15=\frac1{20}$, so $c=20$. If $b=6$, $\frac1c=\frac14-\frac16=\frac1{12}$, so $c=12$. If $b=7$, then $\frac1c=\frac14-\frac17=\frac3{28}$, so there is no solution in this case. If $b=8$, clearly $c=8$ as well. Finally, if $b>8$, then $\frac1b+\frac1b\le\frac29<\frac14$, and there are no further solutions.
The solutions $\langle a,b,c\rangle$ with $a\le b\le c$ are therefore $\langle 4,4,4\rangle$, $\langle 3,3,12\rangle$, $\langle 3,4,6\rangle$, $\langle 2,5,20\rangle$, $\langle 2,6,12\rangle$, and $\langle 2,8,8\rangle$, for a total of six solutions. If you count different permutations of the same integers separately, there are $3\cdot3!+2\cdot3+1=25$ solutions.