Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
1  
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
add comment

1 Answer 1

up vote 17 down vote accepted

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.