If $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{29}{72},\ \ c<b<a<60,\ \ \{a,b,c\}\in\mathbb{N} $.

How many sets of $(a,b,c)$ exists ?


$a.)\ 3 \quad \quad \quad \quad \quad b.)\ 4 \\ c.)\ 5 \quad \quad \quad \quad \quad \color{green}{d.)\ 6} \\ $

by trial and error i found

$$\begin{array}{c|c} 2 & 72 \\ \hline 2 &36 \\ \hline 2 &18 \\ \hline 3 &9 \\ \hline 3 &3 \\ \hline &1\\ \end{array}$$


$$\dfrac{29}{72}=\dfrac{18}{72}+\dfrac{9}{72}+\dfrac{2}{72}= \dfrac{1}{c}+\dfrac{1}{b}+\dfrac{1}{a}\\~\\ \implies (a,b,c)=(36,8,4)$$

This question is from chapter quadratic equations.

I look for a short and simple way.

I have studied maths up to $12$th grade.

  • $\begingroup$ I think easier approach(yet trial and error) , is to write down all the possible dividers and try to reach 29 :$24,18,12,9,8,6,4,3,2,1$. so for example you found $(18,9,2)$ you can see that $(18,8,3)$ is also good. thus $\frac14 + \frac19 + \frac1{24}$ $\endgroup$
    – d_e
    Jul 19, 2015 at 18:50
  • $\begingroup$ The avg time to solve such questions with options given is $1-3$ min , so that approach seems a little time consuming for me. $\endgroup$
    – R K
    Jul 19, 2015 at 18:54
  • $\begingroup$ so I guess I have to return to elementary school again :). anyway it is not that long .. if you really try to use it.(because you know the bigger must be at least 12. because 9+8+6 < 29 $\endgroup$
    – d_e
    Jul 19, 2015 at 18:55
  • $\begingroup$ Also I am not interested in necessarily finding all actual solutions , but the number of solutions.so I think it might have some combinatorics/lcm involved. $\endgroup$
    – R K
    Jul 19, 2015 at 18:59
  • $\begingroup$ but what I wrote above is basically "combinatorics": how many ways you can reach 29 by adding 3 numbers. it should take less than 60 seconds if you really try it. $\endgroup$
    – d_e
    Jul 19, 2015 at 19:17

2 Answers 2


In this type of problem, you have to go through cases, usually on the extreme variables.

Initially, $\dfrac{1}{c} < \dfrac{29}{72}$, so $c > \dfrac{72}{29} =2+\dfrac{14}{29} $, so $c \ge 3$.

In the other direction, since $\dfrac{1}{c}+\dfrac{1}{b}+\dfrac{1}{a} < \dfrac{3}{c} $, $\dfrac{3}{c} > \dfrac{29}{72}$ or $c < \dfrac{216}{29} $ or $c \le 7$.

Looking at $a$, $\dfrac{3}{a} < \dfrac{29}{72}$, or $a \ge 8$.

For each $3 \le c \le 7$, compute $d =\dfrac{72}{29}-\dfrac{1}{c} $. Then $\dfrac{1}{a}+\dfrac{1}{b} = d$, so $\dfrac{2}{a} < d < \dfrac{2}{b}$ or $c+1 \le b < \dfrac{2}{d}$.

For each $c+1 \le b < \dfrac{2}{d}$, compute $\dfrac{1}{d}-\dfrac{1}{b}$ and see if that is of the form $\dfrac{1}{a}$ for $a > b$.

I'll leave the actual computation to you.


This cannot be done without a somewhat clumsy search. The aim then is to keep the number of cases to check as small as possible.

We are looking for integer solutions of $${1\over a}+{1\over b}+{1\over c}={29\over72}$$ with $1\leq c<b<a$.

The conditions $${1\over c}<{29\over 72},\qquad{1\over c}+{1\over c+1}+{1\over c+2}\geq{29\over 72}$$ imply $3\leq c\leq 6$.

Given $c$ in this range, put $$q:={29\over 72}-{1\over c}\ .$$ We then have to solve $${1\over a}+{1\over b}=q\ .$$ The conditions for $b$ are $$b>c,\qquad {1\over b}<q, \qquad {1\over b}>{q\over2}\ ,$$ the third of these in order to guarantee $a>b$. This enforces $$\max\left\{{1\over q},c\right\}<b<{2\over q}\ .$$ Given $b$ in this range we have to check whether the resulting $$ a={b\over bq-1}$$ is integer. If yes, we accept the triple $(a,b,c)$. Mathematica found $7$ such triples, three of which violated the extra condition $a<60$. Here is the output:

enter image description here

  • $\begingroup$ So only 4 satisfy all requirements, which I got also by another long-ish (not clever) method. +1, still wonder if there is some clever way without trying so many cases... $\endgroup$
    – coffeemath
    Jul 22, 2015 at 12:53

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