# Find the number of sets of $(a,b,c)$ for $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{29}{72}$

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 ?

Options

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

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.

• 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 $\frac14 + \frac19 + \frac1{24}$
– d_e
Commented Jul 19, 2015 at 18:50
• The avg time to solve such questions with options given is $1-3$ min , so that approach seems a little time consuming for me.
– R K
Commented Jul 19, 2015 at 18:54
• 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
– d_e
Commented Jul 19, 2015 at 18:55
• 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.
– R K
Commented Jul 19, 2015 at 18:59
• 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.
– d_e
Commented Jul 19, 2015 at 19:17

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:

• 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... Commented Jul 22, 2015 at 12:53
• @coffeemath You can get pretty far (whittling down to six cases) based on $72=2^33^2$. Commented Mar 15, 2023 at 7:52

If $$c the answer is $$4$$.

Here is the method for finding the complete solutions $$(a,b,c)$$ of equation $$\frac{29}{72}=\frac{1}{c}+\frac{1}{b}+\frac{1}{a} ~(1)$$.

Let $$0\frac{1}{c} ⟹c>\frac{72}{29}⟹c≥3$$.

On the other hand $$\frac{1}{72}=\frac{1}{29c}+\frac{1}{29b}+\frac{1}{29a}⟹29c≤3×72$$ (recalling $$~ c)$$⟹c≤\frac{216}{29}$$, so that $$c∈[3,4,5,6,7]$$.

Hence find the equation $$(1)$$ solutions now equivalent to find solutions for below five reduced equations:

$$\frac{29}{72}-\frac{1}{3}=\frac{1}{b}+\frac{1}{a}⟹\frac{5}{72}=\frac{1}{b}+\frac{1}{a};~$$

$$\frac{29}{72}-\frac{1}{4}=\frac{1}{b}+\frac{1}{a}⟹\frac{11}{72}=\frac{1}{b}+\frac{1}{a};~$$

$$\frac{29}{72}-\frac{1}{5}=\frac{1}{b}+\frac{1}{a}⟹\frac{73}{360}=\frac{1}{b}+\frac{1}{a};~$$

$$\frac{29}{72}-\frac{1}{6}=\frac{1}{b}+\frac{1}{a}⟹\frac{17}{72}=\frac{1}{b}+\frac{1}{a};~$$

$$\frac{29}{72}-\frac{1}{7}=\frac{1}{b}+\frac{1}{a}⟹\frac{131}{504}=\frac{1}{b}+\frac{1}{a};~$$

Find above five equations two unit fractions expansion, we have following solutions, becasue $$c, so the the answer is 4, the solutions are $$(a,b,c)=(36,24,3),(36,8,4),(24,9,4),(9,8,6)$$.

• This answer does not add anything useful, not to mention the lack of explanation for how the solutions are obtained. Commented Mar 14, 2023 at 3:05
• @VTand, see above general algorithm, in fact it is a standard algorithm for finding any fraction $\frac{m}{n}$ fixed terms (if have) unit fractions expansion problem. Commented Mar 15, 2023 at 6:42

You can reduce the inevitable case-checking down to 6 cases by thinking about prime powers dividing fraction denominators. First, as with other answers, we can conclude that $$c\leq7$$: $$\frac{29}{72}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}<\frac{3}{c}\implies c<\frac{216}{29}\implies c\leq7$$

Meanwhile, you can't have a denominator of $$72$$ in the sum of three fractions without one of the three original denominators divisible by $$9$$. It must be $$a$$ or $$b$$ since $$c\leq7$$. Likewise, $$a$$ or $$b$$ must be divisible by $$8$$. And since all denominators are less than $$60$$, neither $$a$$ nor $$b$$ is divisible by both $$8$$ and $$9$$. So between $$a$$ and $$b$$, one is divisible by $$9$$ and the other by $$8$$.

So I disregard the ordering $$b and imagine those two denominators to be $$x=9m$$ and $$y=8n$$.

$$\frac{1}{9m}+\frac{1}{8n}+\frac{1}{c}=\frac{29}{72}$$ $$\frac{8}{m}+\frac{9}{n}+\frac{72}{c}=29\tag{*}$$

with $$m\leq6, n\leq7, c\leq 7$$.

Now we cannot have $$n=7$$, or else $$c$$ must also be $$7$$ (because $$7$$ would need to divide at least two of the denominators in (*)). But then $$\frac{81}{7}$$ does not reduce, and $$7$$ would have to also divide $$m$$, which is a contradiction. So $$m\in\{1,2,3,4,5,6\}$$ and $$n\in\{1,2,3,4,5,6\}$$.

We cannot have $$n$$ being even either, or else $$\frac{9}{n}$$ does not reduce to an integer and at lease one of $$m,z$$ would have to be divisible by $$16$$. So $$m\in\{1,2,3,4,5,6\}$$ and $$n\in\{1,3,5\}$$.

Similarly we cannot have $$m$$ divisible by $$3$$, or else $$\frac{8}{m}$$ does not reduce to an integer and at lease one of $$n,z$$ would have to be divisible by $$27$$. So $$m\in\{1,2,4,5\}$$ and $$n\in\{1,3,5\}$$.

And we cannot have either $$m,n$$ equal to $$5$$. If $$n=5$$, at least one of $$m,c$$ would have to equal $$5$$. But neither $$\frac{8}{5}+\frac{9}{5}$$ nor $$\frac{9}{5}+\frac{72}{5}$$ is an integer, so actually all three denominators would have to be $$5$$, and $$\frac{89}{5}\neq29$$.

And if $$m=5$$, at least one of $$n,c$$ would have to equal $$5$$. Even if $$n=5$$, then since $$\frac{8}{5}+\frac{9}{5}$$ is not an integer, $$c$$ would have to also be $$5$$. So either way, $$c=5$$. But just assuming $$m,c=5$$ means $$\frac{9}{n}=13$$, which is not possible.

So $$m\in\{1,2,4\}$$ and $$n\in\{1,3\}$$. Now there are only $$6$$ cases to examine (*) and see if $$z$$ is an integer. Four of the cases pass and two do not. The four that pass are

$$\frac{8}{1}+\frac{9}{1}+\frac{72}{6}=29\quad \frac{8}{1}+\frac{9}{3}+\frac{72}{4}=29\quad \frac{8}{4}+\frac{9}{1}+\frac{72}{4}=29\quad \frac{8}{4}+\frac{9}{3}+\frac{72}{3}=29$$

Recalling $$x=9m$$ and $$y=8n$$, the original triples $$(x,y,z)$$ are $$(9,8,6)$$, $$(9,24,4)$$, $$(36,8,4)$$, and $$(36,24,3)$$.