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

Question

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.

Answer 1

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.

Answer 2

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:

Answer 3

If $c<b<a<60$ 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 $0<c<b<a<60, a,b,c∈N^{+} ⟹ \frac{29}{72}>\frac{1}{c} ⟹c>\frac{72}{29}⟹c≥3$.

On the other hand $\frac{1}{72}=\frac{1}{29c}+\frac{1}{29b}+\frac{1}{29a}⟹29c≤3×72$ (recalling $~ c<b<a<60$)$⟹c≤\frac{216}{29}$, 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 $c<b<a<60$, so the the answer is 4, the solutions are $(a,b,c)=(36,24,3),(36,8,4),(24,9,4),(9,8,6)$.

Answer 4

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 $b<a$ 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 $(36,24,3)$.