How Would The Egyptians Solve it?

Positive rational numbers can always be written as Egyptian fractions using the so-called Greedy Algorithm. Simply stated, this says that, at each stage, pick the largest possible reciprocal. Proving that the algorithm works is a simple matter of induction (and using the divergence of the harmonic series to handle rationals greater than $1$ ).

In this case,

• $\tfrac12$ is the largest reciprocal less than or equal to $\tfrac{9}{10}$ , so choose $x=2$ ,
• $\tfrac13$ is the largest reciprocal less than or equal to $\tfrac{9}{10} - \tfrac12 = \tfrac25$ , so choose $y=3$ ,
• $\tfrac{1}{15}$ is the largest reciprocal less than or equal to $\tfrac25 - \tfrac13 = \tfrac{1}{15}$ , so choose $z=15$ and we are done.

The answer is $2+3+15 = \boxed{20}$ .

Consider $\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}=\dfrac{47}{60}<\dfrac{54}{60}=\dfrac{9}{10}$ .

This shows that $\dfrac{1}{2}$ must be one of the terms. without the loss of generality (WLOG), $x=2$ .

We know have $\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{2}{5}$ . It would be nice to simply have $y=z=5$ , but they have to be distinct.

So, we need to find the answer to the question $\dfrac{2}{5}=\dfrac{(a+b)}{5c}$ where $(a+b)=2c$ , $c$ is a constant, and $a|5c$ and $b|5c$ . So we need a multiple of $2$ who can be written as the sum of the same multiple of $5$ . We know $c>2$ since $2\times2=4<5$ . Using trial and error we try $c=3$ and find that $2\times3=6=5+1$ and $5c=5\times3=15$ which $5|15$ and $1|15$ .

So are two fractions are $\dfrac{1}{15}$ and $\dfrac{5}{15}=\dfrac{1}{3}$ implying $y=3$ and $z=15$ .

Then $x+y+z=2+3+15=20$

P.S. Please post a solution that doesn't use trial and error, I kind of just stumbled upon using $\dfrac{1}{3}$ and figured out that $\dfrac{1}{15}$ would fill in the void.

We can write given equation as $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} + \frac{1}{10} = \frac{1}{1}$ Using Fibonacci's identity regarding Egyptian fractions $\frac{1}{1} = \frac{1}{2} + \frac{1}{2}$ , so maybe $x = 2$ . Which gives $\frac{1}{y} + \frac{1}{z} + \frac{1}{10} = \frac{1}{2}$ Again $\frac{1}{2} = \frac{1}{3} + \frac{1}{6}$ , so maybe $y = 3$ . Which gives $\frac{1}{z} + \frac{1}{10} = \frac{1}{6}$ Hmm, $\frac{1}{6} - \frac{1}{10} = ?$ $\text{BINGO!}\quad z = 15$ I don't know whether this method gives guaranteed solution or if this is just another trial and error method.

Desenvolvendo,

$\\ \mathsf{\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{9}{10}} \\\\\\ \mathsf{\frac{xy + xz + yz}{xyz} = \frac{9}{10}}$

Desse modo, temos que: $\mathsf{\exists \ k \in \mathbb{Z}}$ tal que $\mathsf{\frac{xy + xz + yz}{xyz} = \frac{9k}{10k}}$ .

Com efeito, obtemos $\begin{cases} \mathsf{xy + xz + yz = 9k \quad \quad (i)} \\ \mathsf{x \cdot y \cdot z = 10k \qquad \qquad (ii)} \end{cases}$ .

De $\mathsf{(ii)}$ , $\mathsf{x \cdot y \cdot z = 2 \cdot 5 \cdot k}$ . Portanto, sem perda de generalidade, $\boxed{\mathsf{x = 2}}$ , $\boxed{\mathsf{y = 5}}$ e $\boxed{\mathsf{z = k}}$ ; substituindo em $\mathsf{(i)}$ ,

$\\ \mathsf{xy + xz + yz = 9k} \\\\ \mathsf{2 \cdot 5 + 2 \cdot k + 5 \cdot k = 9k} \\\\ \mathsf{10 = 9k - 7k} \\\\ \boxed{\mathsf{k = 5}}$

Mas, de acordo com o enunciado, x, y e z são inteiros distintos. Ou seja, $\mathsf{z \neq 5}$ . Então,

$\\ \mathsf{\frac{1}{x} + \frac{1}{y} + \frac{1}{z} =} \\\\\\ \mathsf{\frac{1}{2} + \frac{1}{5} + \frac{1}{5} =} \\\\\\ \mathsf{\frac{1}{2} + \frac{2^{\times 3}}{5^{\times 3}} =} \\\\\\ \mathsf{\frac{1}{2} + \frac{6}{15} =}$

$\\ \mathsf{\frac{1}{2} + \left ( \frac{5}{15} + \frac{1}{15} \right ) =} \\\\\\ \mathsf{\frac{1}{2} + \frac{1}{3} + \frac{1}{15}}$

Logo,

$\\ \mathsf{x + y + z = 2 + 3 + 15} \\\\ \boxed{\boxed{\mathsf{x + y + z = 20}}}$

1/x+1/y+1/z=9/10

Therefore, gcd(x,y,z) is a multiple of 10, like 10,20 or 30.

hence 9/10=18/20=27/30=d/n.

and x,y,z must divide the denominator,like 10 or 20 or 30.

d/x+d/y+d/z=n

Now only 2 and 5 divide 10. Thus it has no solution.

2,4,5,10 divide 20. it also does not have any solution.

2,3,4,5,6,10,15 divide 30.

30/2+30/3+30/15=15+10+2=27

This gives 1/2+1/3+1/15=27/30=9/10

Well, you need to just check the first few cases.