Find five positive integers whose reciprocals sum to $1$

Find a positive integer solution $(x,y,z,a,b)$ for which

$$\frac{1}{x}+ \frac{1}{y} + \frac{1}{z} + \frac{1}{a} + \frac{1}{b} = 1\;.$$

Is your answer the only solution? If so, show why.

I was surprised that a teacher would assign this kind of problem to a 5th grade child. (I’m a college student tutor) This girl goes to a private school in a wealthy neighborhood.

Please avoid the trivial $x=y=z=a=b=5$. Try looking for a solution where $ x \neq y \neq z \neq a \neq b$ or if not, look for one where one variable equals to another, but explain your reasoning. The girl was covering “unit fractions” in her class.


The perfect number $28=1+2+4+7+14$ provides a solution:


If they’ve been doing unit (or ‘Egyptian’) fractions, I’d expect some to see that since $\frac16+\frac13=\frac12$, $$\frac16+\frac16+\frac16+\frac16+\frac13=1$$ is a solution, though not a much more interesting one than the trivial solution. The choice of letters might well suggest the solution


A little playing around would show that $\frac14+\frac15=\frac9{20}$, which differs from $\frac12$ by just $\frac1{20}$; that yields the solution


If I were the teacher, I’d hope that some kids would realize that since the average of the fractions is $\frac15$, in any non-trivial solution at least one denominator must be less than $5$, and at least one must be greater than $5$. Say that $x\le y\le z\le a\le b$. Clearly $x\ge 2$, so let’s try $x=2$. Then we need to solve


Now $y\ge 3$. Suppose that $y=3$; then $$\frac1z+\frac1a+\frac1b=\frac16\;.$$

Now $1,2$, and $3$ all divide $36$, and $\frac16=\frac6{36}$, so we can write


and we get another ‘nice’ solution,


Source : Link , Question Author : Low Scores , Answer Author : Aarthi

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