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:

$$\frac1{28}+\frac1{14}+\frac17+\frac14+\frac12=\frac{1+2+4+7+14}{28}=1\;.$$

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

$$\frac16+\frac16+\frac16+\frac14+\frac14\;.$$

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

$$\frac1{20}+\frac15+\frac14+\frac14+\frac14\;.$$

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 

$$\frac1y+\frac1z+\frac1a+\frac1b=\frac12\;.$$

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

$$\frac1{36}+\frac1{18}+\frac1{12}=\frac{1+2+3}{36}=\frac6{36}=\frac16\;,$$

and we get another ‘nice’ solution,

$$\frac12+\frac13+\frac1{12}+\frac1{18}+\frac1{36}\;.$$
Note that $\frac{4!}{4!}=1$. Now write $4!=24$ as $1+2+3+6+12$ (basically the divisors of $24$) Then we have $$1=\frac{4!}{4!}=\frac{1+2+3+6+12}{24}=\frac{1}{24}+\frac{1}{12}+\frac{1}{8}+\frac{1}{4}+\frac{1}{2}$$