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. 

There are many ways to obtain many solutions. Here is one systematic way to obtain solutions.
First look at a class of solutions such that $x \leq y \leq z \leq a \leq b \leq 10$ (other solution can be obtained as permutation of these).
This implies that $2 \leq x \leq 5$. So now start with $x=2$. This now means that $3 \leq y \leq 8$. Choose $y=3$. This now means that $7 \leq z \leq 10$. You will quickly find that no solution exists such that $x \leq y \leq z \leq a \leq b \leq 10$. Then choose $y=4$. Again you will find that no solution exists such that $x \leq y \leq z \leq a \leq b \leq 10$.
Going through this you will find that if we want $x=2$, and $x \leq y \leq z \leq a \leq b \leq 10$, then $$[2, 5, 10, 10, 10]; 
[2, 6, 9, 9, 9]; 
[2, 8, 8, 8, 8]$$ are the only solutions with $x=2$ and $x \leq y \leq z \leq a \leq b \leq 10$.
Now set $x=3$ and get bounds for the remaining variables to see that $$ 
[3, 3, 9, 9, 9]; 
[3, 4, 6, 8, 8]; 
[3, 5, 5, 6, 10]; 
[3, 6, 6, 6, 6]; $$ are the only solutions with $x=3$ and $x \leq y \leq z \leq a \leq b \leq 10$.
Here are the solutions such that $x \leq y \leq z \leq a \leq b \leq 10$
$$[x,y,z,a,b] \in \{[2, 5, 10, 10, 10]; 
[2, 6, 9, 9, 9]; 
[2, 8, 8, 8, 8]; 
[3, 3, 9, 9, 9]; 
[3, 4, 6, 8, 8]; 
[3, 5, 5, 6, 10];\\ 
[3, 6, 6, 6, 6]; 
[4, 4, 4, 8, 8]; 
[4, 4, 5, 5, 10]; 
[4, 4, 6, 6, 6]; 
[5, 5, 5, 5, 5]; \}
$$
There are $114$ distinct solutions i.e. without permutations such that
$$x \leq y \leq z \leq a \leq b \leq 100$$ and can be found here.
