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. 
 A: $$1={1\over 3}+{1\over 2}+{1\over 6}={1\over 3}+\left({1\over 4}+{1\over 4}\right)+{1\over 6}+\left({1\over5}-{1\over 5}\right)={1\over3}+{1\over4}+{1\over5}+{1\over6}+{1\over 20}$$
So we have 
$${1\over3}+{1\over4}+{1\over5}+{1\over6}+{1\over20}=1$$
A: You can try taking $x, y, z, a$ each with value larger than $5$ which may or may not be equal. Now find $b$ such a way that the given relation is satisfied. But this question is way beyond fifth grade student (my opinion).
A: The number of solutions of $$1={1\over x_1}+{1\over x_2}+\cdots+{1\over x_n},\ \ \ 0\lt x_1\le x_2\le\cdots\le x_n$$ is tabulated, as a function of $n$, at OEIS A002966 but only a few terms are given: $1, 1, 3, 14, 147, 3462, 294314, 159330691$. I don't know whether the number of solutions with all denominators distinct has been tabulated at that site. 
A: 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,
$$\frac12+\frac13+\frac1{12}+\frac1{18}+\frac1{36}\;.$$
A: You could connect it to geometry. Cut the square into equally-sized pieces, then take some of those pieces and cut them into even-smaller ones until you get five pieces. e.g.


After that, you could try some different-sized pieces and have a good chance of getting something like this:

A: My first reaction to the question was $\frac12+\frac14+\frac18+\frac1{16}+\frac1{16}$. sorry, but it seems too natural to need explaining, but also not satisfactory because not all denominators are different.
Then other replies remind me of the "5th grade level" equation $1 = \frac12 + \frac13 + \frac16$.
Well, then the pupil will probably try to use the same equation to divide one of the three fractions and realize that the only one that will provide different denominators is the $\frac16$.
That is $1/6 = \frac{1/6}2 + \frac{1/6}3 + \frac{1/6}6 = \frac1{12} + \frac1{18} + \frac1{36}$
Thus building the solution $1 = \frac12 + \frac13 + \frac1{12} + \frac1{18} + \frac1{36}$

Thinking further about it, I assume that anybody confronted with fractions will easily agree with the simple statement: $1 = \frac1n + \frac{n-1}n$. 
Dividing the equation by $n-1$ and arranging differently will produce $\frac1{n-1} = \frac1n + \frac1{n(n-1)}$, of which $\frac12 = \frac13 + \frac16$ is a special case.
once you have a couple of these equations spelled out, like:
$$\frac12 = \frac13 + \frac16 $$
$$\frac13 = \frac14 + \frac1{12} $$
$$\frac14 = \frac15 + \frac1{20} $$
$$\frac15 = \frac16 + \frac1{30} $$
it should be a kid's game to expand
$$1 = \frac12 + \frac14 + \frac14$$
into a solution of the given problem.
you might also want to include the case for $n=2$:
$$1 = \frac12 + \frac12$$
and start with the expansion of $1$.
A: First, without context, there is no way to tell if the problem is suitable for a given grade. It doesn't seem likely, but then maybe they did some work on the Egyptian representation of fractions, which is close enough that they might think of it. For me, that was my first thought, and I then came to:
$$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{16}=1$$
Even with that hint, the question about uniqueness is not trivial. You can play with replacing only part of the expression, for example:
$$\frac{1}{8}+\frac{1}{16} = \frac{1}{6}+\frac{1}{48}$$
which yields:
$$\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{16}+\frac{1}{48}=1$$
but I don't see how 5th grade student can do that other than trial and error.
Other than that, I don't see what “typical fith-grade” reasoning can be used. Perfect numbers seem out of the question.
A: This solution may be too advanced for a fifth-grader, but you can do this problem algorithmically - simply by searching all the possible fractions.
The gist of it is to use a greedy algorithm - start with the biggest fraction, and continue iterating small fractions until you can't anymore. For example, $\displaystyle \frac 1 2 + \frac 1 3 + \frac 1 7 + \frac 1 {43} + \frac 1 {1806}$ would be the first one you find. The next one would be $\displaystyle \frac 1 2 + \frac 1 3 + \frac 1 7 + \frac 1 {44} + \frac 1 {924}$, then $\displaystyle \frac 1 2 + \frac 1 3 + \frac 1 7 + \frac 1 {45} + \frac 1 {630}$, and so on.
Here's the algorithm in more detail:


*

*For the first number, start with $\displaystyle \frac 12$, eventually working your way down to $\displaystyle \frac15$ (which is the smallest that the largest fraction can be, so you can stop there).

*Subtract this fraction from 1, and use the remaining part to determine what the next few numbers will iterate through.

*For each of the subsequent fractions, start with the largest unit fraction smaller than both the "remaining part" that's left and the fraction before it, and work down until you reach the smallest unit fraction larger than $\displaystyle \frac1n$ the "remaining part", where your fraction is the $n$th last fraction, and do the same as above, subtracting the unit fraction from the "remaining part" for the next fraction to use.

*Once you have four fractions, if the "remaining part" for the last fraction can be expressed as a unit fraction, you have a solution. Otherwise, you don't, and continue onwards.


This algorithm will eventually return all the possible unit fraction combinations, starting from $\displaystyle \frac 1 2 + \frac 1 3 + \frac 1 7 + \frac 1 {43} + \frac 1 {1806}$ and ending with $\displaystyle \frac 15 + \frac 15 + \frac 15 + \frac 15 + \frac 15$.

http://joezeng.com/code/fractions/fractions.html $\leftarrow$ This is a list of all of the fractions, dynamically generated using some recursive Javascript that implements the algorithm above. You can view the source code here, which I have MIT-licensed for demonstration purposes.
According to the generator, there are a total of 147 solutions for 5 fractions, and the "minimum unique solution" such that all denominators are distinct and their sum is the lowest possible is $\displaystyle \frac 13 + \frac 14 + \frac 15 + \frac 16 + \frac 1 {20}$. There is another "minimum unique solution", such that the largest denominator is the lowest, which is $\displaystyle \frac 12 + \frac 14 + \frac 1{10} + \frac 1{12} + \frac 1{15}$.
You can also use the generator to generate fraction lists of arbitrary sizes by modifying the initial function call to use 6 levels (or 2, 3, or 4) instead of 5, as well as generate Egyptian fraction expansions for arbitrary fractions (by modifying the first two terms 1, 1 to be other things).
A: 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.
A: $$\frac{1}{2}+\frac{1}{3}+\frac{1}{8}+\frac{1}{30}+\frac{1}{120}=1$$
In  general, if you want $n$ distinct unit fractions to sum up to $1$ , you can use a formula
 (which I found almost by luck) and find a direct solution  
$$\frac{1}{\frac{2!}{1}}+\frac{1}{\frac{3!}{2}}+\frac{1}{\frac{4!}{3}}+...+\frac{1}{\frac{n!}{n-1}}+\frac{1}{n!}=1$$
It is accepted at the Monthly and it will be published.
A: Yet another method would be to start with $${1\over2}+{1\over3}+{1\over 6}=1$$
Divide by two and add $1\over 2$; this yields $${1\over 2}+{1\over 4}+{1\over 6}+{1\over 12}=1$$
Again, divide by two and add $1\over 2$; this yields $${1\over 2}+{1\over 4}+{1\over8}+{1\over 12}+{1\over 24}=1$$
A: Surprised that no-one proposed $1,-1,1,-1,1$ and a whole bunch of variations around it.
A: Note that $1=\dfrac12+\dfrac12.$ Then we can use $\dfrac1n=\dfrac1{(n+1)}+\dfrac1{n(n+1)}$ to expand this up to five terms. In more general,

$$\color{Green}{\dfrac{1}{pqr}=\dfrac{1}{pr(p+q)}+\dfrac{1}{qr(p+q)}.}$$

For distinct 4-tuples, I got that $$1=\dfrac12+\dfrac14+\dfrac16+\dfrac1{12}$$
$$1=\dfrac12+\dfrac13+\dfrac17+\dfrac1{42}$$
$$1=\dfrac12+\dfrac13+\dfrac19+\dfrac1{18}$$
$$1=\dfrac12+\dfrac13+\dfrac1{10}+\dfrac1{15}.$$ Then continue inductively .....
Similarly, $1=\dfrac13+\dfrac13+\dfrac13$ is a special case of

$$\color{Blue}{\dfrac{1}{pqr}=\dfrac{1}{pq(p+q+r)}+\dfrac{1}{qr(p+q+r)}++\dfrac{1}{pr(p+q+r)}.}$$

For example, $\dfrac1{2n}=\dfrac1{2(n+3)}+\dfrac1{n(n+3)}+\dfrac1{2n(n+3)}$ for any positive integer $n.$
A: I am posting another solution since one of my students found this method.
(He is 16 years old but as you see he is gifted)
Start with $\frac{1}{2}+\frac{1}{3}+\frac{1}{6}$. No write $\frac{1}{6}=\frac{3}{18}=\frac{1}{18}+\frac{2}{18}=\frac{1}{9}+\frac{1}{18}$ and you get $\frac{1}{2}+\frac{1}{3}+\frac{1}{9}+\frac{1}{18}=1$. Proceed like this: Write the last fraction (which will have an even denominator) of the sum as $\frac{1}{2n}=\frac{1}{6n}+\frac{2}{6n}=\frac{1}{3n}+\frac{1}{6n}$ and so on!
A: Note that $\frac{4!}{4!}=1$. Now write $4!=24$ as $1+2+3+6+12$ (basically the divisors of $24$) Then we have $$1=\frac{4!}{4!}=\frac{1+2+3+6+12}{24}=\frac{1}{24}+\frac{1}{12}+\frac{1}{8}+\frac{1}{4}+\frac{1}{2}$$
A: If you allow 2 fractions to be equal, as the question allowed, you can just use powers of $2$ to generate a solution for any number of fractions since:
$1/2 +1/2^2+...+1/2^n + 1/2^n = 1$
for all $n>1,$ containing $n+1$ fractions, so we need $n=4$
$1/2 + 1/4 + 1/8+ 1/16 + 1/16 =1$
This may be an easier solution to discover at school.
In addition with this idea you can easily create a solution with distinct fractions from the case of $1/2+1/3+1/6=1$ (which can easily be found by trial and error) using this idea by dividing by $4$ and using the $n=2$ case.
Thus $1/8+1/12+1/24=1/4$
and $1/2+1/4+1/4=1$ giving
$1/2+1/4+1/8+1/12+1/24=1.$
