# Find five positive integers whose reciprocals sum to $1$

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

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

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.

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A $5$-tuple is hardly a ‘pair’! – Brian M. Scott Jan 30 '13 at 9:42
$(2, 4, 8, 16, 16)$ is another solution. – Karolis Juodelė Jan 30 '13 at 9:45
$x\neq y\neq z\neq a\neq b$ is not the same as saying that $x, y, z, a,$ and $b$ are all distinct. For example $1\neq 2\neq 1\neq 2\neq 1$, but they are not distinct. – Michael Albanese Jan 30 '13 at 9:52
I wouldn't be surprised if a bright fifth grader came up with the trivial x=y=z=a=b=5. Then answer the next question with "No": x=y=z=a=6, b=3. That seems do-able for bright grade fives. – Michael J Swart Jan 30 '13 at 13:47
Is there a reason why 1/5 + 1/5 + 1/5 + 1/5 + 1/5 = 1 is a bad answer? Especially if this is grade school math. For a generic fifth grader is seems that this would be a rather logical introduction to fractions. I know you're looking for 'x≠y≠z≠a≠b' but is there some outside reason as to why? The question, as stated, doesn't seem to have that restriction. Occam's Razor and all that. – grauwulf Jan 30 '13 at 20:13

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,

$$\frac12+\frac13+\frac1{12}+\frac1{18}+\frac1{36}\;.$$

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Very nice.${}{}$ – Michael Albanese Jan 30 '13 at 9:47
Oh my gosh Brian. How did you figure that out? It took 5 adults in the room and 50 minutes yet no one got as close as Michael or Karolis! There was a woman in the room who had her B.A. in math from a prestigious university and she couldn't figure this out. Amazing. Is it possible for you to show your reasoning or is this something that "popped" into your head ...like magic? – Low Scores Jan 30 '13 at 9:49
@LowScores: I realized immediately that a perfect number would work if it had exactly $5$ proper divisors. $6$ is too small, and I knew that the next perfect number is $28$, so I tried it, and it worked. – Brian M. Scott Jan 30 '13 at 9:53
Something's wrong with your second equation: $\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{2}\ne 1$. – Joel Reyes Noche Jan 30 '13 at 13:29
I love that we have real professors on here. – Aerovistae Jan 30 '13 at 18:17

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:

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I think this solution is the most impressive, because it seems like the only one that instantly and accessibly conveys to almost everybody that there is an infinite amount of solutions. – gmlime Jan 31 '13 at 9:28
There aren't infinitely many solutions, though. There are 147, as shown in Joe Zeng's answer – Max Jan 31 '13 at 10:41
@gmline This wasn't supposed to be a method for enumerating all the answers or anything like that; just a way to approach it visually that might help for some kids. – Mark Eichenlaub Jan 31 '13 at 17:57
An excellent idea. Cutting out fractions of a unit circle and passing them to the class to put back together in different ways would work really well! – Joshua Shane Liberman Jan 31 '13 at 20:31
This is genius. If I were the teacher, I'd hope for this kind of thinking! – awendt Feb 1 '13 at 10:16

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).

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Brian had already presented the best solution; I was simply presenting a more comprehensive, less appropriate one. – Joe Z. Jan 30 '13 at 17:51
Also, this solution would be to the benefit of that one girl with a B.A. in math who couldn't solve it, as mentioned by @LowScores in one comment on Brian's answer. – Joe Z. Jan 30 '13 at 17:55
There are 3462 solutions for six terms, 2320 with all unique terms (72 solutions for five unique terms). There are 294326 solutions with seven terms (245767 unique); the least solution is 3 4 9 10 12 15 18. Beyond seven I'd want a cleverer search strategy (breadth-first for small values would be fine). – Rex Kerr Jan 30 '13 at 19:36
"Brian had already presented the best solution; I was simply presenting a more comprehensive, less appropriate one." +1 for this charming analysis of your work! – Mark Harrison Feb 1 '13 at 0:06
I’m very glad that someone took the trouble to do the exhaustive analysis! – Brian M. Scott Feb 1 '13 at 21:38

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 $${1\over 2}+{1\over 4}+{1\over8}+{1\over 12}+{1\over 24}=1$$

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one can reach your solution with a similarly simple reasoning: start with 1/2+1/4+1/4 then divide the 1/4 using 1/2+1/3+1/6. – mariotomo Jan 30 '13 at 23:00
I like this method a lot. – Joe Z. Jan 30 '13 at 23:36
This is simple but effective. Very nice. – Ryan Amos Jan 31 '13 at 4:05
I like this method, too. It may be worth noting that your first equation is the result of applying the recursive function to $${1\over3}+{2\over 3}=1$$ – David Kaczynski Feb 1 '13 at 15:36
@DavidKaczynski 2/3 isn't a unit fraction, though. – Joe Z. Feb 1 '13 at 16:47

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 http://oeis.org/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.

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A006585 gives the number of distinct denominator solutions, or $n!$ times that if you can reorder – Henry Feb 1 '13 at 13:58
I think it's less than $n!$ if you still consider $\frac{1}{2} + \frac{1}{2}$ and $\frac{1}{2} + \frac{1}{2}$ to be the same. – asmeurer Feb 12 '13 at 3:29
@asmeurer, Henry is writing about distinct denominator solutions, so I don't think $(1/2)+(1/2)$ is an issue. – Gerry Myerson Feb 12 '13 at 5:10
Ah, missed that. – asmeurer Feb 12 '13 at 19:07

A simple C++ code to find the a-b-c-x-y pairs for the first positive 100 integers.

const double MAX_NUM = 100.0;
const double EPSILON = 0.0000001;
double Sum;
for (double a=1; a<=MAX_NUM; a++)
{
for (double b=1; b<=a; b++)
{
for (double c=1; c<=b; c++)
{
for (double x=1; x<=c; x++)
{
for (double y=1; y<=x; y++)
{
Sum = 1.0/a + 1.0/b + 1.0/c + 1.0/x + 1.0/y;
if (abs(Sum - 1.0) < EPSILON)
{
std::cout << a << "\t" << b << "\t" << c
<< "\t" << x << "\t" << y << std::endl;
}
}
}
}
}
}


Its output is:

5       5       5       5       5
6       6       6       4       4
6       6       6       6       3
8       8       4       4       4
8       8       6       4       3
8       8       8       8       2
9       9       9       3       3
9       9       9       6       2
10      5       5       4       4
10      6       5       5       3
10      10      10      5       2
12      6       4       4       4
12      6       6       4       3
12      8       8       3       3
12      8       8       6       2
12      12      4       4       3
12      12      6       3       3
12      12      6       6       2
12      12      12      4       2
14      7       7       7       2
15      5       5       5       3
15      10      4       4       3
15      10      6       3       3
15      10      6       6       2
15      12      10      4       2
15      15      5       3       3
15      15      6       5       2
16      16      8       4       2
18      9       4       4       3
18      9       6       3       3
18      9       6       6       2
18      12      9       4       2
18      18      18      3       2
20      5       4       4       4
20      6       5       4       3
20      8       8       5       2
20      10      10      4       2
20      12      5       3       3
20      12      6       5       2
20      20      5       5       2
20      20      15      3       2
21      7       7       3       3
21      7       7       6       2
21      21      14      3       2
24      8       4       4       3
24      8       6       3       3
24      8       6       6       2
24      12      8       4       2
24      16      16      3       2
24      24      4       3       3
24      24      6       4       2
24      24      12      3       2
28      14      7       4       2
28      21      4       3       3
28      21      6       4       2
28      21      12      3       2
30      10      5       3       3
30      10      6       5       2
30      15      5       5       2
30      15      15      3       2
30      20      4       3       3
30      20      6       4       2
30      20      12      3       2
30      30      10      3       2
33      22      11      3       2
35      14      5       5       2
35      15      14      3       2
36      9       9       4       2
36      18      4       3       3
36      18      6       4       2
36      18      12      3       2
36      36      9       3       2
40      10      8       4       2
40      24      10      3       2
40      40      5       4       2
42      7       4       4       3
42      7       6       3       3
42      7       6       6       2
42      12      7       4       2
42      14      14      3       2
45      9       5       3       3
45      9       6       5       2
45      30      9       3       2
45      36      5       4       2
48      16      4       3       3
48      16      6       4       2
48      16      12      3       2
48      48      8       3       2
54      27      9       3       2
56      42      8       3       2
60      5       5       4       3
60      12      5       5       2
60      15      4       3       3
60      15      6       4       2
60      15      12      3       2
60      20      10      3       2
60      30      5       4       2
60      40      8       3       2
70      7       7       5       2
70      28      5       4       2
72      9       8       4       2
72      24      9       3       2
72      36      8       3       2
78      13      13      3       2
84      14      4       3       3
84      14      6       4       2
84      14      12      3       2
84      84      7       3       2
88      33      8       3       2
90      18      10      3       2
91      78      7       3       2
96      32      8       3       2
99      22      9       3       2
100     25      5       4       2

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Your code doesn't seem very efficient. – Joe Z. Feb 1 '13 at 16:50
In addition, it won't find unit fraction combinations below a certain granularity, while there is a way to do the question using only integers. – Joe Z. Feb 1 '13 at 19:43
This is like an algorithm that will be used as a bad example of not thinking. – mez Feb 11 '13 at 23:12
This is not a programming community. I wrote the simplest code in shortest time to solve this problem numerically with least effort. I don't care about code efficiency in this particular case. – hkBattousai Feb 14 '13 at 6:43
C++ for 5th graders! – Ahaan S. Rungta Oct 31 '13 at 14:32

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.

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You can edit the code in my answer to preclude answers that have numbers above a certain bound. – Joe Z. Jan 30 '13 at 23:39

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.

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If the goal is simply to find any arbitrary solution, the problem should be solvable if someone starts with the three-number case; it shouldn't take long to find 1/2 + 1/3 + 1/6. From there, one may try to decompose the 1/6 into two parts which are unit fractions. A natural first effort would be to see what's left if one bumps 1/6 to the next smaller unit fraction (1/7). That turns out to leave 1/42--a unit fraction (hooray!) Subtract the next unit fraction smaller than that (1/43) to leave 1/1806. I wouldn't expect students to enumerate the solutions, though. – supercat Jan 30 '13 at 16:26
@supercat: if you decompose the 1/6 in '''three''' parts that are unit fractions, then you get the five fractions you need. see my other answer. according to me easy enough for the fifth grade student. – mariotomo Jan 30 '13 at 22:55

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 $1$.

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Please avoid the trivial $x=y=z=a=b=5$. Try looking for a solution where $x≠y≠z≠a≠b$ 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.

Was this a requirement in the homework problem? The easiest way to prove that there is more than one solution to a problem is to find more than one solution to the problem, even if the additional solution is trivial. A starting point of any arbitrary number of unit fractions that sum to $1$ and that also can be combined into other unit fractions is a very valid starting point.

Case in point: I would propose that this be solved at the fifth-grade level by splitting fractions, with a little special knowledge to produce unique values. $1 = 1/2 + 1/2$. Now we need to come up with three more fractions and differentiate the two we have. We do this by continuing to split, with some clever knowledge of prime factors.

For instance, if a number in the denominator is divisible by both $2$ and $3$, then you can combine two or three of that unit fraction, and when reducing to lowest terms you will get a new unit fraction. $6$ is the simplest example; $1/6 + 1/6 = 1/3$ and $1/6 + 1/6 + 1/6 = 1/2$. This is useful to us: $1/6 + 1/3 = 1/2$, so $1/6 + 1/3 + 1/2 = 1$. Now we just need two more unique fractions. Well, let's split up the largest fraction, $1/2$, into more pieces than we tried before. If we divide it into 12 units, each of those will be $1/24$ of the whole. $2/24 = 1/12, 3/24 = 1/8$, and $6/24 = 1/4$. We can also combine $4$ and $8$ of these, but those two produce the fractions $1/6$ and $1/3$ which we already have. Now, by serendipity (or not), $1 + 2 + 3 + 6 = 12$. That would give us $1/24 + 1/12 + 1/8 + 1/4 = 1/2$. That's four of the five we need, and $1/2$ (the remaining fraction of the whole) isn't spoken for yet and can be produced by recombining $1/3$ and $1/6$, so $1/2 + 1/4 + 1/8 + 1/12 + 1/24 = 1$, and thus $\{2,4,8,12,24\}$ is a valid solution.

This generalizes into the following statement: find five numbers, $j,k,m,n$, and $p$, that are all factors of an arbitrary $z$, and that sum to $z$. Then $1/x=j/z$, $1/y=k/z$, and so on.

A good z-value to try in this case is $100$, which turns the fractions into simple integer percentages. $100$ has the following prime factorization: $2^2*5^2$. Each unique combination of those factors, plus the universal factor $1$, is a factor of $100$ and thus a possible value for the set of $5$ we need. There are 8 factors of $100$, not including $100$; $1, 2, 4, 5, 10, 20, 25$, and $50$. We can now apply a variation of the value-splitting; find numbers in this set that will sum to other numbers in the set. $100=50+50$. $50=25+25$. $25=20+5$, and $5=4+1$. Putting these identities together, $1+4+20+25+50 = 100$. Dividing everything by $100$ and finding lowest terms gives $1/100+1/25+1/5+1/4+1/2 = 1$, so $\{2,4,5,25,100\}$ is another valid solution.

Pretty much any number with $4$ or more factors other than $1$ and itself is a candidate z-value. It helps to have a few extra factors laying around, as was shown from the example above; a number with only $4$ unique factors, say $20$ $(2, 4, 5, 10)$, is unlikely to have those factors all sum to the original number as well (in this case the sum of those four factors is $21$; close but no cigar). It also helps if the prime factorization of the number (a graspable concept for a fifth grader, with multiplication, division and primes under her belt) includes more than one prime factor; this allows you to do the "divide by one, then multiply by the other" trick to produce unique denominators.

As there are an infinite number of natural numbers, and of primes and composites, there are an infinite number of possible Z-values and unique factors of those Z-values, and thus an infinite number of solutions to this problem.

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Surprised that no-one proposed $1,-1,1,-1,1$ and a whole bunch of variations around it.

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Such as N, -N, M, -M, 1 - where N, M are any numbers. A whole infinity of answers. – Andrey Regentov Feb 1 '13 at 8:32
Sorry, missed the "positive" in question... – Andrey Regentov Feb 1 '13 at 8:33

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).

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As long as it’s clearly understood to be a somewhat open-ended and challenging question, I think that it’s a great question at the $5$-th grade level. – Brian M. Scott Jan 30 '13 at 10:03

$$\frac{1}{2}+\frac{1}{3}+\frac{1}{8}+\frac{1}{30}+\frac{1}{120}=1$$
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.

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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 $$1=\dfrac12+\dfrac14+\dfrac16+\dfrac1{12}$$ $$1=\dfrac12+\dfrac13+\dfrac17+\dfrac1{42}$$ $$1=\dfrac12+\dfrac13+\dfrac1{10}+\dfrac1{15}.$$ Then continue.....

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## protected by Asaf KaragilaJan 31 '13 at 17:19

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