Unusual 5th grade problem, how to solve it

Question

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.

Answer 1

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,

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

Answer 2

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:

Answer 3

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 $\displaystyle \frac 15 + \frac 15 + \frac 15 + \frac 15 + \frac 15$.

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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).

Answer 4

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

Answer 5

For 5th grader, I fear the only way to get a 5-tuple answer is providing hint like:

If you have a solution for $1 = \frac{1}{x} + \frac{1}{y}$, say $1 = \frac{1}{2} + \frac{1}{2}$, can you come up a solution for $1 = \frac{1}{x} + \frac{1}{y} + \frac{1}{z}$?

If the kid still has no clue, then tell he/she $2\times2 = 4$ and $1 = \frac{1}{2} + \frac{1}{4} + \frac{1}{4}$.

This is also a good opportunity to teach the kid the possibility to build up new solutions from the old.

Answer 6

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.

Answer 7

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.

Answer 8

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 $1 = \frac12 + \frac13 + \frac1{12} + \frac1{18} + \frac1{36}$

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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 $1$.

Answer 9

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.

Answer 10

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

Answer 11

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.

Answer 12

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

Answer 13

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).