Fractions | A

Question 2

Can you find four different fractional units that add up to 1?

It turns out that this problem has six solutions! Can you find at least one of them? Can you find them all? You can try using similar reasoning as in the cases of two and three fractional units—or find your own method!

Once you find one solution, try to divide a circle into parts like in the figure above to visualise it!

Solution

Understand the Question

A fractional unit (or unit fraction) is a fraction with a numerator of $1$ , written as $\dfrac{1}{n}$ where $n$ is a positive integer.
We want to find four distinct positive integers $a < b < c < d$ such that: $\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} + \dfrac{1}{d} = 1$
By systematically choosing the smallest denominators and solving for the remaining unit fractions, we find all $6$ unique combinations.

Step 1 · Determine the Largest Unit Fraction

Let the four distinct unit fractions be $\dfrac{1}{a}, \dfrac{1}{b}, \dfrac{1}{c}, \dfrac{1}{d}$ with $a < b < c < d$ .

If the smallest denominator is $a = 3$ , the maximum possible sum of four distinct unit fractions starting from $\dfrac{1}{3}$ is:

\begin{aligned} \dfrac{1}{3} + \dfrac{1}{4} + \dfrac{1}{5} + \dfrac{1}{6} &= \dfrac{20}{60} + \dfrac{15}{60} + \dfrac{12}{60} + \dfrac{10}{60} \\[0.6em] &= \dfrac{20 + 15 + 12 + 10}{60} \\[0.6em] &= \dfrac{57}{60} < 1 \end{aligned}

Since this sum is strictly less than $1$ , the largest unit fraction must have $a = 2$ , so the first fraction is $\dfrac{1}{2}$ .

Step 2 · Find Solutions with Second Fraction $\dfrac{1}{3}$

Subtracting $\dfrac{1}{2}$ from $1$ :

\begin{aligned} \dfrac{1}{b} + \dfrac{1}{c} + \dfrac{1}{d} &= 1 - \dfrac{1}{2} = \dfrac{1}{2} \end{aligned}

Choosing $b = 3$ gives:

\begin{aligned} \dfrac{1}{c} + \dfrac{1}{d} &= \dfrac{1}{2} - \dfrac{1}{3} \\[0.6em] &= \dfrac{3}{6} - \dfrac{2}{6} \\[0.6em] &= \dfrac{1}{6} \end{aligned}

Now, test values of $c > 6$ :

For $c = 7$ : $\dfrac{1}{d} = \dfrac{1}{6} - \dfrac{1}{7} = \dfrac{7}{42} - \dfrac{6}{42} = \dfrac{1}{42} \implies d = 42$ $\mathbf{\text{Solution 1: } \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{7} + \dfrac{1}{42} = 1}$
For $c = 8$ : $\dfrac{1}{d} = \dfrac{1}{6} - \dfrac{1}{8} = \dfrac{4}{24} - \dfrac{3}{24} = \dfrac{1}{24} \implies d = 24$ $\mathbf{\text{Solution 2: } \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{8} + \dfrac{1}{24} = 1}$
For $c = 9$ : $\dfrac{1}{d} = \dfrac{1}{6} - \dfrac{1}{9} = \dfrac{3}{18} - \dfrac{2}{18} = \dfrac{1}{18} \implies d = 18$ $\mathbf{\text{Solution 3: } \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{18} = 1}$
For $c = 10$ : $\dfrac{1}{d} = \dfrac{1}{6} - \dfrac{1}{10} = \dfrac{5}{30} - \dfrac{3}{30} = \dfrac{2}{30} = \dfrac{1}{15} \implies d = 15$ $\mathbf{\text{Solution 4: } \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{10} + \dfrac{1}{15} = 1}$

(Note: $c = 11$ gives $\dfrac{5}{66}$ which is not a unit fraction, and $c = 12$ gives $c = d = 12$ , which violates distinctness).

Step 3 · Find Solutions with Second Fraction $\dfrac{1}{4}$

Choosing $b = 4$ gives:

\begin{aligned} \dfrac{1}{c} + \dfrac{1}{d} &= \dfrac{1}{2} - \dfrac{1}{4} \\[0.6em] &= \dfrac{2}{4} - \dfrac{1}{4} \\[0.6em] &= \dfrac{1}{4} \end{aligned}

Now, test values of $c > 4$ with $c \neq 4$ :

For $c = 5$ : $\dfrac{1}{d} = \dfrac{1}{4} - \dfrac{1}{5} = \dfrac{5}{20} - \dfrac{4}{20} = \dfrac{1}{20} \implies d = 20$ $\mathbf{\text{Solution 5: } \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{5} + \dfrac{1}{20} = 1}$
For $c = 6$ : $\dfrac{1}{d} = \dfrac{1}{4} - \dfrac{1}{6} = \dfrac{3}{12} - \dfrac{2}{12} = \dfrac{1}{12} \implies d = 12$ $\mathbf{\text{Solution 6: } \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{6} + \dfrac{1}{12} = 1}$

(Note: $c = 7$ gives $\dfrac{3}{28}$ which is not a unit fraction, and $c = 8$ gives duplicate denominators).

Step 4 · Check $b = 5$ and Higher Values

If $b = 5$ :

\begin{aligned} \dfrac{1}{c} + \dfrac{1}{d} &= \dfrac{1}{2} - \dfrac{1}{5} = \dfrac{3}{10} \end{aligned}

Testing candidates for $c$ :

$c = 4 \implies \dfrac{1}{d} = \dfrac{3}{10} - \dfrac{1}{4} = \dfrac{1}{20}$ (yields the duplicate set $\{2, 4, 5, 20\}$ ).
$c = 6 \implies \dfrac{1}{d} = \dfrac{3}{10} - \dfrac{1}{6} = \dfrac{4}{30} = \dfrac{2}{15}$ (not a unit fraction).

No new solutions exist, confirming there are exactly $6$ solutions.

Step 5 · Visualise a Solution Using a Circle

Diagram 1

Visualising $\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{5} + \dfrac{1}{20} = 1$ by dividing a circle into $20$ equal sectors:

$\dfrac{1}{2} = \dfrac{10}{20}$ ( $10$ sectors)
$\dfrac{1}{4} = \dfrac{5}{20}$ ( $5$ sectors)
$\dfrac{1}{5} = \dfrac{4}{20}$ ( $4$ sectors)
$\dfrac{1}{20} = \dfrac{1}{20}$ ( $1$ sector)

Total $= 10 + 5 + 4 + 1 = 20$ sectors ( $1$ whole circle).

Answer

The $6$ solutions are:

$\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{7} + \dfrac{1}{42} = 1$
$\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{8} + \dfrac{1}{24} = 1$
$\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{18} = 1$
$\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{10} + \dfrac{1}{15} = 1$
$\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{5} + \dfrac{1}{20} = 1$
$\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{6} + \dfrac{1}{12} = 1$

Common Mistakes

Repeated Fractions: Forgetting the condition that all four unit fractions must be distinct (e.g., $\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{8} = 1$ is invalid because $\dfrac{1}{8}$ is repeated).
Non-Unit Fractions: Ending up with a fraction where the numerator is not $1$ after simplification (e.g., $\dfrac{2}{15}$ ) and forgetting to check that the numerator must reduce to $1$ .

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